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 9: Project Scheduling: PERT/CPM
♦ R.C. Coleman
Chapter 2: An Introduction to Linear Programming
♦ Workload Balancing
♦ Production Strategy
♦ Hart Venture Capital
Chapter 10: Inventory Models
♦ Wagner Fabricating Company
♦ River City Fire Department
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 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 7: Integer Linear Programming
♦ Textbook Publishing
♦ Yeager National Bank
♦ Production Scheduling with Changeover Costs
Chapter 16: Markov Processes
♦ Dealer’s Absorbing State Probabilities in
Black Jack
Chapter 8: Nonlinear Optimization Models
♦ Portfolio Optimization with Transaction Costs
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
R1 vs. R2
R3 vs. R4
R5 vs. R6
Week 2
R1 vs. R3
R2 vs. R5
R4 vs. R6
Week 3
R1 vs. R4
R2 vs. R6
R3 vs. R5
Week 4
R1 vs. R5
R2 vs. R4
R3 vs. R6
Week 5
R1 vs. R6
R2 vs. R3
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
R1 vs. W1
R2 vs. W2
R3 vs. W3
Week 7
R1 vs. W2
R2 vs. W3
R3 vs. W1
Week 8
R1 vs. W3
R2 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 3week 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)
(W1, W2, W3, B4, B5, B6)
(W4, W5, W6, R4, R5, R6)
(W1, W2, W3, R4, R5, R6)
(W4, W5, W6, B1, B2, B3)
Weeks 12-14
(R1, R2, R3, B1, B2, B3)
Weeks 15-17
(R1, R2, R3, B4, B5, B6)
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.
Model
DI-910
DI-950
Production Rate
(minutes per printer)
Line 1
Line 2
3
4
6
2
Profit Contribution ($)
42
87
Capacity: 8 hours × 60 minutes/hour = 480 minutes per day
D1 = number of units of the DI-910 produced
D2 = number of units of the DI-950 produced
Let
Max
s.t.
42D1
+ 87D2
3D1 + 6D2
4D1 + 2D2
D 1, D 2 ≥ 0
≤ 480
≤ 480
Line 1 Capacity
Line 2 Capacity
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
Max
s.t.
BP100 = the number of BodyPlus 100 machines produced
BP200 = the number of BodyPlus 200 machines produced
371BP100 +
8BP100
5BP100
2BP100
-0.25BP100
+
+
+
+
461BP200
12BP200
10BP200
2BP200
0.75BP200
≤
≤
≤
≥
600
450
140
0
BP100, BP200 ≥ 0
CP - 4
Machining and Welding
Painting and Finishing
Assembly, Test, and Packaging
BodyPlus 200 Requirement
Solutions to Case Problems
BP200
80
Number of BodyPlus 200
.
70
Assembly, Test, and Packaging
60
50
Machining and Welding
40
BodyPlus 200 Requirement
30
20
10
Painting and Finishing
BP100
30 40 50 60 70 80 90 100
Number of BodyPlus 100
Optimal Solution
0
10
20
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
s.t.
1,800,000S
+
1,600,000M
600,000S
600,000S
250,000S
S
+
+
+
500,000M
350,000M
400,000M
S,M
≥
M
0
≤
≤
≤
≤
≤
800,000
700,000
500,000
1
1
Year 1
Year 2
Year 3
Maximum for S
Maximum for M
The solution obtained using The Management Scientist software package is shown below:
OPTIMAL SOLUTION
Objective Function Value =
2486956.522
Variable
-------------S
M
Value
--------------0.609
0.870
Reduced Costs
-----------------0.000
0.000
Constraint
-------------1
2
3
4
5
Slack/Surplus
--------------0.000
30434.783
0.000
0.391
0.130
Dual Prices
-----------------2.783
0.000
0.522
0.000
0.000
OBJECTIVE COEFFICIENT RANGES
Variable
-----------S
M
Lower Limit
--------------No Lower Limit
No Lower Limit
Current Value
--------------1800000.000
1600000.000
Upper Limit
--------------No Upper Limit
No Upper Limit
Current Value
--------------800000.000
700000.000
500000.000
1.000
1.000
Upper Limit
--------------822950.820
No Upper Limit
No Upper Limit
No Upper Limit
No Upper Limit
RIGHT HAND SIDE RANGES
Constraint
-----------1
2
3
4
5
Lower Limit
--------------No Lower Limit
669565.217
461111.111
0.609
0.870
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.
Security Systems
Market Analysis
Total
Year 1
$365,400
$435,000
$800,400
Year 2
$365,400
$304,500
$669,900
Year 3
$152,250
$348,000
$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 =
Variable
-------------S
M
Constraint
-------------1
2
3
4
5
2550819.672
Value
--------------0.689
0.820
Slack/Surplus
--------------77049.180
0.000
0.000
0.311
0.180
Reduced Costs
-----------------0.000
0.000
Dual Prices
-----------------0.000
2.098
2.164
0.000
0.000
OBJECTIVE COEFFICIENT RANGES
Variable
-----------S
M
Lower Limit
--------------No Lower Limit
No Lower Limit
Current Value
--------------1800000.000
1600000.000
CP - 7
Upper Limit
--------------No Upper Limit
No Upper Limit
Chapter 2
RIGHT HAND SIDE RANGES
Constraint
-----------1
2
3
4
5
4.
Lower Limit
--------------822950.820
No Lower Limit
No Lower Limit
0.689
0.820
Upper Limit
--------------No Upper Limit
802173.913
630555.556
No Upper Limit
No Upper Limit
If an additional $100,000 is made available, the allocation plan would change as follows:
Security Systems
Market Analysis
Total
5.
Current Value
--------------900000.000
700000.000
500000.000
1.000
1.000
Year 1
$413,400
$410,000
$823,400
Year 2
$413,400
$287,000
$700,400
Year 3
$172,250
$328,000
$500,250
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
s.t.
1.65R
+
2.00D
+
2.25H
0.15R
0.25R
0.25R
0.10R
0.25R
R
+
+
+
+
+
0.20D
0.20D
0.20D
0.20D
0.20D
+
+
+
+
+
0.25H
0.15H
0.15H
0.25H
0.20H
D
H
R, D, H ≥ 0
CP - 9
≤
≤
≤
≤
≤
≥
≥
≥
6000
7500
7500
6000
7500
10000
3000
5000
Almonds
Brazil
Filberts
Pecans
Walnuts
Regular
Deluxe
Holiday
Chapter 3
The solution found using The Management Scientist is shown below.
Objective Function Value =
61375.000
Variable
-------------R
D
H
Value
--------------17500.000
10624.999
5000.000
Reduced Costs
-----------------0.000
0.000
0.000
Constraint
-------------1
2
3
4
5
6
7
8
Slack/Surplus
--------------0.000
250.000
250.000
875.000
0.000
7500.000
7624.999
0.000
Dual Prices
-----------------8.500
0.000
0.000
0.000
1.500
0.000
0.000
-0.175
OBJECTIVE COEFFICIENT RANGES
Variable
-----------R
D
H
Lower Limit
--------------1.500
1.892
No Lower Limit
Current Value
--------------1.650
2.000
2.250
Upper Limit
--------------2.000
2.200
2.425
Current Value
--------------6000.000
7500.000
7500.000
6000.000
7500.000
10000.000
3000.000
5000.000
Upper Limit
--------------6583.333
No Upper Limit
No Upper Limit
No Upper Limit
7750.000
17500.000
10624.999
9692.307
RIGHT HAND SIDE RANGES
Constraint
-----------1
2
3
4
5
6
7
8
Lower Limit
--------------5390.000
7250.000
7250.000
5125.000
6750.000
No Lower Limit
No Lower Limit
-0.000
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
-------------R
D
H
Value
--------------11666.667
17916.668
5000.000
Reduced Costs
-----------------0.000
0.000
0.000
Constraint
-------------1
2
3
4
5
6
7
8
Slack/Surplus
--------------416.667
250.000
250.000
0.000
0.000
1666.667
14916.667
0.000
Dual Prices
-----------------0.000
0.000
0.000
5.667
4.333
0.000
0.000
-0.033
OBJECTIVE COEFFICIENT RANGES
Variable
-----------R
D
H
Lower Limit
--------------1.000
1.976
No Lower Limit
Current Value
--------------1.650
2.000
2.250
Upper Limit
--------------1.750
3.300
2.283
Current Value
--------------7000.000
7500.000
7500.000
6000.000
7500.000
10000.000
3000.000
5000.000
Upper Limit
--------------No Upper Limit
No Upper Limit
No Upper Limit
6250.000
7750.000
11666.667
17916.668
15529.412
RIGHT HAND SIDE RANGES
Constraint
-----------1
2
3
4
5
6
7
8
5.
Lower Limit
--------------6583.333
7250.000
7250.000
4210.000
7250.000
No Lower Limit
No Lower Limit
0.002
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)
2)
3)
4)
5)
6)
7)
G + I + M < 800000
.8G -.2I -.2M > 0
.6G -.4I -.4M < 0
-.2G +.8I -.2M > 0
-.5G +.5I -.5M < 0
-.3G -.3I +.7M > 0
.05G + .02I -.04M < 0
Funds Available
Min growth fund
Max growth fund
Min income fund
Max income fund
Min money market fund
Max risk
OPTIMAL SOLUTION
Objective Function Value =
94133.336
Variable
-------------G
I
M
Value
--------------248888.906
160000.000
391111.094
Reduced Costs
-----------------0.000
0.000
0.000
Constraint
-------------1
2
3
4
5
6
7
Slack/Surplus
--------------0.000
88888.898
71111.109
0.000
240000.000
151111.109
0.000
Dual Prices
-----------------0.118
0.000
0.000
-0.020
0.000
0.000
1.167
OBJECTIVE COEFFICIENT RANGES
Variable
-----------G
I
M
Lower Limit
--------------0.150
-0.463
0.015
Current Value
--------------0.180
0.125
0.075
CP - 12
Upper Limit
--------------No Upper Limit
0.145
0.180
Solutions to Case Problems
RIGHT HAND SIDE RANGES
Constraint
-----------1
2
3
4
5
6
7
Lower Limit
--------------0.188
No Lower Limit
-71111.109
-106666.641
-240000.000
No Lower Limit
-8000.000
Current Value
--------------800000.000
0.000
0.000
0.000
0.000
0.000
0.000
Upper Limit
--------------No Upper Limit
88888.898
No Upper Limit
133333.313
No Upper Limit
151111.109
6399.999
Rounding to the nearest dollar, the portfolio recommendation for Langford is as follows.
Fund:
Growth
Income
Money Market
Total
Amount
Invested
$248,889
160,000
391,111
$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
Fund:
Growth
Income
Money Market
Total
Amount
Invested
$293,333
160,000
346,667
$800,000
The portfolio yield becomes 98,800/800,000 = .124 or 12.4%
MAX .18G +.125I +.075M
S.T.
1)
2)
3)
4)
5)
6)
7)
G + I + M < 800000
.8G -.2I -.2M > 0
.6G -.4I -.4M < 0
-.2G +.8I -.2M > 0
-.5G +.5I -.5M < 0
-.3G -.3I +.7M > 0
.045G + .015I-.045M < 0
CP - 13
Chapter 3
OPTIMAL SOLUTION
Objective Function Value =
3.
98800.000
Variable
-------------G
I
M
Value
--------------293333.313
160000.000
346666.656
Reduced Costs
-----------------0.000
0.000
0.000
Constraint
-------------1
2
3
4
5
6
7
Slack/Surplus
--------------0.000
133333.328
26666.666
0.000
240000.000
106666.664
0.000
Dual Prices
-----------------0.124
0.000
0.000
-0.020
0.000
0.000
1.167
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)
2)
3)
4)
5)
6)
7)
G + I + M < 800000
.8G -.2I -.2M > 0
.6G -.4I -.4M < 0
-.2G +.8I -.2M > 0
-.5G +.5I -.5M < 0
-.3G -.3I +.7M > 0
.05G + .02I-.04M < 0
OPTIMAL SOLUTION
Objective Function Value =
Variable
-------------G
I
M
85066.664
Value
--------------160000.016
293333.313
346666.688
CP - 14
Reduced Costs
-----------------0.000
0.000
0.000
Solutions to Case Problems
Constraint
-------------1
2
3
4
5
6
7
4.
Slack/Surplus
--------------0.000
0.000
160000.000
133333.313
106666.688
106666.688
0.000
Dual Prices
-----------------0.106
-0.010
0.000
0.000
0.000
0.000
0.833
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)
2)
3)
4)
5)
6)
7)
8)
G + I + M < 800000
.8G -.2I -.2M > 0
.6G -.4I -.4M < 0
-.2G +.8I -.2M > 0
-.5G +.5I -.5M < 0
-.3G -.3I +.7M > 0
.05G + .02I-.04M < 0
G - I < 0
OPTIMAL SOLUTION
Objective Function Value =
93066.656
Variable
-------------G
I
M
Value
--------------213333.313
213333.313
373333.313
Reduced Costs
-----------------0.000
0.000
0.000
Constraint
-------------1
2
3
4
5
6
7
8
Slack/Surplus
--------------0.000
53333.324
106666.664
53333.324
186666.656
133333.328
0.000
0.000
Dual Prices
-----------------0.116
0.000
0.000
0.000
0.000
0.000
1.033
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.
xij = number of trucks obtained from a short term lease signed in month i for a period of j
Let
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.
