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Chapter 3
APPLICATIONS OF LINEAR
PROGRAMMING
In Chapter 2 you learned to formulate and classify deterministic problems according to the mathematical properties of the variables, objective function, and constraints. One such classification is the
linear programming problem — that is, a problem in which the objective function and all constraints are linear and all the variables are continuous (allowed to assume fractional values). Special
attention is given to linear programming problems because they have wide practical applications in
such diverse areas as allocation of scarce resources, purchasing and manufacturing, diet planning,
portfolio management, blending, and production planning as illustrated with the examples in this
chapter. In Chapters 4 to 6, you will learn how to obtain the solution to such problems efficiently
by computer.

3.1

LINEAR PROGRAMMING MODELS FOR
PRODUCT-MIX DECISIONS

Managers often need to determine how to allocate various scarce resources — such as labor, raw
materials, capital, and so on — to several alternatives that compete for these resources. The final
decision is based on the availability of these resources and on achieving an overall objective for the
organization. For example, in a production setting, the mix of products to manufacture is ultimately
based on an overall corporate objective such as maximizing profits or minimizing total production
costs. Linear programming models are often used to help managers make such decisions. Consider
the problem faced by the management of BlubberMaid, Inc.
Example 3.1: The Product-Mix Problem of BlubberMaid, Inc.
BlubberMaid, Inc., manufactures three rubber-based products: Airtex (a spongy material), Extendex (a stretchy material), and Resistex (a hard material). All three products require the same
three chemical polymers and base. The amount of each ingredient used per pound of final product
is given in Table 3.1.
For the coming week, BlubberMaid has a commitment to produce at least 1000 pounds of Airtex,
500 pounds of Extendex, and 400 pounds of Resistex, but the company management knows it can
sell more of each of the three products. Current inventories of the ingredients are 500 pounds of
Polymer A, 425 pounds of Polymer B, 650 pounds of Polymer C, and 1100 pounds of the Base. Each
pound of Airtex nets the company a profit of $7, each pound of Extendex a profit of $7, and each
1

2

Chapter 3

Table 3.1: Ingredients Used in Producing Airtex, Extendex, and Resistex
Ingredients (ounces per pound of product)
Polymer A Polymer B Polymer C Base
Product
Airtex
Extendex
Resistex

4
3
6

2
2
3

4
2
5

6
9
2

pound of Resistex a profit of $6. As manager of the Production Department, you need to determine
an optimal production plan for this week.
3.1.1

IDENTIFYING THE DECISION VARIABLES

Following the steps of problem formulation from Chapter 2, first identify the decision variables.
Asking yourself what you can control and what information constitutes a production plan, should
lead you to identify the following variables:
A
E
R

3.1.2

=
=
=

the number of pounds of Airtex to produce this week,
the number of pounds of Extendex to produce this week,
the number of pounds of Resistex to produce this week.

IDENTIFYING THE OBJECTIVE FUNCTION

For BlubberMaid, the logical objective is to determine how much of each product to make so as to
maximize total profit. Applying the technique of decomposition leads to:
Total profit

=

Profit from Airtex + Profit from Extendex +
Profit from Resistex.

Since each pound of Airtex nets a profit of $7, A pounds of Airtex yields $7A. Similarly, Extendex
and Resistex contribute $7E and $6R, respectively, to the total profit. In terms of the decision
variables and the profit data, the objective function is:
Maximize 7A + 7E + 6R.
3.1.3

IDENTIFYING THE CONSTRAINTS

Applying the technique of grouping should lead you to identify the following three groups of constraints:
1. Resource constraints to insure that no more of the three polymers and base are used than
are available.
2. Demand constraints to insure that company commitments are met.
3. Logical constraints to specify that all production quantities are nonnegative.

Chapter 3

3

Resource Constraints
This group consists of four constraints: one for each of the three polymers and one for the base. For
the limited availability of 500 pounds of Polymer A:
Amount of Polymer A used ≤ 500 pounds.
Using decomposition leads to:
Amount of Polymer A used

=

Amount used to produce A pounds of Airtex +
Amount used to produce E pounds of Extendex +
Amount used to produce R pounds of Resistex.

To determine the amount of Polymer A used in making each product, work through a specific
example. For instance, set A = 100, E = 300, and R = 200. According to the data in Table 3.1:
Amount of Polymer A used in Airtex
Amount of Polymer A used in Extendex
Amount of Polymer A used in Resistex

=
=
=

4(100)
3(300)
6(200)

=
=
=

400,
900,
1200.

In terms of the decision variables, then, you might think that the appropriate constraint for Polymer
A is:
4A + 3E + 6R ≤ 500.
However, this constraint is not correct. The reason is that the units in the expression on the lefthand side are in ounces (see Table 3.1) but the units on the right-hand side are in pounds. This
discrepency can be corrected by converting the units of either side to those of the other side. For
example, converting the 500 available pounds of Polymer A to 8000 ounces (1 pound equals 16
ounces) results in the following constraint:
4A + 3E + 6R ≤ 8000. (Polymer A)
Following similar logic for the remaining three resources results in these constraints:
2A + 2E
4A + 2E
6A + 9E

+ 3R ≤
6800 (Polymer B)
+ 5R ≤ 10400 (Polymer C)
+ 2R ≤ 17600. (Base)

Demand Constraints
This group consists of three constraints: one for the minimum requirement on the amount of each
of the three products. These constraints are:
A ≥ 1000 (Airtex)
E ≥ 500 (Extendex)
R ≥ 400. (Resistex)
Logical Constraints
Since all production quantities must be nonnegative, the following logical constraints are needed:
A, E, R ≥ 0.

4
3.1.4

Chapter 3
COMPLETE FORMULATION AND SOLUTION OF THE
PRODUCT-MIX PROBLEM OF BLUBBERMAID, INC.

As manager of the Production Department, putting together all the pieces results in the following
mathematical model of the linear programming problem of BlubberMaid, Inc.:
Maximize 7A + 7E
Subject to
Resource Constraints
4A + 3E
2A + 2E
4A + 2E
6A + 9E
Demand Constraints
A
E
Logical Constraints
A ,
E

+ 6R

+
+
+
+

,

6R
3R
5R
2R

≤ 8000
≤ 6800
≤ 10400
≤ 17600



R ≥

1000
500
400

R ≥

0.

(Polymer A)
(Polymer B)
(Polymer C)
(Base)
(Airtex)
(Extendex)
(Resistex)

The optimal solution to this problem that results from using any linear programming software
package is:
A
E
R

= 1000.00
= 533.33
= 400.00

with an objective function value of 13,133.33. In other words, the optimal weekly plan is to produce
1000 pounds of Airtex, 533.33 pounds of Extendex, and 400 pounds of Resistex, resulting in a net
profit of $13,133.33. You will be asked to verify this solution in the exercises following Chapter 6.

3.2

LINEAR PROGRAMMING MODELS FOR
MAKE-OR-BUY DECISIONS

In many production settings, a company may not have enough resources to meet an unexpectedlylarge demand for one or more products. In such cases, the company can supplement its production
capacity by purchasing some of the products from outside suppliers. The central issue in those
situations is for the managers to decide how much of each product to produce versus how much to
purchase from the outside. A linear programming model is often useful in making such decisions, as
illustrated by the following example.
Example 3.2: The Make-or-Buy Problem of the MTV Steel Company
The MTV Steel Company produces three sizes of tubes: A, B, and C, that sell, respectively, for
$10, $12, and $9 per foot. To manufacture each foot of tube A requires 0.5 minutes of processing
time on a particular type of shaping machine. Similarly, each foot of Tube B needs 0.45 minutes
and each foot of Tube C needs 0.6 minutes. After production, each foot of tube, regardless of type,
requires one ounce of welding material. The total production cost is estimated to be $3, $4, and $4
per foot of tubes A, B, and C, respectively.
For the coming week, MTV Steel have received an exceptionally-large order of 2000 feet for Tube
A, 4000 feet for Tube B, and 5000 feet for Tube C. As only 40 hours of machine time are available this

Chapter 3

5

Table 3.2: Data for the Make-or-buy Problem of MTV Steel

Type

Selling
price
($/ft.)

A
B
C

10
12
9

Machine
Demand
time
(feet)
(min./ft.)

2000
4000
5000

Amount
Available

0.50
0.45
0.60

Welding
material
(oz./ft.)

Production Purchase
cost
cost
($/ft.)
($/ft.)

1
1
1

3
4
4

6
6
7

40 hrs. 5500 ounces

week, and only 5500 ounces of welding material are in inventory, the Production Department will not
be able to meet these demands which require a total of 97 hours of machine time and 11,000 ounces
of welding material. Since this high level of demand is not anticipated to continue, rather than
expanding the capacity of the production facilities, the management of MTV Steel is considering
purchasing some of these tubes from suppliers in Japan at a delivered cost of $6 per foot of Tube
A, $6 per foot of Tube B, and $7 per foot of Tube C. These various data are summarized in Table
3.2. As manager of the Production Department, you have been asked to make recommendations as
to how much of each tube type to produce and how much to purchase from Japan so as to meet the
demands and maximize the company’s profits.
3.2.1

IDENTIFYING THE DECISION VARIABLES

In this problem you are free to choose how many feet of each type of tube to produce and how many
feet to purchase from Japan. This gives rise to the following six decision variables:
AP
BP
CP

=
=
=

the number of feet of tube Type A to produce,
the number of feet of tube Type B to produce,
the number of feet of tube Type C to produce,

AJ
BJ
CJ

=
=
=

the number of feet of tube Type A to buy from Japan,
the number of feet of tube Type B to buy from Japan,
the number of feet of tube Type C to buy from Japan.

3.2.2

IDENTIFYING THE OBJECTIVE FUNCTION

As stated in the problem description, the overall objective is to maximize total profits. Applying
decomposition results in:
Total profits

=

(Profits from production) +
(Profits from products purchased from Japan).

Applying decomposition to the profits from production yields:

6

Chapter 3

Profits from production

=

Profits from producing Type A tubes +
Profits from producing Type B tubes +
Profits from producing Type C tubes.

Each of these profits, in turn, is computed as the revenue minus the cost per foot. For example,
since tubes of Type A sell for $10 per foot but cost $3 to produce, the net profit is $7 per foot.
Thus, the profit for producing AP feet of Type A tube is 7AP. A similar computation for Type B
and C tubes results in:
Profits from production = 7AP + 8BP + 5CP.
Applying a similar decomposition and logic to the products purchased from Japan results in:
Profits from products purchased from Japan = 4AJ + 6BJ + 2CJ.
As you would expect, each foot of tube produced results in a higher profit than when purchased
from the outside supplier. Combining these two profit components results in the following overall
objective function:
Maximize 7AP + 8BP + 5CP + 4AJ + 6BJ + 2CJ.
3.2.3

IDENTIFYING THE CONSTRAINTS

Applying the technique of grouping should lead you to identify the following three groups of constraints:
1. Resource constraints to insure that the amount of machine time and welding material used
does not exceed available supplies.
2. Demand constraints to insure that the demand for each type of tube is met.
3. Logical constraints.
Resource Constraints
To produce these tubes requires two resources: machine time and welding material. Since these resources are limited, two constraints are needed to insure that the available supplies are not exceeded.
The machine-time constraint is:
Total machine time used should not exceed 40 hours.
Appplying decomposition leads to:
Total machine time used
= (Machine time used to produce Type A tube) +
(Machine time used to produce Type B tube) +
(Machine time used to produce Type C tube).
Recall from Table 3.2 that each foot of Tube A requires 0.5 minutes of machine time. Thus, to
produce AP feet requires 0.5AP minutes. Similarly, each foot of Tube B requires 0.45 and each foot
of Tube C needs 0.6 minutes. You would thus write the constraint:
0.5AP + 0.45BP + 0.6CP ≤ 40.

Chapter 3

7

However, observe that the quantity on the left-hand side is expressed in minutes while that on the
right-hand side is expressed in hours. One way to correct this inconsistency is to convert 40 hours
into 40 ∗ 60 = 2400 minutes:
0.5AP + 0.45BP + 0.6CP ≤ 2400. (Machine time)
Turning to the availability of the welding material, the associated constraint is:
Total welding material used should not exceed 5500 ounces.
Applying decomposition and recalling that each foot of tube, regardless of type, requires one ounce
of welding material, this resource constraint is:
AP + BP + CP ≤ 5500. (Welding material)
Demand Constraints
This group consists of three constraints, one for the demand associated with each type of tube. For
Tube A,
Total number of feet of Type A tube = 2000 feet.
Applying decomposition,
Total number of feet of Type A tube
=

(Number of feet of Type A produced) +
(Number of feet of Type A purchased from Japan)

=

AP + AJ.

