An Introduction to Management Science: Quantitative Approaches to Decision Making, Revised, 13th Edition solutions manual and test bank David R. Anderson  Dennis J. Sweeney  Thomas A. Williams  Jeffrey D. Camm  R. Kipp Martin
An Introduction to Linear Programming
Case Problem 1: Workload Balancing
1.
Production Rate (minutes per printer)  
Model  Line 1  Line 2  Profit Contribution ($) 
DI910  3  4  42 
DI950  6  2  87 
Capacity: 8 hours60 minutes/hour = 480 minutes per day
Let D_{1} = number of units of the DI910 produced
D_{2} = number of units of the DI950 produced
Max  42D_{1}  +  87D_{2}  
s.t.  
3D_{1}  +  6D_{2}  £  480  Line 1 Capacity  
4D_{1}  +  2D_{2}  £  480  Line 2 Capacity 
D_{1}, D_{2} ³ 0
The optimal solution is D_{1} = 0, D_{2} = 80. The value of the optimal solution is $6960.
Management would not implement this solution because no units of the DI910 would be produced.
2. Adding the constraint D_{1} ³ D_{2} and resolving the linear program results in the optimal solution D_{1} = 53.333, D_{2} = 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 T_{1} = total time spent on Line 1
T_{2} = total time spent on Line 2
Whatever the value of T_{2} is,
T_{1} £ T_{2} + 30
T_{1} ³ T_{2}  30
Thus, with T_{1} = 3D_{1} + 6D_{2 }and T_{2} = 4D_{1} + 2D_{2}
3D_{1} + 6D_{2 }£ 4D_{1} + 2D_{2} + 30
3D_{1} + 6D_{2 }³ 4D_{1} + 2D_{2}  30
Hence,
1D_{1} + 4D_{2 }£ 30
1D_{1} + 4D_{2 }³ 30
Rewriting the second constraint by multiplying both sides by 1, we obtain
1D_{1} + 4D_{2 }£ 30
1D_{1}  4D_{2 }£ 30
Adding these two constraints to the linear program formulated in part (2) and resolving we obtain the optimal solution D_{1} = 96.667, D_{2} = 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 D_{1} = 106.667, D_{2} = 26.667. The total profit contribution is
42(106.667) + 87(26.667) = $6800
Comparing the solutions to part (4) and part (5), maximizing the number of printers produced (106.667 + 26.667 = 133.33) has increased the production by 133.33  (96.667 + 31.667) = 5 printers but has reduced profit contribution by $6815  $6800 = $15. But, this solution results in perfect workload balancing because the total time spent on each line is 480 minutes.
Case Problem 2: Production Strategy
1. Let BP100 = the number of BodyPlus 100 machines produced
BP200 = the number of BodyPlus 200 machines produced
Max  371BP100  +  461BP200  
s.t.  
8BP100  +  12BP200  £  600  Machining and Welding  
5BP100  +  10BP200  £  450  Painting and Finishing  
2BP100  +  2BP200  £  140  Assembly, Test, and Packaging  
0.25BP100  +  0.75BP200  ³  0  BodyPlus 200 Requirement 
BP100, BP200 ³ 0
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 highend 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.
Case Problem 3: Hart Venture Capital
1. Let S = fraction of the Security Systems project funded by HVC
M = fraction of the Market Analysis project funded by HVC
Max  1,800,000S  +  1,600,000M  
s.t.  
600,000S  +  500,000M  £  800,000  Year 1  
600,000S  +  350,000M  £  700,000  Year 2  
250,000S  +  400,000M  £  500,000  Year 3  
S  £  1  Maximum for S  
M  £  1  Maximum for M  
S,M  ³  0 
The solution obtained is shown below:
OPTIMAL SOLUTION
2486956.52174  
Variable  Value  Reduced Cost 
S  0.60870  0.00000 
M  0.86957  0.00000 
Constraint  Slack/Surplus  Dual Value 
1  0.00000  2.78261 
2  30434.78261  0.00000 
3  0.00000  0.52174 
4  0.39130  0.00000 
5  0.13043  0.00000 
Objective  Allowable  Allowable 
Coefficient  Increase  Decrease 
1800000.00000  120000.00000  800000.00000 
1600000.00000  1280000.00000  100000.00000 
RHS  Allowable  Allowable 
Value  Increase  Decrease 
800000.00000  22950.81967  60000.00000 
700000.00000  Infinite  30434.78261 
500000.00000  25000.00000  38888.88889 
1.00000  Infinite  0.39130 
1.00000  Infinite  0.13043 
Thus, the optimal solution is S = 0.609 and M = 0.870. In other words, approximately 61% of the Security Systems project should be funded by HVC and 87% of the Market Analysis project should be funded by HVC.
