Taylor Chap 3 Answers
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Chapter Three: Linear Programming: Computer Solution and Sensitivity Analysis PROBLEM SUMMARY
35. Model formulation
1. QM for Windows
36. Graphical solution; sensitivity analysis (3–35)
2. QM for Windows and Excel
37. Computer solution; sensitivity analysis (3–35)
3. Excel
38. Model formulation; computer solution
4. Graphical solution; sensitivity analysis
39. Sensitivity analysis (3–38)
5. Model formulation
40. Model formulation; computer solution
6. Graphical solution; sensitivity analysis (3–5)
41. Sensitivity analysis (3–40)
7. Sensitivity analysis (3–5)
42. Model formulation
8. Model formulation
43. Computer solution; sensitivity analysis (3–42)
9. Graphical solution; sensitivity analysis (3–8)
44. Model formulation
10. Sensitivity analysis (3–8)
45. Computer solution; sensitivity analysis (3–44)
11. Model formulation
46. Model formulation
12. Graphical solution; sensitivity analysis (3–11)
47. Computer solution; sensitivity analysis (3–46)
13. Computer solution; sensitivity analysis (3–11)
48. Model formulation
14. Model formulation
49. Computer solution; sensitivity analysis (3–48)
15. Graphical solution; sensitivity analysis (3–14)
50. Computer solution
16. Computer solution; sensitivity analysis (3–14)
PROBLEM SOLUTIONS
17. Model formulation
1.
18. Graphical solution; sensitivity analysis (3–17)
x2
19. Computer solution; sensitivity analysis (3–17)
110
20. Model formulation
100
21. Graphical solution; sensitivity analysis (3–20) 90
22. Computer solution; sensitivity analysis (3–20) 80
23. Model formulation 24. Graphical solution; sensitivity analysis (3–23)
70
25. Computer solution; sensitivity analysis (3–23)
60
26. Model formulation
50
27. Graphical solution; sensitivity analysis (3–26)
A 40
28. Computer solution; sensitivity analysis (3–26)
30
optimal point: x1 = 15.29 x2 = 38.24 Z = 4,205.88
B
29. Model formulation
20
30. Graphical solution; sensitivity analysis (3–29)
C
Z 10
31. Computer solution; sensitivity analysis (3–29) 32. Model formulation
0
33. Model formulation; computer solution 34. Computer solution; sensitivity analysis
17
10
20
30
40
50
60
70
80
90
100
110
x1
2.
6.
QM for Windows establishes a “template” for the linear programming model based on the user’s specification of the type of objective function, the number of constraints and number of variables, then the model parameters are input and the problem is solved. In Excel the model “template” must be developed by the user.
x2
*A: x1 = 0 x2 = 160 Z = 2,560
300 250
B: x1 = 128.5 x2 = 57.2 Z = 2457.2
200
A 150
C : x1 = 167 x2 = 0 Z = 2,004
100
3.
4.
Changing cells: B10:B12 Constraints: B10:B12 0 G6 F6 G7 F7 Profit: B10 * C4 B11 * D4 B12 * E4
B
50
Z 0
50
100
Point A is optimal
150 C 200
250
300
350
x1
(a) A: 3(0) + 2(160) + s1 = 500 s1 = 180 4(0) + 5(160) + s2 = 800 s2 = 0 B: 3(128.5) + 2(57.2) + s1 = 500 s1 = 0 4(128.5) + 2(57.2) + s2 = 800 s2 = 0
The slope of the constraint line is –70/60. The optimal solution is at point A where x1 = 0 and x2 = 70. To change the solution to B, c1 must increase such that the slope of the objective function is at least as great as the slope of the constraint line, –c1/50 = –70/60 c1 = 58.33
C: 2(167) + 2(0) + s1 = 500 s1 = 0 4(167) + 5(0) + s2 = 800 s2 = 132
Alternatively, c1 must decrease such that the slope of the objective function is at least as great as the slope of the constraint line, –30/c2 = –70/60 c2 = 25.71
(b) Z = 12x1 + 16x2 and,
Thus, if c1 increases to greater than 58.33 or c2 decreases to less than 25.71, B will become optimal.
x2 = Z/16 – 12 x1/16 The slope of the objective function, –12/16, would have to become steeper (i.e., greater) than the slope of the constraint line 4x1 + 5x2 = 800, for the solution to change.
5. (a) x1 = no. of basketballs x2 = no. of footballs maximize Z = 12x1 + 16x2 subject to
The profit, c1, for a basketball that would change the solution point is,
3x1 + 2x2 ≤ 500 4x1 + 5x2 ≤ 800 x1, x2 ≥ 0
4/5 = –c1/16 5c1 = 64 c1 = 12.8
(b) maximize Z = 12x1 + 16x2 + 0s1 + 0s2 subject to
Since $13 > 12.8 the solution point would change to B where x1 = 128.5, x2 = 57.2. The new Z value is $2,585.70.
3x1 + 2x2 + s1 = 500 4x1 + 5x2 + s2 = 800 x1, x2, s1, s2 ≥ 0
For a football, –4/5 = –12/c2 4c2 = 60 c2 = 15 Thus, if the profit for a football decreased to $15 or less, point B will also be optimal (i.e., multiple optimal solutions). The solution at B is x1 = 128.5, x2 = 57.2 and Z = $2,400. 18
point A where x1 = 0, x2 = 160,
(c) If the constraint line for rubber changes to 3x1 + 2x2 = 1,000, it moves outward, eliminating points B and C. However, since A is the optimal point, it will not change and the optimal solution remains the same, x1 = 0, x2 = 160 and Z = 2,560. There will be an increase in slack, s1, to 680 lbs.
3x1 + 2x2 = q1 3(0) + 2(160) = q1 q1 = 320 For q2 the upper limit is at the point where the rubber constraint line (3x1 + 2x2 = 500) intersects with the leather constraint line (4x1 + 5x2 = 800) along the x2 axis, i.e., x1 = 0, x2 = 250,
If the constraint line for leather changes to 4x1 + 5x2 = 1,300, point A will move to a new location, x1 = 0, x2 = 250, Z = $4,000.
4x1 + 5x2 = q2 4(0) + 5(250) = q2 q2 = 1,250
7. (a) For c1 the upper limit is computed as –4/5 = –c1/16 5c1 = 64 c1 = 12.8
The lower limit is 0 since that is the lowest point on the x2 axis the constraint line can decrease to.
and the lower limit is unlimited.
Summarizing,
For c2 the lower limit is,
320 ≤ q1 ≤ ∞ 0 ≤ q2 ≤ 1,250
–4/5 = –12/c2 4c2 = 60 c2 = 15
(b)
and the upper limit is unlimited.