Cost
$4000 + $2000 = $6000
2 ($3700) + $2000 = $9400
3 ($3225) + $2000 = $11,675
4 ($3040) + $2000 = $14,160
Decision Variables
x11, x21, x31, x41
x12, x22, x32
x13, x23
x14
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)
2)
3)
4)
5)
6)
7)
8)
X11 + X12
X21 + X22
X31 + X32
X41 + X32
Y1 < 1
Y2 < 2
Y3 < 3
Y4 < 1
+
+
+
+
X13
X23
X23
X23
+
+
+
+
X14
X14
X22
X14
+
+
+
+
Y1
X13
X14
Y4
CP - 16
=
+
+
=
10
X12 + Y2 = 12
X13 + Y3 = 14
8
Solutions to Case Problems
Objective Function Value =
151660.000
Variable
-------------X11
X12
X13
X14
X21
X22
X23
X31
X32
X41
Y1
Y2
Y3
Y4
Value
--------------0.000
0.000
3.000
6.000
0.000
0.000
1.000
1.000
0.000
0.000
1.000
2.000
3.000
1.000
Reduced Costs
-----------------3515.000
3725.000
0.000
0.000
2810.000
210.000
0.000
0.000
915.000
3515.000
0.000
0.000
0.000
0.000
Constraint
-------------1
2
3
4
5
6
7
8
Slack/Surplus
--------------0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
Dual Prices
------------------2485.000
-3190.000
-6000.000
-2485.000
485.000
1190.000
4000.000
485.000
OBJECTIVE COEFFICIENT RANGES
Variable
-----------X11
X12
X13
X14
X21
X22
X23
X31
X32
X41
Y1
Y2
Y3
Y4
Lower Limit
--------------2485.000
5675.000
10760.000
13950.000
3190.000
9190.000
10485.000
3190.000
8485.000
2485.000
No Lower Limit
No Lower Limit
No Lower Limit
No Lower Limit
Current Value
--------------6000.000
9400.000
11675.000
14160.000
6000.000
9400.000
11675.000
6000.000
9400.000
6000.000
2000.000
2000.000
2000.000
2000.000
CP - 17
Upper Limit
--------------No Upper Limit
No Upper Limit
11885.000
15075.000
No Upper Limit
No Upper Limit
11885.000
6915.000
No Upper Limit
No Upper Limit
2485.000
3190.000
6000.000
2485.000
Chapter 3
RIGHT HAND SIDE RANGES
Constraint
Lower Limit
-------------------------1
4.000
2
11.000
3
13.000
4
2.000
5
0.000
6
1.000
7
0.000
8
0.000
Current Value
--------------10.000
12.000
14.000
8.000
1.000
2.000
3.000
1.000
Upper Limit
--------------11.000
13.000
No Upper Limit
11.000
7.000
3.000
4.000
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)
2)
3)
4)
5)
6)
7)
8)
9)
10)
1T1<10
1R1<15
1N1<20
10000T1+10000T2+3000R1+3000R2+1000N1+1000N2<279000
4000T1+1500T2+2000R1+1200R2+1000N1+800N2>100000
-2T1-2T2+1R1+1R2>0
1T1+1T2<20
10000T1+10000T2>140000
3000R1+3000R2<99000
1000N1+1000N2>30000
OPTIMAL SOLUTION
Objective Function Value =
Variable
-------------T1
T2
R1
R2
N1
N2
2160.000
Value
--------------10.000
5.000
15.000
18.000
20.000
10.000
CP - 19
Reduced Costs
-----------------0.000
0.000
0.000
0.000
0.000
0.000
Chapter 4
Constraint
-------------1
2
3
4
5
6
7
8
9
10
Slack/Surplus
--------------0.000
0.000
0.000
0.000
27100.000
3.000
5.000
10000.000
0.000
0.000
Dual Prices
-----------------35.000
5.000
5.000
0.006
0.000
0.000
0.000
0.000
0.001
0.000
OBJECTIVE COEFFICIENT RANGES
Variable
-----------T1
T2
R1
R2
N1
N2
Lower Limit
--------------55.000
No Lower Limit
20.000
16.500
5.000
No Lower Limit
Current Value
--------------90.000
55.000
25.000
20.000
10.000
5.000
Upper Limit
--------------No Upper Limit
66.667
No Upper Limit
25.000
No Upper Limit
5.500
Current Value
--------------10.000
15.000
20.000
279000.000
100000.000
0.000
20.000
140000.000
99000.000
30000.000
Upper Limit
--------------15.000
33.000
30.000
294000.000
127100.000
3.000
No Upper Limit
150000.000
109000.000
40000.000
RIGHT HAND SIDE RANGES
Constraint
-----------1
2
3
4
5
6
7
8
9
10
1.
Lower Limit
--------------0.000
0.000
0.000
269000.000
No Lower Limit
No Lower Limit
15.000
No Lower Limit
93375.000
20000.000
Summary of the Optimal Solution
T1 + T2 = 10 + 5 = 15 Television advertisements
R1 + R2 = 15 + 18 = 33 Radio advertisements
N1 + N2 = 20 + 10 = 30 Newspaper advertisements
Advertising Schedule:
Media
Television
Radio
Newspaper
Totals
Number of Ads
15
33
30
78
Total Exposure Rating:
Total New Customers Reached:
Budget
$150,000
99,000
30,000
$279,000
2,160
127,100 (Surplus constraint 5)
CP - 20
Solutions to Case Problems
2.
The dual price shows that total exposure increases 0.006 points for each one dollar increase in the
advertising budget. Right Hand Side Ranges show this dual price applies for a budget increase of up
to $294,000 - $279,000 = $15,000. Thus the dual price applies for the $10,000 increase.
Total Exposure Rating would increase by 10,000(0.006) = 60 points
A $10,000 increase in the advertising budget is a 3.6% increase. But, it only provides a 2.8%
increase in total exposure. Management may decide that the additional exposure is not worth the
cost. This is a discussion point.
3.
The ranges for the exposure rating of 90 for the first 10 television ads show that the solution remains
optimal as long as the exposure rating is 55 or higher. This indicates that the solution is not very
sensitive to the exposure rating HJ has provided. Indeed, we would draw the same conclusion after
reviewing the next four ranges. We could conclude that Flamingo does not have to be concerned
about the exact exposure rating. The only concern might be the newspaper exposure rating of 5. A
rating of 5.5 or better can be expected to alter the current optimal solution.
4.
Remove constraint #5 for the linear programming model and use it to develop the objective function:
MAX 4000T1+1500T2+2000R1+1200R2+1000N1+800N2
Solving provides the following Optimal Solution
T1 + T2 = 10 + 4 = 14 Television advertisements
R1 + R2 = 15 + 13 = 28 Radio advertisements
N1 + N2 = 20 + 35 = 55 Newspaper advertisements
Advertising Schedule:
Media
Television
Radio
Newspaper
Totals
Number of Ads
14
28
55
97
Total New Customers Reached:
Budget
$140,000
83,000
55,000
$279,000
139,600
Total Exposure Rating
90(10) + 55(4) + 25(15) + 20(13) + 10(20) + 5(35) = 2130
5.
The solution with the objective to maximize the number of potential new customers reached looks
attractive. The total number of ads is increased from 78 to 97 (24%) and the number of potential
new customers reached is increased by 139,600 – 127,100 = 12,500 (9.8%).
Maximizing total exposure may seem to be the preferred objective because it is a more general
measure of advertising effectiveness. Exposure includes issues of image, message recall and appeal
to repeat customers. However, in this case, many more potential new customers will be reached
with the objective of maximizing reach, and the total exposure is only reduced by 2160 – 2130 = 30
points (1.4%).
At this point, we would expect some discussion concerning which solution is preferred: the one
obtained by maximizing total exposure or the one obtained by maximizing potential new customers
reached. Expect students to have differing opinions on the final recommendation. Basically, there
are two good media allocation solutions for this problem.
CP - 21
Chapter 4
Case Problem 2: Phoenix Computer
1.
The monthly cost of a new employee is $2,250 = $27,000/12. The training program is 3 months for
a new employee and costs $1,500. So the total cost of hiring a new employee and training her/him
to be a laptop specialist is $8,250 = 3($2,250) + $1,500.
2.
The training program is only 2 months for current employees and costs $1,000. So a replacement
employee will only need to be hired 2 months before the laptop specialist is needed in this case. So
the incremental cost of putting a current employee through the training program is $5,500 =
2($2,250) + $1,000.
3.
It is clear that 100 new employees will need to be hired either to become laptop specialists
themselves or to replace current employees who will become laptop specialists. So the company's
monthly payroll cost will increase by $225,000 = 100($2,250) in September over January.
A linear program can be formulated and solved to minimize the cost of hiring and training over the
period from February to August. The following variable definitions are used:
Feb 3
•
•
•
Jun 3
Mar 2
•
•
•
Jul 2
= No. of new employees entering the 3-month program in February
= No. of new employees entering the 3-month program in June
= No. of current employees entering the 2-month program in March
= No. of current employees entering the 2-month program in July
IdleMay = No. of trained laptop specialists in excess of those needed in May
•
•
•
IdleSep = No. of trained laptop specialists in excess of those needed in September
There are 16 constrains needed. The first 5 deal with meeting the needs for laptop specialists.
Constraints (6) through (10) restrict the number of current employees that may enter the program
while Constraints (11) through (16) deal with the capacity of the training center.
The following output from The Management Scientist shows a model and solution that can be used
to answer the questions in part 3.
LINEAR PROGRAMMING PROBLEM
MIN
8250Feb3+5500Mar2+2250IdleMay+8250Mar3+5500Apr2+2250IdleJun+8250Apr3+55
00May
2+2250IdleJul+8250May3+5500Jun2+2250IdleAug+8250Jun3+5500Jul2+2250IdleS
ep
CP - 22
Solutions to Case Problems
S.T.
1)
2)
3)
4)
5)
1Feb3+1Mar2-1IdleMay=20
1Feb3+1Mar2+1Mar3+1Apr2-1IdleJun=30
1Feb3+1Mar2+1Mar3+1Apr2+1Apr3+1May2-1IdleJul=85
1Feb3+1Mar2+1Mar3+1Apr2+1Apr3+1May2+1May3+1Jun2-1IdleAug=85
1Feb3+1Mar2+1Mar3+1Apr2+1Apr3+1May2+1May3+1Jun2+1Jun3+1Jul2-IdleSep
=100
6) 1Mar2<15
7) 1Mar2+1Apr2<35
8) 1Mar2+1Apr2+1May2<35
9) 1Mar2+1Apr2+1May2+1Jun2<40
10) 1Mar2+1Apr2+1May2+1Jun2+1Jul2<50
11) 1Feb3<25
12) 1Mar2+1Mar3<25
13) 1Apr2+1Apr3<25
14) 1May2+1May3<25
15) 1Jun2+1Jun3<25
16) 1Jul2<25
OPTIMAL SOLUTION
Objective Function Value =
Variable
-------------Feb3
Mar2
IdleMay
Mar3
Apr2
IdleJun
Apr3
May2
IdleJul
May3
Jun2
IdleAug
Jun3
Jul2
IdleSep
698750.000
Value
--------------10.000
10.000
0.000
15.000
0.000
5.000
25.000
25.000
0.000
0.000
0.000
0.000
0.000
15.000
0.000
CP - 23
Reduced Costs
-----------------0.000
0.000
2250.000
0.000
2250.000
0.000
0.000
0.000
4500.000
2250.000
0.000
2250.000
0.000
0.000
10500.000
Chapter 4
Constraint
-------------1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Slack/Surplus
--------------0.000
0.000
0.000
0.000
0.000
5.000
25.000
0.000
5.000
0.000
15.000
0.000
0.000
0.000
25.000
10.000
Dual Prices
-----------------0.000
2250.000
-2250.000
0.000
-8250.000
0.000
0.000
0.000
0.000
2750.000
0.000
0.000
2250.000
2250.000
0.000
0.000
Case Problem 3: Textile Mill Scheduling
Let X3R = Yards of fabric 3 on regular looms
X4R = Yards of fabric 4 on regular looms
X5R = Yards of fabric 5 on regular looms
X1D = Yards of fabric 1 on dobbie looms
X2D = Yards of fabric 2 on dobbie looms
X3D = Yards of fabric 3 on dobbie looms
X4D = Yards of fabric 4 on dobbie looms
X5D = Yards of fabric 5 on dobbie looms
Y1 = Yards of fabric 1 purchased
Y2 = Yards of fabric 2 purchased
Y3 = Yards of fabric 3 purchased
Y4 = Yards of fabric 4 purchased
Y5 = Yards of fabric 5 purchased
Profit Contribution per Yard
Fabric
1
2
3
4
5
Manufactured
0.33
0.31
0.61
0.73
0.20
Purchased
0.19
0.16
0.50
0.54
0.00
1
2
3
4
5
Regular
—
—
0.1912
0.1912
0.2398
Dobbie
0.21598
0.21598
0.1912
0.1912
0.2398
Production Times in Hours per Yard
Fabric
Model may use a Max Profit or Min Cost objective function.
CP - 24
Solutions to Case Problems
Max
0.61X3R + 0.73X4R + 0.20X5R
+ 0.33X1D + 0.31X2D + 0.61X3D + 0.73X4D + 0.20X5D
+ 0.19Y1 + 0.16Y2 + 0.50Y3 + 0.54Y4
or
Min
0.49X3R + 0.51X4R + 0.50X5R
+ 0.66X1D + 0.55X2D + 0.49X3D + 0.51X4D + 0.50X5D
+ 0.80Y1 + 0.70Y2 + 0.60Y3 + 0.70Y4 + 0.70Y5
Regular Hours Available
30 Looms x 30 days x 24 hours/day = 21600
Dobbie Hours Available
8 Looms x 30 days x 24 hours/day = 5760
Constraints:
Regular Looms:
0.192X3R + 0.1912X4R + 0.2398X5R ≤ 21600
Dobbie Looms:
0.21598X1D + 0.21598X2D + 0.1912X3D + 0.1912X4D + 0.2398X5D ≤ 5760
Demand Constraints
X1D + Y1
X2D + Y2
X3R + X3D + Y3
X4R + X4D + Y4
X5R + X5D + Y5
= 16500
= 22000
= 62000
= 7500
= 62000
OPTIMAL SOLUTION
Objective Function Value =
Variable
-------------X3R
X4R
X5R
X1D
X2D
X3D
X4D
X5D
Y1
Y2
Y3
Y4
Y5
62531.91797
Value
--------------27711.29297
7500.00000
62000.00000
4669.13672
22000.00000
0.00000
0.00000
0.00000
11830.86328
0.00000
34288.70703
0.00000
0.00000
CP - 25
Reduced Costs
-----------------0.00000
0.00000
0.00000
0.00000
0.00000
0.01394
0.01394
0.01748
0.00000
0.01000
0.00000
0.08000
0.06204
Chapter 4
Constraint
-------------1
2
3
4
5
6
7
Slack/Surplus
--------------0.00000
0.00000
0.00000
0.00000
0.00000
0.00000
0.00000
Dual Prices
-----------------0.57531
0.64821
0.19000
0.17000
0.50000
0.62000
0.06204
OBJECTIVE COEFFICIENT RANGES
Variable
-----------X3R
X4R
X5R
X1D
X2D
X3D
X4D
X5D
Y1
Y2
Y3
Y4
Y5
Lower Limit
--------------0.50000
0.71606
0.18252
0.31426
0.30000
No Lower Limit
No Lower Limit
No Lower Limit
0.18000
No Lower Limit
0.48606
No Lower Limit
No Lower Limit
Current Value
--------------0.61000
0.73000
0.20000
0.33000
0.31000
0.61000
0.73000
0.20000
0.19000
0.16000
0.50000
0.54000
0.00000
Upper Limit
--------------0.62394
No Upper Limit
No Upper Limit
0.34000
No Upper Limit
0.62394
0.74394
0.21748
0.20574
0.17000
0.61000
0.62000
0.06204
Current Value
--------------21600.00000
5760.00000
16500.00000
22000.00000
62000.00000
7500.00000
62000.00000
Upper Limit
--------------28156.00000
8315.23047
No Upper Limit
26669.13672
No Upper Limit
35211.29297
84095.07813
RIGHT HAND SIDE RANGES
Constraint
-----------1
2
3
4
5
6
7
Lower Limit
--------------16301.60059
4751.55957
4669.13672
10169.13672
27711.29297
0.00000
34660.54688
Production/Purchase Schedule (Yards)
Regular
Looms
Fabric
1
2
3
4
5
27711
7500
62000
Projected Profit: $62,531.92
CP - 26
Dobbie
Looms
4669
22000
Purchased
11831
34289
Solutions to Case Problems
Value of 9th Dobbie Loom
Dual Price (Constraint 2) = 0.64821 per hour dobbie
Monthly Value of 1 Dobbie Loom
(30 days)(24 hours/day)($0.64821) = $466.71
Note: This change is within the Right-Hand Side Ranges for Constraint 2.
Discussion of Objective Coefficient Ranges
For example, fabric one on the dobbie loom shares ranges of 0.31426 to 0.34 for the profit
maximization model or 0.64426 to 0.67 for the cost minimization model.
Note here that since demand for the fabrics is fixed, both the profit maximization and cost
minimization models will provide the same optimal solution. However, the interpretation of the
ranges for the objective function coefficients differ for the two models. In the profit maximization
case, the coefficients are profit contributions. Thus, the range information indicates how price per
unit and cost per unit may vary simultaneously. That is, as long as the net changes in price per unit
and cost per unit keep the profit contributions within the ranges, the solution will remain optimal. In
the cost minimization model, the coefficients are costs per unit. Thus, the range information
indicates that assuming price per unit remains fixed how much the cost per unit may vary and still
maintain the same optimal solution.
Case Problem 4: Workforce Scheduling
1.
