2015
Inteegraated Loggistiics
Beullenns P.
Univerrsity of South
hampton
2/1/20015
Integrated Logistics
Contents
1
Introduction .............................................................................................. 4
1.1
Defining Supply Chain Management ........................................................... 4
1.2
Understanding Supply Chain Management .................................................... 4
1.3
Measuring Supply Chain Performance ......................................................... 4
1.4
What is Integrated Logistics? ................................................................... 4
2 IL & Finance .............................................................................................. 5
2.1
Strategy and Finance ............................................................................ 5
2.2
Time value of money............................................................................. 6
2.2.1 Net Present Value for discrete interest rates ............................................. 6
2.2.2 Net Present Value for continuous interest rates ......................................... 8
2.2.3 Annuity Stream ............................................................................... 10
2.2.4 Linear approximations of NPV and AS ..................................................... 10
2.2.5 NPV and AS of a few useful cases .......................................................... 11
2.3
Economic Order Quantity (EOQ) – Harris (1913) ............................................ 14
2.4
Economic Production Quantity (EPQ) – Taft (1918) ........................................ 18
2.5
EOQ with batch demand – Grubbström (1980) .............................................. 21
2.6
Lot-for-lot production at finite rate – Monahan (1984) .................................... 24
2.7
EOQ for batch demand – Goyal (1976) ....................................................... 26
2.8
EPQ for batch demand – Joglekar (1988) .................................................... 28
2.9
Vending machine ................................................................................ 29
2.10 Payment structures ............................................................................. 30
2.11 Consignment arrangements .................................................................... 32
2.12 The profit function of the integrated supply chain ........................................ 34
2.13 Using the NPV framework to include other cost components ............................ 35
3 IL in Buyer-Supplier Supply Chains .................................................................. 38
3.1
One-to-one shipping............................................................................. 38
3.1.1 Shortest Path Problem ....................................................................... 38
3.1.2 Economic Transport Quantity (ETQ) ....................................................... 41
3.1.3 Maximum Economic Haulage Radius (MEHR) ............................................. 45
3.1.4 Optimal policy to order N different items from a Cross-Docking Facility (CDF).... 46
3.2
One-to-many shipping .......................................................................... 48
3.2.1 Length of an optimal TSP tour visiting many customers ............................... 48
3.2.2 Expected length of an optimal TSP tour visiting a few customers ................... 51
3.2.3 Continuous approximation of an optimal vehicle routing solution ................... 52
3.2.4 One-to-many ETQ ............................................................................ 54
4 Strategy in the supply chain: Alliances versus Leaders .......................................... 61
4.1
Alliances .......................................................................................... 61
4.1.1 Cooperative game theory ................................................................... 61
4.1.2 Alternative methods ......................................................................... 71
4.2
Two-party alliances in the supply chain ..................................................... 71
4.2.1 Two-party alliance: example ............................................................... 71
4.2.2 Two-party alliance: example 2 ............................................................. 79
4.3
Perfect coordination ............................................................................ 82
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4.3.1 Perfect coordination schemes: examples ................................................ 83
4.4
Leaders and followers .......................................................................... 88
4.4.1 Leaders and followers: Stackelberg games ............................................... 88
4.4.2 Stackelberg leader in the supply chain: examples ...................................... 90
5 Stochastic Models ...................................................................................... 96
5.1
When is the assumption of a constant demand rate valid? ............................... 96
5.2
(r, Q) reorder point models .................................................................... 97
5.2.1 Backorder case ................................................................................ 98
5.2.2 Lost sales case .............................................................................. 103
5.2.3 Service level approach .................................................................... 103
5.2.4 Standard Tables ............................................................................ 105
5.2.5 Exercises ..................................................................................... 110
5.3
News vendor problems ........................................................................ 111
5.3.1 Example ...................................................................................... 111
5.3.2 Cost minimisation .......................................................................... 111
5.3.3 Profit maximisation ........................................................................ 112
5.3.4 Regret minimisation ....................................................................... 113
5.3.5 Exercises ..................................................................................... 114
6 MRP, JIT, and Bottlenecks .......................................................................... 116
6.1
Materials Requirements Planning ........................................................... 116
6.1.1 MRP inputs ................................................................................... 116
6.1.2 MRP outputs ................................................................................. 118
6.1.3 Lot sizing policies .......................................................................... 122
6.1.4 Remarks on MRP ............................................................................ 125
6.2
Just-In-Time .................................................................................... 130
6.2.1 Motivation ................................................................................... 131
6.2.2 Push and pull systems ..................................................................... 131
6.2.3 The Kanban System ........................................................................ 131
6.2.4 How many cards? ........................................................................... 132
6.2.5 Subcontractors .............................................................................. 134
6.2.6 Fluctuations in demand ................................................................... 134
6.2.7 Comparison JIT and MRP .................................................................. 136
6.3
Bottleneck scheduling ........................................................................ 139
6.3.1 Optimised Production Technology (OPT) ............................................... 139
6.3.2 Theory Of Constraints (TOC).............................................................. 143
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Integrated Logistics
1 Introduction
For any questions, e‐mail:
[email protected]
See also the material on blackboard.
This text follows the lectures given, but I may have arranged the sequence of a few topics in order to
arrive at a more logical flow.
1.1 Defining Supply Chain Management
See slides on blackboard and read the following article (on blackboard):
MENTZER J T, DE WITT W, KEEBLER J S, MIN S, NIX N W, SMITH C D, AND ZACHARIA Z G. 2001.
DEFINING SUPPLY CHAIN MANAGEMENT. JOURNAL OF BUSINESS LOGISTICS 22 (2), 1‐25.
1.2 Understanding Supply Chain Management
See slides on blackboard and read the following article (on blackboard):
CHEN IJ AND PAULRAJ A. 2004. UNDERSTANDING SUPPLY CHAIN MANAGEMENT: CRITICAL RESEARCH
AND A THEORETICAL FRAMEWORK. INTERNATIONAL JOURNAL OF PRODUCTION RESEARCH 42 (1), 131‐
163.
1.3 Measuring Supply Chain Performance
See slides on blackboard and read the following article (hand‐out):
SHEPHERD C AND GÜNTER H. 2006. MEASURING SUPPLY CHAIN PERFORMANCE: CURRENT RESEARCH
AND FUTURE DIRECTIONS. INTERNATIONAL JOURNAL OF PRODUCTIVITY AND PERFORMANCE
MANAGEMENT 55 (3/4), 242‐258.
1.4 What is Integrated Logistics?
See the slides on blackboard.
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Integrated Logistics
2 IL & Finance
2.1 Strategy and Finance
Source: SILVER A., PYKE D.F., AND PETERSON R. 1998. INVENTORY MANAGEMENT AND PRODUCTION
PLANNING AND SCHEDULING, THIRD ED. JOHN WILEY & SONS, NEW YORK, CHAPTER 2: STRATEGIC
ISSUES, P. 14.
Integrated Logistics (IL) should be linked to the corporate and business strategy. The most important
objective of any firm is arguably long term profitability. In this context the operating profit is defined as:
IL can affect both terms on the right hand side. By reducing e.g. aggregate inventory levels in the firm,
the investment cost can be reduced. By allocating proper inventory levels among different items in an
improved way, sales revenue may increase.
One common aggregate performance measure in IL and inventory management is the inventory
turnover:
£
£
An increase in sales without a corresponding increase in inventory will increase the inventory
turnover, as will a decrease in inventory without a decline in sales. Turnover can be a useful measure to
compare divisions of a firm or firms in an industry.
A higher turnover ratio for the same level of sales means a more profitable business, as less money is
tied up in inventories. The danger is that a reduction of inventory levels may also negatively affect sales.
When it is not known with certainty how much demand there is per period, an amount of safety stock is
needed to make sure enough products are in stock in case demand would be somewhat higher than
expected. Furthermore, the right figure of inventory turnover for your firm depends on the level of in‐
house production versus out‐sourcing. A firm that does everything in‐house will need a lower inventory
turnover than a firm that is completely based on outsourcing of the production. Indeed, the first firm will
also have a stock of raw materials and work‐in‐process, while the second will only have end products in
inventory. These considerations are important when comparing inventory turnover figures of different
firms.
Other useful performance measures for IL include those that measure all sorts of: costs; average and
variability of lead‐times (often seen as the time between initiation of some sales order and realisation of
the sales, sometimes also just the time is takes for a product to move through a particular part of the
supply chain); product and service quality; customer satisfaction; and innovativeness (see also Section
1.3).
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2.2 Time value of money
Source (on blackboard): GRUBBSTRÖM, R.W. 1980. A PRINCIPLE FOR DETERMINING THE CORRECT
CAPITAL COSTS OF WORK‐IN‐PROGRESS AND INVENTORY. INT. J. OF PRODUCTION ECONOMICS 18(2),
259‐271.
2.2.1
Net Present Value for discrete interest rates
If I invest V (£) into a project now, the project having a rate of return =0.2 after one year, what will be
the amount of money a (£) that I will receive one year later?
Cash‐flows
a
0
1
Time
V
The answer:
1
1.2
If I keep the investment running not for one but for T years in total, what will be my reward at the
end?
Cash‐flows
a
0
1
...
2
T
Time
V
The answer:
1
1
… 1
1
We can turn this around and ask the question: what is the current value V (£) of receiving a (£) within
T years time from now? Algebraic manipulation of the above equation gives:
1
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V is called the Net Present Value of a. It thus all depends on the value of , called the discount factor,
interest rate, or internal rate of return. Companies typically use values of within the range 0.1 to 0.3.
Consider a sequence of n payments of different amounts ,
1, … , at equidistant points in time
with cycle time T (see Figure below). What is the Net Present Value of these payments, if is the rate of
return over one period T?
T
T
a2
Cash‐flows
V
0
an
a1
1
...
2
n
Time
The answer:
1
The rate of return is a function of the time period over which it is defined. For example, if is the
rate of return over a one year period, what is the equivalent rate of return defined over a period of one
month (take this to be 1/12th of a year) that would give the same Net Present Value?
Answer: Let us call the equivalent rate of return over one month, then with T = 12:
1
1
1
1
Maths refresher
ln
ln 1
1
ln 1
T
ln 1
ln 1
Thus, if the annual interest rate is 0.2, the equivalent monthly interest rate would be given by:
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1
ln 1
12
0.2
ln 1
.
1 0.0153
Note that if the interest rate would be defined over a small time period, then ln 1
2.2.2
.
Net Present Value for continuous interest rates
In general, the conversion of interest rates corresponding to different lengths of period obeys the
formula:
1
,
ln 1
where is the interest rate pertaining to a period of length and is the continuous interest rate
corresponding to the limit length zero.
The definition of Net Present Value in the continuous case becomes:
,
where
is the cash‐flow at time plus some multiple of Dirac’s Delta function at points at
which there are finite payments .
We will typically need to solve only special cases, given below.
Example 1 – A one‐off lump sum (£) received at future time L
Cash‐flows
a
V
0
L
Time
Thus the NPV of a cash‐flow received L time units in the future is the cash‐flow multiplied with its
“delay” factor
.
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Example 2 – A finite number of cash revenues, each received their own future moment
T1
T2
a2
Cash‐flows
an
a1
V
0
1
...
2
n
Time
Example 3 – An infinite number of equal cash revenues received at equidistant moments
T
T
Cash‐flows
...
a
0
a
a
1
2
...
Time
Maths refresher
If | | 1 , then
∑
For
, | | 1
Thus:
1
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Example 4 ‐ A continuous cash‐flow at a rate of a (£/year) received for eternity.
Cash‐flows
a
....
0
Time
2.2.3
Annuity Stream
The Annuity Stream AS of a series of cash‐flows is the Net Present Value V of these cash‐streams times
the rate of return :
From example 4, it is clear that the annuity stream is that continuous stream of cash yielding the same
net present value as the original series of cash‐flows.
2.2.4
Linear approximations of NPV and AS
Maths refresher
Maclaurin expansion of an exponential function
∑
1
and converges for ∞
Using this result for
!
!
⋯
∞
, it is easy to see that:
1
2
,
and it can also be proven (after lengthy algebraic manipulation):
1
2
12
1
This can be used to derive linear or quadratic approximations in of NPV or AS functions, as
illustrated in the next section.
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2.2.5
NPV and AS of a few useful cases
Case 1 ‐ A one‐off lump sum a (£) received at future moment L
Cash‐flows
a
0
L
Time
1
1
For a one‐off lump sum, both the linear approximation of the NPV and the quadratic approximation in
of the AS are acceptable, and indeed would be of the same accuracy. However, the linear
approximation of AS would be insufficiently accurate – as indeed the above shows, cannot be the AS
as a itself is not its NPV. The linear approximation of the AS would thus neglect the delay effect
altogether.
Case 2 ‐ An infinite series of lump‐sum payments a (£) received with cycle time T.
T
T
Cash‐flows
...
a
0
a
a
1
2
...
Time
1 1
1
2
1
2
1
12
12
1
1
2
For an infinite series of cash‐flows, the linear approximation of AS is acceptable.
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Case 3 ‐ A continuous cash‐flow at rate a (£/year), starting at time L, and forever lasting.
L
Cash‐flows
a
....
0
Time
∞
It can be observed that this result could have been obtained by combining example 4 and case 3 from
above. Indeed the NPV at time L of the continuous cash‐flow a is a/ (example 4), and then accounting
for the delay with time L (case 1) gives the above result.
1
2
1
Case 4 ‐ A continuous cash‐flow at rate a (£/year) received in future time period T, starting at
time L.
L
T
Cash‐flows
a
0
Time
1
This result can also be obtained by viewing the temporary continuous cash‐flow a as the sum of two
infinite continuous cash‐flows +a and –a (starting a time T later), and thus by applying the result of case 3
twice:
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1
To find the linear approximation, it is safe to take the quadratic terms first into consideration and
approximate later on:
1
2
⋯
1
1—
2
⋯
2
1
Case 5 ‐ An infinite series of continuous cash‐flows at rate a (£/year) with cycle time L and each
time received for a length of time T (T ≤ L).
L
Cash‐flows
T
a
L
T
....
a
0
Time
The result of case 4 (without delay) can first be applied to find the NPV of the cash‐flow of every
period L. Using the approach as in case 2 then gives:
1
1
1
To find the linear approximation of AS, it is again safe to first consider the quadratic terms as well:
1
1
1
1
⋯
⋯
2!
2
2
2
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2.3 Economic Order Quantity (EOQ) – Harris (1913)
A retailer sells to customers a type of product at the price p per product. Demand for the product can
be assumed to occur at a constant rate according to an annual demand of products.
The retailer has to purchase from an external supplier at price per product, and also has to pay a
fixed order cost for each order placed at this supplier (this could be the fixed cost of transport plus the
fixed cost of administration to place an order).
retailer
w(£/product)
s(£/order)
y(products/year)
p(£/product)
Q (products/order)?
T (cycle time) ?
Inventory
retailer
T
...
Q
r
L
Time
Figure 1. Lot-size model (EOQ).
When the retailer would place an order or size Q (products/order) with cycle time T, the inventory
level over time at the retailer will follow the classical saw‐tooth pattern as illustrated in the Figure above.
It can be observed that:
If there is a non‐zero lead‐time , and we want to make sure that the order arrives when the
inventory drops to zero, we must order in advance when the inventory level is . This is called the
reorder point.
The traditional approach to deriving the optimal policy
We consider a deterministic system in which all relevant parameters are constant and shortages are not
allowed. The policy used is (r, Q). Although the aim is to find optimal values for both r and q, the optimal
value for is easily determined. The problem therefore reduces to finding the optimal lot size Q. In this
classic economic lot size system the following assumptions are made:
1. constant annual demand rate y (items / year);
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Integrated Logistics
2.
3.
4.
5.
6.
7.
constant infinite replenishment rate R = ;
constant unit holding cost h(£ / item, year);
constant unit order cost s (£ / order);
no shortages allowed;
constant lead time L = 0, easily extended to constant L 0 (years).
constant replenishment quantity Q (items/order).
The inventory fluctuations in this system are illustrated in Figure 1. It is clear that we place the order
at exactly that moment so that replenishments arrive when on‐hand inventory reaches zero: if lead time
L = 0, we order when the inventory level I(t) = r* = 0; if L 0 (and L < T), we order when I(t) = r* = D L.
Shortages are therefore not possible and shortage costs do not need to be considered. Total costs will
hence consist of inventory holding costs and procurement costs. Increasing the lot size Q will reduce the
number of orders to make per year and hence the annual procurement costs. However, it will increase
the average inventory on‐hand and therefore the annual holding costs. The optimal lot size Q* will be
that quantity that minimises the sum of both annual costs.
The average amount of inventory on‐hand being E(I) = Q/2, the holding cost per year is:
hQ
1
h (Q)
2
As y/Q represents the number of orders per year, the order cost per year is given by:
p (Q)
2
sy
Q
Per definition we want no backorders or lost sales:
s (Q) 0
3
It follows that the total cost equals:
(Q) h (Q) p (Q) s (Q)
hQ s y
Q
2
4
To find the optimal lot size we take the derivative of (Q) with respect to Q:
d(Q) h s y
2 0
dQ
2 Q
5
which gives:
Q*
2s y
h
6
Note that we will always denote with superscript * the optimal value of a decision variable; i.e. Q* is
the optimal lot size. The corresponding optimal cost is:
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(Q)
Total cost
Holding cost
Order cost
0
Q*
Q
Figure 2. Global cost function of the EOQ model.
Q* 2 s h y
7
And since
y = Q/T,
8
Q*
2s
y
hy
9
the optimal cycle time is:
T*
Graphically, the cost equations can be described as in Figure 2. At optimum, annual holding costs
equal annual replenishment costs.
Using the NPV framework to derive the optimal policy
The NPV framework can also be used to derive the optimal values of Q and T. Therefore, the cash‐flows
for the retailer have to be determined. The following assumptions are adopted: Since demand is
occurring at a constant rate , the customers pay a continuous cash‐flow of to the retailer. The
to the
retailer will pay the set‐up cost upon receiving every batch , as well as the amount
supplier. This is illustrated in Figure 3.
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Integrated Logistics
T
Cash‐flows
py
....
0
....
Time
s+wyT
Figure 3. Cash-flows representation of the EOQ model.
The annuity stream of the profit function for the retailer is:
1
The linear approximation
in is thus:
1
2
Since
2
2
:
2
2
The optimal value for Q can be obtained by taking the derivative of this profit function to Q, and
setting this equal to 0:
2
∗
2
0
Thus the same result is obtained, and in addition it becomes very clear how the holding cost (£ per
product per year), used in the traditional inventory framework, needs to be calculated:
Thus, the holding cost for the retailer is related to the amount of money invested per product, i.e. .
We will later encounter examples were the holding costs per product per year will be different.
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2.4 Economic Production Quantity (EPQ) – Taft (1918)
In this section we will analyse a lot size system in which the replenishment rate is not necessarily infinite
as has been assumed in the previous section. Specifically, the system has a uniform replenishment rate R
(items/year), where it is obviously necessary that R y. This type of replenishing generally occurs when
the demand has to be met by a manufacturing department inside the company. The inventory
fluctuations can then be described graphically as in Figure 4.
producer
c(£/product)
s(£/order)
R(products/yr)
y(products/year)
w(£/product)
Q (products/order)?
T (cycle time) ?
A producer sells to customers a type of product at the price w per product. Demand for the product
can be assumed to occur at a constant rate according to an annual demand of products.
The producer has to make the products at a variable production cost per product, and also has to
pay a fixed set‐up cost for each run of these products. Production occurs at a constant finite production
rate equivalent to an annual rate of (products per year).
The traditional approach to deriving the optimal policy
I(t)
q
y
a
b
R
c
0
Lr
t
T
Figure 4. Manufacturing lot size system.
The average amount of inventory equals
E(I) = |bc|/2
10
We have the following relationships:
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bc ac ab
(geometrical relationship)
11
y
ab
R
ac
12
(demand rate)
r
L
Q
Lr
r
L
(replenishment rate)
13
Hence the average amount in inventory can be rewritten as a function of the known parameters y and
R and the variable Q:
1
y Q
y
E ( I ) Q Q 1
2
R 2 R
14
The number of replenishments per year equals y/Q. The total system cost is the sum of holding costs
and replenishment costs:
15
y hQ y
y
(Q) hE( I ) s
Q
1 s
2 R Q
The value Q* which minimises total costs can be obtained as follows:
d(Q) h
y sy
1 2 0
dQ
2 R Q
16
Q*
where Q
*
EOQ
2 sy
*
QEOQ
y
h 1
R
17
1
y
1
R
refers to the economic order quantity of the basic EOQ model. The corresponding cost
is given by substitution of
∗
:
(Q* )
h
y
1
2 R
2 sy
s
y
h1
R
y
2 sy
y
h1
R
18
Rearranging:
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Integrated Logistics
(Q* )
19
y
y
shy1
shy1
R
R
2
2
20
y
y
(Q * ) 2 shy 1 *EOQ 1
R
R
where *EOQ refers to the cost obtained in basic EOQ model. It is easy to show that the EOQ model is
a special case of the continuous rate EOQ model by simply substituting R = .
Example.
Let y = 1000 items/year, R = 2000 units/year, h= 1.6 £/year, and s = 200 £/year. Then:
2 sy
y
h1
R
Q*
21
2 (1000)(2000)
500000 707 units/order.
1.61 0.5
Using the NPV framework to derive the optimal policy
The following cash‐flows are assumed:
T
Cash‐flows yT/R
wy
....
0
....
cR
Time
s
Then, using Case 5 of Section 2.2.5 for the variable production costs:
1
1
2
2
2
2
1
2
The optimal order quantity is derived in the usual manner from the first order conditions:
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2
∗
1
The same result is thus obtained as in the traditional derivation if:
2.5 EOQ with batch demand – Grubbström (1980)
We consider the same system as in the previous section but with two modifications:
1. The production rate is set such that
;
2. Sales occurs in batches of size
producer
c(£/product)
s(£/order)
R(products/yr)
y(products/year)
w(£/product)
Q (products/order)?
T (cycle time) ?
The inventory level as a function of time looks like a mirror image of the EOQ sawtooth pattern:
T
Inventory
....
0
Time
The traditional approach to deriving the optimal policy
This would be exactly the same model as the EOQ of Harris and would produce:
(Q) hE( I ) s
∗
y hQ
y
s
Q 2
Q
Using the NPV framework to derive the optimal policy
First NPV solution. This is how Grubbström (1980) derived a solution:
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Integrated Logiistics
on:
Hencce the linear aapproximatio
1
1
1
2
2
2
2
Thereefore:
∗
2
This m
means that th
he holding co
osts in the traaditional fram
mework are to be based oon the sales p
price of
the prod
duct and not tthe purchase price :
delivery of orrders to the ccustomer
Noticce, however, that with the increase off we are delaying the d
while thee start of pro
oduction is ke
ept fixed at ccurrent time 0. We can in
nterpret this as a model w
where we
have placed a boundaary condition
n at time 0. Thhis is not neccessarily always the best aassumption.
ution. Beullen
ns and Jansseens (2011) in
ntroduced another model where this b
boundary
Alternattive NPV solu
condition is placed att the momen
nt in time wh en the first b
batch to the ccustomer hass to be delive
ered. This
assumpttion would make
m
more sense if the ccustomer waants this to happen,
h
becaause otherwise if the
momentt of first delivvery is not fixxed, the custtomer may e
either run out of stock or
r receive the products
too earlyy and has un
nnecessary stock too earrly. A produccer under the
ese circumsttances would
d need to
derive an
n optimal pollicy from the following NPPV calculation
ns:
This can be re‐written
n as:
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d constant, w
we need to ffind the optim
mal policy fro
om the functtion inside th
he square
And sincce is a fixed
bracketss:
Giving, when
3
2
2
2
:
∗
2
2
e latter case, the produce
er has an ince
entive to makke the lot‐size
e as large
and ∗ → ∞ otheerwise. (In the
as alloweed by the cusstomer and su
uch that ∗
.
The h
holding costs in this case ccould hence bbe taken as:
2
Alternatively, we can look at to derive tw
wo terms:
products, justt like in the ttraditional EO
OQ, to be
1. TThe producerr will have a holding cost of keeping p
vvalued at his own invesstment costss and this holding costt correspondds to the trraditional
in
nterpretation
n of being bassed on investtments made
e into the product placed iin stock:
;
2. TThe producerr will also havve a positive eeffect from th
he batch deliveries to the
e customer givving a so‐
ccalled unit supplier’s rew
ward
(B
Beullens and Janssens, 22011). Note that this
p
produces a reevenue, not a cost!
nction:
We can tthen rewrite the profit fun
3
2
2
2
usion, it is in
n some mode
els therefore important how
h
to set th
he boundary condition in the NPV
In conclu
framewo
ork. This boun
ndary conditiion is called tthe Anchor Po
oint (Beullens and Janssenns, 2011).
23
Integrated Logistics
2.6 Lot‐for‐lot production at finite rate – Monahan (1984)
We can extend the previous model of batch demand by assuming that the producer runs every batch
at some finite production rate
. The producer would hence only start a production run
some time / earlier relative to the delivery of a batch to the customer.
T
Inventory
yT/R
....
0
Time
The classic cost function of this system (Monahan, 1984) is given by:
Φ
2
A first solution using NPV is found from setting the Anchor Point at start of production at time 0,
assuming the cash‐flows as in the figure below.
T
Cash‐flows yT/R
wyT
....
0
....
cR
Time
s
If the start of production has to occur at time 0, the annuity stream is:
1
1
1
1
2
2
2
2
2
2
2
2
Comparing with the classic cost function, we find that
2
and we have as an extra term
the supplier’s reward with
.
In the special case that we retrieve the solution from the previous section:
2
2
24
Integrated Logistics
and the holding cost is then
. Note, however, that we can always write
2
, so instead of adopting a special interpretation
, we see that we can equally adopt the
more general interpretation (since valid for any
of having
2
and an extra term, the
.
supplier’s reward, with
When setting the Anchor Point at the start of sales, we find:
1
The function in between the square brackets is re‐written as:
1 2
2
2
2
Thus we can take
and identify, again, the supplier’s reward as an extra term with
. The special case of is as seen in the previous section and can use the same
interpretations for and .
Using the traditional inventory modeling approach, one would only find the first part . This is in fact
what happened in the literature that has followed Monahan’s (1984) model, and hence there are models
in the literature for which it may not be easy to see whether they will lead to inventory policies which will
also maximise the NPV of the future profits of the firm. See also Beullens (2014).
We finish this section by providing some intuition behind the supplier’s reward. As shown above, it
indicates that there is a positive revenue term in the producer’s linearised AS profit function:
2
The supplier’s reward arises from the fact that the customer orders in batch rather than one product
at a time. An intuitive explanation is the following:
Case A. Suppose you have two options to receive income: either receiving £1,200 at the start of every
year, or £100 at the start of every month, what would you choose? The logical answer would be to
choose the first option.
Case B. Suppose you have two options to pay expenses: either paying £1,200 at the start of every
year, or £100 at the start of every month, what would you choose? The logical answer this time is to
choose the latter option.
The fact that a customer orders in batch will cause inventory costs for this customer. The
disadvantage for the customer to order in batch is extra inventory holding costs to be valued at invested
cost , but at the same time this creates an advantage for the supplier as he receives his revenues
earlier. The supplier’s reward term incorporates this advantage into the supplier’s profit
function.
Homework
Derive the optimal order quantity using the classic cost function Φ
in which
and then derive
the optimal order quantity using the linearised annuity stream function under the two assumptions of
the Anchor Point. Compare the three results.
25
Integrated Logistics
2.7 EOQ for batch demand – Goyal (1976)
We consider a model of a distributor who needs to deliver orders of batch size to customers with an
average demand rate . Customers pay per product to the distributor but the distributor incurs a
delivery cost per delivery. The distributor can place orders of size to its own supplier and has to pay
per product and has an order cost of per order.
distributor
c(£/product)
(£/order)
y(products/year)
w(£/product)
(products/order)
(£/order)
(products/order)?
(cycle time) ?
Inventory
....
0
Time
It can be proven that it is optimal for the distributor to have
for some positive integer.
See also the above figure. The classic derivation of the cost function is again based on trigonometry and
produces the result (Goyal, 1976):
Φ
1
2
To derive the AS profit function, we use the following cash‐flows:
Cash‐flows
wyT
wyT
....
0
s
....
Time
Note that placing the Anchor Point at start of sales to the customers or placing the Anchor Point at
start of the first batch arriving from the supplier produces the same boundary condition at time 0. The AS
function is:
26
Integrated Logistics
1
2
2
Comparison of this result with the classic cost function shows that
.
extra term, the supplier’s reward, with
2
, but also that there is an
Exercise
Derive the optimal order quantity ∗ ∗ using the classic cost function Φ
in which
then derive the optimal order quantity using the linearised annuity stream function .
We illustrate the procedure for the classic function. We need to minimise:
Φ
and
1
2
by choosing and optimal integer value, say . This value must satisfy two conditions:
Φ
1
Φ
Φ
1
Φ
From the first condition we can derive:
1
2
1
1
1
2
Hence
1
1
1
Or
2
1
This inequality is a quadratic function in
2
.
. The non‐negative root is given by:
1
1
2
1
8
But the solution has to be integer: (=the largest integer not larger than
quadratic inequality.
From the second condition we derive similarly that:
2
1
.
And we can proceed as for the first condition to derive the same
. Hence:
together imply ∗
∗
∗
1
1
2
1
8
always satisfies the
as always feasible. Both conditions
27
Integrated Logiistics
ptimal ∗ fo r can be d
derived from this result foor the classicc function
Note thaat the derivattion of the op
by substtitution of
. The reason
r
is thaat the supplier’s reward is a constantt term that does not
depend o
on .
2.8 E
EPQ for ba
atch dema
and – Jogleekar (198
88)
We intro
oduce in thee previous model
m
a finitee production
n rate (with
h
). In this model, is the
production cost per p
product and is the set‐uup cost for a production run. The decission variable is the
or a single pro
oduction run..
production lot‐size fo
producer
c(£/productt)
(£/order)
R (products//year)
(prooducts/order)?
(cyccle time) ?
y(products/year)
w(£/prroduct)
(prod
ducts/order)
(£/orrder)
If we assume thatt
a
as in the prevvious model, the inventorry level over time may look like in
n example).
the following figure (tthis is just an
unction is derived in Joglekkar (1988):
The cclassic cost fu
Φ
1
2
2
Settin
ng the Ancho
or Point at sta
art of sales a s in the figurre above, and
d when makinng assumptio
ons about
the timing of cash‐fflows as in the
t previouss sections, produces the following li nearised AS function
ns, 2014):
(Beullenss and Janssen
2
1
1
2
2
2
2
Hencce,
.
, an
nd again therre is the extraa term with
ng the Ancho
or Point at start of the firsst production
n run will pro
oduce a diffeerent funcction. We
Settin
do not derive the result here.
28
Integrated Logistics
Note that the model of Joglekar incorporates previously considered models:
For → ∞, we get Goyal’s model;
For
1, we get Monahan’s model;
For
1 and
, the EOQ model with batch demand of Grubbström.
Homework
in which
and then derive
Derive the optimal order quantity using the classic cost function Φ
the optimal order quantity using the linearised annuity stream function . Compare the two results.