Z = 2560.000
Summarizing,
Variable
Value
Reduced Cost
x1
0.00
0.800
x2
160.000
0.000
Constraint
Slack/Surplus
Shadow Price
c1
180.00
0.00
c2
0.00
3.20
∞ ≤ c1 ≤ 12.8 15 ≤ c2 ≤ ∞ For q1 the upper limit is ∞ since no matter how much q1 increases the optimal solution point A will not change. The lower limit for q1 is at the point where the constraint line 3x1 + 2x2 = q1 intersects with Objective Coefficient Ranges Variables x1 x2
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
No limit 15.000
12.000 16.000
12.800 No limit
0.800 No limit
No limit 1.000
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
500.000 800.000
No limit 1250.000
No limit 450.000
180.000 800.000
Right Hand Side Ranges Constraints c1 c2
Lower Limit 320.000 0.000
19
(b) The constraint line 12x1 + 4x2 = 60 would move inward resulting in a new location forpoint B at x1 = 2, x2 = 4, which would still be optimal.
(c) The shadow price for rubber is $0. Since there is slack rubber left over at the optimal point, extra rubber would have no marginal value.
(c) In order for the optimal solution point to change from B to A the slope of the objective function must be at least as flat as the slope of the constraint line, 4x1 + 8x2 = 40, which is –1/2. Thus, the profit for product B would have to be,
The shadow price for leather is $3.20. For each additional ft.2 of leather that the company can obtain profit would increase by $3.20, up to the upper limit of the sensitivity range for leather (i.e., 1,250 ft.2).
–9/c2 = –1/2 c2 = 18
8.(a) x1 = no. of units of A x2 = no. of units of B maximize Z = 9x1 + 7x2
If the profit for product B is increased to $15 the optimal solution point will not change, although Z would change from $57 to $81.
subject to 12x1 + 4x2 ≤ 60 4x1 + 8x2 ≤ 40 x1,x2 ≥ 0
If the profit for product B is increased to $20 the solution point will change from B to A, x1 = 0, x2 = 5, Z = $100.
(b) maximize Z = 9x1 + 7x2 + 0s1 + 0s2 subject to
10.(a) For c1 the upper limit is computed as,
12x1 + 4x2 + s1 = 60 4x1 + 8x2 + s2 = 40 x1, x2, s1, s2 ≥ 0
–c1/7 = –3 c1 = 21 and the lower limit is, –c1/7 = –1/2 c1 = 3.50
9. x2
A: x1 = 0 x2 = 5 Z = 35
30
For c2 the upper limit is,
25 20 15
and the lower limit is,
C: x1 = 5 x2 = 0 Z = 45
10 5
–9/c2 = –1/2 c2 = 18
*B: x1 = 4 x2 = 3 Z = 57
A
Summarizing,
Point B is optimal
B C 0
5
–9/c2 = –3 c2 = 3
10
15
20
25
30
35
40
3.50 ≤ c1 ≤ 21 –3 ≤ c2 ≤ 18
x1
(b)
(a) A: 12(0) + 4(5) + s1 = 60 s1 = 40 4(0) + 8(5) + s2 = 40 s2 = 0 B:
Z = 57.000
12(4) + 4(3) = 60 s1 = 0 4(4) + 8(3) + s2 = 40 s2 = 0
C: 12(5) + 4(0) + s1 = 60 s1 = 0 4(5) + 8(0) + s2 = 40 s2 = 20
20
Variable
Value
Reduced Cost
x1
4.000
0.000
x2
3.000
0.000
Constraint
Slack/Surplus
Shadow Price
c1
0.000
0.550
c2
0.000
0.600
Objective Coefficient Ranges Variables
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1 x2
3.500 3.000
9.000 7.000
21.000 18.000
12.000 11.000
5.500 4.000
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
60.000 40.000
120.000 120.000
60.000 80.000
40.000 20.000
Right Hand Side Ranges Lower Limit
Constraints c1 c2
20.000 20.000
(c) The shadow price for line 1 time is $0.55 per hour, while the shadow price for line 2 time is $0.60 per hour. The company would prefer to obtain more line 2 time since it would result in the greatest increase in profit.
(a)
11.(a) x1 = no. of yards of denim x2 = no. of yards of corduroy maximize Z = $2.25x1 + 3.10x2 subject to
(b) In order for the optimal solution point to change from B to C the slope of the objective function must be at least as great as the slope of the constraint line, 3.0x1 + 3.2x2 = 3,000, which is –3/3.2. Thus, the profit for denim would have to be, –c1/3.0 = –3/3.2 c1 = 2.91 If the profit for denim is increased from $2.25 to $3.00 the optimal solution would change to point C where x1 = 1,000, x2 = 0, Z = 3,000. Profit for corduroy has no upper limit that would change the optimal solution point.
5.0x1 + 7.5x2 ≤ 6,500 3.0x1 + 3.2x2 ≤ 3,000 x2 ≤ 510 x1, x2 ≥ 0 (b)maximize Z = $2.25x1 + 3.10x2 + 0s1 + 0s2 + 0s3 subject to 5.0x1 + 7.5x2 + s1 = 6,500 3.0x1 + 3.2x2 + s2 = 3,000 x2 + s3 = 510 x1, x2, s1, s2, s3 ≥ 0 12.
5.0(456) + 7.5(510) + s1 = 6,500 s1 = 6,500 – 6,105 s1 = 395 lbs. 3.0(456) + 3.2(510) + s2 = 3,000 s2 = 0 hrs. 510 + s3 = 510 s3 = 0 therefore demand for corduroy is met.
(c) The constraint line for cotton would move inward as shown in the following graph where point C is optimal.
x2
A: x1 = 0 x2 = 510 Z = $1,581
1600
x2 1600
1400
*B: x1 = 456 x2 = 510 Z = $2,607
1200 1000
1400 1200
C: x1 = 1,000 x2 = 0 Z = $2,250
800
1000
C, optimal x1 = 1,000 x2 = 0 Z = $2,250
800 600 A
600 A
B
400
400
B
C
400
600
200 200
C 0
200
400
600
800 1000 1200 1400 1600
x1
D 0
21
200
800 1000 1200 1400 1600
x1
13. Z = 2607.000 Variable
Value
Reduced Cost
x1 x2
456.000 510.000
0.000 0.000
Constraint c1 c2 c3
Slack/Surplus
Shadow Price
395.000 0.000 0.000
0.000 0.750 0.700
Objective Coefficient Ranges Variables
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1 x2
0.000 2.400
2.250 3.100
2.906 No limit
0.656 No limit
2.250 0.700
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
6015.000 1632.000 0.000
6500.000 3000.000 510.000
No limit 3237.000 692.308
No limit 237.000 182.308
395.000 1368.000 510.000
Right Hand Side Ranges Constraints c1 c2 c3
(a) The company should select additional processing time, with a shadow price of $0.75 per hour. Cotton has a shadow price of $0 because there is already extra (slack) cotton available and not being used so any more would have no marginal value. (b)
14.
