Chapter Summary of an Introduction to Management Science: Quantitative Approaches to Decision-Making

Solutions to Case Problems Manual to Accompany

An Introduction To

Management Science

Quantitative Approaches

To Decision Making

Twelfth Edition

David R. Anderson
University of Cincinnati

Dennis J. Sweeney
University of Cincinnati

Thomas A. Williams
Rochester Institute of Technology

R. Kipp Martin

University of Chicago

South-Western

Cincinnati, Ohio

Contents

Preface

Chapter 1: Introduction
♦ Scheduling a Golf League

Chapter 2: An Introduction to Linear Programming

♦ Workload Balancing ♦ Production Strategy ♦ Hart Venture Capital

Chapter 3: Linear Programming: Sensitivity Analysis

and Interpretation of Solution
♦ Product Mix

♦ Investment Strategy ♦ Truck Leasing Strategy

Chapter 4: Linear Programming Applications in Marketing, Finance and Operations Management

♦ Planning an Advertising Campaign ♦ Phoenix Computer

♦ Textile Mill Scheduling ♦ Workforce Scheduling

♦ Duke Energy Coal Allocation

Chapter 6: Distribution and Network Models
♦ Solution Plus
♦ Distribution Systems Design

Chapter 7: Integer Linear Programming

♦ Textbook Publishing ♦ Yeager National Bank
♦ Production Scheduling with Changeover Costs

Chapter 8: Nonlinear Optimization Models
♦ Portfolio Optimization with Transaction Costs

Chapter 9: Project Scheduling: PERT/CPM
♦ R.C. Coleman

Chapter 10: Inventory Models

♦ Wagner Fabricating Company ♦ River City Fire Department

Chapter 11: Waiting Line Models
♦ Regional Airlines
♦ Office Equipment, Inc.

Chapter 12: Simulation
♦ Tri-State Corporation

♦ Harbor Dunes Golf Course ♦ County Beverage Drive-Thru

Chapter 13: Decision Analysis

♦ Property Purchase Strategy ♦ Lawsuit Defense Strategy

Chapter 14: Multicriteria Decision Problems
♦ EZ Trailers, Inc.

Chapter 15: Forecasting
♦ Forecasting Sales
♦ Forecasting Lost Sales

Chapter 16: Markov Processes
♦ Dealer’s Absorbing State Probabilities in

Black Jack

Chapter 21: Dynamic Programming
♦ Process Design

Preface

The purpose of An Introduction to Management Science is to provide students with a sound conceptual understanding of the role management science pays in the decision-making process. The text emphasizes the application of management science by using problem situations to introduce each of the management science concepts and techniques. The book has been specifically designed to meet the needs of nonmathematicians who are studying business and economics.

The Solutions to Case Problems Manual contains solutions to the case problems.

Note: The solutions to the end-of-chapter problems and learning objectives for each chapter are included in the Solutions Manual.

Acknowledgements

We would like to provide a special acknowledgement to Catherine J. Williams for her efforts in preparing the Instructor's Manual. We are also indebted to our acquisitions editor Charles E. McCormick, Jr. and our developmental editor Alice C. Denny for their support during the preparation of this manual.

David R. Anderson

Dennis J. Sweeney

Thomas A. Williams

R. Kipp Martin

Chapter 1
Introduction

Case Problem: Scheduling a Golf League

Note to Instructor: This case problem illustrates the value of the rational management science approach. The problem is easy to understand and, at first glance, appears simple. But, most students will have trouble finding a solution. The solution procedure suggested involves decomposing a larger problem into a series of smaller problems that are easier to solve. The case provides students with a good first look at the kinds of problems where management science is applied in practice. The problem is a real one that one of the authors was asked by the Head Professional at Royal Oak Country Club for help with.

Solution: Scheduling problems such as this occur frequently, and are often difficult to solve. The typical approach is to use trial and error. An alternative approach involves breaking the larger problem into a series of smaller problems. We show how this can be done here using what we call the Red, White, and Blue algorithm.

Suppose we break the 18 couples up into 3 divisions, referred to as the Red, White, and Blue divisions. The six couples in the Red division can then be identified as R1, R2, R3, R4, R5, R6; the six couples in the White division can be identified as W1, W2,…, W6; and the six couples in the Blue division can be identified as B1, B2,…, B6. We begin by developing a schedule for the first 5 weeks of the season so that each couple plays every other couple in its own division. This can be done fairly easily by trial and error. Shown below is the first 5-week schedule for the Red division.

