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Chapter 3(3) Supply Chain Logistics and Operations Management

This Week’s Agenda 0) Review of HW1/2 & Inventory Control Theory 1) Risk Pooling, Extensions 2) IP/LP Modeling Revisited 3) Case Discussion: Sport Obermeyer

Inventory Management
• Independent Demand Inventory
– Dependent vs. independent demand – Basic Economic Order Quantity (EOQ) model – Multi-period models: demand / lead time variability
• No fixed ordering costs: the base-stock model (S) • Fixed ordering costs: the base-stock model (s,S)

– Risk pooling – Inventory management in multiple locations – ABC classification scheme

• Sport Obermeyer case

Why do companies hold inventory? Why might they avoid doing so?
• WHY?
– To meet anticipated customer demand – To account for differences in production timing (smoothing) – To protect against uncertainty (demand surge, price increase, lead time slippage) – To maintain independence of operations (buffering) – To take advantage of economic purchase order size

• WHY NOT?
– Requires additional space – Opportunity cost of capital – Spoilage / obsolescence

Independent vs. Dependent Demand
IndependentDemand Demand(Demand (Demandnot notrelated relatedto toother other Independent itemsor orthe thefinal finalend-product) end-product) items
Ford Taurus

Body Assy.

Wheel Assy. (4) Tire (1)

E(1)

Wheel (1)

Dependent Dependent Demand Demand (Deriveddemand demand (Derived itemsfor for items componentparts, parts, component subassemblies, subassemblies, rawmaterials, materials, raw etc.) etc.)

The Context of the Problem Independent Demand Inventory
• Typical applications of independent demand models
– In a manufacturing company, to control inventory for enditems; e.g., number of A-type washers in warehouse X. – In manufacturing companies, to control inventory for standard parts, such as screws, bolts, etc. – In service firms, to control demand for goods that are provided with the service (e.g., raw materials at McDonald’s) – In retailer service companies

Two Decisions in Inventory Management
• When is it time to reorder?

• If it is time to reorder, how much?

Economic Order Quantity Model: Where it all started….
Time Between Orders (Cycle Time) T = Q/D

On-hand Inventory

Q

Demand Rate, D
Average Cycle Inventory, Q/2 Reorder Point, R

Q/2

Place Order

Receive order

Time

Lead Time, L

Basic EOQ Assumptions
• • • • Constant Demand Rate Constant Lead Time Orders received in full after lead-time. Constant Unit Price (no discounts)

Economic Order Quantity Cost Model: Constant Demand, No Shortages
TC D Q K h = = = = = total annual inventory cost annual demand (units / year) order quantity (units) cost of placing an order or setup cost ($) annual inventory carrying cost ($ / unit /year)

Total Annual Inventory = Cost

Annual Ordering Cost

Annual + Holding Cost

TC

=

(D / Q) K + (Q / 2) h

Cost Relationships for Basic EOQ
(Constant Demand, No Shortages)
TC – Annual Cost

Total Cost Carrying Cost

Ordering Cost

Q*

Order Quantity (how much)

EOQ balances carrying costs and ordering costs in this model.

Trade-off in EOQ Model: Inventory Level vs. Number of Orders
Many orders, low inventory level
On-hand Inventory

Q

Time

Q

Few orders, high inventory level

On-hand Inventory

Time

EOQ Results (How Much to Order)
(Constant Demand, No Shortages)
Economic Order Quantity = Q* = 2DK h Number of Orders per year = D / Q* Length of order cycle T = Q* / D Total cost = TC = (D / Q*) K + (Q* / 2) h

Determining When to Reorder
• Quantity to order (how much…) determined by EOQ • Reorder point (when…)determined by finding the inventory level that is adequate to protect the company from running out during delivery lead time • With constant demand and constant lead time, (EOQ assumptions), the reorder point is exactly the amount that will be sold during the lead time.
Example:

EOQ Example
D = 1,000 units per year BE CAREFUL! K = $20 per order h = $8.33 per unit per month

HOW MUCH TO ORDER? WHEN TO ORDER? Number of orders per year = Length of order cycle = T = Total cost =

Exercise
Question: What if the company can only order in multiples of 12? (That is, order either 0 or 12 or 24 or 36, etc…)?

