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
Let
D1 = number of units of the DI-910 produced
D2 = number of units of the DI-950 produced
Max
s.t.
42D1
+ 87D2
3D1 + 6D2
4D1 + 2D2
D1, D2  0
 480
 480
Line 1 Capacity
Line 2 Capacity
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 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
20 30 40 50 60 70 80 90 100
Number of BodyPlus 100
Optimal Solution
0
10
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.
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 Excel Solver.
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 is shown below:
OPTIMAL SOLUTION
Optimal Objective Value
2486956.52174
Variable
S
M
Constraint
1
2
3
4
5
Value
Reduced Cost
0.60870
0.00000
0.86957
0.00000
Slack/Surplus
0.00000
30434.78261
0.00000
0.39130
0.13043
Dual Value
2.78261
0.00000
0.52174
0.00000
0.00000
Objective
Allowable
Coefficient
Increase
1800000.00000 120000.00000
1600000.00000 1280000.00000
Allowable
Decrease
800000.00000
100000.00000
RHS
Value
800000.00000
700000.00000
500000.00000
1.00000
1.00000
Allowable
Increase
22950.81967
Infinite
25000.00000
Infinite
Infinite
CP - 6
Allowable
Decrease
60000.00000
30434.78261
38888.88889
0.39130
0.13043
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.
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 follows:
OPTIMAL SOLUTION
Optimal Objective Value
2550819.67213
Variable
S
M
Constraint
1
2
3
4
5
Value
Reduced Cost
0.68852
0.00000
0.81967
0.00000
Slack/Surplus
77049.18033
0.00000
0.00000
0.31148
0.18033
Dual Value
0.00000
2.09836
2.16393
0.00000
0.00000
Objective
Allowable
Coefficient
Increase
1800000.00000 942857.14286
1600000.00000 1280000.00000
Allowable
Decrease
800000.00000
550000.00000
RHS
Value
Allowable
Increase
CP - 7
Allowable
Decrease
Chapter 2
900000.00000
700000.00000
500000.00000
1.00000
1.00000
4.
77049.18033
110000.00000
135714.28571
0.31148
0.18033
If an additional $100,000 is made available, the allocation plan would change as follows:
Security Systems
Market Analysis
Total
5.
Infinite
102173.91304
45833.33333
Infinite
Infinite
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.
CP - 8
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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
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6000
7500
7500
6000
7500
10000
3000
5000
Almonds
Brazil
Filberts
Pecans
Walnuts
Regular
Deluxe
Holiday
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Chapter 3
Optimal Objective Value
61375.00000
Variable
R
D
H
Value
17500.00000
10625.00000
5000.00000
Reduced Cost
0.00000
0.00000
0.00000
Slack/Surplus
0.00000
250.00000
250.00000
875.00000
0.00000
7500.00000
7625.00000
0.00000
Dual Value
8.50000
0.00000
0.00000
0.00000
1.50000
0.00000
0.00000
-0.17500
Objective
Coefficient
1.65000
2.00000
2.25000
Allowable
Increase
0.35000
0.20000
0.17500
Allowable
Decrease
0.15000
0.10769
Infinite
RHS
Value
6000.00000
7500.00000
7500.00000
6000.00000
7500.00000
10000.00000
3000.00000
5000.00000
Allowable
Increase
583.33333
Infinite
Infinite
Infinite
250.00000
7500.00000
7625.00000
4692.30769
Allowable
Decrease
610.00000
250.00000
250.00000
875.00000
750.00000
Infinite
Infinite
5000.00000
Constraint
1
2
3
4
5
6
7
8
3.
From the dual values 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 value shows that each pound is worth $8.50; the dual value is
applicable for increases up to 583.33 pounds.
CP - 10
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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.