The net present value of the investment is approximately $2,486,957.
2.
Year 1  Year 2  Year 3  
Security Systems  $365,400  $365,400  $152,250 
Market Analysis  $435,000  $304,500  $348,000 
Total  $800,400  $669,900  $500,250 
Note: The totals for Year 1 and Year 3 are greater than the amounts available. The reason for this is that rounded values for the decision variables were used to compute the amount required in each year.
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  Value  Reduced Cost 
S  0.68852  0.00000 
M  0.81967  0.00000 
Constraint  Slack/Surplus  Dual Value 
1  77049.18033  0.00000 
2  0.00000  2.09836 
3  0.00000  2.16393 
4  0.31148  0.00000 
5  0.18033  0.00000 
Objective  Allowable  Allowable 
Coefficient  Increase  Decrease 
1800000.00000  942857.14286  800000.00000 
1600000.00000  1280000.00000  550000.00000 
RHS  Allowable  Allowable 
Value  Increase  Decrease 
900000.00000  Infinite  77049.18033 
700000.00000  102173.91304  110000.00000 
500000.00000  45833.33333  135714.28571 
1.00000  Infinite  0.31148 
1.00000  Infinite  0.18033 
4. If an additional $100,000 is made available, the allocation plan would change as follows:
Year 1  Year 2  Year 3  
Security Systems  $413,400  $413,400  $172,250 
Market Analysis  $410,000  $287,000  $328,000 
Total  $823,400  $700,400  $500,250 
5. Having additional funds available in year 1 will increase the total net present value. The value of the objective function increases from $2,486,957 to $2,550,820, a difference of $63,863. But, since the allocation plan shows that $823,400 is required in year 1, only $23,400 of the additional $100,00 is required. We can also determine this by looking at the slack variable for constraint 1 in the new solution. This value, 77049.180, shows that at the optimal solution approximately $77,049 of the $900,000 available is not used. Thus, the amount of funds required in year 1 is $900,000  $77,049 = $822,951. In other words, only $22,951 of the additional $100,000 is required. The differences between the two values, $23,400 and $22,951, is simply due to the fact that the value of $23,400 was computed using rounded values for the decision variables.
Chapter 2
An Introduction to Linear Programming
Learning Objectives
1. Obtain an overview of the kinds of problems linear programming has been used to solve.
2. Learn how to develop linear programming models for simple problems.
3. Be able to identify the special features of a model that make it a linear programming model.
4. Learn how to solve two variable linear programming models by the graphical solution procedure.
5. Understand the importance of extreme points in obtaining the optimal solution.
6. Know the use and interpretation of slack and surplus variables.
7. Be able to interpret the computer solution of a linear programming problem.
8. Understand how alternative optimal solutions, infeasibility and unboundedness can occur in linear programming problems.
9. Understand the following terms:
problem formulation feasible region
constraint function slack variable
objective function standard form
solution redundant constraint
optimal solution extreme point
nonnegativity constraints surplus variable
mathematical model alternative optimal solutions
linear program infeasibility
linear functions unbounded
feasible solution
Solutions:
1. a, b, and e, are acceptable linear programming relationships.
c is not acceptable because of
d is not acceptable because of
f is not acceptable because of 1AB
c, d, and f could not be found in a linear programming model because they have the above nonlinear terms.
2. a.
b.
c.
3. a.
b.
c.
4. a.
b.
c.
5.
6. 7A + 10B = 420 is labeled (a)
6A + 4B = 420 is labeled (b)
4A + 7B = 420 is labeled (c)
7.
8.
9.
10.
A  +  2B  =  6  (1)  
5A  +  3B  =  15  (2)  
(1) × 5  5A  +  10B  =  30  (3) 
(2)  (3)    7B  =  15  
B  =  15/7 
From (1), A = 6  2(15/7) = 6  30/7 = 12/7
11.
12. a.
b.
c. There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3).
13. a.
b. The extreme points are (5, 1) and (2, 4).
c.
14. a. Let F = number of tons of fuel additive
S = number of tons of solvent base
Max  40F  +  30S  
s.t.  