15. x2 8
A: x1 = 0 x2 = 6 Z = 42,000
7
A
6
0 ≤ c1 ≤ 2.906 2.4 ≤ c2 ≤ ∞
6,105 ≤ q1 ≤ ∞ 1,632 ≤ q2 ≤ 3,237 0 ≤ q3 ≤ 692.308 The demand for corduroy can decrease to zero or increase to 692.308 yds. without changing the current solution mix of denim and corduroy. If the demand increases beyond 692.308 yds., then denim would no longer be produced and only corduroy would be produced.
*C: x1 = 4 x2 = 0 Z = 24,000
B: x1 = 1 x2 = 3 Z = 27,000
5 4
B
3 2 1
C
x1 = no. of days to operate mill 1 x2 = no. of days to operate mill 2 minimize Z = 6,000x1 + 7,000x2 subject to
0
6x1 + 2x2 ≥ 12 2x1 + 2x2 ≥ 8 4x1 + 10x2 ≥ 5 x1, x2 ≥ 0 22
1
2
3
4
5
6
7
8
x1
(a) 6(4) + 2(0) – s1 = 12 s1 = 12 2(4) + 2(0) – s2 = 8 s2 = 0 4(4) + 10(0) – s3 = 5 s3 = 11
does not change the optimal variable mix. B remains optimal but moves to a new location, x1 = 0.5, x2 = 3.5, Z = $27,500. 16. Z = 24000
(b) The slope of the objective function, –6000/7,000 must become flatter (i.e., less) than the slope of the constraint line,
Variable
Value
x1
4.000
x2
0.000
Constraint
Slack/Surplus
Shadow Price
c1
12.000
0.000
c2 c2
0.000 11.000
–3000.000 0.000
2x1 + 2x2 = 8, for the solution to change. The cost of operating Mill 1, c1, that would change the solution point is, –c1/7,000 = –1 c1 = 7,000 Since $7,500 > $7,000, the solution point will change to B where x1 = 1, x2 = 3, Z = $28,500. (c) If the constraint line for high-grade aluminum changes to 6x1 + 2x2 = 10, it moves inward but Objective Coefficient Ranges Variables
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1
0.000
6000.000
7000.000
1000.000
6000.000
x2
6000.000
7000.000
No limit
No limit
1000.000
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
Right Hand Side Ranges Constraints
Lower Limit
c1
No limit
12.000
24.000
12.000
No limit
c2
4.000
8.000
No limit
No limit
4.000
c3
No limit
5.000
16.000
11.000
No limit
(a) There is surplus high-grade and low-grade aluminum so the shadow price is $0 for both. The shadow price for medium-grade aluminum is $3,000 indicating that for every ton that this constraint could be reduced, cost will decrease by $3,000. (b)
0 ≤ c1 ≤ 7,000 6,000 ≤ c2 ≤ ∞
17.
x1 = no. of acres of corn x2 = no. of acres of tobacco maximize Z = 300x1 + 520x2 subject to x1 + x2 ≤ 410 105x1 + 210x2 ≤ 52,500 x2 ≤ 100 x1, x2 ≥ 0
∞ ≤ q1 ≤ 24 4 ≤ q2 ≤ ∞ ∞ ≤ q3 ≤ 16
(c) There will be no change.
23
18.
x2
A: x1 = 0 x2 = 100 Z = 52,000
600 500
B: x1 = 300 x2 = 100 Z = 142,000
400
The profit for corn must be greater than $520 for the Bradleys to plant only corn.
*C: x1 = 320 x2 = 90 Z = 142,800
(c) If the constraint line changes from x1 + x2 = 410 to x1 + x2 = 510, it will moveoutward to a location which changes the solution to the point where 105x1 + 210x2 = 52,500 intersects with the axis. This new point is x1 = 500, x2 = 0, Z = $150,000.
D: x1 = 410 x2 = 0 Z = 123,000
300 200 100
0
(a)
(d) If the constraint line changes from x1 + x2 = 410 to x1 + x2 = 360, it moves inward to a location which changes the solution point to the intersection of x1 + x2 = 360 and 105x1 + 210x2 = 52,500. At this point x1 = 260, x2 = 100 and Z = $130,000.
A
100
200
B
C
300
400D 500
Point C is optimal 600
700
800
x1
x1 = 320, x2 = 90 320 + 90 + s1 = 410 s1 = 0 acres uncultivated 90 + s3 = 100 s3 = 10 acres of tobacco allotment unused
19. Z = 142800.000
(b) At point D only corn is planted. In order for point D to be optimal the slope of the objective function will have to be at least as great (i.e., steep) as the slope of the constraint line, x1 + x2 = 410, which is –1. Thus, the profit for corn is computed as, –c/520 = –1 c1 = 520
Variable
Value
x1
320.000
x2
90.000
Constraint
Slack/Surplus
Shadow Price
c1
0.000
80.000
c2 c3
0.000 10.000
2.095 0.000
Objective Coefficient Ranges Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1
260.000
300.000
520.000
220.000
40.000
x2
300.000
520.000
600.000
80.000
220.000
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
Variables
Right Hand Side Ranges Constraints
Lower Limit
c1
400.000
410.000
500.000
90.000
10.000
c2
43050.000
52500.000
53550.000
1050.000
9450.000
c3
90.000
100.000
No limit
No limit
10.000
24
(a)x1 = 300, x2 = 100, Z = $230 .10(300) + s1 = 30 s1 = 0 left over sausage .15(100) + s2 = 30 s2 = 15 lbs. left over ham .01(300) + .024(100) + s4 = 6 s4 = 0.6 hr.
(a) No, the shadow price for land is $80 per acre indicating that profit will increase by no more than $80 for each additional acre obtained. The maximum price the Bradley’s should pay is $80 and the most they should obtain is at the upper limit of the sensitivity range for land. This limit is 500 acres, or 90 additional acres. Beyond 90 acres the shadow price would change.
(b) The slope of the objective function, –6/5, must become flatter (i.e., less) than the slope of the constraint line, .04x1 + .04x2 = 16, for the solution to change. The profit for ham, c2, that would change the solution point is,
(b) The shadow price for the budget is $2.095. Thus, for every $1 dollar borrowed they could expect a profit increase of $2.095. If they borrowed $1,000 it would not change the amount of corn and tobacco they plant since the sensitivity range has a maximum allowable increase of $1,050. 20.