Let tij = number of temporary employees hired under option i (i = 1, 2, 3) in month j (j = 1 for
January, j = 2 for February and so on)
The following table depicts the decision variables used in this case problem.
Option 1
Option 2
Option 3
Jan.
t11
t21
t31
Feb.
t12
t22
t32
Mar.
t13
t23
t33
Apr.
t14
t24
t34
May
t15
t25
June
t16
Costs: Contract cost plus training cost
Option
1
2
3
Contract Cost
$2000
$4800
$7500
Training Cost
$875
$875
$875
Min. 2875(t11 + t12 + t13 + t14 + t15 + t16)
+ 5675(t21 + t22 + t23 + t24 + t25)
+ 8375(t31 + t32 + t33 + t34)
One constraint is required for each of the six months.
CP - 27
Total Cost
$2875
$5675
$8375
Chapter 4
Constraint 1: Need 10 additional employees in January
t11 = number of temporary employees hired under Option 1 (one-month contract) in January
t21 = number of temporary employees hired under Option 2 (two-month contract) in January
t31 = number of temporary employees hired under Option 3 (three-month contract) in January
t11 + t21 + t31 = 10
Constraint 2: Need 23 additional employees in February
t12 , t22 and t32 are the number of temporary employees hired under Options 1, 2 and 3 in
February.
But, temporary employees hired under Option 2 or Option 3 in January will also be available to
satisfy February needs.
t21 + t31 + t12 + t22 + t32 = 23
Note: The following table shows the decision variables used in this constraint
Jan.
Option 1
Option 2
Option 3
t21
t31
Feb.
t12
t22
t32
Mar.
Apr.
May
June
Constraint 3: Need 19 additional employees in March
Option 1
Option 2
Option 3
Jan.
Feb.
t31
t22
t32
Mar.
t13
t23
t33
Apr.
May
June
t31 + t22 + t32 + t13 + t23 + t33 = 19
Constraint 4: Need 26 additional employees in May
Jan.
Option 1
Option 2
Option 3
Feb.
Mar.
t32
t23
t33
Apr.
t14
t24
t34
May
June
t32 + t23 + t33 + t14 + t24 + t34 = 26
Constraint 5: Need 20 additional employees in May
Jan.
Feb.
Option 1
Option 2
Option 3
Mar.
Apr.
t33
t24
t34
t33 + t24 + t34 + t15 + t25 = 20
CP - 28
May
t15
t25
June
Solutions to Case Problems
Constraint 6: Need 14 additional employees in June
Jan.
Feb.
Mar.
Option 1
Option 2
Option 3
Apr.
May
June
t16
t25
t34
t34 + t25 + t16 = 14
Optimal Solution: Total Cost = $313,525
Option 1
Option 2
Option 3
Jan.
0
3
7
Feb.
1
0
12
Mar.
0
0
0
Apr.
0
0
14
May
6
0
June
0
2.
Option
1
2
3
3.
Number Hired
7
3
33
Total:
Contract Cost
$14,000
$14,400
$247,500
$275,900
Training Cost
$6,125
$2,625
$28,875
$37,625
Total Cost
$20,125
$17,025
$276,375
$313,525
Hiring 10 full-time employees at the beginning of January will reduce the number of temporary
employees needed each month by 10. Using the same linear programming model with the righthand sides of 0, 13, 9, 16, 10 and 4, provides the following schedule for temporary employees:
Option 1
Option 2
Option 3
Option
1
2
3
Total:
Jan.
0
0
0
Feb.
4
0
9
Number Hired
7
3
13
23
Mar.
0
0
0
Apr.
0
3
4
May
3
0
Contract Cost
$14,000
$14,400
$97,500
June
0
Training Cost
$6,125
$2,625
$11,375
Total Cost
$20,125
$17,025
$108,875
$146,025
Full-time employees cost:
Training cost: 10($875) = $8,750
Salary: 10(6)(168)($16.50) = $166,320
Total Cost = $146,025 + $8750 + $166,320 = $321,095
Hiring 10 full-time employees is $321,095 - $313,525 = $7,570 more expensive than using
temporary employees. Do not hire the 10 full-time employees. Davis should continue to contract
with WorkForce to obtain temporary employees.
CP - 29
Chapter 4
4.
With the lower training costs, the costs per employee for each option are as follows:
Option
1
2
3
Cost
$2000
$4800
$7500
Training Cost
$700
$700
$700
Total Cost
$2700
$5500
$8200
Resolving the original linear programming model with the above costs indicates that Davis should
hire all temporary employees on a one-month contract specifically to meet each month's employee
needs. Thus, the monthly temporary hire schedule would be as follows: January - 10; February - 23;
March - 19; April - 26; May - 20; and June - 14. The total cost of this strategy is $302,400. Note
that if training costs were any lower, this would still be the optimal hiring strategy for Davis.
Case Problem 5: Duke Energy Coal Allocation
A linear programming model can be used to determine how much coal to buy from each of the
mining companies and where to ship it. Let
xij = tons of coal purchased from supplier i and used by generating unit j
The objective function minimizes the total cost to buy and burn coal. The objective function
coefficients, cij , are the cost to buy coal at mine i, ship it to generating unit j, and burn it at
generating unit j. Thus, the objective function is ∑ ∑ cij xij . In computing the objective function
coefficients three inputs must be added: the cost of the coal, the transportation cost to the generating
unit, and the cost of processing the coal at the generating unit.
There are two types of constraints: supply constraints and demand constraints. The supply
constraints limit the amount of coal that can be bought under the various contracts. For the fixedtonnage contracts, the constraints are equalities. For the variable-tonnage contracts, any amount of
coal up to a specified maximum may be purchased. Let Li represent the amount that must be
purchased under fixed-tonnage contract i and Si represent the maximum amount that can be
purchased under variable-tonnage contract i. Then the supply constraints can be written as follows:
∑x
ij
= Li
for all fixed-tonnage contracts
≤ Si
for all variable-tonnage contracts
j
∑x
ij
j
The demand constraints specify the number of mWh of electricity that must be generated by each
generating unit. Let aij = mWh hours of electricity generated by a ton of coal purchased from
supplier i and used by generating unit j, and Dj = mWh of electricity demand at generating unit j.
The demand constraints can then be written as follows:
∑a x
ij ij
= Dj
for all generating units
i
Note: Because of the large number of calculations that must be made to compute the objective
function and constraint coefficients, we developed an Excel spreadsheet model for this problem.
Copies of the data and model worksheets are included after the discussion of the solution to parts (a)
through (f).
CP - 30
Solutions to Case Problems
1.
The number of tons of coal that should be purchased from each of the mining companies and where
it should be shipped is shown below:
Miami Fort #
5
Miami Fort # 7
Beckjord
East Bend
Zimmer
0
0
61,538
288,462
0
Peabody
217,105
11,278
71,617
0
0
American
0
0
0
0
275,000
Consol
0
0
33,878
0
166,122
Cyprus
0
0
0
0
0
Addington
0
200,000
0
0
0
Waterloo
0
0
98,673
0
0
RAG
The total cost to purchase, deliver, and process the coal is $53,407,243.
2.
The cost of the coal in cents per million BTUs for each generating unit is as follows:
Miami Fort #5
111.84
3.
Miami Fort #7
136.97
Beckjord
127.24
East Bend
103.85
Zimmer
114.51
The average number of BTUs per pound of coal received at each generating unit is shown
below:
Miami Fort #5
13,300
Miami Fort #7
12,069
Beckjord
12,354
East Bend
13,000
Zimmer
12,468
4.
The sensitivity report shows that the shadow price per ton of coal purchased from American Coal
Sales is -$13 per ton and the allowable increase is 88,492 tons. This means that every additional ton
of coal that Duke Energy can purchase at the current price of $22 per ton will decrease cost by $13.
So even paying $30 per ton, Duke Energy will decrease cost by $5 per ton. Thus, they should buy
the additional 80,000 tons; doing so will save them $5(80,000) = $400,000.
5.
If the energy content of the Cyprus coal turns out to be 13,000 BTUs per ton the procurement plan
changes as shown below:
Miami Fort # 5 Miami Fort # 7 Beckjord
Zimmer
0
0
61,538
288,462
0
Peabody
36,654
191,729
71,617
0
0
American
0
0
0
0
275,000
Consol
0
0
33,878
0
166,122
Cyprus
0
0
85,769
0
0
Addington
200,000
0
0
0
0
Waterloo
0
0
0
0
0
RAG
6.
East Bend
The shadow prices for the demand constraints are as follows:
Miami Fort #5
21
Miami Fort #7
20
Beckjord
20
CP - 31
East Bend
18
Zimmer
19
Chapter 4
The East Bend unit is the least cost producer at the margin ($18 per mWh), and the allowable
increase is 160,000 mWh. Thus, Duke Energy should sell the 50,000 mWh over the grid. The
additional electricity should be produced at the East Bend generating unit. Duke Energy’s profit will
be $12 per mWh.
The Excel data and model worksheets used to solve the Duke Energy coal allocation problem are as
follows:
Duke Energy Coal Allocation Model (Data)
Duke Energy Coal Allocation Model (Solution)
CP - 32
Chapter 6
Distribution and Network Models
Case Problem 1: Solution Plus
1.
This case can be formulated as a transportation problem with the Cincinnati and Oakland production
facilities as the origins. The locations of the railway stations are the destinations. Each objective
function coefficient is the sum of the production cost at an origin and the freight cost to ship from
the origin to a destination. The minimum cost solution has a value of $1,318,985 and 773,522
gallons of cleaning fluid are produced and shipped. So, the average cost per gallon of producing the
cleaning fluid and shipping it to the railway stations is $1.705168. Shown below is the optimal
shipping plan.
Santa Ana
El Paso
Pendleton
Houston
Kansas City
Los Angeles
Glendale
Jacksonville
Little Rock
Bridgeport
Sacramento
Cincinnati
0
6800
39636
100447
24570
0
0
68486
148586
111475
0
500000
Total
Oakland
22418
0
40654
0
0
64761
33689
0
0
0
112000
273522
From the shipping plan it can be seen that the Cincinnati plant is at capacity. The dual price for the
Cincinnati capacity constraint indicates that if more capacity can be made available the total cost can
be decreased by $.11 for every additional gallon up to 40,654.
2.
Since it costs Solutions Plus $1.705168 per gallon to produce and deliver the cleaning fluid to the
railroad, this is the breakeven point. If Mr. Miller bids any less, Solutions Plus will lose money on
the contract.
3.
A 15% profit margin corresponds to a price of 1.15($1.705168) = $1.9609432 per gallon. Rounding
up, we would recommend the president bid $1.97 per gallon if a 15% profit margin is desired.
4.
If management of Solutions Plus expects oil prices to go up, they can also expect freight rates to go
up. The breakeven point found by solving the minimum cost transportation problem is using freight
rates for the first year. So, management should perhaps bid a bit higher to allow for an increase in
freight rates in the second year of the contract. You might offer to resolve the problem assuming,
say, a 10% increase in freight rates to see what the breakeven point would be in this situation. On
the other hand, if freight rates are expected to go down in the second year, management might want
to bid a bit lower to increase the chances of winning the contract.
CP - 33
Chapter 6
Solutions Plus
Production Cost
Santa Ana
Origin
Cincinnati
10
Oakland
0.22
22418
Demand
CP - 34
Santa Ana
Origin
Cincinnati
11.2
Oakland
1.87
22418
Demand
Cincinnati Oakland
1.2
1.65
El Paso
Pendleton
0.84
0.83
0.74
0.49
6800
80290
FREIGHT COST
Destination
Houston
Kansas City Los Angeles Glendale Jacksonville Little Rock Bridgeport Sacramento Supply
0.45
0.36
10
10
0.34
0.34
0.34
10 500000
10
10
0.22
0.22
10
10
10
0.15 500000
100447
24570
64761
33689
68486
148586
111475
112000
El Paso
Pendleton
2.04
2.03
2.39
2.14
6800
80290
FREIGHT + PRODUCTION COST
Destination
Houston
Kansas City Los Angeles Glendale Jacksonville Little Rock Bridgeport Sacramento Supply
1.65
1.56
11.2
11.2
1.54
1.54
1.54
11.2 500000
11.65
11.65
1.87
1.87
11.65
11.65
11.65
1.8 500000
100447
24570
64761
33689
68486
148586
111475
112000
Model
Total
Per Gallon
Min Cost 1318984.93 1.705167959
Santa Ana
Origin
0
Cincinnati
Oakland
22418
22418
Total
=
22418
El Paso
Pendleton
6800
39636
0
40654
6800
80290
=
=
6800
80290
Destination
Houston
Kansas City Los Angeles Glendale Jacksonville Little Rock Bridgeport Sacramento
100447
24570
0
0
68486
148586
111475
0
0
0
64761
33689
0
0
0
112000
100447
24570
64761
33689
68486
148586
111475
112000
=
=
=
=
=
=
=
=
100447
24570
64761
33689
68486
148586
111475
112000
Total
500000 <= 500000
273522 <= 500000
773522
Solutions to Case Problems
Case Problem 2: Distribution Systems Design
Three related linear programming models were developed and solved to answer the questions in this case.
First, we developed a linear programming formulation of the network model shown on the following page;
we refer to this network model and the corresponding linear program as the Part 2 Network Model and the
Part 2 Linear Program, respectively. Variations of the Part 2 Linear Program were then used to answer the
questions in parts 1 and 3. For instance, to answer the question concerning the existing system in part 1 we
added constraints to the Part 2 Linear Program which set the flow equal to 0 over the distribution centercustomer arcs that Darby does not currently use. For Part 3, we added three plant to customer arcs to the
Part 2 Network Model and corresponding linear program: El Paso - San Antonio, San Bernardino - Los
Angeles, and San Bernardino - San Diego.
The decision variables used in each of the linear programs use 4 letters to describe the amount of flow over
each arc. The first two letters in each variable name identify the “from” node and the second two letters
identify the “to” nodes. For instance, EPFW represents the amount shipped from El Paso to Ft. Worth and
LVDE represents the amount shipped from Las Vegas to Denver.
A description of the LP models that provide the basis for answering the questions in the managerial report
follows the network model for the questions in part 2.
CP - 35
Chapter 6
Part 2 Network Model:
Dallas
6300
San
Antonio
4880
Wichita
2130
Kansas
City
1210
Denver
6120
Salt
Lake
City
4830
Phoenix
2750
0.3
2.1
Ft. Worth
3.1
4.4
6.0
3.20
30,000
5.4
El Paso
2.20
4.5
5.2
6.0
4.20
2.7
Santa Fe
4.7
3.90
20,000
3.4
3.3
2.7
San
Bernardino
1.20
5.4
3.3
Las Vegas
2.4
2.1
2.5
Los
Angeles
San Diego
CP - 36
8580
4460
Solutions to Case Problems
LP Model and Solution for Part 2
MIN 13.7EPFW + 12.7EPSF + 14.7EPLV + 13.9SBSF + 11.2SBLV + .3FWDA +
2.1FWSA + 3.1FWWI + 4.4FWKC + 6.0FWDE + 5.2SFDA + 5.4SFSA + 4.5SFWI +
6.0SFKC + 2.7SFDE + 4.7SFSL + 3.4SFPH + 3.3SFLA + 2.7SFSD + 5.4LVDE +
3.3LVSL + 2.4LVPH + 2.1LVLA + 2.5LVSD
S.T.
