Taylor Chap 3 Answers

0 10 20 30 40 50 60 70x

x

1

optimal point:

Z

C

B

A

10

20

30

40

50

60

2

x1 = 15.29x2 = 38.24Z = 4,205.88

80 90 100 110

70

80

90

100

110

1.

35. Model formulation

36. Graphical solution; sensitivity analysis (3–35)

37. Computer solution; sensitivity analysis (3–35)

38. Model formulation; computer solution

39. Sensitivity analysis (3–38)











50. Computer solution

PROBLEM SOLUTIONS

PROBLEM SUMMARY

1. QM for Windows

2. QM for Windows and Excel

3. Excel

4. Graphical solution; sensitivity analysis






























34. Computer solution; sensitivity analysis

17

Chapter Three: Linear Programming: Computer Solution and Sensitivity Analysis

18

(a) A: 3(0) + 2(160) + s1 = 500s1 = 180

4(0) + 5(160) + s2 = 800s2 = 0

B: 3(128.5) + 2(57.2) + s1 = 500s1 = 0

4(128.5) + 2(57.2) + s2 = 800s2 = 0

C: 2(167) + 2(0) + s1 = 500s1 = 0

4(167) + 5(0) + s2 = 800s2 = 132

(b) Z = 12x1 + 16x2

and,

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.

The profit, c1, for a basketball that would change the solution point is,

4/5 = –c1/165c1 = 64

c1 = 12.8

Since $13 > 12.8 the solution point would change to B where x1 = 128.5, x2 = 57.2. Thenew Z value is $2,585.70.

For a football,

–4/5 = –12/c24c2 = 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 isx1 = 128.5, x2 = 57.2 and Z = $2,400.

6.2. QM for Windows establishes a “template” forthe linear programming model based on theuser’s specification of the type of objectivefunction, the number of constraints and numberof variables, then the model parameters areinput and the problem is solved. In Excel themodel “template” must be developed by theuser.

3. Changing cells: B10:B12Constraints: B10:B12 � 0

G6 � F6G7 � F7

Profit: � B10 * C4 � B11 * D4 � B12 * E4

4. The slope of the constraint line is –70/60. Theoptimal solution is at point A where x1 = 0 andx2 = 70. To change the solution to B, c1 mustincrease such that the slope of the objectivefunction is at least as great as the slope of theconstraint line,

–c1/50 = –70/60c1 = 58.33

Alternatively, c1 must decrease such that theslope of the objective function is at least as greatas the slope of the constraint line,

–30/c2 = –70/60c2 = 25.71

Thus, if c1 increases to greater than 58.33 or c2decreases to less than 25.71, B will becomeoptimal.

5. (a) x1 = no. of basketballsx2 = no. of footballsmaximize Z = 12x1 + 16x2subject to

3x1 + 2x2 ≤ 5004x1 + 5x2 ≤ 800

x1, x2 ≥ 0

(b) maximize Z = 12x1 + 16x2 + 0s1 + 0s2subject to

3x1 + 2x2 + s1 = 5004x1 + 5x2 + s2 = 800

x1, x2, s1, s2 ≥ 0

0 50 100 150 200 250 300 350x

x

1

Point A is optimalZ

C

B

A

50

100

150

200

250

300

2

*A: x1 = 0x2 = 160Z = 2,560

B: x1 = 128.5x2 = 57.2Z = 2457.2

C: x1 = 167x2 = 0Z = 2,004

19

(c) If the constraint line for rubber changes to 3x1 + 2x2 = 1,000, it moves outward,eliminating points B and C. However, since A isthe optimal point, it will not change and theoptimal solution remains the same,x1 = 0, x2 = 160 and Z = 2,560. There will be anincrease in slack, s1, to 680 lbs.

If the constraint line for leather changes to 4x1 + 5x2 = 1,300, point A will move to a newlocation, x1 = 0, x2 = 250, Z = $4,000.

7. (a) For c1 the upper limit is computed as

–4/5 = –c1/165c1 = 64c1 = 12.8

and the lower limit is unlimited.

For c2 the lower limit is,

–4/5 = –12/c24c2 = 60c2 = 15

and the upper limit is unlimited.

Summarizing,

∞ ≤ c1 ≤ 12.815 ≤ c2 ≤ ∞

For q1 the upper limit is ∞ since no matter howmuch q1 increases the optimal solution point Awill not change.

The lower limit for q1 is at the point where theconstraint line 3x1 + 2x2 = q1 intersects with

point A where x1 = 0, x2 = 160,

3x1 + 2x2 = q13(0) + 2(160) = q1

q1 = 320

For q2 the upper limit is at the point where therubber constraint line (3x1 + 2x2 = 500)intersects with the leather constraint line (4x1 + 5x2 = 800) along the x2 axis, i.e., x1 = 0,x2 = 250,

4x1 + 5x2 = q24(0) + 5(250) = q2

q2 = 1,250

The lower limit is 0 since that is the lowest point on the x2 axis the constraint line candecrease to.

Summarizing,

320 ≤ q1 ≤ ∞0 ≤ q2 ≤ 1,250

(b)

Z = 2560.000

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

Objective Coefficient Ranges

Lower Current Upper Allowable AllowableVariables Limit Values Limit Increase Decrease

x1 No limit 12.000 12.800 0.800 No limitx2 15.000 16.000 No limit No limit 1.000

Right Hand Side Ranges

Lower Current Upper Allowable AllowableConstraints Limit Values Limit Increase Decrease

c1 320.000 500.000 No limit No limit 180.000c2 0.000 800.000 1250.000 450.000 800.000

20

(b) The constraint line 12x1 + 4x2 = 60 would moveinward resulting in a new location forpoint B atx1 = 2, x2 = 4, which would still be optimal.

(c) In order for the optimal solution point to changefrom B to A the slope of the objective functionmust be at least as flat as the slope of theconstraint line, 4x1 + 8x2 = 40, which is –1/2.Thus, the profit for product B would have to be,

–9/c2 = –1/2c2 = 18

If the profit for product B is increased to $15 theoptimal solution point will not change, althoughZ would change from $57 to $81.