Week 1
Week 2
Week 3
Week 4
Week 5
R1 vs. R2
R1 vs. R3
R1 vs. R4
R1 vs. R5
R1 vs. R6
R3 vs. R4
R2 vs. R5
R2 vs. R6
R2 vs. R4
R2 vs. R3
R5 vs. R6
R4 vs. R6
R3 vs. R5
R3 vs. R6
R4 vs. R5

Similar 5-week schedules can be developed for the White and Blue divisions by replacing the R in the above table with a W or a B.

To develop the schedule for the next 3 weeks, we create 3 new six-couple divisions by pairing 3 of the teams in each division with 3 of the teams in another division; for example, (R1, R2, R3, W1, W2, W3), (B1, B2, B3, R4, R5, R6), and (W4, W5, W6, B4, B5, B6). Within each of these new divisions, matches can be scheduled for 3 weeks without any couples playing a couple they have played before. For instance, a 3-week schedule for the first of these divisions is shown below:

Week 6
Week 7
Week 8
R1 vs. W1
R1 vs. W2
R1 vs. W3
R2 vs. W2
R2 vs. W3
R2 vs. W1
R3 vs. W3
R3 vs. W1
R3 vs. W2

A similar 3-week schedule can be easily set up for the other two new divisions. This will provide us with a schedule for the first 8 weeks of the season.

For the final 9 weeks, we continue to create new divisions by pairing 3 teams from the original Red, White and Blue divisions with 3 teams from the other divisions that they have not yet been paired with. Then a 3-week schedule is developed as above. Shown below is a set of divisions for the next 9 weeks.

CP - 1

Chapter 1

Weeks 9-11

(R1, R2, R3, W4, W5, W6)
(W1, W2, W3, B1, B2, B3)
(R4, R5, R6, B4, B5, B6)
Weeks 12-14

(R1, R2, R3, B1, B2, B3)
(W1, W2, W3, B4, B5, B6)
(W4, W5, W6, R4, R5, R6)
Weeks 15-17

(R1, R2, R3, B4, B5, B6)
(W1, W2, W3, R4, R5, R6)
(W4, W5, W6, B1, B2, B3)

This Red, White and Blue scheduling procedure provides a schedule with every couple playing every other couple over the 17-week season. If one of the couples should cancel, the schedule can be modified easily. Designate the couple that cancels, say R4, as the Bye couple. Then whichever couple is scheduled to play couple R4 will receive a Bye in that week. With only 17 couples a Bye must be scheduled for one team each week.

This same scheduling procedure can obviously be used for scheduling sports teams and or any other kinds of matches involving 17 or 18 teams. Modifications of the Red, White and Blue algorithm can be employed for 15 or 16 team leagues and other numbers of teams.

CP - 2

Chapter 2
An Introduction to Linear Programming

Case Problem 1: Workload Balancing

1.

Production Rate

(minutes per printer)

Model
Line 1
Line 2
Profit Contribution ($)
DI-910
3
4
42
DI-950
6
2
87

Capacity: 8 hours 60 minutes/hour = 480 minutes per day

Let D1 = number of units of the DI-910 produced
D2 = number of units of the DI-950 produced

Max 42D1
+
87D2

s.t.

3D1
+
6D2
≤
480
Line 1 Capacity
4D1
+
2D2
≤
480
Line 2 Capacity
D1, D2 ≥ 0

Using The Management Scientist, the optimal solution is D1 = 0, D2 = 80. The value of the optimal solution is $6960.

Management would not implement this solution because no units of the DI-910 would be produced.

2. Adding the constraint D1 ≥ D2 and resolving the linear program results in the optimal solution D1 = 53.333, D2 = 53.333. The value of the optimal solution is $6880.

3. Time spent on Line 1: 3(53.333) + 6(53.333) = 480 minutes Time spent on Line 2: 4(53.333) + 2(53.333) = 320 minutes

Thus, the solution does not balance the total time spent on Line 1 and the total time spent on Line 2. This might be a concern to management if no other work assignments were available for the employees on Line 2.