1)
2)
3)
4)
EPFW
SBSF
FWDA
SFDA
+
+
+
+
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)
LVDE
FWDA
FWSA
FWWI
FWKC
FWDE
SFSL
SFPH
SFLA
SFSD
+
+
+
+
+
+
+
+
+
+
SFSD
EPSF
SBLV
FWSA
SFSA
LVSL
SFDA
SFSA
SFWI
SFKC
SFDE
LVSL
LVPH
LVLA
LVSD
+ EPLV < 30000
< 20000
+ FWWI + FWKC + FWDE - EPFW = 0
+ SFWI + SFKC + SFDE + SFSL + SFPH + SFLA +
EPSF - SBSF = 0
+ LVPH + LVLA + LVSD - EPLV - SBLV = 0
= 6300
= 4880
= 2130
= 1210
+ LVDE = 6120
= 4830
= 2750
= 8580
= 4460
Objective Function Value =
Variable
-------------EPFW
EPSF
EPLV
SBSF
SBLV
FWDA
FWSA
FWWI
FWKC
FWDE
SFDA
SFSA
SFWI
SFKC
SFDE
SFSL
SFPH
SFLA
SFSD
LVDE
LVSL
600942.000
Value
--------------14520.000
6740.000
0.000
0.000
20000.000
6300.000
4880.000
2130.000
1210.000
0.000
0.000
0.000
0.000
0.000
6120.000
0.000
0.000
0.000
620.000
0.000
4830.000
CP - 37
Reduced Costs
-----------------0.000
0.000
1.800
2.900
0.000
0.000
0.000
0.000
0.000
4.300
3.900
2.300
0.400
0.600
0.000
1.200
0.800
1.000
0.000
2.900
0.000
Chapter 6
Variable
-------------LVPH
LVLA
LVSD
Value
--------------2750.000
8580.000
3840.000
Reduced Costs
-----------------0.000
0.000
0.000
LP Model and Solution for Part 1 Questions (Existing System)
The following constraints were added to the above LP Model so that each customer can only be served by
the distribution center it is currently served by.
15)
16)
17)
18)
19)
20)
21)
22)
23)
24)
SFDA
SFSA
SFWI
SFKC
SFLA
SFSD
LVDE
FWDE
LVSL
LVPH
=
=
=
=
=
=
=
=
=
=
0
0
0
0
0
0
0
0
0
0
The optimal solution shows that the total cost (manufacturing plus distribution) of the existing system is
$620,770. Details of this solution and the modifications suggested in questions 2 and 3 are contained in the
“Summary of Optimal Solutions.”
LP Model and Solution for Part 3 (Plant-Customer Shipments)
The network model for the part 2 questions must be modified by adding 3 arcs: San Bernardino - Los
Angeles, San Bernardino - San Diego, and El Paso - San Antonio. The corresponding linear program must
also be modified to incorporate the new variables.
CP - 38
Solutions to Case Problems
A summary of the optimal solutions suggested by questions (1), (2), and (3) follows.
Summary of Optimal Solutions
Part 1
Part 2
Costs
Existing
System
Any Distribution
Center
Part 3
Plant Customer
Shipments
Manufacturing Cost
$426,710
$423,230
$423,230
194,060
177,712
130,304
$620,770
$600,942
$553,534
EPFW
14520
14520
9640
EPSF
13700
6740
6740
EPLV
0
0
0
SBSF
0
0
0
SBLV
13040
20000
6960
FWDA
6300
6300
6300
FWSA
4880
4880
0
FWWI
2130
2130
2130
FWKC
1210
1210
1210
FWDE
0
0
0
SFDA
0
0
0
SFSA
0
0
0
SFWI
0
0
0
SFKC
0
0
0
SFDE
6120
6120
6120
SFSL
4830
0
0
SFPH
2750
0
620
SFLA
0
0
0
SFSD
0
620
0
LVDE
0
0
0
LVSL
0
4830
4830
LVPH
0
2750
2130
Shipping Cost
Total Cost
Decision Variables
CP - 39
Chapter 6
Part 1
Part 2
Existing
System
Any Distribution
Center
Part 3
Plant Customer
Shipments
LVLA
8580
8580
0
LVSD
4460
3840
0
EPSA
4880
SBLA
8580
SBSD
4460
Discussion
1.
With the current system of assigning customers to distribution centers, the minimum total cost
is $620,770; the total cost is the sum of $426,710 for manufacturing and $194,060 for shipping.
One might suspect that restricting customers to a single distribution center is not optimal.
2.
Allowing customers to be serviced by any distribution center reduces total cost to $600,942; the
total cost consists of $423,230 of manufacturing cost and $177,712 of shipping cost. This is a
3.19% decrease in total cost and an 8.42% decrease in shipping cost. Only one customer, San
Diego, is served by more than one distribution center.
3.
Allowing some direct shipments from plants to customers causes a substantial reduction in total
cost. The total cost is $553,534, a 10.83% reduction over the current plan. Note, however,
there is a 32.85% reduction in shipping cost. Again, one customer, Phoenix, receives shipments
from 2 distribution centers.
4.
Total capacity at the two plants (50,000 units) is adequate to handle the growth in demand.
However, the company might want to consider expanding the more efficient San Bernardino
plant. The dual price from Part 3 output (not shown here) indicates that each unit of added
capacity will reduce distribution costs by $2.50. However, the range of feasibility shows this
dual price is only applicable for the net 620 units of capacity. Additional computer runs with
higher capacity levels for San Bernardino are recommended.
CP - 40
Chapter 7
Integer Linear Programming
Case Problem 1: Textbook Publishing
An integer programming model can be used advantageously to assist in developing recommendations.
Let x i =
book i is scheduled for publication
{ 10 ifotherwise
The subscripts correspond to the books as follows:
Book
Business Calculus
Finite Math
General Statistics
Mathematical Statistics
Business Statistics
Finance
Financial Accounting
Managerial Accounting
English Literature
German
i
1
2
3
4
5
6
7
8
9
10
An integer programming model for maximizing projected sales (thousands of units) subject to the
restrictions mentioned is given.
Max 20x1 + 30x2 + 15x3
s.t.
30x1 + 16x2 + 24x3
40x1 + 24x2
30x3
x3
x1
+
x2
+ 10x4 +
25x5 +
+ 20x4 +
10x5
+ 24x4 +
x4 +
+
18x6 + 25x7 + 50x8 + 20x9 + 30x10
+ 40x9
≤ 60 John
+ 24x7 + 28x8 + 34x9 + 50x10 ≤ 40 Susan
16x5 + 14x6 + 26x7 + 30x8 + 30x9 + 36x10 ≤ 40 Monica
x5
≤ 2 No. of Stat Books
x7 +
x8
≤ 1 Account Book
= 1 Math Book
xi = 0, 1 for all i
The optimal solution (x2 = x5 = x6 = 1) calls for publishing the finite math, the business statistics and the
finance books. Projected sales are 73,000 copies.
(1)
If Susan can be made available for another 12 days, the optimal solution is x2 = x8 = 1. This calls
for publishing the finite math and managerial accounting texts for projected sales of 80,000 copies.
(2)
If Monica is also available for 10 more days, a big improvement can be made. The new optimal
solution calls for producing the finite math book, the business statistics book, and the managerial
accounting book. Projected sales are 105,000 copies.
CP - 41
Chapter 7
(3)
The solution in (2) above does not include any new books. In the long run this would appear to be a
bad strategy for the company. A variety of modifications can be made to the model to examine the
short run impact of postponing a revision. For instance, a constraint could be added to require
publication of at least one new book.
Case Problem 2: Yeager National Bank
A mixed integer linear programming (MILP) model can be used advantageously to assist in preparing a
report for Mr. Wolff. Since the annual fixed costs of operating the lockboxes are not known exactly we
formulate a model that can be used without that information. Then determine if information more precise
than the $20,000 - $30,000 estimate is needed to make a final decision. We formulate a model that can be
solved to find the best set of locations with a given number of lockboxes. It is then solved 4 times to find
the best set of locations with 1 lockbox, up to 2 lockboxes, up to 3 lockboxes, and up to 4 lockboxes. We
then can address the question of whether annual operating cost information is needed from some or all of
the potential lockbox locations.
We use the following variable definitions:
P = 1 if a lockbox is in Phoenix; 0 otherwise
S = 1 if a lockbox is in Salt Lake City; 0 otherwise
A = 1 if a lockbox is in Atlanta; 0 otherwise
B = 1 if a lockbox is in Boston; 0 otherwise
PNW = 1 if the Northwest region is assigned to Phoenix; 0 otherwise
PSW = 1 if the Southwest region is assigned to Phoenix; 0 otherwise
.
.
.
BSE = 1 if the Southeast region is assigned to Boston; 0 otherwise
The MILP model has 24 variables and 26 constraints. The objective function calls for minimizing lost
interest income. To see how the objective function coefficients are computed, consider assigning the
Northwest region to Phoenix. Daily collections are $80,000 and it takes 4 days to receive and process
payments. At a 15% rate Yeager could save $48,000 = (.15)(4)($80,000) annually over this assignment if
the Northwest collections could be credited to their account instantaneously. A similar calculation is made
for all the other potential assignments. The objective function coefficients for P, S, A, and B are zero since
we are not including the cost of operating the lockboxes at this stage.
Constraints (1) to (5) ensure that each region is assigned a lock box. Constraints (6) to (25) ensure that a
region is only assigned to a lockbox location that is open. And, constraint (26) is a limitation on the
number of lockbox locations that may be chosen. The right-hand-side of that constraint will be varied from
1 to 4 as the problem is solved 4 times. Shown below is the MILP model that is solved to find the optimal
solution when up to 2 lockboxes may be used.
CP - 42
Solutions to Case Problems
INTEGER LINEAR PROGRAMMING PROBLEM
MIN
48PNW+27PSW+112.5PCE+135PNE+60PSE+24SNW+40.5SSW+67.5SCE+108SNE+90SSE+48
ANW+5
4ASW+67.5ACE+81ANE+30ASE+48BNW+81BSW+90BCE+54BNE+45BSE
S.T.
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)
15)
16)
17)
18)
19)
20)
21)
22)
23)
24)
25)
26)
1PNW+1SNW+1ANW+1BNW=1
+1PSW+1SSW+1ASW+1BSW=1
+1PCE+1SCE+1ACE+1BCE=1
+1PNE+1SNE+1ANE+1BNE=1
+1PSE+1SSE+1ASE+1BSE=1
1PNW-1P<0
+1PSW-1P<0
+1PCE-1P<0
+1PNE-1P<0
+1PSE-1P<0
+1SNW-1S<0
+1SSW-1S<0
+1SCE-1S<0
+1SNE-1S<0
+1SSE-1S<0
+1ANW-1A<0
+1ASW-1A<0
+1ACE-1A<0
+1ANE-1A<0
+1ASE-1A<0
+1BNW-1B<0
+1BSW-1B<0
+1BCE-1B<0
+1BNE-1B<0
+1BSE-1B<0
+1P+1S+1A+1B<2
OPTIMAL SOLUTION
Objective Function Value = 231.000
CP - 43
Chapter 7
Variable
-------------PNW
PSW
PCE
PNE
PSE
SNW
SSW
SCE
SNE
SSE
ANW
ASW
ACE
ANE
ASE
BNW
BSW
BCE
BNE
BSE
P
S
A
B
Value
--------------0.000
0.000
0.000
0.000
0.000
1.000
1.000
1.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.000
1.000
0.000
1.000
0.000
1.000
Constraint
-------------1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
Slack/Surplus
--------------0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
1.000
1.000
0.000
0.000
0.000
0.000
0.000
1.000
1.000
1.000
0.000
0.000
0.000
CP - 44
Solutions to Case Problems
With 2 lockbox locations, the lost interest income is $231,000. Resolving with values of 1, 3, and 4 on the
right-hand-side of constraint 26 provides the following:
No. of
Lockboxes
1
2
3
4
Locations
Atlanta
Salt Lake City & Boston
Salt Lake City, Atlanta & Boston
Salt Lake City, Phoenix, Atlanta & Boston
Lost Interest
Income ($)
280,500
231,000
216,000
202,500
From this information, it is clear that if the fixed cost per lockbox location is anywhere between $20,000
and $30,000 two locations (Salt Lake City & Boston) should be chosen. Mr. Wolff can be told that he only
needs to collect information and negotiate on the cost to choose a site in those 2 locations.
Case Problem 3: Production Scheduling with Changeover Costs
A mixed integer programming model can be used advantageously to assist in developing recommendations.
We describe such a model here; it has 48 decision variables and 64 constraints. We show here how to use
Microsoft Excel to formulate and solve the problem. The spreadsheet at the end shows how we set up the
problem and the optimal solution. We describe the model now.
Variables
There are variables for production, inventory, setup, and changeover in each week.
Pi = number of P-Heads produced in week i
Hi = number of H-Heads produced in week i
IPi = number of P-Heads in inventory at end of week i
IHi = number of H-Heads in inventory at end of week i
SPi = 1 if line is setup for P in week i, 0 if setup for H
Changei = 1 if a changeover occurs at the beginning of week i, 0 otherwise
Constraints
There are constraints for production capacity, inventory balance, maintenance of safety stock, and
enforcement of changeovers. Also, Excel requires that you identify the 0-1 (binary) variables in the Solver
dialog box. The constraints as specified in the Excel Solver dialog box are as follows (references are to
cells of the spreadsheet):
B20:C27 ≤ B34:C41
production capacity, or nonnegativity of slack
G20:G27 ≥ H34:H41
G20:G27 ≥ I34:I41
forces Changei to 1 when a changeover occurs
D20:E27 ≥ D6:E13
ending inventory ≥ safety stock
D34:E41 = B6:C13
beg. inv. + production - end inv. = demand
F18:F25 = Bin
setup variables must be binary
CP - 45
Chapter 7
Note that even though the Changei variable must also be integer it is not necessary to require it because
minimization will never let it be any bigger than it has to be. And, the second set of constraints force it to
be ≥ 1 whenever the setup variable changes from 1 to 0 or from 0 to 1.
Objective
We want to minimize total cost which is represented by cell J23 in the spreadsheet. It is the sum of
production cost, inventory cost, and changeover cost.
The Spreadsheet
The first 14 rows of the spreadsheet contain the data for the problem; information on demand, safety stock,
various costs and beginning inventories are given. Cells B20:G27, as shown contain the optimal solution to
the problem. Before solving, those cells were empty.
The spreadsheet formulation and solution are shown.
A
1
2
3
B
C
D
Product Demand
P
H
P
E
4
5
Week
6
7
8
1
2
3
55
55
44
38
38
30
44
35.2
0
9
10
4
5
6
7
8
9
0
45
45
36
35
35
0
48
48
58
57
58
36
36
28.8
28
28
11
12
13
14
F
G
H
I
J
Production Scheduling
Safety Stock
P
H
H
Production Cost
225
310
30.4
24
0
Max Weekly Rate
Changeover Cost
Weekly Inv. Rate
100
500
0.00375
80
500
0.00375
38.4
38.4
46.4
45.6
46.4
Weekly Inv. Cost
Beginning Inv.
0.84375
125
1.1625
143
15
16
17
18
19
Model
Week
P
H
Inv. P
Inv. H
Setup P
Changeover
20
1
18.00
0.00
88.00
105.00
1.00
0.00
Prod. Cost
117374
21
22
23
2
3
4
100.00
100.00
0.00
0.00
0.00
1.40
133.00
189.00
189.00
67.00
37.00
38.40
1.00
1.00
0.00
0.00
0.00
1.00
Inv. Cost
Changeover Cost
Min Total Cost
1280.35125
500
119154.3513
24
25
5
6
0.00
0.00
48.00
56.00
144.00
99.00
38.40
46.40
0.00
0.00
0.00
0.00
26
27
7
8
0.00
0.00
57.20
57.80
63.00
28.00
45.60
46.40
0.00
0.00
0.00
0.00
28
29
30
31
32
33
Inventory Balance
Production Capacity
P
H
100
100
Beginning Inv. + Prod.
