Solutions Manual Introduction to Management Science 11th Edition Bernard W. Taylor III

Instant download and all chapters Solutions Manual Introduction to Management Science 11th Edition Bernard W. Taylor III https://testbankdata.com/download/solutions-manual-introduction-managementscience-11th-edition-bernard-w-taylor-iii/ Chapter Four: Linear Programming: Modeling Examples PROBLEM SUMMARY

4-1

36.

Assignment (minimization), sensitivity analysis

37.

Transportation (minimization)

38.

Scheduling (minimization)

39.

Production line scheduling (maximization)

40.

College admissions (maximization)

41.

Network flow (minimization)

42.

Blend (maximization)

43.

Personal scheduling (maximization)

44.

Employee allocation (minimization)

45.

Trim loss (minimization)

46. 47.

Multiperiod investment (maximization) Multiperiod sales and inventory (maximization)

48.

Multiperiod production and inventory (minimization)

49.

Employee assignment (maximization)

50.

Data envelopment analysis

51.

Data envelopment analysis

52.

Network flow (maximization)

53.

Multiperiod workforce planning (minimization)

54. 55.

Integer solution (4-53) Machine scheduling (maximization), sensitivity analysis

56.

Cargo storage (maximization)

57.

Broadcast scheduling (maximization)

58.

Product mix (maximization)

59.

Product mix/advertising (maximization)

60.

Scheduling (minimization)

61.

Consultant project assignment (minimization)

62.

Multiperiod workforce planning (minimization)

63.

Multiperiod workforce (4-62)

64.

Coal transportation (minimization)

65.

Soccer field assignment (minimization)

66.

Data envelopment analysis

67.

Airline crew scheduling (maximization)

68.

Product flow/scheduling (minimization)

69.

Transshipment (minimization)

70.

Assignment (minimization)

PROBLEM SOLUTIONS 1.

Many different combinations of maximum servings of each of the 10 food items could be used. As an example, limiting the four hot and cold cereals, x1, x2, x3, and x4, to four cups, eggs to three, bacon to three slices, oranges to two, milk to two cups, orange juice to four cups, and wheat toast to four slices results in the following solution:

Since the profit values would change, the shadow prices would no longer be effective. Also, the sensitivity analysis provided in the computer output does not provide ranges for constraint parameter changes. Thus, the model would have to be resolved.

x3 = 2 cups of oatmeal

The reformulated model would have unit costs increased by 10 percent. This same amount would be subtracted from unit profits. The individual processing times would be reduced by 10 percent. This would result in a new, lower solution of $43,310. Thus, the suggested alternative should not be implemented.

x4 = 1.464 cups of oat bran x5 = .065 eggs x8 = 1.033 cups of milk x10 = 4 slices of wheat toast

Probably not. The t-shirts are a variable cost and any additional t-shirts purchased by Quick-Screen would likely reduce unit profit, which would change the current shadow price for blank t-shirts. The shadow price is effective only if the profit is based on costs that would be incurred without regard to the acquisition of additional resources.

Z = $0.828 Further limiting the servings of the four hot and cold cereals to two cups, xl + x2 + x3 + x4 < 2, results in the following solution: x3 = 2 cups of oatmeal x6 = .750 slices of bacon

The new requirement is that

x8 = 2 cups of milk

x1 = x2 = x3 = x4

x9 = .115 cups of orange juice

This can be achieved within the model by creating three additional constraints, x1 = x2

x10 = 4 slices of wheat toast Z = $0.925 3.

x1 = x3 x1 = x4 If x1 equals x2, x3 and x4 then x2, x3 and x4 must also equal each other. These constraints are changed to, x1 - x2 = 0 x1 - x3 = 0 x1 - x4 = 0 The new solution is x1 = x2 = x3 = x4 = 112.5. 2.