4. Let T1 = total time spent on Line 1
T2 = total time spent on Line 2

Whatever the value of T2 is,

T1 ≤ T2 + 30
T1 ≥ T2 - 30

Thus, with T1 = 3D1 + 6D2 and T2 = 4D1 + 2D2

3D1 + 6D2 ≤ 4D1 + 2D2 + 30 3D1 + 6D2 ≥ 4D1 + 2D2 − 30

CP - 3

Chapter 2

Hence,

−1D1 + 4D2 ≤ 30 −1D1 + 4D2 ≥ −30

Rewriting the second constraint by multiplying both sides by -1, we obtain

−1D1 + 4D2 ≤ 30 1D1 − 4D2 ≤ 30

Adding these two constraints to the linear program formulated in part (2) and resolving using The Management Scientist, we obtain the optimal solution D1 = 96.667, D2 = 31.667. The value of the optimal solution is $6815. Line 1 is scheduled for 480 minutes and Line 2 for 450 minutes. The effect of workload balancing is to reduce the total contribution to profit by $6880 - $6815 = $65 per shift.

5. The optimal solution is D1 = 106.667, D2 = 26.667. The total profit contribution is

42(106.667) + 87(26.667) = $6800

Comparing the solutions to part (4) and part (5), maximizing the number of printers produced (106.667 + 26.667 = 133.33) has increased the production by 133.33 - (96.667 + 31.667) = 5 printers but has reduced profit contribution by $6815 - $6800 = $15. But, this solution results in perfect workload balancing because the total time spent on each line is 480 minutes.

Case Problem 2: Production Strategy

1.
Let
BP100
= the number of BodyPlus 100 machines produced

BP200
= the number of BodyPlus 200 machines produced

Max
371BP100
+
461BP200

s.t.

8BP100
+
12BP200 ≤ 600
Machining and Welding

5BP100
+
10BP200
≤
450
Painting and Finishing

2BP100
+
2BP200
≤
140
Assembly, Test, and Packaging

-0.25BP100
+
0.75BP200
≥
0
BodyPlus 200 Requirement

BP100, BP200 ≥ 0

CP - 4

Solutions to Case Problems

BP200

80

200
70

Assembly, Test, and Packaging

of BodyPlus
60

50

Machining and Welding

40

Number
30

BodyPlus 200 Requirement

20

10

Optimal Solution

Painting and Finishing

BP100

0
10
20
30
40
50
60
70
80
90
100

Number of BodyPlus 100

Optimal solution: BP100 = 50, BP200 = 50/3, profit = $26,233.33. Note: If the optimal solution is rounded to BP100 = 50, BP200 = 16.67, the value of the optimal solution will differ from the value shown. The value we show for the optimal solution is the same as the value that will be obtained if the problem is solved using a linear programming software package such as The Management Scientist.

2. In the short run the requirement reduces profits. For instance, if the requirement were reduced to at least 24% of total production, the new optimal solution is BP100 = 1425/28, BP200 = 225/14, with a total profit of $26,290.18; thus, total profits would increase by $56.85. Note: If the optimal solution is rounded to BP100 = 50.89, BP200 = 16.07, the value of the optimal solution will differ from the value shown. The value we show for the optimal solution is the same as the value that will be obtained if the problem is solved using a linear programming software package such as The Management Scientist.

3. If management really believes that the BodyPlus 200 can help position BFI as one of the leader's in high-end exercise equipment, the constraint requiring that the number of units of the BodyPlus 200 produced be at least 25% of total production should not be changed. Since the optimal solution uses all of the available machining and welding time, management should try to obtain additional hours of this resource.

CP - 5

Chapter 2

Case Problem 3: Hart Venture Capital

1.
Let S = fraction of the Security Systems project funded by HVC

M = fraction of the Market Analysis project funded by HVC

Max
1,800,000S
+
1,600,000M

s.t.