- Ending Inv.
P
H
0
55
0
55
34
35
Week
1
2
36
37
38
3
4
5
100
0
0
0
80
80
39
40
6
7
0
0
41
8
0
Changeover Def.
To P if 1
To H if 1
0.00
0.00
0.00
0.00
38
38
Week
1
2
44
0
45
30
0
48
3
4
5
0.00
-1.00
0.00
0.00
1.00
0.00
80
80
45
36
48
58
6
7
0.00
0.00
0.00
0.00
80
35
57
8
0.00
0.00
CP - 46
Solutions to Case Problems
Solution Comments
The spreadsheet contains the optimal solution. The minimum total cost is $119,154.35. The components
of that cost are production: $117,374, inventory: $1280.35 and changeover: $500. From cells F20:F27 we
see that the line will be setup to produce P-Heads in weeks 1-3 and H-Heads in weeks 4-8. Cell G23 shows
that there will be a changeover from producing P-Heads to H-Heads at the beginning of week 4.
By adjusting the data for this problem (e.g. beginning inventories and the various costs) a number of
variations of this problem can be created with the same basic model. Also, one might want to vary the
safety stock requirements and the number of weeks in the planning horizon to create other variations of the
problem.
CP - 47
Chapter 8
Nonlinear Optimization Models
Case Problem: Portfolio Optimization with Transaction Costs
1.
If $41,268.51are spent purchasing the Intermediate-Term Bond fund, and the transaction cost is 1
percent (i.e. one cent on the dollar), then the transaction fee is
.01 ($41, 268.51) = $412.6851
2.
The total dollar amount spent on transaction costs is $1090.311.
3.
After rebalancing, Ms. Delgado has $100,000 - $1090.311 = $98,909.689
4.
To calculate the expected amount in the Intermediate-Term bond fund at year end, figure out the
year-end amount in each scenario and then weight each scenario by a factor of 1/5 = .2. Since Ms.
Delgado started with $51,268.51 after rebalancing, this gives
$51,268.51 + (.2)(.1764)( $51,268.51) + (.2)(.0325)( $51,268.51)
+ (.2)(.0751)( $51,268.51) - (.2)(.0133)( $51,268.51) +(.2)(.0736)( $51,268.51)
= $51,268.51 + $3530.35 = $54,798.86
5.
From the LINGO solution we see that Ms. Delgado can expect an average return of $10,000 on her
investment. After rebalancing (see question 3) Ms. Delgado has $98,909.689 in her portfolio to start
the year, so she is getting a return of greater than 10 percent on the amount of money she starts
with after rebalancing! However, this is not what she wants or expects. Since she starts the year with
$98,909.689 after rebalancing and earns (in expectation) $10,000 she can expect $98,909.689 +
$10,000 = $108,909.689 in her portfolio. However, this does not give a return of 10 percent on her
original portfolio of $100,000 which is what she expected. In order to end up with a return of at least
10 percent on her original portfolio she must have at least $110,000.00 in her portfolio at the end of
the year.
6.
Here is the formulation that provides an expected value at least $110,000 at year end.
MIN = (1/5)*((R1 - RBAR)^2 + (R2 - RBAR)^2 + (R3 - RBAR)^2 + (R4 RBAR)^2 + (R5 - RBAR)^2);
! THE SCENARIO RETURNS;
1.1006*FS + 1.1764*IB + 1.3241*LG + 1.3236*LV + 1.3344*SG + 1.2456*SV =
R1;
1.1312*FS + 1.0325*IB + 1.1871*LG + 1.2061*LV + 1.1940*SG + 1.2532*SV =
R2;
1.1347*FS + 1.0751*IB + 1.3328*LG + 1.1293*LV + 1.0385*SG + (1.0670)*SV = R3;
1.4542*FS + (1 - .0133)*IB + 1.4146*LG + 1.0706*LV + 1.5868*SG +
1.0543*SV = R4;
(1-.2193)*FS + 1.0736*IB + (1- .2326)*LG + (1- .0537)*LV
+ (1.0902)*SG + 1.1731*SV = R5;
! PORTFOLIO AVERAGE RETURN;
(1/5)*(R1 + R2 + R3+ R4 + R5) = RBAR;
! UNITY CONSTRAINT;
FS + IB + LG + LV + SG + SV + TRANS_COST = 100000;
CP - 48
Solutions to Case Problems
RBAR > RMIN;
RMIN = 10000 + 100000;
! DEFINE BUY AND SELL QUANTITIES;
FS = FS_START + FS_BUY - FS_SELL;
IB = IB_START + IB_BUY - IB_SELL;
LG = LG_START + LG_BUY - LG_SELL;
LV = LV_START + LV_BUY - LV_SELL;
SG = SG_START + SG_BUY - SG_SELL;
SV = SV_START + SV_BUY - SV_SELL;
! DEFINE TOTAL TRANSACTION COSTS;
TRANS_COST = TRANS_FEE*(FS_BUY + FS_SELL + IB_BUY + IB_SELL +
LG_BUY + LG_SELL + LV_BUY + LV_SELL + SG_BUY + SG_SELL + SV_BUY +
SV_SELL);
FS_START = 10000;
IB_START = 10000;
LG_START = 10000;
LV_START = 40000;
SG_START = 10000;
SV_START = 20000;
TRANS_FEE = 0.01;
@FREE(R1);
@FREE(R2);
@FREE(R3);
@FREE(R4);
@FREE(R5);
CP - 49
Chapter 6
7.
Here is the solution for the formulation in Question 7.
Local optimal solution found.
Objective value:
Total solver iterations:
Variable
R1
RBAR
R2
R3
R4
R5
FS
IB
LG
LV
SG
SV
TRANS_COST
RMIN
FS_START
FS_BUY
FS_SELL
IB_START
IB_BUY
IB_SELL
LG_START
LG_BUY
LG_SELL
LV_START
LV_BUY
LV_SELL
SG_START
SG_BUY
SG_SELL
SV_START
SV_BUY
SV_SELL
TRANS_FEE
0.3380209E+08
46
Value
119713.7
110000.0
112235.3
105190.3
109673.8
103186.9
10000.00
44769.45
10870.66
0.000000
719.8605
32664.18
975.8443
110000.0
10000.00
0.000000
0.000000
10000.00
34769.45
0.000000
10000.00
870.6613
0.000000
40000.00
0.000000
40000.00
10000.00
0.000000
9280.140
20000.00
12664.18
0.000000
0.1000000E-01
CP - 50
Reduced Cost
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000
327.7706
0.000000
0.000000
0.000000
0.000000
0.000000
41.03809
106.0720
0.000000
0.000000
147.1100
0.000000
0.000000
147.1100
0.000000
147.1100
0.000000
0.000000
147.1100
0.000000
0.000000
0.000000
147.1100
0.000000
Chapter 9
Project Scheduling: PERT/CPM
Case Problem: R.C. Coleman
1.
R.C. Coleman's Project Network
D
A
Start
C
B
E
F
G
Activity
A
B
C
D
E
F
G
H
I
J
K
Activity
A
B
C
D
E
F
G
H
I
J
K
I
Earliest
Start
0
0
9
13
13
23
13
29
29
35
39
K
H
J
Expected Time
6
9
4
12
10
6
8
6
7
4
4
Latest
Start
3
0
9
17
13
23
21
29
32
35
39
Earliest
Finish
6
9
13
25
23
29
21
35
36
39
43
CP - 51
Finish
Variance
0.44
2.78
0.44
7.11
1.00
0.44
7.11
0.44
2.78
0.11
0.44
Latest
Finish
9
9
13
29
23
29
29
35
39
39
43
Slack
3
0
0
4
0
0
8
0
3
0
0
Critical
Activity
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Chapter 9
The expected project completion time is 43 weeks. The critical path activities are B-C-E-F-H-J-K.
The variance of the critical path is 5.67.
z=
40 − 43
5.67
= −1.26
Area = 0.3962
P(T ≤ 40) = 0.5000 - 0.3962 = 0.1038
Given the above calculations, we can conclude that there is about a 10% chance that the project can
be completed in 40 weeks or less. Coleman should consider crashing project activities.
2.
20%
80%
Planned project
completion time
Desires 40-week
completion time
For 80% chance,
z = +0.84
Thus
40 − E (T )
5.67
= 0.84
Solve for E(T) = 38 weeks.
R.C. Coleman should crash activities to reduce the expected project completion time to 38 weeks.
3.
In this section, we will use expected activity times as normal times and use a linear programming
model based on expected times to make the crashing decisions.
Let
xi = the completion time for activity i
yi = the amount of crash time for activity i
CP - 52
Solutions to Case Problems
Min 450yA + 400yB + 600yC + 300yD + 1000yE + 550yF + 750yG + 700yH + 800yI + 400yJ +
500yK
s.t.
xA + y A ≥ 6
xK ≤ 38
xB + yB ≥ 9
yA ≤ 2
xC + yC – xA ≥ 4
yB ≤ 2
xC + yC – xB ≥ 4
yC ≤ 2
xD + yD – xC ≥ 12
yD ≤ 4
xE + yE – xC ≥ 10
yE ≤ 3
xF + yF – xE ≥ 6
yF ≤ 2
xG + yG – xC ≥ 8
yG ≤ 3
xH + yH – xF ≥ 6
yH ≤ 2
xH + yH – xG ≥ 6
yI ≤ 3
xI + y I – xD ≥ 7
yJ ≤ 1
xI + y I – xF ≥ 7
yK ≤ 1
xJ + yJ – xH ≥ 4
All xi,yi ≥ 0
xK + yK – xI ≥ 4
xK + yK – xJ ≥ 4
The optimal crashing decisions are as follows:
Crash Activity
B
F
J
K
Weeks
2
1
1
1
Total
CP - 53
Cost
800
550
400
500
2250
Chapter 9
A revised activity schedule based on these crashing decisions is as follows:
Activity
A
B
C
D
E
F
G
H
I
J
K
Earliest
Start
0
0
7
11
11
21
11
26
26
32
35
Latest
Start
1
0
7
14
11
21
18
26
28
32
35
Earliest
Finish
6
7
11
23
21
26
19
32
33
35
38
Latest
Finish
7
7
11
26
21
26
26
32
35
35
38
Slack
1
0
0
3
0
0
7
0
2
0
0
Critical
Activity
Yes
Yes
Yes
Yes
Yes
Yes
Yes
The student should comment on the fact that the crashing decisions may alter the variance in the
project completion time. By defining revised optimistic, most probable, and pessimistic times for
crashed activities B, F, J, and K, a revised variance in the project completion time can be found.
Using this result, a revised probability of a 40-week completion time can be computed.
CP - 54
Chapter 10
Inventory Models
Case Problem 1: Wagner Fabricating Company
1.
2.
Holding Cost
Cost of capital
Taxes/Insurance (24,000/600,000)
Shrinkage (9,000/600,000)
Warehouse overhead (15,000/600,000)
Annual rate
Ordering Cost
2 hours at $28.00
Other expenses (2,375/125)
Cost per order
3.
14.0%
4.0%
1.5%
2.5%
22.0%
$56.00
19.00
$75.00
Set-up Cost
8 Hours at $50.00
$400 per set-up
4. & 5.
a. Order from Supplier - EOQ model
Ch = IC = 0.22 ($18.00) = $3.96
Q* =
2 DCo
=
Ch
2(3200)75
= 348.16 units
3.96
Number of orders = D/Q = 9.19/year
Cycle time = 250(Q) / D = 250(348.16) / 3200 = 27.2 days
Reorder Point:
P(Stockout) = 1 / 9.19 = 0.1088
Stockout
64
CP - 55
z = 1.24
Chapter 10
r = 64 + 1.24(10) = 76.4
Safety stock = 76.4 - 64 = 12.4
Maximum inventory = Q + 12.4 = 360.56
Average inventory = Q/2 + 12.4 = 186.48
Annual holding cost = 186.48(3.96) = $738.46
Annual ordering cost = 9.19(75) = $689.35
Purchase cost = 3200($18) = $57,600
Total annual cost = $59,027.81
b. Manufacture - Production lot size model
Ch = IC = 0.22($17.00) = $3.74
P = 1000(12) = 12,000/year
Note: The five-month capacity of 5,000 units is sufficient to handle annual demand of 3,200 units.
Q* =
2 DCo
=
(1 − D / P )Ch
2(3200)(400)
= 966.13
(1 − 3200 /1200)3.74
Number of production runs = D/Q = 3.31/year
Cycle Time = 250(Q) / D = 250(966.13) / 3200 = 75.48 days
Reorder point:
P(Stockout) = 1 / 3.31 = 0.3021
Stockout
128
r = 128 + 0.52(20) = 138.4
Safety stock = 138.4 - 128 = 10.4
Maximum inventory = (1 - 3200/12000)966.13 + 10.4 = 718.89
Annual holding cost = (354.25 + 10.4)(3.74) = $1363.79
Annual set up cost = 3.31(400) = $1363.79
Manufacturing cost = 3200($17) = $54,400
Total Annual Cost = $57,088.67
CP - 56
z = 0.52
Solutions to Case Problems
6.
Recommend manufacturing the part
Savings: $59,027.81 - 57,088.67 = $1,939.14 (3.3%)
Case Problem 2: River City Fire Department
1.
During a three-week scheduling period, a unit is scheduled seven days and must have at least 186
firefighters on duty each day. Thus, each unit must provide staffing for
7 × 186 = 1302 firefighter days
During a three-week scheduling period, each firefighter provides six days of service. Thus, the base
number of firefighters per unit must be
1302
= 217 firefighters
6
2.
The single-period inventory model can be used to determine the number of additional firefighters to
be added to the unit to cover daily absences.
Let Q* = minimum cost number of additional firefighters
If Demand < Q*, Q* has overestimated demand.
The cost of overestimating demand is the daily wage rate (d) for a firefighter.
co = d
If Demand > Q*, Q* has underestimated demand and overtime is needed.
The cost of underestimating demand is the overtime wage (1.55d) minus the wage rate (d) that
would have been required if we had hired enough additional firefighters.
cu = 1.55d - d = 0.55d
Using the single period inventory model,
P(Demand < Q*) =
cu
0.55d
0.55
=
=
= 0.3548
cu + co 0.55 + d 1.55
Using the normal distribution, area = 0.5000 - 0.3548 = 0.1452. z = -0.37
Q* = µ + zσ = 20 − 0.37(5) = 18.14
Use 18 additional firefighters.
CP - 57
Chapter 10
3.
From part 2, this is the probability demand will exceed 18.14 is 1 - 0.3548 = 0.6452. This is an
acceptable approximation of the probability overtime will be needed.
A more complete analysis would include the fact that we have rounded the number of firefighters
needed to 18. Then allowing for continuity correction, overtime would only be needed if demand
exceeded 18.5 firefighters. Thus a revised estimate of the probability of overtime is:
z=
18.5 − 20
= −.30 Area = .1179
5
P(Overtime) = .5000 + .1179 = .6179
4.
Firefighters for each unit
217 + 18 additional = 235
Firefighters for the department (3 units)
3 × 235 = 705 firefighters
CP - 58
Chapter 11
Waiting Line Models
Case Problem 1: Regional Airlines
The analysis that follows is based upon the assumptions of Poisson arrivals and exponential
service times. With one call every 3.75 minutes, we have an average arrival rate of
λ = 60/3.75 = 16 calls per hour
Similarly, with an average service time of 3 minutes, we have a service rate of
µ = 60/3 = 20 calls per hour
1.
Current System with no waiting allowed
With 1 reservation agent:
P0 =
(λ / µ )0 / 0!
k
∑ (λ / µ )
i
=
(16 / 20)0 / 0!