Thus, the demand constraint for Type A tube is:
AP

+ AJ

= 2000. (Demand for Type A)

A similar logic results in the following demand constraints for Types B and C tubes:
BP
CP

+ BJ
+ CJ

= 4000 (Demand for Type B)
= 5000. (Demand for Type C)

Logical Constraints
The only logical constraints in this problem are that all variables be nonnegative.
3.2.4

COMPLETE FORMULATION AND SOLUTION OF THE MAKE-OR-BUY
PROBLEM OF THE MTV STEEL COMPANY

Putting together the pieces results in the following linear programming model for the problem of
the MTV Steel Company:

8

Chapter 3

Maximize
7AP +
8BP +
Subject to
Demand Constraints
AP
BP

5CP + 4AJ + 6BJ + 2CJ

+

AJ
+

BJ

CP

+

Resource Constraints
0.5AP + 0.45BP + 0.6CP
AP +
BP +
CP
Logical Constraints
AP ,

BP ,

CP ,

= 2000
= 4000
CJ = 5000

≤ 2400
≤ 5500
AJ ,

BJ ,

CJ ≥

(Demand for Type A)
(Demand for Type B)
(Demand for Type C)

(Machine time)
(Welding material)

0.

The optimal solution to this problem that results from using any linear programming software
package is:
AP
BP
CP
AJ
BJ
CJ

=
=
=
=
=
=

2000.000
0.000
2333.333
0.000
4000.000
2666.667

with a net profit of $55,000. In other words, MTV Steel should produce 2000 feet of Type A tube
and 2333.333 feet of Type C tube while importing 4000 feet of Type B tube and 2666.667 feet of
Type C tube from Japan. You will be asked to verify this solution in the exercises following Chapter
6.

3.3

LINEAR PROGRAMMING MODELS FOR DIET
PROBLEMS

Linear programming model can also be applied in the planning of diets. In particular, given a
number of food alternatives, each of which yields a known amount of a needed nutrient, you want to
determine how much of each food type to include in a diet to insure minimum nutrient requirements
while achieving an overall objective. Such a problem is illustrated in the following example.
Example 3.3: The Diet Problem of the Mountain View General Hospital
The Nutrition Department of the Mountain View General Hospital prepares 30 dinner menus,
one for each day of the month. In preparing one meal consisting of spaghetti, turkey, scalloped
potatoes, spinach, and apple strudel, the director of the Nutrition Department has determined that
this meal must provide 63000 milligrams (mg.) of protein, 10 mg. of iron, 15 mg. of niacin, 1 mg.
of thiamin, and 50 mg. of vitamin C. Each 100 grams of these foods provides the amount of each
nutrient and fat indicated in Table 3.3.
To avoid too much of one type of food, not more than 300, 300, 200, 100, and 100 grams
of each of the foods as listed in Table 3.3 should be included in the meal. As director of the
Nutrition Department, you want to determine the composition of a meal that meets the nutritional
requirements, and provides the least amount of fat.

Chapter 3

9

Table 3.3: Nutrients Provided by the Various Foods
Nutrients (in mg. per 100 gm.)
Protein Iron Niacin Thiamin Vitamin C Fat
5000
29300
5300
3000
4000

Spaghetti
Turkey
Potatoes
Spinach
App. Strud.

3.3.1

1.1
1.8
0.5
2.2
1.2

1.4
5.4
0.9
0.5
0.6

0.18
0.06
0.06
0.07
0.15

0.0
0.0
10.0
28.0
3.0

5000
5000
7900
300
14300

IDENTIFYING THE DECISION VARIABLES

In this problem, you can control how much of each of the five foods to include in the meal, leading
you to define the following five variables:
SP AG
T U RK
P OT A
SP IN
AP P L

=
=
=
=
=

the
the
the
the
the

number
number
number
number
number

of
of
of
of
of

100
100
100
100
100

grams
grams
grams
grams
grams

of
of
of
of
of

spaghetti to include,
turkey to include,
potatoes to include,
spinach to include,
apple strudel to include.

For convenience, the units of the variables are chosen to be in hundreds of grams because those are
the units used in Table 3.3.
3.3.2

IDENTIFYING THE OBJECTIVE FUNCTION

As stated in the problem description, the overall objective is to minimize the total fat content of the
diet. Applying decomposition results in:
Total fat content

=

(Fat
(Fat
(Fat
(Fat
(Fat

contributed
contributed
contributed
contributed
contributed

by
by
by
by
by

spaghetti) +
turkey) +
potatoes) +
spinach) +
apple strudel).

Using the data in the last column of Table 3.3, and working through a specific example should lead
you to identify the following overall objective:
Minimize 5000SP AG + 5000T U RK + 7900P OT A + 300SP IN + 14300AP P L.

3.3.3

IDENTIFYING THE CONSTRAINTS

Applying the technique of grouping leads you to the following three groups of constraints:
1. Nutrient constraints to insure that the diet provides the minimum amount of each nutrient.

10

Chapter 3
2. Bound constraints to insure that not too much of one type of food is included (for example,
asking a patient to eat 1000 grams of spinach).
3. Logical constraints to insure that all variables are nonnegative.

Nutrient Requirements
This group consists of five constraints, one to insure the minimum amount of each of the five
nutrients. Consider the protein requirement:
Total amount of protein in the diet ≥ 63000 mg.
Applying decomposition yields:
Total amount of protein in the diet
=

(Amount
(Amount
(Amount
(Amount
(Amount

of
of
of
of
of

protein
protein
protein
protein
protein

from
from
from
from
from

spaghetti) +
turkey) +
potatoes) +
spinach) +
apple strudel).

Refer to the first column of Table 3.3. Each 100 grams of spaghetti contains 5000 mg. of protein.
Thus, SP AG hundred grams of this food provides 5000SP AG mg. of protein to the meal. Similarly,
using the remaining data in the first column of Table 3.3 results in the following constraint for
protein:
5000SP AG + 29300T U RK + 5300P OT A + 3000SP IN + 4000AP P L ≥ 63000. (Protein)
Although the units of the variables are expressed in hundreds of grams, the units on both sides of
the constraint above are in milligrams.
Using the next four columns of data in Table 3.3 results in the following similar constraints for
each of the remaining four nutrients:
1.1SP AG + 1.8T U RK + 0.5P OT A
1.4SP AG + 5.4T U RK + 0.9P OT A
0.18SP AG + 0.06T U RK + 0.06P OT A
10P OT A

+ 2.2SP IN
+ 0.5SP IN
+ 0.07SP IN
+ 28SP IN

+ 1.2AP P L
+ 0.6AP P L
+ 0.15AP P L
+
3AP P L

≥ 10
≥ 15
≥ 1
≥ 50.

(Iron)
(Niacin)
(Thiamin)
(Vitamin C)

Bound Constraints
These restrictions limit the maximum amount of each food type in the meal. Keeping in mind that
the units of the variables are in hundreds of grams, the following five bound constraints arise:
SP AG
T U RK
P OT A
SP IN
AP P L

≤ 3
≤ 3
≤ 2
≤ 1
≤ 1.

Chapter 3

11

Logical Constraints
The only logical constraints in this problem are that all variables be nonnegative.
3.3.4

COMPLETE FORMULATION AND SOLUTION OF THE DIET PROBLEM
OF THE MOUNTAIN VIEW GENERAL HOSPITAL

Putting together all this information results in the following linear programming model for the
problem of the Mountain View General Hospital:

Minimize 5000SP AG+ 5000T U RK+7900P OT A+ 300SP IN +14300AP P L
Subject to
Nutrient Constraints
5000SP AG+29300T U RK+5300P OT A+3000SP IN + 4000AP P L ≥ 63000
1.1SP AG+ 1.8T U RK+ 0.5P OT A+ 2.2SP IN + 1.2AP P L ≥
10
1.4SP AG+ 5.4T U RK+ 0.9P OT A+ 0.5SP IN + 0.6AP P L ≥
15
0.18SP AG+ 0.06T U RK+ 0.06P OT A+ 0.07SP IN + 0.15AP P L ≥
1
10P OT A+ 28SP IN +
3AP P L ≥
50
Bound Constraints
SP AG
T U RK
P OT A
SP IN

Logical Constraints
SP AG ,

T U RK ,

P OT A ,

SP IN ,





AP P L ≤

3
3
2
1
1

AP P L ≥

0.

(Protein)
(Iron)
(Niacin)
(Thiamin)
(Vitamin C)

The optimal solution to this problem that results from using any linear programming software
package is:
SP AG
T U RK
P OT A
SP IN
AP P L

=
=
=
=
=

3.000
2.833
2.000
1.000
0.667

with a total fat content of 54,800 milligrams. In other words, the meal should consist of 300 grams
of spaghetti, 283.3 grams of turkey, 200 grams of potatoes, 100 grams of spinach, and 66.7 grams of
apple strudel. You will be asked to verify this solution in the exercises following Chapter 6.

3.4

LINEAR PROGRAMMING MODELS FOR
PORTFOLIO MANAGEMENT

Recall the portfolio-management problem of High Tech in Example 2.3 in Section 2.2. The decision
in that problem is to determine which investments to select. That problem required making a
“no/yes” decision that resulted in an integer programming model with 0 − 1 variables. As you will

12

Chapter 3

Table 3.4: Risk and Expected Rate of Return for Six Mutual Funds
Fund

Price ($/share)
Expected return (%)
Risk category

1

2

3

4

5

6

45
30
High

76
20
High

110
15
High

17
12
Med.

23
10
Med.

22
7
Low

see now, an investment decision may also require determining how much to invest in each available
alternative. A linear programming model can often be formulated for a problem of this nature.
The overall objective of an investor is to gain the highest possible return. But high return
comes at a price: risk. An investor must balance return against risk. One can often formulate a
linear programming model to design an investment strategy that achieves the maximum return while
satisfying certain risk requirements. Consider the problem faced by the general managers of Pension
Planners, Inc.
Example 3.4: The Investment Problem of Pension Planners, Inc.
The Portfolio Manager of Pension Planners, Inc., has been asked to invest $1,000,000 of a large
pension fund. The Investment Research Department has identified six mutual funds with varying
investment strategies resulting in different potential returns and associated risks, as summarized in
Table 3.4.
One way to control the risk is to limit the amount of money invested in the various funds. To
that end, the management of Pension Planners, Inc., has specified the following guidelines:
1. The total amount invested in high-risk funds must be between 50% and 75% of the portfolio.
2. The total amount invested in medium-risk funds must be between 20% and 30% of the
portfolio.
3. The total amount invested in low-risk funds must be at least 5% of the portfolio.
A second way to control risk is to diversify, that is to spread the risk by investing in many
different alternatives. In this case, the management of Pension Planners, Inc., has specified that the
amount invested in the high-risk Funds 1, 2, and 3 should be in the ratio 1:2:3, respectively, while
the amount invested in the medium-risk Funds 4 and 5 should be 1:2.
With these guidelines, what portfolio should the manager recommend so as to maximize the
expected rate of return?
3.4.1

IDENTIFYING THE DECISION VARIABLES

In this problem, you can control how much to invest in each of the six mutual funds, thus giving rise
to six decision variables. As usual, you must specify the units associated with each variable. For
example, for the first fund you could define any one of the following variables:

Chapter 3

F1
F1
F1

=
=
=

13

the number of shares of Fund 1 to buy,
the number of dollars to invest in Fund 1,
the fraction of the portfolio to invest in Fund 1.

Each choice leads to a different, but equivalent, mathematical model. Here, the last choice is used.
In the exercises following this chapter you are asked to develop the appropriate models corresponding
to the other two choices. So, for each of the remaining funds, define:
F2
F3
F4
F5
F6

3.4.2

=
=
=
=
=

the
the
the
the
the

fraction
fraction
fraction
fraction
fraction

of
of
of
of
of

the
the
the
the
the

portfolio
portfolio
portfolio
portfolio
portfolio

to
to
to
to
to

invest
invest
invest
invest
invest

in
in
in
in
in

Fund
Fund
Fund
Fund
Fund

2,
3,
4,
5,
6.

IDENTIFYING THE OBJECTIVE FUNCTION

As stated in the problem description, the overall objective is to maximize the expected rate of return,
that is,
Expected rate of return =

Expected total return
Amount invested

Applying decomposition to the numerator leads to
Expected total return

=

Expected
Expected
Expected
Expected
Expected
Expected

return
return
return
return
return
return

from
from
from
from
from
from

Fund
Fund
Fund
Fund
Fund
Fund

1+
2+
3+
4+
5+
6.

To determine the expected return from Fund 1, work through a specific example in which 10% of
the portfolio is invested in Fund 1, that is, F1 = 0.10. In this case, 0.10 ∗ 1, 000, 000 = $100, 000
dollars of the portfolio are invested in Fund 1. According to the data in Table 3.4, these monies are
expected to return 30%, or 0.30 ∗ 100, 000 = $30, 000. Thus, in terms of F1 ,
Expected return from Fund 1

=
=
=

(Amount invested in Fund 1) ∗
(Rate of return from Fund 1)
(F1 ∗ 1, 000, 000) ∗ 0.30
300000F1 .

Using a similar logic for the remaining five funds leads to:
Expected total return = 300000F1
120000F4

+ 200000F2
+ 100000F5

+ 150000F3 +
+ 70000F6 .

Dividing this by the total investment of $1,000,000 provides the rate of return and hence the following
objective function:
Maximize 0.30F1 + 0.20F2 + 0.15F3 + 0.12F4 + 0.10F5 + 0.07F6 .