2.9 Vending machine
We illustrate the usefulness of the NPV framework for deriving the profit function of a vending machine
operator. We consider a single product sold at a price in a vending machine of capacity . The product
has a cost price for the vendor. The vendor has a set‐up cost for delivering a batch of products to the
vending machine. Upon the delivery of products to the vending machine, the operator collects the coins
of the customers. Assume a constant demand rate .
The inventory level over time follows the EOQ saw‐tooth pattern. The cash‐flows however differ from
that in the EOQ model and are given in the figure below.
T
Cash‐flows
pyT
....
0
....
Time
s+wyT
2
2
The holding cost for a vending machine is to be based on
, i.e. based on the sum of cost
price and sales price!
The reason why this result differs from the EOQ model is that the coins put into a vending machine by
customers is cash that is not yet accessible to the vendor operator. Only upon collection of these coins
can the vendor have access to this capital for reinvestment. It is as if the customers only exchange the
cash with the operator the moment the operator empties the vending machine’s coins register.
It can be proven that the optimal lot‐size is
∗
min
,
.
This example illustrates that the profit function in the NPV framework will depend on the assumptions
we make about when cash is exchanged. We call this the payment structure and it is further discussed in
the next section.
29
Integrated Logiistics
2.10 P
Payment sstructuress
Considerr the transfer of a batch of productts between aa buyer and a supplier. Leet be the price per
product that the buyyer has to pay the supplieer. In the pre
evious sections, we have consistently assumed
that the total amountt is paid ffor the momeent that the
e batch arrive
es at the buyeer.
If thee payment occurs in full at the time of delivery then it is said that the payment strructure is
Conventtional (C). It is not the on
nly reasonab le assumptio
on we can make. In realitty, it may be
e that the
buyer paays at differen
nt moments in time, incluuding:
t buyer paays at some time
t
before the delivery is made, sayy at time
Cash‐In‐Advaance (CIA): the
, with
0.
, with
Credit (CR): tthe buyer can
n pay the suppplier later th
han the time of delivery, ssay at time
0
Theree are furtherm
more the following two coonsiderationss:
n instalmentss. The buyer may pay, forr example, a fraction of t he amount d
due
,
Payments in
with 0
1, as CIA (this could be aa deposit paid the momen
nt when the bbuyer places his order
at the supplier), and the rremainder 1
witth a CR arrangement.
ansaction dellays. Due to inefficiencies and costs chharged by the
e financial
Transaction costs and tra
used, the sup
pplier may recceive a differrent amount and at a diffferent time re
elative to
instrument u
the amount aand time the buyer has m
made a payme
ent.
The payment of the amou
unt is CIA, of is C, annd of is CR
R and has
The ffigure below illustrates. T
transactiion costs an
nd delays. We
W call the first two paayments
andd the third payment
.
will hencefortth assume th
hat transactioon costs and
d delays are zero or so sm
mall that the
ey can be
We w
ignored. All payments further considered are hhence
.
It is not d
difficult to ad
dapt all previously consideered models for situations where paym
ments occur CIA or CR
in cases where the tim
me is speciffied. An exam
mple follows.
EOQ mo
odel with CR
R and CIA pay
yment strucctures
Assume that all custo
omers pay with a CR of tiime but th
hat you have to pay the ssupplier with CIA with
ow diagram is given below
w. Since we have to pay the supplier in advance, we must
time . The cash‐flo
assume tthat the first delivery will occur at som
me future time with
.
30
Integrated Logistics
L
T
Cash‐flows
py
....
0
.... Time
s
wyT
This gives:
2
2
2
∗
We also get:
Φ
∗
∗
∗
2
2
and
∗
2
2
Note. It is easy to see that if customers would pay CIA we can use the above model but consider negative
values for . Likewise, we can study the impact of receiving a credit period from the supplier by taking in
the above model negative values for .
Numerical example
The table below illustrates the impact for an example with
£25/
,
£150/
,
2000
/
,
0 and
£50/
. We take
0.2. <Note. We only evaluate
the part of the profit AS function that is within the square brackets.>
The % gap is a common measure for getting insight in relative differences between a scenario and a
base case scenario. In the table it is calculated as the percentage difference relative to the base case
scenario of a conventional payment of the supplier i.e. for
0. The % gap for measure is:
∗
∗
0
,
%
100
∗ 0
where in the above table is, respectively, , Φ, and . It can be observed that the increase in
logistics costs is much smaller than the corresponding profit loss. For assessing the impact of different
timings of payments it is hence much safer to consider the profit function.
31
Integrated Logistics
∗
(months)
(products/order)
% gap
Φ∗
% gap
(-)
(£/year)
(-)
∗
(£/year)
% gap
(-)
0
346
0.00
1732
0.00
48253
0.00
1
344
‐0.83
1747
0.84
47398
‐1.77
2
341
‐1.65
1761
1.68
46529
‐3.57
3
338
‐2.47
1776
2.53
45646
‐5.40
4
335
‐3.28
1791
3.39
44747
‐7.27
5
332
‐4.08
1806
4.25
43834
‐9.16
6
330
‐4.88
1821
5.13
42906
‐11.08
2.11 Consignment arrangements
Consignment arrangements are popular payment structures in some industries between suppliers and
buyers. These buyers are companies themselves and may be retailers or production companies. The stock
held at a buyer under this arrangement is called the consignment stock. It is hence inventory that is
physically held at the premises of the buyer, but financially it is still under the (partial) ownership of the
supplier. Only when a product is removed from this consignment stock, will the buyer have to complete
the payment for the product to the supplier.
Assume that the price for a product that a buyer needs to pay to the supplier is .
We can distinguish between the following three common consignment arrangements:
Full Consignment (FC): the supplier retains ownership of the inventory at the buyer and this is
implemented by letting the buyer pays the supplier the price for a product only at the moment
that this product is actually taken out of the consignment stock at the buyer. This product is then
to be used in the buyer’s production process, or in the case of the buyer being a retailer, when
the buyer actually sells the product to one of its own customers.
Partial Consignment (PC): the supplier is paid an amount for each product when the product is
delivered to the buyer’s consignment inventory, but the remainder
is only paid out the
moment the product is taken out of the consignment stock by the buyer (for reasons as
explained above for the case of full consignment). The price can in principle be any agreed
amount; it does not need to be the supplier’s own cost price.
Grace period (GP(z)): This is somewhat similar to PC in that the buyer may first pay an amount
the moment that products are delivered to its consignment stock, but the remainder
is
then to be paid back according to a credit arrangement that depends on the average cycle time
between deliveries. This credit period is called the grace period and its moment of payment is in
general as follows:
,
where is a suitably chosen constant. The idea is that when a buyer orders in larger lot‐sizes,
that the credit period the supplier is willing to offer also becomes longer.
We illustrate the three payment structures with the following example.
32
Integrated Logistics
A buyer‐supplier model with consignment arrangements
We consider a buyer having an EOQ problem and the supplier delivering to this buyer then to have the
EOQ problem with batch demand. Both models were discussed previously. The AS function of the buyer,
under conventional payment structures, was:
2
2
,
where the subscript is used to make clear that it concerns the buyer. The supplier’s AS function was:
1
2
2
2
.
We now introduce a generalised payment structure which has the above discussed three variations of
consignment in it:
An amount is paid for a product when it is delivered to the buyer;
An amount is paid for a product when the buyer sells the product to its own customers;
is paid for with a grace period time units after the
The remainder
delivery of the product to the buyer.
Hence, for
0 ,
0 and
this corresponds to FC; for
0 but
0 this
corresponds to PC; and with
0 but
0 to GP(z). The more general case has all three amounts
non‐zero. Note that
.
Customers of the buyer still pay according to the conventional structure C, as in the EOQ model.
The buyer’s AS function changes to:
We linearise the exponential terms in the decision variable and find:
1
2
2
2
Note that the holding cost for the buyer is based on
1 2 . For
, the
holding cost for the component is zero. For
1, the holding cost for the component becomes a
negative cost i.e. will increase the buyer’s profit function.
The supplier’s AS function changes to:
2
The supplier’s reward is affected and now to be based on
1
1
2
1
2
2
2
.
.
33
Integrated Logistics
2.12 The profit function of the integrated supply chain
If firms in a supply chain want to identify the optimal way of coordinating, then a commonly used
benchmark is assuming that the firms would operate as one virtual organisation. This virtual organisation
represents the integrated supply chain. To derive its optimal policy, we arguably need to have a profit
function as well. We illustrate with an example what can be done, and where there are still some open
problems.
Consider the buyer‐supplier model from the previous section, but assume conventional payment
structures. The profit functions of both firms are, respectively:
2
2
1
2
,
2
2
.
It is tempting to postulate that the profit function of the integrated supply chain is found as their sum:
This gives:
1
2
2
2
However, this function now makes use of two different opportunity costs of capital, and . Would
an integrated firm not be in a position to use one common opportunity cost of capital instead? If we
would assume that both capital cost rates are replaced by one common rate we would find:
2
1
2
2
2
2
Such a result would also be found if we started from the cash‐flow diagrams of both firms, and then
recognising that cash‐exchanged between the firms would cancel each other out.
This conundrum has not yet been adequately addressed in the literature. We will assume henceforth
; in this case the problem does not present itself. Under this assumption, we can
that
indeed sum the profit functions of individual firms to find the profit function of their integrated supply
chain.
Exercise 1
Derive for the above buyer‐supplier model with conventional payment structures the order quantities ∗
.
and ∗ ∗ ∗ that will maximise
We take the partial derivative of
, to for a constant , and from setting this to zero we
find:
∗
Substitution of this result into
2
:
34
Integrated Logistics
2
2
2
The optimal positive integer value is hence
∗
2
∗
1. Therefore, we also get
2
and
∗
2
2
.
Exercise 2
Consider the buyer‐supplier model with consignment payment structures derived previously. Repeat the
above analysis for this case, i.e:
a. Find the profit function of the integrated supply chain;
b. Find the values of ∗ and ∗ ∗ ∗ that maximise this function.
You will find that the profit function of the integrated supply chain is exactly the same, and therefore
also the optimal values derived above apply. This gives an important insight: the payment structures
between the firms in a supply chain do not affect their integrated profit function!
Homework
Consider the set of models we have looked at so far, and make up your own feasible combination of a
single supplier – single buyer model and repeat the above analysis, i.e. determine the integrated supply
chain profit function, and derive the optimal policy that will maximise this function.
2.13 Using the NPV framework to include other cost components
The NPV framework also allows for the consideration of other cost components. We illustrate with a few
examples, using the EOQ model as our base case scenario.
Material handling and insurance
A variable material handling cost e (£/product) for each product placed in inventory, paid upon
receiving the batch
A cost directly proportional to the average inventory level (e.g. the cost of insurance against fire for
the average amount of inventory held), to be paid as e.g. a continuous stream at the rate of /2
(£/year).
The AS function for the EOQ model is adapted to:
2
2
1
35
Integrated Logistics
1
2
2
2
2
Thus the optimal lot‐size becomes:
2
∗
and the holding costs are to be calculated as follows:
In general, we can hence identify two different components of holding costs:
a. The financial cost of keeping stock, arising from investments made into products stored, here the
term
;
b. Out‐of‐pocket holding costs, arising as real expenses to be made when placing products in stock;
here . Note that these costs need to be a function of the stock position i.e. the corresponding
annual cost needs to vary with .
It is generally accepted that in many practical situations the financial holding costs are much larger
than the out‐of‐pocket holding costs. That is also why in many models out‐of‐pocket holding costs are
simply ignored.
Price elastic demand functions
Final demand for a product could be a function of price . For example, one possible function is (for
0):
where and are positive constants. If the retailer with an EOQ model could decide on the price
, then there is the practical constraint:
as otherwise the retailer would not make any profits.
The retailer thus has the AS profit function:
,
2
2
One way to solve this problem is taking partial derivatives to and and solve the non‐linear system
of two equations. A more pragmatic approach would be to apply the following algorithm:
1. Input: Values for , , , and a function
with maximum price
,
2 ,…,
2. For a sequence of prices ∈ ,
2.1. Determine
2.2. Use this to calculate
∗
2.3. Calculate
, ∗
3. Retain that price that gives the highest value for
2
,
∗
.
36
Integrated Logistics
Credit elastic demand functions
Sometimes firms will offer a delay of payments in order to boost demand. In the simplest case, the
complete amount has to be paid for time units after the purchase. Assume that the function
is
known. The AS function for the retailer now becomes:
1
1
,
2
,
2
2
To find the optimal values for L and Q, the following pragmatic algorithm could be used:
1. Input: Values for , , , and a function
with maximum credit period
2. For a sequence of delay values ∈ 0, , 2 , … ,
2.1. Determine
2.2. Use this to calculate
2
∗
2.3. Calculate
, ∗
3. Retain that L that gives the highest value for
,
∗
.
.
References
GRUBBSTRÖM, R.W. 1980. A PRINCIPLE FOR DETERMINING THE CORRECT CAPITAL COSTS OF WORK‐IN‐
PROGRESS AND INVENTORY. INT. J. OF PRODUCTION ECONOMICS 18(2), 259‐271.
BEULLENS, P. AND JANSSENS, G.K. 2011. HOLDING COSTS UNDER PUSH OR PULL CONDITIONS ‐ THE
IMPACT OF THE ANCHOR POINT. EUROPEAN JOURNAL OF OPERATIONAL RESEARCH 215, (1), 115‐125.
BEULLENS, P. 2014. REVISITING FOUNDATIONS IN LOT SIZING ‐ CONNECTIONS BETWEEN HARRIS,
CROWTHER, MONAHAN, AND CLARK. INT. J. OF PRODUCTION ECONOMICS 155, 68‐81.
BEULLENS, P. AND JANSSENS, G.K. 2014. ADAPTING INVENTORY MODELS FOR HANDLING VARIOUS
PAYMENT STRUCTURES USING NET PRESENT VALUE EQUIVALENCE ANALYSIS. INT. J. OF PRODUCTION
ECONOMICS 157, 190‐200.
37
Integrated Logistics
3 IL in Buyer-Supplier Supply Chains
3.1 One‐to‐one shipping
This strategy involves shipping products from a location A to a location B, by letting a vehicle (truck, taxi,
train, ship, airplane) taking a feasible and optimal route, typically the cheapest or fastest.
Let us assume that the vehicle is dedicated to the transaction, i.e. during its trip it will not make
detours or perform other transportation duties. It is then also called Direct Shipping (DS) or line‐haul
shipping.
A
B
Figure 5: One-to-one shipping (Direct shipping, line-haul shipping, FTL shipping)
The transit time is the time the goods are underway from the source to the destination. It is in general
the smallest relative to other shipping strategies (discussed later).
Because a vehicle is dedicated to this single transportation job, it is in general only justifiable from a
cost perspective to transport small amounts when the product transported has a high value density (£/m3
or £/kg) or when it needs to arrive fast (emergency shipment). It is not unusual, for example, to dispatch a
private jet or helicopter for the transfer of human transplant organs in order to save a life, or to deliver a
single spare part by taxi in order to prevent excessive delays for a passenger air flight.
For goods with lower value density, direct shipping is only cost effective when a vehicle can transport
a large enough amount. Therefore this strategy is sometimes also referred to as Full‐Truck‐Load (FTL)
shipping, although it is not restricted to road transport, nor is it always optimal to ship in quantities that
fill up the vehicle’s capacity.
3.1.1
Shortest Path Problem
Finding the optimal route from A to B can be modelled as a Shortest Path Problem (SPP). The SPP calls for
finding the shortest path from an origin node to a destination node in a connected directed graph G=(N,
A) with node set N and arc set A and where every arc a A has a non‐negative length. Shortest path
problems can be efficiently solved using Dijkstra’s algorithm.
2
3
4
2
4
2
6
1
2
3
3
3
5
Figure 6: A shortest path problem
38
Integrated Logistics
Figure 6 shows an example of a shortest path problem. There are six nodes in the graph numbered
from 1 to 6, and seven arcs, where each arc’s direction and length is also indicated. The head of an arc is
the node adjacent to the arc’s arrow head, and the other adjacent node of an arc is called the arc’s tail.
Dijkstra’s Algorithm
To find the shortest path from in such graph from some origin node to some destination node, we can
use Dijkstra’s algorithm:
Dijkstra’s
Algorithm
1. Associate with all nodes a temporary label with value
2. Start at the origin node by changing its label to 0 and make the label permanent. Call it
the current node.
3. For every arc going out of the current node that has a head node with a temporary label,
replace the value of the temporary label of this head node with the value:
min
,
4. Among all temporary labelled nodes, select one with the smallest label value and make
the label permanent. If this node is the destination node, stop, else call this node the
current node and go back to Step 3 (iterate).
The optimal total length is now equal to the label value of the destination node. To determine
the optimal path, start at the destination node and work backwards in the graph by finding
the arc which cost corresponds to the difference of the labels of its head and tail nodes. There
may be more than one optimal path.
Applied to the problem in Figure 6, the algorithm gives the following sequence of label values for each
node (note that permanent label values are indicated with a*, and the position in the sequence
corresponds to the node number):
[ ]
[0* ]
[0* 4 3 ]
[0* 4 3* ]
[0* 4 3* 6 ]
[0* 4* 3* 6 ]
[0* 4* 3* 7 6 ]
[0* 4* 3* 7 6* ]
[0* 4* 3* 7 6* 8]
[0* 4* 3* 7* 6* 8]
[0* 4* 3* 7* 6* 8*]
The length of the optimal path is thus 8. The optimal path is derived starting from node 6 and going
back to node 5 since the length 2 of that arc is equal to the difference of its head label and tail labels, 2 =
8 – 6. It is then either possible to go to node 2, since the arc’s length 2 = 6 – 4, or to node 3 since its arc’s
length 3 = 6 – 3. From node 2 we go back to node 1. From node 3 we would also go back to node 1. There
are thus two optimal paths: either the sequence of nodes 1‐2‐5‐6 or the sequence of nodes 1‐3‐5‐6.
Note that all nodes in this examples have permanent labels at the end. This is not always so; in
general there could be nodes that still carry temporary labels. Note also that the algorithm still works if
there there are multiple arcs having the same tail and head nodes, as long as we select in Step 3 of the
algorithm the shortest arc in the formula.
39
Integrated Logistics
The algorithm finds the optimal solution since every time we make a label of some node j permanent,
we will have found the shortest path from the origin node to that node j and its permanent label value is
the length of this optimal path. The optimality of this path from origin to j is not depending on any
decisions we need to make later on in the algorithm, and vice versa: if the optimal path from origin to
destination would pass node j, then the optimal path from the origin to j will be completely part of the
optimal solution independent of the decision made in the path from j to the destination. We call this the
principle of optimality. (We will further discuss conditions under which this principle is no longer true and
therefore the algorithm would not be applicable.)
Different objective functions
Different objective functions can be used – minimising total length, time or cost – by letting the length of
each arc in the SPP correspond to its distance, expected transport time to cross the arc, or total cost to
use it, respectively. It is easy to incorporate any fixed costs or fixed delays encountered on arcs, such as
on toll roads, at toll bridges, and at toll tunnels, or at border crossings (administration, border control).
Tachograph legislation
For truck transport over longer distances, attention has to be paid to the so‐called tachograph legislation
which requires drivers to take breaks and limits the total number of driving hours per day. Typical
constraints may include the following:
1. no more than 2 hours of consecutive driving is allowed;
2. 45 minutes of rest time needs to be taken either after 2 hours of constant driving or during the 2
hours of driving (in breaks of minimum 15 minutes each);
3. total number of driving hours per driver and day is restricted to 8 hours.
Companies may have the choice between using two drivers reducing total route time or using a single
driver with longer total travel time. To minimise to total costs, one can run the SPP algorithm on the
graph using only travel time related costs for one driver and then, also knowing the total travel time, add
the costs of the extra driver, required breaks, and rest times to find the total cost and real total time. The
option that is the cheapest can then be retained.
Time‐dependent travel times
The problem of minimising total travel time in graphs with time‐dependent expected travel times on arcs
can be adequately modelled as a SPP if the start time at the origin node is given. The algorithm now
needs to use in Step 3 the expected travel time of an arc based on the calculated expected arrival time at
its tail node.
Under the assumption that vehicles that arrive later at the tail node of an arc can never arrive earlier
at the head of the arc then vehicles of the same type that arrived earlier at the tail node, the algorithm
finds the optimal path. The assumption is in general realistic for queue induced travel delays on roads
since vehicles of some type at the end of a queue will find it very difficult to beat vehicles of the same
type at the top of the queue. The assumption is known as the “no‐overtaking” property. Note that this
property implies that waiting at any node in the graph is never optimal. This approach also works for
when a desired arrival time at the destination node is given by letting the algorithm start at the
destination node, working back to the origin node, and subtracting times during the search.
To find minimum cost solutions for time‐dependent travel times for which the no‐overtaking property
holds, the travel cost must be monotone increasing with travel time and a given units of driving time
must cost the same as the same units of waiting time on nodes. In that case there is always an optimal
solution in which no waiting occurs, and thus the algorithm will find an optimal solution.
If waiting (i.e. resting) is less costly than driving, it may be optimal to wait for times with less
congestion on roads in order to minimise costs. However, waiting at some tail node means that
opportunities for faster driving may be lost on arcs closer to the destination. Thus, the principle of
optimality no longer holds and other algorithms are to be used.
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Integrated Logistics
Multi‐model transportation
Multi‐modal transportation problems can be modelled as SPPs by associating with each arc in the graph
the data relevant for a specific mode and vehicle type; multi‐model transfer nodes are introduced in the
graph which connect the graphs of the different transportation modes. Each transfer node will add its
transfer cost or expected time of transfer to the relevant arc leaving the node. The normal SPP algorithm
will then be applicable.
However, there may be other constraints in practice making this approach less realistic. Ships, trains,
and airplanes typically travel according to pre‐specified schedules and routes and some ships or trains are
cheaper than others. It may thus be cheaper to wait at a transfer node for the cheapest vehicle (train,
ship) or route. This however, violates the principle of optimality.
Note. For calculating SPP on road networks, various internet‐based resources can now be used. For
example, Google maps has a function to allow you to seek for the quickest route from A to B, where you
can specify your starting time and which takes into account time‐dependent travel times.
3.1.2
Economic Transport Quantity (ETQ)
Consider the situation that a buyer needs regular supply of some good. We construct a cost model for the
situation of direct shipping of the product from a supplier to this buyer. We assume that the optimal
travel route has been determined (as e.g. an SPP) and that its cost and total travel time are independent
of the time of the year at which it is undertaken. We denote by the total transit time.
We assume a buyer with an EOQ model and a supplier producing lot‐for‐lot at a finite production rate.
We now also consider the intermediate stage where the products are on a vehicle in transit. We place the
Anchor Point at the delivery of the first batch to the buyer at arbitrary time .
T
Inventory
supplier
yT/R
....
Time
0
T
Inventory
in transit
L
....
Time
0
T
Inventory
buyer
0
L
....
Time
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Integrated Logistics
T
Cash‐flows
supplier
L
yT/R
wyT
....
0
Time
cR
s
T
L
Cash‐flows
3PL
....
0
Time
Cash‐flows
buyer
T
L
....
0
Time
Figure 7: Inventory positions and cash-flows in the direct shipping model
We assume that the transport is undertaken by a 3PL and that the buyer pays this company a fixed
transport cost per shipment and a variable transport cost that depends on the lot‐size being shipped. The
buyer pays the supplier for the products the moment that a batch is delivered. We assume that next to
set‐up costs for loading and unloading the vehicle, the 3PL also incurs a transport cost at a rate for the
duration of the journey. See also Figure 7. The capacity of the vehicle is .
We proceed by deriving the AS profit functions of the three firms involved from their cash‐flow
functions. For the buyer, we have:
2
2
For the 3PL, we find:
1
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Integrated Logistics
Define:
1
Then:
2
2
For the supplier, finally, we derive:
1
This leads to, after some algebraic manipulation:
1
2
2
2
2
Optimal three‐firm solution
If the three firms are interested in determining the optimal shipping strategy for their integrated supply
chain, we can derive this from their supply chain profit function. This function is found from the
). This produces:
summation of their profit functions. (As always, assuming
Define:
1
1
1
2
Then:
This produces an optimal unrestricted lot‐size
∗∗
∗∗
2
2
:
2
1
1
However, since the vehicle capacity is , the optimal feasible Economic Transport Quantity (ETQ) is:
∗
min ∗∗ ,
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Integrated Logistics
Determining the value of
The parameter is an annuity stream cost rate and hence needs to be expressed in (£/year). We show
how to incorporate two relevant components in case that the vehicle is a road vehicle: driver wages and
fuel costs. Driver wages are typically given in (£/hr), so if a driver costs (£/hr), then it has to be
converted to a cost rate (£/year) as follows:
8760 ,
since there are approximately 24 365
8760 hours in one year (i.e. not counting years with an
extra day in February). Fuel costs are typically expressed in (£/km). Therefore, a fuel cost (£/km) has to
be converted into a rate (£/year) by assuming an average speed of the vehicle of (km/hr):
8760 .
If these are the only relevant costs, it would hence produce
.
Exercise
Determine the optimal lot‐size when the buyer would independently be able to determine ∗ .
∗∗
. This produces ∗ min
, where:
In that case, the buyer would aim to maximise
2
∗∗
Note. Observe that the unrestricted optimal lot‐size for the integrated three‐firm solution is a function of
the transit lead‐time. However, for the above solution for the buyer it is independent of this lead‐time.
Optimal solution when transport is outsourced
If the 3PL works independently and the supplier and buyer want to determine the optimal shipping
strategy for themselves as two firms, we can derive this from:
Redefine:
1
2
Then:
This produces an optimal unrestricted lot‐size
∗∗
1
2
:
2
∗∗
2
1
However, since the vehicle capacity is , the optimal feasible result is
∗
min
∗∗
,
.
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Integrated Logistics
Note
In international transport by sea it is often that case that buyer and supplier need to rely on a 3PL for the
shipping. Here, could be the maximum load of the product into one container. The 3PL may charge
different rates between shipping a full container versus shipping a fraction of a container load.
Homework
Derive the optimal lot‐size when: (1) buyer and 3PL would seek to find their integrated optimal solution;
(2) supplier and 3PL would seek to find their integrated optimal solution.
Note
The situation of lot‐for‐lot at an infinite rate can be retrieved from the above functions by considering the
case → ∞. The case that the supplier produces according to an EOQ with batch demand is retrieved by
setting
.
3.1.3
Maximum Economic Haulage Radius (MEHR)
There is a limit as to how far one can reasonably transport a certain type of product using direct shipping.
In general, the higher the value density of a product the further it can be transported in small quantities.
This makes it reasonable to use dedicated small transport vehicles such as small vans, taxis, or expensive
vehicles such as aeroplanes.
The lower the value density of a product, the shorter the distance over which it can be economically
transported. To cover longer distances would need shipments in large quantities such as large trucks with
a second trailer or bulk transport in a barge (inland waterways), or container transport on trains (rail) or
ships (sea and ocean transport) where transport costs can be shared with other goods from other
companies.
Using the profit function to derive the MEHR
Knowledge of the AS profit function allows us to calculate the maximum distance over which a product
can be transported. Since
, only strictly positive value for will also produce strictly
positive values for NPV. As any project is not considered worthwhile by a company if its respective NPV
would become negative, the boundary condition would be that:
,
0
Note that when
,
0, this does not mean that the company would not produce a positive
profit in accounting terms. It simply means that it would not produce more profit than from the next best
available alternative! If
0.20 then the firm would still gain a respective 20% of profits from this
activity.
We sketch the approach with the following example. Take the three‐firm integrated profit function
derived previously:
,
2
1
2
It is sufficient to focus on the function in between the square brackets. Furthermore, we can
substitute ∗ into this function, producing the boundary condition:
∗
∗
2
1
0,
2
where
1
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Integrated Logiistics
1
1
2
now need to cconsider two cases.
We n
Case ∗
.
The lot‐ssize is in otheer words a co
onstant. It caan be observe
ed from the a
above functioons that we have two
types off terms in th
he boundary condition: tterms that arre not a function of , aand terms th
hat are a
function of
. W
Working out this boundaryy condition, w
we will hence find a formuula of the sortt:
,
wherre working ou
ut the actual vvalues for aand is left aas an exercise
e to the read er. This mean
ns that:
1
The right‐hand side
s
of this inequality w
would be the
e maximum economic h aulage radiu
us
expresseed in (years). To convert th
his to a distannce, you need
d to considerr the type of vvehicle used.
∗∗
.
Case ∗
Substituttion of the reesult for
∗∗
derived prreviously wou
uld produce tthe conditionn:
2
2
1
0
he maximum possible vallue of . Ho
owever, a
It maay be difficult to derive an analytic result for th
practicall approach would
w
be to use
u an algoritthm where you
y would increase unttil the right‐h
hand side
reduces to zero. (Similar to the alggorithms pressented at the
e end of Section 2.13).
3.1.4
er N differen
nt items from
m a Cross‐D
Docking Faciility (CDF)
Optimal policy to orde
Cross‐D
Docking Facility
Inbound transport
2
1
Outbound transport
Figurre 8: Cross-Do
ocking Facility
y
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Integrated Logistics
A Cross‐Docking Facility (CDF) is typically located near the boundaries of a populated area. It receives
goods in (large) vehicles from various suppliers; these incoming goods are separated and mixed as
required at the CDF, and subsequently sent out in vehicles without being held in storage to different
destinations in the local area.
The cross‐docking operations may require large areas in the warehouse where inbound materials are
sorted, consolidated, and stored until the outbound shipment is complete and ready to ship. If this takes
several days or even weeks it is not considered a CDF but a warehouse. In most CDFs, goods do not stay
longer than 48 hours.
Optimal order policy for N items received from a CDF
Consider the problem where your firm represents stocking point 2 in Figure 8. Your firm has a demand for
N different types of items (i = 1, ..., N). For each item the uniform annual demand rate of your
customers is yi and your cost price is . You order each of these items from the CDF. Each time a
vehicle from the CDF visits your firm, however, you have to pay a fixed transport cost . What would be
the optimal lot‐size for ordering each item?
Since you have constant annual demand for each item, you have an EOQ‐type problem for each item.
Your AS profit function for item , when each item is ordered separately, is then arguably of the following
form:
2
2
For each delivery, the CDF charges a set‐up cost , and that is why in the above function we have
. Your total profit function would then be the sum over all items:
2
2
Could there be a better way of ordering? Is it perhaps worthwhile to order some items together into
one trip? This would safe on transportation charges .
Property. An optimal policy contains scheduling periods Ti of equal length, i.e. Ti = T for all i N.
Proof. Suppose the theorem does not hold. Let T designate the smallest scheduling period which
happens to be for item k, i.e. T = Tk < Ti ( i k). Consider now an item j for which Tj > T and let the
average inventory carried of this item be E(Ij)
.
Now suppose we decrease its scheduling period to Tj`= T. The average inventory will decrease from
E(Ij) to E(Ij`). Since the replenishment cost is independent of the quantity ordered, no additional
replenishment cost is incurred for replenishing item type j every Tj` = T units of time. Hence the global
cost will decrease.