= .5556
(16 / 20)0 / 0!+ (16 / 20)1 /1!
=
(16 / 20)1 /1!
= .4444
(16 / 20)0 / 0!+ (16 / 20)1 /1!
/ i!
i =0
P1 =
(λ / µ )1 /1!
k
∑ (λ / µ )
1
/ i!
i =1
The probability a caller will get a busy signal and be blocked is P1 = .4444.
With 2 reservation agents:
P0 =
(λ / µ )0 / 0!
k
∑ (λ / µ )
i
=
(16 / 20)0 / 0!
= .4717
(16 / 20) / 0!+ (16 / 20)1 /1!+ (16 / 20)2 / 2!
=
(16 / 20)1 /1!
= .3774
(16 / 20) / 0!+ (16 / 20)1 /1!+ (16 / 20)2 / 2!
=
(16 / 20) 2 / 2!
= .1509
(16 / 20) / 0!+ (16 / 20)1 /1!+ (16 / 20)2 / 2!
/ i!
0
i =0
P1 =
(λ / µ )1 /1!
k
∑ (λ / µ )
i
/ i!
0
i=0
P2 =
(λ / µ ) 2 / 2!
k
∑ (λ / µ )
i
/ i!
0
i=0
The probability a caller will get a busy signal and be blocked is P2 = .1509.
Conclusion: The current system will answer approximately 85% of the calls immediately if two
agents are employed. But, 15% of callers will receive a busy signal and be turned away. The ones
turned away may not call back.
CP - 59
Chapter 11
2.
Expanded System with waiting allowed
With 1 reservation agent:
 λ
16
= 0.20
P0 = 1 −  = 1 −
µ
20


Lq =
λ2
162
=
= 3.2
µ ( µ − λ ) 20(20 − 16)
λ
16
= 3.2 +
=4
20
µ
L = Lq +
Wq +
Lq
=
λ
1
W = Wq +
Pw =
3.2
= 0.20 hours (12 minutes)
16
µ
= 0.20 +
1
= 0.25 hours (15 minutes)
20
λ 16
=
= 0.80
µ 20
Operating the expanded reservation service with only one reservation agent appears
unacceptable. With 80% of incoming calls waiting (Pw = 0.80) and an average waiting time of
12 minutes (Wq = 12), the company clearly needs to consider using two or more agents.
Since Regional's management team agreed that an acceptable service goal was to immediately
answer and process at least 85% of the incoming calls, the probability of waiting must be 15%
or less. Computing Pw for k = 2 agents and k = 3 agents provides the following.
Pw = 1 λ
k! µ
k
kµ P
0
kµ - λ
For k = 2
Pw = 1 16
2! 20
2
2(20)
(0.4286) = 0.2286
2(20) - 16
3
3(20)
(0.4472) = 0.0520
3(20) - 16
For k = 3
Pw = 1 16
3! 20
Based on the value of Pw, 3 reservation agents will be required to meet the service goal of
answering 85% of the calls answered immediately. Other operating characteristics of the system
with 3 reservation agents are as follows:
P0 = 0.4472
CP - 60
Solutions to Case Problems
Lq = 0.0189
L = 0.8189
Wq = 0.0012 hours = 0.07 minutes
W = 0.0512 hours = 3.07 minutes
3.
The analysis in parts (1) and (2) will provide discussion as to which is the best system. If meeting the
service goal of answering 85% of the calls immediately is the only consideration, the current system
is preferred. Only 2 reservation agents are needed. But, the drawback is that 15% of the callers will
be turned away. The callers will either call back later or, perhaps, become lost business.
If the expanded system is employed, 3 reservation agents will be needed to meet the service goal of
answering 85% of the calls immediately. But, the advantage is that all calls will be answered and the
average waiting time is only .0012 hours, or approximately 4 seconds. An extra $20 per hour will be
required with the third agent, but better overall service is provided and no customer has to call back.
Further analysis of the system with 3 reservation agents will provide additional information for
management. For example, the probability of n units in the system is as follows:
Units in the System
0
P0 = .4472
1
P1 =
(λ / µ )1
(16 / 20)1
P0 =
(0.4472) = 0.3577
1!
1!
2
P2 =
(λ / µ ) 2
(16 / 20) 2
(0.4472) = 0.1431
P0 =
2!
2!
Thus, the probability that there are three or more customers in the system:
1 – (0.4472 + 0.3577 + 0.1431) = 0.0520.
The third agent would be busy when there are three or more customers in the system. The probability
the third agent would be needed is 0.0520. This is equivalent to saying the agent would be utilized
5.2% of the time. During a one hour period, this would be 0.0520(60 minutes) = 3.12 minutes.
Management should be advised that the third agent will be underutilized. Is it worth $20 an hour to
have an agent who needs to be available for 3.12 minutes? On an 8-hour basis, this is a cost of $160
for approximately 25 minutes of work. This analysis suggests that management should reconsider
the system with two reservation agents. The operating characterizes of this system are as follows:
P0 = 0.4286
Lq = 0.1524
L = 0.9524
Wq = 0.0095 hours = 34 seconds
CP - 61
Chapter 11
W = 0.0595 hours = 3.57 minutes
Pw = 0.2286
The operating characteristics of this 2-agent system are still very good. With Pw = 0.2286, 1 - Pw =
0.7714, or approximately 77% of the calling customers will receive immediate service. But this is
only 8% less than the target service goal of 85%. More importantly, the average waiting time is only
34 seconds, which is not unreasonable with today’s telephone systems. This solution saves $20 per
hour, or $160 per day, over the system with the underutilized third agent.
Look for some good discussion of the pros and cons of the alternative systems in this case. Each of
three systems may be considered the best depending upon the criteria used by the analyst. A
recommendation that provides management with the key summary statistics for all three systems and
the advantages of each may be the best possible write-up for this case.
4.
We would need to know the average arrival rate for each hourly period throughout the day. An
analysis similar to the one above would determine the recommended number of reservation agents
each hour. This information could then be used to develop full-time and part-time shift schedules
which would meet the service goals.
Case Problem 2: Office Equipment, Inc.
1.
λ = 1 call/50 hours = 0.02 calls per hour
2.
Mean service time = travel time + repair time = 1 + 1.5 = 2.5 hours
µ = 1 / 2.5 hours = 0.4 customers per hour
3.
The travel time is 1 hour. While this is considered part of the service time it actually means that the
customer will be waiting during the first hour of the service time. Thus, travel time must be added to
the time spent in line as predicted model in order to determine the total customer waiting time.
4.
Using output from The Management Scientist, we have the following:
Probability that no customers are in the system
Average number of customers waiting
Average number of customers in the system
Average time a customer spends in the waiting line
Average time until the machine is back in operation
Probability of a wait more than one hour
Hours a week the technician is not on service calls
(0.5380) x 40 hours = 21.5 hours
Total cost per hour for the service operation
0.5380
0.2972
0.7593
1.6082 hours*
4.1082 hours
0.4620
$155.93
*The average time a customer spends in the waiting line is 1.6082 hours. This is the average time
for the service technician to complete all previous service call commitments and be ready to travel to
the new customer. Since the average travel time is 1 hour for the service technician to reach the new
customer's office, the total customer waiting time is 1.6082 + 1 = 2.6082 hours. Thus, the one
technician is able to meet the company's 3-hour service guideline. The total cost is $155.93 per
hour.
Note that the waiting line model indicates the probability that a customer has to wait is 0.4620.
Since all customers wait an average of 1-hour of travel time whenever the service technician is free,
this probability is actually the probability that a customer will have to wait more than 1-hour for a
service technician to arrive.
CP - 62
Solutions to Case Problems
5.
If the company continues to use one technician when the customer base expands to 20 customers, the
average time in the waiting line will increase to 6.9454 hours. With an average travel time of 1
hour, the average total waiting time will be 6.9454 + 1 = 7.9454 hours. The total cost will be
$397.78 per hour. This average total waiting time is too long and a second technician is definitely
necessary. Using output from The Management Scientist, two service technicians provide the
following:
Probability that no customers are in the system
0.3525
Average number of customers in the waiting line
0.2104
Average number of customers in the system
1.1527
Average time a customer spends in the waiting line
0.5581 hours*
Average time until the machine is back in operation
3.0581 hours
Probability of a wait more than one hour
0.2949
Hours a week the technicians are not on service calls
P0 = 0.3525 (0.3525) x 2 technicians x 40 hours = 28.2 hours
P1 = 0.3525 (0.3525) x 1 technician x 40 hours = 14.1 hours
Total = 42.3 hours
Total cost per hour of service operation
$275.27
*The average time a customer spends in the waiting line is 0.5581 hours. This is the average time
for the service technician to complete all previous service call commitments and be ready to travel to
the new customer. Since the average travel time is 1-hour for the service technician to reach the new
customer's office, the total customer waiting time is 0.5581 + 1 = 1.5581 hours. Thus, two
technicians are needed to meet the company's 3-hour service guideline when the company reaches 20
customers. The total cost is $275.27 per hour.
6.
A comparison of two and three technicians with 30 customers shows that the average total waiting
time with two technicians will be 2.6895 hours and the average total waiting time with three
technicians will be 1.2626 hours. The hourly cost with two technicians is $391.94 and the hourly
cost with three technicians is $397.08. While three technicians provide a smaller waiting time, two
technicians are able to meet the 3-hour service guideline for a total lower cost. Thus, the company
should continue to use two technicians when the customer base expands to 30 customers. Using
output from The Management Scientist, two service technicians provide the following:
Probability that no customers are in the system
0.1760
Average number of customers in the waiting line
0.9353
Average number of customers in the system
2.3194
Average time a customer spends in the waiting line
1.6895 hours*
Average time until the machine is back in operation
4.1895 hours
Probability of a wait more than one hour
0.5600
Hours a week the technicians are not on service calls
P0 = 0.1760 (0.1760) x 2 technicians x 40 hours = 14.08 hours
P1 = 0.2640 (0.2640) x 1 technician x 40 hours = 10.56 hours
Total = 24.64 hours
Total cost per hour of service operation
$391.94
*The average time a customer spends in the waiting line is 1.6895 hours. While the average travel
time is 1-hour for the service technician to reach the new customer's office, the average total
customer waiting time is 1.6895 + 1 = 2.6895 hours.
7.
The OEI planning committee’s proposal anticipated that three technicians would be needed at a total
cost of $397.08 per hour. Thus, the recommendation to stay with two technicians has as annual
savings of (397.08 – 391.94) x 8 hours/day x 250 days/year = $10,280.
CP - 63
Chapter 12
Simulation
Case Problem 1: Tri-State Corporation
With the specific financial analysis data input into cells D3:D8, the formulas used to develop the portfolio
projection worksheet are shown below. The rows are copied to extend the worksheet to the desired 30-year
projection.
A
10
11
12
13
14
15
16
Year
1
2
3
4
5
1.
B
Age
=D3
=B12+1
=B13+1
=B14+1
=B15+1
C
D
Beginning
Portfolio
=D5
=G12
=G13
=G14
=G15
E
Salary
=D4
=(1+$D$6)*D12
=(1+$D$6)*D13
=(1+$D$6)*D14
=(1+$D$6)*D15
New
Investment
=$D$7*D12
=$D$7*D13
=$D$7*D14
=$D$7*D15
=$D$7*D16
F
Portfolio
Earnings
=$D$8*(C12+0.5*E12)
=$D$8*(C13+0.5*E13)
=$D$8*(C14+0.5*E14)
=$D$8*(C15+0.5*E15)
=$D$8*(C16+0.5*E16)
G
Ending
Portfolio
=C12+E12+F12
=C13+E13+F13
=C14+E14+F14
=C15+E15+F15
=C16+E16+F16
Increasing the annual investment rate in cell D7 will generate the following 30-year portfolio
projections:
Projected Portfolio
$ 721,667
815,397
909,127
1,002,857
Rate
5%
6%
7%
8%
A 1% increase in the annual investment rate increases the projected 30-year portfolio by $93,730. The
annual investment rate would have to be increased to 8% in order to achieve the $1,000,000 goal.
2.
The simulation worksheet that we developed placed the simulated annual salary growth rate in column
D and the simulated annual portfolio growth rate in column G. The revised data input section and the
cell formulas used to develop the simulation worksheet are as follows.
Data Input
A
1
2
3
4
5
6
7
8
9
10
11
12
B
C
D
Financial Analysis - Portfolio Projection
Age
Current Salary
Current Portfolio
Annual Salary Growth Rate
Minimum Rate
Maximum Rate
Annual Investment Rate
Annual Portfolio Growth Rate
Mean Rate
Standard Deviation
CP - 64
25
$34,000
$14,500
0%
10%
8%
10%
5%
Solutions to Case Problems
Formula Worksheet – Columns D to I
14
15
16
17
18
19
20
D
Salary
Growth %
E
=RAND()*($D$8-$D$7)
=RAND()*($D$8-$D$7)
=RAND()*($D$8-$D$7)
=RAND()*($D$8-$D$7)
Salary
=D4
=(1+D17)*E16
=(1+D18)*E17
=(1+D19)*E18
=(1+D20)*E19
F
New
Investment
=$D$9*E16
=$D$9*E17
=$D$9*E18
=$D$9*E19
=$D$9*E20
G
Portfolio
Growth %
=NORMINV(RAND(),$D$11,$D$12)
=NORMINV(RAND(),$D$11,$D$12)
=NORMINV(RAND(),$D$11,$D$12)
=NORMINV(RAND(),$D$11,$D$12)
=NORMINV(RAND(),$D$11,$D$12)
H
Portfolio
Earnings
=G16*(C16+0.5*F16)
=G17*(C17+0.5*F17)
=G18*(C18+0.5*F18)
=G19*(C19+0.5*F19)
=G20*(C20+0.5*F20)
I
Ending
Portfolio
=C16+F16+H16
=C17+F17+H17
=C18+F18+H18
=C19+F19+H19
=C20+F20+H20
The simulated 30-year portfolio amounts will vary considerably from trial to trial. The uncertainty
associated with the annual salary growth rate and the annual portfolio growth rate will show that there
is uncertainty in reaching the $1,000,000 even if the annual investment rate is increased to 8%.
A few simulation trials will point out that the 30-year portfolio variability and the uncertainty of
reaching the $1,000,000 goal. However, repeating the simulation numerous times will be necessary to
provide an objective basis for estimating the probability of reaching $1,000,000.
We preformed the simulation of 1000 trials. The probability of reaching $1,000,000 was estimated to
be 0.48. Thus, the simulation conclusion is that there is less than a 50% chance of reaching $1,000,000
even if an 8% investment rate is used. During the 1000 trials, the 30-year portfolio varied from
$550,000 to $1,900,000. There was a 30% chance that the portfolio would not reach $900,000.
3.
The simulation model suggests additional strategies should be considered to obtain a reasonably high
probability of reaching the $1,000,000 portfolio goal. Increasing the annual investment rate to 9% or
10% may be worth considering. If this rate is getting too high, then extending the 30-year period by
adding years may be the best strategy for reaching the $1,000,000 goal.
4.
The longer 35-year period is definitely a good idea. Expanding the simulation spreadsheet by five
years will show that almost every simulated 35-year portfolio exceeded $1,000,000. 1000 simulated
trials with the 35-year period showed better than a 99% chance that the portfolio would exceed
$1,000,000.
In fact, the extra five years increased the expected portfolio from $1,002,857 to $1,697,622, or almost
$700,000. This result demonstrates the long-term advantages of investing.
5.
The simulation worksheet can be used for any employee. The employee may enter his/her age, current
salary, current portfolio, and any assumptions he/she cares to make about salary growth rate,
investment rate, and portfolio growth rate. These assumptions in cells D3:D8 would be different for
each employee and rows would be added to the worksheet to reflect the number of years appropriate
for the employee.