14

Chapter 3

3.4.3

IDENTIFYING THE CONSTRAINTS

Applying the technique of grouping should lead you to identify the following three groups of constraints:
1. Investment limitations to control the amount invested in each of the three risk categories.
2. Diversification constraints to spread the investment within each risk category.
3. Logical constraints.
Investment-Limitation Constraints
This group consists of three subgroups of constraints, one for each category of risk, namely:
1. The total amount invested in high-risk funds must be between 50% and 75% of the portfolio.
Since F1 , F2 , and F3 respresent the fraction of the portfolio to invest in the high-risk funds,
the fraction of the total portfolio invested in high-risk funds is F1 +F2 +F3 . These constraints
are:
+ F2
+ F2

F1
F1

+ F3
+ F3

≥ 0.50 (Min. in high-risk)
≤ 0.75. (Max. in high-risk)

2. The total amount invested in medium-risk funds must be between 20% and 30% of the
portfolio. Since F4 and F5 represent the fraction of the portfolio to invest in the mediumrisk funds, the fraction of the total portfolio invested in medium-risk funds is F4 + F5 . These
constraints are:
F4
F4

+ F5
+ F5

≥ 0.20 (Min. in medium-risk)
≤ 0.30. (Max. in medium-risk)

3. The total amount invested in low-risk funds must be at least 5% of the portfolio. Since F6
is the fraction of the portfolio invested in low-risk funds, this constraint is:
F6 ≥ 0.05. (Min. in low-risk)
Diversification Constraints
This group of constraints is used to control risk by insuring that the amount invested in each fund
within a given risk category is in the specified ratio, as follows:
1. The amount invested in the high-risk Funds 1, 2, and 3 should be in the ratio 1:2:3. This
constraint specifies that the amount invested in Fund 2 be twice the amount invested in
Fund 1:
F2 = 2F1 .
Rearranging this so that all variables are on the left-hand side:
−2F1 + F2 = 0.

(Ratio of F1 to F2 )

Similarly, the amount invested in Fund 3 is to be three times that invested in Fund 1:
F3 = 3F1 , or
−3F1 + F3 = 0. (Ratio of F1 to F3 )

Chapter 3

15

2. The amount invested in the medium-risk Funds 4 and 5 should be in the ratio of 1:2, that
is, the amount invested in Fund 5 should be twice that in Fund 4:
F5 = 2F4 .
Rearranging this so that all variables are on the left-hand side:
−2F4 + F5 = 0.

(Ratio of F4 to F5 )

Logical Constraints
Of course one set of logical constraints is that each variable be nonnegative. Since it is possible to
buy fractional shares of a mutual fund, these variables are allowed to have any fractional value, thus
resulting in a linear programming problem. However, another logical constraint is needed to insure
that the total portfolio of precisely $1,000,000 is invested. Since the decision variables represent the
fraction of this portfolio to invest in the various funds, this constraint is:
Total fraction of $1,000,000 invested should equal 1, or
F1 + F2 + F3 + F4 + F5 + F6 = 1.0. (Total portfolio)
3.4.4

COMPLETE FORMULATION AND SOLUTION OF THE
INVESTMENT PROBLEM OF PENSION PLANNERS, INC.

Putting together the pieces results in the following linear programming model for the general partners
of Pension Planners, Inc.:
Maximize 0.30F1 + 0.20F2 + 0.15F3 + 0.12F4 + 0.10F5 + 0.07F6
Subject to
Investment-Limitation Constraints
F1 + F2 + F3
≥ 0.50 (Min. in high-risk)
F1 + F2 + F3
≤ 0.75 (Max. in high-risk)
F4 + F5
≥ 0.20 (Min. in medium-risk)
F4 + F5
≤ 0.30 (Max. in medium-risk)
F6 ≥ 0.05 (Min. in low-risk)
Diversification Constraints
−2F1 + F2
−3F1 +
F3

− 2F4 + F5

Logical Constraints
F1 + F2 + F3 +
F1 , F2 , F3 ,

=
=
=

F4 + F5 + F6 =
F4 , F5 , F6 ≥

0
0
0

(Ratio of F1 to F2 )
(Ratio of F1 to F3 )
(Ratio of F4 to F5 )

1.0 (Total portfolio)
0.

The optimal solution to this problem that results from using any linear programming software
package is:
F1
F2
F3
F4
F5
F6

=
=
=
=
=
=

0.1250
0.2500
0.3750
0.0667
0.1333
0.0500

16

Chapter 3

with a rate of return of 0.168583. In other words, the amount of money invested in each of the six
funds is:
Amount
Amount
Amount
Amount
Amount
Amount

in
in
in
in
in
in

Fund
Fund
Fund
Fund
Fund
Fund

1
2
3
4
5
6

= 0.1250 ∗ 1, 000, 000 = $
= 0.2500 ∗ 1, 000, 000 = $
= 0.3750 ∗ 1, 000, 000 = $
= 0.0667 ∗ 1, 000, 000 = $
= 0.1333 ∗ 1, 000, 000 = $
= 0.0500 ∗ 1, 000, 000 = $

TOTAL INVESTMENT

125, 000
250, 000
375, 000
66, 700
133, 300
50, 000

= $1, 000, 000

with an expected rate of return of 16.86% (or $168,600 total return). You are asked to verify this
solution in the exercises following Chapter 6.
Recall that the decision variables are defined as the fraction of the portfolio to invest, rather
than the number of dollars. This approach has a distinct advantage. Should the dollar amount of
the portfolio change — a likely event — the current model remains unchanged. You simply need need
to multiply the fractions obtained in the solution above by the new portfolio size to determine the
new amounts to invest in each of the six funds.

3.5

LINEAR PROGRAMMING MODELS FOR
BLENDING PROBLEMS

Another example of the use of a linear programming model is in the blending of various components
to produce a final product. For example, how does a refinery blend and process crude oils to
produce gasoline? How does a metal-making business blend alloys to produce a new alloy with
certain properties? In such problems, each component contains amounts of certain ingredients, such
as sulfur in the crude oils, or iron in the alloys, that may either be required or restricted to be
present in certain amounts in the final blend. The objective in blending problems is to determine
the amounts of each component in the blend that yield the desired product at a minimum cost.
Consider the gasoline-blending problem facing the managers at the Hexxon Oil Company.
Example 3.5: The Gasoline-Blending Problem of the Hexxon Oil
Company
The Hexxon Oil Company obtains three types of crude oils from its wells in Mississippi, New
Mexico, and Texas. The gasoline obtained from these crude oils are blended together with two
additives to obtain the final product. These crude oils and additives contain sulfur, lead, and
phosphorus, as shown in Table 3.5. The cost of each component is also presented. However, due to
by-products and impurities, each gallon of Mississippi crude oil results in only 0.35 gallons of the
final product that contains 7% sulfur. Similarly, each gallon of New Mexico crude yields 0.40 gallons
of the final product containing 8% sulfur and each gallon of Texas crude results in 0.30 gallons of
the final product that contains 10% sulfur.
Since the blend contains sulfur, lead, and phosphorus, each gallon of the resulting gasoline must
meet the following specifications:
1. At most 0.07% sulfur.
2. Between 1.25 and 2.5 grams of lead per gallon.
3. Between 0.0025 and 0.0045 grams of phosphorus per gallon.

Chapter 3

17

Table 3.5: Composition and Cost of the Blending Components

Sulfur (%)
Lead (gm/gal)
Phos. (gm/gal)
Cost ($/gal)

Miss.
0.07


0.55

Crude oils
New Mex.
0.08


0.47

Texas
0.10


0.33

Additives
1
2


7
6
0.025 0.02
0.08 0.12

4. The total amount of the additives cannot exceed 19% of the blend.
As Production Manager, determine a blending plan that yields an acceptable gasoline at the least
cost.
3.5.1

IDENTIFYING THE DECISION VARIABLES

You can control how much of each crude and additive to blend when producing a gallon of gasoline.
This leads to the following five decision variables:
xM

=

xN

=

xT

=

A1

=

A2

=

3.5.2

the number
making one
the number
making one
the number
making one

of gallons of Mississippi crude oil used in
gallon of gasoline,
of gallons of New Mexico crude oil used in
gallon of gasoline,
of gallons of Texas crude oil used in
gallon of gasoline,

the number
making one
the number
making one

of gallons of Additive 1 used in
gallon of gasoline,
of gallons of Additive 2 used in
gallon of gasoline.

IDENTIFYING THE OBJECTIVE FUNCTION

As stated in the problem description, the overall objective is to minimize the cost of the components
used in making each gallon of gasoline. Applying decomposition leads to:
Total cost

=

Cost
Cost
Cost
Cost
Cost

of
of
of
of
of

Mississipi crude oil +
New Mexico crude oil +
Texas crude oil +
Additive 1 +
Additive 2.

Using the variables and the associated costs in Table 3.5 results in the following objective function:
Minimize 0.55xM + 0.47xN + 0.33xT + 0.08A1 + 0.12A2 .

18

Chapter 3

3.5.3

IDENTIFYING THE CONSTRAINTS

Applying the technique of grouping should lead you to identify the following three groups of constraints:
1. A production constraint to insure that one gallon of gasoline is produced, since the blending
plan is for each gallon, a
2. Blend-composition constraints to insure that the resulting gasoline meets the sulfur, lead,
phosphorus, and additive requirements.
3. Logical constraints.
Production Constraint
This constraint insures that precisely one gallon of gasoline is produced:
Amount of gasoline produced = 1 gallon.
Applying decomposition leads to
Amount of gasoline produced

=

Amount
Amount
Amount
Amount
Amount

produced from Miss. crude +
produced from New Mexico crude +
produced from Texas crude +
of Additive 1 +
of Additive 2.

Recall that each gallon of Mississippi crude yields only 0.35 gallons of gasoline. Thus, xM gallons
of this crude yield 0.35xM gallons of gasoline. Similarly, since each gallon of New Mexico crude oils
yield 0.40 gallons of gasoline and each gallon of Texas crude oil results in 0.30 gallons of gasoline,
this constraint is:
0.35xM + 0.40xN + 0.30xT + A1 + A2 = 1.0. (Production)
Blend-Composition Constraints
This group consists of three sets of constraints, one for each of the sulfur, lead, and phosphorus
restrictions on the final blend. For example, for sulfur:
Proportion of sulfur in the blend ≤ 0.0007 (that is, ≤ 0.07%.)
Applying decomposition,
Proportion of sulfur in the blend = Amount of sulfur in the blend .
Total amount of the blend
However, from the production constraint above, the total amount of the blend is precisely one gallon
so, all that need be computed is the amount of sulfur in the blend. Applying decomposition,
Amount of sulfur in the blend
=

Amount
Amount
Amount
Amount
Amount

of
of
of
of
of

sulfur
sulfur
sulfur
sulfur
sulfur

from
from
from
from
from

Mississippi crude oil +
New Mexico crude oil +
Texas crude oil +
Additive 1 +
Additive 2.

Chapter 3

19

According to Table 3.5, each gallon of Mississippi crude oil yields 0.35 gallons of gasoline containing
0.07% sulfur. Thus, xM gallons of this crude oil yield 0.35xM gallons that contains 0.07% sulfur.
So,
Amount of sulfur from Mississippi crude oil

= 0.0007 ∗ 0.35xM
= 0.000245xM .

Noting that the additives contribute no sulfur and applying similar logic to the other two crude oils
results in the following sulfur constraint:
0.35 ∗ 0.0007xM + 0.40 ∗ 0.0008xN + 0.30 ∗ 0.001xT ≤ 0.0007,

or

0.000245xM + 0.00032xN + 0.0003xT ≤ 0.0007. (Sulfur)
There are both lower and upper limits on the amounts of lead and phosphorus in the final blend.
Applying the same reasoning as used in developing the sulfur constraint results in the following four
constraints for lead and phosphorus:
7A1
7A1
0.025A1
0.025A1

+
+

6A2
6A2

+ 0.02A2
+ 0.02A2




2.50
1.25

(Upper limit on lead)
(Lower limit on lead)

≤ 0.0045 (Upper limit on phosphorus)
≥ 0.0025. (Lower limit on phosphorus)

Finally, there is a restriction that the blend contain at most 19% of additives. Thus the total of
A1 and A2 must be at most 0.19 gallons, resulting in the following constraint:
A1 + A2 ≤ 0.19. (Upper limit on additives)
Logical Constraints
The only logical constraints are that all variables be nonnegative.
3.5.4

COMPLETE FORMULATION AND SOLUTION OF THE
BLENDING PROBLEM OF THE HEXXON OIL COMPANY

Putting together this information results in the following linear programming model for the Production Manager of the Hexxon Oil Company:
Minimize 0.55xM + 0.47xN + 0.33xT + 0.08A1 + 0.12A2
Subject to
Production Constraint
0.35xM + 0.40xN + 0.30xT + A1 + A2 =
1.0
Blend-Composition Constraints
0.000245xM + 0.00032xN + 0.0003xT
7A1 +
6A2 ≤
2.50
7A1 +
6A2 ≥
1.25
0.025A1 + 0.02A2 ≤ 0.0045
0.025A1 + 0.02A2 ≥ 0.0025
A1 +
A2 ≤
0.19
Logical Constraint
xM , xN , xT , A1 , A2 ≥ 0.