600,000S
+
500,000M
≤
800,000
Year 1

600,000S
+
350,000M
≤
700,000
Year 2

250,000S
+
400,000M
≤
500,000
Year 3

S

≤
1
Maximum for S

M
≤
1
Maximum for M

S,M
≥
0

The solution obtained using The Management Scientist software package is shown below:

OPTIMAL SOLUTION

Objective Function Value =
2486956.522

Variable

Value
Reduced Costs
--------------
---------------
------------------
S

0.609
0.000
M

0.870
0.000
Constraint
Slack/Surplus
Dual Prices
--------------
---------------
------------------
1

0.000
2.783
2

30434.783
0.000
3

0.000
0.522
4

0.391
0.000
5

0.130
0.000

OBJECTIVE COEFFICIENT RANGES

Variable
Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
S
No Lower Limit
1800000.000
No Upper Limit
M
No Lower Limit
1600000.000
No Upper Limit
RIGHT HAND SIDE RANGES

Constraint
Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
1
No Lower Limit
800000.000
822950.820
2
669565.217
700000.000
No Upper Limit
3
461111.111
500000.000
No Upper Limit
4
0.609
1.000
No Upper Limit
5
0.870
1.000
No Upper Limit

CP - 6

Solutions to Case Problems

Thus, the optimal solution is S = 0.609 and M = 0.870. In other words, approximately 61% of the Security Systems project should be funded by HVC and 87% of the Market Analysis project should be funded by HVC.

The net present value of the investment is approximately $2,486,957.

2.

Year 1
Year 2
Year 3
Security Systems
$365,400
$365,400
$152,250
Market Analysis
$435,000
$304,500
$348,000
Total
$800,400
$669,900
$500,250

Note: The totals for Year 1 and Year 3 are greater than the amounts available. The reason for this is that rounded values for the decision variables were used to compute the amount required in each year. To see why this situation occurs here, first note that each of the problem coefficients is an integer value. Thus, by default, when The Management Scientist prints the optimal solution, values of the decision variables are rounded and printed with three decimal places. To increase the number of decimal places shown in the output, one or more of the problem coefficients can be entered with additional digits to the right of the decimal point. For instance, if we enter the coefficient of 1 for S in constraint 4 as 1.000000 and resolve the problem, the new optimal values for S and D will be rounded and printed with six decimal places. If we use the new values in the computation of the amount required in each year, the differences observed for year 1 and year 3 will be much smaller than we obtained using the values of S = 0.609 and M = 0.870.

3. If up to $900,000 is available in year 1 we obtain a new optimal solution with S = 0.689 and M = 0.820. In other words, approximately 69% of the Security Systems project should be funded by HVC and 82% of the Market Analysis project should be funded by HVC.

The net present value of the investment is approximately $2,550,820.

The solution obtained using The Management Scientist software package follows:

OPTIMAL SOLUTION

Objective Function Value =
2550819.672

Variable

Value
Reduced Costs
--------------
---------------
------------------
S

0.689
0.000
M

0.820
0.000
Constraint
Slack/Surplus
Dual Prices
--------------
---------------
------------------
1

77049.180
0.000
2

0.000
2.098
3

0.000
2.164
4

0.311
0.000
5

0.180
0.000

OBJECTIVE COEFFICIENT RANGES

Variable
Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
S
No
Lower
Limit
1800000.000
No
Upper
Limit
M
No
Lower
Limit
1600000.000
No
Upper
Limit

CP - 7

Chapter 2

RIGHT HAND SIDE RANGES

Constraint
Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
1
822950.820
900000.000
No Upper Limit
2
No Lower Limit
700000.000
802173.913
3
No Lower Limit
500000.000
630555.556
4
0.689
1.000
No Upper Limit
5
0.820
1.000
No Upper Limit

4. If an additional $100,000 is made available, the allocation plan would change as follows:

Year 1
Year 2
Year 3
Security Systems
$413,400
$413,400
$172,250
Market Analysis
$410,000
$287,000
$328,000
Total
$823,400
$700,400
$500,250

5. Having additional funds available in year 1 will increase the total net present value. The value of the objective function increases from $2,486,957 to $2,550,820, a difference of $63,863. But, since the allocation plan shows that $823,400 is required in year 1, only $23,400 of the additional $100,00 is required. We can also determine this by looking at the slack variable for constraint 1 in the new solution. This value, 77049.180, shows that at the optimal solution approximately $77,049 of the $900,000 available is not used. Thus, the amount of funds required in year 1 is $900,000 - $77,049 = $822,951. In other words, only $22,951 of the additional $100,000 is required. The differences between the two values, $23,400 and $22,951, is simply due to the fact that the value of $23,400 was computed using rounded values for the decision variables. The value of $22,951 is computed internally in The Management Scientist output and is not subject to this rounding. Thus, the most accurate value is $22,951.