Robustness of EOQ model
Annual Cost

Very Flat Curve - Good!!
Total Cost

∆TC

Q*-∆Q

Q*

Q*+∆Q

Order Quantity

Would have to mis-specify Q* by quite a bit before total annual inventory costs would change significantly. Example

Example: EOQ Robustness
• Suppose that in the last problem, you have misspecified the order costs by 100% and the holding costs by 50%. That is,
– K used in the computation = $40/order (actual cost = $20 / order) – h used in computation = $150 / unit / year (actual = $ 2(1 ,000)40 100) Q' = = 23.1 150 – Then, using these wrong costs, you would have gotten
Your actual TC (computed substituting Q’ into TC, using correct costs of K = $20, and h = $100: 1 ,000 23 TC = 20 + 100 = $2,019 Only 1% above minimum 23 2

TC!

Variations of EOQ
Some assumptions so far
• Constant price • Certain and constant demand rate • Instantaneous replenishment

Some variations of EOQ
• EOQ with quantity discounts …(BMGT 734) • EOQ with uncertainty (demand and/or lead time)

Effects of Demand / Lead Time Variability on Reorder Point (When)
Variable demand Expected demand at average demand rate d s
Safety Stock level

QUESTION: How much inventory is needed during lead time L?

Place order L

Receive order

KEY POINT: s is larger when there is uncertainty about demand or L

Safety Stock
• Stock carried to provide a level of protection against costly stockouts due to uncertainty of demand during lead time • Stock outs occur when
– Demand over the lead time is larger than expected
Inventory Level

s= ROP

Expected demand

Time

Safety Stock

• Safety Stock Criterion:
– stockouts occur when demand during lead time (DL) … – service level (1 - α )100%.

• DL is a random variable. What kind of probability distribution?

Computing s …
Assumption: Demand Demand over over lead-time lead-timeis isnormally normallydistributed distributed Assumption: Probability {Demand over lead-time < s} = 1- α

Probabilitydistribution distribution Probability ofdemand demandover overLL of

ServiceLevel Level Service 1- α

α
s

Computing s: Normal Distribution
Probabilitydistribution distribution Probability ofdemand demandover overL: L: of Mean= =µ µ StdDev Dev= =σ σ Mean ; ;Std

1- α
µ

α
s

1 - α .90 z 1.28

.95 1.65

.98 2.05

.99 2.33

.999 3.09
1- α

z=
α

s−µ

σ

⇒ s = µ + zσ

From normal table or, in Excel, use: =normsinv (0.90)

0

z

Computation of Variance for Demand over Lead Time: Variability Comes From Two Sources
1: Suppose only demand di in day i is variable; lead time is constant at AVGL

DL = d1 + d 2 + ... + d AVGL
AVGL times

Var {DL } = Var {d1 + d 2 + ... + d AVGL } = Var {d1 } + Var {d 2 } + ... + Var {d AVGL } = AVGL ⋅ STD 2
di’s are independent di’s are identically distributed

2: Now, suppose only lead time is variable; daily demand is constant at AVG

DL = L ⋅ AVG

Var {DL } = Var {L ⋅ AVG } = AVG 2 ⋅ Var {L} = AVG 2 ⋅ STDL2
3: Adding the two terms, we get to our result

Var {DL } = AVGL ⋅ STD 2 + AVG 2 ⋅ STDL2

More specifically….
Mean demand over LT Safety factor (std normal table) Standard deviation of demand over LT Safety stock SS

s = AVG⋅ AVGL + z ⋅ STDL2 ⋅ AVG2 + STD2 ⋅ AVGL
Note: •If lead time is constant, STDL = 0 • If demand is constant, STD = 0
Note: This is a very good approximation even when demand is not normally distributed.