≤ 0.0007

(Production)

(Sulfur)
(Upper limit on lead)
(Lower limit on lead)
(Upper limit on phosphorus)
(Lower limit on phosphorus)
(Upper limit on additives)

20

Chapter 3

The optimal solution to this problem that results from using any linear programming software
package is:
xM
xN
xT
A1
A2

=
=
=
=
=

0.0000
1.3750
0.8667
0.1400
0.0500

with an objective function value of 0.94945. In other words, each gallon of final product is made by
blending and processing 1.3750 gallons of New Mexico crude oil, and 0.8667 gallons of Texas crude
oil with 0.14 gallons of Additive 1 and 0.05 gallons of Additive 2, at a total cost of 94.945 cents.
You are asked to verify this solution in the exercises following Chapter 6.

3.6

LINEAR PROGRAMMING MODELS FOR
AGGREGATE PRODUCTION PLANNING

Another application of linear programming is in the area of production planning. Managers in
production planning must determine how many of one or more items to produce and how many
to use from existing inventories to meet anticipated demands for a specific period of time. Any
leftover items are stored in inventory. The overall objective is to minimize total costs, made up of
production, inventory, and other costs. Consider the problem facing the management of the National
Steel Corporation.
Example 3.6: The Production-Planning Problem of the National Steel
Corporation
The National Steel Corporation (NSC) produces a special-purpose steel used in the aircraft and
aerospace industries. The Sales Department of NSC has received orders of 2400, 2200, 2700, and
2500 tons of steel for each of the next four months. NSC can meet these demands by producing the
steel, by drawing from its inventory, or by using any combination of the two alternatives.
The production costs per ton of steel during each of the next four months are projected to be
$7400, $7500, $7600, and $7650. Because costs are rising each month — due to inflationary pressures
— NSC might be better off producing more steel than it needs in a given month and storing the
excess. Production capacity, though, cannot exceed 4000 tons in any one month. The monthly
production is finished at the end of the month, at which time the demand is met. Any remaining
steel is then stored in inventory at a cost of $120 per ton for each month that it remains there.
These data are summarized in Table 3.6.
If the production level is increased from one month to the next, then the company incurs a cost
of $50 per ton of increased production to cover the additional labor and/or overtime. Each ton of
decreased production incurs a cost of $30 to cover the benefits of unused employees.
The production level during the previous month was 1800 tons and the beginning inventory is
1000 tons. If the inventory at the end of the fourth month must be at least 1500 tons, formulate a
production plan for NSC that minimizes the total costs over the next four months.
3.6.1

IDENTIFYING THE DECISION VARIABLES

In this problem, you are free to choose how many tons of steel to produce each month to meet
demand. Four variables arise:

Chapter 3

21

Table 3.6: Data for the Production-Planning Problem of NSC
Month
1
2
3
4
2400 2200 2700 2500
7400 7500 7600 7650

Demand (tons)
Production cost ($/ton)
Inventory cost
($ per ton per month)

x1
x2
x3
x4

=
=
=
=

the
the
the
the

number
number
number
number

of
of
of
of

tons
tons
tons
tons

of
of
of
of

steel
steel
steel
steel

to
to
to
to

120

produce
produce
produce
produce

120

during
during
during
during

120

month
month
month
month

120

1,
2,
3,
4.

At first glance, you might think that these are all the variables needed. With these variables, you
can always determine the amount in inventory. For example, looking at the schematic diagram in
Figure 3.1, the inventory at the end of the first month is:
Inventory at the end of month 1 = Beginning inventory +
Production amount −
Demand
= 1000 + x1 − 2400.
However, writing the inventory at the end of the second, third, and subsequent months is more
complicated. For example, for month 2:
Inventory at the end of month 2 = Beginning inventory +
Production amount −
Demand
= (1000 + x1 − 2400) + x2 − 2200.
To simplify, it is expedient to create five more variables to represent the inventory levels at the
beginning of each month:
I1
I2
I3
I4
I5

3.6.2

=
=
=
=
=

inventory
inventory
inventory
inventory
inventory

in
in
in
in
in

tons
tons
tons
tons
tons

at
at
at
at
at

the
the
the
the
the

beginning
beginning
beginning
beginning
beginning

of
of
of
of
of

month
month
month
month
month

1,
2,
3,
4,
5.

IDENTIFYING THE OBJECTIVE FUNCTION

As stated in the problem description, the overall objective is to minimize the total costs over the
four-month planning horizon. Applying decompostion to identify three different cost components
leads to:

22

Chapter 3

Production Quantity
(x1 )
?
Beginning Inventory
(I1 = 1000)

-

Month 1

- Ending Inventory
(I2 )

?
Demand
(D1 = 2400)

Figure 3.1: Relationship Among Inventory Levels, Production, and Demand

Total costs

=

Production costs + Inventory costs +
Change-in-production costs.

Production Costs
Applying decomposition again identifies the production costs as the sum of the production costs in
each of the four months. Using the production variables x1 , x2 , x3 , and x4 together with the per-ton
production costs in Table 3.6 yields:
Production costs = 7400x1 + 7500x2 + 7600x3 + 7650x4 .
Inventory Costs
A similar decomposition yields a total inventory cost as the sum of the inventory costs during each
of the four months. Since the inventory levels change only at the end of the month, all inventories
at the beginning of the month incur a cost of $120 per ton for that month. Using the variables I1 ,
I2 , I3 , and I4 yields:
Inventory costs = 120I1 + 120I2 + 120I3 + 120I4 .
Observe that I5 is not included in this portion because the objective is to minimize total costs over
the next four months only, and I5 incurs cost during the fifth month.
Change-In-Production Costs
To determine the change-in-production costs from one month to the next, work through a specific
example in which, say, x1 = 100 and x2 = 300. In this case, there is an increase of 300 − 100 = 200
tons of steel from month 1 to month 2. Thus, at a cost of $50 per ton of increase,
Change-in-production cost = (300 − 100) ∗ 50 = $10, 000.
Using this example, you might write the following general expression:

Chapter 3

23

Change-in-production cost = (x2 − x1 ) ∗ 50.
However, what if x1 = 300 and x2 = 100? That is, what if the production level decreases. In this
case, the expression above results in a cost of (100−300)∗50 = −$10, 000, that is, a profit of $10,000,
which makes no sense. Instead, at a cost of $30 per ton of decrease, the correct expression is:
Change-in-production cost = (300 − 100) ∗ 30
= $6, 000.
In general, when the production level decreases from month 1 to month 2, the correct expression is:
Change-in-production cost = (x1 − x2 ) ∗ 30.
Combining the expressions for an increase and a decrease results in the following change-in-production
costs from month 1 to month 2:

Change-in-production costs

=


⎨ 50(x2 − x1 ) if x2 ≥ x1


30(x1 − x2 ) if x1 > x2

(increase)
(decrease).

Since the values of x1 and x2 are as yet unknown, the issue is how to combine these two cases into
a single expression.
One approach is to create additional decision variables whose values are precisely the amounts
of increased and decreased production from one month to the next. That is,
S1
D1

=
=

the number of tons of increased production in month 1,
the number of tons of decreased production in month 1,

S2
D2

=
=

the number of tons of increased production in month 2,
the number of tons of decreased production in month 2,

S3
D3

=
=

the number of tons of increased production in month 3,
the number of tons of decreased production in month 3,

S4
D4

=
=

the number of tons of increased production in month 4,
the number of tons of decreased production in month 4.

The values of these variables depend on the production levels. For example, when x2 = 300 and
x1 = 100, you want S2 to be 200 and D2 to be 0. If x2 = 100 and x1 = 300, you want S2 to be 0
and D2 to be 200. The constraints that insure the proper relationships among these variables are
identified in Section 3.6.3.
With these new variables, when S1 is positive, D1 must be 0. Similarly, when D1 is positive, S1
must be 0. Thus, the change-in-production costs for the first month are 50S1 + 30D1 . Hence the
total change-in-production costs are:

24

Chapter 3

Change-in-prodcution costs

=

Change-in-production
Change-in-production
Change-in-production
Change-in-production

cost
cost
cost
cost

in
in
in
in

month
month
month
month

=

(50S1 + 30D1 ) + (50S2 + 30D2 )+
(50S3 + 30D3 ) + (50S4 + 30D4 ).

1+
2+
3+
4

Complete Objective Function
Combining the three cost components results in the following overall objective function:
Minimize total costs = 7400x1 + 7500x2 + 7600x3 + 7650x4 +
120I1 + 120I2 + 120I3 + 120I4 +
50S1 + 30D1 + 50S2 + 30D2 +
50S3 + 30D3 + 50S4 + 30D4 .
3.6.3

IDENTIFYING THE CONSTRAINTS

Applying the technique of grouping should lead you to identify the following six groups of constraints:
1. Initial and final inventory constraints to insure the proper beginning and ending inventory
levels.
2. Production-bound constraints to insure that the production in any given month not exceed
4000 tons.
3. Inventory-balance constraints to insure the proper relationship between the production and
inventory variables.
4. The change-in-production constraints to insure the proper relationship between the production and change-in-production variables.
5. Demand constraints to insure that demands are met in each month.
6. Logical constraints to insure all variables are nonnegative.
Initial and Final Inventory Constraints
In words, the two constraints in this group are:
Initial inventory level is 1000 tons,

Final inventory level must be at least 1500 tons.
Since I1 and I5 represent the initial and final inventories at the beginning and end of the fourmonth planning period, respectively, these constraints are:
I1
I5

= 1000 (Beginning inventory)
≥ 1500. (Ending inventory)

Chapter 3

25

Production-Bound Constraints
Since the production in any month cannot exceed 4000 tons, the four constraints in this group are:
≤ 4000
≤ 4000
≤ 4000
≤ 4000.

x1
x2
x3
x4

(Bound
(Bound
(Bound
(Bound

in
in
in
in

month
month
month
month

1)
2)
3)
4)

Inventory-Balance Constraints
This group consists of four constraints to insure the proper relationship between the hypothetical
production and inventory amounts illustrated in Figure 3.2. For example, for month 1,
Inventory at the end of month 1

=

Inventory at the beginning of month 1 +
Amount produced in month 1 −
Demand for month 1.

Since the inventory at the end of month 1 is precisely the beginning inventory in month 2,
I2 = I1 + x1 − 2400, or
−I2 + I1 + x1 = 2400. (Inv. balance in month 1)
A similar constraint is needed for each of the remaining three months, resulting in:
−I3
−I4
−I5

+ I2
+ I3
+ I4

+ x2
+ x3
+ x4

= 2200 (Inv. balance in month 2)
= 2700 (Inv. balance in month 3)
= 2500. (Inv. balance in month 4)

Change-In-Production Constraints
This group of constraints insures the proper relationship between the production and change-inproduction variables. For instance, consider the change in production from month 1 to month 2.
After working through several specific examples, you might conclude that:
Production in month 2

=

Production in month 1 +
(Increase in production in month 2 −
Decrease in production in month 2).

Using the variables x2 , x1 , S2 , and D2 leads to:
x2 = x1 + (S2 − D2 ), or
x2 − x1 − S2 + D2 = 0. (Change in month 2)
For example, when x2 = 300 and x1 = 100, the values of S2 = 200 and D2 = 0 satisfy this constraint,
as they should. However, what insures that if S2 is positive, then D2 is in fact 0? For instance,
when x2 = 300 and x1 = 100, the values of S2 = 250 and D2 = 50 also satisfies the above constraint.
Likewise, if x2 is less than x1 , what insures that D2 is positive and S2 is 0? It would appear that
an additional constraint is necessary, for example:

26

Chapter 3

Figure 3.2: Inventory Levels of NSC

Chapter 3

27

S2 ∗ D2 = 0.
Including such constraints results in a nonlinear model which is substantially more difficult to solve
than a linear programming model. It is indeed fortunate that in this particular problem these
nonlinear constraints are not needed. This is because the objective function serves the same purpose.
To see how, consider the numerical example just used:
x2

x1

S2

300 100 200
300 100 250

D2

Change-in-production costs

0 (200 ∗ 50) + (0 ∗ 30) = 10000
50 (250 ∗ 50) + (50 ∗ 30) = 14000

Since the objective is to achieve minimum costs, it is always less expensive to make one of the two
variables S2 or D2 have value 0. As a result, no additional constraints are needed to insure this
relationship.
A similar constraint for each of the first, third, and fourth months yields:
x1
x1

− 1800 − S1
− S1

+ D1
+ D1

=
0, or
= 1800
(Change in month 1)

x3
x4




+ D3
+ D4

=
=

x2
x3

− S3
− S4

0
0.

(Change in month 3)
(Change in month 4)

Demand Constraints
To insure that the demands are met, consider month 1. The appropriate constraint is:
Beginning inventory
in month 1

+

Amount produced
in month 1



Demand
in month 1

.

Using the decision variables results in:
I1 + x1 ≥ 2400, or
I1 + x1 − 2400 ≥ 0.
However, observe from the inventory balance constraint for month 1 that:
I2 = I1 + x1 − 2400.
Thus, the demand constraint for month 1 can also be written as:
I2 ≥ 0.
In other words, requiring that each inventory variable be nonnegative insures that the demand for
the previous month is met. Thus, these demand constraints can be included as the logical constraints
below.
Logical Constraints
The only logical constraints are that each production, inventory, and change-in-production variable
be nonnegative.