If the profit for product B is increased to $20 thesolution point will change from B to A, x1 = 0, x2= 5, Z = $100.

10.(a) For c1 the upper limit is computed as,

–c1/7 = –3c1 = 21

and the lower limit is,

–c1/7 = –1/2c1 = 3.50

For c2 the upper limit is,

–9/c2 = –1/2c2 = 18

and the lower limit is,

–9/c2 = –3c2 = 3

Summarizing,

3.50 ≤ c1 ≤ 21–3 ≤ c2 ≤ 18

(b)

Z = 57.000


x1 4.000 0.000

x2 3.000 0.000


c1 0.000 0.550

c2 0.000 0.600

(c) The shadow price for rubber is $0. Since there isslack rubber left over at the optimal point, extrarubber would have no marginal value.

The shadow price for leather is $3.20. For eachadditional ft.2 of leather that the company can obtain profit would increase by $3.20, up to theupper limit of the sensitivity range for leather (i.e., 1,250 ft.2).

8.(a) x1 = no. of units of Ax2 = no. of units of Bmaximize Z = 9x1 + 7x2

subject to

12x1 + 4x2 ≤ 604x1 + 8x2 ≤ 40

x1,x2 ≥ 0

(b) maximize Z = 9x1 + 7x2 + 0s1 + 0s2

subject to

12x1 + 4x2 + s1 = 604x1 + 8x2 + s2 = 40

x1, x2, s1, s2 ≥ 0

9.

(a) A: 12(0) + 4(5) + s1 = 60s1 = 40

4(0) + 8(5) + s2 = 40s2 = 0

B: 12(4) + 4(3) = 60s1 = 0

4(4) + 8(3) + s2 = 40s2 = 0

C: 12(5) + 4(0) + s1 = 60s1 = 0

4(5) + 8(0) + s2 = 40s2 = 20

0 5 10 15 20 25 30 35x

x

1

Point B is optimalB

C

A5

10

15

20

25

30

2

A: x1 = 0x2 = 5Z = 35

*B: x1 = 4x2 = 3Z = 57

C: x1 = 5x2 = 0Z = 45

40

21



x1 3.500 9.000 21.000 12.000 5.500x2 3.000 7.000 18.000 11.000 4.000



c1 20.000 60.000 120.000 60.000 40.000c2 20.000 40.000 120.000 80.000 20.000

(c) The shadow price for line 1 time is $0.55 perhour, while the shadow price for line 2 time is$0.60 per hour. The company would prefer toobtain more line 2 time since it would result inthe greatest increase in profit.

11.(a) x1 = no. of yards of denimx2 = no. of yards of corduroymaximize Z = $2.25x1 + 3.10x2

subject to

5.0x1 + 7.5x2 ≤ 6,5003.0x1 + 3.2x2 ≤ 3,000

x2 ≤ 510x1, x2 ≥ 0

(b)maximize Z = $2.25x1 + 3.10x2 + 0s1 + 0s2 + 0s3

subject to

5.0x1 + 7.5x2 + s1 = 6,5003.0x1 + 3.2x2 + s2 = 3,000

x2 + s3 = 510x1, x2, s1, s2, s3 ≥ 0

12.

(a) 5.0(456) + 7.5(510) + s1 = 6,500s1 = 6,500 – 6,105s1 = 395 lbs.

3.0(456) + 3.2(510) + s2 = 3,000s2 = 0 hrs.

510 + s3 = 510s3 = 0

therefore demand for corduroy is met.

(b) In order for the optimal solution point to changefrom B to C the slope of the objective functionmust be at least as great as the slope of the con-straint line, 3.0x1 + 3.2x2 = 3,000, which is –3/3.2.Thus, the profit for denim would have to be,

–c1/3.0 = –3/3.2c1 = 2.91

If the profit for denim is increased from $2.25 to$3.00 the optimal solution would change topoint C where x1 = 1,000, x2 = 0, Z = 3,000.

Profit for corduroy has no upper limit thatwould change the optimal solution point.

(c) The constraint line for cotton would move inward as shown in the following graph wherepoint C is optimal.

0 200 400 600 800 1000 1200 1400x

x

1

B

C

A

200

400

600

800

1000

1200

1400

1600

2

A: x1 = 0x2 = 510Z = $1,581

*B: x1 = 456x2 = 510Z = $2,607

C: x1 = 1,000x2 = 0Z = $2,250

1600

0 200 400 600 800 1000 1200 1400x

x

1

B C

D

A

200

400

600

800

1000

1200

1400

1600

2

x1 = 1,000x2 = 0Z = $2,250

1600

C, optimal

22

13.

Z = 2607.000


x1 456.000 0.000

x2 510.000 0.000


c1 395.000 0.000

c2 0.000 0.750

c3 0.000 0.700



x1 0.000 2.250 2.906 0.656 2.250x2 2.400 3.100 No limit No limit 0.700



c1 6015.000 6500.000 No limit No limit 395.000c2 1632.000 3000.000 3237.000 237.000 1368.000c3 0.000 510.000 692.308 182.308 510.000

(a) The company should select additionalprocessing time, with a shadow price of $0.75per hour. Cotton has a shadow price of $0because there is already extra (slack) cottonavailable and not being used so any more wouldhave no marginal value.

(b) 0 ≤ c1 ≤ 2.906 6,105 ≤ q1 ≤ ∞2.4 ≤ c2 ≤ ∞ 1,632 ≤ q2 ≤ 3,237

0 ≤ q3 ≤ 692.308The demand for corduroy can decrease to zeroor increase to 692.308 yds. without changing thecurrent solution mix of denim and corduroy. Ifthe demand increases beyond 692.308 yds., thendenim would no longer be produced and onlycorduroy would be produced.