Since an optimal policy contains periods T of equal length, we have:
∀
And we get:
2
2
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Integrated Logistics
We find:
2
∗
And:
∑
1
∗
∗
∗
2
2
3.2 One‐to‐many shipping
3.2.1
Length of an optimal TSP tour visiting many customers
Draw on a piece of paper a square. Call this your “service region.” Indicate the x and y axis (see
Figure 9). We arbitrarily take the length of the sides equal to 1 m, so that the service region covers an area
of 1 x 1 = 1 m2.
1
(x1, y1)
0
(x1, y1)
(x1, y1)
(x1, y1)
x
0
1
Figure 9: Randomly distributed points in a rectangle.
Now draw a random number from the interval [0, 1] and call this x1. Draw a second random number
from the interval [0, 1] and call this y1. Use these coordinates to draw a point (x1, y1) in your service
region. Repeat this procedure. Now you will have a second point located at some coordinates (x2, y2).
Continue until you have generated n points in your service region. Figure 9 shows the exercise at four
different stages, i.e. for n = 5, n = 10, n = 15 and n = 20. We call the set of n point obtained Xn:
X n {( x1 , y1 ), ( x2 , y2 ),...., ( xn , yn )}
Draw a tour through your n points, visiting each point only once and returning to the first city where
you started as in Figure 10. Note that each time you leave a point you have to decide which point to go to
next. You can thus construct several different tours, all having a different total distance. In fact, there are
(n‐1)!/2 different possible tours through n points.
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Integrated Logistics
Figure 10: A TSP tour through 9 points.
Suppose we are interested in that tour of which the total distance travelled is minimal. If n grows
large, the number of possible tours becomes excessively large and we can’t just find the best tour by
trying all possible tours and retain the best one.
We call this problem of finding the shortest tour through n points the Travelling Salesman Problem
(TSP). The TSP is one of the classic problems in Operational Research.
Without knowing the optimal tour itself, let us call the length of the optimal TSP tour T*(Xn) in our
example of randomly distributed points in a square of area 1. Then Beardwood et al. (1959) proved that
when you make n very large, the ratio T*(Xn)/ n becomes constant. In other words, for very large n:
T *(X n ) n
where the constant is believed to be 0.7124.
In fact, this also holds for a service region of more irregular shapes (as in Figure 11). If A is the area
(m2) of a service region of any finite shape, then
T *(X n )
An
Figure 11: Locations randomly distributed in a service region.
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Integrated Logistics
Finally, the result is even more general than that. The points do not need to be drawn from a uniform
distribution across the service area, any distribution will do. The Theorem of Beardwood et al. (1959) is
given in Figure 12.
Theorem BBH. (Beardwood et al., 1959). If T*(Xn) is
the length of the optimal travelling salesman tour
through n points which are independently drawn from
an identical distribution over a bounded region a
of the Euclidean plane, then there exists a constant
such that with probability one
lim
n
T *(X n )
n
d
,
a
where the integration is with respect to the
Lebesgue (area) measure. For the uniform
distribution over [0,1]2, the integration term is
equal to one.
Note: 0.7124 (Johnson et al., 1996, Percus and Martin, 1996).
Figure 12: Theorem of Beardwood et al. (1959).
Example 1
A printed circuit board of 20 cm by 10 cm needs 1000 little holes drilled in it. Drilling is performed by a
moving pin. What is the expected distance that the pen needs to travel? Assume that the holes are
uniformly distributed across the board.
Using the Theorem, we get for A = 200 cm2, n = 1000
T *(X n )
An 0.71 200(1000) 318 cm
Example 2
In his small van, an express courier has small packages destined for 250 customers located across
Hampshire. Assume that Hampshire covers 22500 km2 and customers are uniformly distributed.
a) Estimate the distance to be travelled to deliver all mail.
b) Give a reasonable estimate of how long it is going to take the courier to deliver the mail.
c) What will be the result of dividing the work up in five vans, each covering an equal part of Hampshire?
a) Using the Theorem, we get for A = 22500 km2, n = 250
T * ( X n ) An 0.71 22500(250) 1690 km
b) Taking an average driving speed of 50 km/hour, total travel time = 1690 km /(50 km/hr) = 33.8 hours.
In addition, it may take our driver at least one minute for every customer to make the delivery, which
adds another 250 min = 4.2 hours to the total delivery time.
c) The distance travelled by each driver, since now A = 22500/5 km2 and n = 250/5
T * ( X n ) An 0.71 (22500 / 5)(250 / 5) 1690/5 = 338 km
Taking an average driving speed of 50 km/hour, total travel time = 338 km /(50 km/hr) = 6.76 hours. A
driver now visits about 50 customers, which should keep total working time per driver within 8 hours.
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Integrated Logistics
3.2.2
Expected length of an optimal TSP tour visiting a few customers
Computational tests for uniformly generated random customers in a square area find that the formula of
Beardwood et al. also serves as a good prediction for the expected or average optimal TSP tour length
even when n is relatively small (Eilon et al., 1971, see also Haimovich and Rinnooy Kan, 1985).
Figure 13: Few customers uniformly distributed in a square.
Figure 13 shows three examples. The first row shows four different instances for n = 5. The length of
the optimal TSP tour in all four will be different, say Ti * (i = 1, …,4). The average length, however, can be
expected to be
1 4 *
Ti
4 i 1
An
A5 0.71 5
Example
A pizza takeaway also makes home deliveries using scooters. On average there are 10 home delivery
orders per hour, randomly located in a 50 km2 service region located around the pizza restaurant.
a) Estimate the average total distance a scooter will travel to deliver 3 orders and return to the
restaurant.
b) Estimate the average total time needed for a scooter to deliver 3 orders and return to the pizza
restaurant.
c) Estimate the minimum number of scooters needed when a scooter will deliver on average to 3
customers per trip.
d) Estimate the time when the scooter arrives at the third customer on its trip (relative to its starting time
at the restaurant).
a) With three customers plus the restaurant, n = 3 + 1, and A = 50, the formula gives us an estimated
average time of visiting three customers in the best sequence as to minimise travel distance:
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Integrated Logistics
An 0.71 50(4) 10 km
b) Assuming the average speed of a scooter is around 35 km/hour, total travel time is 10/35 = 0.28 hour =
17 min. Assuming it takes 4 minutes to deliver at a customer location, it will take in total on average 17 +
12 = 29 minutes or about half an hour to make the delivery trip.
c) It takes one scooter half an hour to deliver to 3 customers and in the same time 10/2 = 5 new orders
arrive. Therefore one needs minimum 2 scooters. (In order to be able to perform all deliveries with one
scooter, the average number of delivery orders needs to drop to 6 per hour, or smaller.)
d) The total travel time of 17 min per trip is divided over 4 “legs”, and the third customer is reached at
the end of the third “leg”. The scooter is therefore expected to arrive after travelling a total time of ¾
(17) = 12.75 min. However, we have to add to this the time is takes for delivering the first and second
order, in total 8 min. Therefore the time of arrival at the third customer is estimated to be 12.75 + 8 21
min. after leaving the restaurant.
3.2.3
Continuous approximation of an optimal vehicle routing solution
In physical distribution, goods need to be delivered using vehicles of limited capacity. In the one‐to‐many
shipping mode, a vehicle will deliver to more than one customer during one trip. In the so‐called
Capacitated Vehicle Routing Problem (CVRP) (Dantzig and Ramser, 1959), all vehicles have the same
capacity. The CVRP consists of finding a set of vehicle routes of minimum cost such that: every customer
is serviced exactly by one vehicle, each route starts and ends at the depot and the total demand serviced
by a route does not exceed vehicle capacity. The CVRP, like TSP, is a classic problem in Operational
Research.
To find an optimal set of routes for the CVRP is a difficult problem that can require a lot of
computational time for large problems. One often makes use of heuristics i.e. methods that can find in a
relatively short time a good, but not necessarily optimal, solution. Good (near‐optimal) solutions typically
look like in Figure 14 a: the total service region is divided into a number of districts; and customers
belonging to one district are served by one and the same vehicle, as indicated in Figure 14 b.
A good approximation for the distance travelled by a vehicle serving a district i is (Daganzo, 1984):
Di* 2ri (ni ) Ai ni
where Ai is the size of the district, ri is the average distance from a customer in district i to the depot,
ni is the number of customers in the district, and the dimensionless factor (ni) depends on the metric
and the number of customers in the district. For ni 6 and the Euclidean metric, (ni) is a constant =
0.57, and for smaller values of ni, the parameter becomes larger and goes to about 0.73 for ni = 2 (see
Campbell, 1993). The first term 2ri represents the line‐ and backhaul distance travelled by the vehicle
from the depot to the centre of district i and back. The second term
(ni ) Ai ni
represents the local distance travelled to service all customers in the district. The complete formula
*
for Di is accurate when the density of points is (nearly) uniform over the service region; the density
should not vary much over a district (Daganzo, 1984, Campbell, 1993).
In particular, assuming identical vehicles and unit loads and a uniform density of the locations over
the area A of the service region, the total route distance over all districts is (taking the sum):
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Integrated Logistics
D * 2r
n
(q) A n
q
where r is the average distance between a customer and the depot, q is the number of customers
*
serviced per vehicle, and n is the total number of customers. The above formulas for Di* and D also
hold for situations where the depot would be located outside the service area.
a
Area A
b
Area Ai
Back‐haul ri
Line‐haul ri
Depot
Figure 14: a) Service area, 18 districts, b) A vehicle serving district i
Example
The data are given in Table 1. If vehicles can serve about 5 customers on one tour, the total distance
travelled D * is approximately (taking = 0.57)
D * 2r
n
40
(q) A n 2 (48.62)
0.57 10000 (40) 1138.47 km
q
5
This solution needs n / q = 40 / 5 = 8 tours (or vehicles).
When vehicles are bigger or smaller quantities are delivered so that 10 customers can be served in
one tour by a vehicle, then the total distance travelled becomes
D * 2r
n
40
(q ) A n 2 (48.62)
0.57 10000 (40) 759.75 km
q
10
and 4 tours are needed.
Approximation when customers order different quantities
Consider the situation where customers
1, … , each order a quantity and the vehicles each
have a capacity . Campbell (1993) found that Daganzo’s result is fairly robust if the distribution of order
quantities is reasonably centred around its average
∑
:
∗
2
√
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Integrated Logistics
Table 1: Data for example
2
Service area (km )
No. customers
Customer no.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Total distance
Average distance
3.2.4
10000.00
40.00
Distance to depot
(km)
55.90
48.32
31.70
89.85
71.73
18.45
65.19
53.83
32.59
13.81
74.65
6.16
8.94
47.71
39.65
48.19
32.44
10.40
62.04
15.55
95.75
22.29
47.97
68.17
62.55
61.73
54.90
31.16
54.74
36.30
52.74
90.40
99.25
90.75
13.14
19.65
47.14
39.08
65.69
64.42
1944.92
48.62
One‐to‐many ETQ
We model the distribution from a supplier to buyers. The general optimal distribution pattern is a
difficult problem for which a general analytical solution is unknown. We assume that the buyers’ orders
are small relative to the vehicle capacity, so that an optimal integrated solution would make use of a
CVRP‐type solution. We assume that the total time in transit is relatively small. If, for example, the total
transport time is one to a few days only for a single distribution round, the effect in the NPV approach
would remain fairly small so that this lead‐time effect from transport can be easily ignored.
If the buyers are located in close vicinity to each other relative to the supplier’s location, then the
transportation cost is largely determined from the line‐haul distance between the supplier’s location and
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Integrated Logistics
the centre of these buyers, and this then resembles somewhat the problem we have seen before of
ordering N items from a CDF with a common transport cost. We recall from the problem of ordering N
items from a CDF that in that case it is optimal to order all items with the same cycle time .
This motives us to develop the following model as depicted in the figure below.
T
yT/R
Inventory
supplier
....
Time
0
T
Inventory
Buyer
....
Time
0
T
Inventory
Buyer
....
Time
0
T
Cash‐flows
supplier
....
0
Time
s
cR
s
cR
T
Inventory
Buyer
0
....
Time
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Integrated Logistics
We further make the simplifying assumptions that all buyers order the same product from the
supplier and pay the price .
We have for buyer :
∑
For the supplier, where
2
:
1
1
1
2
2
2
2
The integrated profit function across supplier and all buyers:
1
1
One must now be aware of the fact that
solution from the above function.
2
1
2
2
is a function of , so we cannot yet derive the optimal
Determination of
This is the cost (£) of a CVRP‐solution to deliver from the supplier to the buyers. It arguably consists of:
a. The cost for loading a vehicle at the supplier, (£/vehicle);
b. The cost for unloading goods at each buyer , (£/stop);
c. The cost for delivering with a fleet of vehicles.
/ vehicles on the road, the total cost of this is:
a. Since there are
b. Since there are stops to be made, the total cost of this is:
c.
The cost for delivering a fleet of vehicles are mainly arising from driver wages and fuel costs. The
total distance travelled in an optimal CVRP solution is approximately:
∗
2
0.57 √
2
0.57 √
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Integrated Logistics
With (km/hr) the average vehicle speed,
we can hence specify the travel cost as:
∗
(£/hr) the cost for a driver, and (£/km) the fuel cost,
2
0.57
√
We hence find:
2
0.57
√
Substition of this result into the profit function:
1
2
2
∑
0.57
2
√
0.57
1
√
2
1
2
After algebraic manipulation this gives:
2
1
∑
2
0.57
0.57
√
1
√
2
1
2
2
And therefore:
∗
2
∑
1
0.57
∗
√
2
∗
Note that this solution will not be found using classic inventory theory. The reason is that in the classic
inventory theory, the opportunity cost of the set‐up cost would not be accounted for, and this proves
in the above derivation of crucial importance to retrieve the correct holding costs in this model, which
are:
2
1
.
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Integrated Logistics
Example
Consider an integrated supply chain of a producer of sports articles and a set of retailers, with the
following characteristics:
The production cost of sports articles is 1000 £/m3
Production rate is 20,000 m3/year and set‐up cost is £500/production run
There are 45 retailer located across the service area of size 80000 km2
Annual demand at each retail shop is on average 300 m3/year
The cost of a stop at a retailer is on average £25
The cost of dispatching a vehicle (loading at the central depot) is £40
All vehicles have a capacity of 20 m3
The total variable transport cost (from fuel and driver wages) is 0.5 £/km
Average distance between the production facility and a retailer is 160 km
The value of sports articles at consumer sales prices is 1500 £/m3
The opportunity cost of capital is 20%.
In an optimal distribution pattern for the integrated supply chain, what would be, approximately:
a) The number of times per year a retail shop will be supplied?
b) The shipment size per retailer delivery and the number of visits one vehicle will make on one round‐
trip?
c) The yearly net profit?
d) The total logistics cost of distribution per year?
Do not only provide a numerical answer, but show which analytical formulas you have used, and state the
assumptions on which they are based.
Solution
We adopt the model developed in LN p.57. (LN: Lecture Notes.) Hence the main assumptions are:
1. Use of NPV profit functions with Anchor Point at time of delivery of first batch to retailers
2. Conventional and symmetric payment structures throughout
3. Production model: lot‐for‐lot as in Monahan (1984) with one lot corresponding to the total freight
load over all vehicles used in one delivery round
4. A delivery round as the optimal solution to a CVRP with vehicles loaded at full capacity, in every
delivery each retailer is visited
5. A continuous approximation model as in Daganzo (1984) to estimate the cost of the CVRP
solution, for retail locations approximately uniformly distributed over the service region
6. Lead‐time in transit is not in the model since it is assumed small compared to average time a
product spends waiting at the production site and at the retailer before it is sold to customers
(needs to be checked afterwards – see note on p. 60) and then its impact on holding costs can be
ignored
7. Each retailer has an EOQ model as in Harris (1913)
a) We first determine the time between two delivery rounds from:
∗
2
∑
1
0.57
√
2
∗
2 500
0.2 1,000 1
1,125 0.57 0.5 √3,600,000
13,500
40 2 160
0.5 13,500
20,000
20
20
0.03083
11.3
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Integrated Logistics
32.4
/
Therefore each retailer is visited about
.
b) It will depend on each retailer’s annual demand rate since:
∗
∗
The expected quantity delivered per stop, for an average retailer, is
∗
300 0.0308
9.23 /
The number of stops each vehicle will make depends on the annual demand rate of the set of
retailers it visits, as this determines the lot‐size delivered to each. A rough estimate, assuming all retailers
have a demand rate close to the average:
20
2.17
/
9.23
c) The linear approximation of the AS profit function for the integrated supply chain is as on LN p. 57,
and is an approximation of the net profit per year:
2
1
∑
2
0.57
0.57
√
1
√
2
1
2
2
and a numerical value is found from evaluating the above equation using the optimal distribution
cycle time ∗ . This is left to the reader.
Note that the value of is an unspecified arbitrary constant, so you can only find a numerical result
for the equation in between the square brackets. It is, however, possible to determine the smallest
feasible value for :
∗
.
,
0.208
76
And therefore, if your evaluation of the equation within the square brackets leads to a numerical
value of , the annual profits can’t be larger than
£0.96 /
.
d) The logistics costs for the integrated supply chain are found from subtracting from the result in part
c) the annual marginal profit:
Since
1,500 £/
, the annual marginal profit is equal to:
1,500
1,000 45 300
6,750,000 £/
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Integrated Logistics
The numerical evaluation is left to the reader. If this finds a logistics cost of £/
as in part c) above.
corrected with factor
, it should also be
Note
Is it reasonable to neglect the impact of the lead‐time in transit on holding costs in the above example?
Lead‐time in transit can be estimated from three components:
1. An estimate for the time of loading a vehicle at the depot. At £40 dispatch costs, assuming driver
is paid at full cost of around £20/hour, it will take no longer than 2 hours.
2. The total distance travelled as in LN p.56:
∗
2
0.57 √
2
0.57 √
We need to estimate, however, the time it takes for one single vehicle as this is what counts for every
single product’s time on the road.
Since there are
vehicles on the road, the average time spend on the line‐haul part of a single
160 . At a conservative speed of
50 /
this part takes 3.2 hours.
vehicle route is
The average local distance travelled by one single vehicle is the total local distance travelled,
0.57 √
, divided by the number of vehicles,
:
0.57√
0.57√3,600,000
52.1
45 9.23
20
At an average speed (on city roads with some congestion) of
35 /
, this will take about
1.5
. The average product will spend about half this time in the vehicle.
3. An estimate for the time of unloading at each retailer. At £25 for unloading, driver wages of
£20/hour plus one extra person at retail shop at same rate, this wil take about 38 minutes
0.6
.
In total, we arrive at an approximation of the time a product spends in transit by summation of the
above results:
1.5
0.6 6.55
2 3.2
2
The average cycle time is 11.3
(for a product, half of this time is spend in the production site,
and half of this time is spend at the retailer). So the assumption 6 on p. 58 is reasonable.
Homework
Consider the data of the example on p. 58. Now consider an individual customer located at a distance
from the production site. Use a direct shipping ETQ model (LN. p 41) to anwser the following
questions:
a) In an integrated supply chain of production site, 3PL, and this retailer only, what is the optimal
delivery quantity to this retailer as a function of ? (Hints: (1) Insert the numerical values of all
known parameters in an appropriate formula; (2) be careful to not confuse distance with lead‐time,
i.e.
.)
b) Which of the following two statements is correct, and why: (X) If the distance is smaller than some
critical value then it is optimal to not fill the vehicle at full capacity, i.e. the optimal delivery
quantity is lower than the vehicle capacity; or (Y) If the distance is larger than then the optimal
delivery quantity is lower than the vehicle capacity? (Hint: Use the result obtained from answering
part a) to determine a numerical value for at which point it becomes optimal to fill the vehicle up
to capacity.)
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Integrated Logistics
4 Strategy in the supply chain: Alliances
versus Leaders
4.1 Alliances
An alliance is a partnership between two or more companies in which both the risks and rewards of the
cooperation are shared. Alliances may be formed when it is thought to lead to long term strategic
benefits for all partners involved.
One problem with alliances is to reach an agreement about how exactly the risks and rewards are
shared. Such an agreement should be (considered) “fair” by each partner, otherwise the alliance might
break down. However, it is difficult to define “fairness”: there are many ways in which fairness can be
defined in a seemingly objective way. Therefore a single universal method to devise such a fair
agreement does not seem to exist.
In this section, we discuss two solution concepts to determine a way for sharing risks or rewards
among partners: the core and the Shapley value. Before we do this, we need to introduce some concepts
from cooperative game theory.
4.1.1
Cooperative game theory
Source: WINSTON, L.W. 1994. OPERATIONS RESEARCH: APPLICATIONS AND ALGORITHMS (3RD ED.).
DUXBURY PRESS, CALIFORNIA. ISBN 0534209718. (CHAPTER 15)
Let there be a set of n players N ={1, 2, …, n}. A game between these players is specified by its
characteristic function. The characteristic function v(S), for every subset S N, is defined as the total
reward that the members of S can be sure of receiving if they act together and form a coalition.
Example 1: The Drug Game
Joe Willie has invented a new drug. Joe cannot manufacture the drug
himself, but he can sell the drug’s formula to company 2 or company 3. The
lucky company will split a $1 mio profit with Joe Willie. Find the
characteristic function of this game.
Solution
Letting Joe Willie be player 1, company 2 be player 2, and company 3 be
player 3:
v({})=v({1})= v({2})= v({3})= v({2, 3})=$0
v({1, 2}) = v({1, 3})= v({1, 2, 3}) = $1000000
Example 2: The Garbage Game
Each of 4 property owners has one bag of garbage and must dump it on
somebody’s property. If b bags of garbage are dumped on the coalition of
property owners, the coalition receives a reward of –b. Find the
characteristic function of this game.
Solution
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Integrated Logistics
The best that the members of any coalition S can do is to dump all of their
garbage on the property of owners who are not in S. With |S| denoting the
number of players in S, the characteristic function
v({S})= - 4 + |S|
v({S})= - 4
(if |S| < 4)
(if |S| = 4)
Example 3: The Land Development Game
Player 1 owns a piece of land and values the land at $10000. Player 2 is a
subdivider who can develop the land and increase its worth to $20000.
Player 3 is a subdivider who can develop the land and increase its worth to
$30000. There are no other prospective buyers. Find the characteristic
function for this game.
Solution
Any coalition that does not contain player 1, who owns the land, has a
worth or value of $0. Any other coalition has a value equal to the maximum
value that a member of a coalition places on the piece of land:
v({})= v({2})= v({3})= v({2, 3})=$0
v({1}) = $10000; v({1, 2})=$20000; v({1, 3})=$30000
v({1, 2, 3})=$30000
Consider any two subsets A and B of N such that A and B have no players in common (A B = ). In
many games, the following inequality holds:
v({A, B}) v({A}) + v({B}).
This property of the characteristic function is called superaddivitity. It is often a valid assumption,
because if the players in A B form a coalition, one of their options (but not necessarily their only
option) is to let players A fend for themselves and let players B fend for themselves. This would result in
the coalition receiving an amount v({A}) + v({B}). Thus v({A, B}) must at least be as large as v({A}) + v({B}).
Assume that we know the characteristic function. The problem now is to find a reasonable and
acceptable way to divide the reward v(N) between the n partners. Often, the solution concept is build
around finding a reward vector x = { x1, x2, …, xn}, in which xi specifies the reward for player i of the
coalition. The minimum conditions for x to be reasonable are the following:
(Group Rationality)
v(N) = x1 + x2 + …+ xn
(Individual Rationality)
xi v({i}) (for each i N)
Group Rationality says that the sum of individual rewards given to each player need to sum up to the
total value that can be attained by the supercoalition consisting of all players. Individual Rationality
means that each player needs to be at least as best of as when not participating at all within any
coalition, because otherwise this player will indeed play on its own and earn v({i}). If these two conditions
hold, the reward vector is also called an imputation.
Example 4
Let there be two players, N = {1, 2}, with the following characteristic
function:
v({1}) = 100, v({2}) = -50, and v({1,2}) = 70.
Questions
a) Does superadditivity hold?
b) Find an imputation.
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Integrated Logistics
Solution
a) Yes because v({1,2}) v({1}) + v({2}) since 70 100 -50
b) For x = { x1, x2} to be an imputation, we need
1) group rationality i.e. x1 + x2 = 70
2) individual rationality i.e.
x1 100 and x2 -50
An imputation therefore is e.g. x1 = 110 and x2 = 70 - x1 = - 40
Example 3: The Land Development Game
x
($10000, $10000, $10000)
($5000, $20000, $5000)
($10000, $20000, $0)
($10000, $0, $20000)
($30000, $0, $0)
($12000, $19000, -$1000)
($11000, $11000, $11000)
Is x an imputation?
Yes
No, x1 < v({1})
Yes
Yes
Yes
No, x3 < v({3})
No, v(N) < x1 + x2 + x3
A solution concept for an n‐person game is the specification of how to choose some subset of the set
of imputations (possibly empty) as the solution to the game. We discuss two solution concepts: the core
and the Shapley value.
Core
The core of the game is the set of all imputations which are “undominated”. A more practical definition
of the core follows:
An imputation x is in the core if and only if:
S N : xi v S
iS
The core is the set of imputations for which also “Subgroup Rationality” holds. Taking a reward vector
from the core will avoid the situation in which some players would like to form a subgroup and fend for
themselves because they could be better off by excluding some other players.
Example 3: The Land Development Game
x
($10000, $10000, $10000)
Is x an imputation?
Yes
($10000, $20000, $0)
Yes
($10000, $0, $20000)
($30000, $0, $0)
Yes
Yes
Is x in the core?
No, because {1, 3} could
do better by themselves
No, because {1, 3} could
do better by themselves
Yes
Yes
Example 1: The Drug Game
Find the core of the game.
Solution
For this game, x = { x1, x2, x3} will be an imputation if and only if:
x1
x2
x3
x1
+
0
0
0
x2 + x3 = 1000000
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Integrated Logistics
For this game, x = { x1, x2, x3} will be in the core if and only if in
addition:
x1
x1
x2
x1
+
+
+
+
x2
x3
x3
x2
+
1000000
1000000
0
x3 1000000
The unique solution is ($1000000, 0, 0). The core emphasises the importance
of player 1. The core contains a single solution.
Remark: Consider the game with only two players and show that the core of
the game now puts less emphasis on player 1.
Example 2: The Garbage Game
Find the core of the game.
Solution
Requirements for imputation:
x1 -3
x2 -3
x3 -3
x4 -3
x1 + x2 + x3
In addition:
x1 + x2 -2
x1 + x3 -2
x1 + x4 -2
x2 + x3 -2
x2 + x4 -2
x3 + x4 -2
x1 + x2 + x3
x1 + x2 + x4
x1 + x3 + x4
x2 + x3 + x4
+ x4 = -4
-1
-1
-1
-1
The system of equalities has no solution. Indeed, note that the last four
inequalities can be added together, which gives:
3(x1 + x2 + x3 + x4) -4
But this can never holds since in order to be an imputation,
x4 = -4. Therefore the core of this game is empty!
x1 + x2 + x3 +
Remark: Consider the game with only two players and show that the core is
not empty and equal to (-1, -1) !
Example 3: The Land Development Game
Find the core of the game.
Solution
For this game, x = { x1, x2, x3} will be an imputation if and only if:
x1
x2
x3
x1
+
$10000
$0
$0
x2 + x3 = $30000
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Integrated Logistics
For this game, x = { x1, x2, x3} will be in the core if and only if in
addition:
x1
x1
x2
x1
+
+
+
+
x2
x3
x3
x2
+
$20000
$30000
$0
x3 $30000
The solution is the set of imputations ($20000 x1 $30000, $0, $(30000x1)). The core contains an infinite number of solutions.
In conclusion, the core can be empty, contain one unique solution, or have an infinite number of
solutions. As a result, the core leaves still room on the negotiation table.
Shapley Value
The Shapley value given, in contrast to the core, always a unique reward vector x = { x1, x2, …, xn} where:
xi
pn (S)v S {i } v S
all S for which
i is not in S
where
pn S
S ! n S 1!
n!
and |S| is the number of players in S, and for n 1, n! = n (n‐1) … 2 (1)
Although the equation seems complex, it has a simple interpretation. Suppose that players 1, 2, …, n
arrive in random order. Then, any of the n! permutations of 1, 2, …, n has a 1/n! chance of being the
order in which the players arrive.
Suppose that when player i arrives, he of she finds that the players in the set S have already arrived. If
player i forms a coalition with the players who are present when he arrives, player i adds
v S {i } v S to the coalition S. It can be shown that the probability that when player i arrives the
players in the coalition S are present is pn S . The Shapley value implies that player’s i reward should be
the expected amount that player i adds to the coalition made up of players who are present when he or
she arrives.
Example 1: The Drug Game
Find the Shapley value.
Solution
Player 1. First, list all coalitions S for which player 1 is not a member.
For each of these coalitions, we compute v S {i } v S and pn S .
S
pn S
vS {i} vS
{ }
2/6
$0
{2}
1/6
$1000000
{3}
1/6
$1000000
{2, 3}
2/6
$1000000
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Integrated Logistics
Since player 1 adds on average
(2/6)(0)+(1/6)($1000000)+(2/6)($1000000)+(1/6)($1000000)=$4000000/6
the Shapley value concept recommends that player 1 receive a reward of
$4000000/6.
Player 2. First, list all coalitions S for which player 2 is not a member.
For each of these coalitions, we compute vS {i} vS and pn S
vS {i} vS
pn S
S
{ }
2/6
$0
{1}
1/6
$1000000
{3}
1/6
$0
{1, 3}
2/6
$0
The Shapley value concept recommends that player 2 receive a reward of
$1000000/6.
Since v({1, 2, 3}) = $1000000, Player 3 will receive a reward x3 =
$10000000 – x1 – x2=$1000000/6.
Note that, compared to the core in which all the money went to player 1,
the Shapley value gives a more equitable solution.
Alternative solution method for Shapley Value ***RECOMMENDED**
For games with a few players, it may be easier to compute each player’s Shapley value by using the fact
that player i should receive the expected amount that he or she adds to the coalition present when he or
she arrives.
For a three player game, the general approach is to fill in this table:
Order
of
Cost Added by Player’s Arrival
Arrival
Player 1
Player 2
Player 3
1,2,3
v({1})
v({1,2})-v({1})
v({1,2,3}-v({1,2})
1,3,2
v({1})
v({1,2,3}-v({1, 3})
v({1,3})-v({1})
2,1,3
v({1,2})-v({2})
v({2})
v({1,2,3}-v({1,2})
2,3,1
v({1,2,3})-v({2,3})
v({2})
v({2,3})-v({2})
3,1,2
v({1,3})-v({3})
v({1,2,3})-v({1,3})
v({3})
3,2,1
v({1,2,3})-v({2,3})
v({2,3})-v({3})
v({3})
The value assigned to each player is now the sum of its column divided by the number of rows.
Example 1: The Drug Game
Find the Shapley value.
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Integrated Logistics
Order
Amount Added by Player’s Arrival
of Arrival
Player 1
Player 2
Player 3
1,2,3
$0
$1000000
$0
1,3,2
$0
$0
$1000000
2,1,3
$1000000
$0
$0
2,3,1
$1000000
$0
$0
3,1,2
$1000000
$0
$0
3,2,1
$1000000
$0
$0
Since each of the six orderings of arrivals are equally likely, the
expected amount added by each player is the average of its column.