28
3.6.4

Chapter 3
COMPLETE FORMULATION AND SOLUTION OF THE
PRODUCTION-PLANNING PROBLEM OF NSC

Putting together the pieces results in the following linear programming model for the productionplanning problem of the National Steel Corporation:
Minimize

7400x1 + 7500x2 + 7600x3 + 7650x4 +
120I1 + 120I2 + 120I3 + 120I4 +
50S1 + 30D1 + 50S2 + 30D2 +
50S3 + 30D3 + 50S4 + 30D4

Subject to
Initial and Final Inventory Constraints
I1 = 1000
I5 ≥ 1500
Production Bound
x1 ≤
x2 ≤
x3 ≤
x4 ≤

(Beginning inventory)
(Ending inventory)

Constraints
4000
4000
4000
4000

Inventory Balance Constraints
−I2 + I1 + x1
−I3 + I2 + x2
−I4 + I3 + x3
−I5 + I4 + x4
Change-In-Production
x1
x2 − x1
x3 − x2
x4 − x3

(Bound
(Bound
(Bound
(Bound

=
=
=
=

2400
2200
2700
2500

Constraints
− S1 + D1
− S2 + D2
− S3 + D3
− S4 + D4

(Inv.
(Inv.
(Inv.
(Inv.

= 1800
=
0
=
0
=
0

in
in
in
in

month
month
month
month

balance
balance
balance
balance

(Change
(Change
(Change
(Change

in
in
in
in

in
in
in
in

1)
2)
3)
4)

month
month
month
month

month
month
month
month

1)
2)
3)
4)

1)
2)
3)
4)

Logical Constraints
x1 , x2 , x3 , x4 , I1 , I2 , I3 , I4 , I5 ,
S1 , S2 , S3 , S4 , D1 , D2 , D3 , D4 ≥ 0.
The optimal solution to this problem that results from using any linear programming software
package is:
x1
x2
x3
x4

=
=
=
=

1800.00
1800.00
2700.00
4000.00

I1
I2
I3
I4
I5

= 1000.00
= 400.00
=
0.00
=
0.00
= 1500.00

S1
S2
S3
S4

=
0.00
=
0.00
= 900.00
= 1300.00

D1
D2
D3
D4

=
=
=
=

0.00
0.00
0.00
0.00

with an objective function value of 78,218,000. In other words, the optimal production plan can be
summarized as:

Chapter 3

29

Beginning inventory
Amount produced
Demand
Ending inventory

1
1000
1800
2400
400

Month
2
3
400
0
1800
2700
2200
2700
0
0

4
0
4000
2500
1500

with a total cost of $78,218,000. You are asked to verify this solution in the exercises following
Chapter 6.
In Sections 3.1 through 3.6 you have seen several examples of how linear programming models
are formulated to help managers make decisions. These models are formulated using the techniques
and guidelines given in Chapter 2. Unlike the examples presented so far, most real-world problems
involve hundreds or thousands of variables and a similar number of constraints. Formulating these
large, complex models is accomplished using these same techniques, as you will see in the case study
presented in Section 3.7.

3.7

CASE STUDY: GLOBAL PLANNING FOR THE
AMERICAN STEEL COMPANY

All the problems discussed so far have involved only one functional area of an organization. For
instance, the transportation problem in Example 2.2 in Section 2.2 involves only the distribution of
the finished computers given that the production plan has already been decided. Moreover, these
models have required very few variables and constraints. Most real-world problems involve many
functional areas of the company (sales, production, distribution, marketing, and so on). Formulating
such problems requires a large number of variables and constraints — perhaps tens of thousands of
variables and thousands of constraints. Although teams of people may be involved in formulating
these models, the basic process is the same as the one you have already learned: identify the decision
variables, the data, the objective function, and the constraints.
Due to the large-scale nature of these problems, additional techniques are sometimes helpful in
making the formulation manageable. Some of these techniques are illustrated in this section.
3.7.1

THE PROBLEM OF THE AMERICAN STEEL COMPANY

American Steel has received annual orders from four countries — Japan, Korea, Taiwan, and Mexico
— for two different types of steels it produces: high grade and low grade. These steels are produced at
its two plants, located in Pittsburgh and Youngstown, using iron ores supplied by two mining companies — Butte Minerals and Cheyenne Mines. The management of American Steel needs an overall
annual purchase/production/distribution plan to minimize total costs. Various departments have
collected the necessary data regarding the sales commitment, ore availability and cost, production
characteristics, and distribution costs of ore and finished steel.
American Steel can obtain up to 1000 tons of Grade-A iron ore from Butte Minerals and up to
2000 tons of Grade-B iron ore from Cheyenne Mines. American Steel can specify how much of each
ore is to be shipped to each of its two steel mills. The associated purchase cost and shipping charge
per ton is given in Table 3.7.
Each of American Steel’s two mills can produce high-grade steel and/or low-grade steel. Highgrade steel requires blending Grade-A and Grade-B iron ore in a ratio of 1 to 2. Low-grade steel
requires a ratio of 1 to 3. The Youngstown mill can process up to 1500 tons of iron ore and the
Pittsburgh facility can handle at most 700 tons. The mill at Pittsburgh is a modern facility and has

30

Chapter 3

Table 3.7: Ore Purchase and Shipping Costs
Purchase
Shipping cost to
cost
Pittsburgh Youngstown
($/ton)
($/ton)
($/ton)
Butte Minerals

130

10

13

Cheyenne Mines

110

14

17

a lower processing cost per ton of steel produced than does the facility at Youngstown, as indicated
in Table 3.8.
Table 3.8: Processing Costs
Processing cost ($/ton)
Pittsburgh Youngstown
High-grade steel

32

39

Low-grade steel

27

32

The finished steel is shipped to Japan, Korea, Taiwan, and Mexico. The International Sales
Division of American Steel has received orders for each type of steel, given in Table 3.9. This table
also includes the shipping costs per ton for each type of steel.
As manager of the Management Science Group of American Steel, you have been asked to
make recommendations on the purchasing, processing, and shipping functions with the objective of
minimizing the total annual cost.
3.7.2

MATHEMATICAL FORMULATION

Your first step is to formulate the mathematical model which, once classified, can be solved by
computer using an appropriate algorithm to be discussed in subsequent chapters. While the problem
and all its data may at first appear overwhelming, you can formulate this problem using the same
techniques you learned in Section 2.2. Some additional techniques are also introduced to help make
the formulation more manageable.
Identifying the Decision Variables
One technique for simplifying the process of identifying all the decision variables is to decompose
the overall problem into a collection of smaller, more manageable subproblems. While there are no
specific rules for doing this, one typical approach is to focus on the individual functional areas, as
will be done in this example. To that end, consider the sequence of operations that are performed:
1. The purchase of the iron ores from the two mining companies and the shipment of the ores
to the two steel mills.

Chapter 3

31

Table 3.9: Demands and Unit Shipping Costs of Steel

Country

Steel type

Demand Shipping cost ($/ton) from
(tons)
Pittsburgh
Youngstown

Japan

High grade
Low grade

400
200

110
100

115
110

Korea

High grade
Low grade

200
100

140
130

150
145

Taiwan

High grade
Low grade

200
100

130
125

135
127

Mexico

High grade
Low grade

150
50

80
80

90
85

2. The processing of the ores at the two mills to produce the two grades of steel.
3. The shipping of the finished steel to the four countries.
Having decomposed the original problem into these three subproblems, you can now proceed to
identify the decision variables associated with each function.
Decision Variables for Purchasing and Shipping Iron Ore
Asking yourself the usual question of what you can control should lead you to identify the following
decision variables pertaining to this functional area:
IBP

=

IBY

=

ICP

=

ICY

=

The number of tons of Grade A iron ore to buy from Butte Minerals
and ship to the Pittsburgh mill.
The number of tons of Grade A iron ore to be bought from Butte
Minerals and ship to the Youngstown mill.
The number of tons of Grade B iron ore to buy from Cheyenne
Mines and ship to the Pittsburgh mill.
The number of tons of Grade B iron ore to buy from Cheyenne
Mines and ship to the Youngstown mill.

The variables corresponding to this portion of the problem are illustrated in the schematic diagram of Figure 3.3. Schematic diagrams are used throughout this formulation to help identify the
relationships among the decision variables for the various functional areas.
Decision Variables for Producing Steel
Recall that the production of low-grade and high-grade steel at the mills is the second functional
area. For this phase, ask yourself what decisions you are free to make. What can you control? One
answer is the number of tons of each type steel to produce at each mill. Another answer is the

32

Chapter 3

'

¡

¡

¡ @

@ ¡

BUTTE @

¡
@




@ −
Q
CHEYENNE
Q
¡

¡

¡

PITTSBURGH

$

¡¡µ

¡IBP



IBY

@


@ −
−@
@


−ICP

−Á
− &

'

%
YOUNGSTOWN

$

@@R

¡ @

QICY
Q

Q

Q

QQs
&

%

Figure 3.3: Schematic Diagram for American Steel: Ore Purchase

Chapter 3

33

amount of each grade of iron ore to blend. It is possible to develop the model by using either set of
decision variables. In this case, the formulation is more manageable using both sets. So let

HP

=

LP

=

HY

=

LY

=

OBPH

=

OCPH

=

OBPL

=

OCPL

=

OBYH

=

OCYH

=

OBYL

=

OCYL

=

The number of tons of high-grade steel to produce at the Pittsburgh
mill.
The number of tons of low-grade steel to produce at the Pittsburgh
mill.
The number of tons of high-grade steel to produce at the
Youngstown mill.
The number of tons of low-grade steel to produce at the
Youngstown mill.
The number of tons of iron ore from Butte Minerals used to produce
high-grade steel at the Pittsburgh mill.
The number of tons of iron ore from Cheyenne mines used to produce high-grade steel at the Pittsburgh mill.
The number of tons of iron ore from Butte Minerals used to produce
low-grade steel at the Pittsburgh mill.
The number of tons of iron ore from Cheyenne mines used to produce low-grade steel at the Pittsburgh mill.
The number of tons of iron ore from Butte Minerals used to produce
high-grade steel at the Youngstown mill.
The number of tons of iron ore from Cheyenne mines used to produce high-grade steel at the Youngstown mill.
The number of tons of iron ore from Butte Minerals used to produce
low-grade steel at the Youngstown mill.
The number of tons of iron ore from Cheyenne mines used to produce low-grade steel at the Youngstown mill.

Combining these decision variables with the variables corresponding to the purchase of the iron ore
in Figure 3.3 results in the schematic diagram of Figure 3.4.

Decision Variables for the Distribution of Steel

Recall that the last functional area pertains to the shipping of the final steel products from the two
mills to each of the four countries. You might recognize the similarity between this component of the
problem and the transportation problem in Section 2.2.1. Think of each mill as a supply point and
each country as a demand point. The distribution can be viewed as two transportation problems —
one for each type of steel — leading to the following decision variables:

34

Chapter 3

'

¡

¡ @

@ ¡
BUTTE @

¡
@

¡
¡IBP

IBY

@

¡

¡





@ −
−@
@


¡µ¡

PITTSBURGH

OBPH
a a
!
a a
!
!
aOBPL
a a
! !
!
a !a
a a
! !
!
a a
!OCPH
!
a a
!
!
!
-

Á−
− &

'

-

HP

½
¾
LP

%
YOUNGSTOWN

½
¾

¼
»

$
-

−ICP

HY

½
¾
-

¼
»
LY

OCYL

&

¼
»

-

OCPL

OBYH
@@R
H H
©*
©
H H
©

¡ @
HOBYL
© ©
H H
¡
@ −
©
Q
CHEYENNE
H ©
Q ICY
© © H H
Q
©
H H
Q
© ©OCYH
Q
H H
©
Q
jHQQs © ©



»

¾
$

%
½

¼

Figure 3.4: Schematic Diagram for American Steel: Ore Purchase and Steel
Production

Chapter 3

35

For High-Grade Steel
SHPJ
= The number of tons of high-grade steel to ship from the Pittsburgh
mill to Japan.
SHPK = The number of tons of high-grade steel to ship from the Pittsburgh
mill to Korea.
SHPT = The number of tons of high-grade steel to ship from the Pittsburgh
mill to Taiwan.
SHPM = The number of tons of high-grade steel to ship from the Pittsburgh
mill to Mexico.
SHYJ

=

SHYK

=

SHYT

=

SHYM

=

The number of tons of high-grade steel to ship from the Youngstown
mill to Japan.
The number of tons of high-grade steel to ship from the Youngstown
mill to Korea.
The number of tons of high-grade steel to ship from the Youngstown
mill to Taiwan.
The number of tons of high-grade steel to ship from the Youngstown
mill to Mexico.

For Low-Grade Steel
SLPJ
= The number of tons
mill to Japan.
SLPK
= The number of tons
mill to Korea.
SLPT
= The number of tons
mill to Taiwan.
SLPM = The number of tons
mill to Mexico.
SLYJ

=

SLYK

=

SLYT

=

SLYM

=

of low-grade steel to ship from the Pittsburgh
of low-grade steel to ship from the Pittsburgh
of low-grade steel to ship from the Pittsburgh
of low-grade steel to ship from the Pittsburgh

The number of tons of low-grade steel to ship from the Youngstown
mill to Japan.
The number of tons of low-grade steel to ship from the Youngstown
mill to Korea.
The number of tons of low-grade steel to ship from the Youngstown
mill to Taiwan.
The number of tons of low-grade steel to ship from the Youngstown
mill to Mexico.