14. x1 = no. of days to operate mill 1x2 = no. of days to operate mill 2 minimize Z = 6,000x1 + 7,000x2subject to

6x1 + 2x2 ≥ 122x1 + 2x2 ≥ 8

4x1 + 10x2 ≥ 5x1, x2 ≥ 0

15.

0 1 2 3 4 5 6 7x

x

1

B

C

A

1

2

3

4

5

6

7

8

2

8

A: x1 = 0x2 = 6Z = 42,000

B: x1 = 1x2 = 3Z = 27,000

*C: x1 = 4x2 = 0Z = 24,000

23

(a) 6(4) + 2(0) – s1 = 12s1 = 12

2(4) + 2(0) – s2 = 8s2 = 0

4(4) + 10(0) – s3 = 5s3 = 11

(b) The slope of the objective function, –6000/7,000must become flatter (i.e., less) than the slope ofthe constraint line,

2x1 + 2x2 = 8, for the solution to change. Thecost of operating Mill 1, c1, that would changethe solution point is,

–c1/7,000 = –1c1 = 7,000

Since $7,500 > $7,000, the solution point willchange to B where x1 = 1, x2 = 3, Z = $28,500.

(c) If the constraint line for high-grade aluminumchanges to 6x1 + 2x2 = 10, it moves inward but

does not change the optimal variable mix. Bremains optimal but moves to a new location, x1= 0.5, x2 = 3.5, Z = $27,500.

16.

Z = 24000

Variable Value

x1 4.000

x2 0.000


c1 12.000 0.000

c2 0.000 –3000.000

c2 11.000 0.000



x1 0.000 6000.000 7000.000 1000.000 6000.000

x2 6000.000 7000.000 No limit No limit 1000.000



c1 No limit 12.000 24.000 12.000 No limit

c2 4.000 8.000 No limit No limit 4.000


(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 ∞ ≤ q1 ≤ 246,000 ≤ c2 ≤ ∞ 4 ≤ q2 ≤ ∞

∞ ≤ q3 ≤ 16

(c) There will be no change.

17. x1 = no. of acres of cornx2 = no. of acres of tobaccomaximize Z = 300x1 + 520x2subject to

x1 + x2 ≤ 410105x1 + 210x2 ≤ 52,500

x2 ≤ 100x1, x2 ≥ 0

24

The profit for corn must be greater than $520for the Bradleys to plant only corn.

(c) If the constraint line changes from x1 + x2 = 410to x1 + x2 = 510, it will moveoutward to alocation which changes the solution to the pointwhere 105x1 + 210x2 = 52,500 intersects withthe axis. This new point is x1 = 500, x2 = 0,Z = $150,000.

(d) If the constraint line changes from x1 + x2 = 410to x1 + x2 = 360, it moves inward to a locationwhich changes the solution point to theintersection of x1 + x2 = 360 and 105x1 + 210x2 = 52,500. At this point x1 = 260, x2 = 100and Z = $130,000.

19.

Z = 142800.000

Variable Value

x1 320.000

x2 90.000


c1 0.000 80.000

c2 0.000 2.095

c3 10.000 0.000

18.

(a) x1 = 320, x2 = 90320 + 90 + s1 = 410

s1 = 0 acres uncultivated90 + s3 = 100

s3 = 10 acres of tobacco allotmentunused

(b) At point D only corn is planted. In order forpoint D to be optimal the slope of the objectivefunction 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 forcorn is computed as,

–c/520 = –1c1 = 520

0 100 200 300 400 500 600 700x

x

1

B CA

100

200

300

400

500

600

2

800

A: x1 = 0x2 = 100Z = 52,000

B: x1 = 300x2 = 100Z = 142,000

*C: x1 = 320x2 = 90Z = 142,800

D: x1 = 410x2 = 0Z = 123,000

Point C is optimal

D



x1 260.000 300.000 520.000 220.000 40.000

x2 300.000 520.000 600.000 80.000 220.000



c1 400.000 410.000 500.000 90.000 10.000

c2 43050.000 52500.000 53550.000 1050.000 9450.000


25

(a) No, the shadow price for land is $80 per acreindicating that profit will increase by no morethan $80 for each additional acre obtained. Themaximum price the Bradley’s should pay is $80and the most they should obtain is at the upperlimit of the sensitivity range for land. This limitis 500 acres, or 90 additional acres. Beyond 90acres the shadow price would change.

(b) The shadow price for the budget is $2.095.Thus, for every $1 dollar borrowed they couldexpect a profit increase of $2.095. If theyborrowed $1,000 it would not change theamount of corn and tobacco they plant since thesensitivity range has a maximum allowableincrease of $1,050.

20. x1 = no. of sausage biscuitsx2 = no. of ham biscuitsmaximize Z = .60x1 + .50x2subject to

.10x1 ≤ 30.15 x2 ≤ 30

.04x1 + .04x2 ≤ 160..01x1 + .024x2 ≤ 6

x1, x2 ≥ 0

21.

(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.

(b) The slope of the objective function, –6/5, mustbecome flatter (i.e., less) than the slope of theconstraint line, .04x1 + .04x2 = 16, for thesolution to change. The profit for ham, c2, thatwould change the solution point is,

–0.6/c2 = –1c2 = .60

Thus, an increase in profit for ham of 0.60 willcreate a second optimal solution point at Cwhere x1 = 257, x2 = 143 and Z = $225.70.(Point D would also continue to be optimal, i.e.,multiple optimal solutions.)

(c) A change in the constraint line from, .04x1+ .04x2 = 16 to .04x1 + .04x2 = 18would movethe line outward, eliminating both points C andD. The new solution point occurs at theintersection of 0.01x1 + .024x2 = 6 and .10x =30. This point is x1 = 300, x2 = 125, and Z =$242.50.

22.