CP - 8

Chapter 3
Linear Programming: Sensitivity Analysis and Interpretation of Solution

Case Problem 1: Product Mix

Note to Instructor: The difference between relevant and sunk costs is critical. The cost of the shipment of nuts is a sunk cost. Practice in applying sensitivity analysis to a business decision is obtained. You may want to suggest that sensitivity analyses other than the ones we have suggested be undertaken.

1. Cost per pound of ingredients

Almonds
$7500/6000 = $1.25
Brazil
$7125/7500 = $.95
Filberts
$6750/7500 = $.90
Pecans
$7200/6000 = $1.20
Walnuts $7875/7500 = $1.05

Cost of nuts in three mixes:

Regular mix: .15($1.25) + .25($.95) + .25($90) + .10($1.20) + .25($1.05) = $1.0325

Deluxe mix .20($1.25) + .20($.95) + .20($.90) + .20($1.20) + .20($1.05) = $1.07

Holiday mix: .25($1.25) + .15($.95) + .15($.90) + .25($1.20) + .20($1.05) = $1.10

2. Let R = pounds of Regular Mix produced D = pounds of Deluxe Mix produced

H = pounds of Holiday Mix produced

Note that the cost of the five shipments of nuts is a sunk (not a relevant) cost and should not affect the decision. However, this information may be useful to management in future pricing and purchasing decisions. A linear programming model for the optimal product mix is given.

The following linear programming model can be solved to maximize profit contribution for the nuts already purchased.

Max
1.65R
+
2.00D
+
2.25H

s.t.

0.15R
+
0.20D
+
0.25H
≤
6000
Almonds

0.25R
+
0.20D
+
0.15H
≤
7500
Brazil

0.25R
+
0.20D
+
0.15H
≤
7500
Filberts

0.10R
+
0.20D
+
0.25H
≤
6000
Pecans

0.25R
+
0.20D
+
0.20H
≤
7500
Walnuts

R

≥
10000
Regular

D

≥
3000
Deluxe

H
≥
5000
Holiday
R,
D, H ≥ 0

CP - 9

Chapter 3

The solution found using The Management Scientist is shown below.

Objective Function Value =
61375.000

Variable

Value

Reduced Costs
--------------
---------------
------------------
R

17500.000

0.000
D

10624.999

0.000
H

5000.000

0.000
Constraint

Slack/Surplus
Dual Prices
--------------
---------------
------------------
1

0.000

8.500
2

250.000

0.000
3

250.000

0.000
4

875.000

0.000
5

0.000

1.500
6

7500.000

0.000
7

7624.999

0.000
8

0.000

-0.175
OBJECTIVE COEFFICIENT RANGES

Variable

Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
R

1.500

1.650
2.000
D

1.892

2.000
2.200
H
No Lower Limit

2.250
2.425
RIGHT HAND SIDE RANGES

Constraint

Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
1

5390.000

6000.000
6583.333
2

7250.000

7500.000
No Upper Limit
3

7250.000

7500.000
No Upper Limit
4

5125.000

6000.000
No Upper Limit
5

6750.000

7500.000
7750.000
6

No Lower Limit

10000.000
17500.000
7

No Lower Limit

3000.000
10624.999
8

-0.000

5000.000
9692.307

3. From the dual prices it can be seen that additional almonds are worth $8.50 per pound to TJ. Additional walnuts are worth $1.50 per pound. From the slack variables, we see that additional Brazil nut, Filberts, and Pecans are of no value since they are already in excess supply.

4. Yes, purchase the almonds. The dual price shows that each pound is worth $8.50; the dual price is applicable for increases up to 583.33 pounds.

CP - 10

Solutions to Case Problems

Resolving the problem by changing the right-hand side of constraint 1 from 6000 to 7000 yields the following optimal solution. The optimal solution has increased in value by $4958.34. Note that only 583.33 pounds of the additional almonds were used, but that the increase in profit contribution more than justifies the $1000 cost of the shipment.