Optimal Objective Value
66333.33333
Variable
R
D
H
Value
11666.66667
17916.66667
5000.00000
Reduced Cost
0.00000
0.00000
0.00000
Slack/Surplus
416.66667
250.00000
250.00000
0.00000
0.00000
1666.66667
14916.66667
0.00000
Dual Value
0.00000
0.00000
0.00000
5.66667
4.33333
0.00000
0.00000
-0.03333
Objective
Coefficient
1.65000
2.00000
2.25000
Allowable
Increase
0.10000
1.30000
0.03333
Allowable
Decrease
0.65000
0.02353
Infinite
RHS
Value
7000.00000
7500.00000
7500.00000
6000.00000
7500.00000
10000.00000
3000.00000
5000.00000
Allowable
Increase
Infinite
Infinite
Infinite
250.00000
250.00000
1666.66667
14916.66667
10529.41176
Allowable
Decrease
416.66667
250.00000
250.00000
1790.00000
250.00000
Infinite
Infinite
5000.00000
Constraint
1
2
3
4
5
6
7
8
5.
From the dual values 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
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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 Objective Value
94133.33333
Variable
G
I
M
Value
248888.88889
160000.00000
391111.11111
Reduced Cost
0.00000
0.00000
0.00000
Constraint
1
2
3
4
5
6
7
Slack/Surplus
0.00000
88888.88889
71111.11111
0.00000
240000.00000
151111.11111
0.00000
Dual Value
0.11767
0.00000
0.00000
-0.02000
0.00000
0.00000
1.16667
Allowable
Increase
Infinite
0.02000
0.10500
Allowable
Decrease
0.03000
0.58833
0.06000
Objective
Coefficient
0.18000
0.12500
0.07500
CP - 12
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Solutions to Case Problems
RHS
Value
800000.00000
0.00000
0.00000
0.00000
0.00000
0.00000
0.00000
Allowable
Increase
Infinite
88888.88889
Infinite
133333.33333
Infinite
151111.11111
6400.00000
Allowable
Decrease
800000.00000
Infinite
71111.11111
106666.66667
240000.00000
Infinite
8000.00000
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 value 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
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Chapter 3
Optimal Objective Value
98800.00000
3.
Variable
G
I
M
Value
293333.33333
160000.00000
346666.66667
Reduced Cost
0.00000
0.00000
0.00000
Constraint
1
2
3
4
5
6
7
Slack/Surplus
0.00000
133333.33333
26666.66667
0.00000
240000.00000
106666.66667
0.00000
Dual Value
0.12350
0.00000
0.00000
-0.02000
0.00000
0.00000
1.16667
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 Objective Value
85066.66667
Variable
G
I
M
Value
160000.00000
293333.33333
346666.66667
Reduced Cost
0.00000
0.00000
0.00000
Slack/Surplus
Dual Value
CP - 14
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Solutions to Case Problems
Constraint
1
2
3
4
5
6
7
4.
0.00000
0.00000
160000.00000
133333.33333
106666.66667
106666.66667
0.00000
0.10633
-0.01000
0.00000
0.00000
0.00000
0.00000
0.83333
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 Objective Value
93066.66667
Variable
1
2
3
Value
213333.33333
213333.33333
373333.33333
Reduced Cost
0.00000
0.00000
0.00000
Constraint
1
2
3
4
5
6
7
8
Slack/Surplus
0.00000
53333.33333
106666.66667
53333.33333
186666.66667
133333.33333
0.00000
0.00000
Dual Value
0.11633
0.00000
0.00000
0.00000
0.00000
0.00000
1.03333
0.01200
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.
CP - 15
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Chapter 3
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.
Case Problem 3: Truck Leasing Strategy
1.
Let xij = number of trucks obtained from a short term lease signed in month i for a period of j
months and yi = number of trucks obtained from the long-term lease that are used in month i
Monthly fuel costs are 20 ($100) = $2000.
Monthly Costs for Short-Term Leased Trucks
Note: the costs shown here include monthly fuel costs of $2000.
Decision Variables
x11 , x21 , x31 , x41
x12 , x22 , x32
x13 , x23
x14
Cost
$4000 + $2000 = $6000
2 ($3700 + $2000) = $11,400
3 ($3225 + $2000) = $15,675
4 ($3040 + $2000) = $20,160
Monthly Costs for Long-Term Leased Trucks
Since Reep Construction is committed to the long-term lease and since employees cannot
be laid off, the only relevant cost for the long-term leased trucks is the monthly fuel cost of
$2000.
LINEAR PROGRAMMING PROBLEM
MIN
6000X11+11400X12+15675X13+20160X14+6000X21+11400X22+15675X23+6000X31+
11400X32+6000X41+2000Y1+2000Y2+2000Y3+2000Y4
S.T.