Z = 230.000

Variable Value

x1 300.000

x2 100.000


c1 0.000 1.000

c2 15.000 0.000

c3 0.000 12.500

c4 0.600 0.0000 100 200 300 400 500 600 700

x

x

1

B

C

D

E

A

100

200

300

400

500

600

2

800

A: x1 = 0x2 = 200Z = 100

B: x1 = 120x2 = 200Z = 172

C: x1 = 257x2 = 143Z = 225.70

*D: x1 = 300x2 = 100Z = 230

E: x1 = 300x2 = 0Z = 180

Point D is optimal




x2 0.000 0.500 0.600 0.100 0.500

26

(a) The shadow price for sausage is $1. For everyadditional pound of sausage that can be obtainedprofit will increase by $1. The shadow price forflour is $12.50. For each additional pound offlour that can be obtained, profit will increase bythis amount. There are extra ham and labor hoursavailable, so their shadow prices are zero,indicating additional amounts of those resourceswould add nothing to profit.

(b) The constraint for flour, indicated by the highshadow price.

(c) .50 ≤ c1 ≤ ∞25.714 ≤ q1 ≤ 40

The sensitivity range for profit indicates that theoptimal mix of sausage and ham biscuits willremain optimal as long as profit does not fallbelow $0.50. The sensitivity range for sausageindicates the optimal solution mix will bemaintained as long as the available sausage isbetween 25.714 and 40 lbs.

23. x1 = no. of telephone interviewersx2 = no. of personal interviewersminimize Z = 50x1 + 70x2subject to

80x1 + 40x2 ≥ 3,00080x1 ≥ 1,00040x2 ≥ 800

x1, x2 ≥ 0

24.

(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) thanthe slope of the constraint line, 80x1 + 40x2 =3,000, for the solution point to change from B toA. The cost of a telephone interviewer that wouldchange the solution point is,

–c1/70 = –2c1 = 140

This is the upper limit of the sensitivity range forc1. The lower limit is 0 since as the slope of theobjective function becomes flatter, the solutionpoint will not change from B until the objectivefunction is parallel with the constraint line. Thus,

0 ≤ c1 ≤ 140

Since the constraint line is vertical, it canincrease as far as point B and decrease all theway to the x2 axis before the solution mix willchange. At point B,

80(27.5) = q1q1 = 2,200

At the axis,

80(0) = q1q1 = 0

Summarizing,

0 ≤ q1 ≤ 2,200

(b) At the optimal point, B, x1 = 27.5 and x2 = 20.

80(27.5) – s2 = 1,000s2 = 1,200 extra telephone interviews

40(20) – s3 = 800s3 = 0

(c) A change in the constraint line from 40x2 = 800to 40x2 = 1,200, moves the lineup, but it does notchange the optimal mix. The new solution valuesare x1 = 22.5, x2 = 30, Z = $3,225.



c1 25.714 30.000 40.000 10.000 4.286


c3 12.000 16.000 17.000 1.000 4.000


0 10 20 30 40 50 60 70x

x

1

Point B is optimal

B

A

10

20

30

40

50

60

70

80

2

A: x1 = 12.5x2 = 50Z = 4,125

*B: x1 = 27.5x2 = 20Z = 2,775

27

25.

Z = 2775.000

Variable Value

x1 27.500

x2 20.000


c1 0.000 –0.625

c2 1200.000 0.000

c3 0.000 –1.125



x1 0.000 50.000 140.000 90.000 50.000






c3 0.000 800.000 2000.000 1200.000 800.000

(a) Reduce the personal interview requirement; itwill reduce cost by $0.625 per interview, while atelephone interview will not reduce cost; i.e., ithas a shadow price equal to $0.

(b) 25 ≤ c2 ≤ ∞1,800 ≤ q1 ≥ ∞

26. x1 = no. of gallons of ryex2 = no. of gallons of bourbonmaximize Z = 3x1 + 4x2subject to

x1 + x2 ≥ 400x1 ≥ .4(x1 + x2)x2 ≤ 250x1 = 2x2

x1 + x2 ≤ 500x1, x2 ≥ 0

27.

0 100 200 300 400 500 600 700x

x

1

Point B is optimal

Feasiblesolution line

Z

BA100

200

300

400

500

600

2

A: x1 = 266.7x2 = 133.3Z = 1,333.20

*B: x1 = 333.3x2 = 166.7Z = 1,666

28

(a) Optimal solution at B: x1 = 333.3 and x2 = 166.7

(333.3) + (166.7) – s1 = 400s1 = 100 extra gallons of blended whiskey produced

.6(333.33) – .4(166.7) – s2 = 0s2 = 133.3 extragallons of rye inthe blend

(166.7) + s3 = 250

s3 = 83.3 fewer gallons of bourbon than the maximum

(333.3) + (166.7) + s4 = 500s4 = 100 gallons of blendproduction capacity left over

(b) Because the “solution space” is not really anarea, but a line instead, the objective functioncoefficients can change to any positive value andthe solution point will remain the same, i.e.,point B. Observing the graph of this model, nomatter how flatter or steeper the objectivefunction becomes, point B will remain optimal.

28.

Z = 1666.667

Variable Value

x1 333.333

x2 166.667


c1 100.000 0.000

c2 133.333 0.000

c3 83.333 0.000

c5 0.000 3.333

29. x1 = $ amount invested in landx2 = $ amount invested in cattlemaximize Z = 1.20x1 + 1.30x2subject to

x1 + x2 ≤ 95,000.18x1 + .30x2 ≤ 20,000

x1, x2 ≥ 0

30.

(a) The optimal solution point is at B where x1 = $70,833.33, and x2 = $24, 166.67. The slopeof the objective function, –1.2/1.3, must becomeflatter than the slope of the constraint line, .18x1+ .30x2 = 20,000, for the solution point tochange to A (i.e., only cattle). The return oncattle that will change the solution point is

–1.2/c2 = –.18/30c2 = 2

Thus, the return must be 100% before Alexiswill invest only in cattle.

(b) Yes, there is no slack money left over at theoptimal solution.

(c) Since her investment is $95,000, she couldexpect to earn $21,416.67.