Objective Function Value =
66333.336

Variable

Value

Reduced Costs
--------------
---------------
------------------
R

11666.667

0.000
D

17916.668

0.000
H

5000.000

0.000
Constraint

Slack/Surplus
Dual Prices
--------------
---------------
------------------
1

416.667

0.000
2

250.000

0.000
3

250.000

0.000
4

0.000

5.667
5

0.000

4.333
6

1666.667

0.000
7

14916.667

0.000
8

0.000

-0.033
OBJECTIVE COEFFICIENT RANGES

Variable

Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
R

1.000

1.650
1.750
D

1.976

2.000
3.300
H
No Lower Limit

2.250
2.283
RIGHT HAND SIDE RANGES

Constraint

Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
1

6583.333

7000.000
No Upper Limit
2

7250.000

7500.000
No Upper Limit
3

7250.000

7500.000
No Upper Limit
4

4210.000

6000.000
6250.000
5

7250.000

7500.000
7750.000
6

No Lower Limit

10000.000
11666.667
7

No Lower Limit

3000.000
17916.668
8

0.002

5000.000
15529.412

5. From the dual prices it is clear that there is no advantage to not satisfying the orders for the Regular and Deluxe mixes. However, it would be advantageous to negotiate a decrease in the Holiday mix requirement.

CP - 11

Chapter 3

Case Problem 2: Investment Strategy

1. The first step is to develop a linear programming model for maximizing return subject to constraints for funds available, diversity, and risk tolerance.

Let G = Amount invested in growth fund I = Amount invested in income fund

M = Amount invested in money market fund

The LP formulation and optimal solution found using The Management Scientist are shown.

MAX .18G +.125I +.075M

S.T.

1)
G +
I + M
< 800000
Funds Available
2)
.8G
-.2I
-.2M
> 0
Min growth fund
3)
.6G
-.4I
-.4M
< 0
Max growth fund
4)
-.2G
+.8I
-.2M
> 0
Min income fund
5)
-.5G
+.5I
-.5M
< 0
Max income fund
6)
-.3G
-.3I
+.7M
> 0
Min money market fund

7) .05G + .02I -.04M < 0
Max risk

OPTIMAL SOLUTION

Objective Function Value =
94133.336

Variable

Value

Reduced Costs
--------------
---------------
------------------
G

248888.906

0.000
I

160000.000

0.000
M

391111.094

0.000
Constraint

Slack/Surplus
Dual Prices
--------------
---------------
------------------
1

0.000

0.118
2

88888.898

0.000
3

71111.109

0.000
4

0.000

-0.020
5

240000.000

0.000
6

151111.109

0.000
7

0.000

1.167
OBJECTIVE COEFFICIENT RANGES

Variable

Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
G

0.150

0.180
No Upper Limit
I

-0.463

0.125
0.145
M

0.015

0.075
0.180

CP - 12

Solutions to Case Problems
RIGHT HAND SIDE RANGES

Constraint
Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
1
0.188
800000.000
No Upper Limit
2
No Lower Limit
0.000
88888.898
3
-71111.109
0.000
No Upper Limit
4
-106666.641
0.000
133333.313
5
-240000.000
0.000
No Upper Limit
6
No Lower Limit
0.000
151111.109
7
-8000.000
0.000
6399.999

Rounding to the nearest dollar, the portfolio recommendation for Langford is as follows.

Amount
Fund:

Invested
Growth
$248,889
Income
160,000
Money Market
391,111

Total
$800,000

Yield = 94,133 / 800,000 = .118

The portfolio yield is .118 or 11.8%.

Note that the portfolio yield equals the dual price for the funds available constraint.

2. If Langford’s risk index is increased by .005 that is the same as increasing the right-hand side of constraint 7 by .005 (800,000) = 4000. Since this amount of increase is within the right-hand-side range, we would expect an increase in return of 1.167 (4000) = 4668. The revised formulation and new optimal solution are shown below. Except for rounding, the value has increased as predicted; the new optimal allocation is

Amount
Fund:

Invested
Growth
$293,333
Income
160,000
Money Market
346,667

Total
$800,000

The portfolio yield becomes 98,800/800,000 = .124 or 12.4%

MAX .18G +.125I +.075M

S.T.