Putting together all of the decision variables for this problem results in the schematic diagram in
Figure 3.5 where, due to limited space, the symbolic names of most of the shipping variables are not
included. This diagram illustrates: (1) the three functional areas of this problem, (2) the sequential
nature in which the various operations are performed, and (3) a summary of all the variables used
in the formulation. Since it provides a concise visual summary of the entire process, you will find
this diagram useful throughout the rest of the formulation.

Identifying the Objective Function
As stated in the problem description of Section 3.7.1, the overall objective is to minimize the total
cost. Applying the technique of decomposition according to the functional areas in Figure 3.5, you
have
Total cost

=

Purchase costs

+

Production costs

+

Distribution costs.

36

Chapter 3

'

PITTSBURGH

a a
!
¡¡µ
a a
!
!
aOBPL
¡
a a
! !
!
¡
a !a
a a
¡
! !
!
a a
!OCPH
¡IBP
!
a a
!
¡
!
¡ @
!
¡
@ ¡
OCPL
−Á
BUTTE @
− &
%
@ IBY

@

@ −
YOUNGSTOWN
−@
'
$
@

OBYH
@@R
−ICP
H H
©*

©
H H

© ©
¡ @
HOBYL
©
H H
¡
@ −
©
Q
CHEYENNE
H ©H
©
Q ICY
H H
Q
© ©
Q
©
H H
OCYH
©
Q
©
H Hj
Q
©
QQs ©
OBPH

-

HP

½
¾
LP

½
¾
-

HY

½
¾
-

LY

OCYL

&

»

¾
$

SHPJ
- JAPAN
¡µ£¢¸£±
BA@
¡
BA@
¡ ¢¢£
¼BA @
¡
BA @
¢£
» BA ¡ @ ¢ £
¡
£
¡ BBAA ¢¢@£ @
¡
B ¢A ££ ¡µ@R- ¢¸ KOREA
A@
A@ ¢BB¢ A£A¡ ¡¢
¼ A ¢@ B £¡ A ¢
A @£ ¢
» ¢A ¡£B@ ¢A
B
¢ ¡
A
¢¡ A££A ¢B¢@ @A
¡¢
£ ¢A BB ¡µ@R- AUTAIWAN
@
@ ££¢¢ AA¡B ¡
B
¼ £ ¢@
@ ¡ A B
» £¢ ¡ @ A B
£¢ ¡
@ AAB
£¢¡
@@R- BAUBN
¡¢£
MEXICO
SLYM

%
½

¼

Figure 3.5: Schematic Diagram for American Steel: Ore Purchase, Steel Production, and Distribution

Chapter 3

37

Purchase Costs
Looking at Figure 3.5, you can further decompose the purchase costs into the cost of the iron ore
plus the shipping cost to each of the mills. Use the decision variables IBP , IBY , ICP , and ICY
and the specific data in Table 3.7 to get:
Purchase costs = Iron ore costs + Shipping costs
= (130IBP
( 10IBP

+ 130IBY
+ 13IBY

+ 110ICP
+ 14ICP

+
+

110ICY )
17ICY )

= (140IBP

+ 143IBY

+ 124ICP

+ 127ICY ).

+

Production Costs
In a similar manner, you can decompose the production cost into the production costs at the Pittsburgh mill plus those at the Youngstown mill. Using the variables HP , LP , HY , and LY together
with the data in Table 3.8 leads to:

Production costs

=

(Costs at Pittsburgh)

+

(Costs at Youngstown)

=

( 32HP + 27LP )

+

( 39HY + 32LY ).

Shipping Costs
You can decompose the total shipping costs into the shipping costs from the Pittsburgh mill to each
of the four countries plus the shipping costs from the Youngstown mill to each of the four countries.
Using the appropriate decision variables and the data in Table 3.9 leads to:

Shipping costs

=

(Costs from Pittsburgh) + (Costs from Youngstown)

=

(110SHP J + 140SHP K + 130SHP T + 80SHP M +
100SLP J + 130SLP K + 125SLP T + 80SLP M ) +
(115SHY J + 150SHY K + 135SHY T + 90SHY M +
110SLY J + 145SLY K + 127SLY T + 85SLY M ).

Combining the three individual cost components, the complete objective function is:
Minimize

(140IBP + 143IBY + 124ICP + 127ICY ) +
(32HP + 27LP ) + (39HY + 32LY ) +
(110SHP J + 140SHP K + 130SHP T + 80SHP M +
100SLP J + 130SLP K + 125SLP T + 80SLP M ) +
(115SHY J + 150SHY K + 135SHY T + 90SHY M +
110SLY J + 145SLY K + 127SLY T + 85SLY M ).

38

Chapter 3

Identifying the Constraints
Constraints are needed to reflect the appropriate physical limitations of the problem and to specify
the logical relationships among the decision variables of the different functional areas. To develop
both types of constraints, use the technique of decomposition and follow the flow of information in
Figure 3.5.
Purchasing Constraints
Recalling the physical capacity limitations of the two mines should lead you to identify the following
constraints: the total number of tons of iron ore shipped from each mining company cannot exceed
the available supply. Use the decision variables of this functional area to formulate these constraints
mathematically:
IBP + IBY
ICP + ICY

≤ 1000 (Butte Minerals supply)
≤ 2000 (Cheyenne Mines supply).

(1)
(2)

Production Constraints
Asking yourself what restrictions apply to the production process will lead you to identify the
following groups of constraints:
1. Ore-processing capacities — the total number of tons of ore processed at each mill cannot
exceed the mill’s capacity.
2. The blending requirements for producing each grade of steel — that is, each grade of steel
requires the blending of a certain proportion of each type of iron ore.
Using the decision variables, these constraints are formulated mathematically as follows.
1. Ore-Capacity Constraints

OBP H + OBP L + OCP H + OCP L ≤ 700 (Pittsburgh mill)
OBY H + OBY L + OCY H + OCY L ≤ 1500 (Youngstown mill).

(3)
(4)

2. Blending Requirements
According to the problem description, the iron ore from Cheyenne Mines and Butte Minerals
must be blended in a ratio of 2 to 1 to produce high-grade steel. Thus you are led to the
following mathematical constraints for high-grade steel at the Pittsburgh mill:
OCP H/OBP H = 2/1.

To make this constraint linear, multiply through by OBPH and then subtract 2 OBPH from
both sides to obtain:

OCP H − 2OBP H = 0.

(5)

Chapter 3

39

Likewise, the constraint for low-grade steel, requiring a ratio of 3 to 1, is:

OCP L − 3OBP L = 0.

(6)

Similar constraints are needed for the Youngstown mill, so:

OCY H − 2OBY H
OCY L − 3OBY L

= 0
= 0.

(7)
(8)

In addition to these constraints, there are some implicit relationships between the amounts of
iron ore purchased from the mines and the amounts of these ores used in the production of steel.
For example, from Figure 3.5 you can see that the total amount of ore used in producing the two
steels cannot exceed the amount purchased. These implicit constraints must be made explicit:
3. Production and Purchasing Relationships

OBP H + OBP L
OCP H + OCP L

≤ IBP
≤ ICP

OBY H + OBY L ≤ IBY
OCY H + OCY L ≤ ICY

(Ore from Butte used in Pittsburgh)
(Ore from Cheyenne used in Pittsburgh)

(9)
(10)

(Ore from Butte used in Youngstown)
(Ore from Cheyenne used in Youngstown).

(11)
(12)

Another implicit constraint evident from Figure 3.5 is that the total amount of ore used in
the production process must equal the total amount of steel produced. This must be the
case for each grade of steel at each mill. These implicit constraints can be made explicit as
follows:

OCP H + OBP H
OCP L + OBP L

= HP
= LP

(High-grade steel produced at Pittsburgh)
(Low-grade steel produced at Pittsburgh)

(13)
(14)

OCY H + OBY H
OCY L + OBY L

= HY
= LY

(High-grade steel produced at Youngstown)
(Low-grade steel produced at Youngstown).

(15)
(16)

Distribution Constraints
From Figure 3.5 you still need appropriate constraints to insure that the demands for the amounts
of the two grades of steel supplied by the two mills to each of the four countries are met. You might
recognize that this portion is a transportation problem, as described in Section 2.2.1, and should
lead to the following mathematical constraints:

40

Chapter 3
1. Supplies at the Mills

SHP J + SHP K + SHP T + SHP M

≤ HP

SLP J + SLP K + SLP T + SLP M

≤ LP

SHY J + SHY K + SHY T + SHY M

≤ HY

SLY J + SLY K + SLY T + SLY M

≤ LY

(High-grade steel shipped
from Pittsburgh)
(Low-grade steel shipped
from Pittsburgh)
(High-grade steel shipped
from Youngstown)
(Low-grade steel shipped
from Youngstown).

(17)
(18)

(19)
(20)

2. Demands for Steel in the Four Countries

SHP J + SHY J
SLP J + SLY J

≥ 400 (Demand for high-grade steel in Japan)
≥ 200 (Demand for low-grade steel in Japan)

(21)
(22)

SHP K + SHY K
SLP K + SLY K

≥ 200 (Demand for high-grade steel in Korea)
≥ 100 (Demand for low-grade steel in Korea)

(23)
(24)

SHP T + SHY T
SLP T + SLY T

≥ 200 (Demand for high-grade steel in Taiwan)
≥ 100 (Demand for low-grade steel in Taiwan)

(25)
(26)

SHP M + SHY M
SLP M + SLY M

≥ 150 (Demand for high-grade steel in Mexico)
≥ 50 (Demand for low-grade steel in Mexico).

(27)
(28)

The final set of constraints are nonnegativity constraints on all the variables. These constraints,
together with the objective function and all other constraints formulated above provide a complete
and correct formulation of the problem, which follows.
3.7.3

Mathematical Formulation of the Problem of the American Steel Company
Minimize

(140IBP + 143IBY + 124ICP + 127ICY ) +
(32HP + 27LP ) + (39HY + 32LY ) +
(110SHP J + 140SHP K + 130SHP T + 80SHP M +
100SLP J + 130SLP K + 125SLP T + 80SLP M ) +
(115SHY J + 150SHY K + 135SHY T + 90SHY M +
110SLY J + 145SLY K + 127SLY T + 85SLY M )

Subject to
Purchasing Constraints
IBP + IBY ≤ 1000 (Butte Minerals supply)
ICP + ICY ≤ 2000 (Cheyenne Mines supply)

(1)
(2)

Chapter 3

41

Production Constraints
Ore Capacity Constraints
OBP H + OBP L + OCP H + OCP L ≤ 700 (Pittsburgh mill)
OBY H + OBY L + OCY H + OCY L ≤ 1500 (Youngstown mill)

Blending Requirements
OCP H − 2OBP H =
OCP L − 3OBP L =
OCY H − 2OBY H =
OCY L − 3OBY L =

0
0
0
0

(3)
(4)

(5)
(6)
(7)
(8)

Production and Purchasing Relationships
OBP H + OBP L ≤ IBP (Ore from Butte used in Pittsburgh)
OCP H + OCP L ≤ ICP (Ore from Cheyenne used in Pittsburgh)

(9)
(10)

OBY H + OBY L ≤ IBY
OCY H + OCY L ≤ ICY

(Ore from Butte used in Youngstown)
(Ore from Cheyenne used in Youngstown)

(11)
(12)

Production-Balance Constraints
OCP H + OBP H = HP (High-grade steel produced at Pittsburgh)
OCP L + OBP L = LP (Low-grade steel produced at Pittsburgh)

(13)
(14)

OCY H + OBY H
OCY L + OBY L

(15)
(16)

= HY
= LY

(High-grade steel produced at Youngstown)
(Low-grade steel produced at Youngstown)

Distribution Constraints
Supplies at the Mills
SHP J + SHP K + SHP T + SHP M
SLP J + SLP K + SLP T + SLP M

≤ HP
≤ LP

SHY J + SHY K + SHY T + SHY M

≤ HY

SLY J + SLY K + SLY T + SLY M

≤ LY

(high-grade steel shipped
from Pittsburgh)
(low-grade steel shipped
from Pittsburgh)
(high-grade steel shipped
from Youngstown)
(low-grade steel shipped
from Youngstown)

(17)
(18)

(19)
(20)

42

Chapter 3

Demands for Steel in the Four Countries
SHP J + SHY J
≥ 400 (Demand for high-grade steel in Japan)
SLP J + SLY J
≥ 200 (Demand for low-grade steel in Japan)

(21)
(22)

SHP K + SHY K
SLP K + SLY K

≥ 200 (Demand for high-grade steel in Korea)
≥ 100 (Demand for low-grade steel in Korea)

(23)
(24)

SHP T + SHY T
SLP T + SLY T

≥ 200 (Demand for high-grade steel in Taiwan)
≥ 100 (Demand for low-grade steel in Taiwan)

(25)
(26)

SHP M + SHY M
SLP M + SLY M

≥ 150 (Demand for high-grade steel in Mexico)
≥ 50 (Demand for low-grade steel in Mexico)

(27)
(28)

and all variables nonnegative.