2/5F  +  1/2 S  £  200 Material 1  
1/5 S  £  5 Material 2  
3/5 F  +  3/10 S  £  21 Material 3 
F, S ³ 0


c. Material 2: 4 tons are used, 1 ton is unused.
d. No redundant constraints.
15. a.
b. Similar to part (a): the same feasible region with a different objective function. The optimal solution occurs at (708, 0) with a profit of z = 20(708) + 9(0) = 14,160.
c. The sewing constraint is redundant. Such a change would not change the optimal solution to the original problem.
16. a. A variety of objective functions with a slope greater than 4/10 (slope of I & P line) will make extreme point (0, 540) the optimal solution. For example, one possibility is 3S + 9D.
b. Optimal Solution is S = 0 and D = 540.
c.
Department  Hours Used  Max. Available  Slack 
Cutting and Dyeing  1(540) = 540  630  90 
Sewing  ^{5}/_{6}(540) = 450  600  150 
Finishing  ^{2}/_{3}(540) = 360  708  348 
Inspection and Packaging  ^{1}/_{4}(540) = 135  135  0 
17.
Max  5A  +  2B  +  0S1  +  0S2  +  0S3  
s.t.  
1A    2B  +  1S1  =  420  
2A  +  3B  +  1S2  =  610  
6A    1B  +  1S3  =  125 
A, B, S1, S2, S3 ³ 0
18. a.
Max  4A  +  1B  +  0S1  +  0S2  +  0S3  
s.t.  
10A  +  2B  +  1S1  =  30  
3A  +  2B  +  1S2  =  12  
2A  +  2B  +  1S3  =  10 
A, B, S1, S2, S3 ³ 0
b.
c. S1 = 0, S2 = 0, S3 = 4/7
19. a.
Max  3A  +  4B  +  0S1  +  0S2  +  0S3  
s.t.  
1A  +  2B  +  1S1  =  8 (1)  
1A  +  2B  +  1S2  =  12 (2)  
2A  +  1B  +  1S3  =  16 (3) 
A, B, S1, S2, S3 ³ 0
b.
c. S1 = 8 + A – 2B = 8 + 20/3  16/3 = 28/3
S2 = 12  A – 2B = 12  20/3  16/3 = 0
S3 = 16 – 2A  B = 16  40/3  8/3 = 0
20. a.
Max  3A  +  2B  
s.t.  
A  +  B   S1  =  4  
3A  +  4B  + S2  =  24  
A   S3  =  2  
A    B   S4  =  0 
A, B, S1, S2, S3, S4 ³ 0
b.
c. S1 = (3.43 + 3.43)  4 = 2.86
S2 = 24  [3(3.43) + 4(3.43)] = 0
S3 = 3.43  2 = 1.43
S4 = 0  (3.43  3.43) = 0
21. a. and b.
c. Optimal solution occurs at the intersection of constraints 1 and 2. For constraint 2,
B = 10 + A
Substituting for B in constraint 1 we obtain
5A + 5(10 + A)  = 400 
5A + 50 + 5A  = 400 
10A  = 350 
A  = 35 
B = 10 + A = 10 + 35 = 45
Optimal solution is A = 35, B = 45
d. Because the optimal solution occurs at the intersection of constraints 1 and 2, these are binding constraints.
e. Constraint 3 is the nonbinding constraint. At the optimal solution 1A + 3B = 1(35) + 3(45) = 170. Because 170 exceeds the righthand side value of 90 by 80 units, there is a surplus of 80 associated with this constraint.
22. a.
b.
Extreme Point  Coordinates  Profit 
1  (0, 0)  5(0) + 4(0) = 0 
2  (1700, 0)  5(1700) + 4(0) = 8500 
3  (1400, 600)  5(1400) + 4(600) = 9400 
4  (800, 1200)  5(800) + 4(1200) = 8800 
5  (0, 1680)  5(0) + 4(1680) = 6720 
Extreme point 3 generates the highest profit.
c. Optimal solution is A = 1400, C = 600
d. The optimal solution occurs at the intersection of the cutting and dyeing constraint and the inspection and packaging constraint. Therefore these two constraints are the binding constraints.
e. New optimal solution is A = 800, C = 1200
Profit = 4(800) + 5(1200) = 9200
23. a. Let E = number of units of the EZRider produced
L = number of units of the LadySport produced
Max  2400E  +  1800L  
s.t.  
6E  +  3L  £  2100  Engine time  
L  £  280  LadySport maximum  
2E  +  2.5L  £  1000  Assembly and testing 
E, L ³ 0
b.
c. The binding constraints are the manufacturing time and the assembly and testing time.