–0.6/c2 = –1 c2 = .60 Thus, an increase in profit for ham of 0.60 will create a second optimal solution point at C where x1 = 257, x2 = 143 and Z = $225.70. (Point D would also continue to be optimal, i.e., multiple optimal solutions.)
x1 = no. of sausage biscuits x2 = no. of ham biscuits maximize Z = .60x1 + .50x2 subject to
(c) A change in the constraint line from, .04x1 + .04x2 = 16 to .04x1 + .04x2 = 18would move the line outward, eliminating both points C and D. The new solution point occurs at the intersection of 0.01x1 + .024x2 = 6 and .10x = 30. This point is x1 = 300, x2 = 125, and Z = $242.50.
.10x1 ≤ 30 .15 x2 ≤ 30 .04x1 + .04x2 ≤ 16 0..01x1 + .024x2 ≤ 6 x1, x2 ≥ 0
22.
21.
Z = 230.000
x2 600 500 400
A: x1 = 0 x2 = 200 Z = 100
C: x1 = 257 x2 = 143 Z = 225.70
B: x1 = 120 x2 = 200 Z = 172
*D: x1 = 300 x2 = 100 Z = 230
300 200
A
E: x1 = 300 x2 = 0 Z = 180
B C
100
Point D is optimal
D E 0
100
200
300
400
500
600
700
800
x1
Variable
Value
x1
300.000
x2
100.000
Constraint
Slack/Surplus
Shadow Price
c1
0.000
1.000
c2
15.000
0.000
c3
0.000
12.500
c4
0.600
0.000
Objective Coefficient Ranges Variables
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1
0.500
0.600
No limit
No limit
0.100
x2
0.000
0.500
0.600
0.100
0.500
25
Right Hand Side Ranges Lower Limit
Constraints
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
c1
25.714
30.000
40.000
10.000
4.286
c2
15.000
30.000
No limit
No limit
15.000
c3
12.000
16.000
17.000
1.000
4.000
c4
5.400
6.000
No limit
No limit
0.600
(a) The shadow price for sausage is $1. For every additional pound of sausage that can be obtained profit will increase by $1. The shadow price for flour is $12.50. For each additional pound of flour that can be obtained, profit will increase by this amount. There are extra ham and labor hours available, so their shadow prices are zero, indicating additional amounts of those resources would add nothing to profit.
(a) The optimal point is at B where x1 = 27.5 and x2 = 20. The slope of the objective function –50/70, must become greater (i.e., steeper) than the slope of the constraint line, 80x1 + 40x2 = 3,000, for the solution point to change from B to A. The cost of a telephone interviewer that would change the solution point is, –c1/70 = –2 c1 = 140
(b) The constraint for flour, indicated by the high shadow price. (c)
This is the upper limit of the sensitivity range for c1. The lower limit is 0 since as the slope of the objective function becomes flatter, the solution point will not change from B until the objective function is parallel with the constraint line. Thus,
.50 ≤ c1 ≤ ∞ 25.714 ≤ q1 ≤ 40 The sensitivity range for profit indicates that the optimal mix of sausage and ham biscuits will remain optimal as long as profit does not fall below $0.50. The sensitivity range for sausage indicates the optimal solution mix will be maintained as long as the available sausage is between 25.714 and 40 lbs.
23.
0 ≤ c1 ≤ 140 Since the constraint line is vertical, it can increase as far as point B and decrease all the way to the x2 axis before the solution mix will change. At point B,
x1 = no. of telephone interviewers x2 = no. of personal interviewers minimize Z = 50x1 + 70x2 subject to
80(27.5) = q1 q1 = 2,200 At the axis, 80(0) = q1 q1 = 0
80x1 + 40x2 ≥ 3,000 80x1 ≥ 1,000 40x2 ≥ 800 x1, x2 ≥ 0 24.
Summarizing, 0 ≤ q1 ≤ 2,200
x2
(b) At the optimal point, B, x1 = 27.5 and x2 = 20.
80
A: x1 = 12.5 x2 = 50 Z = 4,125
70 60
*B: x1 = 27.5 x2 = 20 Z = 2,775
A
50
80(27.5) – s2 s2 40(20) – s3 s3
(c) A change in the constraint line from 40x2 = 800 to 40x2 = 1,200, moves the lineup, but it does not change the optimal mix. The new solution values are x1 = 22.5, x2 = 30, Z = $3,225.
40 30
B
20
Point B is optimal
10
0
10
20
30
40
50
60
70
= 1,000 = 1,200 extra telephone interviews = 800 =0
x1
26
25. Z = 2775.000 Variable
Value
x1
27.500
x2
20.000
Constraint
Slack/Surplus
Shadow Price
c1
0.000
–0.625
c2
1200.000
0.000
c3
0.000
–1.125
Objective Coefficient Ranges Variables
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1
0.000
50.000
140.000
90.000
50.000
x2
25.000
70.000
No limit
No limit
45.000
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
Right Hand Side Ranges Lower Limit
Constraints c1
1800.000
3000.000
No limit
No limit
1200.000
c2
No limit
1000.000
2200.000
1200.000
No limit
c3
0.000
800.000
2000.000
1200.000
800.000
27.
(a) Reduce the personal interview requirement; it will reduce cost by $0.625 per interview, while a telephone interview will not reduce cost; i.e., it has a shadow price equal to $0. (b)
x2
A: x1 = 266.7 x2 = 133.3 Z = 1,333.20
600 500
25 ≤ c2 ≤ ∞ 1,800 ≤ q1 ≥ ∞
*B: x1 = 333.3 x2 = 166.7 Z = 1,666
400 300
26.
x1 = no. of gallons of rye x2 = no. of gallons of bourbon maximize Z = 3x1 + 4x2 subject to
200
B A
100
Z
x1 + x2 ≥ 400 x1 ≥ .4(x1 + x2) x2 ≤ 250 x1 = 2x2 x1 + x2 ≤ 500 x1, x2 ≥ 0
0
27
100
Feasible solution line Point B is optimal
200
300
400
500
600
700
x1
30.
(a) Optimal solution at B: x1 = 333.3 and x2 = 166.7 (333.3) + (166.7) – s1 = 400 s1 = 100 extra gallons of blended whiskey produced .6(333.33) – .4(166.7) – s2 = 0 s2 = 133.3 extra gallons of rye in the blend (166.7) + s3 = 250
x2 160 140 120
*B: x1 = 70,833.33 x2 = 24,166.67 Z = $116,416.67
100 80
s3 = 83.3 fewer gallons of bourbon than the maximum (333.3) + (166.7) + s4 = 500 s4 = 100 gallons of blend production capacity left over
C: x1 = 95,000 x2 = 0 Z = $114,000
A 60 40
B 20
(b) Because the “solution space” is not really an area, but a line instead, the objective function coefficients can change to any positive value and the solution point will remain the same, i.e., point B. Observing the graph of this model, no matter how flatter or steeper the objective function becomes, point B will remain optimal.