This section has provided an example of formulating a large and complex problem with many
variables and constraints. (The solution to this problem is given in Chapter 6.) In addition to the
techniques of Section 2.2, you learned to decompose a problem into smaller, more manageable subproblems. This is often done based on the functional areas of an organization, or on a sequence of
operations to be performed. A schematic diagram is useful to summarize this decompostion. Then,
each subproblem is formulated by identifying the decision variables relevant to that subproblem.
After specifying an overall objective function, constraints are needed not only to reflect the restrictions within each subproblem, but also to specify the relationships among the variables from one
subproblem to another. Now that you know how to formulate an appropriate linear programming
model, it is time to learn the algorithm available for solving these problems. That study begins in
Chapter 4 after a brief summary of this chapter in Section 3.8.

3.8

SUMMARY

In this chapter you have seen numerous applications of linear programming problems that arise in
such areas as production planning, financial planning, blending decisions, allocating scarce resources,
and so on. These applications are limited only by your ability to identify and formulate such
problems.
To formulate a linear programming problem, follow these steps that are discussed in Chapter 2:
1. Identify decision variables.
2. Identify an overall objective function.
3. Identify constraints,
and, as necessary, identify data that are needed to help solve the problem. After formulating the
problem, be sure you have a linear programming problem by verifying that:
1. The objective function is linear.
2. All constraints are linear.

Chapter 3

43

3. All variables are continuous (that is, they can assume all fractional values).
Now that you know various applications of linear programming problems and how to formulate
them, the next three chapters provide the details involved in solving such problems — that is, in finding
the values of the variables that provide the best value of the objective function while satisfying all
of the constraints simultaneously. In particular, Chapter 4 shows how to solve graphically linear
programming problems having only two variables. This method provides insight into how such
problems having any number of variables and/or constraints are solved algebraically by computer,
as described in Chapters 5 and 6.

EXERCISES FOR CHAPTER 3
Exercise 3.1 Gasahol, Inc., has 14,000 and 16,000 gallons of a gasoline-alcohol mixture stored at
respective facilities in Fresno and Bakersfield from which it must supply Fresh Food Farms (FFF)
with 10,000 gallons and American Growers (AG) with 20,000 gallons. The cost for shipping one
gallon from each storage facility to each customer is given below:
From/To
Fresno
Bakersfield

FFF
$0.04
$0.05

AG
$0.06
$0.03

Formulate a linear programming model to determine the least-cost shipping plan that meets the
supply and demand constraints.
Exercise 3.2 The HealthNut Company is developing a new peanut butter and chocolate candy
bar. The candy must have at least 5 grams of protein but not more than 5 grams of carbohydrates
and 3 grams of saturated fats. Develop a linear program to determine the amount of each ingredient
to use that meets these nutritional requirements at least total cost, based on the following data:
Cost ($/oz.)
Protein (gm./oz.)
Carbohydrates (gm./oz.)
Saturated fats (gm./oz.)

Peanut butter
0.10
4.00
2.50
2.00

Chocolate
0.18
0.80
1.00
0.50

Exercise 3.3 The HealthNut Company has a machine that grinds raw psyllium seeds at the rate
of 300 pounds per hour into a fine powder. The company also uses the machine to make peanut
butter from roasted peanuts at the rate of 600 pounds per hour. The set-up time to switch the
machine from one product to the other is negligible. The monthly demand and inventory holding
costs for each product are provided in the following table:

May
June
July

Demand (lbs.)
Peanut butter Psyllium
400
600
450
700
500
650

Holding costs ($/lb.)
Peanut butter Psyllium
0.10
0.05
0.10
0.05
0.12
0.05

The initial inventory of each product at the beginning of May is 0 and should also be 0 at the end of
July. At no time can the inventory of psyllium exceed 1000 pounds nor that of peanut butter exceed
500 pounds. Formulate a linear program to determine a production plan for the months of May,
June, and July that minimizes the total holding costs, assuming that demand is met at the end of

44

Chapter 3

each month and inventory holding costs are based on the amount in inventory at the beginning of
the month.
Exercise 3.4 The FMR Company has a machine capable of making large and small pipes for
plumbing contractors. Large pipes are produced at the rate of 200 feet per hour and small pipes
at the rate of 300 feet per hour. However, each hour the machine is used for producing large pipes
typically results in 1.5 jams and 3 jams when producing the small pipes. Each jam requires about
five minutes to reset, during which time the machine cannot produce pipes. Formulate a model
to determine how much of an eight-hour day should be allocated to producing large versus small
pipes, given that the anticipated lengths of both size pipes produced should be equal and as large
as possible. Use the number of hours of machine time to devote to making small and large pipes as
the decision variables.
Exercise 3.5 Repeat Exercise 3.4 using the fraction of eight hours of machine time to devote to
making small and large pipes as the decision variables.
Exercise 3.6 Repeat Exercise 3.4 using the number of feet of small and large pipes to make in
eight hours of machine time as the decision variables.
Exercise 3.7 Sulfur, charcoal, and saltpeter are mixed to produce gunpowder at Explosives, Inc.
The final product must contain at least 10%, but not more than 20%, charcoal by weight, while
the amount of saltpeter cannot exceed 50% of the amount of charcoal used. To avoid inadvertant
explosion, the sum of 50% of the sulfur plus 60% of the charcoal plus 30% of the saltpeter used cannot
exceed 35% of the final product. Since sulfur is by far the most expensive component, formulate a
model to determine how much of each ingredient should be used in making each pound of gunpowder
that satisfies the constraints while requiring the least amount of sulfur.
Exercise 3.8 Case Chemicals dillutes each liter of concentrated sulfuric acid with 20 liters of
distilled water to produce H2-SO4. Similarly, each liter of concentrated hydrochloric acid is dilluted
with 30 liters of distilled water to produce H-CL. These two products, H2-SO4 and H-CL, are each
sold to high schools at $0.10 per 100-milliliter (that is, 0.1 liter) bottle. The company currently has
50,000 empty bottles in inventory. Assume that there is virtually an unlimited amount of distilled
water costing $0.15 per liter and that the following data are available:
Cost ($/liter)
Supply (liters)

Sulfuric acid
12.00
200

Hydrochloric acid
18.00
150

Formulate a model to determine how much of each concentrated acid to dillute to maximize the
total profits. Can you solve this model as a linear program? Why or why not? Discuss.
Exercise 3.9 The ManuMania Company uses a base and two gum products, all in equal amounts,
to make its Gooey Gum. The company can produce a combined total of up to 800 pounds of the
base and two gum products. Alternatively, it can purchase these ingredients on the open market at
the following prices per pound:
Product
Base
GP-1
GP-2

Production cost
$1.75
$2.00
$2.25

Purchase cost
$3.00
$3.25
$3.75

Formulate a model to determine the least-cost production/purchasing plan to meet a demand of

Chapter 3

45

1200 pounds of Gooey Gum.
Exercise 3.10 Each week, Florida Citrus, Inc., uses a single machine for 150 hours to distill orange
and grapefruit juice into a concentrates that are then stored in two separate 1000-gallon tanks before
being frozen. The machine can process 25 gallons of orange juice per hour but only 20 gallons of
grapefruit juice. Each gallon of orange juice cost $1.50 and loses 30% in water content when distilled
into a concentrate that then sells for $6.00 per gallon. Each gallon of grapefruit juice costs $2.00
and loses 25% when distilled into a concentrate that then sells for $8.00 per gallon. Formulate a
linear programming model to determine a production plan to maximize the profit for the coming
week using the variables:
OJ
GJ

=
=

the number of gallons of orange juice to use this week,
the number of gallons of grapefruit juice to use this week.

Exercise 3.11 For the problem in Exercise 3.10, formulate a linear programming model to determine a production plan to maximize the profit for the coming week using the variables:
OC
GC

=
=

number of gallons of orange-juice concentrate to produce this week,
number of gallons of grapefruit-juice concentrate to produce this week.

Exercise 3.12 For the problem in Exercise 3.10, formulate a linear programming model to determine a production plan to maximize the profit for the coming week using the variables:
OT

=

GT

=

the number of hours of machine time to use this week for
distilling orange juice,
the number of hours of machine time to use this week for
distilling grapefruit juice.

Exercise 3.13 Oklahoma Oil, Inc., must transport 100,000 barrels from each of its three oil fields
to its storage tank in Oklahoma City. Oil can be trucked directly from the fields to the storage tank
at a cost of $0.03 per barrel per mile. Up to 150,000 barrels of oil can also be sent from the fields
via a pipeline to a central hub in Tulsa at a cost of $0.02 per barrel per mile, and then trucked
to Oklahoma City for $1 per barrel. Formulate a model to determine the least-cost shipping plan,
given the following distances in miles:
From/To
Oil Field 1
Oil Field 2
Oil Field 3

Oklahoma City
150
170
190

Tulsa
50
65
80

Exercise 3.14 Cajun World mixes six spices to make a product for blackening fish. The following
table provides the cost of each spice together with the minimum and maximum percentages by
weight that can be used in the final product:

46

Chapter 3
Spice
Cayenne
Black pepper
Fennel seeds
Onion powder
Garlic
Oregano

Cost ($/gm.)
0.020
0.025
0.082
0.025
0.028
0.075

Minimum (%)
18
15
12
16
12
14

Maximum (%)
20
18
14
20
15
18

Formulate a linear program to determine the amount of each spice to use in making each kilogram
of product that minimizes the total cost.
Exercise 3.15 The Incredible Indelible Ink Company mixes three additives, A1 , A2 , and A3 , to a
base in different proportions to obtain different colors of ink. Red ink is obtained by mixing A1 , A2 ,
and A3 in the ratio of 3:1:2; blue ink in the ratio of 2:3:4, and green ink in the ratio of 1:2:3. After
mixing these additives, an equal amount of base is added for each color. The company currently has
1000 gallons of A1 , 1500 of A2 , 2000 of A3 , and 4000 gallons of base. Given that the selling price
per gallon for each type of ink is the same, develop a model to determine how these resources should
be used to obtain the maximum revenue.
Exercise 3.16 The Department of Energy of Lilliput is currently in the process of developing a
national energy plan for the next year. Lilliput can generate energy from any one of five sources:
coal, natural gas, nuclear materials, hydroelectric projects, and petroleum. The data on the energy
resources, generation capacities measured in megawatt hours (MW-hr), and unit costs of generation
are given in Table 1.
Table 3.1 1 Generation Capacities and Costs

Energy source
Coal
Natural gas
Nuclear
Hydroelectric
Petroleum

Total capacity
(MW-hr)
45,000
15,000
45,000
24,000
48,000

Cost of generating
($/MW-hr)
6.0
5.5
4.5
5.0
7.0

Lilliput needs 50,000 MW-hr of energy for domestic use but the country has a commitment to
produce 10,000 MW-hr for export. Furthermore, to conserve the energy resources and to protect
the environment, the government has passed the following regulations:
(a) The generation from nuclear materials should not exceed 20% of the total energy generated
by Lilliput.
(b) At least 80% of the capacity of the coal plants should be utilized.
(c) The effluents let off into the atmosphere should not exceed the limits specified in Table 2.
(d) The amount of energy generated from natural gas should be at least 30% of that generated
from petroleum.
Table 3.2 2 Pollution Data for Generating Energy

Chapter 3

47

Energy
source
Coal
Natural gas
Nuclear
Hydroelectric
Petroleum
Max. kg. allowed

Sulfur
dioxide
1.5
0.2


0.4
75

Pollutant (gm/MW-hr)
Carbon
Dust
Solid
monoxide particles waste
1.2
0.7
0.4
0.5


0.1
0.2
0.7



0.8
0.5
0.1
60
30
25

Formulate a linear program to determine a least-cost energy plan.
Exercise 3.17 Fresh Food Farms, Inc., has 50 acres of land on which to plant any amount of
corn, soy beans, lettuce, cotton, and broccoli. The following table shows the relevant information
pertaining to the yield, cost for planting, expected selling price, and water requirements of each
crop:
Crop
Corn
Soy beans
Lettuce
Cotton
Broccoli

Yield in
kgs. per acre
640
500
400
300
350

Cost
per kg.
$1.00
$0.50
$0.40
$0.25
$0.60

Selling price
per kg.
$1.70
$1.30
$1.00
$1.00
$1.30

Required liters
of water per kg.
8.75
5.00
2.25
4.25
3.50

For the coming season, there are 100,000 liters of water available and the company has contracted
to sell at least 5120 kilograms of corn. Formulate a linear program to determine an optimal planting
strategy for Fresh Food Farms, Inc. Use the number of acres of each crop to plant as the decision
variables.
Exercise 3.18 Repeat Exercise 3.17 using the number of kilograms of each crop to produce as the
decision variables.
Exercise 3.19 Formulate the Investment Problem of Pension Planners, Inc., described in Section
3.4, using the following collection of decision variables:
F1
F2
F3
F4
F5
F6

=
=
=
=
=
=

the
the
the
the
the
the

number
number
number
number
number
number

of
of
of
of
of
of

shares
shares
shares
shares
shares
shares

of
of
of
of
of
of

Fund
Fund
Fund
Fund
Fund
Fund

1
2
3
4
5
6

to
to
to
to
to
to

buy,
buy,
buy,
buy,
buy,
buy.