0 20 40 60 80 100 120 140 x

x

1

C

B

A

20

40

60

80

100

120

2

160

140

160

*B: x1 = 70,833.33x2 = 24,166.67Z = $116,416.67

C: x1 = 95,000x2 = 0Z = $114,000

A: x1 = 0x2 = 66,666.7Z = $86,667

29



x1 –2.000 3.000 No limit No limit 5.000x2 –6.000 4.000 No limit No limit 10.000



c1 No limit 400.000 500.000 100.000 No limitc2 No limit 0.000 133.333 133.333 No limitc3 166.667 250.000 No limit No limit 83.333c4 –250.000 0.000 500.000 500.000 250.000c5 400.000 500.000 750.000 250.000 100.000

(a)–2.0 ≤ c1 ≤ ∞–6.0 ≤ c2 ≤ ∞

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.

(b)The shadow price for production capacity is $3.33. Thus, for each gallon increase incapacity 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 withnormal sensitivity analysis. Instead the model

must be solved again on the computer, which results in the following solution output.

Z = 1625.000

Variable Value

x1 375.000

x2 125.000


c1 100.000 0.000

c2 175.000 0.000

c3 125.000 0.000

c5 0.000 3.250



x1 –1.333 3.000 No limit No limit 4.333x2 –9.000 4.000 No limit No limit 13.000



c1 No limit 400.000 500.000 100.000 No limitc2 No limit 0.000 175.000 175.000 No limitc3 125.000 250.000 No limit No limit 125.000c4 –500.000 0.000 500.000 500.000 500.000c5 400.000 500.000 1000.000 500.000 100.000



x1 No limit 1.200 1.300 0.100 No limit





c2 0.000 20000.000 28500.000 8500.000 20000.000

30



x1 0.780 1.200 1.300 0.100 0.420

x2 1.200 1.300 2.000 0.700 0.100



c1 66666.667 95000.000 111111.111 16111.111 28333.333

c2 17100.000 20000.000 28500.000 8500.000 2900.000

(a) The shadow price for invested money is$1.05.Thus, for every dollar of her own moneyAlexis invested she could expect a return of $0.05or 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 shadowprice would change.

(b) This would change the constraint, .18x1 + .30x2= 20,000 to .30x1 + .30x2 = 20,000. In order toassess the effect of this change the problem must

be solved again using the computer, as follows.

Z = 86666.667

Variable Value

x1 0.000

x2 66666.667


c1 28333.33 0.000

c2 0.000 4.333

31.

Z = 116416.667

Variable Value

x1 70833.333

x2 24166.667


c1 0.000 1.050

c2 0.000 0.833

31

32. maximize Z = 140x1 + 205x2 + 190x3 + 0s1 + 0s2+ 0s3 + 0s4subject to

10x1 + 15x2 + 8x3 +s1 = 610x1 – 3x2 + s2 = 0

.6x1 – .4x2 – .4x3 – s3 = 0x2 – x3 – s4 = 0

x1, x2, s1, s2, s3, s4 ≥ 0

Z = 9765.596


x1 22.385 0.000

x2 16.789 0.000

x3 16.789 0.000


c1 0.000 16.009

c2 27.982 0.000

c3 0.000 –33.486

c4 0.000 –48.532



x1 –237.857 140.000 171.739 31.739 377.857

x2 132.000 205.000 325.227 120.227 73.000





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

33. (a) and (b)minimize Z = $400x1 + 180x2 + 90x3subject to

x1 ≥ 200x2 ≥ 300x3 ≥ 100

4x3 – x1 – x2 ≤ 0x1 + x2 + x3 = 1,000x1, x2, x3 ≥ 0

(c)

Z = 206000.000

Variable Value

x1 200.000

x2 600.000

x3 200.000


c1 0.000 –220.000

c2 500.000 0.000

c3 100.000 0.000

c4 0.000 18.000

32




x2 90.000 180.000 400.000 220.000 90.000

x3 No limit 90.000 180.000 90.000 No limit



c1 0.000 200.000 700.000 500.000 200.000



c4 –500.000 0.000 2500.000 2500.000 500.000


34.(a) x1 � 36.7142x2 � 58.6371x3 � 0x4 � 63.5675Z � 9,177.85

(b) 34.6871 � q1 � 61.533543.0808 � q2 � 71.2315� � q3 � 65.468655 � q4 � �529.0816 � c1 � 747.9999350.3345 � c2 � �3,488.554 � c3 � �1,363.636 � c4 � 1,761.476�20.132 � c5 � 64.4643

(c) process 1 time is the most valuable with a dual value of $7.9275

(d) Product 3(x3) is not produced; it would require a profit of $65.4686 to be produced.

35. Maximize Z � $0.50x1 � 0.75x2subject to:

$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

0 2 4 6 8 10 12 14 16 18 20 22

2

4

6

8

10

12

14

16

18

x 2 (1

000s

)

x1 (1000s)

B

CA

Optimal

36.

(a) c2 � .50

(b) s1 � $140

(c) There would be no feasible solution.

B: x1 � 8,000x2 � 10,000Z � $11,500

33

the original solution with a 140,000 ft2 store,thus, given these conditions, Mega-Mart shouldnot purchase the land.

40.(a) Maximize Z � $0.97x1 � 0.83x2 � 0.69x3subject to:

x1 � x2 � x3 � 324 cartonsx3 � x1 � x2

(b) x1 = 54,x2 = 108,x3 = 162,Z 5 $253.80

41.(a) The shadow price for shelf space is $0.78 percarton, however, this is only valid up to 360cartons, the upper limit of the sensitivity rangefor shelf space.

(b)The shadow price for available local dairycartons is $0 so it would not increase profit toincrease the available amount of local dairymilk.

(c) The discount would change the objectivefunction to,

maximize Z � 0.86x1 � 0.83x2 � 0.69x3

and the constraint for relative demand wouldchange to

the resulting optimal solution is,

x1 = 108, x2 = 54, x3 = 162, Z � $249.48

Since the profit declines the discount should notbe implemented.