With a minimum of 500 calories, the three food items remain the same; however, the amount of each and the total cost increases: x3 = 2.995 cups of oatmeal, x8 = 1.352 cups of milk, x10 = 1.005 slices of toast, and Z = $0.586. With a minimum of 600 calories, the food items change to x3 = 4.622 cups of oatmeal and 1.378 cups of milk and Z = $0.683. A change of variables would be expected given that 600 calories is greater than the upper limit of the sensitivity range for calories.

4.

It would have no effect; the entire $70,000 would be invested anyway. Since the upper limit of the sensitivity range for the investment amount is “unlimited,” an increase of $10,000 will not affect the shadow price, which is $0.074. Thus, the total increase in return will be $740 (i.e., $10,000 x .074 = $740). No, the entire amount will not be invested in one alternative. The new solution is x1 = $22,363.636, x3 = $43,636.364, and x4 = $14,000. The shadow price is 1.00; thus for every $1 increase in budget, up to the sensitivity range upper limit of $14,000, audience exposure will increase by 1.00. The total audience increase for a $20,000 budget increase is 20,000. This new requirement results in two new model constraints, 20,000x, = 12,000x2 20,000x1 = 9,000x3 or, 20,000x, - 12,000x2 = 0 20,000x1 - 9,000x3 = 0

4-2

The new solution is = 3.068, x2 = 5.114, x3 = 6.818 and Z = 184,090. This results in approximately 61,362 exposures per type of advertising (with some slight differences due to computer rounding). 5.

4,500 to 4,501 results in an increase in total cost to $76,820. 7. a) Maximize Z = $190xt + 170x2 + 155x3 subject to

The slack variables for the three < warehouse constraints would be added to the constraints as follows:

3.5xj + 5.2x2 + 2.8x3 < 500 1.

2xj + 0.8x2 + 1.5x3 < 240 40xj + 55x2 + 20x3 < 6,500

x1A + x1B + x1C + s1 = 300

xj,x2,x3 > 0

x2A + x2B + x2C + s2 = 200

b) x = 41.27, x2 = 0, x3 = 126.98, Z = $27,523.81

x3A + x3B + x3C + s3 = 200 These three slacks would then be added to the objective function with the storage cost coefficients of $9 for s1, $6 for s2, and $7 for s3. This change would not result in a new solution. The model must be reformulated with three new variables reflecting the shipments from the new warehouse at Memphis (4) to the three stores, x4A, x4B, and x4C. These variables must be included in the objective function with the cost coefficients of $18, $9, and $12 respectively. A new supply constraint must be added, X

4A + X4B + X4C < 200

The solution to this reformulated model is x1C = 200 x2B = 50 x3A = 150 X4B = 200 Z = 6,550 Yes, the warehouse should be leased. The shadow price for the Atlanta warehouse shows the greatest decrease in cost, $6 for every additional set supplied from this source. However, the upper limit of the sensitivity range is 200, the same as the current supply value. Thus, if the supply is increased at Atlanta by even one television set the shadow price will change. 6.

This change would not affect the solution at all since there is no surplus with any of the three constraints. Component 1 has the greatest dual price of $20. For each barrel of component 1 the company can acquire, profit will increase by $20, up to the limit of the sensitivity range which is an increase of 1,700 bbls. or 6,200 total bbls. of component 1. For example an increase of 1 bbl. of component 1 from

1

s = s2 = 0, s3 = 2,309.52 1

8. a) It would not affect the model. The slack apples are multiplied by the revenue per apple of $.08 to determine the extra total revenue, i.e., (2,309.52)($.08) = $184.76. b) This change requires a new variable, x4, and that the constraint for apples be changed from < to =. No, the Friendlys should not produce cider. The new solution would be x = 135, x2 = 0, x3 = 0, x4 = 18.33 and Z = $26,475. This reduction in profit occurs because the requirement that all 6,500 apples be used forces resources to be used for cider that would be more profitable to be used to produce the other products. If the final model constraint for apples is < rather than =, the previous solution in 7(b) results. 1

