# Taylor Chap 3 Answers

#### Description

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

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

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

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

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

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

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

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

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

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

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

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

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

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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