Chapter Four: Linear Programming: Modeling Examples PROBLEM SUMMARY 4-1

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1 Instant download and all chapters Solutions Manual Introduction to Management Science 11th Edition Bernard W. Taylor III PROBLEM SUMMARY Chapter Four: Linear Programming: Modeling Examples 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. Multiperiod investment (maximization) 47. 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. Integer solution (4-53) 55. 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) Assignment (minimization)

2 PROBLEM SOLUTIONS 1. 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. 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. 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. The new requirement is that x 1 = x 2 = x 3 = x 4 This can be achieved within the model by creating three additional constraints, x 1 = x 2 x 1 = x 3 x1 = x4 If x 1 equals x 2, x 3 and x 4 then x 2, x 3 and x 4 must also equal each other. These constraints are changed to, x 1 - x 2 = 0 x 1 - x 3 = 0 x 1 - x 4 = 0 The new solution is x 1 = x 2 = x 3 = x 4 = With a minimum of 500 calories, the three food items remain the same; however, the amount of each and the total cost increases: x 3 = cups of oatmeal, x 8 = cups of milk, x 10 = slices of toast, and Z = $ With a minimum of 600 calories, the food items change to x 3 = cups of oatmeal and cups of milk and Z = $ A change of variables would be expected given that 600 calories is greater than the upper limit of the sensitivity range for calories. 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, x 1, x 2, x 3, and x 4, 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: x 3 = 2 cups of oatmeal x 4 = cups of oat bran x 5 =.065 eggs x 8 = cups of milk x 10 = 4 slices of wheat toast Z = $0.828 Further limiting the servings of the four hot and cold cereals to two cups, x l + x 2 + x 3 + x 4 < 2, results in the following solution: x 3 = 2 cups of oatmeal x 6 =.750 slices of bacon x 8 = 2 cups of milk x 9 =.115 cups of orange juice x 10 = 4 slices of wheat toast Z = $ 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 $ 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 x 1 = $22, , x 3 = $43, , and x 4 = $14, 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 The total audience increase for a $20,000 budget increase is 20,000. This new requirement results in two new model constraints, 20,000x, = 12,000x 2 20,000x1 = 9,000x3 or, 20,000x, - 12,000x 2 = 0 20,000x 1-9,000x 3 = 0 4-2

3 The new solution is = 3.068, x 2 = 5.114, x 3 = and Z = 184,090. This results in approximately 61,362 exposures per type of advertising (with some slight differences due to computer rounding). 5. The slack variables for the three < warehouse constraints would be added to the constraints as follows: x 1A + x 1B + x 1C + s 1 = 300 x 2A + x 2B + x 2C + s 2 = 200 x 3A + x 3B + x 3C + s 3 = 200 These three slacks would then be added to the objective function with the storage cost coefficients of $9 for s 1, $6 for s 2, and $7 for s 3. 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, x 4A, x 4B, and x 4C. 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 + X 4B + X 4C < 200 The solution to this reformulated model is x 1C = 200 x 2B = 50 x 3A = 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 4,500 to 4,501 results in an increase in total cost to $76, a) Maximize Z = $190x t + 170x x 3 3.5xj + 5.2x x 3 < xj + 0.8x x3 < xj + 55x x 3 < 6,500 xj,x 2,x 3 > 0 b) x1 = 41.27, x 2 = 0, x 3 = , Z = $27, s1 = s 2 = 0, s 3 = 2, 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) = $ b) This change requires a new variable, x 4, and that the constraint for apples be changed from < to =. No, the Friendlys should not produce cider. The new solution would be x1 = 135, x 2 = 0, x 3 = 0, x 4 = 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. 9. a) xl = no. of eggs 10. x 2 = no. of bacon strips x 3 = no. of cups of cereal minimize Z = 4x l + 3x 2 + 2x 3 2x t + 4x 2 + x 3 > 16 3xi + 2x 2 + x 3 > 12 x 1, x 2, x 3 > 0 b) x = 2 x 2 = 3 Z = $0.17 a) x t = 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, + 18,000x ,000x 3 = 12,500x t + 8,500x ,700x 3 + 2,800,000 or, 10,500x t + 9,500x ,300x 3 = 2,800,000 minimize Z = 12,500x t + 8,500x ,700x 3 4-3