1)
2)
3)
4)
5)
6)
7)
8)
1X11+1X12+1X13+1X14+1Y1=10
1X12+1X13+1X14+1X21+1X22+1X23+1Y2=12
1X13+1X14+1X22+1X23+1X31+1X32+1Y3=14
1X14+1X23+1X32+1X41+1Y4=8
1Y1<1
1Y2<2
1Y3<3
1Y4<1
CP - 16
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Solutions to Case Problems
Optimal Objective Value
203660.00000
Variable
X11
X12
X13
X14
X21
X22
X23
X31
X32
X41
Y1
Y2
Y3
Y3
Constraint
1
2
3
4
5
6
7
8
Objective
Coefficient
6000.00000
11400.00000
15675.00000
20160.00000
6000.00000
11400.00000
15675.00000
6000.00000
11400.00000
6000.00000
2000.00000
Value
Reduced Cost
0.00000
1515.00000
0.00000
1725.00000
3.00000
0.00000
6.00000
0.00000
0.00000
810.00000
0.00000
210.00000
1.00000
0.00000
1.00000
0.00000
0.00000
915.00000
0.00000
1515.00000
1.00000
0.00000
2.00000
0.00000
3.00000
0.00000
1.00000
0.00000
Slack/Surplus
0.00000
0.00000
0.00000
0.00000
0.00000
0.00000
0.00000
0.00000
Dual Value
4485.00000
5190.00000
6000.00000
4485.00000
-2485.00000
-3190.00000
-4000.00000
-2485.00000
Allowable
Increase
Infinite
Infinite
210.00000
915.00000
Infinite
Infinite
210.00000
915.00000
Infinite
Infinite
2485.00000
Allowable
Decrease
1515.00000
1725.00000
915.00000
210.00000
810.00000
210.00000
1515.00000
810.00000
915.00000
1515.00000
Infinite
CP - 17
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Chapter 3
2000.00000
2000.00000
2000.00000
3190.00000
4000.00000
2485.00000
Infinite
Infinite
Infinite
RHS
Value
10.00000
12.00000
14.00000
8.00000
1.00000
2.00000
3.00000
1.00000
Allowable
Increase
1.00000
1.00000
Infinite
3.00000
6.00000
1.00000
1.00000
6.00000
Allowable
Decrease
6.00000
1.00000
1.00000
6.00000
1.00000
1.00000
3.00000
1.00000
2.
The total cost associated with the leasing plan is $203,660.
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
Leased
1
2
3
4
1
0
0
0
0
Length of Lease (Months)
2
3
0
3
0
1
0
4
7
In addition, in month 2, one of the trucks from the long-term leases was used and in month 3 three
of the trucks from the long-term leases were used. The total cost of this leasing plan is $224,620.
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
Leased
1
2
3
4
1
0
0
1
0
Length of Lease (Months)
2
3
0
3
0
1
0
4
6
The total cost associated with this solution is $226,060. Thus, if Reep maintains their current
policy of no layoffs they will incur an additional cost of $226,060 - $224,620 = $1,440.
CP - 18
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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 are as follows:
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
Optimal Objective Value
2160.00000
Variable
T1
T2
R1
R2
N1
N2
Value
Reduced Cost
10.00000
0.00000
5.00000
0.00000
15.00000
0.00000
18.00000
0.00000
20.00000
0.00000
10.00000
0.00000
Slack/Surplus
CP - 19
Dual Value
Chapter 4
1.
Constraint
1
2
3
4
5
6
7
8
9
10
0.00000
0.00000
0.00000
0.00000
27100.00000
3.00000
5.00000
10000.00000
0.00000
0.00000
35.00000
5.00000
5.00000
0.00550
0.00000
0.00000
0.00000
0.00000
0.00117
-0.00050
Objective
Coefficient
90.00000
55.00000
25.00000
20.00000
10.00000
5.00000
Allowable
Increase
Infinite
11.66667
Infinite
5.00000
Infinite
0.50000
Allowable
Decrease
35.00000
5.00000
5.00000
3.50000
5.00000
Infinite
RHS
Value
10.00000
15.00000
20.00000
279000.00000
100000.00000
0.00000
20.00000
140000.00000
99000.00000
30000.00000
Allowable
Increase
5.00000
18.00000
10.00000
15000.00000
27100.00000
3.00000
Infinite
10000.00000
10000.00000
10000.00000
Allowable
Decrease
10.00000
15.00000
20.00000
10000.00000
Infinite
Infinite
5.00000
Infinite
5625.00000
10000.00000
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
CP - 20
Solutions to Case Problems
Advertising Schedule:
Media
Television
Radio
Newspaper
Totals
Number of Ads
15
33
30
78
Total Exposure Rating:
Total New Customers Reached:
2.