C 0
20
40
60
80
100
120
140
160
Z = 1666.667
–1.2/c2 = –.18/30 c2 = 2
Variable
Value
x1
333.333
x2
166.667
Constraint
Slack/Surplus
Shadow Price
c1
100.000
0.000
c2
133.333
0.000
c3
83.333
0.000
c5
0.000
3.333
x1
(a) The optimal solution point is at B where x1 = $70,833.33, and x2 = $24, 166.67. The slope of the objective function, –1.2/1.3, must become flatter than the slope of the constraint line, .18x1 + .30x2 = 20,000, for the solution point to change to A (i.e., only cattle). The return on cattle that will change the solution point is
28.
29.
A: x1 = 0 x2 = 66,666.7 Z = $86,667
Thus, the return must be 100% before Alexis will invest only in cattle. (b) Yes, there is no slack money left over at the optimal solution. (c) Since her investment is $95,000, she could expect to earn $21,416.67.
x1 = $ amount invested in land x2 = $ amount invested in cattle maximize Z = 1.20x1 + 1.30x2 subject to x1 + x2 ≤ 95,000 .18x1 + .30x2 ≤ 20,000 x1, x2 ≥ 0
28
Objective Coefficient Ranges Variables
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1 x2
–2.000 –6.000
3.000 4.000
No limit No limit
No limit No limit
5.000 10.000
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
No limit No limit 166.667 –250.000 400.000
400.000 0.000 250.000 0.000 500.000
500.000 133.333 No limit 500.000 750.000
100.000 133.333 No limit 500.000 250.000
No limit No limit 83.333 250.000 100.000
Right Hand Side Ranges Constraints c1 c2 c3 c4 c5
(a)–2.0 ≤ c1 ≤ ∞ –6.0 ≤ c2 ≤ ∞
must be solved again on the computer, which results in the following solution output.
Because there is only one effective solution point the objective function can take on any negative (downward) slope and the solution point will not change. Only “negative” coefficients that result in a positive slope will move the solution to point A, however, this would be unrealistic.
Z = 1625.000
(b)The shadow price for production capacity is $3.33. Thus, for each gallon increase in capacity profit will increase by $3.33. (c)This new specification changes the constraint, x1 – 2x2 = 0, to x1 – 3x2 = 0. This change to a constraint coefficient cannot be evaluated with normal sensitivity analysis. Instead the model
Variable
Value
x1
375.000
x2
125.000
Constraint
Slack/Surplus
Shadow Price
c1
100.000
0.000
c2
175.000
0.000
c3
125.000
0.000
c5
0.000
3.250
Objective Coefficient Ranges Variables
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1 x2
–1.333 –9.000
3.000 4.000
No limit No limit
No limit No limit
4.333 13.000
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
No limit No limit 125.000 –500.000 400.000
400.000 0.000 250.000 0.000 500.000
500.000 175.000 No limit 500.000 1000.000
100.000 175.000 No limit 500.000 500.000
No limit No limit 125.000 500.000 100.000
Right Hand Side Ranges Constraints c1 c2 c3 c4 c5
29
31. Z = 116416.667 Variable
Value
x1
70833.333
x2
24166.667
Constraint
Slack/Surplus
Shadow Price
c1
0.000
1.050
c2
0.000
0.833
Objective Coefficient Ranges Variables
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1
0.780
1.200
1.300
0.100
0.420
x2
1.200
1.300
2.000
0.700
0.100
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
c1
66666.667
95000.000
111111.111
16111.111
28333.333
c2
17100.000
20000.000
28500.000
8500.000
2900.000
Right Hand Side Ranges Constraints
(a) The shadow price for invested money is $1.05.Thus, for every dollar of her own money Alexis invested she could expect a return of $0.05 or 5%. The upper limit of the sensitivity range is $111,111.11, thus, Alexis could invest $16,111.11 of her own money before the shadow price would change.
be solved again using the computer, as follows.
Variable
Value
x1
0.000
x2
66666.667
(b) This would change the constraint, .18x1 + .30x2 = 20,000 to .30x1 + .30x2 = 20,000. In order to assess the effect of this change the problem must
Constraint
Z = 86666.667
Slack/Surplus 28333.33
c1 c2
0.000
Objective Coefficient Ranges Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1
No limit
1.200
1.300
0.100
No limit
x2
1.200
1.300
No limit
No limit
0.100
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
c1
66666.667
95000.000
No limit
No limit
28333.333
c2
0.000
20000.000
28500.000
8500.000
20000.000
Variables
Right Hand Side Ranges Constraints
30
Shadow Price 0.000 4.333
32.
maximize Z = 140x1 + 205x2 + 190x3 + 0s1 + 0s2 + 0s3 + 0s4 subject to
Z = 9765.596 Variable
Value
Reduced Cost
x1
22.385
0.000
x2
16.789
0.000
x3
16.789
0.000
Constraint
Slack/Surplus
Shadow Price
10x1 + 15x2 + 8x3 +s1 = 610 x1 – 3x2 + s2 = 0 .6x1 – .4x2 – .4x3 – s3 = 0 x2 – x3 – s4 = 0 x1, x2, s1, s2, s3, s4 ≥ 0
c1
0.000
16.009
c2
27.982
0.000
c3
0.000
–33.486
c4
0.000
–48.532
Objective Coefficient Ranges Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1
–237.857
140.000
171.739
31.739
377.857
x2
132.000
205.000
325.227
120.227
73.000
x3
117.000
190.000
No limit
No limit
73.000
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
c1
0.000
610.000
No limit
No limit
610.000
c2
–27.982
0.000
No limit
No limit
27.982
c3
–21.217
0.000
11.509
11.509
21.217
c4
–20.890
0.000
28.154
28.154
20.890
Variables
Right Hand Side Ranges Constraints
33.