9.

a) x = no. of eggs l

x2 = no. of bacon strips x3 = no. of cups of cereal minimize Z = 4xl + 3x2 + 2x3 subject to 2xt + 4x2 + x3 > 16 3x + 2x2 + x3 > 12 x1, x2, x3 > 0 b) x = 2 x2 = 3 Z i

= $0.17 10. a)

xt = number of boats of type i, i = 1 (bass boat), 2 (ski boat), 3 (speed boat) In order to break even total revenue must equal total cost: 23,000x, + 18,000x2 + 26,000x3 = 12,500xt + 8,500x2 + 13,700x3 + 2,800,000 or, 10,500xt + 9,500x2 + 12,300x3 = 2,800,000 minimize Z = 12,500xt + 8,500x2 + 13,700x3

4-3

13. a) xt = no. of lb of Super Two at Fresno

subject to

x2 = no. of lb of Super Two at Dearborn x3 = no. of

10,500x + 9,500X2 + 12,300x3 = 2,800,000 x, > 70

lb of Green Grow at Fresno x4 = no. of lb of Green

x2 > 50

Grow at Dearborn maximize Z = 7xl + 5x2 + 5x3 +

x3 > 50

4x4 subject to

x1 < 120

2x + 4x2 + 2x3 + 3x4 < 45,000 x + x2 < 6,000 x3 + x4 1

1

x2 < 120

< 7,000 x + x3 < 5,000 x2 + x4 < 6,000 xv x2, x3, x4 > 0 b)

x3 < 120

x = 5,000 x2 = 1,000 x4 = 5,000 Z = $60,000

1

1

14. a) xt = ore i (i = 1,2,3,4,5,6) minimize Z = 27xt + 25X2 + 32x3 + 22X4 + 2()x3 + 24X6

x1,x2,x3 > 0 b) x = 70.00 x2 = 1

subject to

120.00 x3 = 75.203 Z =

.19xt + .43x2 + .17x3 + .20x4 +

.12x6 > .21

.15xt + .10x2 + .12x4 + .24x5 +

.18x6 < .12

$2,925,284.5

.12xt + .25x2

53

x2 = no. of radios x3 = no. of toasters maximize Z = 8x1 + 10x2 + 7x3 subject to

.16x6 < .07

.14x +

.07x2 + .53x3 + .18x4

+ .31x5 +

.25x6 > .30

.14xt +

.07x2 + .53x3 + .18x4

+ .31x5 +

.25x6 < .65

.60xt +

.85X2 + .70x3 + .50x4

+ .65x5 +

.71x6 = 1.00

1

11. a) x1 = no. of clocks

+ .10x5 +

b) x2 = .1153 ton - ore 2 x3 =

7x + 10x2 + 5x3 < 2,000 l

.8487 ton - ore 3 x4 =

2x + 3x2 + 2x3 < 660 x, < i

.0806 ton - ore 4 x5 =

200 x2 < 300 x3 < 150

.4116 ton - ore 5 Z =

xt,x2,x3 > 0

$40.05 15. a) x- j = number of trucks assigned to route from warehouse i to terminal j, where i = 1 (Charlotte), 2 (Memphis), 3 (Louisville) and j = a (St. Louis), b (Atlanta), c (New York) maximize Z = 1,800x1a + 2,100x1t + 1,60C)x1c+ 1,000x2a + 700x2i, + 900X2 + 1,40()x3a + 800x3i, + 2,200x3c

b) x = 178.571 x3

t

= 150.00 Z = $2,478.571 12. a) x = no.

of gallons of Yodel

l

x2 = no.

of gallons of Shotz

x3 = no.

of gallons of Rainwater

C

maximize Z = 1.50xt + 1.60x2 + 1.25x3 subject to x + x2 + x3 = 1,000 1.50xt t

+ .90x2 + .50x3 < 2,000 x1 < 400 x2 < 500 x3 < 300 xv x2, x3 > 0 b) x1 = 400 x2 = 500 x3 = 100 Z = $1,525.00