Exercise 3.20 Formulate the Investment Problem of Pension Planners, Inc., described in Section
3.4, using the following collection of decision variables:
F1
F2
F3
F4
F5
F6

=
=
=
=
=
=

the
the
the
the
the
the

number
number
number
number
number
number

of
of
of
of
of
of

dollars
dollars
dollars
dollars
dollars
dollars

to
to
to
to
to
to

invest
invest
invest
invest
invest
invest

in
in
in
in
in
in

Fund
Fund
Fund
Fund
Fund
Fund

1,
2,
3,
4,
5,
6.

48

Chapter 3

Exercise 3.21 Formulate the Investment Problem of Pension Planners, Inc., described in Section
3.4, using the following collection of decision variables:
H
M
L

=
=
=

the fraction of the portfolio to invest in high-risk funds,
the fraction of the portfolio to invest in medium-risk funds,
the fraction of the portfolio to invest in low-risk funds.

CASE-STUDY FORMULATIONS
Formulate a single mathematical model for each of the following problems. These models can be
linear, integer, or even nonlinear programming problems. Whenever possible, draw a schematic
diagram or network to represent the problem. Check with your instructor for additional information
and guidelines.

CASE STUDY #1
The Blending Problem of the Hexxon Oil Company
Mr. Sam Barton, production manager at the Hexxon Oil Company, has been asked by Mr.
James Arden, the vice president of production, to formulate a new daily production plan for the
three brands of gasoline the company sells: Regular (90 octane), Unleaded (96 octane), and Supreme
(100 octane). In their meeting, Mr. Arden brought with him the data in Table 1 consisting of the
projected daily demands for these three gasolines and their respective selling prices, prepared by
Ms. Jean Farrow in the Accounting Department. When Mr. Arden expressed the desire to achieve
the highest possible daily profit, Mr. Barton said that he would have to meet with Mr. Allen, the
production supervisor, to discuss the availability and costs of the constituents used in making the
three brands of gasoline and would then get back to the vice president with a production plan.
Table 3.3 1 Selling Prices and Demands for Gasoline
Brand of
gasoline
Regular
Unleaded
Supreme

Minimum octane
rating
90
96
100

Selling price
($ per barrel)
16.50
18.00
22.50

Demand per
day (barrel)
2000
4000
3000

When Mr. Barton discussed the problem on the phone with the production supervisor, Mr. Allen
said he would obtain the necessary information pertaining to the four constituents used in making
the three brands of gasoline. At their meeting the next day, he brought the data in Table 2 that
includes, for each constituent: (a) the octane rating, (b) the cost in dollars per barrel, and (c) the
maximum supply available per day.
Table 3.4 2 Data on the Constituents for Blending Gasoline
Blending
constituent
1
2
3
4

Octane rating
per barrel
102
96
93
110

Cost ($ per
barrel)
15.00
12.00
9.00
24.00

Supply per
day (barrels)
2500
3000
3500
2000

Mr. Allen reminded Mr. Barton that each of the three brands of gasoline must meet the minimum
standard for octane rating (see Table 1). When asked precisely how this is accomplished, Mr. Allen
explained that the octane rating of a mixture consisting of x1 , x2 , x3 , and x4 barrels of the four
constituents is the ratio of 102x1 +96x2 + 93x3 + 110x4 to the total number of barrels of the mixture,
namely, x1 + x2 + x3 + x4 .

Chapter 3

49

1. Formulate a production model for Mr. Barton to maximize the daily profits that satisfies
all constraints.
2. Ms. Jean Farrow in the Accounting Department has discovered that the costs of the constituents given in Table 2 were not exactly correct. Formulate the new objective function
on the basis of the following memo she sent to Mr. Barton:
********************************************************
TO:
Mr. S. Barton, Production Manager
FROM:

Ms. J. Farrow, Head of Accounting

RE:

Cost of Blending Constituent 1 to Make Supreme
Gasoline

Please note that the cost of blending x barrels of
Constituent 1 when making Supreme gas should be
13.50 + 0.001x per barrel, instead of $15.00 per barrel.
*******************************************************

CASE STUDY #2
The Production Problem of the ASA Steel Company
ASA is a large steel company that produces each of five different types of iron plates at its eight
factories. At a recent strategic-planning meeting, the management allocated the budgets in Table 1
to each of the eight factories for the next fiscal year. These budgets were based, in part, on the fixed
demands in Table 2 for the five iron plates, provided by the Sales Department. As vice president of
production, you, Mr. Leroy Adams, have been asked to determine a production plan for each of the
eight factories.
Table 3.5 1 The budget assigned to each factory (in thousands of dollars)
Factory
Budget

1
900

2
1050

3
950

4
1050

5
1000

6
1600

7
950

8
1050

Table 3.6 2 The demand in tons for each type of iron plate
Plate type
Demand

1
450

2
800

3
500

4
650

5
180

In preparation to formulate such a plan, you asked your associate, Mr. James Aaron, to obtain
production costs for each of the five iron plates at each of the eight factories. At your next meeting,
Mr. Aaron brought the cost data in Table 3. He also pointed out that, as external suppliers, ASA
has three subcontractors from which they can purchase precisely the amounts of different types of
iron plates at the prices shown in Table 4. Mr. Aaron reminded you that contractual obligations
require that if a particular subcontractor is chosen, ASA must purchase all of its supplies of all five
types of iron plates at the indicated prices. You can, of course, choose to purchase from either none,
one, two, or all three subcontractors. When you relayed this information to Mr. Charles Bentley,
the CEO, he approved the use of subcontractors if it would lower the total costs and informed you
that the cost for buying the iron plates from the subcontractors would not affect the budgets of any
of the factories (see Table 1).
Table 3.7 3 The cost of proucing one ton of each iron plate type i at factory j (in thousands of
dollars)

50

Chapter 3

Factory #
Plate type
1
2
3
4
5

1

2

3

4

5

6

7

8

5
3
7
6
8

3
4
6
6
9

4
6
5
6
8

3
2
8
5
7

3
5
4
3
10

4
3
3
5
9

6
6
4
5
8

4
4
5
4
6

Table 3.8 4 The cost per ton and the amount of each plate type available from each subcontractor
Plate type
1
2
3
4
5

Subcontractor #1
cost* amount**
5
40
5
20
5
30
3
40
5
20

Subcontractor #2
cost
amount
4
10
8
80
6
50
4
20
3
10

Subcontractor #3
cost
amount
5
20
6
40
6
10
5
10
4
50

*cost: The cost to purchase one ton of plate type i from the subcontractor, in thousands of dollars.
**amount: For any subcontractor used, all of its available amounts of all five types of iron plates must
be purchased at the indicated costs.
As production manager, formulate a model to determine a production plan for each of the eight
factories and which subcontractor(s), if any, to buy iron plates from to minimize the total costs
while staying within each factory’s budget and meeting the given demands.

CASE STUDY #3
The Delivery Problem of the Hexxon Oil Company
Mr. John Porter, production manager of the Hexxon Oil Company, just received a call from Mr.
Peter Finch, the vice president of production, regarding the need to get at least 1,500,000 barrels of
oil to the refinery in New Orleans (Refinery 1), 1,600,000 barrels to the refinery in Houston (Refinery
2), and 1,400,000 barrels to the refinery in Mobile (Refinery 3) for the next production quarter. You
told him that you would develop a plan of least cost to transport the oil by tanker from the port in
Mexico and via a pipeline from their facility in Alaska.
Making phone calls to these two facilities, you learned that there are up to 3,500,000 barrels of
oil at the port in Mexico and up to 1,200,000 barrels at the facility in Alaska. To transport the oil
from Mexico to the refineries, you can lease at most three tankers, A, B, and C, each with a capacity
of 1,600,000 barrels. In discussions with Ms. Yuki Ando in the Accounting Department, you learned
that the cost of transporting the oil by tanker consists of an insurance cost against spills and a fixed
cost for using a tanker. The insurance cost is $0.25/barrel/1000 miles traveled plus a surcharge that
depends on the age of the tanker. Ms. Ando provided the following data for each tanker:
Tanker
A
B
C

Base charge
($/barrel/1000 mi.)
0.25
0.25
0.25

Surcharge
($/barrel/1000 mi.)
0.0000
0.1250
0.0625

Fixed Cost
($)
200,000
100,000
150,000

She was also able to provide the approximate distances from the port in Mexico to each of the
refineries and the total cost for pumping a barrel of oil from Alaska to each of the three refineries.
These data follow:

Chapter 3

Refinery 1
Refinery 2
Refinery 3

51

Distance in miles
from the Mexican port
2500
2000
1800

Cost in $/barrel to ship
by pipeline from Alaska
1.40
1.25
1.20

As manager of the Production Department, formulate a single model to determine the least-cost
way to ship all the needed oil from both Alaska and the port in Mexico to the three refineries. Keep
in mind that you can determine which, if any, tankers to use and which refineries to send them to,
but a tanker can go to at most one refinery. (Include an appropriate schematic diagram.)

52

Chapter 3

CASE STUDY #4
The Delivery Problem of Gasahol, Inc.
Gasahol, Inc., produces a special fuel consisting of a mixture of gasoline and alcohol at each of
three plants located in Denver, Oklahoma City, and St. Louis. This fuel is supplied to customers in
Dallas, Kansas City, Phoenix, San Francisco, and Los Angeles. Gashol, Inc., can also subcontract to
GasMix, Inc., a California- based company, to produce and deliver a combined total of up to 18,000
gallons of the same product to the customers in Los Angeles and San Francisco. As vice president of
the company, one of your responsibilities is to determine a least-cost production and shipping plan.
Speaking to the production managers at each plant, you have obtained each of their production
capacity and cost per gallon, as follows:
Plant
Denver
Oklahoma
St. Louis

Capacity (gal.)
14,000
16,000
20,000

Cost per gallon
$0.55
$0.62
$0.48

Mr. Tubb Jones, manager of the Sales Department, has obtained for you next month’s orders for
this fuel from each of the customers. Knowing that to supply these customers, Gasahol, Inc., incurs
a delivery charge at the rate of $0.01/gallon/100 miles traveled, you have used a road map to find
the approximate distances in miles from the plants to the five customers. These distances and the
customer demands are summarized in the following table:

Denver
Oklahoma
St. Louis
Demand (gal.)

Dallas
780
210
660
11000

Distance in miles
Kansas Phoenix
SF
610
810
1260
350
980
1660
260
1480
2120
10000
9000 12000

LA
1030
1340
1860
18000

Checking your records, you find that your subcontractor, GasMix, Inc., charges a combined production and transportation cost based on the amount shipped (regardless of whether to San Francisco
or to Los Angeles) as follows:
Production and
transportation
price per gallon
$0.95
$0.90
$0.85

When the following
amounts (in gallons)
are shipped
up to 7,999
8,000 — 15,999
16,000 — 23,999

As vice president of the Production Department of Gasahol, Inc., formulate a single mathematical
model to determine the optimal production and shipping plan that precisely meets the demands.
Include an appropriate schematic diagram. (Hint: Create an integer variable, Y , whose value is 0 if
0 - 7,999 gallons are shipped by GasMix to San Francisco, 1 if 8000 - 15,999 gallons are shipped by
GasMix to San Francisco, and 2 if 16,000 - 23,999 gallons are shipped by GasMix to San Francisco.
In terms of this variable, the cost for shipping each gallon is 0.95 − 0.05Y . Create a similar variable,
Z, for shipping from GasMix to Los Angeles. Be sure to include constraints to insure that the
amounts shipped by GasMix and the unit shipping costs match those in the table above.)

CASE STUDY #5
The Data-Transmission Problem of Tele Com

Chapter 3

53

Ms. Amy Jenkins, director of communications for Tele Com, has just come from a meeting at which
it was decided that, due to a major new group of clients in Los Angeles and Boston, it is necessary
to increase the existing capacity for transmitting data between the offices in these two cities. The
current communication network has intermediate offices with computers and retransmission capabilities in Salt Lake City, Phoenix, Denver, Albuquerque, Minneapolis, Houston, Chicago, Atlanta,
Cleveland, Washington, D.C., and New York.
As a first step, Ms. Jenkins needs to review the current system. She has obtained from her files
the following list of communication links between certain pairs of these cities and the maximum
number of bits per day that can be sent through that link:
From
Los Angeles
Los Angeles
Salt Lake
Salt Lake
Phoenix
Denver
Albuquerque
Minneapolis
Houston
Chicago
Atlanta
Atlanta
Cleveland
Cleveland
Washington
New York

To
Salt Lake
Phoenix
Denver
Albuquerque
Albuquerque
Minneapolis
Houston
Chicago
Atlanta
Cleveland
Cleveland
Washington
Washington
Boston
New York
Boston

Max. bits per day (billions)
15
12
10
10
12
8
9
15
12
15
12
14
8
12
15
18

Formulate a single mathematical model to determine the maximum number of bits per day that can
be transmitted from the Los Angeles to the Boston office through the existing network. (Include a
schematic diagram.)

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