(c)
(a) and (b) minimize Z = $400x1 + 180x2 + 90x3 subject to
Z = 206000.000
x1 ≥ 200 x2 ≥ 300 x3 ≥ 100 4x3 – x1 – x2 ≤ 0 x1 + x2 + x3 = 1,000 x1, x2, x3 ≥ 0
31
Variable
Value
x1
200.000
x2
600.000
x3
200.000
Constraint
Slack/Surplus
c1
0.000
–220.000
c2
500.000
0.000
c3
100.000
0.000
c4
0.000
18.000
Shadow Price
Objective Coefficient Ranges Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1
180.000
400.000
No limit
No limit
220.000
x2
90.000
180.000
400.000
220.000
90.000
x3
No limit
90.000
180.000
90.000
No limit
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
c1
0.000
200.000
700.000
500.000
200.000
c2
No limit
100.000
600.000
500.000
No limit
c3
No limit
100.000
200.000
100.000
No limit
c4
–500.000
0.000
2500.000
2500.000
500.000
c5
500.000
1000.000
No limit
No limit
500.000
Variables
Right Hand Side Ranges Constraints
34.(a) x1 36.7142 x2 58.6371 x3 0 x4 63.5675 Z 9,177.85
36. 18
B: x1 8,000 x2 10,000 Z $11,500
16 14 12
x2 (1000s)
(b) 34.6871 q1 61.5335 43.0808 q2 71.2315 q3 65.4686 55 q4 529.0816 c1 747.9999 350.3345 c2 3,488.554 c3 1,363.636 c4 1,761.476 20.132 c5 64.4643
Optimal 10
B
8
A
4 2 0
2
4
6
(c) process 1 time is the most valuable with a dual value of $7.9275
8
10 12 14 x1 (1000s)
16
18
(a) c2 .50
(d) Product 3(x3) is not produced; it would require a profit of $65.4686 to be produced. 35.
C
6
(b) s1 $140
Maximize Z $0.50x1 0.75x2 subject to:
(c) There would be no feasible solution.
$0.17x1 0.25x2 $4,000 (printing budget) x1 x2 18,000 (total copies, rack space) x1 8,000 (entertainment guide) x2 8,000 (real estate guide) x1, x2 0
32
20
22
37.
x1 8,000 x2 10,000 Z $11,500
the original solution with a 140,000 ft2 store, thus, given these conditions, Mega-Mart should not purchase the land. 40.(a) Maximize Z $0.97x1 0.83x2 0.69x3 subject to:
(a) The dual value of rack space is 0.75 so an increase in rack space to accommodate an additional 500 copies would result in increased advertising revenue of $375. An increase in rack space to 20,000 copies would be outside the sensitivity range for this constraint and require the problem to be solved again. The new solution is x1 8,000, x2 10,560 and Z $11,920
x1 x2 x3 324 cartons x3 x1 x2 x1 + x 2 + x3 ≤ 324 cartons x3 ≥ x1 + x 2 x3 ≥3 x1 x 2 ≤ 120
(b) 7,000 is within the sensitivity range for the entertainment guide (6,250 q3 10,000). The dual value is $0.25 thus for every unit the distribution requirement can be reduced, revenue will be increased by $0.25, or $250. Thus, Z $11,750
(b) x1 = 54,x2 = 108,x3 = 162,Z 5 $253.80 41.(a) The shadow price for shelf space is $0.78 per carton, however, this is only valid up to 360 cartons, the upper limit of the sensitivity range for shelf space.
38. (a) Maximize Z 4.25x1 5.10x2 4.50x2 5.20x4 4.10x5 4.90x6 3.80x7 subject to:
(b)The shadow price for available local dairy cartons is $0 so it would not increase profit to increase the available amount of local dairy milk.
x1 x2 x3 x4 x5 x6 x7 x8 140,000 xi 15,000, i 1, 2, …,7
(c) The discount would change the objective function to,
xi ≥ 15, 000, i = 1, 2, … , 7 xi ≤ .20, i = 1, 2, … , 7 Σxi
maximize Z 0.86x1 0.83x2 0.69x3 and the constraint for relative demand would change to x3 ≥ 1.5 x1
x8 = .10 x 4 + x6 + x 7 xi ≥ 0
the resulting optimal solution is,
(b) x1 15,000 x2 26,863 x3 20,588.24 x4 26,862.75 x5 15,000 x6 15,000 x7 15,000 x8 5,686 Z $625,083
x1 = 108, x2 = 54, x3 = 162, Z $249.48 Since the profit declines the discount should not be implemented. 42.
x1 = road racing bikes x2 = cross country bikes x3 = mountain bikes maximize Z 600x1 400x2 300x3 subject to
39.(a) A 20,000 ft2 increase in store size to 160,000 ft2 would increase annual profit to $718,316. This is a $93,233 increase in profit. Given the price of the land ($190,000) relative to the increase in profit, it would appear that the cost of the land would be offset in about 2 years, therefore the decision should be to purchase the land.
1,200x1 1,700x2 900x3 $12,000 x1 x2 x3 20 8x1 12x2 16x3 120 x3 2(x1 x2) x1, x2, x3 0
(b) The decrease in profit in all departments would result in a new solution with Z $574,653. This is a reduction of $50,430 annually in profit from
43.
33
x1 = 3 x3 = 6 Z 3,600
(a) More hours to assemble; the dual value for budget and space is zero, while the dual value for assembly is $30/hour.
47.
(b) The additional net sales would be $900. Since the cost of the labor is $300, the additional profit would be $600.
(a) The sensitivity range for x2 is 7,500 c2 8,774.999. Since $7,600 is within this range the values for x1, x2, and x3 would not change, but the profit would decline to $683,333.30 (i.e., less the difference in profit, ($600)(x2 33.3334)
(c) It would have no effect on the original solution. $700 profit for a cross country bike is within the sensitivity range for the objective function coefficient for x2. 44.
(b) One ton of grapes; the dual value is $23,333.35 (c) Grapes: (0.5)($23,333.35) $11,666.68 Casks: (4)($3,833.329) $15,333.32 Production: $0 Select the casks.
Maximize Z $0.35x1 0.42x2 0.37x3 subject to: 0.45x1 0.41x2 0.50x3 960 x1 x2 x3 2,000 x1 200 x2 200 x3 200 x1 x2 x3 x1, x2, x3 0
45.
x1 20 x2 33.3334 x3 26.6667 Z $703,333.40
(d) $6,799; slightly less than the lower band of the sensitivity range for cj. 48.
x1 1,000 x2 800 x3 200 Z $760
(a) Increase vending capacity by 100 sandwiches. There is already excess assembly time available (82 minutes) and the dual value is zero whereas the dual value of vending machine capacity is $0.38. $38 in additional profit.
Minimize Z $37x11 37x12 37x13 46x21 46x22 46x23 50x31 50x32 50x33 42x41 42x42 42x43 subject to: .7x11 .6x21 .5x31 .3x41 400 tons .7x12 .6x22 .5x32 .3x42 250 tons .7x13 .6x23 .5x33 .3x43 290 tons x11 x12 x13 350 tons x21 x22 x23 530 tons x31 x32 x33 610 tons x41 x42 x43 490 tons
49. x13 350 tons x21 158.333 tons x22 296.667 tons x23 75 tons x31 610 tons x42 240 tons Z $77,910
(b) x1 1,000 x2 1,000 Z $770 The original profit is $760 and the new solution is $770. It would seem that a $10 difference would not be worth the possible loss of customer goodwill due to the loss of variety in the number of sandwiches available.