42. x1 = road racing bikesx2 = cross country bikesx3 = mountain bikes

maximize Z � 600x1 � 400x2 � 300x3subject to

1,200x1 � 1,700x2 � 900x3 � $12,000x1 � x2 � x3 � 208x1 � 12x2 � 16x3 � 120x3 � 2(x1 � x2)x1, x2, x3 � 0

43. x1 = 3x3 = 6Z � 3,600

37. x1 � 8,000x2 � 10,000Z � $11,500

(a) The dual value of rack space is 0.75 so anincrease in rack space to accommodate anadditional 500 copies would result in increasedadvertising revenue of $375. An increase in rackspace to 20,000 copies would be outside thesensitivity range for this constraint and requirethe problem to be solved again. The new solutionis x1 � 8,000, x2 � 10,560 and Z � $11,920

(b) 7,000 is within the sensitivity range for theentertainment guide (6,250 � q3 � 10,000). Thedual value is $0.25 thus for every unit thedistribution requirement can be reduced, revenuewill be increased by $0.25, or $250. Thus, Z �$11,750

38. (a) Maximize Z � 4.25x1 � 5.10x2 � 4.50x2 �5.20x4 � 4.10x5 � 4.90x6 � 3.80x7subject to:

x1 � x2 � x3 � x4 � x5 � x6 � x7 � x8 �140,000xi � 15,000, i � 1, 2, …,7

(b) x1 � 15,000x2 � 26,863x3 � 20,588.24x4 � 26,862.75x5 � 15,000x6 � 15,000x7 � 15,000x8 � 5,686Z � $625,083

39.(a) A 20,000 ft2 increase in store size to 160,000 ft2

would increase annual profit to $718,316. This isa $93,233 increase in profit. Given the price ofthe land ($190,000) relative to the increase inprofit, it would appear that the cost of the landwould be offset in about 2 years, therefore thedecision should be to purchase the land.

(b) The decrease in profit in all departments wouldresult in a new solution with Z � $574,653. Thisis a reduction of $50,430 annually in profit from

x ix

xi

i

i

i

≥ = …

≤ = …

15 000 1 2 7

20 1 2 7

, , , , ,

. , , , ,

Σ

x

x x xxi

8

4 6 710

0+ +

=

≥

. x

x3

11 5≥ .

x x xx x xx

xx

1 2 3

3 1 2

3

1

2

3

120

+ + ≤≥ +

≥

≤

324 cartons

34

(a) More hours to assemble; the dual value forbudget and space is zero, while the dual valuefor assembly is $30/hour.

(b) The additional net sales would be $900. Sincethe cost of the labor is $300, the additional profitwould be $600.

(c) It would have no effect on the original solution.$700 profit for a cross country bike is within thesensitivity range for the objective functioncoefficient for x2.

44. Maximize Z � $0.35x1 � 0.42x2 � 0.37x3subject to:

0.45x1 � 0.41x2 � 0.50x3 � 960x1 � x2 � x3 � 2,000x1 � 200x2 � 200x3 � 200x1 � x2 � x3x1, x2, x3 � 0

45. x1 � 1,000x2 � 800x3 � 200Z � $760

(a) Increase vending capacity by 100 sandwiches.There is already excess assembly time available(82 minutes) and the dual value is zero whereasthe dual value of vending machine capacity is$0.38. $38 in additional profit.

(b) x1 � 1,000x2 � 1,000Z � $770

The original profit is $760 and the new solutionis $770. It would seem that a $10 differencewould not be worth the possible loss of customergoodwill due to the loss of variety in the numberof sandwiches available.

(c) Profit would increase to $810 but the solutionvalues would not change. If profit is increased to$0.45 the solution values change to x1 � 1,600,x2 � 200, x3 � 200.

46.(a) Maximize Z � 7,500x1 � 8,200x2 � 10,500x3subject to:

.21x1 � .24x2 � .18x3 � 17x1 � x2 � x3 � 8012x1 � 14.5x2 � 16x3 � 2,500x3 � (x1 � x2)/2x1, x2, x3 � 0

47. x1 � 20x2 � 33.3334x3 � 26.6667Z � $703,333.40

(a) The sensitivity range for x2 is 7,500 � c2 �8,774.999. Since $7,600 is within this range thevalues for x1, x2, and x3 would not change, butthe profit would decline to $683,333.30 (i.e., lessthe difference in profit, ($600)(x2 � 33.3334)

(b) One ton of grapes; the dual value is $23,333.35

(c) Grapes: (0.5)($23,333.35) � $11,666.68Casks: (4)($3,833.329) � $15,333.32Production: $0Select the casks.

(d) $6,799; slightly less than the lower band of thesensitivity range for cj.

48. Minimize Z � $37x11 � 37x12 � 37x13 �46x21 � 46x22 � 46x23 � 50x31 � 50x32 �50x33 � 42x41 � 42x42 � 42x43subject to:.7x11 � .6x21 � .5x31 � .3x41 � 400 tons.7x12 � .6x22 � .5x32 � .3x42 � 250 tons.7x13 � .6x23 � .5x33 � .3x43 � 290 tons

x11 � x12 � x13 � 350 tonsx21 � x22 � x23 � 530 tonsx31 � x32 � x33 � 610 tonsx41 � x42 � x43 � 490 tons

49. x13 � 350 tonsx21 � 158.333 tonsx22 � 296.667 tonsx23 � 75 tonsx31 � 610 tonsx42 � 240 tonsZ � $77,910

Mine 1 � 350 tonsMine 2 � 530 tonsMine 3 � 610 tonsMine 4 � 240 tons

Multiple optimal solutions exist

(a) Mine 4 has 240 tons of “slack” capacity.

(b) The dual values for the 4 constraintsrepresenting the capacity at the 4 mines showthat mine 1 has the highest dual value of $61, soits capacity is the best one to increase.

(c) The sensitivity range for mine 1 is 242.8571 �c1 � 414.2857, thus capacity could be increased

35

by 64.2857 tons before the optimal solutionpoint would change.