Budget
$150,000
99,000
30,000
$279,000
2,160
127,100 (Surplus constraint 5)
The dual value shows that total exposure increases 0.0055 points for each one dollar increase in the
advertising budget. Right Hand Side Ranges show this dual value applies for a budget increase of up
to $15,000. Thus the dual value applies for the $10,000 increase.
Total Exposure Rating would increase by 10,000(0.0055) = 55 points
A $10,000 increase in the advertising budget is a 3.6% increase. But, it only provides a 2.54%
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
CP - 21
Chapter 4
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.
Case Problem 2: Schneider’s Sweet Shop
Papa Jack’s recipe can just be calculated, whereas the optimized and flexible optimized recipes
require linear programming. The following linear programs may be used to calculate the latter two:
Let Xi = amount (in pounds) of ingredient i to use in making 50 pounds of ice cream, i = 1,2,3…14
Min
1.19X1 + .70X2 + 2.32X3 + …… + 1.68X13 + 0.0X14
s.t.
.4X1 + .2X2 + .8X3 + .8X4 + .9X5 + .1X6 + .5X10 + .6X11
= .16(50)
.1X1 +.1X4 + .1X6 + .3X7 + 1X8
= .08(50)
.7X9 + .1X10
= .16(50)
.4X10 + .4X11
= .0035(50)
X12
= .0025(50)
X13
= .0015(50)
.5X1 + .8X2 + .2X3 + .1X4 + .1X5 +.8X6 + .7X7 + .3X9 + X14
= .5925(50)
X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 + X11 + X12 + X13 + X14 = 50
X1 , X2 , X3 , X4 , X5 , X6 , X7 , X8 , X9 , X10 , X11 , X12 , X13 , X14 ≥ 0
And for the flexible recipe:
Min
1.19X1 + .70X2 + 2.32X3 + …… + 1.68X13 + 0.0X14
s.t.
.15(50)
≤ .4X1 + .2X2 + .8X3 + .8X4 + .9X5 + .1X6 + .5X10 + .6X11 ≤ .17(50)
.07(50)
≤ .1X1 +.1X4 + .1X6 + .3X7 + 1X8
≤ .09(50)
.155(50) ≤ .7X9 + .1X10
≤ .165(50)
.003(50) ≤ .4X10 + .4X11
≤ .004(50)
.002(50) ≤
X12
≤ .003(50)
.001(50) ≤
X13
≤ .002(50)
.58(50) ≤
.5X1 + .8X2 + .2X3 + .1X4 + .1X5 +.8X6 + .7X7 + .3X9 + X14
≤ .595(50)
X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 + X10 + X11 + X12 + X13 + X14 = 50
X1 , X2 , X3 , X4 , X5 , X6 , X7 , X8 , X9 , X10 , X11 , X12 , X13 , X14 ≥ 0
CP - 22
Solutions to Case Problems
The recipes and costs are as follows:
Ingredient Name
40% Cream
23% Cream
Butter
Plastic Cream
Butter Oil
4% Milk
Skim Condensed Milk
Skim Milk Powder
Liquid Sugar
Sugared Frozen Fresh Egg Yolk
Powdered Egg Yolk
Stabilizer
Emulsifier
Water
Fat
Serum Solids
Sugar Solids
Egg Solids
Stabilizer
Emulsifier
Water
Ingredient
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Total
Cost
Papa Jack Optimized
16.0%
16.0%
8.0%
8.0%
16.0%
16.0%
0.4%
0.4%
0.2%
0.3%
0.1%
0.2%
59.2%
59.3%
Papa Jack
0
0
0
9.73
0
0
0
3.03
11.37
0.44
0
0.12
0.07
25.24
50
$28.37
Optimized
0
0
0
5.72
0
32.05
0
0.22
11.37
0.44
0
0.13
0.08
0
50
$25.35
Optimized
Flexible
0
0
0
5.12
0
32.15
0
0.47
11.73
0.38
0
0.10
0.05
0
50
$24.04
Optimized
Flexible
15.0%
8.4%
16.5%
0.3%
0.2%
0.1%
59.5%
The shaded areas in the recipe table indicate larger differences. The recipes generated by the
cost-minimizing linear program use no water and instead get the required water from other
ingredients that provide water and other requirements. In fact, the table shows that even though
water is (essentially) free, it is not used because it does not contribute to other requirements. The
optimized recipe saves $28.37 - $25.35 = $3.02 per $50 batch and the optimized flexible recipe
saves an additional $25.35-$24.04 = $1.31 per 50 pound batch. Note that the flexible recipe
increases the amount of sugar solids and that may or may not be accepted well by customers. Since
the optimized recipe follows the original requirements and saves $3.02 per 50 pound batch, we
recommend going with that and maybe test marketing the optimized flexible recipe to better
understand its acceptance by the market.