Mine 1 350 tons Mine 2 530 tons Mine 3 610 tons Mine 4 240 tons
(c) Profit would increase to $810 but the solution values would not change. If profit is increased to $0.45 the solution values change to x1 1,600, x2 200, x3 200.
Multiple optimal solutions exist (a) Mine 4 has 240 tons of “slack” capacity.
46.(a) Maximize Z 7,500x1 8,200x2 10,500x3 subject to:
(b) The dual values for the 4 constraints representing the capacity at the 4 mines show that mine 1 has the highest dual value of $61, so its capacity is the best one to increase.
.21x1 .24x2 .18x3 17 x1 x2 x3 80 12x1 14.5x2 16x3 2,500 x3 (x1 x2)/2 x1, x2, x3 0
(c) The sensitivity range for mine 1 is 242.8571 c1 414.2857, thus capacity could be increased 34
by 64.2857 tons before the optimal solution point would change.
.16x1 + .20x2 + s3 = 40 32.8x1 + 20x2 + s4 = 6,000 x1, x2, s1, s2, s3, s4 ≥ 0
(d) The effect of simultaneous changes in objective function coefficients and constraint quality values cannot be analyzed using the sensitivity ranges provided by the computer output. It is necessary to make both changes in the model and solve it again. Doing so results in a new solution with Z $73,080, which is $4,830 less than the original solution, so Exeter should make these changes.
(c) x2 400 350 300
50.
minimize Z = 8.2x1 + 7.0x2 +6.5x3 + 9.0x4 + 0s1 + 0s2 + 0s3 + 0s4 subject to
clay
250
D: x1 = 136.36 x2 = 76.36 Z = $44,234.80
*B: x1 = 56.70 x2 = 154.64 Z = $47,886.60
E: x1 = 182.93 x2 = 0 Z = $34,756.70
C : x1 = 100 x2 = 120 Z = $47,800
molding
200
A
6x1 + 2x2 + 5x3 + 7x4 – s1 = 820 .7x1 – .3x2 – .3x3 – .3x4 – s2 = 0 –.2x1 + x2 + x3 – .2x4 + s3 = 0 x3 – x1 – x4 –- s4 = 0
150
B C
100
baking 50
D
***** Input Data ***** Max. Z = 8.2x1 + 7.0x2+ 6.5x3 + 9.0x4
0
Subject to c1 c2 c3 c4
A: x1 = 0 x2 = 181.03 Z = $43,447.20
50
100
150 E 200
glazing 250
300
350
400
x1
(d)x1 = 56.70, x2 = 154.64
6x1 + 2x2 + 5x3 + 7x4 ≥ 820 .7x1 – .3x2 – .3x3 – .3x4 ≥ 0 –.2x1 + 1x2 + 1x3 – .2x4 ≤ 0 –1x1 + 1x3 – 1x4 ≥ 0
.30(56.7) + .25(154.64) + s1 = 60 s1 = 4.33 hr. of molding time .27(56.7) + .58(154.64) + s2 = 105 s2 = 0 hr. of baking time .16(56.7) + .20(154.64) + s3 = 40 s3 = 0 hr. of glazing time 32.8(56.7) + 20(154.64) + s4 = 6,000 s4 = 1,047.42 lbs. of clay
***** Program Output ***** Infeasible Solution because Artificial variables remain in the final tableau.
(e)The optimal solution is at point B. For point C to become optimal the profit for a large tile, x1, would have to become steeper, than the constraint line for glazing, .16x1 + .20x2 = 40:
CASE SOLUTION: MOSAIC TILE COMPANY
–c1/240 = .16/.20 c1 = 192
(a)maximize Z = $190x1 + 240x2 subject to
This is the upper limit for c1. The lower limit is at point A which requires an objective function slope flatter than the constraint line for baking,
.30x1 + .25x2 ≤ 60 hr.– molding .27x1 + .58x2 ≤ 105 hr. – baking .16x1 + .20x2 ≤ 40 hr.– glazing 32.8x1 + 20x2 ≤ 6,000 lb. – clay x1, x2, ≥ 0
–c1/240 = .27/.58 c1 = 111.72 Thus, 111.72 ≤ c1 ≤ 192
(b)maximize Z = $190x1 + 240x2 + 0s1 + 0s2 + 0s3 + 0s4 subject to
The same logic is used to compute the sensitivity range for c2. The lower limit is computed as,
.30x1 + .25x2 + s1 = 60 .27x1 + .58x2 + s2 = 105
–190/c2 = –.16/.20 c2 = 237.5 35
For the clay constraint the upper limit is ∞ since the constraint can increase indefinitely. The lower limit is at the point where the constraint line intersects with point B:
The upper limit is, –190/c2 = .27/.58 c2 = 408.15 The sensitivity ranges for the constraint quantity values are determined by observing the graph and seeing where the new location of the constraint lines must be to change the solution point.
At B: 32.8(56.7) + 20(154.64) = q4 q4 = 4,952.56 Thus, 4,952.56 ≤ q4 ≤ ∞
For the molding constraint, the lower limit of the range for q1 is where the constraint line intersects with point B,
(f) The slope of the objective must be flatter than the slope of the constraint that intersects with the x2 axis at point A, which is the baking constraint,
.30(56.7) + .25(154.64) = q1 q1 = 55.67 The upper limit is ∞ since it can be seen that this constraint can increase indefinitely without changing the solution point.
–190/c2 = .27/.58 c2 = $408.14 (g)
Thus, 55.67 ≤ q1 ≤ ∞
Problem Title: Case Problem: Mosaic Tile Company
For the baking constraint the lower limit of the range for q2 is where point C becomes optimal, and the upper limit is where the baking constraint intersects with the x2 axis (x2 = 200).
***** Input Data ***** Max. Z = 190x1 + 240x2 Subject to
At C:
.27(100) + .58(120) = q2 q2 = 96.6 At x2 axis: .27(0) + .58(200) = q2 q2 = 116 Thus,
***** Program Output *****
96.6 ≤ q2 ≤ 116
Final Optimal Solution At Simplex Tableau : 2
For the glazing constraint the lower limit of the range for q3 is at point A, and the upper limit is where the glazing constraint line, .16x1 + .20x2 = 40, intersects with the baking and molding constraints (i.e., x1 = 80.28 and x2 = 143.68). At A:
.30x1 + .25x2 ≤ 60 .27x1 + .58x2 ≤ 105 .16x1 + .20x2 ≤ 40 32.8x1 + 20x2 ≤ 6000
c1 c2 c3 c4
Z = 47886.598
.16(0) + .20(181.03) = q3 q3 = 36.21
At intersection of constraints: .16(80.28) + .20(143.68) = q3 q3 = 41.58 Thus, 36.21 ≥ q3 ≥ 41.58
36
Variable
Value
x1
56.701
x2
154.639
Constraint
Slack/Surplus
Shadow Price
c1
4.330
0.000
c2
0.000
10.309
c3
0.000
1170.103
c4
1047.423
0.000
Objective Coefficient Ranges Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
x1
111.724
190.000
192.000
2.000
78.276
x2
237.500
240.000
408.148
168.148
2.500
Lower Limit
Current Values
Upper Limit
Allowable Increase
Allowable Decrease
c1
55.670
60.000
No limit
No limit
4.330
c2
96.600
105.000
116.000
11.000
8.400
c3
36.207
40.000
41.577
1.577
3.793
c4
4952.577
6000.000
No limit
No limit
1047.423
Variables
Right Hand Side Ranges Constraints
CASE SOLUTION: “THE POSSIBILITY” RESTAURANT –– CONTINUED
(h)Since there is already slack molding hours left over, reducing the time required to mold a batch of tiles will only create more slack molding time. Thus, the solution will not change.