24. a. Let R = number of units of regular model.
C = number of units of catcher’s model.
Max  5R  +  8C  
s.t.  
1R  +  3/2 C  £  900 Cutting and sewing  
1/2 R  +  1/3 C  £  300 Finishing  
1/8 R  +  1/4 C  £  100 Packing and Shipping 
R, C ³ 0
b.
c. 5(500) + 8(150) = $3,700
d. C & S 1(500) + 3/2(150) = 725
F 1/2(500) + 1/3(150) = 300
P & S 1/8(500) + 1/4(150) = 100
e.
Department  Capacity  Usage  Slack 
C & S  900  725  175 hours 
F  300  300  0 hours 
P & S  100  100  0 hours 
25. a. Let B = percentage of funds invested in the bond fund
S = percentage of funds invested in the stock fund
Max  0.06 B  +  0.10 S  
s.t.  
B  ³  0.3  Bond fund minimum  
0.06 B  +  0.10 S  ³  0.075  Minimum return  
B  +  S  =  1  Percentage requirement 
b. Optimal solution: B = 0.3, S = 0.7
Value of optimal solution is 0.088 or 8.8%
26. a. a. Let N = amount spent on newspaper advertising
R = amount spent on radio advertising
Max  50N  +  80R  
s.t.  
N  +  R  =  1000  Budget  
N  ³  250  Newspaper min.  
R  ³  250  Radio min.  
N  2R  ³  0  News ³ 2 Radio  
N, R ³ 0
b.
27. Let I = Internet fund investment in thousands
B = Blue Chip fund investment in thousands
Max  0.12I  +  0.09B  
s.t.  
1I  +  1B  £  50  Available investment funds  
1I  £  35  Maximum investment in the internet fund  
6I  +  4B  £  240  Maximum risk for a moderate investor 
I, B ³ 0
Internet fund $20,000
Blue Chip fund $30,000
Annual return $ 5,100
b. The third constraint for the aggressive investor becomes
6I + 4B £ 320
This constraint is redundant; the available funds and the maximum Internet fund investment constraints define the feasible region. The optimal solution is:
Internet fund $35,000
Blue Chip fund $15,000
Annual return $ 5,550
The aggressive investor places as much funds as possible in the high return but high risk Internet fund.
c. The third constraint for the conservative investor becomes
6I + 4B £ 160
This constraint becomes a binding constraint. The optimal solution is
Internet fund $0
Blue Chip fund $40,000
Annual return $ 3,600
The slack for constraint 1 is $10,000. This indicates that investing all $50,000 in the Blue Chip fund is still too risky for the conservative investor. $40,000 can be invested in the Blue Chip fund. The remaining $10,000 could be invested in lowrisk bonds or certificates of deposit.
28. a. Let W = number of jars of Western Foods Salsa produced
M = number of jars of Mexico City Salsa produced
Max  1W  +  1.25M  
s.t.  
5W  7M  £  4480  Whole tomatoes  
3W  +  1M  £  2080  Tomato sauce  
2W  +  2M  £  1600  Tomato paste 
W, M ³ 0
Note: units for constraints are ounces
b. Optimal solution: W = 560, M = 240
Value of optimal solution is 860
29. a. Let B = proportion of Buffalo's time used to produce component 1
D = proportion of Dayton's time used to produce component 1
Maximum Daily Production  
Component 1  Component 2  
Buffalo  2000  1000 
Dayton  600  1400 
Number of units of component 1 produced: 2000B + 600D
Number of units of component 2 produced: 1000(1  B) + 600(1  D)
For assembly of the ignition systems, the number of units of component 1 produced must equal the number of units of component 2 produced.
Therefore,
2000B + 600D = 1000(1  B) + 1400(1  D)
2000B + 600D = 1000  1000B + 1400  1400D
3000B + 2000D = 2400
Note: Because every ignition system uses 1 unit of component 1 and 1 unit of component 2, we can maximize the number of electronic ignition systems produced by maximizing the number of units of subassembly 1 produced.
Max 2000B + 600D
In addition, B £ 1 and D £ 1.
The linear programming model is:
Max  2000B  + 600D  
s.t.  
3000B  + 2000D  = 2400  
B  £ 1  
D  £ 1  
B, D  ³ 0  

The graphical solution is shown below.