4-4

subject to

b) xtl = 600 xa =

Xa + Xb +

30

=

X

2a + X2b + ** =

180 x13 =

30

1,680 x„i =

30

600 X

Xa + 2a + 3a <

40

x

X

60

x

X

X

3a + 3b + ** = X

1b +

X

X

2b

+ X3b <

X

1C + -*2 + X3C <

50

450

=

m2

480

m3 =

=0 xM = 270 M

>

x

b3 = 240 Z = $10,123.50 18. a) x = acres of crop i planted on plot j, where i = c (corn), p (peas), s (soybeans) and j = 1,2,3

0

b) X1b = 30

v

X

2a = 30

X

30

3C =

maximize Z = 600(xc1 + xc2 + xc3)

Z = $159,000 16.

+ 450(x?1 + x* + V +

a) x1 = no. of sofas

300(x

,1 + xs2 + xJ

x2 = no. of tables x3 subject to

= no. of chairs maximize Z = 400x1 + 275x2 + 190x3 subject to 1

x

+ x,1 + x» 1 + x,1 + x„ 1 + x,2 + x» 2 + xp + x» 2 2 + xp + x» 3 3 + p + x» 3 3 + x»2 + x< 3 + xp + xp 2 3 x » + x»2 + x» 1 3 + xp1 + xs1)

1

c 1x c 1x » 2 x » 2 x < 3x < 3 x » 1x ,1

7x3 < 1,000 6x1 + 9x2 + 5x3 < 240 x1 + x2 + x3 < 650 x1, x2, x3 > 0 b) x1 = 40 Z = $16,000 17.

a) x = lbs. of seed i used in mix j, where i = t (tall fescue), m (mustang fescue), b (bluegrass) and j = 1,2,3. ÿ

minimize Z = 1.70 (x„ + x + x13) a

+ 2.80 (x„1 + x„2 + xJ + 3.25 (xb1 + xb2 + xb3) 800(x

,1

subject to 50x

- 50x

- 50x

. ,1 . m1 . b1 < -

.2°x,1 + .80xm1

-

-

0

.2°xM >

0

- 30x

70x

0

-

.30xQ + .70x^2 — .30x^2 > 0 .80x,2

—

— 20x

.20xm2

. b2 <

800(x

0

< 700 < 900 < 700 < 1,000

=

0

d + xp1 + xs1) - 500(x c3 + xp3 + xs3)

=

0

—

xc3 = 300

»2 =

l0xi3 — .10xm3 + .90xb3 > 0

ij

^0

xc1 = 500 x

100

x?2 = 700

x

,1 + xm1 + xb1 > 1,200 x«2 + m2 + b2 >

> 420

+ xp2 + xs2) = 0 700(xc2

x

b)

900 x

< 800

+ xp3 + xs3)

c3

.30x,3 — .70xm3 — .70xb3 < 0

x

> 480

700(x

.50x,3 — .50xm3 — .50xb3 > 0

x

<

+ xp2 + xs2)

.3 ^2 . m2 + . b2 > — 0x

500(x c2

X

> 300 O O

7x + 5x2 + 4x3 < 2,250 12x +

xs3 = 400

x

,3 + m3 +

Z

x

b3 > 2,400 xij > 0

=

$975,000 19. a) x1 = $ allocated to job training x2 = $ allocated to parks x3 = $ allocated to sanitation x