(d) The effect of simultaneous changes in objectivefunction coefficients and constraint qualityvalues cannot be analyzed using the sensitivityranges provided by the computer output. It isnecessary to make both changes in the modeland solve it again. Doing so results in a newsolution with Z � $73,080, which is $4,830 lessthan the original solution, so Exeter shouldmake these changes.

50. minimize Z = 8.2x1 + 7.0x2 +6.5x3 + 9.0x4 + 0s1 + 0s2 + 0s3 + 0s4subject to

6x1 + 2x2 + 5x3 + 7x4 – s1 = 820.7x1 – .3x2 – .3x3 – .3x4 – s2 = 0

–.2x1 + x2 + x3 – .2x4 + s3 = 0 x3 – x1 – x4 –- s4 = 0

***** Input Data *****

Max. Z = 8.2x1 + 7.0x2+ 6.5x3 + 9.0x4

Subject to

c1 6x1 + 2x2 + 5x3 + 7x4 ≥ 820c2 .7x1 – .3x2 – .3x3 – .3x4 ≥ 0c3 –.2x1 + 1x2 + 1x3 – .2x4 ≤ 0c4 –1x1 + 1x3 – 1x4 ≥ 0

***** Program Output *****

Infeasible Solution

because Artificial variables remain in the final

tableau.

CASE SOLUTION:MOSAIC TILE COMPANY

(a)maximize Z = $190x1 + 240x2subject to

.30x1 + .25x2 ≤ 60 hr.– molding

.27x1 + .58x2 ≤ 105 hr. – baking

.16x1 + .20x2 ≤ 40 hr.– glazing32.8x1 + 20x2 ≤ 6,000 lb. – clay

x1, x2, ≥ 0

(b)maximize Z = $190x1 + 240x2 + 0s1 + 0s2 + 0s3 + 0s4subject to

.30x1 + .25x2 + s1 = 60

.27x1 + .58x2 + s2 = 105

.16x1 + .20x2 + s3 = 4032.8x1 + 20x2 + s4 = 6,000x1, x2, s1, s2, s3, s4 ≥ 0

(c)

(d)x1 = 56.70, x2 = 154.64

.30(56.7) + .25(154.64) + s1 = 60s1 = 4.33 hr. of molding time

.27(56.7) + .58(154.64) + s2 = 105s2 = 0 hr. of baking time

.16(56.7) + .20(154.64) + s3 = 40s3 = 0 hr. of glazing time

32.8(56.7) + 20(154.64) + s4 = 6,000 s4 = 1,047.42 lbs. of clay

(e)The optimal solution is at point B. For point Cto become optimal the profit for a largetile, x1, would have to become steeper, than theconstraint line for glazing, .16x1 + .20x2 = 40:

–c1/240 = .16/.20c1 = 192

This is the upper limit for c1. The lower limit is at point A which requires an objective functionslope flatter than the constraint line for baking,

–c1/240 = .27/.58c1 = 111.72

Thus, 111.72 ≤ c1 ≤ 192

The same logic is used to compute the sensitivity range for c2. The lower limit iscomputed as,

–190/c2 = –.16/.20c2 = 237.5

0 50 100 150 200 250 300 350 x

x

1

C

B

D

A

50

100

150

200

250

300

2

400

350

400

*B: x1 = 56.70x2 = 154.64Z = $47,886.60

C: x1 = 100x2 = 120Z = $47,800

A: x1 = 0x2 = 181.03Z = $43,447.20

E: x1 = 182.93x2 = 0Z = $34,756.70

D: x1 = 136.36x2 = 76.36Z = $44,234.80

glazing

baking

molding

clay

E

36

The upper limit is,

–190/c2 = .27/.58c2 = 408.15

The sensitivity ranges for the constraint quantity values are determined by observing thegraph and seeing where the new location of theconstraint lines must be to change the solutionpoint.

For the molding constraint, the lower limit of the range for q1 is where the constraint lineintersects with point B,

.30(56.7) + .25(154.64) = q1q1 = 55.67

The upper limit is ∞ since it can be seen that this constraint can increase indefinitely withoutchanging the solution point.

Thus,

55.67 ≤ q1 ≤ ∞

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 bakingconstraint intersects with the x2 axis (x2 = 200).

At C: .27(100) + .58(120) = q2q2 = 96.6

At x2 axis: .27(0) + .58(200) = q2q2 = 116

Thus,

96.6 ≤ q2 ≤ 116

For the glazing constraint the lower limit of therange for q3 is at point A, and the upper limit iswhere the glazing constraint line, .16x1 + .20x2= 40, intersects with the baking and moldingconstraints (i.e., x1 = 80.28 and x2 = 143.68).

At A: .16(0) + .20(181.03) = q3q3 = 36.21

At intersection of constraints:

.16(80.28) + .20(143.68) = q3q3 = 41.58

Thus,

36.21 ≥ q3 ≥ 41.58

For the clay constraint the upper limit is ∞since the constraint can increase indefinitely.The lower limit is at the point where theconstraint line intersects with point B:

At B: 32.8(56.7) + 20(154.64) = q4q4 = 4,952.56

Thus,

4,952.56 ≤ q4 ≤ ∞

(f) The slope of the objective must be flatter than the slope of the constraint that intersects withthe x2 axis at point A, which is the bakingconstraint,

–190/c2 = .27/.58c2 = $408.14

(g)

Problem Title: Case Problem: Mosaic Tile Company

***** Input Data *****

Max. Z = 190x1 + 240x2

Subject to

c1 .30x1 + .25x2 ≤ 60c2 .27x1 + .58x2 ≤ 105c3 .16x1 + .20x2 ≤ 40c4 32.8x1 + 20x2 ≤ 6000

***** Program Output *****

Final Optimal Solution At Simplex Tableau : 2

Z = 47886.598

Variable Value

x1 56.701

x2 154.639


c1 4.330 0.000

c2 0.000 10.309

c3 0.000 1170.103

c4 1047.423 0.000

37



x1 111.724 190.000 192.000 2.000 78.276

x2 237.500 240.000 408.148 168.148 2.500




c2 96.600 105.000 116.000 11.000 8.400

c3 36.207 40.000 41.577 1.577 3.793


(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 moldingtime. Thus, the solution will not change.