Optimal Solution: B = .8, D = 0
Optimal Production Plan
Buffalo  Component 1 .8(2000) = 1600
Buffalo  Component 2 .2(1000) = 200
Dayton  Component 1 0(600) = 0
Dayton  Component 2 1(1400) = 1400
Total units of electronic ignition system = 1600 per day.
30. a. Let E = number of shares of Eastern Cable
C = number of shares of ComSwitch
Max  15E  +  18C  
s.t.  
40E  +  25C  £  50,000  Maximum Investment  
40E  ³  15,000  Eastern Cable Minimum  
25C  ³  10,000  ComSwitch Minimum  
25C  £  25,000  ComSwitch Maximum 
E, C ³ 0
b.
c. There are four extreme points: (375,400); (1000,400);(625,1000); (375,1000)
d. Optimal solution is E = 625, C = 1000
Total return = $27,375
31.
Objective Function Value = 13
32.



Extreme Points  Objective Function Value  Surplus Demand  Surplus Total Production  Slack Processing Time 
(A = 250, B = 100)  800  125  —  — 
(A = 125, B = 225)  925  —  —  125 
(A = 125, B = 350)  1300  —  125  — 
33. a.


Optimal Solution: A = 3, B = 1, value = 5
b.
(1)  3 + 4(1) = 7  Slack = 21  7 = 14 
(2)  2(3) + 1 = 7  Surplus = 7  7 = 0 
(3)  3(3) + 1.5 = 10.5  Slack = 21  10.5 = 10.5 
(4)  2(3) +6(1) = 0  Surplus = 0  0 = 0 
c.


Optimal Solution: A = 6, B = 2, value = 34
34. a.


b. There are two extreme points: (A = 4, B = 1) and (A = 21/4, B = 9/4)
c. The optimal solution is A = 4, B = 1
35. a.
Min  6A  +  4B  +  0S1  +  0S2  +  0S3  
s.t.  
2A  +  1B    S1  =  12  
1A  +  1B    S2  =  10  
1B  +  S3  =  4 
A, B, S1, S2, S3 ³ 0
b. The optimal solution is A = 6, B = 4.
c. S1 = 4, S2 = 0, S3 = 0.
36. a. Let T = number of training programs on teaming
P = number of training programs on problem solving
Max  10,000T  +  8,000P  
s.t.  
T  ³  8  Minimum Teaming  
P  ³  10  Minimum Problem Solving  
T  +  P  ³  25  Minimum Total  
3 T  +  2 P  £  84  Days Available 
T, P ³ 0
b.
c. There are four extreme points: (15,10); (21.33,10); (8,30); (8,17)
d. The minimum cost solution is T = 8, P = 17
Total cost = $216,000
37.
Regular  Zesty  
Mild  80%  60%  8100 
Extra Sharp  20%  40%  3000 
Let R = number of containers of Regular
Z = number of containers of Zesty
Each container holds 12/16 or 0.75 pounds of cheese
Pounds of mild cheese used = 0.80 (0.75) R + 0.60 (0.75) Z
= 0.60 R + 0.45 Z
Pounds of extra sharp cheese used = 0.20 (0.75) R + 0.40 (0.75) Z
= 0.15 R + 0.30 Z
Cost of Cheese = Cost of mild + Cost of extra sharp
= 1.20 (0.60 R + 0.45 Z) + 1.40 (0.15 R + 0.30 Z)
= 0.72 R + 0.54 Z + 0.21 R + 0.42 Z
= 0.93 R + 0.96 Z
Packaging Cost = 0.20 R + 0.20 Z
Total Cost = (0.93 R + 0.96 Z) + (0.20 R + 0.20 Z)
= 1.13 R + 1.16 Z
Revenue = 1.95 R + 2.20 Z
Profit Contribution = Revenue  Total Cost
= (1.95 R + 2.20 Z)  (1.13 R + 1.16 Z)
= 0.82 R + 1.04 Z
Max  0.82 R  +  1.04 Z  
s.t.  
0.60 R  +  0.45 Z  £  8100  Mild  
0.15 R  +  0.30 Z  £  3000  Extra Sharp 
R, Z ³ 0
Optimal Solution: R = 9600, Z = 5200, profit = 0.82(9600) + 1.04(5200) = $13,280
38. a. Let S = yards of the standard grade material per frame
P = yards of the professional grade material per frame
Min  7.50S  +  9.00P  
s.t.  