4 10,500x + 9,500X2 + 12,300x 3 = 2,800,000 x, > 70 x 2 > 50 x 3 > 50 x 1 < 120 x 2 < 120 x 3 < 120 x 1,x 2,x 3 > 0 b) x1 = x 2 = x 3 = Z = $2,925, a) x 1 = no. of clocks x 2 = no. of radios x 3 = no. of toasters maximize Z = 8x x 2 + 7x 3 7xl + 10x 2 + 5x 3 < 2,000 2xi + 3x 2 + 2x 3 < 660 x, < 200 x 2 < 300 x 3 < 150 x t,x 2,x 3 > 0 b) x = x 3 = Z = $2, a) xl = no. of gallons of Yodel x 2 = no. of gallons of Shotz x 3 = no. of gallons of Rainwater maximize Z = 1.50x t x x a) x t = no. of lb of Super Two at Fresno x 2 = no. of lb of Super Two at Dearborn x 3 = no. of lb of Green Grow at Fresno x 4 = no. of lb of Green Grow at Dearborn maximize Z = 7x l + 5x 2 + 5x 3 + 4x 4 2x1 + 4x 2 + 2x 3 + 3x 4 < 45,000 x1 + x 2 < 6,000 x 3 + x 4 < 7,000 x1 + x 3 < 5,000 x 2 + x 4 < 6,000 x v x 2, x 3, x 4 > 0 b) x1 = 5,000 x 2 = 1,000 x 4 = 5,000 Z = $60, a) x t = ore i (i = 1,2,3,4,5,6) minimize Z = 27x t + 25X x X 4 + 2()x X 6.19x t +.43x x x x 6 >.21.15x t +.10x x x x 6 <.12.12x t +.25x x x 6 <.07.14x1 +.07x x x x x 6 >.30.14x t +.07x x x x x 6 <.65.60x t +.85X x x x x 6 = 1.00 b) x 2 =.1153 ton - ore 2 x 3 =.8487 ton - ore 3 x 4 =.0806 ton - ore 4 x 5 =.4116 ton - ore 5 Z = $ a) x-tj = 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 = 1,800x 1a + 2,100x 1t + 1,60C)x 1c+ 1,000x 2a + 700x 2i, + 900X 2C + 1,40()x 3a + 800x 3i, + 2,200x 3c xt + x 2 + x 3 = 1, x t +.90x x 3 < 2,000 x1 < 400 x 2 < 500 x 3 < 300 x v x 2, x 3 > 0 b) x1 = 400 x 2 = 500 x 3 = 100 Z = $1,