Example:
• Consider inventory management for a certain SKU at Home Depot. Supply lead time is variable (since it depends on order consolidation with other stores) and has a mean of 5 days and std deviation of 2 days. Daily demand for the item is variable with a mean of 30 units and c.v. of 20%. Find the reorder point for 95% service level.

AVG = 30;

STD = 0.2(30) = 6
z = 1.64

AVGL = 5;

STDL = 2

95% service level

s = AVG⋅ AVGL + z ⋅ STDL2 ⋅ AVG2 + STD2 ⋅ AVGL = 30⋅ 5 + 1.64 22 ⋅ 302 + 62 ⋅ 5 = 150 + 100.8 = 251

The (s,S) Policy: Fixed Ordering Costs
Inventory

S
Average demand during lead time L Order placed Order arrives Safety Stock

R s

Time

s should be set to cover the lead time demand and together with a safety stock that insures the stock out probability is within the specific limit (When to reorder). S depends on the fixed order cost – EOQ (How much)

• Order when: inventory position (IP) drops below s • Order how much: bring IP to S

The (s,S) Policy: Fixed Ordering Costs

• Compute s exactly as in the base-stock model:
s = AVG⋅ AVGL + z ⋅ STDL2 ⋅ AVG2 + STD2 ⋅ AVGL

• Compute Q using the EOQ formula, using mean demand D = AVG (be careful about units…):
Q= 2 ⋅ K ⋅ AVG h

• Set S = s + Q

Example: (s,S) Model
• Consider previous Home Depot example, however, there are fixed ordering costs, which are estimated at $50. Assume that holding costs are 15% of the product cost ($80) per year. Also, assume that the store is open 360 days a year.

s = 251

(from previous calculations)

h = (.15)80/360 = 0.0333;
Q=

K = 50;

AVG = 30
S = s + Q = 251+ 300 = 551

2⋅ AVG⋅ K 2(30)50 = = 300 h 0.0333

Summary of Inventory Models
Use EOQ •How much: EOQ formula •When: d*L

yes Is demand rate and lead time constant?

Are there fixed yes ordering no costs?

Use (s, S) policy •How much: Q = S – s (Q is from EOQ formula) •When: IP drops below s (base-stock policy formula)

no

Use base stock (s) policy •When: IP drops below s •How much: necessary to bring IP back to s

Exercise
• Consider previous Home Depot example, however, there are fixed ordering costs, which are estimated at $50. Assume that holding costs are 15% of the product cost ($80) per year. Also, assume that the store is open 360 days a year.

Risk Pooling
• (safety) stock based on standard deviation
– square root law: stock for combined demands usually less than the combined stocks (depends on what?)

• Example: independent demand

σ

2 X +Y

= σ +σ
2 X 2 X

2 Y 2 Y

σ X+Y = σ + σ

HP Example: Benefits of a Universal Product
Because of a different power supplies, HP had two laser printers, one for Europe and one for N. America. A universal product (with a universal power supply) has been proposed, but costs $30 extra. Is it worthwhile?
N. America N(200,60) Europe N(150,50)

Consider z = 2 (98% of service level)
What is the difference is safety stocks required? assume independent demand seen by HP (NA and Europe)

Exercise from Quiz/HW1 revisited
Calculate the stock levels for the following (12) cases : (a) centralized vs. decentralized (b) independent vs. negatively correlated vs. positively correlated (c) 75% and 100% service levels

Inventory Management in Multiple Locations: Echelon Inventory
Supplie r Warehouse echelon lead time

Now,both bothwarehouse warehouseand and Now, retailershold holdinventory. inventory. retailers Howto tomanage manageinventory inventory How inthis thissupply supplychain? chain? in

Warehou se

Retailer s

ABC Inventory Classification: Where To Focus Management Attention
• GOAL: Determine the small # of items that account for the majority of the cost (exploiting “80/20 rule”) which need a tighter inventory control • Items classified as A, B, or C items based on the percentage of annual sales that they command
– A items: very tight control, complete records, regular review – B items: less tightly controlled, good records, regular review – C items: simple controls, minimal records, large inventories, periodic review

• Typical split between classifications: No more than 20% of the items as Class A; no less than 50% of the items as Class C.