Example 5: Runway
Suppose three types of planes (Piper Cubs, DC-10s, and 707s) use an
airport. A Piper Cub requires a 100-yd runway, a DC-10 requires a 250-yd
runway, and a 707 requires a 400-yd runway. Suppose the cost (in $) of
maintaining a runway for one year is equal to the length of the runway.
Since 707s land at the airport, the airport will have a 400-yd runway. How
much of the $400 annual maintenance cost should be charged to each plane?
Solution
Let player 1 = Piper Cub, player 2 = DC-10, player 3 = 707. We can now
define a 3-player game in which the value to a coalition is the cost
associated with the runway length needed to service thee largest plane in
the coalition. The characteristic function therefore is:
v({})=$0
v({1})=-$100
v({2})= v({1, 2})= -$150
v({3})= v({1, 3})= v({2, 3})= v({1, 2, 3})=-$400
Order
of
Cost Added by Player’s Arrival
Arrival
Player 1
Player 2
Player 3
1,2,3
$100
$50
$250
1,3,2
$100
$0
$300
2,1,3
$0
$150
$250
2,3,1
$0
$150
$250
3,1,2
$0
$0
$400
3,2,1
$0
$0
$400
The Shapley value concept suggests that Piper Cub pay $33.33, the DC-10
pay $58.33, and the 707 pay $308.33.
Note. In general, it has been shown that for more than one plane of each
type, the Shapley value for the airport problem allocates runway operating
costs as follows: all planes that use a portion of the runway should divide
equally the cost of that portion of the runway. Thus, all planes should
cover the cost of the first 100 yard of runway, the DC-10s and 707s should
pay for the next 150-100=50 yard of runway, and the 707 should pay for the
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Integrated Logistics
last 400-150 = 250 yard runway. If there were ten Piper Cub landings, five
DC-10 landings, and two 707 landings, the Shapley value concept would
recommend that each Piper Cub pay 100/(10+5+2) = $5.88 in landing fees,
each DC-10 pay $5.88+50/(5+2) = $13.03, and each 707 pay $13.03 + 250/2 =
$138.03.
Exercise 1
Three companies consider a strategic alliance which has the following characteristic cost function: v(1)
=10, v(2) = 20, v(3) = 30, v(1,2) = 25, v(1,3) = 32, v(2,3) = 45, v(1,2,3) = 50.
a. How much of the total cost would each company carry when they would adopt the Shapley value
method?
b. Check the collaboration for stability – are there any companies or subcoalitions who might wish to
break from the grand coalition and why (not)?
Solution
a.
arrival
123
132
213
231
312
321
cost
Player 1
10
10
5
5
2
5
6.17
Player 2
15
18
20
20
18
15
17.67
Player 3
25
22
25
25
30
30
26.17
b. Individual rationality = ok; Subgroup rationality: v(1.2) > 6.17 + 17.67 = 23.84; v(1,3) < 6.17 + 26.17 =
32.34 (slightly); v(2,3)> 17.67 + 26.17 = 43,.. Subgroup (1,3) causes instability and might break‐off (a small
incentive only).
Exercise 2
Three companies (A, B, and C) have their depot in the harbour of Rotterdam, from which they supply
their customers in the Netherlands (an area of 10 km2): Company A has 300 customers, Company B 200
customers, and Company C 100 customers. All three companies use similar vehicles of 10‐ton capacity.
The average demand for company A is 0.5 ton/customer, for B 1 ton/customer, and for C 0.5
ton/customer. Assume that all customers are uniformly distributed over the area and that the average
distance from the depot to a customer is 220 km. We ignore the cost of loading and unloading and use a
cost of £0.5/km for the transport cost.
a) Estimate the distance travelled for delivery to all customers if company A, B, and C do not cooperate
with each other.
b) Estimate the distance travelled when company A, B, and C would cooperate and share the same
vehicles for distributing to their customers.
c) Use the Shapley value to allocate the total transport cost under cooperation to each of the
companies A, B, and C.
Solution
a) Using Daganzo’s approximation formula for a CVRP solution (LN p. 53):
∗
2
√
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Integrated Logistics
For A:
∗
2 220
For B:
∗
2 220
For B:
∗
2 220
300 0.5
10
200 1
10
100 0.5
10
0.57 10 300
0.57 10 200
0.57 10 100
b) For {A, B, C} (calculate the average load in the linehaul distance term):
3 0.5
2 1
1 0.5
600
6
∗
2 220
0.57 10 600
, ,
10
c) Transport cost for coalition is 0.5 ∗ . The characteristic function value for grand coalition can be
found from part b), and for the individual firms from part a). For the subcoalitions, apply the same
approach:
3 0.5
2 1
500
5
∗
2 220
0.57 10 500
,
10
3 0.5
1 0.5
400
4
∗
2 220
0.57 10 400
,
10
2 1
1 0.5
300
3
∗
2 220
0.57 10 300
,
10
Work out the above formulas to find the numerical values of the costs for each (sub)coalition. Then fill
in the table for the Shapley value as in the above examples. This is left to the reader.
Homework
The net benefit from cooperation between players can be defined as:
1, 2, . . ,
<for cost games, use negative values for . >
Prove that for any cooperative game with two players, the Shapley value will assign half of the net
benefit obtained from cooperation to each player.
(Hint: adapt the table format developed for the three player situation on LN p. 66 to a two‐player
situation and then show that each player gets its original value
plus half the net benefit.)
<Note. This is in general not true for games with more than two players.>
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Integrated Logistics
Exercise 3
Three companies A, B, and C wish to explore how they could benefit from sharing rented warehouse
space. It costs £1 per m2 of floor space and per month if rented on a yearly basis, while £1.5 per m2 and
per month if rented on a monthly basis. Table 1 lists monthly space requirements for each of the three
companies.
Table 1: Warehouse space requirements (m2 floor space needed per month)
Month
A
B
C
Total
January
100
50
10
160
February
100
50
10
160
March
100
50
10
160
April
100
50
10
160
May
50
100
10
160
June
50
100
10
160
July
50
100
10
160
August
50
100
10
160
September
0
0
200
200
October
0
0
200
200
November
0
0
200
200
December
0
0
200
200
Total
600
600
880
2080
a) Find the characteristic function.
b) Using the Shapley value, determine how these companies could split the costs of renting warehouse
space together as one virtual company.
c) Is the coalition stable, or will some companies or subcoalitions try to break off from this coalition?
Justify your answer.
Solution
a) Let A=1, B=2, C=3.
Method: If company 1 rents for one year is has to rent 100 m2 for 12 months at 1£/m2, thus costing
£1200 per year. If it rents per month, it would rent 100m2 for 4 months, and 50m2 for another 4 months
at 1.5£/m2, thus paying £900 per year. Thus v(1) = min{1200,900} = 900.
Similarly:
v(2) = v(1) = 900,
v(3) = min{2400,1320}=1320,
v(1,2) = min{1800,1800}=1800,
v(1,3)=min{2400,2220}=2220,
v(2,3)=v(1,3)=2220,
v(1,2,3)=min{2400,3120}=2400.
b) Shapley value table:
arrival
A=1
B=2
C=3
123
900
900
600
132
900
180
1320
213
900
900
600
231
180
900
1320
312
900
180
1320
321
180
900
1320
cost
660
660
1080
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Integrated Logistics
c) Checking stability means checking whether each subcoalition needs to pay less than if it would go by
itself. Thus: v(1)>=600, v(2)>=660, v(3)>= 1080, v(1,3)>=660+1080, v(1,2)>=660+660,
v(2,3)>=660+1080. Therefore, the coalition is stable.
(The three companies should rent together a space of 200m2 for a whole year. The extra capacity in
the first 8 months means they also have the flexibility to use more space if needed.)
4.1.2
Alternative methods
There exists a wealth of other methods in the literature besides the core and Shapley value approaches.
A quite popular approach in practise is the so‐called proportional allocation method. This approach aims
to let each partner in the cooperation pay costs that are proportional to this partner’s use of the shared
resources, while making sure that all the cost in the system are allocated. This seems very intuitive but
may cause problems of the coalition being unstable. Indeed, the method does not give any assurance
that partners working by themselves or with only a few of the partners in the coalition are better off
using the cooperative system. If partners break away from the coalition it is not a good thing because, for
super‐additive games, the grand coalition can achieve globally more efficient outcomes.
There exist also a variety of methods that are based on cooperative game theory but use other
mechanisms to find the distribution of the rewards of cooperation. Most of these techniques, such as e.g.
the nucleolus, are however more mathematically challenging that the Shapley value.
Finally, there is a specific strand of research that investigates the efficiency‐fairness trade‐off, see e.g.
Bertsimas et al. (2012) article “On the efficiency‐fairness trade‐off” published in Management Science 58
(12).
4.2 Two‐party alliances in the supply chain
In order to apply the cooperative game theory framework for a situation of a supplier and its buyer ,
we will need to find the characteristic functions of the following coalitional sets:
;
;
,
What makes this situation somewhat special compared to the previous examples in this chapter is that
these parties, even when `working alone’, remain in a relationship in which they exchange physical
goods and cash with each other. Even if these parties would act `independently’, their actions can’t be
completely independent and at least one of them is going to have to consider which actions are
undertaken by the other party. Care should hence be taken when defining the characteristic function
value of a player acting alone how we specify the assumptions of this working relationship and the
order of decision making events in such a `status quo’ scenario. The following example illustrates.
4.2.1
Two‐party alliance: example
We consider an EOQ‐type buyer and its supplier as in the model of Goyal (1976). We will use the profit
functions of the players under consignment arrangements, functions we have developed for this case in
Section 2.11. In addition, we incorporate a material handling cost per unit of product as in Section 2.13
for the buyer, paid upon receiving a batch to an outside third party (i.e. not paid to the supplier). We also
assume a maximum capacity on the warehouse at the buyer, .
2
1
2
2
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Integrated Logistics
1
2
1
2
2
2
.
2
2
Status‐quo scenario
To determine
and
, we assume that the buyer, given a payment structure with the supplier,
∗
|
. The supplier will then determine the
that maximises
will first determine the lot size
∗
∗
∗ ∗
|
.
that will maximise
,
lot size
We find: If
1 2
0, then the unconstrained optimal lot‐size is
2
∗∗
1
else
∗∗
→ ∞. Therefore,
∗
∗∗
min
,
2
, and:
∗|
For the supplier, given ∗ , the problem of identifying the optimal value for is as was previously derived
in the exercise of Section 2.7:
1
1
2
∗
8
1
∗
and hence:
∗ | ∗
,
Note that the total profits made in this supply chain under the status‐quo scenario are:
∗|
∗ | ∗
∗
,
, ∗|
∗
∗
2
∗
∗
2
Alliance
The value for
, is found from considering the alliance between buyer and supplier in which the
, |
.
values of and are simultaneously derived as to maximise
We start by following the approach of exercise 1 in Section 2.12:
2
∗∗
but the capacity constraint implies that
∗
min
∗∗
,
.
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Integrated Logistics
Case 1: The hypothesis is that
∗
∗∗
, |
. Substituting this result into
:
2
2
2
2
Hence, in the case that
0, ∗ 1. In the case that
Section 2.7. We seek an integer value such that:
The first condition:
0, we apply the technique presented first in
1
1
2
2
2
2
1
1
is simplified to:
2
2
1
1
We now take the square of both sides, and simplify to obtain:
1
1
Rearranging terms:
1
The procedure now follows the same logic as before in Section 2.7, and we find:
∗
1
1
2
1
4
It can now be tested whether the assumption made holds. If it does, then the profit in the integrated
supply chain is:
∗
,
, ∗ ∗ |
Case 2: The hypothesis is that ∗
.
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Integrated Logistics
2
2
The derivation of the optimal value for is very similar to the procedure applied in LN Section 2.7:
∗
1
1
2
8
1
And:
∗
,
Shapley value
The net benefit of cooperation is
∗
∗
, ∗|
, ∗|
∆
,
In a two‐player cooperative game, the Shapley value will assign half of the net benefit to each player (see
Homework LN p. 69).
,
, this means that the allocation should be:
With reward vector
∆
2
∆
2
,
Exercise 1
Consider the supply chain of an EOQ‐type buyer and its EOQ‐with‐batch‐demand‐type wholesaler. We
have the following characteristics:
Annual demand rate at buyer is 5,000 products/year;
Sales price charged by buyer to customers is £40/product;
Capacity of warehouse at the buyer is 800 products;
Wholesale price charged by wholesaler to buyer is £30/product;
Material handling cost at buyer is £4/product;
Set‐up cost at buyer per lot‐size received from wholesaler is £100/order;
Set‐up cost for wholesaler to deliver a lot‐size to buyer is £150/order;
Acquisition price charged to wholesaler by its supplier is £20/product;
Set‐up cost for wholesaler to receive a batch from its supplier is £1000/order.
Opportunity cost of capital is 20%.
Determine the value of an alliance for supplier and buyer, and use the Shapley value to find the profits
each of the parties should earn under the alliance. Do this for the following two cases:
A. In a status‐quo scenario where the payment structure between buyer and supplier is conventional;
B. In a status‐quo scenario where the payment structure between buyer and supplier is full
consignment.
In each status‐quo scenario, the buyer first determines its own optimal pattern of ordering from the
wholesaler, and knowing this, the wholesaler subsequently determines its own optimal ordering pattern
from its supplier.
74
Integrated Logistics
Do not only provide a numerical answer, but show which analytical formulas you have used, and state the
assumptions on which they are based.
Solution
We adopt the model developed in LN p. 71‐73. Hence the main assumptions are:
1. <this is left to the reader…>
2. <this is left to the reader…>
3. …
A. Conventional payment structure.
The buyer has to pay the wholesaler the full price when a product is delivered. This means that, in the
scheme as developed in LN Section 2.11, that
, and
0.
Since
1 2
30 4 0
we have for the buyer:
∗
2
min
1
,
2
∗
2 100 5,000
, 800
0.2 30 4
min
min 383.5, 800
383.5
/
This gives:
∗|
383.5|0.2
∗
∗
4 5,000
100
2
40
30
30,000
1,303.8
1
5,000
383.5
10
0.2
100
2
1,303.9
2
0.2 30
£27,382.3/
2
4
383.5
2
For the wholesaler:
∗
1
1
2
8
1
∗
1
1
2
1
1
2
8 1,000 5,000
0.2 20 383.5
1
√1
67.995
4.65
4
Therefore:
∗ | ∗
9|383.5,0.2
,
∗
∗
2
∗
1
∗
2
1
2
∗
2
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Integrated Logistics
30
20 5,000
1,000
9
0.2 30
5,000
383.5
383.5
20
2
150
0.2
1,000 150
2
0.2 20 9
1
383.5
2
50,000 3,404.32 115 6,136.0 383.5 £40,728.2/
For an alliance between buyer and supplier, we first test for which value of the buyer’s lot‐size is
unconstrained:
2
∗
800 ?
Numerical evaluation shows that for this to hold, it must be that
3. Since for
0, the value
using the unconstrained lot‐size for produces ∗ 1, and given that is small relative to the other
cost parameters in the model, we think it will be unlikely that we are in Case 1. Indeed, if we calculate the
value of for the unconstrained lot‐size we get:
1
1
2
∗
4
1
1
1
2
1
1
1
2
√1
4 1,000 4
100 150 20
3.2
1.52
We see that this would produce an unconstrained value
800:
Case 2, with ∗
1
1
2
∗
∗
1
1
1613.7 ≫ . We hence calculate under
8
1
1
1
2
8 1,000 5,000
0.2 20 800
1
2.54
2
,
∗
40
20
4 5,000
100
0.2 20 2
2
80,000
150
4 800
4,687.5
125
1,000 5,000
800
2
3,520
2
0.2
100
£71667.5/
150
2
∗
2
1,000
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Integrated Logistics
The net benefit from forming the alliance is hence:
∆
,
£71667.5 £40,728.2 £27,382.3 £3557/
The Shapley value will hence recommend that this net benefit is shared equally. Therefore the net profits
of buyer and supplier become, respectively:
£3,557
£27,382.3
£29,160.8/
2
£3,557
£40,728.2
£42,506.7/
2
The benefits of forming the alliance in terms of profit increase achieved might not seem much when
compared to the overall profit figures of the firms in status‐quo. However, note that this benefit has been
realised solely by adopting a logistics strategy that works better for the integrated supply chain. It is
hence more insightful to compare this logistics benefit relative to the logistics costs of the firms in the
status‐quo scenario. This is what we do next.
The logistics costs can be found from subtracting the marginal profit term from a firm’s profit function.
For example, for the buyer the logistics costs in the status‐quo scenario are:
∗
£2,617.7/
The relative gain is then:
3,557
2
100%
67.9%,
2,617.7
and thus the buyer saves as a result 67.9% on its logistics costs. Similarly, we find that the logistics costs
for the wholesales in the status‐quo scenario are £9,271.8/
, and then:
3,557
2
19.2%,
100%
9,271.8
or the wholesaler saves 19.2% on its logistics costs.
B. Full consignment.
The buyer has to pay the wholesaler the full price only the moment that a product is sold to its
customers. This means that, in the scheme as developed in LN Section 2.11, that
0 and
.
Since
1 2
4 0
we have for the buyer:
∗
min
2
1
2
,
∗
min
2 100 5,000
, 800
0.2 4
min 1118.0, 800
800
/
This gives:
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Integrated Logistics
∗|
383.5|0.2
∗
∗
2
40
30
4 5,000
100
30,000
625
1
10
5,000
800
0.2
320
100
2
2
2
0.2 4
£29,045.0/
800
2
For the wholesaler:
1
1
2
∗
8
1
∗
1
1
2
8 1,000 5,000
0.2 20 800
1
2.53
2
Therefore:
∗ | ∗
9|383.5,0.2
,
∗
30
1,000
2
20 5,000
0.2
20
∗
∗
2
150
800
2
5,000
800
0.2
1
∗
2
1,000 150
2
1
0.2 20 2
2
∗
2
1
800
2
50,000 4,062.5 115 1,600 1,600 £42,622.5/
For an alliance between buyer and supplier, we note that the payment structure adopted between them
does not influence the integrated supply chain profit function. We therefore find the same optimal result
800 and ∗ 2 and gives a total supply chain
as in Scenario A. We recall that this also takes ∗
profit of £71667.5/
. The net benefit of an alliance would hence be:
£71667.5 £42,622.5 £29,045.0 £0/
In other words, if the firms have adopted the full consignment scheme in status‐quo, they derive no
benefit from forming an alliance.
Note 1. Under a conventional payment structure, both firms do worse in status‐quo that if they would
use a full consignment arrangement in status‐quo. If you compare what the firms would get when
adopting the alliance under a conventional payment structure, then the buyer would gain a bit more than
working under the full consignment scheme in status‐quo, and also the supplier would gain relatively less
than what he could get using a full consignment arrangement in status‐quo.
In conclusion, starting from a status‐quo scenario of conventional payments, the buyer would prefer to
form an alliance and use the Shapley value criterion to distribute the net benefits of the alliance, whereas
78
Integrated Logistics
the wholesaler would rather want to propose the adoption of a full consignment payment structure,
without forming an alliance.
Note 2. From a practical point of view, changing the payment structure from conventional to full
consignment is much easier than implementing the alliance and using the Shapley value.
In the alliance option, someone has to calculate the optimal scenario of the supply chain first, which
requires a party having knowledge of the integrated supply chain function. This party then needs to
enforce the buyer and wholesaler to select their decision variables and accordingly. This may be
difficult to achieve if the logistics managers in both firms are used to select the lot‐sizes that optimises
the profit functions of their own firms. It may seem to them that the decisions imposed as to arrive to the
optimal supply chain result is not what optimises their own functions. (Indeed it often does not, prior to
establishing the net benefits distribution.) This may cause internal friction in the firms between general
management and the logistics division. Then the problem arises of calculating the net benefit of the
supply chain and to redistribute these savings to both parties. In fact, the net benefit of £3,557/year is an
annuity stream value, which means that both parties in effect need to receive payments that are
equivalent to such a continuous cash flow throughout the year. One may wonder, rightfully, how this is
going to be established, since in practise there will not be a `central party’ that has such a source of
funding available…
Adopting the full consignment payment structure in status‐quo is, on the other hand, much more in line
with what will be common practise in the firms. The logistics manager in the buyer’s firm will keep
selecting the optimal lot‐size that maximises the profit for his firm, and then the logistics manager in the
wholesaler’s firm will use this to calculate the optimal value of that will maximise the wholesaler’s
profit function. Under full consignment, furthermore, both firms realise that this process of sequential
and independent local decision making, although not optimal in general, would in this case indeed also
be optimal for their integrated supply chain! There is furthermore no need for a central party to keep
track of net benefits and to redistribute these benefits between the parties, because under the full
consignment policy they will both fair better in a status‐quo scenario than when using the conventional
payment structure under a status‐quo scenario.
4.2.2
Two‐party alliance: example 2
Consider the supply chain of an EOQ‐type buyer and a lot‐for‐lot producer. We have the following
characteristics:
Annual demand rate at buyer is 5,000 products/year;
Sales price charged by buyer to customers is £40/product;
Capacity of warehouse at the buyer is 800 products;
Transfer price charged by producer to buyer is £30/product;
Material handling cost at buyer is £4/product;
Set‐up cost at buyer per lot‐size received from producer is £100/order;
Set‐up cost for producer to deliver a lot‐size to buyer is £0/order;
Production rate at producer is £10,000 products/year;
Production cost at producer is £20/product;
Set‐up cost for producer to initiate production is £500/run.
Opportunity cost of capital is 20%.
A. Determine the value of an alliance for producer and buyer, and use the Shapley value to find the
profits each of the parties should earn under the alliance. Do this in a case of a conventional payment
structure between buyer and producer.
B. Now assume that buyer and producer keep paying each other according to a conventional payment
structure when adopting the solution that maximises the value of the alliance. Compare the profits
each of these firms earn with your answer to part A.
79
Integrated Logistics
Do not only provide a numerical answer, but show which analytical formulas you have used, and state the
assumptions on which they are based.
Solution
We adopt the ETQ model developed in LN Section 3.1.2, but do not consider the 3PL in the middle. Hence
the main assumptions are:
1. <this is left to the reader…>
2. <this is left to the reader…>
3. …
A. Conventional payment structure.
2
2
1
2
2
2
2
1
2
2
where:
1
2
In status‐quo, the buyer will choose:
2
∗
2 100 5000
0.2 30 4
383.5
/
This gives:
∗
40
30
4 5,000
100
5,000
383.5
0.2
100
2
0.2 30
4
383.5
2
£27,382.3/
∗
1
∗
2
2
∗
2
∗
2
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Integrated Logistics
30
20 5,000
5,000
0
383.5
500
500 1
0.2
2
383.5
2
0.2 30
20
50,000
6,518.9
2
1
2
0
0.2 20
1 383.5
2 2
100
383.5
383.5
£43,381.1/
In the alliance, the lot‐size that maximises the integrated supply chain’s profit function is:
2
∗
2 100
1
500
0.2 20 1
0 5,000
1
4
2
939.3
/
∗
,
∗
∗
40
20
4 5,000
600
5,000
939.3
1
2
0.2
1,100
2
0.2 20 1
1
2
2
4
939.3
2
80,000
3193.9
110
3193.9
£73,502.2/
The net benefit is therefore:
∆
73,502.2
43,381.1
27,382.3
£2,738.8/
The Shapley value awards the players the following profits when adopting the alliance:
2,738.8
27,382.3
£28,751.7/
2
2,738.8
43,381.1
£44,750.5/
2
B. Using the conventional payment structure when adopting ∗ , the buyer and producer will get,
respectively:
∗
∗
40
30
4 5,000
2
5,000
939.3
0.2
532.3
10
100
2
100
2
0.2 30
4
939.3
2
30,000
3193.6
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Integrated Logistics
£26264.1/
≪
≪
30
20 5,000
500
1
2
∗
5,000
939.3
∗
2
0.2
1,000
2
0.2 20
∗
2
1 939.3
2 2
0.2 30
20
2
939.3
2
50,000
2661.6
100
939.3
939.3
£47,238.4
≫ ≫
Note that the sum of profits earned by both players is still equal to the supply chain profit of £73,502.2/
.
However, the producer would reap much more of the benefits. The buyer would earn even less than its
earnings obtained in the status‐quo scenario, and therefore this solution violates the condition of
individual rationality for the buyer. It is very unlikely that this could be adopted (unless the supplier could
enforce it.)
4.3 Perfect coordination
We have seen how for number of firms in the supply chain we can calculate the logistics policy that
would be optimal for their integrated supply chain. We do this by deriving the integrated supply chain
profit function and use it to determine the optimal values of all decision variables simultaneously.
We have seen in previous section how we can distribute the net benefits of full cooperation across the
firms, based on cooperative game theory solution concepts such as the Shapley value. If this finds a
, ,…,
that satisfies the conditions of group rationality, individual
reward distribution vector
rationality, and subgroup rationality (i.e. if the reward vector is a solution in the core) then this reward
scheme is stable. In that case, no individual firm nor any subcoalition would do financially better by
breaking off from the grand coalition and acting on their own.
This is not always possible. There are games where the core is empty, and hence no stable reward vector
exists that can offer this stability. These situations can be handled if the firms in the coalition would be
able to enforce a contract whereby a firm which breaks off from the coalition would have to pay a fine. If
the fine is large enough, the net financial benefit of breaking off would become negative. However, we
do not further explore this here.
The two examples in the previous section illustrated that the implementation of a reward distribution
vector can be difficult in practise. The reason is that the firms in a supply chain, even if they wanted to
adopt the optimal solution for their supply chain, remain individual firms with their own identity. They
will likely not agree to set‐up an overarching entity that would coordinate their whole supply chain. This
entity would not only have to dictate the optimal logistical decisions to be taken by each firm, but will
have to oversee the financial flows exchanged between them so as to end up with the targeted reward
distribution vector that will end up distributing the net benefits fairly between the firms.
It would be much easier to find a solution whereby the firms will retain their individual decision making
power, but such that their individual decisions would lead to the logistics policy that is optimal for the
integrated supply chain. (One such approach was adopting a full consignment scheme for the situation in
Section 4.2.1.)
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Integrated Logistics
The following two definitions are adapted from Chen et al. (2001).
Coordination mechanism
If the firms in a supply chain maintain decentralised decision making, a coordination mechanism is an
approach by which the structure of costs and revenues of each individual firm are adapted so as to find a
better alignment of each firm’s decisions to the integrated supply chain’s objective function.
Perfect coordination
If the coordination mechanism achieves the result that the decentralised decisions also maximise the
integrated supply chain’s objective function, the coordination mechanism is said to be perfect.
4.3.1
Perfect coordination schemes: examples
We consider the situation of Section 4.2.2. The payment structure between buyer and producer is
conventional in the status quo scenario. Recall the profit functions:
2
2
1
2
2
2
2
1
1
2
2
2
Solution 1
In Chen et al. (2001) the coordination mechanism presented is based on:
1. The producer charging the buyer its set‐up cost whenever the buyer places an order
2. To modify the transfer price
We now explore how this work. In the status‐quo scenario, the buyer determines the lot‐size
maximising its profit function. The terms that are relevant for this are:
Π
2
∗
by
When we observe the supply chain profit function, then the buyer would arrive at the correct lot‐size
∗
that maximises the supply chain’s profit function if these terms were:
Π
1
Take the difference (as always, assume
Π
Π
2
:
2
1
2
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Integrated Logistics
For the buyer to arrive at the correct lot‐size ∗ ≡ ∗ , we can restructure the buyer’s profit function so
that the above difference becomes zero.
1. The producer charging his set‐up costs
every time the buyer places an order
Call the new profit function of the buyer Π . We see that this will produce the result that
Π .
will drop out of Π′
2. Adapt the transfer price :
After implementing step 1, the difference remaining is set to zero to calculate the value of :
Π′
Π
1
2
∗
1
0
2
Critical evaluation
If Solution 1 is adopted, what would be the profits of the firms? We keep assuming a conventional
payment structure between them.
In order to arrive at the correct profit functions of the firms, we need to make assumptions about when
the producer charges his set‐ups costs to the buyer. According to the conventional payment structure,
this would happen the moment an order is delivered.
After an analysis of the new cash‐flow functions of the players, and the derivation of linearised annuity
stream functions, we should find that the profit functions are now:
1
2
1
1
2
2
2
2
1
2
The supply chain profit function is the same as presented in Section 4.2.2. Recall that ∗ 939.3. (You
∗
. Hence:
can verify that ∗
∗
1
5,000
600
1
939.3
40 20 1
4 5,000
600
0.2
0.2 20 1
4
2
939.3
2
2
2
30,000 3,193.9 60 3,193.6
£23552.5/
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∗
20 1
1
2
500
1 939.3
0.2 20
2
2 2
939.3
20
2
20 5,000
0.2 20 1
1
2
0.2
50,000
50
939.3
939.3
£49950.0/
Recall from Section 4.2.2:
£27,382.3/
£43,381.1/
27,382.3
2,738.8
2
£28,751.7/
43,381.1
2,738.8
2
£44,750.5/
It can be observed that the buyer would not be happy to adopt this perfect coordination scheme since
under the status‐quo scenario he could make more profits. To achieve the equal distribution of net
profits when adopting this perfect coordination scheme, the supplier would still need to pay to the buyer
an amount of:
49950.0 44,750.5
£5199.2/
28,751.7 23552.5
Coordination mechanisms that require such so‐called side‐payments are not ideal because it is difficult to
arrange for such side‐payments to be implemented in the real world.
In conclusion, perfect coordination should ideally not only look to ensure that decentralised decision
making leads to the logistics policy that is optimal for the integrated supply chain, it should at the same
time ensure that all parties are automatically financially rewarded without the need to resort to any side‐
payments.
Solution 2
This solution is similar to Solution 1. The set‐up cost payments are the same as in Solution 1. However,
instead of changing the transfer price itself, we will change the payment structure between buyer and
supplier so that its impact on the holding cost term of the buyer has the same effect as to align it to that
in the supply chain profit function.
Recall from Section 2.11 that by the introduction of prices , , and parameter we can find profit
functions that reflect the impact of consignment arrangements on the firms profit functions. If we use
this scheme for this supply chain, the profit functions are as follows:
2
1
2
2
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Integrated Logistics
1
1
2
2
2
2
2
1
1
2
2
2
Therefore, an alternative implementation of Solution 1 is to keep the original price but select a
combination of
1 2 such that:
1
2
1
1
as the amount to be
This can be done in an infinite number of ways. For example, choose
be
paid by buyer to producer the moment a product is delivered, and let the remainder
paid when the buyer sells the product to its customers.
In general, such an implementation alters the profit functions of buyer and producer less because in their
marginal profit term we still retain the original price. This is likely to produce results requiring, in most
situations, less side‐payments. However, it can on itself not eliminate that side‐payments may be needed.
Note. When applying this scheme to the numerical example, we have in this case the same result simply
because
20 1
£30/
£30/
happens, by sheer coincidence, to be equal to
∗
1
.
Solution 3
This solution will adapt Solution 1 and also make use of Solution 2’s strategy to change the payment
structure.