1) G + I + M < 800000

2) .8G -.2I -.2M > 0
3) .6G -.4I -.4M < 0
4) -.2G +.8I -.2M > 0

5) -.5G +.5I -.5M < 0
6) -.3G -.3I +.7M > 0

7) .045G + .015I-.045M < 0

CP - 13

Chapter 3

OPTIMAL SOLUTION

Objective Function Value =
98800.000
Variable

Value
Reduced Costs
--------------
---------------
------------------
G

293333.313
0.000
I

160000.000
0.000
M

346666.656
0.000
Constraint
Slack/Surplus
Dual Prices
--------------
---------------
------------------
1

0.000
0.124
2

133333.328
0.000
3

26666.666
0.000
4

0.000
-0.020
5

240000.000
0.000
6

106666.664
0.000
7

0.000
1.167

3. Since .16 is in the objective coefficient range for the growth fund return, there would be no change in allocation. However, the return would decrease by (.02) ($248,889) = $4978.

A decrease to .14 is outside the objective function coefficient range forcing us to resolve the problem. The new formulation and optimal solution is as follows.

MAX .14G +.125I +.075M

S.T.

1) G + I + M < 800000
2) .8G -.2I -.2M > 0
3) .6G -.4I -.4M < 0

4) -.2G +.8I -.2M > 0
5) -.5G +.5I -.5M < 0

6) -.3G -.3I +.7M > 0
7) .05G + .02I-.04M < 0

OPTIMAL SOLUTION

Objective Function Value =
85066.664
Variable

Value
Reduced Costs
--------------
---------------
------------------
G

160000.016
0.000
I

293333.313
0.000
M

346666.688
0.000

CP - 14

Solutions to Case Problems
Constraint
Slack/Surplus
Dual Prices
--------------
---------------
------------------
1
0.000
0.106
2
0.000
-0.010
3
160000.000
0.000
4
133333.313
0.000
5
106666.688
0.000
6
106666.688
0.000
7
0.000
0.833

4. Since the current optimal solution has more invested in the growth fund than the income fund, adding this requirement will force us to resolve the problem with a new constraint. We should expect a decrease in return as is shown in the following optimal solution.

MAX .18G +.125I +.075M

S.T.

1) G + I + M < 800000
2) .8G -.2I -.2M > 0
3) .6G -.4I -.4M < 0

4) -.2G +.8I -.2M > 0
5) -.5G +.5I -.5M < 0

6) -.3G -.3I +.7M > 0
7) .05G + .02I-.04M < 0
8) G - I < 0

OPTIMAL SOLUTION

Objective Function Value =
93066.656
Variable

Value
Reduced Costs
--------------
---------------
------------------
G

213333.313
0.000
I

213333.313
0.000
M

373333.313
0.000
Constraint
Slack/Surplus
Dual Prices
--------------
---------------
------------------
1

0.000
0.116
2

53333.324
0.000
3

106666.664
0.000
4

53333.324
0.000
5

186666.656
0.000
6

133333.328
0.000
7

0.000
1.033
8

0.000
0.012

Note that the value of the solution has decreased from $94,133 to $93,067. This is only a decrease of 0.2% inyield. Since the yield decrease is so small, Williams may prefer this portfolio for Langford.

5. It is possible a model such as this could be developed for each client. The changed yield estimates would require a change in the objective function coefficients and resolving the problem if the change was outside the objective coefficient range.

CP - 15

Chapter 3

Case Problem 3: Truck Leasing Strategy

1. Let xij = number of trucks obtained from a short term lease signed in month i for a period of j months

yi = number of trucks obtained from the long-term lease that are used in month i

Monthly fuel costs are 20 ($100) = $2000.

Monthly Costs for Short-Term Leased Trucks

Note: the costs shown here include monthly fuel costs of $2000.

Decision Variables
Cost

x11, x21, x31, x41
$4000 + $2000 = $6000 x12, x22, x32
2 ($3700) + $2000
= $9400 x13, x23
3 ($3225) + $2000
=
$11,675 x14 4 ($3040) + $2000
=
$14,160

Monthly Costs for Long-Term Leased Trucks

Since Reep Construction is committed to the long-term lease and since employees cannot be laid off, the only relevant cost for the long-term leased trucks is the monthly fuel cost of $2000.

MIN 6000X11 + 9400X12 + 11675X13 + 14160X14 + 6000X21 + 9400X22 + 11675X23 + 6000X31 + 9400X32 + 6000X41 + 2000Y1 + 2000Y2 + 2000Y3 + 2000Y4

S.T.