By varying the inputs and projecting the ultimate portfolio value, the employees should be able to
make investment observations such as the following:
•
•
•
•
Begin early with an investment program. The more years the better the ending portfolio value.
Make new contributions to an investment program at the highest possible rate. Increasing the rate
1% or 2% will have a significant impact on the long-term portfolio.
Resist the temptation to use assets in investment programs for short-term personal expenditures.
Early withdraws can be shown to have a major impact the long-term portfolio value.
Be patient. Benefits of investment programs may not be significant for the first 5 to 10 years. It is
really in the later years where a consistent investment program pays its biggest dividends.
CP - 65
Chapter 12
Case Problem 2: Harbor Dunes Golf Course
The Crystal Ball simulation worksheet that we used is as follows:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
A
B
C
Harbor Dunes Golf Course Simulation - Option 1
Available Tee Times
Regular Green Fee
Replay Green Fee
Cart Fee
D
E
20
$160
$25
$20
Number of
Advance
Reservations
8
9
10
11
12
13
14
15
16
17
18
19
20
Probability
0.01
0.04
0.06
0.08
0.10
0.11
0.12
0.15
0.10
0.09
0.07
0.05
0.02
B
C
Option 1: $25 Green Fee + Cart Fee
Number of
Replay Requests
Probability
0
0.01
1
0.03
2
0.05
3
0.05
4
0.11
5
0.15
6
0.17
7
0.15
8
0.13
9
0.09
10
0.06
Cell Formulas:
25
26
27
28
29
30
31
32
33
34
35
36
37
1.
A
Crystal Ball Model
D
E
Number of Advance Reservations
Number of Times Available
Number of Requests for a Replay
Number of Replays
Crystal Ball Custom Distribution with Data B11:C23
=B3-C27
Crystal Ball Custom Distribution with Data D11:E21
=MIN(C28,C29)
Revenue from Advance Reservations
Revenue from Replays
=4*(B4+B6)*C27
=4*(B5+B6)*C30
Total Revenue
Forecast
Cell
=C32+C33
Selected statistical results:
CP - 66
Solutions to Case Problems
Simulation results for various. Below is one set of possible results:
Statistics
Mean
Median
Standard Deviation
Minimum
Maximum
Option 1
$11,062
$11,160
$ 1,802
$ 5,940
$14,400
Option 2
$11,173
$11,360
$ 1,838
$ 5,760
$14,400
Using the mean, the options are similar, but Option 2 is preferred with a daily revenue advantage of
$11,173 - $11,062 = $111.
2.
Go with Option 2: The $50 per replay option.
3.
Without the replay option, Harbor Dunes reported $10,240 daily revenue. Thus the Option 2 replay
policy is estimated to generate an additional revenue of $11,173 - $10,240 = $933 per day. Over a 90day spring season, the estimated revenue increase is 90 * $933 = $83,970. The replay policy is
definitely worthwhile.
4.
One suggestion is that Harbor Dunes wait until mid-morning to determine the afternoon replay option.
If by mid-morning, the afternoon tee time reservations are relatively low, Harbor Dunes may want to
offer the Option 1 replay policy which generates more demand for the afternoon tee times. However, if
by mid-morning, the afternoon tee time reservations were relatively high, Harbor Dunes would want to
continue the Option 2 replay policy. Simulation runs could help determine a level of advance afternoon
tee time reservations that would indicate whether to implement the Option 1 or Option 2 replay policy.
CP - 67
Chapter 12
Case Problem 3: County Beverage Drive-Thru
Excel worksheets patterned after those used for the ATM one- and two- channel simulations in Figure
15.15 and 15.17 may be used to solve this case problem. We recommend using one workbook with three
worksheets, one for each of the following Drive-Thru designs: 1 channel with 1 clerk, 1 channel with 2
clerks and 2 channels with 2 clerks.
The 1 Channel with 1 Clerk Design
Use the format of the Hammondsport 1 ATM simulation model and set up the data for the 1 channel with 1
clerk design as follows:
A
1
B
C
D
County Beverage Drive-Thru with One Channel One Clerk
2
3
4
Interarrival times (Exponential Distribution)
Mean
6
5
6
7
8
9
10
11
12
13
14
15
16
Service Time Distribution
Lower
Random No.
0.00
0.24
0.44
0.59
0.73
0.85
0.93
0.98
Upper
Random No.
0.24
0.44
0.59
0.73
0.85
0.93
0.98
1.00
Service
Time
2
3
4
5
6
7
8
9
The interarrival time for customer 1 would be provided by the cell formula
=-$B$4*LN(RAND())
The service time for customer 1 would be provided by the cell formula
=VLOOKUP(RAND(),$A$9:$C$16,3)
The summary statistics formulas should be provided at the end of the simulation data. The =COUNTIF
function can be used to count the number of customers who wait more than 6 minutes and more than 10
minutes.
CP - 68
Solutions to Case Problems
The 1 Channel with 2 Clerks Design
Use the format of the Hammondsport 1 ATM simulation model and set up the data for the 1 channel with 2
clerks design as follows:
A
1
B
C
D
County Beverage Drive-Thru with One Channel Two Clerks
2
3
4
Interarrival times (Exponential Distribution)
Mean
6
5
6
7
8
9
10
11
12
13
Service Time Distribution
Lower
Random No.
0.00
0.20
0.55
0.85
0.95
Upper
Random No.
0.20
0.55
0.85
0.95
1.00
Service
Time
1
2
3
4
5
The service time for customer 1 would be provided by the cell formula
=VLOOKUP(RAND(),$A$9:$C$13,3)
The summary statistics formulas should be provided at the end of the simulation data. The =COUNTIF
function can be used to count the number of customers who wait more than 6 minutes and more than 10
minutes.
The 2 Channels with 2 Clerks Design
Use the format of the Hammondsport 2 ATMs simulation model and set up the data as previously shown
for the 1 channel with 1 clerk design. Customer interarrival times and service times are the same as shown
for the 1 channel with 1 clerk design.
This part of the case is optional in that significant spreadsheet modeling skills are required to duplicate the
2-channel simulation model. Selected cell formulas area as follows:
Cell I16
Cell J16
Cell I17
Cell J17
=
=
=
=
G16
0
IF(I16=MIN(I16,J16),G17,I16)
IF(J16=MIN(I16,J16),G17,J16)
Cells I17 and J17 can be copied to fill columns I and J.
The summary statistics formulas should be provided at the end of the simulation data. The =COUNTIF
function can be used to count the number of customers who wait more than 6 minutes and more than 10
minutes.
Each design was tested with a 1000 customer simulation run. Data on the first 100 customers was
discarded and summary statistics collected for a total of 900 customers.
CP - 69
Chapter 12
Simulation results will vary but the approximate results are as follows:
Characteristic
Number Waiting
Probability of Waiting
Average Waiting Time
Maximum Waiting Time
Utilization of Drive Thru
Number Waiting > 6 Minutes
Probability Waiting > 6 Minutes
Number Waiting > 10 Minutes
Probability Waiting > 10 Minutes
1 Channel 1 Clerk
640
0.71
6.1
37.8
0.72
322
0.36
191
0.21
1 Channel 2 Clerks
377
0.42
1.0
11.7
0.42
26
0.03
4
0.01
2 Channels 2 Clerks
167
0.19
0.4
9.3
0.36
7
0.01
1
0.00
The 1 channel with 1 clerk system appears unacceptable. The mean waited time is over 6 minutes which
exceeds the company guideline of 1.5 minutes. In addition, over 300 customers waiting over 6 minutes and
almost 200 customers waited over 10 minutes. The company must do something to improve the service
characteristics of its drive-thru operation or face the loss of substantial business.
The 1 channel with 2 clerks system appears to be the best design., The mean waiting time of approximately
1 minute is with the company’s guideline of 1.5 minutes. Relatively few customers experienced the 6 to 10
minute waiting times. The performance of the 2 channel with 2 clerks system is the best overall, but the
added cost may not justify the expansion to the two channel operation.
CP - 70
Chapter 13
Decision Analysis
Case Problem 1: Property Purchase Strategy
The decision tree for the Oceanview decision problem is shown below. Note that the final outcome of
whether the zoning change is approved or rejected by the voters occurs of Oceanview decides to bid and
then wins the bid. Otherwise, the outcome of the vote on the zoning change does not enter into the
problem.
Payoff*
Highest Bid
Bid
Predicts
Approval
Zoning
Approved
1,985,000
Zoning
Rejected
-515,000
9
6
Not Highest Bid
3
-15,000
Do Not Bid
Market
Research
-15,000
Highest Bid
2
Bid
Predicts
Rejection
Zoning
Approved
1,985,000
Zoning
Rejected
-515,000
10
7
Not Highest Bid
4
-15,000
Do Not Bid
1
-15,000
Highest Bid
Bid
No M arket
Research
Zoning
Approved
2,000,000
Zoning
Rejected
-500,000
11
8
Not Highest Bid
5
Do Not Bid
0
0
* Payoff values are the profit computed from the given revenue, property cost, construction expenses,
property cost deposit and marketing research cost given in the problem.
CP - 71
Chapter 13
The decision tree with branch probabilities and expected values is as follows:
Approved
.6585
Highest Bid
.2
EV = 1,131,250
Rejected
.3415
Bid
Predicts
Approval
.41
9
6
Not Highest Bid
.8
3
-15,000
Approved
.0508
2
Highest Bid
.2
EV = 78,993
10
7
-15,000
Do Not Bid
1
-15,000
Approved
.3
Highest Bid
.2
No M arket
Research
11
8
-500,000
EV = 50,000
Not Highest Bid
.8
5
2,000,000
EV = 250,000
Rejected
.7
Bid
-515,000
EV = -89,600
Not Highest Bid
.8
4
1,985,000
EV = -388,000
Rejected
.9492
Bid
Predicts
Rejection
.59
-515,000
EV = 214,250
Do Not Bid
Market
Research
1,985,000
Do Not Bid
0
0
If the market research is not conducted, Oceanview should go ahead and bid. The expected value of this
action is $50,000 as shown at node 8 of the decision tree. However, the analysis shows that the
recommended strategy is to conduct the market research first. the value of the market research exceeds its
cost. The optimal strategy is as follows:
If the market research predicts approval of the zoning change, bid.
If the market research predicts rejection of the zoning change, do not bid.
The expected value of this strategy, including the cost of conducting the market research, is $78,993. The
market research has a value of $78,993 - $50,000 = $28,993 over and above the cost of the research only.
CP - 72
Solutions to Case Problems
Case Problem 2: Lawsuit Defense Strategy
1. Decision Tree
Allied Accepts John's
Offer of $750,000
$750
John Accepts
Allied's Counteroffer
0.10
$400
Allied Loses
Damages of $1,500,000
0.3
$1500
1
Allied Counteroffers
with $400,000
John Rejects Allied's
Allied Loses Damages
Counteroffer
of $750,000
2
3
0.4
0.5
Allied W ins
No Damages
0.2
Allied Accepts
John's Counteroffer
$750
$0
$600
Allied Loses
Damages of $1,500,000
0.3
John Counteroffers
with $600,000
4
0.5
Allied Rejects
John's Counteroffer
Allied Loses Damages
of $750,000
5
0.5
Allied W ins
No Damages
0.20
To find the optimal decision strategy, we must fold the tree back computing expected values at the chance
nodes and choosing the least cost alternative at the decision nodes. Shown below are the values computed
at each node.
Node
1
2
3
4
5
Value
670
670
825
600
825
CP - 73
$1500
$750
$0
Chapter 13
Shown below is the decision tree with all nonoptimal branches eliminated.
John Accepts
Allied's Counteroffer
0.10
$400
Allied Loses
Damages of $1,500,000
0.3
$1500
1
Allied Counteroffers
with $400,000
Allied Loses Damages
John Rejects Allied's
of $750,000
Counteroffer
2
3
0.4
0.5
Allied W ins
No Damages
0.2
Allied Accepts
John's Counteroffer
$750
$0
$600
John Counteroffers
with $600,000
4
0.5
2.
Allied should not accept John's offer to settle for $750,000. The strategy associated with a
counteroffer of $400,000 has an expected value of $670,000.
3.
If John accepts Allied's counteroffer of $400,000, no further action is required. If John rejects
Allied's counteroffer and elects to have a jury decide the settlement amount, Allied must prepare for
a trial. If John counteroffers with $600,000, Allied should accept John's counteroffer.
4.
To develop the risk profile for the optimal strategy, we simply multiple the branch probabilities on
all paths to the end points of the decision tree. The risk profile is given below in tabular form.
Settlement Amount
($1000s)
0
400
600
750
1500
Probability
0.08
0.10
0.50
0.20
0.12
1.00
CP - 74
Chapter 14
Multicriteria Decision Problems
Case Problem: EZ Trailers, Inc.
Let
x11
x12
x21
x22
s11
s12
s21
s22
=
=
=
=
=
=
=
=
number of EZ-190 trailers produced in March
number of EZ-190 trailers produced in April
number of EZ-250 trailers produced in March
numb r of EZ-250 trailers produced in April
EZ-190 ending inventory in March
EZ-190 ending inventory in April
EZ-250 ending inventory in March
EZ-250 ending inventory in April
P1 Goal: Meet demand for the EZ-250:
March:
April:
300 + x21 - s21 - d1+ + d1− = 1000
(1)
s21 + x22 - s22 - d 2+ + d 2− = 1200
(2)
P2 Goal: Meet demand for the EZ-190:
March:
April:
200 + x11 - s11 - d 3+ + d 3− = 800
(3)
s11 + x12 - s12 - d 4+ + d 4− = 600
(4)
P3 Goal: Limit labor fluctuations from month to month to at most 1000:
March:
5300 ≤ 4 x11 + 6 x21 ≤ 7300
4 x11 + 6 x21 - d 5+ + d 5− = 5300
+
6
(5)
−
6
4 x11 + 6 x21 - d + d = 7300
(6)
April:
(4 x11 + 6 x21) - 1000 ≤ 4 x12 + 6 x22 ≤ (4 x11 + 6 x21) + 1000
4 x12 + 6 x22 = [(4 x11+ 6 x21) - 1000] + d 7+ + d 7−
+
7
−
7
4 x12 + 6 x22 - 4 x11 - 6 x21 - d + d = -1000
or
(7)
4 x12 + 6 x22 = [(4 x11 + 6 x21) + 1000] + d 8+ + d8−
or
4 x12 + 6 x22 - 4 x11 - 6 x21 - d 8+ + d8− = 1000
CP - 75
(8)
Chapter 14
The complete goal programming model is
( )
( )
( )
( )
( )
( )
( )
( )
Min P1 d1− + P1 d 2− + P2 d3− + P2 d 4− + P3 d5− + P3 d 6+ + P3 d 7− + P3 d8+
s.t.
x 21 – s21 – d1+ + d –1 = 800
+
–
x22 + s 21 – s 22 – d 2 + d 2 = 1200
+
–
+
–
x11 – s 11 – d3 + d 3 = 600
x12 + s 11 – s 12 – d4 + d 4 = 600
+
–
+
–
4 x11 + 6 x21 – d 5 + d 5 = 5300
4 x11 + 6 x21 – d 6 + d 6 = 7300
+
–
–4 x11 + 4 x12 – 6 x21 +6x22 – d 7 + d 7 = – 1000
–4 x11 + 4 x12 – 6 x21 +6x 22 – d +8 + d –8 = 1000
all variables ≥ 0
Information Needed for the Managerial Report
1.
EZ-190
EZ-250
2.