The solution is, x1 40 x2 20 Z $800
(i) Additional clay will have no effect on the solution since there is already slack clay left. Thus, Mosaic should not agree to the offer of extra clay.
(A)The question regarding a possible advertising expenditure of $350 per day requires that the sensitivity range for q1 be computed. q1: s3: 20 + 7 0 s4: 14 – 1.1 0 7 20 1.1 14 7 2.86 1.1 12.72 s1: 40 – 2 0 s2: 20 – 0 2 40 20 7 20 20
(j) Although an additional hour of glazing has the highest shadow price of $1,170.103, the upper limit of the sensitivity range for glazing hours is 41.577. Thus, with an increase of only 1.577 hours the solution will change and a new shadow price will exist. In order to assess the full impact of a 20 hour increase in glazing hours the problem should be solved again using the computer with this change. This new solution results in a profit of $49,732.39 an increase in profit of only $1,845.79. The reason for this small increase can be observed in the graphical solution; as the glazing constraint increases it quickly becomes a “non-binding” constraint with a new solution point.
Summarizing, –20 2.86 12.72 20 and, 2.86 12.72 Since q1 = 60 + , = q1 – 60. Therefore, –26 ≤ q1 – 60 ≤ 12.72 57.14 ≤ q1 ≤ 72.72
(k)A reduction of 3 hours is within the sensitivity range for kiln hours. However, the shadow price for kiln hours is $1,170.103 per hour. Thus, a loss of 3 kiln hours will reduce profit by (3)(1,170.103) = $3,510.31.
Thus, an increase of 10 meals does not affect the shadow price for mean demand, which is $800. An increase of 10 meals will result in increased profit of ($8)($10) = $80, which exceeds the advertising expenditure of $30. The ad should be purchased.
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CASE SOLUTION: JULIA’s FOOD BOOTH
(B)The reduction in kitchen staff from 20 to 15 hours requires the computation of the sensitivity range for q2. q2: s3: 20 – 10 0 s4: 14 + 4 0 – 20 20 4 14 70 1 4 –3.5 s1: 40 – 40 0 s2: 20 + 4 0 – 40 40 4 20 70 1 4 –5
(A)x1 = pizza slices, x2 = hot dogs, x3 = barbeque sandwiches The model is for the first home game, maximize Z $0.75x1 1.05x2 1.35x3 subject to: $0.75x1 .0.45x2 0.90x3 1,500 24x1 16x2 25x3 55,296 in2 of oven space. x1 x2 + x3
Summarizing, –5 3.5 1 10 and,
x2
≥ 2.0 x3
3.5 1 Since q2 = 20 + , = q2 – 20. Therefore,
x1, x2, x3 0
–3.5 ≤ q2 – 20 ≤ 1 16/5 ≤ q2 ≤ 21
*Note that the oven space required for a pizza slice was determined by dividing the total space required by a pizza, 14 x 14 = 196 in2, by 8, or approximately 24 in2 per slice. The total space available is the dimension of a shelf, 36 in. x 48 in. = 1,728 in2, multiplied by 16 shelves, 27,648 in2, which is multiplied by 2, the times before kickoff and halftime the oven will be filled = 55,296 in2.
A reduction of 5 hours to 15 hours would exceed the lower limit of the sensitivity range. This would result in a change in the solution mix and the shadow price, so the impact could not be totally ascertained from the optimal simplex tableau. solving the model again with q2 = 15 results in the following new solution. s1 5.45 s3 81.82 x1 49.09 x2 5.45 Z $676.36
Solution: x1 = 1,250 pizza slices x2 = 1,250 hot dogs x3 = 0 barbecue sandwiches Z = $2,250
Notice that simply using the shadow price of $16 for staff time (hr) would have indicated a loss in profit of only (5hr)(16) = $80, or Z = $720. The actual reduction in profit to $676.36 is greater. The final question concerns an increase in the coefficient for c1 from $12 to $14. This requires the computation of the range for c1.
Julia should receive a profit of $2,250 for the first game. Her lease is $1,000 per game so that leaves her with $1,250. Her cost of leasing a warming oven is $100 per game, thus she will make a little more than what she needs to, i.e., $1,000, for it to be worth her while to lease the booth.
(C)The final question concerns an increase in the coefficient for c1 from $12 to $14. This requires the computation of the range for c1.
A “tricky” aspect of the model formulation is the $1,500 used to purchase the ingredients. Since the objective function reflects net profit, the $1,500 is recouped and can be used for the next home game to purchase food ingredients; thus, it’s not necessary for Julia to use any of her $1,150 profit to buy ingredients for the next game.
c1, basic: –8 –2 0 –8 –2 8 –8 –2 –4 –4 4
– 16 + 4 0 – 16 + 4 16 – 16 + 4 4
Since c1 = 12 + , = c1 – 12. Therefore,
(B) Yes, she would increase her profit; the dual value is $1.50 for each additional dollar. The upper limit of the sensitivity range for budget is $1,658.88, so she should only borrow approximately $158. Her additional profit would be $238.32 or a total profit of $2,488.32.
–4 c1 – 12 4 –8 c1 16 Since c1 = $14 is within this range the price increase could be implemented without affecting Pierre’s meal plans. 38
(C) Yes, she should hire her friend. It appears impossible for her to prepare all of the food items given in the solution in such a short period of time. The additional profit she would get if she borrowed more money as indicated in part B would offset this additional expenditure. (D) The biggest uncertainty is the weather. If the weather is very hot or cold, fans might eat less. Also, if it is rainy weather for a game or games, the crowd might not be as large, even though the games are all sellouts. The model results show that Julia will reach her goal of $1,000 per game - if everything goes right. She has little slack in her profit margin, thus it seems unlikely that she will achieve $1,000 for each game.
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