0.10S  +  0.30P  ³  6  carbon fiber (at least 20% of 30 yards)  
0.06S  +  0.12P  £  3  kevlar (no more than 10% of 30 yards)  
S  +  P  =  30  total (30 yards) 
S, P ³ 0
b.
c.
Extreme Point  Cost 
(15, 15)  7.50(15) + 9.00(15) = 247.50 
(10, 20)  7.50(10) + 9.00(20) = 255.00 
The optimal solution is S = 15, P = 15
d. Optimal solution does not change: S = 15 and P = 15. However, the value of the optimal solution is reduced to 7.50(15) + 8(15) = $232.50.
e. At $7.40 per yard, the optimal solution is S = 10, P = 20. The value of the optimal solution is reduced to 7.50(10) + 7.40(20) = $223.00. A lower price for the professional grade will not change the S = 10, P = 20 solution because of the requirement for the maximum percentage of kevlar (10%).
39. a. Let S = number of units purchased in the stock fund
M = number of units purchased in the money market fund
Min  8S  +  3M  
s.t.  
50S  +  100M  £  1,200,000 Funds available  
5S  +  4M  ³  60,000 Annual income  
M  ³  3,000 Minimum units in money market 
S, M, ³ 0



Optimal Solution: S = 4000, M = 10000, value = 62000
b. Annual income = 5(4000) + 4(10000) = 60,000
c. Invest everything in the stock fund.
40. Let P1 = gallons of product 1
P2 = gallons of product 2
Min  1P1  +  1P2  
s.t.  
1P1  +  ³  30 Product 1 minimum  
1P2  ³  20 Product 2 minimum  
1P1  +  2P2  ³  80 Raw material 
P1, P2 ³ 0
Optimal Solution: P1 = 30, P2 = 25 Cost = $55
41. a. Let R = number of gallons of regular gasoline produced
P = number of gallons of premium gasoline produced
Max  0.30R  +  0.50P  
s.t.  
0.30R  +  0.60P  £  18,000  Grade A crude oil available  
1R  +  1P  £  50,000  Production capacity  
1P  £  20,000  Demand for premium 
R, P ³ 0
b.
Optimal Solution:
40,000 gallons of regular gasoline
10,000 gallons of premium gasoline
Total profit contribution = $17,000
c.
Constraint  Value of Slack Variable  Interpretation 
1  0  All available grade A crude oil is used 
2  0  Total production capacity is used 
3  10,000  Premium gasoline production is 10,000 gallons less than the maximum demand 
d. Grade A crude oil and production capacity are the binding constraints.

42.


43.


44. a.


b. New optimal solution is A = 0, B = 3, value = 6.
45. a.




b. Feasible region is unbounded.
c. Optimal Solution: A = 3, B = 0, z = 3.
d. An unbounded feasible region does not imply the problem is unbounded. This will only be the case when it is unbounded in the direction of improvement for the objective function.
46. Let N = number of sq. ft. for national brands
G = number of sq. ft. for generic brands
Problem Constraints:
N  +  G  £  200  Space available  
N  ³  120  National brands  
G  ³  20  Generic 
Extreme Point  N  G 
1  120  20 
2  180  20 
3  120  80 
a. Optimal solution is extreme point 2; 180 sq. ft. for the national brand and 20 sq. ft. for the generic brand.
b. Alternative optimal solutions. Any point on the line segment joining extreme point 2 and extreme point 3 is optimal.
c. Optimal solution is extreme point 3; 120 sq. ft. for the national brand and 80 sq. ft. for the generic brand.


Alternative optimal solutions exist at extreme points (A = 125, B = 225) and (A = 250, B = 100).
Cost = 3(125) + 3(225) = 1050
or
Cost = 3(250) + 3(100) = 1050
The solution (A = 250, B = 100) uses all available processing time. However, the solution
(A = 125, B = 225) uses only 2(125) + 1(225) = 475 hours.
Thus, (A = 125, B = 225) provides 600  475 = 125 hours of slack processing time which may be used for other products.
48.
Possible Actions:
i. Reduce total production to A = 125, B = 350 on 475 gallons.
ii. Make solution A = 125, B = 375 which would require 2(125) + 1(375) = 625 hours of processing time. This would involve 25 hours of overtime or extra processing time.
iii. Reduce minimum A production to 100, making A = 100, B = 400 the desired solution.
49. a. Let P = number of fulltime equivalent pharmacists
T = number of fulltime equivalent physicians
The model and the optimal solution are shown below:
MIN 40P+10T
S.T.