March
775
800
No changes since the ending inventories for the optimal production schedule are as follows:
EZ-190
EZ-250
3.
April
425
1200
March
175
0
April
0
0
The following constraints must be added to the model:
April ending inventory: s12 ≥ 100, s22 ≥ 100
Maximum storage of 300 units in each month: s11 ≤ 300, s12 ≤ 300, s21 ≤ 300, s22 ≤ 300
The new optimal production schedule is as follows:
EZ-190
EZ-250
March
900
800
April
400
1300
March
300
0
April
100
100
The corresponding ending inventories are:
EZ-190
EZ-250
CP - 76
Solutions to Case Problems
4.
The new optimal production schedule is:
EZ-190
EZ-250
March
625
800
April
275
1200
March
25
0
April
0
0
The corresponding ending inventories are:
EZ-190
EZ-250
CP - 77
Chapter 15
Forecasting
Case Problem 1: Forecasting Sales
1.
Month 1 corresponds to January for year 1, month 2 corresponds to February for year 1, and so on. A
graph of the time series is shown below:
300
250
Sales
200
150
100
50
0
5
10
15
20
25
Month
2.
Analysis of seasonality:
Month
January
February
March
April
May
June
July
August
September
October
November
December
Seasonal-Irregular
Component Values
1.445
1.441
1.301
1.297
1.344
1.343
1.047
1.034
1.044
1.054
.779
.801
.882
.834
.857
.848
.618
.638
.725
.675
.843
.862
1.137
1.180
Seasonal Index
1.44
1.30
1.34
1.04
1.05
.80
.83
.85
.63
.70
.85
1.16
CP - 78
30
35
40
Chapter 15
The deseasonalized time series is shown below:
t
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
Deseasonalized Sales
168.06
180.77
173.13
171.15
175.24
175.00
174.70
178.82
174.60
185.71
178.82
177.59
182.64
183.08
184.33
185.58
183.81
186.25
Deseasonalized Sales
189.16
189.41
193.65
185.71
196.47
198.28
195.83
196.15
197.76
197.12
200.00
200.00
200.00
204.71
200.00
211.43
203.53
202.59
t
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
The trend line fitted to the deseasonalized time series is
T t = 169.499 + 1.02 t
3.
Sales forecasts
Forecast for Year 4
Using T t = 169.499 + 1.02 t
Month
January
February
March
April
May
June
July
August
September
October
November
December
4.
Trend
Forecast
207.239
208.259
209.279
210.299
211.319
212.339
213.359
214.379
215.399
216.419
217.439
218.459
Seasonal
Index
1.44
1.30
1.34
1.04
1.05
.80
.83
.85
.63
.70
.85
1.16
Monthly
Forecast
298.424
270.737
280.434
218.711
221.885
169.871
177.088
182.222
135.701
151.493
184.823
253.194
Forecast error = $295,000 - $298,424 = -$3,424
The forecast we developed over predicted by $3,424; this represents a very small error.
5.
The analysis can be easily updated each month, especially if a computer software package is
used to perform the analysis.
CP - 79
Solutions to Case Problems
Case Problem 2: Forecasting Lost Sales
1.
The data used for the forecast is the Carlson sales data for the 48 months preceding the storm. Using
the trend and seasonal method, the seasonal indexes and forecasts of sales assuming the hurricane
had not occurred are as follows:
Month
January
February
March
April
May
June
July
August
September
October
November
December
2.
Month
September
October
November
December
Forecast ($ million)
2.16
2.54
3.06
4.60
The data used for this forecast is the total sales for the 48 months preceding the storm for all
department sores in the county. Using the trend and seasonal method, the seasonal indexes and
forecasts of county-wide department store sales assuming the hurricane had not occurred are as
follows:
Month
January
February
March
April
May
June
July
August
September
October
November
December
3.
Seasonal Index
0.957
0.819
0.907
0.929
1.011
0.937
0.936
0.974
0.797
0.936
1.119
1.677
Seasonal Index
0.773
0.813
0.976
0.935
0.989
0.924
0.901
1.017
0.861
0.907
1.141
1.763
Month
September
October
November
December
Forecast ($ million)
50.55
53.20
66.78
103.11
By comparing the forecast of county-wide department store sales with actual sales, one can determine
whether or not there are excess storm-related sales. We have computed a "lift factor" as the ratio of
actual sales to forecast sales as a measure of the magnitude of excess sales.
Forecast Sales ($ million)
50.55
53.20
66.78
103.11
273.64
Actual Sales ($ million)
69.0
75.0
85.2
121.8
351.0
Lift Factor
1.365
1.410
1.276
1.181
1.283
From the analysis a strong case can be made for excess storm related sales. For each month, actual
sales exceed the forecast of what sales would have been without the hurricane. For the 4-month
total, actual sales exceeded the forecast by 28.3%.
CP - 80
Chapter 15
The explanation for the increase is that people had to replace real and personal property damaged by
the storm. In addition, the additional construction workers, the disaster relief teams, and so on,
created additional commercial activity in the area.
4.
One approach would be to use the forecast of what sales would have been without the hurricane and
then multiply by the lift factor to account for the excess storm-related sales. Such an estimate of lost
sales is developed below:
Forecast ($ million)
Lift Factor
2.16
2.54
3.06
4.60
1.365
1.410
1.276
1.181
Lost Sales ($ million)
Total
2.948
3.581
3.905
5.433
15.867
Based on this analysis, Carlson Department Stores can make a case to the insurance company for a
business interruption claim of $15,867,000.
Another approach would be to use the 48 months of historical data to compute a market share for
Carlson. That is, compute Carlson’s sales as a fraction of county-wide department store sales. Then
you could develop a forecast of Carlson’s market share for September through December. Finally,
an estimate of lost sales for each of the four months can be obtained by multiplying the forecasts of
market share by the actual department store sales.
CP - 81
Chapter 16
Markov Processes
Case Problem: Dealer’s Absorbing State Probabilities in Black Jack
1.
The first step is to create the transition probability matrix for the dealer’s hand. The states are
defined by the value of the dealer’s hand after the up card is dealt, and the values the dealer’s hand
may assume after the down card is revealed, and after taking hits. The absorbing states are 17, 18,
19, 20, 21, and bust. According to the house rules, once the dealer’s hand takes on one of these
values she quits taking hits and either pays or collects the bets that are on the table. The states S12,
S13, S14, S15, and S16 represent what are called soft hands; they include an ace that may be played
as 1 or 11. There are no S17, S18, S19, S20, and S21 states because the dealer plays these the same
as hard hands of the same value.
There are 27 states so, due to space restrictions, the transition matrix that is shown below is printed
in 3 parts. All of the rows and a subset of the columns are shown in each part.
A
S12
S13
S14
S15
S16
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Bust
A
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
S12
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
S13
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
S14
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
S15
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
S16
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
CP - 82
2
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
3
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
4
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
5
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
6
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
Solutions to Case Problems
A
S12
S13
S14
S15
S16
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Bust
7
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
8
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
9
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
10
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
11
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
12
13
14
15
16
0.3077
0.0769
0.0769
0.0769
0.0769
0.3077
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
0.3077
0.0769
0.0769
0.0769
0.0000
0.3077
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0
0
0
0
0
0
0.3077
0.0769
0.0769
0.0000
0.0000
0.3077
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0
0
0
0
0
0
0.3077
0.0769
0.0000
0.0000
0.0000
0.3077
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0
0
0
0
0
0
0.3077
0.0000
0.0000
0.0000
0.0000
0.3077
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0
0
0
0
0
0
CP - 83
17
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0769
0.3077
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
1
0
0
0
0
0
Chapter 16
A
S12
S13
S14
S15
S16
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Bust
18
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.3077
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0
1
0
0
0
0
19
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.3077
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0
0
1
0
0
0
20
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.3077
0.0769
0.0769
0.0769
0.0769
0.0769
0.0769
0
0
0
1
0
0
21
0.3077
0.0769
0.0769
0.0769
0.0769
0.0769
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0769
0.3077
0.0769
0.0769
0.0769
0.0769
0.0769
0
0
0
0
1
0
Bust
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.3077
0.3846
0.4615
0.5385
0.6154
0
0
0
0
0
1
CP - 84
Solutions to Case Problems
The Q matrix is defined as the first 21 rows (through state 16) and the first 21 columns (through state
16) of the transition matrix. We must now compute I – Q and (I – Q)-1. We did this using Excel and
Excel’s matrix inversion procedure described in the chapter appendix. To find the absorption
probabilities we multiply (I – Q)-1 times the R matrix. The R matrix is a 21 x 6 matrix. It is
identified as the first 21 rows and the last 6 columns of the transition matrix. We show (I – Q)-1R
below.
Probability of Dealer’s Finishing State Given Starting State
A
S12
S13
S14
S15
S16
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
0.1308
0.1510
0.1455
0.1400
0.1346
0.1292
0.1398
0.1350
0.1305
0.1223
0.1654
0.3686
0.1286
0.1200
0.1114
0.1114
0.1035
0.0961
0.0892
0.0828
0.0769
18
0.1308
0.1510
0.1455
0.1400
0.1346
0.1292
0.1349
0.1305
0.1259
0.1223
0.1063
0.1378
0.3593
0.1200
0.1114
0.1114
0.1035
0.0961
0.0892
0.0828
0.0769
19
0.1308
0.1510
0.1455
0.1400
0.1346
0.1292
0.1297
0.1256
0.1214
0.1177
0.1063
0.0786
0.1286
0.3508
0.1114
0.1114
0.1035
0.0961
0.0892
0.0828
0.0769
20
0.1308
0.1510
0.1455
0.1400
0.1346
0.1292
0.1240
0.1203
0.1165
0.1131
0.1017
0.0786
0.0694
0.1200
0.3422
0.1114
0.1035
0.0961
0.0892
0.0828
0.0769
21
0.3616
0.1510
0.1455
0.1400
0.1346
0.1292
0.1180
0.1147
0.1112
0.1082
0.0972
0.0741
0.0694
0.0608
0.1114
0.3422
0.1035
0.0961
0.0892
0.0828
0.0769
Bust
0.1153
0.2450
0.2725
0.3000
0.3272
0.3541
0.3536
0.3739
0.3945
0.4164
0.4232
0.2623
0.2447
0.2284
0.2121
0.2121
0.4827
0.5196
0.5539
0.5858
0.6154
This matrix shows the probabilities for the ending values of the dealer’s hand given any of the
starting states identified by each row. For instance, if the dealer has a 6 as an up card, the
probability of finishing with 17 is .1654, the probability of finishing with 18 is .1063 and so on. The
probability of the dealer busting is .4232. For this reason, black jack players recommend not taking
a hit with a hand of 12 or better. Why should the player take a chance of busting when the dealer
has a high probability of busting?
CP - 85
Chapter 16
2.
For this situation another row and column must be added to the transition matrix to account for the
soft 17 state, S17. When the dealer has to stay on S17, there was no reason to differentiate soft 17
and hard 17. So only the state, 17 was used. For space reasons, we do not show the new transition
matrix here. But, after computing (I – Q)-1R, the absorbing state probabilities are given below.
Probability of Dealer's Finishing State Given Starting State
A
S12
S13
S14
S15
S16
S17
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
0.0575
0.0829
0.0823
0.0813
0.0801
0.0786
0.3422
0.1301
0.1263
0.1224
0.1184
0.1148
0.3686
0.1286
0.1200
0.1114
0.1114
0.1035
0.0961
0.0892
0.0828
0.0769
18
0.1432
0.1625
0.1562
0.1500
0.1438
0.1377
0.1114
0.1365
0.1320
0.1273
0.1229
0.1148
0.1378
0.3593
0.1200
0.1114
0.1114
0.1035
0.0961
0.0892
0.0828
0.0769
19
0.1432
0.1625
0.1562
0.1500
0.1438
0.1377
0.1114
0.1313
0.1271
0.1228
0.1184
0.1148
0.0786
0.1286
0.3508
0.1114
0.1114
0.1035
0.0961
0.0892
0.0828
0.0769
20
0.1432
0.1625
0.1562
0.1500
0.1438
0.1377
0.1114
0.1257
0.1218
0.1179
0.1138
0.1103
0.0786
0.0694
0.1200
0.3422
0.1114
0.1035
0.0961
0.0892
0.0828
0.0769
21 Bust
0.3740
0.1389
0.1625
0.2669
0.1562
0.2929
0.1500
0.3189
0.1438
0.3448
0.1377
0.3704
0.1114
0.2121
0.1196
0.3567
0.1162
0.3767
0.1126
0.3971
0.1089
0.4177
0.1057
0.4395
0.0741
0.2623
0.0694
0.2447
0.0608
0.2284
0.1114
0.2121
0.3422
0.2121
0.1035
0.4827
0.0961
0.5196
0.0892
0.5539
0.0828
0.5858
0.0769
0.6154
This matrix shows the probabilities for the ending values of the dealer’s hand given any of the
starting states identified by each row for the case when the dealer hits soft 17. For instance, if the
dealer has a 6 as an up card, the probability of finishing with 17 is .1148, the probability of finishing
with 18 is .1148, and so on. The probability of the dealer busting is .4395.
3.
Mathematicians have shown that when the dealer stays on soft 17 it is better for the player. We can
provide the following argument using the finishing values for the dealer’s hand shown in part 2
above. Note that when the dealer gets S17 and hits it, the finishing values and probabilities for the
dealer’s hand are
Ending Value
17
18
19
20
21
Bust
Probability
.3422
.1114
.1114
.1114
.1114
.2121
CP - 86
Solutions to Case Problems
If the player has 17, the probability of a tie is .3422, the probability of losing is 4(.1114) = .4456,
and the probability of winning is .2121. So, when the player has 17, she will do better (a tie) if the
dealer stays on S17. Similarly, if the player has 18, 19, 20, or 21, she will do much better if the
dealer stays on S17. She will always win.
But, if the player has 12, 13, 14, 15, or 16, she will be better off if the dealer hits soft 17. This is
because when the dealer busts she will win (if the dealer does not hit S17 she loses). And, this will
happen with probability .2121. But, the player will lose with probability 1 - .2121 = .7879 when the
dealer hits S17. This small probability of the player winning with 12, 13, 14, 15, and 16 when the
dealer hits S17 is not enough to offset the large increase in the probability of the player winning with
17, 18,19, 20, and 21 when the dealer stays on S17.
CP - 87
Chapter 21
Dynamic Programming
Case Problem: Process Design
The optimal solution is to operate the heater at a temperature of 800°, use catalyst C2 for the reactor, and purchase
separator S1. This results in a weekly profit of $9,580 calculated as follows.
Weekly revenue (2700 lbs. x $6.00)
Weekly costs
$16,200
Payback on heater (12,000/100)
Payback on reactor ( 50,000/100)
Payback on separator (20,000/100)
Operate heater
Operate reactor
Operate separator (2700 x 0.10)
Raw material (4500 x 1.00)
Net Weekly Profit
120
500
200
380
650
270
4,500
6,620
$ 9,580
Calculations
Let stage 1 be separation, stage 2 reactor, and stage 3 heating.
Stage 1
d1
x1
s1
s2
d1*
f1(d1)
900 lb.
$5,110
$5,170
s2
$5,170
1800 lb.
10,420
10,390
s1
10,420
2700 lb.
15,730
15,610
s1
15,730
Stage 2
d2
x2
c1
c2
d 2*
f2(d2)
x1
4500 lb., 700°
$4,220
$9,270
c2
$9,270
1800
4500 lb., 800°
9,470
14,580
c1
14,580
2700
Stage 3
d3
x3
700°
800°
d 3*
f3(d3)
x1
4500 lb.
$4,370
$9,580
800°
$9,580
(4500, 800°)
CP - 88