We start by investigating what determines the optimal lot‐size ∗ . We realise this is given by the ratio
under the square root:
2
1
Consider a non‐zero constant , then we achieve the same ∗ if the ratio would be:
2
1
Property: When
0:
Maths refresher
An affine transformation over real numbers is a function
: → :
,
where and are constants.
is a function of that reaches its maximum in ∗ if and only if
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Integrated Logistics
is a function that reaches its maximum in ∗ .
Therefore, we will also achieve perfect coordination by making the difference Π′
Π equal to zero:
′
Π
′
1
0
Π′
2
2
This leads to the following coordination scheme:
1. The buyer pays the supplier for every order the set‐up cost at the time the order is delivered:
∗
∗
1
2. The buyer pays the producer per product upon delivery the price
∗
1
∗
1
,
the moment that the product is sold by the
and in addition, will pay the remainder
buyer to its customers.
Note that for
1, this scheme is exactly the same as Solution 1, safe for the fact that we change the
payment structure instead of changing the transfer price itself. We know from the critical analysis that
the buyer would be £5199.2/
short if implementing Solution 1.
3. Hence we determine the value of ∗ such that:
∗
1
5199.2
Critical evaluation
We work out the impact of implementing Solution 3.
After an analysis of the new cash‐flow functions of the players, and the derivation of linearised annuity
stream functions, we should find that the profit functions are now:
1
2
2
1
1
Recall that
∗
1
939.3. (You can verify that
40
30
4 5,000
600
∗
5,000
939.3
2
1
∗
2
2
2
2
. Hence:
0.2
600
2
0.2 20 1
1
2
4
939.3
2
30,000
3,193.9
60
3,193.6
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Where we require this to equal
£28,751.7/
20 5,000
0.2 20
. It follows:
∗
1248.3
6447.5
1
0.1936 600
We can check:
30
1 939.3
2 2
0.2 0.1936 20 1
0.1936
5,000
939.3
0.2 1
1
2
1
0.1936 4
48.2
50
939.3
20
0.1936
600
2
939.3
2
0.2
500
2
50,000
2575.6
1636.03
£44750.4
≡
Hence, this coordination scheme is perfect and will also provide the firms with the financial profits they
would receive when adopting the Shapley value distribution of net benefits from adopting the optimal
solution of their alliance. It does not require that firms give up their independent distributed decision
making processes, and it does not require side‐payments.
Note that we do not necessarily need to divide the net benefit equally between the firms – this is only a
requirement that comes from the application of the Shapley value to a two‐player game. By adapting
we can also find solutions whereby the net benefit is split in some pre‐specified but non‐equal split, e.g.
1/3 of ∆ goes to one player, and 2/3 of ∆ to the other player.
4.4 Leaders and followers
The previous sections have looked into the application of cooperative game theory in the supply chain.
This led to seeking coordination mechanisms that are perfect and require no additional side‐payments for
fairly distributing the net benefits when adopting the coordination mechanism.
In the real world, not every situation can be handled with what we have seen so far. There are situations
in which one of the parties has bargaining power. This leads this firm to take on the role of a leader in the
supply chain, and it will set out conditions (i.e. a contract) under which the other firms need to operate.
While leaders thus have the power to dictate to some degree how other firms should behave, it is
generally accepted that a framework of leadership cannot ignore the fact that each firm retains its own
decentralised decision making power. That is, each firm will look at its own profit function and make
decisions about those variables it controls as to maximise this function. A firm can also step away from a
proposal made by a leader and therefore not engage in the working relationship with this leader but try
its chances with other firms instead.
4.4.1
Leaders and followers: Stackelberg games
The framework of a Stackelberg game takes the above considerations of individual decentralised decision
making power into account. We illustrate in subsequent sections how Stackelberg leaders in a supply
chain would formulate their decision problem.
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Integrated Logistics
The key idea is that the Stackelberg leader anticipates how other firms will take decisions that optimise
their individual profit functions, and knowing this, the Stackelberg leader will set out conditions such that
those decisions made by other firms under these conditions will maximise the profits of the Stackelberg
leader.
There are two important considerations we will further explore through investigating some examples:
1. How will a leader achieve that other firms are willing to adopt the contract? What are the
minimal concessions a leader has to offer other firms?
2. What is the best possible outcome for a leader? Is it different or equal to the logistics policy that
is found from maximising the integrated supply chain profit function? Is the profit achievable in
an integrated supply chain an upper bound on the profit that a Stackelberg leader can achieve?
We will first derive a general model for one firm as a Stackelberg leader and another firm as the follower.
be the profit function of the leader, and
that of the follower.
Let
Assume that the Stackelberg leader has power over determining the conditions of working together. Let
these conditions be called ∈ , where is the set (or domain) of all possible ways in which the leader
can specify these conditions. The follower, given the conditions from the leader as expressed through ,
will aim to optimise its own profit function:
, ,
max
∈
where is the set of all possible ways in which the follower can set its own decisions. Let this optimal set
of decisions that the buyer can make be represented by
∈ . We can use the following
mathematical notation for this:
,
arg max
∈
Note that according to the NPV criterion the buyer as a firm should engage in activities when the NPV of
this activity is non‐negative. Hence we have an additional constraint that:
,
0
So this is the minimal condition that the leader has to offer the follower!
The leader therefore, has to find the conditions under which the follower’s profit remains non‐negative
and such that the decisions , maximise the profits of the leader.
It is also a requirement that under , the leader’s profit must itself remain non‐negative.
Hence the leader’s problem can be written as follows:
|
max
∈ , ∈
,
arg max
∈
,
0
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,
0
4.4.2
Stackelberg leader in the supply chain: examples
We will use the example used previously in Section 4.3.1 and also in Section 4.2.2.
Producer is Stackelberg leader
We consider the situation that the producer is the Stackelberg leader and the buyer is the follower.
You may recall from previous sections that is the only logistical decision in this model and it is set by
the buyer. So what is, therefore, the power of the leader?
Contract menu
We will consider that the leader can determine the financial arrangements of the working relationships,
including setting the price for a product and the type of payment structure adopted with the buyer.
We in particular assume that the ingredients of a contract the leader will offer, i.e. , is given as follows:
1. The producer will specify a price ∗ per unit of product that the buyer has to pay as set‐out in
the condition 4. below;
2. The producer will specify a non‐zero value for ∗ that governs what the buyers will pay for and
how, as set out in conditions 3. and 4 below;
3. The buyer is to pay the supplier for every order the set‐up cost at the time the order is delivered:
∗
∗
1
4. The buyer pays the producer per product upon delivery the price
∗
1
∗
1
,
∗
the moment that the product is sold by
and in addition, will pay the remainder
the buyer to its customers.
In short, the leader has to choose values for ∗ and ∗ . The buyer will choose the optimal lot‐size ∗ .
Solution
We note that the above contract scheme is as what was used in Solution 3 in Section 4.3.1. We hence can
specify the profit functions of the players as follows:
,
∗
,
∗
∗
∗
∗
∗
2
1
2
, ,
∗
1
∗
2
1
∗
2
2
2
∗
1
1
2
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Integrated Logistics
To solve this game, we start by deriving the optimal decision for the follower, given some values
∗
. The solution that maximises
, ∗ , ∗ is:
2
∗
∗
,
,
∗
∗
,
∗
and
1
We substitute this into
∗
:
∗
∗
2
2
1
Define the square root term as
(“logistics costs” of buyer).
Intuition says that an optimal solution for the leader will aim to extract all profits from the buyer, or in
,
0 is binding, i.e. the an optimal set ∗ , ∗ will make
other words that the condition
∗ ∗
0. This means that if we solve for ∗ , we find the following condition:
,
∗
∗
2
An interpretation for this is that the leader will set its price equal to the difference between the revenue
per product received by the buyer and the total logistics costs per product experienced by the buyer.
With the numerical data of this example, we have
£40/
, and for ∗ 1, we would get
∗
40
1.2775 4
0.012 £34.71/
.
We now look at the leader’s profit function. We take account of the result found for ∗ . We choose
∗
1. We then get:
∗
∗
1 ,
∗
∗
2
∗
2
∗
1
2
2
which simplifies to:
∗
We substitute the result for
∗
∗
2
2
and find:
∗
2
2
2
2
1
Notice that the producer’s profits are now equal to optimal solution of the integrated supply chain profit
as given on
function. (Recall that ∗ is the optimal solution for the alliance, and substitute ∗ into
LN Section 4.2.1, Solution 2). If you use the numerical data, you will hence find that under this contract:
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Integrated Logistics
£0/
£73,502.2/
∗
≡
So, if there is value in this activity from the viewpoint of the integrated supply chain, as is clearly the case
,
0 is automatically also satisfied. Furthermore, intuition
in this example, then the condition
says that this must indeed be an optimal solution for the Stackelberg leader: there is no more value in the
supply chain then by adopting a policy that is optimal for the integrated supply chain so this must be an
upper bound.
For any other choice for ∗ , it can be shown that the same conclusion can be reached.
General conclusion
It seems that we can conclude that a Stackelberg leader will extract maximum value by ensuring the
logistics policy implemented is that which maximises the integrated supply chain profit function, and by
further specifying the right mix of transfer price and payment structure such that all the profit achieved in
the integrated supply chain goes to the leader.
We have merely used an example to illustrate this, but it is thought to hold in general.
Buyer is Stackelberg leader
We consider the case that the buyer is the leader. The buyer will not only specify ∗ but will offer a
contract setting out the payment conditions. We use the same contract scheme as in the previous case,
but now ∗ and ∗ are specified by the buyer too.
The producer as follower has no decision to take but to consider either to accept the contract or to
walk away.
We proceed as before, but now when it comes to determining ∗ and ∗ , we use the condition:
∗ ∗
0
, , ∗
That is:
∗
,
∗
,
∗
∗
∗
If we choose
∗
∗
1
∗
2
1
∗
2
2
2
∗
∗
1
∗
1
2
0
1 this reduces to:
∗
∗
2
2
0
Therefore:
∗
2
2
An interpretation for this is that the leader will set the price it is willing to pay to the producer to be equal
to the producer unit cost per product plus all other relevant logistics costs per product, which are in this
case only the financial opportunity cost of his set‐ups. With
£20/
, the price offered is just a
!
fraction higher, i.e. ∗ £20.01/
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Substitution of ∗ and this value for ∗ into the buyer’s profit function shows that the buyer’s profit
now equal the maximum profit obtainable from the integrated supply chain profit function:
∗
,
∗
2
2
2
2
1
Numerical evaluation thus shows that adopting this contract leads to:
£0/
£73,502.2/
≡
This is again in support of the general conclusion.
∗
Critical evaluation
Don’t rely on a single powerful buyer or a single powerful supplier
A Stackelberg leader game is much more `unfair’ for an outsider who would care for both firms equally
than a cooperative game. Firms who are under threat from finding themselves in a follower’s role should
anticipate this potential development well in advance and seek business opportunities with other firms. It
is one of the reasons why good intentions to develop a `strategic relationship’ with one powerful supplier
or with one powerful buyer may turn sour. It is good advice that for strategic components of your
business you seek to develop relationships with more firms than just one.
Avoid getting trapped ‐ when it requires investment to break away from a relationship with a leader
Imagine that your firm is a supplier of a particular component to a big manufacturer. The relationship has
gone so well that your firm is now developing a bespoke component for this manufacturer. As a result,
the component is of high value for the manufacturer so you still get a good price. However, your firm had
to configure your own manufacturing facility to a highly specialised degree, so you can’t produce
components for other potential clients anymore.
If your firm wants to switch to producing for another client, your firm will need to invest in new
(it will be a negative value.)
technology. Let the annuity stream value of this investment cost be
Now if the current client manufacturer realises that your firm has locked itself into being too specialised,
it can extort your firm even more, since the condition for your firm to reject any contract from the
current manufacture (see the general model in Section 4.4.1) now becomes:
0
In other words, if
£10,000/
, then the manufacturer can re‐set the boundary condition
in his leader’s optimisation model to:
,
10,000
So the leader can extort your firm even more!
This is not entirely sustainable because your firm may have to lend money to compensate for liquidity
problems and in the long term your firm may go bankrupt. The leader may in fact not care for this if the
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Integrated Logistics
leader has at the same time relationships with other suppliers. Only if you as a supplier are vital to your
client manufacturer will you probably get a slightly better deal so that your firm doesn’t go bankrupt. In
any case, it is then still not an entirely satisfactory way for your firm.
It is for your firm hence not sufficient to stress that your firm is vital to the manufacturer (when is that
really the case if the manufacturer is a leader?). Your firm must make sure it can somehow create the
condition under which your firm could switch to supplying another firm, and make sure that this is known
to the current manufacturer as a credible threat, i.e. an action that your firm can and is willing to take if
necessary.
An accounting perspective
You may be sceptical about the Stackelberg games we have analysed in the supplier‐buyer models above.
In particular, you may find that using the boundary condition:
,
0,
seems unrealistic. However, this only means that the annuity stream profit function is set to zero. You
may be more convinced of the reality of this if we analyse the accounting profits of the follower. We will
show that under these conditions the follower will typically still make a positive accounting profit.
We must first realise the following important difference:
1. The profit functions we have used are not those used in traditional accounting. Our profit
functions reflect the net benefit today of future cash‐flows derived from an activity, i.e. it is to be
used to make optimal decisions about the future directions for your firm under the realisation
that there is a next best available alternative activity that the firm could invest its money in
instead.
2. Traditional accounting looks on the other hand into the past, and records which cash‐flows have
gone out and come in over a fiscal year, without consideration of the actual timing of these cash‐
flows nor of making a comparison with what would be the next best available alternative to the
firm.
If we use the same example as before, the accounting profit over a past year will be found from our AS
functions, before substitution of any optimal policy (!), by setting the opportunity cost of capital
0.
(This is in fact a rough approximation of accounting profit, since e.g. we then do not account for any
inventory at the end of the fiscal year, depreciation of assets, etc.)
If we consider the game where the producer is the Stackelberg leader, the accounting profit
for the
buyer with ∗ 1 is:
∗
With
∗
939.3
40
∗
/
and
34.71
4 5,000
∗
£34.71/
600
5,000
939.3
, this evaluates to:
6450
3193.9
£3256.1/
0, in accounting terms the follower will still end up with a (small) profit.
Hence while
If we consider the game where the buyer is the Stackelberg leader, the accounting profit
supplier with ∗ 1 is:
∗
20.01 20 5,000 £50/
for the
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Quite small indeed, but not zero!
Including all relevant costs
There is also another reason which may help sceptical readers look more favourably on using the
,
0.
boundary condition
In our model we have not considered fixed overhead costs of a business. That is, to be able to execute an
activity some managers and other employees need to devote some of their time to this activity. This is
not a direct variable cost because they are paid monthly salaries. These fixed costs are in part to be
allocated to the current activity nevertheless in order to really establish the profits obtained.
They are typically ignored in logistics models because these costs do not determine optimal logistical
parameters such as a lot‐size. However, in financial terms they would actually need to be included. If we
would adapt our AS function accordingly, then a Stackelberg leader, from consideration of the boundary
condition:
,
0
would arrive at solutions whereby the transfer price ∗ will not be set so sharply as those value we found
in our examples in previous sections.
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Integrated Logistics
5 Stochastic Models
Source: PETERSON, R. AND SILVER, E. 1985. DECISION SYSTEMS FOR INVENTORY MANAGEMENT AND
PRODUCTION PLANNING. WILEY, NEW‐YORK.
5.1 When is the assumption of a constant demand rate valid?
Demand is often “irregular” or “lumpy.” This may be caused by seasonality or other factors. If demand is
irregular, the assumption of having a constant demand rate, used in all the EOQ models, might perhaps
be inaccurate?
Suppose that d1, d2, …, dn has been the observed demand during each of n periods of time. We
assume that this is representative for future demand patterns.
To decide whether demand is sufficiently regular to justify the use of EOQ model assumptions, the
following test is reported in Peterson and Silver (1985):
1. Determine the estimate E(D) of average demand per period, given by:
E(D)
1
n
n
d
i
i 1
2. Determine an estimate of the variance 2 of the per‐period demand from:
1
n
2 di2 E(D)2
n i 1
3. Determine an estimate of the relative variability of demand, called CV, the coefficient of variability:
CV
2
E(D)2
Note that if all di are equal, then 2 = 0, and also CV = 0. Hence the smaller the value of CV, the more it
indicates that demand follows a constant demand rate closely. Research has indicated that if CV < 0.20
then the use of the EOQ‐assumption of a constant demand rate is reasonable.
Example
We use a small example to illustrate the method. <You would want to use more than four observations if
available.> Demands during four quarters of the past year have been as follows: 80 units, 100 units, 130
units, and 90 units. If future demand is known to follow a similar pattern, is the assumption of a constant
demand rate appropriate?
E ( D)
1 n
1
di (80 100 130 90) 100
n i1
4
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1
1 n 2
2
d i E ( D) (80 2 1002 130 2 90 2 ) 1002 350
4
n i1
2
CV
2
E ( D)
2
350
0.035 0.2
1002
In conclusion, the test indicates that it would be appropriate to use the assumption of a constant
demand rate.
5.2 (r, Q) reorder point models
In this section the assumption is made that demand is a random variable following a known normal
(Gaussian) probability distribution. This distribution is characterised by a mean value
and a
standard deviation
. This probability distribution is symmetrical around its mean.
Note
If represents the demand per year, then
, where is the demand rate we have been using in
our EOQ‐type models so far.
The (r, Q) reorder point model represents a widely used inventory system. It is also known as the “Two‐
Bin System.” Its origins can be traced back to the practice of storing the stock of an item in two storage
bins, as a visual inventory control device. Emptying the first bin is the signal to the stock keeper to
reorder a fixed reorder quantity Q, and the contents of the second bin serve to meet the demand until
the reordered amount has been received.1 Therefore, the amount of stock in this second bin should cover
for the demand that arises during the lead‐time, that is the time that elapses between the moment an
order is placed and the moment the goods are received. Upon receipt of the order, the second bin is first
filled up to its original level, and the rest of the order is placed in the first bin.
Thus, whenever the first bin has been emptied, we know that the reorder point has been reached: the
reorder point corresponds to the number of items we keep in our second bin. A mathematical two‐bin
inventory control system is composed of two parts:
Two‐bin
System
Part 1. A method of determining how much to reorder when replenishment is necessary. This
part of the system is taken to be deterministic. That is, it uses a single value for each of the
variables it incorporates; and uses the mean or average annual demand E(D). The order quantity
Q is thus determined following an EOQ model:
∗
2
Part 2. A method for deciding when to reorder, that is, determining the stock level at which a
replenishment order should be placed. The method for determining that reorder point s is
probabilistic and considers the item demand as a random variable that may have any one of
several values.
1
In fact, the Kanban system and the basic ideas of the Just-In-Time philosophy by Toyota in Japan supposedly have been
developed from observing such a bin system used in U.S. supermarkets.
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Integrated Logistics
There are two issues to consider:
Part 1: Is the EOQ order quantity a good approximation for the optimal order quantity?
Part 2: To establish a reorder point so as to account for the proper amount of stock needed to
cover the demand during the lead‐time.
Note that because demand is probabilistic, shortages (i.e. stock‐outs) may still occur during the period
of lead‐time demand. We will consider two cases:
A stock‐out is made up, i.e. the customer is promised the item(s) but will have to wait until the
next order has arrived. This situation is called the backorder case;
A stock‐out results in a lost sale, i.e. the customer leaves and will not purchase the item(s) at your
shop (warehouse,…).
5.2.1
Backorder case
We investigate the (r, Q) model making the following assumptions:
1. annual demand D (items / year) is a continuous random variable with a mean E(D), a variance
2(D), and a standard deviation (D);
2. constant infinite replenishment rate R = ;
3. constant unit holding cost (£ / item, year);
4. constant unit replenishment cost (£ / order);
5. constant unit shortage cost cs (£ / item) which does not depend on how long it takes to make up
the stockout;
6. constant non‐zero lead time L 0;
7. constant replenishment quantity (items/order) and shortages are made up (backorders);
8. constant reorder point .
Possible inventory fluctuations in the (r, Q) reorder point model are depicted in Figure 15. Note that
the lead time L is constant, but the cycle time T, as a consequence of variability of demand, is no longer
constant. Also note that the amount of items left in inventory when the order arrives varies. This is
caused by the variability of the demand during the lead time.
I(t)
L
L
Q
L
Q
Q
s
r
E(X)
‐
T
= Average demand rate
t
= Real demand rate
Figure 15: The (r, Q) reorder point model; no backorders occur
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Integrated Logistics
Let X denote a random variable representing the demand rate during lead time L, with an average
value a mean E(X), a variance 2(X), and a standard deviation (X). If we assume that the demands at
different points in time are independent, then it can be shown that:
2 ( X ) L 2 ( D)
E ( X ) L E ( D)
( X ) L ( D)
Therefore, we can expect that the inventory level during lead time L will drop from the reorder point
to a level:
r E( X )
We call
the safety stock inventory. The safety stock is used to buffer against the uncertainty
on demand during the lead time. It is that part of inventory in the second bin that we will only use when
demand during lead time is larger than the expected amount E(X). We cannot be sure that we will never
run out of stock. Figure 16 depicts a situation where the variability of demand during lead time has
caused stock‐outs, which are then backordered.
I(t)
L
L
L
Q
Q
Q
s
r
E(X)
‐
backorders
t
Figure 16: The (r, Q) reorder point model; backorders occur
The expected total annual cost of the system consists of expected replenishment costs, expected
holding costs, and expected shortage costs.
1. We start with the replenishment cost per year. Since eventually all demand will be met during a
year, we will place on average E(D)/Q orders. Thus the annual replenishment cost is:
2. The expected annual holding cost is considered next. The expected annual holding costs are:
where E(I) is the expected average inventory level throughout the year.
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Integrated Logistics
To calculate E(I), we will make the assumption that backorders occur very infrequently. That means
we can assume inventory levels to be always positive for the calculation of the expected average
inventory level E(I). The expected inventory value I(t) at the start of a cycle is then r – E(X) +Q; the
expected value of I(t) at the end of a cycle is r – E(X); and in between it is expected to follow a linear
downward slope. Therefore the expected total amount in inventory during a cycle of length T is:
1
QT (r E ( X ))T
2
and thus the expected average inventory level per year is obtained by dividing by T:
1
E(I ) Q r E( X )
2
The expected average cost is hence approximately:
1
h Q r E ( X )
2
3. We next compute the expected average shortage cost per year. Let us assume that X follows a
probability density function f(x). The expected number of stock‐out items during a cycle is therefore:
( x r ) f ( x)dx
r
The expected number of cycles per year is equal to the expected number of time we place a
replenishment order, which is E(D)/q. Also remember that we assumed that cs represents the shortage
cost per unit short (£/item) and that is does not depend on the time it takes to make up the shortage.
Therefore the annual shortage cost can be written as:
E ( D)
cs
( x r ) f ( x ) dx
Q r
The total annual cost is the sum of all three cost components:
(r , Q) s
E ( D) 1
E ( D)
h Q r E ( X ) cs
( x r ) f ( x)dx
Q
Q r
2
We could now use the well trodden path of taking the partial derivatives, and setting these equal to
zero to find the optimal values r* and Q*. In most cases when backorders occur very infrequently, it has
been found that Q* is very close to the EOQ value:
∗
2
If we assume this given value of Q, which is not dependent on r, the total cost formula as a function of
the reorder point r can be simplified to:
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Integrated Logistics
(r ) s
E ( D) 1
E ( D)
h Q * r E ( X ) cs
( x r ) f ( x)dx
Q*
Q * r
2
where Q* is now a constant. To find the optimal value of r, we take the derivative:
d ( r ) d E ( D )
E ( D)
1 *
h
Q
r
E
(
X
)
c
( x r ) f ( x ) dx 0
s
s
*
dr
dr Q
Q* r
2
Since the annual replenishment cost term is not a function of r this can be simplified to:
d ( r ) d
dr
dr
1 *
E ( D)
h
Q
r
E
(
X
)
c
( x r ) f ( x ) dx 0
s
Q* r
2
Further:
d ( r )
E ( D) d
h cs
( x r ) f ( x ) dx 0
dr
Q * dr r
d(r )
E ( D)
h cs
f ( x)dx 0
dr
Q * r
Note that the probability that a shortage occurs during a cycle equals:
P( X r ) f ( x)dx
r
Therefore:
P( X r*)
hQ*
cs E ( D )
This result is not an explicit but an implicit specification of the reorder point r. In general, we want to
choose such a value for the reorder point r so that the probability of a stock‐out occurring during a cycle,
P(X r), is very small.
In order to make such calculations, we need to have information about the density function f(x). Often,
demand during lead time is assumed to follow a normal distribution.
Example
Given:
s $24 / order
h $3 / item, year
cs $4 / item
E ( D) 10000 items/year
X normally distributed random variable during lead time L
E ( X ) 300 items
( X ) 100 items
Compute the optimal values for order quantity and reorder point.
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Integrated Logistics
1. Order quantity:
Q*
2 sE ( D )
h
2 ( 24) (10000 )
400 items/order
3
2. Reorder point:
P( X r*)
hQ*
3(400)
0.03
cs E ( D ) 4(10000)
This means a 3% probability of having a backorder in a cycle. Thus the reorder point r should be set
such that the probability of a stock‐out occurring during a cycle equals 0.03. Standardising the above
formula with respect to X, we obtain
P(
X E( X ) r * E( X )
) 0.03
(X )
(X )
P( Z
r * E( X )
) 0.03
(X )
From a Standard Normal Cumulative Probabilities Table, we can get that
P( Z 1.88) 0.97
Then
P( Z 1.88) 0.03
Hence
r * E( X )
1.88
(X )
This gives
r* 1.88 ( X ) E ( X ) 488 items
The (r, Q) reorder point inventory control policy for this item is now established. An order quantity for
400 units is placed when the inventory position reaches 488 units. The annual costs for this system are
shown below to be £1180. This follows from
1. reorder costs: sE( D) / Q 24(10000) / 400 $600
2. lot size inventory holding costs: hQ / 2 3(400) / 2 $600
3. buffer inventory holding costs: hr E ( X ) 3(488 300) $564
4. stock‐out costs:
E ( D)
cs ( x r ) f ( x)dx 10000
4 (100)(0.0116) $116
400
Q
r
(See Partial Expectation for Standard Normally Distributed Random Variable Table)
Note. ( x r ) f ( x )dx , i.e. the expected number of items stock out during one cycle, is also called
r
the partial expectation. The partial expectation for a standard normally distributed random variable can
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Integrated Logistics
be found in a Table; the partial expectation for any normally distributed variable can be found by
simply multiplying the corresponding value with the standard deviation (X ) of the random variable.
5.2.2
Lost sales case
In the lost sales case, similar assumptions hold as in the backorders case, except the following
differences:
The stock out cost cs includes the lost profit as well as any penalty costs.
When replenishment stock arrives the inventory is increases by the full amount of the order i.e. by
Q, as no backorders exist to be filled.
From the second point, it follows that the expected inventory level is the expected inventory level of
the backorders case, plus the expected number of items short per cycle. The expected inventory level is
therefore:
1
E ( I ) Q r E ( X ) ( x r ) f ( x ) dx
2
r
All the other terms stay equal. Following the same approach as in the backorders case gives:
(r ) s
1 *
E ( D)
E ( D)
h
Q
r
E
(
X
)
(
x
r
)
f
(
x
)
d
x
c
2
s Q * ( x r ) f ( x )d
Q*
r
r
d ( r )
d
E ( D) d
h h ( x r ) f ( x ) d x cs
( x r ) f ( x ) dx 0
dr
dr r
Q * dr r
d(r )
E ( D)
h h f ( x)dx cs
f ( x ) dx 0
dr
Q * r
r
Which leads to
P( X r*)
hQ*
cs E ( D) hQ *
Note. The shortage cost cs is now the penalty cost of a stock out plus the opportunity cost (lost profit)
of a lost sale.
5.2.3
Service level approach
It may be very difficult to determine accurately the cost of being one unit short. For this reason,
managers often decide to control shortages by meeting a specified service level. Service level can be
defined in different ways.
Definition 1. The service level is the expected number of cycles per year during which a shortage
occurs.
Definition 2. The service level is the expected fraction of all demand that is met on time.
Assume that all shortages are backordered (backlogged).
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Integrated Logistics
Definition 1
Given a reorder point r, the fraction of all cycles per year that will lead to a stock out is equal to:
P( X r )
Since E(D)/Q cycles per year will occur, the expected number of cycles per year during which a
shortage occurs equals
P( X r )
E ( D)
Q*
According to Definition 1, we can write
P( X r )
E ( D)
s1
Q*
where s1 is the allowed number of expected stock outs per year. This also gives
P ( X r*) s1
Q*
E ( D)
Definition 2
Given a reorder point r, the expected number of items short per cycle is equal to:
( x r ) f ( x)dx
r
Since E(D)/Q cycles per year will occur, the expected number items short per year equals
( x r ) f ( x)dx
r
E ( D)
Q*
The expected fraction of items short per year is obtained by dividing the latter equation by E(D),
which gives
1
( x r ) f ( x)dx
Q* r
According to Definition 2, we can write
1
( x r ) f ( x)dx 1 s2
Q* r
where s2 is the expected fraction of items per year that can be met on time. Rearranged:
( x r*) f ( x)dx Q (1 s )
*
2
r*
The left‐hand side is also called the Partial Expectation E ( x r*)
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Integrated Logistics
E ( x r*) ( x r*) f ( x)dx
r*
For a random variable Z following a standard normal distribution, Partial Expectations E ( Z z ) in
function of Z have been tabulated, Section 5.2.4. If X follows a normal distribution with expected value
E(X) and standard deviation (X), then
E (Z z )
E ( X r*) Q* (1 s2 )
(X )
(X )
Therefore, from such a table we can derive z. The reorder point can then be found as
∗
5.2.4
Standard Tables
Standard Normal Cumulative Probability Table
If
denotes the probability distribution of random variable , then
P( X r ) f ( x)dx
r
is the probability that a random draw from the distribution will produce a value of that will be larger
or equal to a given constant .
Given a normally distributed variable with mean
and standard deviation
, then the
random variable , defined as:
will be standard normally distributed. That is,
0 and
1. We have:
We define:
The first two tables on the following pages let you solve any of the following two problems:
1. Given , what is
?
2. Given
, what is ?
Examples
Let
‐2.15. Then from the first table, the row with 2.1 and column with 0.05 intersects at
the value 0.0158. Therefore,
2.15
0.0158. “The probability that a random trial
from a normal distribution produces a value not higher than 2.15 is 1.58%.”
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Integrated Logistics
Let
1.23. Then from the second table, the row with 1.2 and column with 0.03 intersect at
the value
1.23
0.8907. “The probability that a random trial from a normal distribution
produces a value not higher than 1.23 is 89.07%.”
Let
= 0.9500. In the second table, we identify this value in between the values 0.9495
and 0.9505. They are both in the row of 1.6, hence z = 1.6x. To find the value of x, we see from
the column headers that it must fall in between 0.04 and 0.05. In aim to be as accurate as
possible, we could (linearly) interpolate between 1.64 and 1.65, leading in this case to z = 1.645.
However, for the sake of keeping calculations to a minimum, it is adequate to simply take the x
from the column of which its
is closest to 0.9500. Hence both z = 1.64 or z = 1.65 are
good values.
Note that
1
.