1) P+T >=250
2) 2PT>=0
3) P>=90
Optimal Objective Value  
5200.00000  
Variable  Value  Reduced Cost 
P  90.00000  0.00000 
T  160.00000  0.00000 
Constraint  Slack/Surplus  Dual Value 
1  0.00000  10.00000 
2  20.00000  0.00000 
3  0.00000  30.00000 
The optimal solution requires 90 fulltime equivalent pharmacists and 160 fulltime equivalent technicians. The total cost is $5200 per hour.
b.
Current Levels  Attrition  Optimal Values  New Hires Required  
Pharmacists  85  10  90  15 
Technicians  175  30  160  15 
The payroll cost using the current levels of 85 pharmacists and 175 technicians is 40(85) + 10(175) = $5150 per hour.
The payroll cost using the optimal solution in part (a) is $5200 per hour.
Thus, the payroll cost will go up by $50
50. Let M = number of Mount Everest Parkas
R = number of Rocky Mountain Parkas
Max  100M  +  150R  
s.t.  
30M  +  20R  £  7200 Cutting time  
45M  +  15R  £  7200 Sewing time  
0.8M    0.2R  ³  0 % requirement 
Note: Students often have difficulty formulating constraints such as the % requirement constraint. We encourage our students to proceed in a systematic stepbystep fashion when formulating these types of constraints. For example:
M must be at least 20% of total production
M ³ 0.2 (total production)
M ³ 0.2 (M + R)
M ³ 0.2M + 0.2R
0.8M  0.2R ³ 0
The optimal solution is M = 65.45 and R = 261.82; the value of this solution is z = 100(65.45) + 150(261.82) = $45,818. If we think of this situation as an ongoing continuous production process, the fractional values simply represent partially completed products. If this is not the case, we can approximate the optimal solution by rounding down; this yields the solution M = 65 and R = 261 with a corresponding profit of $45,650.
51. Let C = number sent to current customers
N = number sent to new customers
Note:
Number of current customers that test drive = .25 C
Number of new customers that test drive = .20 N
Number sold = .12 ( .25 C ) + .20 (.20 N )
= .03 C + .04 N
Max  .03C  +  .04N  
s.t.  
.25 C  ³  30,000  Current Min  
.20 N  ³  10,000  New Min  
.25 C    .40 N  ³  0  Current vs. New  
4 C  +  6 N  £  1,200,000  Budget 
C, N, ³ 0
52. Let S = number of standard size rackets
O = number of oversize size rackets
Max  10S  +  15O  
s.t.  
0.8S    0.2O  ³  0  % standard  
10S  +  12O  £  4800  Time  
0.125S  +  0.4O  £  80  Alloy 
S, O, ³ 0
53. a. Let R = time allocated to regular customer service
N = time allocated to new customer service
Max  1.2R  +  N  
s.t.  
R  +  N  £  80  
25R  +  8N  ³  800  
0.6R  +  N  ³  0 
R, N, ³ 0
b.
Optimal Objective Value  
90.00000  
Variable  Value  Reduced Cost 
R  50.00000  0.00000 
N  30.00000  0.00000 
Constraint  Slack/Surplus  Dual Value 
1  0.00000  1.12500 
2  690.00000  0.00000 
3  0.00000  0.12500 
Optimal solution: R = 50, N = 30, value = 90
HTS should allocate 50 hours to service for regular customers and 30 hours to calling on new customers.
54. a. Let M1 = number of hours spent on the M100 machine
M2 = number of hours spent on the M200 machine
Total Cost
6(40)M1 + 6(50)M2 + 50M1 + 75M2 = 290M1 + 375M2
Total Revenue
25(18)M1 + 40(18)M2 = 450M1 + 720M2
Profit Contribution
(450  290)M1 + (720  375)M2 = 160M1 + 345M2
Max  160 M1  +  345M2  
s.t.  
M1  £  15  M100 maximum  
M2  £  10  M200 maximum  
M1  ³  5  M100 minimum  
M2  ³  5  M200 minimum  
40 M1  +  50 M2  £  1000  Raw material available 
M1, M2 ³ 0
b.
Optimal Objective Value  
5450.00000  
Variable  Value  Reduced Cost 
M1  12.50000  0.00000 
M2  10.00000  145.00000 
Constraint  Slack/Surplus  Dual Value 
1  2.50000  0.00000 
2  0.00000  145.00000 
3  7.50000  0.00000 
4  5.00000  0.00000 
5  0.00000  4.00000 
The optimal decision is to schedule 12.5 hours on the M100 and 10 hours on the M200.