(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.

(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.

(k)A reduction of 3 hours is within the sensitivity range for kiln hours. However, the shadow pricefor kiln hours is $1,170.103 per hour. Thus, aloss of 3 kiln hours will reduce profit by(3)(1,170.103) = $3,510.31.

CASE SOLUTION:“THE POSSIBILITY” RESTAURANT ––CONTINUED

The solution is,

x1 � 40x2 � 20Z � $800

(A)The question regarding a possible advertisingexpenditure of $350 per day requires that thesensitivity range for q1 be computed.q1:

s3: 20 + 7� � 0 s4: 14 – 1.1� � 07� � 20 1.1� � 147� � 2.86 1.1� � 12.72

s1: 40 – 2� � 0 s2: 20 – � � 02� � 40 � � 207� � 20 � � 20

Summarizing,

–20 � 2.86 � � � 12.72 � 20

and,

2.86 � � � 12.72

Since q1 = 60 + �,� = q1 – 60. Therefore,

–26 ≤ q1 – 60 ≤ 12.7257.14 ≤ q1 ≤ 72.72

Thus, an increase of 10 meals does not affect theshadow price for mean demand, which is $800. Anincrease of 10 meals will result in increased profitof ($8)($10) = $80, which exceeds the advertisingexpenditure of $30. The ad should be purchased.

38

(B)The reduction in kitchen staff from 20 to 15hours requires the computation of thesensitivity range for q2.q2:

s3: 20 – 10� � 0 s4: 14 + 4� � 0– 20� � 20 4� � 14

70� � 1 4� � –3.5s1: 40 – 40� � 0 s2: 20 + 4� � 0

– 40� � 40 4� � 2070� � 1 4� � –5

Summarizing,

–5 � 3.5 � � � 1 � 10

and,

3.5 � � � 1

Since q2 = 20 + �,� = q2 – 20. Therefore,

–3.5 ≤ q2 – 20 ≤ 116/5 ≤ q2 ≤ 21

A reduction of 5 hours to 15 hours would exceedthe lower limit of the sensitivity range. This wouldresult in a change in the solution mix and theshadow price, so the impact could not be totallyascertained from the optimal simplex tableau.solving the model again with q2 = 15 results in thefollowing new solution.

s1 � 5.45s3 � 81.82x1 � 49.09x2 � 5.45Z � $676.36

Notice that simply using the shadow price of $16for staff time (hr) would have indicated a loss inprofit of only (5hr)(16) = $80, or Z = $720. Theactual reduction in profit to $676.36 is greater. Thefinal question concerns an increase in thecoefficient for c1 from $12 to $14. This requiresthe computation of the range for c1.

(C)The final question concerns an increase in thecoefficient for c1 from $12 to $14. This requiresthe computation of the range for c1.

c1, basic:–8 –2� � 0 – 16 + 4� � 0–8 –2� � 8 – 16 + 4� � 16–8 –2� � –4 – 16 + 4� � 4–4 �� 4

Since c1 = 12 + �,� = c1 – 12. Therefore,

–4 � c1 – 12 � 4–8 � c1 � 16

Since c1 = $14 is within this range the priceincrease could be implemented without affectingPierre’s meal plans.

CASE SOLUTION:JULIA’s FOOD BOOTH

(A)x1 = pizza slices, x2 = hot dogs,x3 = barbeque sandwiches

The model is for the first home game,

maximize Z � $0.75x1 � 1.05x2 � 1.35x3subject to:

$0.75x1 � .0.45x2 � 0.90x3 � 1,50024x1 � 16x2 � 25x3 � 55,296 in2 of ovenspace.x1 � x2 + x3

≥ 2.0

x1, x2, x3 � 0

*Note that the oven space required for a pizzaslice was determined by dividing the totalspace required by a pizza, 14 x 14 = 196 in2,by 8, or approximately 24 in2 per slice. Thetotal space available is the dimension of ashelf, 36 in. x 48 in. = 1,728 in2, multiplied by16 shelves, 27,648 in2, which is multiplied by2, the times before kickoff and halftime theoven will be filled = 55,296 in2.

Solution: x1 = 1,250 pizza slicesx2 = 1,250 hot dogsx3 = 0 barbecue sandwichesZ = $2,250

Julia should receive a profit of $2,250 for thefirst game. Her lease is $1,000 per game so thatleaves her with $1,250. Her cost of leasing awarming oven is $100 per game, thus she willmake a little more than what she needs to, i.e.,$1,000, for it to be worth her while to lease thebooth.

A “tricky” aspect of the model formulation isthe $1,500 used to purchase the ingredients.Since the objective function reflects net profit,the $1,500 is recouped and can be used for thenext home game to purchase food ingredients;thus, it’s not necessary for Julia to use any ofher $1,150 profit to buy ingredients for the nextgame.

(B) Yes, she would increase her profit; the dualvalue is $1.50 for each additional dollar. Theupper limit of the sensitivity range for budgetis $1,658.88, so she should only borrowapproximately $158. Her additional profitwould be $238.32 or a total profit of $2,488.32.

x2x3

39

(C) Yes, she should hire her friend. It appearsimpossible for her to prepare all of the fooditems given in the solution in such a shortperiod of time. The additional profit she wouldget if she borrowed more money as indicatedin part B would offset this additionalexpenditure.

(D) The biggest uncertainty is the weather. If theweather is very hot or cold, fans might eat less.Also, if it is rainy weather for a game orgames, the crowd might not be as large, eventhough the games are all sellouts. The modelresults show that Julia will reach her goal of$1,000 per game - if everything goes right. Shehas little slack in her profit margin, thus itseems unlikely that she will achieve $1,000 foreach game.

Documents

Taylor Chap 3 Answers