Standard Normal Loss Function Table
Per definition,
if
0, and
random variable , then
0 if
0. If
denotes a probability distribution of
E ( X r ) ( x r ) f ( x)dx
r
denotes the expected value of the random variable
We also have:
E (Z z)
, where is a given constant.
E ( X r )
(X )
When is normally distributed, will be standard normally distributed, and the Standard Normal
Loss Table given below can be used to solve any of the following two problems:
?
1. Given , what is
2. Given
, what is ?
Examples
Let
1.64. We look up the value in the table in the row with header 1.6 and column with
header 0.04 to find
1.6611
0.0075. In the table, we identify this value at the row with value 2.4 and in
Let
between the columns with values 0.04 and 0.05. Hence z = 2.445 is the value obtained from linear
interpolation. In practise, we are happy with taking either z = 2.44 or z = 2.45.
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Integrated Logiistics
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5.2.5
Exercises
1. Each year, a computer store sells an average of 1000 boxes of disks. Annual demand for boxes of disks
is normally distributed with a standard deviation of 40.8 boxes. The store orders disks from a regional
distributor. Each order is filled in two weeks. The cost of placing each order is $50, and the annual cost of
holding one box of disks in inventory is $10. The per‐unit stock out cost (because of loss of goodwill and
the cost of placing a special order) is assumed to be $20. The store is willing to assume that all demand is
backlogged. Determine for the computer store the proper
a) order quantity
b) reorder point
c) safety stock level
d) What is the probability that stock out occurs during lead time?
Additional questions:
e) If the lead‐time increases to 4 weeks, investigate how this affects your answers to the previous
questions, a to d?
f) What is the expected number of boxes per year that are not delivered on time from available stock
when implementing the reorder point you calculated in part b?
g) What is the percentage of annual demand met on time from available stock when implementing
the reorder point calculated in part b?
2. The same computer store as in Question 1 now doubts whether shortages result in backorders.
Suppose that each box of disks sells for $50 and costs the store $30. Answer the same questions a to d,
for the case that shortages all result in lost sales! <Hint: you need to estimate a value you will need to use
for the shortage cost .>
3. Again, the same computer store as in Question 1 and Question 2. It now doubts whether the values for
shortage cost used were realistic! Answer the same questions, a to d, for the case that the computer
store wants the average number of stock out cycles per year to be at most equal to 3.
4. Yep, the same computer store as in Question 1, 2, and 3. The store manager now thinks it might be
best to aim for satisfying at least 95% of annual demand. Answer the same questions, a to d. <Note: this
example illustrates that a negative safety stock value is possible.>
5. A hospital orders its blood from a regional blood bank. Each year, the hospital uses an average of 1040
pints of type O blood. Each order placed with the regional blood bank incurs a cost of $20. The lead time
for each order is one week. It costs the hospital $20 to hold 1 pint of blood in inventory for a year. The
per‐pint stock out cost is estimated to be $50. Annual demand for type O blood is normally distributed
with standard deviation of 43.26 pints. Assume that 52 weeks = 1 year and that all demand is backlogged.
a) Determine the optimal order quantity, reorder point, and safety stock level.
b) To use the two bin system with backorders, what “unrealistic” assumptions must be made?
6. Furnco sells secretarial chairs. Annual demand is normally distributed with mean of 1040 chairs and
standard deviation of 50.99 chairs. Furnco orders its chairs from its flagship store. It costs $100 to place
an order and the lead time is two weeks. Furnco estimates that each stock out causes a loss of $50 in
future goodwill. Furnco pays $60 for each chair and sells it for $100. The annual cost of holding a chair in
inventory is 30% of its purchase price.
a) Assuming that all demand is backlogged, what are the reorder point and the safety stock level?
b) Assuming that all stock outs result in lost sales, determine the optimal reorder point and the safety
stock level.
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Integrated Logistics
7. FOOTD Inc. sells an average of 1000 food processors each year. Each order for food processors placed
by FOOTD costs $50. The lead time is one month. It costs $10 to hold a food processor in inventory for
one year. Annual demand for food processors is normally distributed with a standard deviation of 69.28.
Determine the reorder point for each of the following values of the service level:
a) 95% of annual demand met on time
b) 99%
c) 99.9%
5.3 News vendor problems
In this section, we consider single‐period or static inventory problems. These are problems in which one
ordering decision has to be made at one moment in time.
Such problems are often called news vendor problems. Consider a vendor who must decide how
many newspapers should be ordered each day from the newspaper plant. If the vendor orders too many
papers, he or she will be left with many worthless newspapers at the end of the day. On the other hand,
the vendor who orders too few newspapers will loose profit (and disappoint customers) that could have
been earned if enough newspapers to meet customer demand had been ordered. The news vendor must
order the number of papers that properly balances these two costs.
More formally, a decision maker is faced with the problem of determining the order quantity q to
purchase an item based on estimates of some future demand for that item. Only after q has been
determined and the order has been placed, the real value of the demand D becomes known. Depending
on the values of q and D, the decision maker incurs a revenue R(q, D), a cost (q, D), and a profit P = R –
.
5.3.1
Example
A student has obtained the concession for picnic lunches at local football games. Records indicate the
following demand distribution for lunches (Table 2):
Table 2: Demand distribution for lunches at the local football club
Demand
Pr(x) = Probability that demand = X
X
0
0.00
1
0.10
2
0.10
3
0.30
4
0.30
5
0.10
6
0.10
The lunches cost £3 and are sold at £5. Any lunches left over can be disposed of for £0.5. The problem
is to determine the order quantity.
Three different approaches to solve this problem will be given in the following sections.
5.3.2
Cost minimisation
Once the decision has been made to place an order, the fixed procurement costs do not influence the
ordering decision. Fixed costs per order are therefore irrelevant in this case. The cash flows influenced by
the order quantity decision are the costs of ordering too less or the cost of ordering too much.
The cost of ordering too much may include the purchase price of the item plus the cost of disposing
any unsold items. The cost of ordering too less includes the purchase price of the item and the cost of lost
sales.
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Integrated Logistics
Let q represent the order quantity and x the realisation of the demand, p the unit purchase price of
the item, w the salvage value per unit for disposing of it, and r the unit selling price for selling an item to
our customers. We can then write:
pq w(q x)
q, x
pq r(x q)
(x q)
(x q)
where p = £3/unit, w = £0.5/unit, and r = £5/unit.
The expected cost, for a given order quantity q, is then given by:
6
x 0
x 1
E (q, x) q, x Pr(x) q, x Pr(x)
where Pr(x) is the probability that a demand of x units materialises.
We want to select this order quantity q that will minimise the expected cost. It is helpful to show the
calculations in a matrix, as presented in Table 3. It can be seen that the best decision would be to order 3
picnic lunches, with an expected cost of £128.5.
Table 3: Conditional cost matrix
Order
quantity q
1
2
Demand x
3
4
Expected
5
6
cost
E(
(q,x))
1
30
80
130
180
230
280
155.0
2
55
60
110
160
210
260
139.5
3
80
85
90
140
190
240
128.5
4
105
110
115
120
170
220
131.0
5
130
135
140
145
150
200
147.0
6
155
160
165
170
175
180
167.5
Pr(x)
0.10
0.10
0.30
0.30
0.10
0.10
5.3.3
Profit maximisation
For problems involving major changes in revenue in which profit maximisation is the objective, profits
may be used as a measure of performance rather than costs. Profit equals revenue minus cost.
In case of ordering too much, the profit comes from the sales of units but there is a cost related to the
disposal of unsold items. In case of ordering too few units, the sole profits arise from sales. Let P(q, x)
denote the profit, then:
(r p)x (p w )(q x)
P(q, x)
(r p)q
(x q)
(x q)
The expected profit, for a given order quantity q, is:
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Integrated Logistics
6
x 0
x 1
E P(q, x) P q, x Pr( x) P q, x Pr(x)
We like to select this order quantity to maximise the expected profit. Results are displayed in Table 4.
The optimal order quantity is again 3 units.
Table 4: Conditional profit matrix
Order
Demand x
Expected
quantity q
1
2
3
4
5
6
profit
E(P(q,x))
1
20
20
20
20
20
20
20.0
2
-5
40
40
40
40
40
35.5
3
-30
15
60
60
60
60
46.5
4
-55
-10
35
80
80
80
44.0
5
-80
-35
10
55
100
100
28.0
6
-105
-60
-15
30
75
120
7.5
Pr(x)
0.10
0.10
0.30
0.30
0.10
0.10
5.3.4
Regret minimisation
A third approach to the example utilises the same basic methodology for decision making, but the
measure of performance can no longer be obtained simply by examining the relevant cash flows
associated with a given order quantity and demand level. The regret measure is obtained by replacing the
outcome in each cell by the value obtained by subtracting the profit of the cell from the maximum profit
for any cell in that cell’s column. This simple concept is intuitively evident. For each state of nature (i.e.
realised demand level), there is an optimum decision (yielding the maximum profit). The regret
associated with making the best decision for a given state of nature is zero. The regret of each of the
other cells in the column simply indicates the difference in profits between the decision indicated and the
best decision for the state of nature that occurred. The regret matrix with respect to our example is given
in Table 5.
Table 5: Conditional regret matrix (in 0.1£)
Order
Demand x
quantity q
Expected
regret
1
2
3
4
5
6
1
0
20
40
60
80
100
50.5
2
25
0
20
40
60
80
34.5
3
50
25
0
20
40
60
23.5
4
75
50
25
0
20
40
26.0
5
100
75
50
25
0
20
42.0
6
125
100
75
50
25
0
62.5
Pr(x)
0.10
0.10
0.30
0.30
0.10
0.10
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Integrated Logistics
Note that expected regret plus expected profit sum to £7 for each order quantity. Since this is true for
each order quantity (q = 1, 2, …6), selecting the alternative that minimises regret is identical to selecting
the alternative that maximises profit. The sum of the two expected values (in this case £7) represents the
expected profit if demand x were known before making decisions. That is, it is the sum of the highest
profit outcome for each state of nature times the probability of the state. The expected profit given
perfect knowledge is:
6
x 0
x 1
E P(q, x) P q, x Pr( x) P q, x Pr(x)
2(0.1) 4(0.1) 6(0.3) 8(0.3) 10(0.1) 12(0.1) £7
The expected regret can therefore be interpreted as the cost of uncertainty.
The cost of uncertainty of the optimal decision represents the maximum value of perfect forecast
information to the decision maker. For this example, the maximum price the decision maker would pay
for perfect information is £2.35.
5.3.5
Exercises
1. In August, Walton Bookstore must decide how many of next year’s nature calendars should be
ordered. Each calendar costs the bookstore $2 and is sold for $4.50. After January 1, any unsold calendars
are returned to the publisher for a refund of $0.75 per calendar. Walton believes that the number of
calendars sold by January 1 follows the probability distribution shown in the Table below. Walton wants
to maximise the expected net profit from calendar sales. How many calendars should the bookstore
order in August?
5.3.5.1.1 No.
of
Calendars
sold
100
150
200
250
300
Probabili
ty
.3
.2
.3
.15
.05
2. Every four years, Blockbuster Publishers revises its textbooks. It has been three years since the
best‐selling book, The Joy of OR, has been revised. At present, 2000 copies of the book are in stock, and
Blockbuster must determine how many copies of the book should be printed for the next year. The sales
department believes that sales during the next year are governed by the distribution in the Table below.
Each copy of the book during the next brings the publisher $35 in revenues. Any copies left at the end of
the next year cannot be sold at full price but can be sold for $5 to Bonds Ennoble and Gitano’s
Bookstores. The cost of a printing of the book is $50000 plus $15 per book printed. How many copies of
The Joy should be printed? Would the answer change if 4000 copies were currently in stock?
Copies Demanded
5000
6000
7000
8000
Probability
.3
.2
.4
.1
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Integrated Logistics
1. Solution: Expected profit is maximised for q* = 250 items/run.
profit maximisation
acquisition cost
salvage value
revenue
p
w
r
order quantity
q
demand x
0
0
-125
-187.5
-250
-312.5
-375
0
0
0
100
150
200
250
300
0
p(x)
2
0.75
4.5
profit per unit sold
loss per unit unsold
2.5
1.25
expected profit
100
0
250
187.5
125
62.5
0
0
0.3
150
0
250
375
312.5
250
187.5
0
0.2
200
0
250
375
500
437.5
375
0
0.3
250
0
250
375
500
625
562.5
0
0.15
300
0
250
375
500
625
750
0
0.05
0
0
-125
-187.5
-250
-312.5
-375
0
0
0
231.25
328.125
387.5
390.625
346.875
0
2. Answer: Order one print run for 5000 units of the book, estimated profit is 65500‐50000=$15500.
profit maximisation
acquisition cost
salvage value
revenue
p
w
r
order quantity
q
demand x
0
0
-30000
-40000
-50000
-60000
0
0
0
0
3000
4000
5000
6000
0
0
p(x)
15
5
35
profit per unit sold
loss per unit unsold
20
10
expected profit
3000
0
60000
50000
40000
30000
0
0
0.3
4000
0
60000
80000
70000
60000
0
0
0.2
5000
0
60000
80000
100000
90000
0
0
0.4
6000
0
60000
80000
100000
120000
0
0
0.1
0
0
-30000
-40000
-50000
-60000
0
0
0
0
0
-30000
-40000
-50000
-60000
0
0
0
0
46500
59000
65500
63000
0
0
If 4000 books in stock, then it does not seem worthwhile to print another edition as shown below.
acquisition cost
salvage value
revenue
p
w
r
order quantity
q
demand x
0
0
-10000
-20000
-30000
-40000
0
0
0
0
1000
2000
3000
4000
0
0
p(x)
15
5
35
profit per unit sold
loss per unit unsold
20
10
expected profit
1000
0
20000
10000
0
-10000
0
0
0.3
2000
0
20000
40000
30000
20000
0
0
0.2
3000
0
20000
40000
60000
50000
0
0
0.4
4000
0
20000
40000
60000
80000
0
0
0.1
0
0
-10000
-20000
-30000
-40000
0
0
0
0
0
-10000
-20000
-30000
-40000
0
0
0
0
15500
28000
34500
32000
0
0
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Integrated Logistics
6 MRP, JIT, and Bottlenecks
6.1 Materials Requirements Planning
Materials Requirements Planning (MRP) is a computer‐based production planning and inventory control
system. It is also known as “time‐phased requirements planning”, as it recognises the time needed to
produce (or purchase) items. MRP also recognises the relationship between the demand for the final
product and the components to make it. These relationships are then used to determine the amount that
should be produced (or purchased) of each final product, assembly, subassembly, and component during
each period of a planning horizon.
6.1.1
MRP inputs
The three major inputs of an MRP system are the master production schedule, the production structure
records, and the inventory status records (Figure 17).
Customer orders
Bills
of materials
Purchase orders
Master
Production
Schedule
Forecast
demand
Inventory
records
MRP
Materials plans
Work orders
Figure 17. MRP environment
Master Production Schedule (MPS)
The Master Production Schedule (MPS) outlines the production plan for all end items. It expresses how
much of each end item is wanted and when it is wanted. Typically, the MPS works with time buckets of
one week (month) and the planning horizon of the MPS covers for the next 1 – 6 months (1 – 2 years).
The MPS is developed from end item customer orders and forecasts.
The end item customer orders or sales orders represent some contractual commitment on behalf of
the customer. Customers, however, may change their minds about what they require after having placed
their orders. Considering that each of several hundred customers may make changes to their sales
orders, not once, but possibly several times after order placement, it is evident that managing the sales
order book is a complex and dynamic process. An organisation may have to make decisions about how
much flexibility they will afford to customers and at what stage their customers have to accept liability for
the implications of their changes.
Not all operations have the same degree of forward “visibility” in terms of known customer orders.
Supermarkets, for example, have no advance notice when a customer will arrive and what and how much
will be purchased. Many businesses also do not have sufficient time to respond “just‐in‐time” when
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Integrated Logistics
customers place orders, because there would be insufficient time left to buy the required materials,
perform the required manufacturing operations on those materials, and then deliver the product to the
customer. Many businesses therefore have to rely on forecasts, i.e. predictions about likely future end
item demands to ensure they have all raw materials, components, assemblies and end item products
available to meet delivery times once a sales order is received.
A combination of known orders and forecasted orders is used to represent end item demand in many
businesses. Important to note is that forecasted orders do not equal sales targets. Sales targets are given
to the sales force and to give them some incentive, these are typically set optimistically high. The
forecasted orders should be the best estimate available at any time of what reasonably could be
expected to happen.
A characteristic of demand management is that the further ahead you look into the future, the less
certainty there is about demand. In the short term, typically, the demand will largely be based on placed
orders. However, as few customer orders are placed well into the future, the demand in the longer term
will be largely based on forecasts i.e. market information from sales operatives as well as historical sales
figures. As time goes by, the fraction of forecast demand for a certain future period will diminish and be
replaced by newly arriving sales orders.
The mix of known orders and forecast orders will also be determined by the type of operation. A
concept sometimes used here is the Order Penetration Point (OPP). The OPP is that storage point in the
supply chain where downstream operations are largely triggered by “make‐to‐order” demand and
upstream operations largely by “make‐to‐stock” decisions. For example, the OPP for a supermarket lies at
the retail point itself, while for a jobbing builder the OPP lies upstream as materials (and other resources)
will only be acquired when the builder has been awarded a specific contract. Many businesses have to
operate in a context where the OPP is located somewhere within their operations or downstream the
supply chain. In that case they need to operate mainly using forecasts. In general, the information of
sales orders combined with forecast demand of end items forms the major input to the Master
Production Schedule.
Note that MPS and MRP can also be applied to a service environment. For example, in a hospital
theatre there is an MPS which contains statements of which operations are planned and when. This
drives provisioning of materials for the operations, such as sterile instruments, blood and dressings. It
also governs the scheduling of staff for operations, including anaesthetists, nurses, and surgeons, and
other facilities such as recovery rooms and beds.
An example of a MPS is given in Table 1. End item demand for the next 9 weeks is given. We start by
having 30 units of the item on hand at the beginning of week 1, of which 10 can be used to fulfil demand
for week 1, and 20 remain available inventory. At the end of week 2, available inventory is reduced to 10
units, just sufficient to satisfy demand for week 3. Therefore, the MPS schedule needs to make available
at least 10 units at the end of week 3 to satisfy demand for week 4.
Table 6. Example of MPS chase strategy.
Week number
1
2
3
4
5
6
7
8
9
Demand
10 10 10
10
15
15
15
20
20
Available 30 20 10 0
0
0
0
0
0
MPS
0
0
10
15
15
15
20
20
The chase strategy used in Table 1 is to make available only those amounts needed to satisfy
immediate future demand. As a result, inventory levels are eventually reduced to zero. Chasing demand
involves adjusting the provision of resources, which may not always be desirable or possible. Level
scheduling, on the other hand, (see Table 2) involves averaging the amount required to be completed to
smooth out peaks and troughs. The average projected inventory of the end item, however, is now much
higher. In practice, companies may use strategies which fall between the chase and the level strategy.
Table 7. Example of MPS level strategy.
Week number
1
2
3
4
5
6
7
8
9
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Integrated Logistics
Demand
10
10
10
10
15
15
15
20
20
Available 30
33
36
39
42
40
38
36
29
MPS
13
13
13
13
13
13
13
13
While the demand for the end items needs to be derived from projected sales orders and forecasts,
the demand for all components necessary to produce these end items can be calculated by the MRP
system from the MPS data.
Bill of Material (BOM)
To work out the volume and timing of assemblies, sub‐assemblies and materials required to meet the
MPS, we need information about the list of assemblies, sub‐assemblies and components needed to
produce the end items. This is provided by the product structure records or Bills of Materials (BOMs).
A Bill of Material (BOM) contains the information on every item required to produce an end item: it
lists all information necessary, in particular the part identification and the quantity needed per assembly.
An example of a simple BOM is displayed in Figure 18.
A
B(1)
E(1)
Level 0
C(3)
F(2)
J(1)
D(1)
G(2)
K(1)
H(1)
Level 1
I(1)
Level 2
Level 3
Figure 18 Typical BOM. The letters represent assemblies, subassemblies and parts. The numbers in
parentheses are the quantities required for assembly.
In Figure 18, item A is the end item and it requires 1 subassembly B, 1 subassembly D, and 3 parts C.
One part E and 2 subassemblies F make up 1 subassembly B. One part I, 1 part H, and 2 parts G make up
the subassembly D. One part J and 1 part K make up 1 subassembly F.
A BOM should display all components, assemblies and sub‐assemblies in levels representing the way
in which they are actually manufactured: from raw materials to components to subassemblies to
assemblies to end items. The BOM for MRP may be different from a product structure plan constructed
by design engineers, as a product may not be assembled the way it was designed.
Inventory status records
These records contain the status of all items in inventory. All inventory items must be uniquely
identified. These records must be kept up to date, with each receipt, disbursement, or withdrawal
documented to maintain record integrity. They should also contain information on lead times, lot sizes, or
other item peculiarities.
Quantities of items in inventory available for the next period are referred to as “on‐hand”; quantities
that are expected to become available are “scheduled receipts.”
6.1.2
MRP outputs
MRP takes the MPS for end items, BOM and inventory status records, and determines what components
and subassemblies are required, how many, and when work orders for subassemblies should be released
to the manufacturing operation and when orders for parts should be placed with suppliers.
For purchased components, the lead time is the time interval between placement of the purchase
order and the availability of these components in inventory. After such an order has been placed, it
changes from “planned” order to “open” or “on order.”
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Integrated Logistics
For manufactured subassemblies, the lead time is the interval between release of the work order to
the manufacturing operation and the availability of the produced subassemblies after completion.
The actual order quantity released to either the manufacturing floor or the supplier may be the net
requirements or adjusted to any suitable lot size.
Example 1: simplified MRP
Question: You are to produce 100 units of product A in period 8. If no stock is on‐hand or scheduled to
arrive, determine when to release orders for each component and the size of each order. The BOM and
Lead time information is displayed in Figure 19.
Solution: Product A is made from components B and C; C is made from components D and E. By simple
computation we calculate quantities required:
Component B: (1) (number of A’s) = 1 (100) = 100;
Component C: (2) (number of A’s) = 2 (100) = 200;
Component D: (1) (number of C’s) = 1 (200) = 200;
Component E: (2) (number of C’s) = 2 (200) = 400.
Now we must consider the time element for all the items. Table 8 below create a simplified MRP
based on the demand for A, the knowledge of how A is made, and the time needed to obtain each
component.
A
Lead time = 4
B(1)
Lead time = 3
C(2)
Lead time = 2
D(1)
Lead time = 1
E(2)
Lead time = 1
Figure 19. BOM/Lead time information of example 1
Table 8. Simplified MRP matrix with solution for example 1.
Item: A
Lead time = 4
Period
1
2
Gross requirements
Planned order releases
Item: B
Lead time = 3
Gross requirements
Planned order releases
Item: C
Lead time = 2
Gross requirements
Planned order releases
4
5
6
7
8
100
100
100
100
Gross requirements
Planned order releases
Item: D
Lead time = 1
Gross requirements
Planned order releases
Item: E
Lead time = 1
3
200
200
200
200
400
400
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Integrated Logistics
Planned order releases of a parent item are used to determine gross requirements for its component
items. Planned order releases of a parent item generate a gross requirement in the same time period for
its lower level components. Planned order release dates of the components are simply obtained by
offsetting the lead times.
The MRP in Table 8 shows which items are needed, how many are needed, and when they are
needed; in order to complete 100 units of product A in period 8, it is necessary to release order for 100
units of B in period 1, 200 units of C in period 2, 200 units of D in period 1, and 400 units of E in period 1.
MRP Matrix
The previous example was a simplified case. In a typical MRP table, we will instead encounter the
following entries (see Table 9):
I. Gross requirements: the total anticipated demand for the item. For end items the quantity is
obtained from the MPS; for components and subassemblies it is derived from the “planned
order releases” of its parents;
II. Scheduled receipts: material that has been ordered previously and is expected to arrive in
this period;
III. Projected on hand: expected quantity in inventory at the end of the period, available for
demand in subsequent periods. It is calculated by subtracting the “gross requirements” for
the period from the “scheduled receipts” and “planned order receipts” for the same period as
well as the “projected on hand” from the previous period. When “safety stock” is maintained
and/or units are “allocated” to future orders, these amounts must be added to the gross
requirements before calculating the “projected on hand”;
IV. Net requirements: the reduction of “gross requirements” by the “scheduled receipts” in the
period plus the “projected on hand” in the previous period. This indicates the net number of
items that must be provided to satisfy the parent or master schedule requirements;
V. Planned order receipts: the size of the planned order (the order has not yet been placed) and
when it is needed. This appears in the same period as “net requirements”, but its size is
modified by the appropriate lot sizing policy (see also next section). If the lot sizing policy is
not lot‐for‐lot, the planned order quantity will generally exceed the “net requirements.” Any
excess beyond the “net requirements” goes into “projected on hand” inventory. With lot‐for‐
lot ordering, the “planned order receipts” is always the same as the “net requirements”;
VI. Planned order releases: when the order should be placed (released) so the items are
available when needed by the parent. This is the same as the “planned order receipts” offset
for lead times. “Planned order releases” at level j generate material requirements are the
lower levels j + 1, etc. When the order is placed, it is removed from the “planned order
receipts” and “planned order releases” rows and entered in the “scheduled receipts” row.
“Planned order releases” show the what, how many, and when of MRP.
Table 9. Typical MRP matrix.
Item:
Level:
Lead time =
On hand =
Period
PD
1
Safety stock =
2
3
Allocated =
4
Lot
5
sizing policy =
6
…
Gross requirements
Scheduled receipts
Projected on hand
Net requirements
Planned order receipts
Planned order releases
Example 2: MRP
Given: There are orders for 103 units of product A in period 8 and 200 units of product Q in period 7. The
on hand inventory levels for each item are A = 18, Q = 6, B = 10, C = 20, D = 0, and E = 30. A safety stock of
five units is maintained on product A and six units on product Q with no safety stock on other
components. Additionally, ten units of the 18 units on hand of product A are already allocated to
particular customers. There are no open orders (scheduled receipts) on any item. The lot size for items A,
120
Integrated Logistics
Q, B, and C is the same as the net requirements (lot‐for‐lot ordering), while the lot size for D is 200 units
and for E is 500 units.
Question: Develop an MRP plan for products A and Q with the product structures (BOM) given below.
What should be the size of the orders for each item, and when should orders be released?
End item A
Lead time = 4
B(1)
C(2)
Lead time = 3
Lead time = 2
D(1)
E(2)
Lead time = 1
Lead time = 1
End item Q
Lead time = 2
E(1)
C(1)
Lead time = 1
Lead time = 2
D(1)
E(2)
Lead time = 1
Lead time = 1
Figure 20. BOM/Lead time information for example 2.
Solution
Since item E appears at level 1 and level 2 in product Q and also at level 2 in product A, it is best to assign
it the lowest level of 2 in order to avoid that we will calculate a material requirements plan for it more
than once. The MRP plan for 103 product A and 200 product Q is shown in Table 10. The table was
developed in the following manner:
The first step is to establish the gross requirements for items A and Q, which are given as 103 and 200
units, respectively.
The planned order releases for A and Q in periods 4 and 5 are exploded (multiplied by use quantities
of items B, C, and E) and accumulated as gross requirements for items B, C, and E. Only items B and C are
then calculated.
The planned order releases for B and C in periods 1, 2, and 3 are exploded and accumulated as gross
requirements for items D and E. Item E already has a gross requirement of 200 units in period 5 from
item Q’s explosion.
Note that we have introduced in example 2 safety stocks. Safety stocks can be used within MRP, but it
does not consider them available for regular use. Safety stock is recommended at the end item level but
not the component level in MRP. The need for safety stocks of components is reduced by MRP since it
calculates the quantities of components needed and when they are needed.
With MRP logic, it is possible to have low level items scheduled in a period that is no longer feasible,
i.e. it should have happened in the past. It is then needed to either revise the MPS and shift end item
demand to later periods, or otherwise to compress lead time by expediting the item, or to reduce lead
time for the lot by allowing overtime in the manufacturing shop.
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Integrated Logistics
Table 10. MPR plan for example 2.
Item:A Level:0
lot-for-lot
Lead time =4
PD
On hand =18
1
2
Safety stock =5
3
4
Gross requirements
Scheduled receipts
Projected on hand
3
3
3
3
3
Net requirements
Planned order receipts
Planned order releases
100
Item:Q Level:0 Lead time =2 On hand =6 Safety stock =6
lot-for-lot
PD
1
2
3
4
Gross requirements
Scheduled receipts
Projected on hand
0
0
0
0
0
Net requirements
Planned order receipts
Planned order releases
Item:B Level:1 Lead time =3 On hand =10 Safety stock =
lot-for-lot
PD
1
2
3
4
Gross requirements
Scheduled receipts
Projected on hand
10
Net requirements
Planned order receipts
Planned order releases
Item:C Level:1
Lead time =2
policy: lot-for-lot
PD
Gross requirements
Scheduled receipts
Projected on hand
20
Net requirements
Planned order receipts
Planned order releases
Item:D Level:2 Lead time =1
policy: 200
PD
Gross requirements
Scheduled receipts
Projected on hand
0
Net requirements
Planned order receipts
Planned order releases
Item:E Level:2
Lead time =1
policy: 500
PD
Gross requirements
Scheduled receipts
Projected on hand
Net requirements
Planned order receipts
Planned order releases
6.1.3
30
Allocated =10 Lot sizing policy:
5
6
7
8
103
3
3
Allocated =0
5
6
3
0
100
100
Lot sizing policy:
7
8
200
0
0
200
Allocated = 0
5
6
0
200
200
Lot sizing policy:
7
8
100
10
10
90
On hand =20
1
20
2
20
10
0
90
90
Safety stock =0
3
20
4
5
200
200
0
180
180
0
200
200
180
On hand =0
200
Safety stock =0
1
2
3
180
200
0
20
180
200
200
200
On hand =30
20
180
200
1
2
3
360
400
170
330
500
500
270
230
500
30
500
4
5
Safety stock =0
4
5
Allocated =0
6
7
Allocated =0
6
7
Allocated =0
6
7
Lot sizing
8
Lot sizing
8
Lot sizing
8
200
270
70
Lot sizing policies
Several methods called “lot sizing policies” are commonly used to determine the planned order releases
for each period. We have already encountered the simple (and widely used) lot‐for‐lot policy, which
basically says that planned order receipts are set equal to net requirements. We now discuss some other
often used lot sizing policies: period order quantity (POQ), economic order quantity (EOQ), and the part‐
period balancing method (PBB).
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Integrated Logistics
To illustrate the various methods, we assume that the average demand for a component X is 225 units
per period, the setup costs of producing (or purchasing) the component X is £225, and the cost of holding
one unit of X in inventory for one period is £0.5. We use Table 11 as the base case, in which the lot‐for‐lot
approach is used.
Table 11. Example to illustrate various lot sizing policies.