A Typical ABC Curve
% of Total Dollar Value 100% 90% 60% B A 20% 50% 100% % of SKU C

Example: ABC Inventory Classification
SKU F-1 F-3 F-7 H-1 H-3 H-4 H-6 J-1 J-3 K-3 Ann. Sales (units) 40,000 195,000 30,000 100,000 3,000 50,000 10,000 30,000 150,000 6,000 Unit Cost 0.06 0.10 0.09 0.05 0.15 0.06 0.07 0.12 0.05 0.09 Annual Sales ($) 2,400 19,500 2,700 5,000 450 3,000 700 3,600 7,500 540 $45,390 Sales % 5.3 43.0 6.0 11.0 1.0 6.6 1.5 7.9 16.5 1.2 Rank 7 1 6 3 10 5 8 4 2 9 ABC Class

Two Network Design Problems
• Problem 1: Given facility locations (plants, warehouses), find the best distribution strategy from plants to warehouses to markets. • Problem 2: Given a set of candidate locations, find the best locations for warehouses and best distribution strategy from plants to warehouses to markets.

Approaches to Use: Heuristics and Exact Algorithms

Problem 1: Finding Best Distribution Strategy (Given Facility Locations)
• Single product • Two plants p1 and p2 -- Plant p1 has unlimited capacity, p2 has an annual capacity of 60,000 units -- The two plants have the same production costs. • Two warehouses w1 and w2 with identical warehouse handling costs, both having unlimited capacity • Three market areas c1, c2 and c3 with annual demands of 50,000, 100,000 and 50,000, respectively • Unit distribution costs: w1 w2 p1 p2 c1 c2 0 5 4 2 3 2 4 1 c3 5 2

Problem:Find Finda adistribution distributionstrategy strategythat thatspecifies specifiesthe theflow flowof ofproducts productsfrom from Problem: plantsto towarehouses warehousesto tomarkets marketswith withminimum minimumtotal totaldistribution distributioncost cost plants Example 2-3, text pp.36-37

Problem 1 Network
markets demand capacity unlimited plants P1 5 0 warehouses 3 W1 5 2 2 W2 C2 1 2 C3 50K 100K 4 C1 50K

4 60K P2

A Heuristic for Problem 1
For each market, choose the cheapest warehouse, and for each warehouse, choose the cheapest plant

markets C1 50K

plants unlimited P1 140K 5

0

warehouses W1

3 4 5

4 60K P2 60K
For every market, W2 is picked

2 2 W2 200K

C2 1 2 C3

100K

50K

Total cost = 5×140 + 2×60 + 2×50 + 1×100 + 2×50 = 1120K

Linear (or Integer) Programming
• Three components:

• What is a feasible solution?

Example of LP Application in Finance: Investment Problem
Kathleen Allen has $70,000 to invest in 4 alternatives with known annual returns: Municipal bonds, 8.5% Certificates of deposit (CDs), 5% Treasury bills, 6.5% Growth stock fund, 13% She has established some guidelines for diversifying her investments: (i) No more than 20% of the total investment should be in municipal bonds (ii) The amount invested in CD should not exceed that invested in the other 3 alternatives (iii) At least 30% of the money should be invested in treasury bills and CDs (iv) More should be invested in treasury bills and CDs than municipal bonds and growth stock fund by a ratio of at least 1.2 to 1. Kathleen wants to invest the entire $70,000 to maximize the total return.

Example of LP Application in Marketing: Advertising Budget Allocation
• A department store chain has hired an advertising firm to determine the types and amount of advertising it should have for its stores. The three types of advertising available are: radio, TV, newspaper. The retail chain wants to know the number of each type of advertisement it should purchase in order to maximize exposure. It is estimated that each ad will reach the following potential audience and cost the following amount: Each TV commercial: exposure = 20,000 people, cost = $15,000 Each radio commercial: exposure = 12,000 people, cost = $6,000 Each newspaper ad: exposure = 9,000 people, cost = $4,000 The company must consider the following resource constraints 1. The budget limit for advertising is $100,000. 2. The TV station has time available for 4 commercials. 3. The radio station has time available for 10 commercials. 4. The newspaper has space available for 7 ads 5. The advertising agency has time and staff available for producing no more than a total of 15 commercials and ads.