4

= $ allocated to library maximize Z =

.02x1 + .09x2 + .06x3 + .04x4

4-5

subject to

+ 9.2x23 + 6.3x24 + 8.4x31 + 8.1x32 + 9.0x33 + 5.8x34

x1 + x2 + x3 + x4 = 4,000,000 x1 <

subject to

1,600,000 x2 < 1,600,000 x3 <

35xt1 + 40x21 + 38x31 < 9,000 41x12

1,600,000 x4 < 1,600,000 x2 - x3 - x4

+ 36x22 + 37x32 < 14,400 34x13 +

< 0 x1 - x3 > 0 x1, x2, x3, x4 > 0 b) x1 = 800,000

32x23 + 33x33 < 12,000 39x14 +

x2 = 1,600,000 x3

43x24 + 40x34 < 15,000

= 800,000 x4 =

x11 + x12 + x13 + x14 = 400

800,000

x21 + x22 + x23 + x24 = 570 x31 + x32 + x33 + x34 =

Z = 240,000 20.

320 31 32 33 x3 4 > 0 b) x11 i j

a) Minimize Z = .80x1 + 3.70x2 + 2.30x3 +.90x4

= 15.385 x14 =

+ .75x5 + .40x6 + .83x7 subject to

384.615 x22 = 400.00 520x1 + 500x2 + 860x3 + 600x4 + 50x5 + 460x6 + 240x7 > 1,500 520x1 + 500x2 + 860x3 + 600x4

x23 = 170.00 x31 = 121.212

+ 50x5 + 460x6 + 240x7 < 2,000

x33 = 198.788

4.4xt + 3.3x2 + .3x3 + 3.4x4

Z = $11,089.73 22. a) Minimize Z = 69xu + 71x12 + 72x13 +74x14 + 76x21 + 74x22 + 75x23 + 79x24 + 86x31 + 89x32 + 80x33 + 82x34

+.5x5 + 2.2x6 + .2x7 > 5 30x1 + 5x2 + 75x3 + 3x4 + 10x7 > 20 30x1 + 5x2 + 75x3 + 3x4 +10x7 < 60 17xl + 85x2 + 82x3 + 10x4 + 6x5

subject to

+ 10x6 + 16x7 > 30

x11 + x12 + x13 + x14 < 220 x21

30x4 + 70x6 + 22x7 > 40 180x1 +

+ x22 + x23 + x24 < 170 x31 +

90x2 + 350x3 + 20x7 < 30

x32 + x33 + x34 < 280 x11 + x21

xi > 0 b)

+ x31 = 110 x12 + x22 + x32 =

x4 = 1.667 x6

160 x13 + x23 + x33 = 90 x14 + x24 + x34 = 180

= 0.304 x7 = 1.500

Z

=

Ash: .03x11 - .01x21 - .02x31 < 0 .04x12 -

$2.867 c) The model becomes infeasible and cannot be solved. Limiting each food item to one- half pound is too restrictive. In fact, experimentation with the model will show that one food item in particular, dried beans, is restrictive. All other food items can be limited except dried beans. 21. a) x = number of units of products i (i = 1,2,3) produced on machine j (j = 1,2,3,4) maximize Z = $7.8xn + 7.8x12 + 8.2x13 + 7.9x14 + 6.7x21 + 8.9x22

.0x22 - .01x32 < 0 .04x13 - .0x23 .01x33 < 0 .03x14 - .01x24 - .02x34 < 0 Sulfur: .01x11 -.01x21- .02x31 < 0 .01x12 -.01x22-.02x32 < 0 .01x13 -.03x23-.04x33 < 0 .0x14 .02x24-.03x34 < 0 14 24

v

b)

x11 = 42 x13 = 18 x

14 = 72

x21 = 10

4-6

x

34 ij

>0

subject to xnc + xnw + xns = 700 x

sc + xsw + xss = 300 xec +

x

ew + xes = 900 x + x + x = 600 wc ww ws xcc + xcw + xsc = 500 nc + xsc + x„ + xwc + xcc < 1,200 xnw + xsw

x

+ xew + xww + xcw < 1,200 xns + xss + xes + x

ws + xcs < 1,200 xj > 0

b) x„c = 700 xss = 300 xes = 900 xww = 600 xxwcwc == 600 500cc

Z = 14,600 25. a) Add the following 3 constraints to the original formulation: xss < 150 xww