Item:X Level:
Lead time =2
On hand =150
policy: lot-for-lot
PD
1
2
3
Gross reqrmnts
Scheduled rcpts
Projected on hand
Net requirements
Pl. order rcpts
Pl. order relses
150
100
50
100
150
140
Safety stock =0
4
5
6
Allocated =0
7
Lot sizing
8
9
10
11
12
240
100
340
100
240
100
340
140
140
100
100
340
340
340
100
100
100
100
240
240
100
100
340
340
340
100
240
The total cost is the sum of holding costs and setup costs:
Holding costs = 0.5(150+150+100)=£200
Setup costs = 7 (225) = £1575
Total costs = £200 + £1575 = £1775
POQ method
This method simply sets, in a period in which production is scheduled (nonzero planned order releases)
the production quantity equal to the sum of P positive net requirements, where P is a fixed number.
Table 12. Example using POQ lot sizing method.
Item:X Level:..
policy: POQ
Lead time =2
PD
Gr.reqrmnts
Sched.rcpts
Proj.on hand
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
150
On hand =150
1
2
3
4
240
100
150
100
50
100
440
140
580
340
580
5
Safety stock =0
6
340
340
440
7
Allocated =0
8
9
10
100
240
100
340
100
440
100
Lot sizing
11
12
340
340
340
340
Table 12 illustrates POQ applied to the example of Table 11 assuming P = 3. Note that we must
produce in period 1 to meet period 3’s net requirements. Since P = 3, out planned order releases for
period 1 should meet total net requirements for periods 3, 4, and 6. To meet net requirements in period
8, we need to produce in period 6, and the quantity produced equals total net requirements of periods 8,
9, and 10.
The total cost is the sum of holding costs and setup costs:
Holding costs = 0.5(150+100+440+340+340+340+100)=£905
Setup costs = 3 (225) = £675
Total costs = £905 + £675 = £1580
EOQ method
In this method, the batch size is set equal to the well‐known economic order quantity EOQ =
2 (setup cost)(average demand per period)
Lot size
(unit holding cost per period)
22
If this results in a failure to meet any period’s net requirement, we then produce the smallest multiple
of the EOQ (i.e. lot size = 2 EOQ, or 3 EOQ, and so on) that will avoid a shortage.
This gives in the example of Table 11:
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Integrated Logistics
Lot size
23
2 (225)(225)
450 units
(0.5)
Table 13. Example using EOQ as lot sizing method.
Item: X Level:..
policy: EOQ
Lead time =2
PD
Gr.reqrmnts
Sched.rcpts
Proj.on hand
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
150
On hand =150
1
2
3
4
240
100
150
100
50
100
310
140
450
210
450
Safety stock =0
5
6
7
340
210
320
130
450
450
320
Allocated = 0
8
9
10
100
240
100
220
430
20
450
330
450
Lot sizing
11
12
340
330
440
10
450
450
We begin by producing 450 units during period 1; this will arrive in time to meet period 3
requirements. These 450 units will enable us to meet all requirements through to period 5, so our next
production of 450 units will be during period 4. In period 9, we will need more units of X, so we produce
450 units during period 7. Finally, we need to produce 450 units during period 10.
The total cost incurred:
Holding costs = 0.5(150+100+…+ 440)=£1685
Setup costs = 4 (225) = £900
Total costs = £1685 + £900 = £2585
Many companies adopt a fixed order quantity lot sizing policy. With this method, each production run
(or purchase order) is of the same size, not necessarily the EOQ.
PBB method
In this method, we make each lot sizing decision by producing the number of periods of net requirements
that makes the holding cost associated with the production batch as close as possible to the setup cost
for producing (purchasing) the batch. This idea is based on the fact that the EOQ minimises costs at an
order quantity for which setup and holding costs are equal.
The calculations when applied to our example from Table 11 are shown in the following three tables.
We first need to determine the lot size for the first production run which is to produce good for period 3
at least. We see from Table 14 that the holding costs will be closest to setup costs when producing for
periods 3 and 4.
Table 14. Determining the lot size for the first production run.
Produced for periods
Setup costs (£)
Holding costs (£)
3
3,4
3,4,6
225
225
225
0
0.5(100) = 50*
0.5(3(340)+100) = 560
As we produce in the first batch for periods 3 and 4, we need to start producing the next batch in
period 4 to meet (at least) requirements of period 6. Calculations are then shown in Table 15. This gives
as solution to produce for periods 6 and 8. The next production run therefore needs to start with
producing for period 9 at least; calculations are shown in Table 16.
Table 15. Determining the lot size for the second production run.
Produced for periods
Setup costs (£)
Holding costs (£)
6
6,8
6,8,9
225
225
225
0
2(0.5)(100) = 100*
2(0.5)(100) +3(0.5)(240) = 460
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Table 16. Determining the lot size for the third production run.
Produced for periods
Setup costs (£)
Holding costs (£)
9
9, 10
9,10,12
225
225
225
0
0.5(100) = 50*
0.5(3(340)+100) = 560
The last batch to produce is in period 10 to meet period 12’s requirements. The complete solution is
displayed in Table 17.
Table 17. Example using PPB as lot sizing method.
Item: X Level: ..
sizing policy: PPB
PD
Gr.reqrmnts
Sched.rcpts
Proj.on hand
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
150
Lead time = 2
On hand = 150
Safety stock =
7
1
2
3
4
240
100
150
100
50
100
240
100
140
240
5
6
340
100
340
440
440
100
340
0
Allocated = 0
8
9
10
100
240
100
11
100
240
340
Lot
12
340
340
340
The total cost incurred:
Holding costs = 0.5(150+100+…+ 100)=£325
Setup costs = 4 (225) = £900
Total costs = £1225
In this example, PPB produced the lowest total cost. For other examples, however, POQ or EOQ might
outperform PPB. The only method guaranteed to yield a lot sizing solution that minimises total costs is
the Wagner‐Whitin method. Unfortunately, the Wagner‐Whitin method is more complex than most
practitioners would like, so it is rarely used. Another method that gives a good performance in practise is
called the Silver‐Meal method. Both the Wagner‐Whitin optimal method and the Silver‐Meal
approximate method is described in Winston (1994).
6.1.4
Remarks on MRP
MRP has been come into use with the rise of computers in the late 1970s.
Even a slight change in the requirements for an end item may cause substantial changes in the timing
and lot sizes for subassemblies, components, and parts. This phenomenon is known as the
nervousness of the MRP system.
An important decision is the frequency with which MRP records are updated. In a regenerative MPR
system, the entire record for each end item, subassembly, component, etc is updated periodically. In
our examples, the record might be updated every period. Every period, we only place orders for the
first period. The approach is sometimes also referred to as a rolling planning horizon.
The MPS should be realistic in terms of what the production facility can achieve. MRP and MPS do
not take capacity restrictions explicitly into account and this may cause practical problems, overtime,
etc.
Assuming everything works as planned, the MRP will result in no shortages or stock outs.
Unfortunately, four types of uncertainty may cause shortages to occur:
Lead time uncertainty.
End item demand time uncertainty.
End item demand size uncertainty.
Production yield uncertainty.
Two approaches are used to avoid shortages causes by these four types of uncertainty:
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Integrated Logistics
Plan orders so they arrive earlier than needed. This is called the safety lead time approach.
This works best for dealing with lead time and demand time uncertainty.
Keep more end items in inventory than what is needed (and increase production batches in
case this inventory is being depleted). This is the safety stock approach. This works best for
demand size and production yield uncertainty.
References
Winston, W.L. 1994. Operations Research: Applications and Algorithms, 3th edition. Duxbury Press. See Chapter 20:
Deterministic Dynamic Programming.
Winston, L.W. 1994. Operations Research: Applications and Algorithms (3rd ed.). Duxbury Press, California. ISBN 0534209718.
Chapter 18, Section 1.
Exercises
1. Given: Table 18 gives gross requirements for final products A and B during the next six weeks. Each unit
of A uses two units of C, and each unit of B uses three units of C. Each unit of C uses four units of D.
Assume that A and B can be produced in zero time once enough C is available. The lead time for both C
and D is one week. At the beginning of week 1, 120 units of C and 160 units of D are on hand. Fifty units
of C and 80 units of D are scheduled to be received at the beginning of week 2. The average weekly
demand is 100 units. The cost per production run is £200, and the cost of holding one unit in inventory
for a week is £4.
Questions:
1. Using the lot‐for‐lot method, determine the MRP records for C and D.
2. Determine the MRP record for C if POC (P = 2) is used.
3. Determine the MRP record for C if the EOQ lot sizing method is used.
4. Determine the MRP record for C if the PPB lot sizing method is used.
Table 18. Data for question 1.
Week
1
20
10
A
B
2
0
30
3
30
0
4
10
10
5
20
5
6
0
10
2. Given: End item A consists of 3 units B, 1 unit C, and 2 units D. B consists of 2 units E and 1 units D.
C is assembled from 1 unit B and 2 units E. Every unit E consists of a unit F. Components B, C, E and F have
a lead time of 1 week. A and D’s lead time is 2 weeks. A, B, and F are produced lot‐for‐lot. For C, D, and E,
a fixed order quantity is used of 50, 100, and 250 units, respectively. Components C, E, and F have 10,
150, and 300 units on hand, respectively. All other components have no on hand inventory. The
scheduled receipts for A are 10 in week 6, 50 units of E and F in week 4, and 100 units D in week 1 and
again in week 2. There are no other scheduled receipts. The MPS states a gross requirement of 30 units A
in week 5 and 30 units A in week 8.
Question: Determine the MRP plan for all items.
3. We presently have 18 units of product A on hand. It takes 2 week to produce a unit of product A;
POQ (P = 4 weeks) is being used to determine planned production. The gross requirements for product A
for the next seven weeks are shown in the table below.
Questions:
I. Determine planned production for each week.
II. Suppose week 2’s gross requirements is reduced by 1 (to 13 units). How does the planned
production for each week change? How does this illustrate nervousness?
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Integrated Logistics
Table 19. Data for question 3.
Week
1
2
2
14
3
3
4
6
5
2
6
1
7
45
Solutions
1. i. Using the lot‐for‐lot method, we obtain the following MRP records for C and D:
Item: C Level: 1
Lead time = 1
sizing policy: lot-for-lot
PD
1
Gr.reqrmnts
70
Sched.rcpts
Proj.on hand
120
50
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
Item: D Level: 2
Lead time = 1
sizing policy: lot-for-lot
PD
1
Gr.reqrmnts
Sched.rcpts
Proj.on hand
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
160
160
On hand =
120
Safety stock = 0
2
3
4
5
6
90
50
10
60
50
55
30
50
50
55
Safety
55
30
55
30
30
stock = 0
50
50
50
50
On hand = 160
2
3
4
5
200
80
40
200
220
120
160
160
220
220
220
120
120
120
160
Allocated =0
Lot
Allocated =0
Lot
Allocated =0
Lot
6
ii. Using the POQ (P = 2) method, we obtain the following MRP record for C:
Item: C Level: 1
Lead time = 1
sizing policy: POQ (P=2)
PD
1
Gr.reqrmnts
Sched.rcpts
Proj.on hand
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
70
120
50
On hand =
120
Safety stock = 0
2
3
4
5
6
90
50
10
60
50
55
30
50
50
100
(50)
30
55
85
(30)
100
85
iii. The EOQ for C, assuming that the average weekly demand of 100 units is related to C items, is
equal to (2 (200)100/4) = 100 units. We obtain the following MRP record for C:
Item: C Level: 1
sizing policy: EOQ
Lead time = 1
PD
Gr.reqrmnts
Sched.rcpts
Proj.on hand
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
120
On hand =
120
Safety stock = 0
1
2
3
4
5
6
70
90
50
10
60
50
55
30
50
50
100
(50)
45
55
100
15
(30)
50
100
Allocated =0
Lot
100
iv. Using PPB, we first have to calculate the lot sizes to be used. The first period for which a net
requirement arises is period 3. Therefore:
Produced for weeks
3
3,4
3,4,5
Produced for weeks
5
5,6
Setup cost
200
200
200
Setup cost
200
200
Holding cost
0
4(50)=200*
4(50 + 2 (55))=640
Holding cost
0
4(30)=120*
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The MRP record therefore is:
Item: C Level: 1
sizing policy: PPB
Lead time = 1
PD
Gr.reqrmnts
Sched.rcpts
Proj.on hand
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
120
On hand =
120
Safety stock = 0
1
2
3
4
5
6
70
90
50
10
60
50
55
30
50
50
100
(50)
30
55
85
(30)
50
100
Allocated =0
Lot
85
By coincidence, this record is equal to the record obtained with the POQ (P = 2) method.
2. The BOM can be drawn as follows:
A
B(3)
D
Level 0
C
E(2)
B
D(2)
Level 1
Level 2
E(2)
Level 3
F
D
E(2)
F
Level 4
F
Notice that items of the same type arise on more than one level in the BOM. Indeed, B is on level 1 as
well as level 2, E on level 2 and level 3, D on level 1 and 3, and F on level 3 and 4. The easiest approach is
to first bring all items down to their lowest level before the MRP calculations start. Therefore we modify
the BOM to:
A
Level 0
C
Level 1
B(3)
D
B
E(2)
F
D
Level 2
E(2)
E(2)
F
F
D(2)
Level 3
Level 4
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Item: A Level: 0
Lead time = 2
sizing policy: lot-for-lot
PD
1
Gr.reqrmnts
Sched.rcpts
Proj.on hand
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
Item: C Level: 1
sizing policy: 50
3
4
5
Allocated =0
6
7
Lead time = 1
1
50
Lead time = 2
2
30
10
3
30
20
50
50
On hand =
0
10
30
30
10
Safety stock = 0
60
50
90
50
90
90
On hand = 0
60
60
90
100
60
60
On hand =
2
100
150
3
250
8
10
10
4
5
Allocated =0
90
50
100
50
7
Lot
20
50
3
50
6
3
Gr.reqrmnts
100
280
Sched.rcpts
Proj.on hand
150
50
20
20
Net reqrmnts
230
Pl.ord.rcpts
250
Pl.ord.relses
250
Item: F Level: 4
Lead time = 1
On hand = 300
sizing policy: lot-for-lot
PD
1
2
3
300
20
20
2
2
1
5
30
10
8
20
Safety stock = 0
Allocated =0
4
1
Lead time = 1
PD
On hand =
Lot
30
30
30
PD
Gr.reqrmnts
Sched.rcpts
Proj.on hand
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
Safety stock = 0
10
10
Gr.reqrmnts
Sched.rcpts
Proj.on hand
10
10
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
Item: B Level: 2
Lead time = 1
sizing policy: lot-for-lot
PD
1
Gr.reqrmnts
Sched.rcpts
Proj.on hand
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
Item: E Level: 3
sizing policy: 250
2
0
30
PD
Gr.reqrmnts
Sched.rcpts
Proj.on hand
Net reqrmnts
Pl.ord.rcpts
Pl.ord.relses
Item: D Level: 3
sizing policy: 100
On hand =
6
60
Safety stock = 0
4
7
6
60
40
40
60
100
(40)
Safety stock = 0
4
5
8
Allocated =0
5
7
Lot
Lot
8
Allocated =0
6
7
8
200
200
200
Lot
120
50
70
200
50
250
250
Safety stock = 0
4
5
6
Allocated =0
7
Lot
8
250
50
50
50
150
150
150
3.
i.
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Item: A Level: 0
Lead time = 2
sizing policy: POQ (P=4)
PD
1
Gross requirements
Scheduled receipts
Projected on hand
Net requirements
Planned order receipts
Planned order releases
18
On hand =
18
Safety stock = 0
Allocated =0
2
3
4
5
6
7
2
14
3
6
2
1
45
16
2
9
1
10
3
6
1
2
1
45
45
10
Lot
45
ii.
Item: A Level: 0
Lead time = 2
sizing policy: POQ (P=4)
PD
1
Gross requirements
Scheduled receipts
Projected on hand
Net requirements
Planned order receipts
Planned order releases
18
On hand =
18
Safety stock = 0
Allocated =0
2
3
4
5
6
7
2
13
3
6
2
1
45
16
3
48
6
54
46
2
45
1
45
Lot
54
A slight change in gross requirements results in a large change in production levels as well as
inventory levels! Small changes on MPS data may therefore result in large changes down the MRP
explosion.
The amplification arises especially from the lot sizing technique used, which combines the positive net
requirements from 4 future periods. If we would use lot‐for‐lot instead, the resulting MRP would show
much less nervousness:
Item: A Level: 0
Lead time = 2
sizing policy: lot-for-lot
PD
1
Gross requirements
Scheduled receipts
Projected on hand
Net requirements
Planned order receipts
Planned order releases
18
Safety stock = 0
Allocated =0
3
4
5
6
7
2
14
3
6
2
1
45
16
2
1
1
2
6
6
1
2
2
45
1
1
45
45
1
18
18
2
Item: A Level: 0
Lead time = 2
sizing policy: POQ (P=4)
PD
1
Gross requirements
Scheduled receipts
Projected on hand
Net requirements
Planned order receipts
Planned order releases
On hand =
6
On hand =
18
Safety stock = 0
Allocated =0
2
3
4
5
6
7
2
13
3
6
2
1
45
16
3
2
2
45
1
1
45
45
2
6
6
1
6
Lot
Lot
6.2 Just‐In‐Time
Based on the example of many successful Japanese companies (most notably Toyota), many companies
worldwide have attempted to reduce inventory levels by implementing a Just‐In‐Time (JIT) approach to
production and purchasing. The JIT approach consists of producing products or obtaining products from
suppliers at the moment they are required. Since JIT strives to reduce inventory levels, JIT is also known
as “stockless production” or “zero inventories”.
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Integrated Logistics
6.2.1
Motivation
The main motivation behind JIT is that inventory represents inefficiency. Just as high water levels in a
river hide dangerous rocks, high inventory levels in a company hide sources of inefficiency and causes of
poor product quality.
Advocates of JIT believe that companies often underestimate the holding cost ch used in the
traditional inventory models such as the Economic Order Quantity (EOQ) model. The advocates of JIT
argue that in the long run, the “fixed” costs of a warehouse and the cost of shop storage should be
considered as a variable cost and therefore included in the value of the holding cost ch. This would, of
course, reduce optimal lot size and average inventory levels.
Another important parameter in inventory models is the replenishment cost cp. Often a major
component of it is the time required to setup a machine for a production run. A reduction of setup cost cp
will typically increase optimal replenishment frequency, reduce optimal lot size, and reduce inventory
levels. Much of the success of implementing JIT comes from the success with which one can achieve a
reduction in setup times.
Another traditional argument to hold inventory is that safety stock must be held to reduce shortages
caused by demand uncertainty and supply uncertainty (such as variability in lead times). JIT reduces the
impact of demand uncertainty on inventory levels by “levelling” the production schedule. For example, if
2000 cars and 1000 lorries per month must be produced and a plant is open 20 days per month, the plant
should aim for producing 2000/20 = 100 cars per day and 1000/20 = 50 lorries per day, and the sequence
in which the vehicles roll from the production line should also reflect the relative demand: car, lorry, car,
car, lorry, car, car, lorry, etc. For JIT to succeed, it is critical that the quantity of a product produced each
day varies by at most 10% from the average daily production level.
By reducing safety stock levels, the causes of supply variability (such as unreliable suppliers, frequent
machine breakdowns, product quality problems) are exposed. By reducing or eliminating the causes of
supply variability, there is less need to keep large amounts of safety stocks.
6.2.2
Push and pull systems
To understand how JIT differs from the traditional approach, we must understand the difference
between push and pull (production) systems. Consider a manufacturing process in which a product must
pass through four workstations before being completed, beginning at station 1 and ending at 4. Think of
the product as flowing down a river. Then the first workstation 1 is farthest upstream and station 4 is
farthest downstream.
In a “push” system, materials push their way through the system from station 1 to 4. When, for
example, 100 units of a product are completed at station 1, they are pushed toward station 2, causing
work for the employees at station 2. Once 20 units, for example, are finished on station 2, they are
pushed to station 3, and so on. Traditional production systems are often push systems, with relatively
large setup costs, so large production lot sizes. This can cause large work‐in‐process inventories to
accumulate. Furthermore, if for example at some moment in time station 3 breaks down, then station 1
and 2 may still produce with stock accumulating before station 3, unless someone tells them to stop.
In contrast, JIT is a pull system, a system in which an upstream station does not produce anything
unless production is authorised by the immediately succeeding downstream station. For example, station
2 cannot work on any material unless authorised by station 3. This also means that if at some point in
time station 3 breaks down, all stations upstream will soon also stop and the build‐up of extra inventory
is avoided.
6.2.3
The Kanban System
The best known method used to implement JIT is Toyota’s Kanban system. The following three types of
items are vital to implementing JIT through the Kanban system:
I. Containers. Each container holds a standard number of parts – usually less than 10% of the daily
requirement for the part. This keeps lot sizes relatively small.
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Integrated Logistics
II. Move cards (also known as withdrawal or conveyance cards). Move cards are used to authorise
movement of a container between two consecutive operations – for example, from station 4 to
station 3 and back to station 4. A container cannot move unless a move card is attached to it.
III. Production cards. A production card is used to authorise production of a container of parts.
Workers cannot produce the parts needed to fill a container unless authorised by a production
card.
There is a separate set of containers, move cards, and production cards for each possible flow of
material. For example, there is one set of containers, move cards, and production cards for movement
between workstation 3 and 4, one other set for movement between 2 and 3, and yet one other set for
movement between 1 and 2.
For example, let’s focus on the movement between station 3 and 4. Suppose we are starting to work
on a container of parts that arrived at station 4. This container will have a move card attached to it.
Remove that card and attach it to an empty container at station 4 (call this container 1). Now container 1
(with its move card) can be transported back to station 3. At station 3, find a full container of parts (call it
container 2). Container 2 will still have its production card attached to it. Remove that production card
and put it in station 3’s production card file box. Replace the production card on container 2 with the
move card from container 1. You can now transport container 2 to station 4. Production workers from
station 3 can use the production card that you have just placed in the file box to attach it to the empty
container 1 and can therefore start production to fill up this container.
This shows how station 4 “pulls” material downstream through the system. The beauty of the
approach is its simplicity: unlike MRP, for example, no complicated paperwork or computer printouts are
needed to keep track of the inventory status at each station. The cards do the job automatically. If we
reduce the number of cards between two workstations, we will automatically reduce the inventory
between the two stations. If we have too few cards, however, shortages may occur frequently. Only trial
and error can determine the “right” number of cards.
The rules of the Kanban system can be summarised as follows:
I.
II.
III.
IV.
At all times, each container must have a card (either a move or a production card) attached to it.
Never move a container unless it has a move card attached to it.
Never begin producing a container without a production card.
A container should always contain a standard number of parts.
If station 4 breaks down, for example, then this will eventually result in production stopping at all
upstream stations (1 – 3). At first glance, this might appear to reduce the plant’s productivity. In reality,
however, frequent plant shutdowns force workers to concentrate more on quality and plant efficiency. In
the long term, the shutdown will become less frequent, and a more efficient (and higher quality)
production process will result.
6.2.4
How many cards?
Here’s a way how to calculate (in theory) how many move and production cards are needed to enable the
plant to meet daily demand without holding unnecessary inventory. Let us again consider the material
movement between station 3 and 4. Define:
c = number of parts in each container (less than 10% of daily requirement);
Tm = time (in days) for worker to go from station 4 to get a container at station 3 and return with it
to station 4;
Tp = time (in days) required to produce a container with c parts at station 3;
D = daily demand for parts produced at station 3;
M = number of move cards;
P = number of production cards.
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Integrated Logistics
How many move cards are needed? We can make 1/Tm trips per day to get containers, and on each
trip we can get M containers with c parts per container. Since we need D parts per day, in theory we can
meet daily requirements without accumulating excess inventory if:
1
M c D
Tm
or
M
DTm
c
How many production cards are needed? With P cards, we can produce up to P containers (each with
c parts) and this 1/Tp times per day. Since we must produce D units per day, we can in theory meet daily
requirements without accumulating excess inventory if
1
P c
Tp
D
or
P
DTp
c
For these formulas to be valid, we must assume that there is no variability in move and production
times. The validity of the formulas also requires that the containers will be ready to pick up whenever a
worker from station 4 arrives at station 3. To allow for some slack in the system, the following formulas
are often used to determine the required number of move and production cards:
M
DTm (1 k)
c
P
D Tp (1 k)
c
where k is a safety factor. If JIT is working well, k should be near 0. Toyota strives to keep k as close as
possible to 0.10. In reality, a company may begin with k = 0.50. If shortages are infrequent, a card may be
removed from the system. This will reduce inventory levels. If shortages are still rare, another card may
be removed. This process continues until satisfactory results, i.e. infrequent shortages and low inventory
levels, have been obtained.
Example
Assume k = 0. Station 4 must obtain 200 parts per day from station 3. Each container contains 10 parts,
and it takes Tp = 0.10 day at station 3 to fill a container. Going from station 4 to station 3 with an empty
container, pick up a full container at station 3 and return to station 4 takes a total of Tm = 0.05 day. Then:
M = 200 (0.05) (1+0)/10 = 1 move card
P = 200 (0.10) (1+0)/10 = 2 production cards
The maximum inventory level is (M + P) c = 30 items.
Let us see how 1 move card and 2 production cards would enable the system to keep up with demand
without accumulating excess inventories. Observe that station 4 will “use up” a container in 0.05 day.
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Suppose that at time 0 (the beginning of a work day) station 4 has one full and one empty container, and
station 3 has two full containers. At time 0, we begin using parts at station 4, and so a worker takes the
empty container (with a move card) to station 3. The worker arrives at station 3 at time 0.025. There, the
worker picks up a full container and removes the production card. At station 3, the detached production
card is used to start production for filling the empty container. After attaching the move card to the full
container, the worker heads back to station 4 and arrives at time 0.05, just in time for the full container
of parts to be used when the full container already at station 4 is used up at time 0.05. At time 0.05,
another worker leaves station 4 and arrives at station 3 at 0.075. Production begins at station 3 on
another container. (This is okay, because we have two production cards). We arrive back at station 4 with
a full container (again, just in time) at time 0.10. At time 0.10, a worker again leaves for station 3, and a
time 0.125, the worker arrives at station 3 and can pickup the full container whose production began at
time 0.25. In summary:
1. At times 0; 0.5; 0.10; 0.15; …, a worker leaves station 4 for station 3.
2. At times 0.25; 0.75; 0.125; …., a worker arrives at station 3 and picks up a full container. Station 3
also simultaneously begins producing a container a parts.
3. At times 0.05; 0.10; 0.15, a worker arrives at station 4 with a full container (just in time)
The maximum inventory level will be 30 parts. With more cards, the maximum inventory level would
increase, but there would be slack in the system to handle unexpected shutdowns, shortages, or delays.
Also note that a levelled production schedule (i.e. always producing 200 parts per day, for example) is
critical to the success of JIT. Section 5 describes how Toyota deals with fluctuations in demand.
6.2.5
Subcontractors
When a company such as Toyota implements JIT, the subcontractors also have to adopt JIT delivery to the
Toyota plants. It is usually the subcontractor who delivers its products to the company. As JIT implies
frequent deliveries in small lot sizes, the inventory levels at both the Toyota plant and the subcontractors
can remain low. However, the distribution costs will rise.
To counter the high distribution costs, these suppliers often collaborate with each other and use the
so‐called “round‐tour mixed‐loading” transportation system. Suppose, for example, that four
subcontracted companies A, B, C, and D are all located in the eastern area of the Toyota plant. Suppose
further that they have to bring their products four times during the day‐shift in small lot sizes. The first
delivery at 9am could be made by subcontractor A, also picking up on the way products for companies B,
C, and D in A’s truck. The second delivery at 11am could be made by company B similarly picking up the
products of A, C, and D on the way. The third delivery at 2pm would be made by company C and the
fourth delivery at 4am would be made by company D.
The subcontractors themselves would also adopt the JIT system in their production plants to keep
inventory levels of work in process down, to increase product and process quality, and lower costs. This
will be beneficial to Toyota as the supplies can be offered at lower prices, and ultimately benefit the final
customers as well.
6.2.6
Fluctuations in demand
The JIT system works only if production is smoothed (or “levelled”, see also Section 1) and therefore if
demand is fairly constant. Within reasonable limits, however, the JIT production system of Toyota is able
to deal with various fluctuations in demand. There are three types of demand changes possible:
Case 1: No change in daily total production load, the only changes are in the kinds of end products (for
example, cars in the Toyota situation), their delivery dates, and their quantities.
The solution adopted by Toyota is to only revise the production schedule for the final production line
where the cars are finalised. Then the schedules for all the preceding processes will be automatically
revised by transferring the Kanbans.
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Case 2: Short term small fluctuations in the daily production load, although the monthly total load is
the same.
For this case, Toyota does not increase the or decrease the number of Kanbans, but simply increases
or decreases the number of Kanban movements. The assembly line of Toyota has a two‐shift system. The
day shift is from 8am to 6pm, and the night shift starts at 9pm, ending at 6am. By inserting early
attendance and overtime before and after these shifts, the line always has the same number of workers
at any time the line is in operation. It is thereby able to produce as many units as a three shift system if
necessary. If, on the other hand, demand is lower, the frequency of Kanban movements is reduced which
results in idle capacity or idle time for the workers. Simultaneously, time spent in carrying inventory is
prolonged. If the lead time is short, however, the loss associated with carrying inventory will not be very
large.
Case 3: Seasonal changes in demand, or when the actual monthly demand is larger or smaller than the
predetermined load or the preceding month’s load.
In this case, the number of Kanbans must be decreased or increased, and at the same time, all the
production lines must be rearranged.
To better understand what to do in Case 3, we first introduce the concept of cycle time. Take the
example of Section 1. If 2000 cars and 1000 lorries per month must be produced and a plant is open 20
days per month, the plant should aim for producing 2000/20 = 100 cars per day and 1000/20 = 50 lorries
per day, and the sequence in which the vehicles roll from the production line should also reflect the
relative demand: car, lorry, car, car, lorry, car, car, lorry, etc. The cycle time is the time between two
vehicles leaving the production line:
cycle time
operating hours per day
necessary output per day
21
150
0.14 hours/vehicle
Therefore every 0.14 hours = 8.4 minutes a vehicle should leave the production line. If the total
number of vehicles needed per month is now decreased from 3000 to 2500, for example, then the daily
output necessary is 2500/20 = 125 and the cycle time becomes:
cycle time
operating hours per day
necessary output per day
21
125
0.168 hours/vehicle
or every 10.08 minutes a vehicle should leave the production line.
How to best achieve this change? The basic idea is to lower the production rate by reducing the
number of operators necessary to do the work. The production layout that is found to be most flexible to
allow for such adjustment is the U‐turn format, as depicted in the figures below. The top figure shows, for
example, the configuration under high monthly demand: five operators are used, each operating two
machines.
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entrance
1
exit
entrance
1
exit
Figure 21. U-form layout and worker allocation under high (top) and low (bottom) monthly demand.
The bottom figure would then be the configuration used when monthly demand is lower: three
operators, two of them operating thee machines and one four machines. The work on each machine will
go slower as an operator now has more machines under control. As a result, the output will be reduced.
In order for this to work, operators have to be multi‐skilled workers, as they may have to change work
from time to time. Each job on each machine has to be perfectly standardised so that different workers
will always follow the same procedures on each machine so that product quality remains constant.
6.2.7
Comparison JIT and MRP
Experience with JIT and MRP has shown, not surprisingly, that MRP tends to outperform JIT when
demand is highly variable and setup costs are high or setup times are long. JIT tends to outperform MRP
when demand is stable and setup times are small.