Back to Problem 1: Solution by LP
capacity unlimited plants P1 0 5 4 60K P2 2 W2 warehouses W1 3 4 5 1 2 C3 50K Define Decision Variables:
X(P1,W1) = amount sent from P1 to X W1 (W1,C1) = amount sent from W1 to C1 X(P1,W2) = amount sent from P1 to X W2 (W1,C2) = amount sent from W1 to C2 X(P2,W1) = amount sent from P2 to X W1 (W1,C3) = amount sent from W1 to C3 X(P2,W2) = amount sent from P2 to X W2 (W2,C1) = amount sent from W2 to C1 X(W2,C2) = amount sent from W2 to C2 X(W2,C3) = amount sent from W2 to C3

markets demand C1 50K

2

C2 100K

LP Formulation for Problem 1
minZ = 0x(p1 ,w1) + 5x(p1 ,w2) + 4x(p2,w1) + 2x(p2,w2) + 3x(w1 ,c1) + 4x(w1 ,c2) + 5x(w1 ,c3) + 2x(w2,c1) + x(w2,c2) + 2x(w2,c3) subject to:
x(p2,w1) + x(p2,w2) ≤ 60,000
Total Distributi on Cost

x(p1 ,w1) + x(p2,w1) = x(w1 ,c1) + x(w1 ,c2) + x(w1 ,c3) x(w1 ,c1) + x(w2,c1) = 50,000 x(w1 ,c2) + x(w2,c2) = 100,000 x(w1 ,c3) + x(w2,c3) = 50,000
All flows greater than or equal to zero

x(p1 ,w2) + x(p2,w2) = x(w2,c1) + x(w2,c2) + x(w2,c3)

Capacity constraint at plant 2 Flow conservation at warehouse 1 (flow in = flow out) Flow conservation at warehouse 2

Flows to customer 1 has to be equal to its demand Flows to customer 2 has to be equal to its demand Flows to customer 3 has to be equal to its demand Non-negativity constraints

Optimal Solution
The solution to the problem can be obtained via Excel Solver (see Excel File)

Optimal Solution Facility p1 Warehouse w1 140,000 w2 p2 c1 0 50,000 0 c2 40,000 60,000 c3 50,000 0

0 60,000

Optimal Total Cost: $740,000
Recall: Total cost by Heuristic = $1120,000 LPhas hashuge hugeadvantage advantage LP overheuristic!! heuristic!! over

Extensions
• nonlinear objective function • integer decision variables • binary decision variables (yes or no) • uncertainty • stochastic programming • simulation • commercial software packages • CAPS logistics (specialized for SCM) • CPLEX, GAMS, IBM OSL (general purpose)

Problem 2: Finding Best Warehouse Locations & Distribution Strategy
capacity 0 unlimited P1 6 4 60K P2 5 0 2 2 0 W2 W1 2 1 2 W3 36 6 C3 50K C2 100K 3 4 5 demand C1 50K

2 7 W4 5

Warehouse cost

W1 W2 W3 W4 600K 500K 400K 400K

Problem: Pick two warehouses and find a distribution strategy such that total warehousing and distribution cost is minimum

A Heuristic for Problem 2: A Two-Step Approach
Step 1: Pick two least-cost warehouses
600K

cost

capacities 0 unlimited P1 6 4 60K P2 5 0 2 2 0

W1
500K

3 4 5 2 1 2 36 6

demands C1 50K

W2
400K

C2 100K

W3
400K

2 7 W4 5

C3 50K

W3 & W4 are picked Total warehouse cost = 400K + 400K = 800K

Step 2: Given the warehouses, apply the heuristic used for Problem 1 (for each market, choose the cheapest warehouse; for each warehouse, choose the cheapest plant) 3 4 5 2 W2 1 2 36 6 C3 50K C2 100K demands C1 50K

capacities 0 unlimited P1
100K 40K

W1 5

6 4

0 2 2 0

60K

W3
100K

P2
60K

2 7 W4 5

For C1, pick W4 For C2, pick W3 For C3, pick W4

100K

Total distribution cost = 40×6 + 50×2 + 100×6 + 50×5 = 1190K Total cost = 800 + 1190 = 1990K