CP - 23
Chapter 4
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.
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
CP - 24
Solutions to Case Problems
Constraints:
Regular Looms:
0.1912X3R + 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
Optimal Objective Value
62531.49090
Variable
X3R
X4R
X5R
X1D
X2D
X3D
X4D
X5D
Y1
Y2
Y3
Y4
Y5
Constraint
1
2
3
4
5
6
7
Objective
Value
Reduced Cost
27707.80815
0.00000
7500.00000
0.00000
62000.00000
0.00000
4668.80000
0.00000
22000.00000
0.00000
0.00000
-0.01394
0.00000
-0.01394
0.00000
-0.01748
11831.20000
0.00000
0.00000
-0.01000
34292.19185
0.00000
0.00000
-0.08000
0.00000
-0.06204
Slack/Surplus
0.00000
0.00000
0.00000
0.00000
0.00000
0.00000
0.00000
Allowable
CP - 25
Dual Value
0.57530
0.64820
0.19000
0.17000
0.50000
0.62000
0.06204
Allowable
Chapter 4
Coefficient
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
Increase
0.01394
Infinite
Infinite
0.01000
Infinite
0.01394
0.01394
0.01748
0.01575
0.01000
0.11000
0.08000
0.06204
Decrease
0.11000
0.01394
0.01748
0.01575
0.01000
Infinite
Infinite
Infinite
0.01000
Infinite
0.01394
Infinite
Infinite
RHS
Value
21600.00000
5760.00000
16500.00000
22000.00000
62000.00000
7500.00000
62000.00000
Allowable
Increase
6556.82444
2555.33477
Infinite
4668.80000
Infinite
27707.80815
22092.07648
Allowable
Decrease
5297.86007
1008.38013
11831.20000
11831.20000
34292.19185
7500.00000
27341.95794
Production/Purchase Schedule (Yards)
Regular
Looms
Fabric
1
2
3
4
5
27711
7500
62000
Projected Profit: $62,531.49
Value of 9th Dobbie Loom
Dual Value (Constraint 2) = 0.6482 per hour dobbie
CP - 26
Dobbie
Looms
4669
22000
Purchased
11831
34289
Solutions to Case Problems
Monthly Value of 1 Dobbie Loom
(30 days)(24 hours/day)($0.6482) = $466.70
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
Total Cost
$2875
$5675
$8375
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.
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
CP - 27
Chapter 4
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
May
t15
t25
June
t33 + t24 + t34 + t15 + t25 = 20
Constraint 6: Need 14 additional employees in June
Jan.
Option 1
Option 2
Option 3
Feb.
Mar.
Apr.
May
t25
t34
t34 + t25 + t16 = 14
CP - 28
June
t16
Solutions to Case Problems
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.
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.
CP - 29
Chapter 4
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
S i
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
i
D j
for all generating units
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).
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
CP - 30
Solutions to Case Problems
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 dual value 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 dual values for the demand constraints are as follows:
Miami Fort #5
21
Miami Fort #7
20
Beckjord
20
East Bend
18
Zimmer
19
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.
CP - 31
Chapter 4
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