Clearly, the smaller lot sizes associated with JIT will increase the number of setups required. Consider
a company producing many different products. If such a company uses JIT, high setup costs will probably
be incurred because it must frequently change from producing one product to another, due to small lot
size of JIT. This explains why JIT is most commonly used in a repetitive manufacturing environment where
most setup times are small and very few products are made.
Exercises
1. Given: In a Kanban production environment, 100 parts per day are required, 5 parts per container are
used, and Tm = Tp = 0.20 days. Assume that material is being transported from station 1 to station 2.
Questions:
a. Determine the number of move and production cards needed.
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b. At what instants will a person arrive at station 2 with full containers?
c. At what instants will production on containers begin at station 1?
d. What is the maximum number of parts in inventory?
2. Given: In a Kanban production environment, 200 parts per day are needed, 20 parts per container
are used, and Tm = 0.10 days and Tp = 0.30 days. Assume that material is being transported from station 1
to station 2.
Questions:
a. Determine the number of move and production cards needed.
b. At what instants will a person arrive at station 2 with full containers?
c. At what instants will production on containers begin at station 1?
d. What is the maximum number of parts in inventory?
e. What is the cycle time?
3. Given: Consider a BOM as displayed in Figure 22.
A
B(2)
C
D
E
Figure 22. BOM for question 3.
A, B, and C are produced at our own factory. A is assembled in station 2, and components B and C are
both produced in station 1. The setup times for the machine in station 1 are displayed in Table 20.
Table 20. Data for question 3.
Setup time (days)
From B
From C
To B
0
0.02
To C
0.01
0
Production time to produce one B on station 1 is 0.01. Production time to produce one C on station 1
is 0.01. Production is controlled according to the JIT approach using a Kanban system. Containers used
between station 1 and station 2 store 4 B items and 2 C items. The time to bring an empty container from
station 2 to station 1 and bring a full container back is 0.05 days.
To produce one item B, one kit of parts D is needed. To produce one item C, one kit of parts E is
needed. D kits are produced by supplier 1, and E kits are produced by supplier 2. Containers used to
transport D kits from supplier 1 to station 1 contains 16 D kits at a time. The same containers can be used
to transport 8 E kits from supplier 2 to station 1.
Supplier 1 needs a time of 0.2 days to produce a container of D kits. Supplier needs a time of 0.15 days
to produce a container of E kits. To transport a full container, and bring an empty one back, supplier 1
needs 0.2 days. To transport a full container and bring an empty container back, supplier 2 also needs 0.2
days. If supplier 1 and supplier 2 would cooperate and adopt the “round‐tour mixed‐loading” system in
which one container from each is picked up, transport time would be 0.25 days for one roundtrip. The
daily production requirement is 100 A items.
Questions:
a. Determine the number of move and production cards between station 2 and station 1.
b. Determine the number of move cards between supplier 1 and station 1 in case of separate
transport
c. Determine the number of move cards between supplier 2 and station 1 in case of separate
transport.
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d. Determine the number of move cards between suppliers 1 and 2 and station 1 in case of the
round‐tour mixed‐loading approach.
e. Determine the number of production cards for each supplier.
f. Determine the maximum inventory level in the supply chain under the separate transport option
g. Determine the maximum inventory level in the supply chain under the round‐tour mixed‐loading
approach.
Solutions
1. Solution:
a. M = 100 (0.2)/5 = 4 move cards; P = 100 (0.2)/5 = 4 production cards
b. Arrive at 0.20; 0.25; 0.30; 0.35; 0.40; etc. (because there have to be 4 movements in a period of 0.2.)
Note: An alternative way to calculate b: cycle time is 1/100 = 0.01 i.e. every 0.01 days an item has to
roll from the production line, which means that a new container of 5 parts has to arrive every 5 (0.01)
= 0.05 days. Assuming the first walk from station 2 to station 1 starts at 0.0, the person will arrive at
station 4 with first container at time 0,20, and then every 0.05 days the following container has to
arrive at station 4. Hence the sequence is: 0.20; 0.25; 0.30; 0.35; 0.40; etc.
c. Begin at 0.1; 0.15; 0.20; 0.25; 0.30; etc (4 containers in production in a period of 0.2) Note: An
alternative way to calculate c: cycle time is 0.01 days and 5 parts per container means every 0.05 days
production to fill a new container has to commence. The first empty container arrives from station 2
at 0.1, which is the time the production to fill this container starts; then at time 0.15 the next empty
containers from station 2 arrives and production starts for this second container; etc.
d. Begin with 4 full containers at each station. Then max inventory level is 8 (5) = 40 items.
2. Solution:
a. M = 200 (0.1)/20 = 1 move card; P = 200 (0.3)/20 = 3 production cards
b. Cycle time is 1/200 = 0.005 days between two items rolling from the production line. A container has
20 items, therefore a container needs to arrive at station 2 every 20(0.005) = 0.1 days. The first
container will arrive at 0.1; the second at 0.2; then 0.3; etc.
c. Similarly, every 0.1 days a full container has to be ready at station 1, and therefore production to fill a
new container needs to start every 0.1 days. Production can begin at 0.05; then production for the
next container can start at 0.15; then at 0.25, etc. The first container for which production started at
0.05 will be ready at 0.35, just in time for pickup by a worker from station 4.
d. Cycle time = time in a day/ items needed per day = 1 / 200 = 0.005 days. This is the time between two
items rolling from the production line.
3. Solution: (Note. In real life, one will have uncertainties in lead times such as those caused by traffic
jams and therefore one has to introduce safety factors)
a. Between station 2 and station 1: Station 1 would have to produce in the following order: B, B, C, B, B,
C, etc (see BOM). A container therefore contains 4 B and 2 C items. Total demand per day in order to
produce 100 A items per day is therefore 200 B + 100 C = 300 items per day. Thus M = 300 (0.05)/6 =
2.5 => 3 move cards. At station 1, the time to fill one container is the time to produce: B, B, C, B, B, C
and set the machine back in a condition to start production on a B item. Total production time is
hence:
Production time: 6 (0.01) = 0.06 days
Setup time: 2 (0.02) + 2 (0.01) = 0.06 days
Total production time to fill container = 0.06 + 0.06 = 0.12 days
Therefore: P = 300 (0.12)/6 = 6 production cards.
b. M = 200 (0.2)/16 = 2.5 => 3 move cards.
c. M = 100 (0.2)/8 = 2.5 => 3 move cards.
d. M = 300 (0.25)/24 = 3.125 = 4 move cards.
e. For supplier 1: P = 200 (0.2)/16 = 2.5 => 3 production cards. For supplier 2: P = 200 (0.15)/8 = 1.875 =>
2 production cards.
f. 32 B items and 16 C items in containers in station 1 or somewhere between station 1 and station 2; 48
D kits at supplier 1; 96 D kits somewhere between supplier 1 and station 1; 16 E kits at supplier 2 and
40 E kits somewhere between supplier 2 and station 1;
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g. Like in f., but only 64 D kits and 32 E kits somewhere between supplier 1 or 2 and station 1.
References
See also e.g. Chapter 18, Section 2 of: Winston, L.W. 1994. Operations Research: Applications and Algorithms (3rd ed.).
Duxbury Press, California. ISBN 0534209718.
6.3 Bottleneck scheduling
Optimised Production Technology (OPT) is a technique that derives its name from a computer software
system for planning and scheduling in an operational facility in the context of variability. The first versions
of this system were introduced as early as 1979.
The internal algorithms of the OPT software are proprietary, but the general rules have been
explained in Goldratt and Cox’s book “The Goal”, a book in which the message is presented as an
adventure story around a manager facing the closure of his factory unless a major improvement in
productivity (more output at lower costs) is obtained within a few days. As a technique, OPT provides yet
another perspective on planning and scheduling compared to MRP and JIT, although it is more a
philosophy than a real technique.
OPT is closely linked with TOC, yet another acronym which stands for the Theory of Constraints,
explained in several other books with again Goldratt as one of the authors.
What these two approaches lead to is the view that modelling can help to identify problems and/or to
optimise the Supply Chain. The optimisation model of a Supply Chain is a successful approach
exemplified by for example the acceptance of Linear Programming.
6.3.1
Optimised Production Technology (OPT)
The basic idea of OPT is that in a production environment in which different machines or stations should
work together to produce end products, some of these machines or stations are heavily used and others
are not. The first and main attention for planning and scheduling should go towards keeping these heavily
used machines, the so‐called bottlenecks, working at high utilisation levels. The planning and scheduling
for the other stations, the non‐bottlenecks stations, comes second. These non‐critical resources are
scheduled below capacity to allow for a safety capacity to exist at all times.
The general rules of OPT are summarised here and will be further discussed in the sections:
1.
Do not balance capacity – balance the flow.
2.
The level of utilisation of a non‐bottleneck resource is determined not by its own potential but by
some other constraint in the system.
3.
Activating a resource (making it work) is not synonymous with utilising a resource effectively.
4.
An hour lost at a bottleneck is an hour lost for the entire system.
5.
An hour saved at a non‐bottleneck is a mirage.
6.
Bottlenecks govern both throughput and inventory in the system.
7.
The transfer batch may not and many times should not be equal to the process batch.
8.
The process batch should be variable both along its route and in time.
9.
Priorities can only be set by simultaneously examining all of the system’s constraints. Lead time
is a derivative of the schedule.
OPT investigates five areas of the production environment: variability, bottlenecks, setups, lot sizes,
and priorities.
Variability
Most scheduling methods attempt to balance resources. The number of workers will be constantly
adjusted to balance machine capacity with worker capacity, and the plans for the subsequent work will
utilise these resources at the capacity required to meet demand. It has been shown through
mathematical simulation that a balanced plant cannot exist in the presence of variability. Variability in
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dependent work centres will cause schedules to be missed, resulting in a decreased throughput,
increased inventory, and increased operating costs.
OPT rules are based on the realisation that the constraints on an operation often exist outside the
operation. The variability within an operation not only affects that operation but all subsequent
operations, especially with bottleneck resources. Therefore, OPT’s first rule is: 1. Do not balance capacity
– balance the flow.
Bottlenecks and Non‐bottlenecks
A bottleneck is a restriction or constraint in the flow of material through a resource. A resource, in this
case, is any element needed to make a product, whether it is a machine, a person, or space. A resource is
considered a bottleneck when it is required to operate at 100 % capacity to meet the present schedule.
As the schedule changes or the product mix changes, different resources may become bottlenecks.
To illustrate the interaction between bottlenecks and non‐bottlenecks, four different cases are
presented in Figure 23; they represent virtually all the possible combinations of interaction. In the
examples, a certain product is manufactured that requires the use of two resources, X and Y. This is the
simplest case and can, of course, be expanded for multiple resources. Demand for this product places
different time requirements on the two resources, and an imbalance occurs. X denotes a bottleneck
resource that has a market demand of 100 hours per week; it also has a capacity of 100 hours per week. Y
denotes a non‐bottleneck resources that has a market demand of 75 hours per week and a capacity of
100 hours per week. These two resources can only interact in four ways.
X
Bottleneck resource
Capacity = 100 hours/week
Demand = 100 hours/week
Y
Non‐bottleneck resource
Capacity = 100 hours/week
Demand = 75 hours/week
Case 1
Case 2
Y
X
Case 3
A
X
X
Y
Case 4
Dependent demand
Y
Independent demand
Customer
X
Y
Figure 23. Interaction between resources
In case 1, all product flows from X to Y. In this case, resource X has a requirement of 100 hours per
week and is fully utilised, but resource Y has only 75 hours of work and is therefore underutilised. The
output from X starves Y.
In case 2, all product flows from Y to X and X can again be utilised 100 % of the time. If Y, however, is
activated 100 % of the time, too much product will be produced and flow to resource X; X will not be able
to process all these products. This will increase work‐in‐process inventory, consume resources, and not
increase throughput. To balance the flow, Y needs to be activated only 75 % of the time.
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In case 3, X and Y both feed a common assembly. Having worked through cases 1 and 2, we can see
that X will be utilised 100 % of the time to meet market demand; however, activating Y 100 % of the time
will exceed the demand for this resource and will again build work‐in‐process inventory in front of the
assembly station.
In case 4, X and Y both feed independent customer demand and are not interrelated. Activating X 100
% of the time will exactly meet customer demand. Activating Y 100 % of the time will exceed customer
demand; this will build inventory and not increase throughput. Y needs only be activated 75 % of the
time. Note that in this case the market has become the constraint.
In all four cases the same results were obtained. X was always active, a good indication of a
bottleneck. Y could not be activated more than 75% of the time, or one or more manufacturing goals
were contravened. From this example comes OPT’s second and third rules: 2. The level of utilisation of a
non‐bottleneck resource is determined not by its own potential but by some other constraint in the
system; 3. Activating a resource (making it work) is not synonymous with utilising a resource effectively.
All non‐bottleneck resources must therefore be scheduled not on the basis of their own constraints
but on the basis of the system’s constraints.
Setups
The available time at a bottleneck resource can be divided into two categories: processing time and setup
time. If an hour of setup is saved at a bottleneck resource, an hour of processing time is gained. Taking
the four case example of the previous section, resource X’s capacity has increased to 101 hours per week.
It means that the entire system is now able to satisfy a higher demand per week, i.e. the throughput of
the system has increased. This leads to OPT’s fourth rule: 4. An hour lost at a bottleneck is an hour lost
for the entire system.
At non‐bottlenecks, however, there may be no gain at all in avoiding setups. Taking again the
examples of the X and Y machines, increasing Y’s capacity from 100 to 101 hours per week would not give
us any increase in throughput. In particular, in cases 1‐3, we would never gain from having Y at a capacity
of 101 hours per week, even if machine X’s capacity would be 101 hours per week and demand for X
would likewise have increased to 101 hours per week. We can realise a throughput equivalent to 101
hours per week of demand for X whether Y’s capacity is 100 or 101 hours per week. In case 4, extending
machine Y’s capacity does also make no sense with a weekly demand that requires only about 75% of its
capacity. Therefore OPT’s fifth rule states: 5. An hour saved at a non‐bottleneck is a mirage.
Lot sizes
It is important that setups be saved at the bottlenecks, since time spent not producing affects the entire
system. OPT therefore advises to produce in larger batches on the bottleneck resource than traditional
lot sizing methods (see e.g. Chapter 1, Section 3) would indicate. Indeed, the cost of a setup at a
bottleneck resource is not just the cost associated with the time and work required at this bottleneck
station, but it should be associated with the loss in throughput for the whole system and therefore it is
very high! Therefore, OPT’s sixth rule reads as follows: 6. Bottlenecks govern both throughput and
inventory in the system.
Since non‐bottleneck resources have actually “spare time” available, it is possible to produce in
smaller lot sizes on such resources. Smaller lot sizes means more setups and setup time, but time is
available and the cost of such setups may only involve the direct labour cost: there is no opportunity cost
of throughput loss as the one associated with bottleneck resources.
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Setup
time
X
Run time
Bottleneck resource
Time saved in setup becomes runtime
Additional run time results in additional hour of
throughput for the whole system
Setup
time
Y
Run time
Idle
time
Non‐bottleneck resource
Any time saved in setup becomes idle time
An additional hour of idle time is worth nothing
Figure 24. Resource activities
The benefit of producing in smaller lot sizes on non‐bottleneck stations are system wide advantages in
that smaller lot sizes may reduce lead times and can help to keep bottleneck resources working all the
time. Indeed, in case 2 of the examples presented before in which machine Y feeds the bottleneck X,
smaller lot sizes on Y may be better able to keep X working all the time. (This is especially relevant in
when there is variability in the system).
As a batch moves through a facility, it encounters both bottleneck and non‐bottleneck resources, and
the question arises as to whether a batch should be subdivided or remain a single entity. Two different
types of batches must be considered:
Transfer batch: the lot size viewed from the standpoint of the part.
Process batch: the lot size viewed from the standpoint of the resource.
The Economic Order Quantity (EOQ) model maintains a balance between holding and setup costs. In
the case of a bottleneck, the setup cost is viewed from the perspective not only of the resource but of the
entire system. An hour lost at the bottleneck is an hour lost for the whole system. One should make sure
the bottleneck station needs relatively few setups and has always a buffer inventory in front of it.
Therefore: 7. The transfer batch may not and many times should not be equal to the process batch.
Since a batch moving through manufacturing will encounter both bottleneck and non‐bottleneck
operations, all with varying setup and processing times, the parameters used in establishing lot size for
the batch must be examined for validity. Typically, batches launched on the shop floor are split,
combined, and overlapped to meet the demands of the schedules. The simplicity of the Economic Order
Quantity (EOQ) formula cannot accommodate the complexities and reality of multiple work centres in a
manufacturing environment, and it should not be expected that one batch size for a product moving
through many work centres should always be the same: 8. The process batch should be variable both
along its route and in time.
Priorities
Many rules exist to determine the sequence in which orders should be processed. The most widely
accepted rules consider the amount of time remaining to complete the order and the time available to
complete the order, or the time to due date.
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MRP logic looks at the priorities of an item based on the required due date and the lead time offset of
components. This initial rough cut does not consider capacity constraints and may result in a plan that
requires reworking until feasible. In effect, MRP looks at priorities and then sees if they fit the capacity.
Invariably, however, conflicting priorities result, since two or more different jobs will require processing
at the same work centre at the same time. To satisfy capacity constraints, one job must be processed first
and the subsequent job(s) delayed. Since delay has occurred, the lead time has been affected.
The lead time of a job is thus affected not only by the capacity of the various work centres but by the
priority of the other jobs. This leads to the ninth and final rule: 9. Priorities can only be set by
simultaneously examining all of the system’s constraints. Lead time is a derivative of the schedule.
6.3.2
Theory Of Constraints (TOC)
Step 1. Identify the system’s constraint(s).
Calculate the process load.
All for breakdowns.
Determine the number of setups that can be done.
Step 2. Decide how to exploit the system’s constraint(s),
i.e. how to make the best use of the constraint.
If no bottleneck exists, then make a predetermined schedule for the constraints.
If a bottleneck does exist, then decide on the product mix before making
a predetermined schedule.
= “The Drum Concept.”
Step 3. Subordinate everything else to the decisions made in Step 2.
Release material into the plant according to the needs of the constraints,
allowing sufficient time for the material to arrive.
= “The Buffer and Rope Concept.”
Step 4. Elevate the system’s constraint(s).
After having made the best possible use of the existing constraint(s),
the next step is to reduce its (their) limitations on the system’s performance.
Example
Assume we operate a simple plant as depicted in Figure 1. We have a product called P which sells for £90
per unit, and the market will take 100 units per week. In order to make that product, we have to put
together an assembly. We have resource D, who does all our assembly work, and it requires 10 minutes
to make one product P. The assembly operation involves putting together a purchased part that costs £5
and putting together a few couple of manufactured parts. Those manufactured parts, of course, also are
made from raw materials, and one particular part goes through department A, where the resource takes
15 minutes to work on it, and then it goes on to department C, where 10 minutes are taken. Another part
starts in department B, where 15 minutes are required, then it goes to department C, where 5 minutes
are required. Finally our assembler puts them together, and we have a finished product.
We have a second product, product Q, which sells for £100 per unit. It is priced higher, so we can sell
only 50 of them per week. It is also done by the assembly department, but the assembly only takes 5
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minutes. It uses two parts, one made from raw material costing which goes through departments A and
then B, and the other part is a type of part also used in product P.
In this particular plant, we have one A, one B, one C, and one D, and they are all working 2400
minutes a day – a 40‐hour day. It costs us £6000 a week to run this plant. What is the maximum amount
of money we can make in this plant?
A,B,C,D: 1 each
Available time:
2400 min/week
Operating
expenses:
£6000/week
Q £100/unit
50
P £90/unit
100
D 5
D 10
Purcha
se part:
£5/unit
C 10
C 5 min
B 15
A 15
B 15
A 10
RM1
£20/
RM2
£20/
RM3
£20/
Figure 1: Simple plant example.
How much profit can we make? Let us first calculate what net profit per week can be, i.e. what the
potential of the plant would be if there were no capacity constraints. The results are shown in the Table 1
below.
Table 1: Net week profit potential for this plant
Product
Total
P
Q
Market potential
100
50
Selling price
£90
£100
Raw material costs
£45
£40
Contribution per unit
£45
£60
Total contribution
£4500
£3000
£7500
Operating expenses
£ (6000)
Net profit per week
£1500
However, we cannot be sure that we can produce everything. Let us check whether what the
workload would be on all of our resources if we would indeed try to produce 100 P products and 50 Q
products. These calculations are shown in the Table 2.
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Table 2: Process load assuming full market demand can be produced
Reso
Process Load
Available
Workload
urce
/ week
time / week
/ week
A
P: 15 x 100 = 1500
2000
2400
83 %
Q: 10 x 50 = 500
B
P: 15 x 100 = 1500
3000
2400
125 %
Q: 15 x 50 + 15 x 50 = 1500
C
P: 10 x 100 + 5 x 100 = 1500
1750
2400
73 %
Q: 5 x 50 = 250
D
P: 10 x 100 = 1000
1250
2400
52 %
Q: 5 x 50 = 250
So there is a problem: we cannot make everything. Workloads are fine for A, C, and D, but resource B
should be used more than 100 % and this is not possible.
Now we have a different kind of problem. Because we cannot make everything, what should we
make? Let us first approach this from an accounting perspective in which we wish to calculate the
profitability of a product, as shown in the Table 3.
Table 3: Conventional approach based on product profitability
Product
P
Q
Market potential
100 units
50 units
Selling price
£90
£100
Raw material costs
£45
£40
Contribution per unit
£45
£60
Direct labour time per unit
55 min
50 min
Contribution per direct labour minute
£0.82
£1.20
=> Product Q is preferable
As product Q is apparently the most profitable product, we will produce all 50 products Q on resource
B, which leaves us with 900 min free on resource B which we will use to produce 900/15 = 60 products P.
The net profits we can expect from the plant per week are therefore:
Table 4: Net profit per week expected under the conventional approach
Product
Total
P
Q
Quantity produced
60
50
Selling price
£90
£100
Raw material costs
£45
£40
Contribution per unit
£45
£60
Total contribution
£2700
£3000
£5700
Operating expenses
£ (6000)
Net profit per week
£ (300)
Unfortunately, we will make a loss of £300 per week. Is this really the best we can do?
We now are going to apply the philosophy of TOC, and see if leads to better results. We know that
resource B is the bottleneck of the plant. What would happen if we tried to establish how to make best
use of this bottleneck resource; what would be the solution if we tried to make sure that every minute of
processing time on the bottleneck will generate us the most profits? Would we then still prefer to
produce Q? The type of calculations needed are displayed in the Table 5.
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Integrated Logistics
Table 5: TOC approach based on bottleneck profitability
Product
P
Q
Market potential
100
50
Selling price
£90
£100
Raw material costs
£45
£40
Contribution per unit
£45
£60
Time on bottleneck per unit
15 min
30 min
Contribution per bottleneck minute
£3.00
£2.00
=> Product P is preferable !
This time product P is preferable. If we would produce all 100 products of P, we would have 900 min
left on resource B to make 900/30 = 30 products Q. The net profit expected is now:
Table 6: Net profit per week expected under the TOC approach
Product
Total
P
Q
Quantity produced
100
30
Selling price
£90
£100
Raw material costs
£45
£40
Contribution per unit
£45
£60
Total contribution
£4500
£1800
£6300
Operating expenses
£ (6000)
Net profit per week
£ 300
We would make a profit of £300 per week!
Think about this: with the conventional method, we need to have accurate estimates of how much
processing time is needed on all work stations to produce a unit of product (i.e. of the whole supply
chain). With the TOC method, we only need to know the time needed on the bottleneck resource. It
seems we can achieve better results with less effort.
Efficiencies
As a result of the plan to produce 100 products P and 30 products Q, we can calculate the actual
workloads for each resource. The results are displayed in Table 7 below.
Table 7: Process load under the TOC planning schedule
Reso
Process Load
Available
Workload
urce
/ week
time / week
/ week
A
P: 15 x 100 = 1500
1800
2400
75 %
Q: 10 x 30 = 300
B
P: 15 x 100 = 1500
2400
2400
100 %
Q: 15 x 30 + 15 x 30 = 900
C
P: 10 x 100 + 5 x 100 = 1000
1150
2400
48 %
Q: 5 x 30 = 150
1150
2400
48 %
D
P: 10 x 100 = 1000
Q: 5 x 30 = 150
We can see that these results are similar to the examples discussed in Section 1 on OPT. It is no use to
run resources A, C, and D at 100% efficiency, as this would not allow us to sell anything more but would
just build up inventory.
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Integrated Logistics
Elevate the system’s constraints
Suppose an engineer at the plant asks permission to spend £5000 in tooling and fixtures that will allow:
The process time at resource C to be increased from 5 to 7 minutes;
The process time at resource B to be decreased from 15 to 14 minutes.
Would you say this is a good proposal, or do you think we need to fire the engineer?
At first sight, the idea may seem strange. Remark that if we would implement this idea, the processing
time to make the part from raw material RM1 goes up from 25 to 27 minutes, the processing time to
make the part from raw material RM2 goes up from 20 to 21 minutes, and only the processing time to
make the part from material RM3 goes down from 25 to 23 minutes. All in all, we do not seem to gain a
lot in processing times, and we have to pay £5000 for it!
Let’s examine the situation, however, from the viewpoint of TOC. The proposal has one good aspect:
it reduces the time needed to produce on the bottleneck resource; processing time goes down from 15 to
14 minutes for P and from 30 to 28 minutes for Q. This gives:
Table 8: TOC approach based on bottleneck profitability
Product
P
Q
Market potential
100
50
Selling price
£90
£100
Raw material costs
£45
£40
Contribution per unit
£45
£60
Time on bottleneck per unit
14 min
28 min
Contribution per bottleneck minute
£3.21
£2.14
=> Product P is still preferable !
This time product P is still preferable. If we would produce all 100 products of P, we would have 1000
min left on resource B to make 1000/28 = 35.7 35 products Q. The net profit expected is now:
Table 9: Net profit per week expected under the TOC approach
Product
Total
P
Q
Quantity produced
100
35
Selling price
£90
£100
Raw material costs
£45
£40
Contribution per unit
£45
£60
Total contribution
£4500
£2100
£6600
Operating expenses
£ (6000)
Net profit per week
£ 600
We would make a profit of £600 per week, an extra £300 per week! Therefore, the investment of
£5000 would pay itself back in less than 5000/300 17 weeks, or in less than 5 months.
A new look at standard costing
OPT and TOC provide new perspectives on the standard approaches to production costing, measuring
production efficiency in work centres, make or buy decisions, and overall on how investors should invest
their money.
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Integrated Logistics
Linear Programming
One of the most powerful developments is the use of operational research methods. For example, the
problem of the simple plant presented in previous section can be easily translated into a Linear
Programming problem:
Let xp be the units of P produced per week, and xq be the units Q produced per week. Then our
objective function, to maximise profits = revenues – costs, can be expressed as:
Max z 90 x p 100 x q 6000 5 x p 20 x p 20 x p 20 x q 20 x q
The constraints in the model are:
1) Demand constraints for P and Q:
x p 100
x q 50
2) Time constraints for A, B, C, and D:
15 x p 10 x q 2400
15( x p x q ) 15 x q 2400
10 x p 5( x p x q ) 2400
10 x p 5 x q 2400
3) Sign restrictions
x p , xq 0
Exercise
1. A company produces two types of product: X and Y. The market potential for X is 50 units per week;
the market potential for product Y is 100 units per week. One unit of X sells for $100, and one unit of Y
sells for $80. The company has three expensive machines: A, B, and C. Each machine can be used for 2400
minutes per week. The fixed costs per week are $4000.
One unit of product X is made from one unit of raw material F and two units of raw material G; one
unit of product Y is made from one unit of raw material G and one unit of raw material H. The unit
purchase price of F is $20; of G is $20; and of H is $30.
Raw materials F and H are first processed on machine B; to process one unit on B requires 10 minutes.
Raw material G is first processed on machine C; it takes 15 minutes to process one unit. The processed
raw materials then flow to machine A to make the final products X and Y. It takes 10 minutes on machine
A to produce one unit of X and 15 minutes to produce one unit of Y.
a) Use the Theory Of Constraints’ four step framework to propose a production plan for the company
to maximise its weekly profits (=revenues – costs).
b) Suppose it takes 20 minutes on resource B to process one unit. Which resource is now the
bottleneck? Write an LP model of this planning problem.
c) Solve the LP problem constructed in b) with Excel Solver.
Answers:
a)
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Integrated Logistics
Table : Net week profit potential for this plant
X
Market potential
50
Selling price
$100
Raw material costs
$60
Contribution per unit
$40
Total contribution
$2000
Operating expenses
Net profit per week
Product
Y
100
$80
$50
$30
$3000
Total
$5000
$(4000)
$1000
Table : Process load assuming full market demand can be produced
Resour
Process Load
Available
ce
/ week
time / week
A
X: 10 x 50 = 500
2000
2400
Y: 15 x 100 = 1500
B
X: 10 x 50 = 500
1500
2400
Y: 10 x 100 = 1000
C
X: 2 x 15 x 50 = 1500
3000
2400
Y: 15 x 100 = 1500
Workload
/ week
83.3%
62.5%
125 %
Table : TOC approach based on bottleneck profitability
Product
X
Y
Market potential
50
100
Selling price
$100
$80
Raw material costs
$60
$50
Contribution per unit
$40
$30
Time on bottleneck C per unit
30 min
15 min
Contribution per bottleneck C minute
$1.33
$2.00
=> Product Y is preferable !
To produce 100 Y products corresponds with 1500 minutes on the bottleneck C, leaving 900 minutes
on C to produce 900/30 = 30 units
Table : Process load under the TOC planning schedule
Resour
Process Load
ce
/ week
A
X: 10 x 3 = 30
130
Y: 15 x 100 = 1500
B
X: 10 x 3 = 30
1300
Y: 10 x 100 = 1000
C
X: 2 x 15 x 3 = 900
2400
Y: 15 x 100 = 1500
Table : Net week profit potential for this plant
Product
Available
time / week
2400
Workload
/ week
54.2%
2400
54.2%
2400
100%
Total
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Integrated Logistics
Market potential
Selling price
Raw material costs
Contribution per unit
Total contribution
Operating expenses
Net profit per week
X
30
$100
$60
$40
$1200
Y
100
$80
$50
$30
$3000
$4200
$(4000)
$200
b)
Table : Process load assuming full market demand can be produced
Resour
Process Load
Available
ce
/ week
time / week
A
X: 10 x 50 = 500
2000
2400
Y: 15 x 100 = 1500
B
X: 20 x 50 = 1000
3000
2400
Y: 20 x 100 = 2000
C
X: 2 x 15 x 50 = 1500
3000
2400
Y: 15 x 100 = 1500
Workload
/ week
83.3%
125%
125%
References
Goldratt, E., and J. Cox. 2004. The Goal: a Process of Ongoing Improvement. Gower Publishing Ltd.,
ISBN 0566086654.
Goldratt, E.M., E. Schragenheim, and C.A. Ptak. 2000. Necessary But Not Sufficient: A Theory of
Constraints Business Novel. North River Press. ISBN 0884271706.
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