Optimization Approach

Step 1

Step 2

Analyze Analyze intangible intangible aspects aspects

List of of List potential potential sites sites

Optimization: Optimization:
•Selectionof ofone one •Selection ormore more sites sites or •Allocationof of •Allocation demandto tosites sites demand

Our focus here

Finding Optimal Solution for Problem 2: Integer Programming
600K

cost

capacities 0 unlimited P1 6 4 60K P2 5 0 2 2 0

W1
500K

3 4 5 2 1 2 36 6

demands C1 50K

W2
400K

C2 100K

W3
400K

2 7 W4 5

C3 50K

Involve discrete choices, cannot use LP! Have to use IP Define Decision Variables:
Y(j) = 1 if warehouse Wj is picked, 0 otherwise, for j = 1, 2, 3, 4 X(i,j) = amount sent from location i to j, for all the links (i, j) in the network

IP Formulation for Problem 2
minimize s.t. capacity constraints demand constraints (3) warehouse capacity constraints (4) pick two warehouses flow conservation at each warehouse (total in flow = total out flow) (4)

Y variables are binary, X variables are nonnegative

The Optimal Solution for Problem 2
600K

cost

capacities 0 unlimited P1
150K

W1 5
150K
500K

3 4 5 2 1 2 36 6

demands C1 50K

6 4

0 2 2 0

W2
400K

C2 100K

60K

W3
400K

P2
50K

2 6 W4 6
50K

C3 50K

Total cost = 1750K
Recall: Total cost of the heuristic = 1990K

Sport Obermeyer Case
1. Identification of major issues in the supply chain. 2. Recommendation on ordering units of each style during initial phase of production. (using sample data in Exhibit 10; assume all ten styles in the sample problem are made in Hong Kong, and that Obermeyer's initial production commitment must be at least 10,000 units; ignore price differences among styles in your initial analysis.) 3. Recommendation of operational changes to lower risk and improve performance. 4. How should Obermeyer management think (both short-term and long-term) about sourcing in Hong Kong vs. China?

Place your vote (sales rate: 3:1, 2:1, 1:1, 1:2, 1:3)
Number Sold: 4,000 Number Sold: 4

Sport Obermeyer, Ltd
• Product lines:

• Product variety within a line:

• Problems with ???

The Supply Chain
Shell Shell Fabric Fabric Lining Lining Fabric Fabric Insulation Insulation Materials Materials Snaps Snaps Zippers Zippers Others Others

Cut/Sew

Distribution Center

Retailers

Textile Suppliers

Obersport

Obermeyer

Retailers

Hong Kong vs. China
Parameter Efficiency *Wages ($7/unit) Speed / lead time Minimum order size Quota restrictions Quality Labor skill / versatility HK China

Total wage benefit: ($7/unit)(200,000 units) = $1.4 mil !

What makes it difficult for Sport Obermeyer to be effective in managing its supply chain?
• Long ___________ • • • • Early ___________ Large _________ Short __________ Uncertain __________
– __________ product: one season – __________ data useless – Late demand signal

high ___________

– Global supply chain (distance / time)

• Limited ___________

Order planning cycle 93-94 line
Time Obermeyer Obersport

Feb-92 Jul-92 Sep-92 Nov-92 Mar-93 Jun-93 Sep-93

Design begins Sketches done Design finalized 1st order placed Las Vegas More retail orders Selling season

Order fabric Prototype production Start work Full scale production Ship to US

Speculative vs. Reactive Capacity
Speculative Production Capacity Reactive Production Capacity

Initial Forecast

Las Vegas Orders

End of Season Problem
Scenario Outcome
•Loss of 8%/unit ~ $9

Excess inventory

•Limited capacity effects (could have used that capacity to produce something that stocked out