55. Mr. Krtick’s solution cannot be optimal. Every department has unused hours, so there are no binding constraints. With unused hours in every department, clearly some more product can be made.
56. No, it is not possible that the problem is now infeasible. Note that the original problem was feasible (it had an optimal solution). Every solution that was feasible is still feasible when we change the constraint to lessthanorequalto, since the new constraint is satisfied at equality (as well as inequality). In summary, we have relaxed the constraint so that the previous solutions are feasible (and possibly more satisfying the constraint as strict inequality).
57. Yes, it is possible that the modified problem is infeasible. To see this, consider a redundant greaterthanorequal to constraint as shown below. Constraints 2, 3, and 4 form the feasible region and constraint 1 is redundant. Change constraint 1 to lessthanorequalto and the modified problem is infeasible.
Original Problem:
Modified Problem:
58. It makes no sense to add this constraint. The objective of the problem is to minimize the number of products needed so that everyone’s top three choices are included. There are only two possible outcomes relative to the boss’ new constraint. First, suppose the minimum number of products is <= 15, then there was no need for the new constraint. Second, suppose the minimum number is > 15. Then the new constraint makes the problem infeasible.
Chapter 2—An Introduction to Linear Programming
MULTIPLE CHOICE
1. The maximization or minimization of a quantity is the
a.  goal of management science. 
b.  decision for decision analysis. 
c.  constraint of operations research. 
d.  objective of linear programming. 
ANS: D PTS: 1 TOP: Introduction
2. Decision variables
a.  tell how much or how many of something to produce, invest, purchase, hire, etc. 
b.  represent the values of the constraints. 
c.  measure the objective function. 
d.  must exist for each constraint. 
ANS: A PTS: 1 TOP: Objective function
3. Which of the following is a valid objective function for a linear programming problem?
a.  Max 5xy 
b.  Min 4x + 3y + (2/3)z 
c.  Max 5x^{2 }+ 6y^{2} 
d.  Min (x_{1} + x_{2})/x_{3} 
ANS: B PTS: 1 TOP: Objective function
4. Which of the following statements is NOT true?
a.  A feasible solution satisfies all constraints. 
b.  An optimal solution satisfies all constraints. 
c.  An infeasible solution violates all constraints. 
d.  A feasible solution point does not have to lie on the boundary of the feasible region. 
ANS: C PTS: 1 TOP: Graphical solution
5. A solution that satisfies all the constraints of a linear programming problem except the nonnegativity constraints is called
a.  optimal. 
b.  feasible. 
c.  infeasible. 
d.  semifeasible. 
ANS: C PTS: 1 TOP: Graphical solution
6. Slack
a.  is the difference between the left and right sides of a constraint. 
b.  is the amount by which the left side of a £ constraint is smaller than the right side. 
c.  is the amount by which the left side of a ³ constraint is larger than the right side. 
d.  exists for each variable in a linear programming problem. 
ANS: B PTS: 1 TOP: Slack variables
7. To find the optimal solution to a linear programming problem using the graphical method
a.  find the feasible point that is the farthest away from the origin. 
b.  find the feasible point that is at the highest location. 
c.  find the feasible point that is closest to the origin. 
d.  None of the alternatives is correct. 
ANS: D PTS: 1 TOP: Extreme points
8. Which of the following special cases does not require reformulation of the problem in order to obtain a solution?
a.  alternate optimality 
b.  infeasibility 
c.  unboundedness 
d.  each case requires a reformulation. 
ANS: A PTS: 1 TOP: Special cases
9. The improvement in the value of the objective function per unit increase in a righthand side is the
a.  sensitivity value. 
b.  dual price. 
c.  constraint coefficient. 
d.  slack value. 
ANS: B PTS: 1 TOP: Righthand sides
10. As long as the slope of the objective function stays between the slopes of the binding constraints
a.  the value of the objective function won't change. 
b.  there will be alternative optimal solutions. 
c.  the values of the dual variables won't change. 
d.  there will be no slack in the solution. 
ANS: C PTS: 1 TOP: Objective function
11. Infeasibility means that the number of solutions to the linear programming models that satisfies all constraints is
a.  at least 1. 
b.  0. 
c.  an infinite number. 
d.  at least 2. 
ANS: B PTS: 1 TOP: Alternate optimal solutions
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