<

300 xcc < 250 xnc = 700 xnw = 0 xSw = 150 xss = 150 xes = 900 xwc = 250 xww

=

300 xws = 50 xcc = 250 xcw = 250 Z = 20,400 b) Change the 3 demand constraints in the (a) formulation from < 1,200 to = 1,000 xnc = 400 xnw = 300 xSw = 150 xss = 150 xec = 50 xes = 850 xwc = 300 xww = 300 xcc = 250

4-7

subject to xa1 > 30,000 x„1 > 40,000 xp1 > 50,000 x

a1 + xm1 + xp1 + S1 = 500,000

x

a2 + xm2 + -*>2 + S2 = S1 + 1'2xa1

x

a3 + xm3 + xs3 = S2 + 1-2xa2 + 1-3x 13x m

l

m

x 1 a4

+ xs4 = S3 + 1-2xa3 +

+

x

j, Sj > 0

Note: Since it is assumed that any amount of funds can be invested in each alternative—i.e., there is no minimum investment required—and funds can always be invested in as short a period as one year yielding a positive return, it is apparent that the s

j

variables for uninvested funds will be driven to zero in every period. Thus, these variables could be omitted from the model formulation for this problem. b) xa1 = 410,000

xm1 = 40,000

xa2 = 492,000

xp1 = 50,000

xa3 = 642,400

Z = $1,015,056

xa4 = 845,880 29. a) x1 = no. of homeowner’s policies x2 = no. of auto policies x3 = no. of life policies maximize Z = 35x1 + 20x2 + 58x3 subject to 14x1 + 12x2 + 35x3 < 35,000 6x1 + 3x2 + 12x3 < 20,000 x1, x2, x3 > 0 b)

x1 = 2,500 Z = $87,500

30. a) x1 = no. of issues of Daily Life x = no. of issues of Agriculture Today x3 = 2

no. of issues of Surf’s Up maximize Z = 2.25x1 + 4.00x2 + 1.50x3 subject to x1 + x2 + x3 > 5,000 .01x1 + .03x2 + .02x3 < 120 .2x1 + .5x2 + .3x3 < 3,000 x1 < 3,000 x2 < 2,000 x3 < 6,000 x1, x2, x3 > 0

4-8

b) x

os = 60 Special: x„s + xcs + xms = 200 lbs. xcs = 80 Dark: xbd + x

md = 72 lbs. xms = 60 Regular: xb[ + x0r + xmr = 138 lbs. x,d = 64.8 xmd = 7.2 xbr = 45.2 xor = 10 xm = 82.8 Z = $1,296 33. a) x1 = $ amount borrowed for six months in July y.t = $ amount borrowed in month i (i = 1, 2, ..., 6) for one month ci = $ amount carried over from month i to i + 1 6

minimize Z = .11x1 + .05^ yt i =1

subject to July:

x1 + y1 + 20,000 - c1 = 60,000

August:

c1 + y2 +30,000 - c2 = 60,000 + y1

September: c2 +y3+ 40,000 - c3 = 80,000 + y2 October: c3+y4 + 50,000 - c4 = 30,000 + y3 November: c4 + y5 + 80,000 - c5 = 30,000 + y4 December: c5 + y6 + 100,000 - c6= 20,000 + y5 End: x1 + y6 < c6 x1, y, c.t > 0 b)

Solution x1 = 70,000 y3 = 40,000 y4 = 20,000 y1 = y2 = y5 = y6 = 0 c1 = 30,000 c5 = 30,000 c6 = 110,000 Z =

c)

$10,700 Changing the six-month interest rate to 9% results in the following new solution: x1 = 90,000 y3 = 20,000 c1 = 50,000 c2 = 20,000 c5 = 50,000 c6 = 130,000 Z = $9,100

4-9

4-10