Stock out

•Loss of profit (24%/unit) ~ $27

What can Obermeyer do? (Operational changes)
• Short term: setting _______________ • Longer term operational changes:
– – – – – – Increase __________ capacity Reduce _____________________________ Get better data earlier than Las Vegas Commit later (__________________) Reduce component __________ (e.g., zippers) ______________ to consumers

Increasing Reactive Capacity
Base case
1st Prod’n Order Speculative production capacity Material LT New Info / 2nd prod’n order Original Reactive capacity Prod’n LT Additional Reactive capacity Delivery LT

Increase Total Capacity Decrease Lead Times

Additional Reactive capacity

Obtain Information Earlier
New Info

Additional Reactive capacity

Nov/92

Mar/93

Jun/93

Aug/93

Sept/93

The Forecast Process: Input for Production Planning

• Independent versus consensus forecasts • Aggregation of expert estimates
– Average of expert estimates is a proxy for the mean of the demand distribution – Standard deviation among expert estimates is a proxy for half the standard deviation of demand distribution

• Forecast updates

How should Wally think about how much of each style he should order in November?
G ail Isis En tice A ssa ult Te ri E le ctra Step han ie S ed uced A nita Daph ne Total 1,00 0 1,00 0 1,00 0 1,00 0 1,00 0 1,00 0 1,00 0 1,00 0 1,00 0 1,00 0 10,00 0

Is this a good plan?

Which Units are Safest to Build First?
• Highest demand
– More likely that unit will sell

• Less variable (lower σ/µ) • Less expensive
– Lower overage costs – In speculative capacity, you are worried about being over – being under not a problem, because you can always use reactive capacity

IP Formulation for Problem 2
Minimize Z = 600Y(w1) + 500Y(w2) + 400Y(w3) + 400Y(w4) + 5X(p1,w2) + 6X(p1,w4) + 4X(p2,w1) + 2X(p2,w2) + 2X(p2,w3) + 3X(w1,c1) + 4X(w1,c2) + 5X(w1,c3) + 2X(w2,c1) + X(w2,c2) + 2X(w2,c3) + 3X(w3,c1) + 6X(w3,c2) + 6X(w3,c3) + 2X(w4,c1) + 7X(w4,c2) + 5X(w4,c3) Subject to (1) Capacity at p2 X(p2,w1) + X(p2,w2) + X(p2,w3) + X(p2,w4) <= 60,000 (2) Demand constraints X(w1,c1) + X(w2,c1)+ X(w3,c1) + X(w4,c1) = 50,000 X(w1,c2) + X(w2,c2)+ X(w3,c2) + X(w4,c2) = 100,000 X(w1,c3) + X(w2,c3)+ X(w3,c3) + X(w4,c3) = 50,000 (3) Warehouse capacity constraints X(p1,w1) + X(p2,w1)<= 200,000Y(w1) X(p1,w2) + X(p2,w2)<= 200,000Y(w2) X(p1,w3) + X(p2,w3)<= 200,000Y(w3) X(p1,w3) + X(p2,w3)<= 200,000Y(w4)
continued on next slide

(4) Pick two warehouses Y(w1) + Y(w2) + Y(w3) + Y(w4) = 2 (5) Flow conservation at each warehouse (total in flow = total out flow) X(p1,w1) + X(p2,w1) = X(w1,c1) + X(w1,c2) + X(w1,c3) X(p1,w2) + X(p2,w2) = X(w2,c1) + X(w2,c2) + X(w2,c3) X(p1,w3) + X(p2,w3) = X(w3,c1) + X(w3,c2) + X(w3,c3) X(p1,w4) + X(p2,w4) = X(w4,c1) + X(w4,c2) + X(w4,c3) (6) Y variables are binary, X variables are nonnegative Y(j) = 0 or 1, for all j = w1, w2, w3, w4 X(i,j) >= 0, for all link (i, j)

SeeExcel Excelfile fileto toknow knowhow howit itis issolved solvedby byExcel ExcelSolver Solver See

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