1) X11 + X12 + X13 + X14 + Y1 = 10

2) X21 + X22 + X23 + X14 + X13 + X12 + Y2 = 12

3) X31 + X32 + X23 + X22 + X14 + X13 + Y3 = 14

4) X41 + X32 + X23 + X14 + Y4 = 8

5) Y1 < 1

6) Y2 < 2

7) Y3 < 3

8) Y4 < 1

CP - 16

Solutions to Case Problems
Objective Function Value =
151660.000
Variable

Value
Reduced Costs
--------------
---------------
------------------
X11

0.000
3515.000
X12

0.000
3725.000
X13

3.000
0.000
X14

6.000
0.000
X21

0.000
2810.000
X22

0.000
210.000
X23

1.000
0.000
X31

1.000
0.000
X32

0.000
915.000
X41

0.000
3515.000
Y1

1.000
0.000
Y2

2.000
0.000
Y3

3.000
0.000
Y4

1.000
0.000
Constraint
Slack/Surplus
Dual Prices
--------------
---------------
------------------
1

0.000
-2485.000
2

0.000
-3190.000
3

0.000
-6000.000
4

0.000
-2485.000
5

0.000
485.000
6

0.000
1190.000
7

0.000
4000.000
8

0.000
485.000

OBJECTIVE COEFFICIENT RANGES

Variable
Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
X11
2485.000
6000.000
No Upper Limit
X12
5675.000
9400.000
No Upper Limit
X13
10760.000
11675.000
11885.000
X14
13950.000
14160.000
15075.000
X21
3190.000
6000.000
No Upper Limit
X22
9190.000
9400.000
No Upper Limit
X23
10485.000
11675.000
11885.000
X31
3190.000
6000.000
6915.000
X32
8485.000
9400.000
No Upper Limit
X41
2485.000
6000.000
No Upper Limit
Y1
No Lower Limit
2000.000
2485.000
Y2
No Lower Limit
2000.000
3190.000
Y3
No Lower Limit
2000.000
6000.000
Y4
No Lower Limit
2000.000
2485.000

CP - 17

Chapter 3

RIGHT HAND SIDE RANGES

Constraint
Lower Limit
Current Value
Upper Limit
------------
---------------
---------------
---------------
1
4.000
10.000
11.000
2
11.000
12.000
13.000
3
13.000
14.000
No Upper Limit
4
2.000
8.000
11.000
5
0.000
1.000
7.000
6
1.000
2.000
3.000
7
0.000
3.000
4.000
8
0.000
1.000
7.000

2. The total cost associated with the leasing plan is $151,660.

3. If Reep Construction is willing to consider the possibility of layoffs, we need to include driver costs of $3200 per month. Replacing the coefficients for y1, y2, y3, and y4 in our previous linear program with $5200 and resolving resulted in the following leasing plan:

Month
Length of Lease (Months)

Leased
1
2
3
4

1
0
0
4
6

2
0
0
2
_

3
0
0
_
_

4
0
_
_
_

In addition, in month 3, two of the trucks from the long-term leases were used. The total cost of this leasing plan is $165,410.

To see what effect a no layoff policy has, we can set y1 = 1, y2 = 2, y3 = 3, y4 = 1 and resolve the linear program using objective coefficients of $5200 for y1, y2, y3, and y4. The new optimal solution forces us to use all the available trucks from the long-term lease; the optimal leasing plan is shown below.

Month
Length of Lease (Months)

Leased
1
2
3
4

1
0
0
3
6

2
0
0
1
_

3
1
0
_
_

4
0
_
_
_

The total cost associated with this solution is $174,060. Thus, if Reep maintains their current policy of no layoffs they will incur an additional cost of $174,060 - $165,410 = $8,650.

CP - 18

Chapter 4
Linear Programming Applications in Marketing, Finance and
Operations Management

Case Problem 1: Planning an Advertising Campaign

The decision variables are as follows:

T1 = number of television advertisements with rating of 90 and 4000 new customers T2 = number of television advertisements with rating of 40 and 1500 new customers R1 = number of radio advertisements with rating of 25 and 2000 new customers R2 = number of radio advertisements with rating of 15 and 1200 new customers

N1 = number of newspaper advertisements with rating of 10 and 1000 new customers N2 = number of newspaper advertisements with rating of 5 and 800 new customers

The Linear Programming Model and solution using The Management Scientist follow:

MAX 90T1+55T2+25R1+20R2+10N1+5N2

S.T.

1) 1T1