5 O O Xa + Xb + = 30 X 2a + X 2b + ** = 30 X 3a + X 3b + ** = 30 Xa + X 2a + X 3a < 40 X 1b + X 2b + X 3b < 60 X 1 C + -*2 + X 3 C < 50 > 0 b) X1b = 30 X 2a = 30 X 3 C = 30 Z = $159, a) x 1 = no. of sofas x 2 = no. of tables x 3 = no. of chairs maximize Z = 400x x x 3 subject to 7x1 + 5x 2 + 4x 3 < 2,250 12x1 + 7x 3 < 1,000 6x 1 + 9x 2 + 5x 3 < 240 x 1 + x 2 + x 3 < 650 x 1, x 2, x 3 > 0 b) x1 = 40 Z = $16, a) xÿ = lbs. of seed i used in mix j, where i = t (tall fescue), m (mustang fescue), b (bluegrass) and j = 1,2,3. minimize Z = 1.70 (x + xa + x 13). 50x, x m x b1 < x, x m1. 80x, (x 1 + x 2 + x J (x b1 + x b2 + x b3 ) -.2 x M >.3 0x^ x m x b2 > 0.30x Q +.70x^2.30x^2 > 0.2 0x m2. 50x, 3. 50x m 3. 50x b 3 > 0.30x, 3. 70x m 3. 70x b 3 < 0. 20x b2 < 0 l 0x i3. 10x m x b3 > 0 x,1 + x m1 + x b1 > 1,200 x «2 + x m2 + x b2 > 900 x,3 + x m3 + x b3 > 2,400 x ij > 0 0 b) x tl = 600 x a = 180 x 13 = 1,680 x i = 600 X m2 = 450 x m3 = 480 x M = 0 x M = 270 x b3 = 240 Z = $10, a) xv = acres of crop i planted on plot j, where i = c (corn), p (peas), s (soybeans) and j = 1,2,3 maximize Z = 600(x c1 + x c2 + x c3) + 450(x?1 + x* + V + 300(x,1 + x s2 + x J x c + x,1 + x» > x c + x,1 + x < 1 1 x» + x,2 + x» > x» + x p + x» < x < + x p + x» > x < + X p + x» < x» + x»2 + x < < x,1 + x p + x p < x» 1 800(x,1 + x p1 + x s1 ) + x p2 + x s2 ) + x»2 + x» 3 < 1, (x c2 + x p2 + x s2 ) = 0 700(x c2-800(x c3 + x p3 + x s3 ) = 700(x d + x p1 + x s1 ) b) x c1 = 500 x»2 = - 500(x c3 + x p3 + x s3 ) = 100 x c3 = 300 x?2 = 700 x s3 = 400 Z = $975,000 x ij ^ a) x 1 = $ allocated to job training x 2 = $ allocated to parks x 3 = $ allocated to sanitation x 4 = $ allocated to library maximize Z = x x x x 4 4-5

6 x 1 + x 2 + x 3 + x 4 = 4,000,000 x 1 < 1,600,000 x 2 < 1,600,000 x 3 < 1,600,000 x 4 < 1,600,000 x 2 - x 3 - x 4 < 0 x 1 - x 3 > 0 x 1, x 2, x 3, x 4 > 0 b) x 1 = 800,000 x2 = 1,600,000 x3 = 800,000 x4 = 800,000 Z = 240, a) Minimize Z =.80x x x x 4 b) x 4 = x x x x 7 520x x x x x x x 7 > 1, x x x x x x x 7 < 2, x t + 3.3x 2 +.3x x 4 +.5x x 6 +.2x 7 > 5 30x 1 + 5x x 3 + 3x x 7 > 20 30x 1 + 5x x 3 + 3x 4 +10x 7 < 60 17x l + 85x x x 4 + 6x x x 7 > 30 30x x x 7 > x x x x 7 < 30 = x 7 = Z = $2.867 xi > 0 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) xv = number of units of products i (i = 1,2,3) produced on machine j (j = 1,2,3,4) maximize Z = $7.8x n + 7.8x x x x x 22 x 31 + x 32 + x 33 + x 34 = x 3 4 i j > 0 b) x x x x x x x 34 35x t1 + 40x x 31 < 9,000 41x x x 32 < 14,400 34x x x 33 < 12,000 39x x x 34 < 15,000 x11 + x12 + x13 + x14 = 400 x21 + x22 + x23 + x24 = 570 = x 14 = x 22 = x 23 = x31 = x33 = Z = $11, a) Minimize Z = 69x u + 71x x x x x x x x x x x 34 x 11 + x 12 + x 13 + x 14 < 220 x 21 + x 22 + x 23 + x 24 < 170 x 31 + x 32 + x 33 + x 34 < 280 x 11 + x 21 + x 31 = 110 x 12 + x 22 + x 32 = 160 x 13 + x 23 + x 33 = 90 x14 + x24 + x 34 = 180 Ash:.03x x x 31 < 0.04x 12 - b) x11 = 42 x 13 = 18 x 14 = 72.0x x 32 < 0.04x x x 33 < 0.03x x x 34 < 0 Sulfur:.01x x x 31 < 0.01x x x 32 < 0.01x x x 33 < 0.0x x x 34 < x 34 ij > 0 x21 =

7 b) x c = 700 xss = 300 x es = 900 x ww = 600 x x w c w c = = xnc + xnw + xns = 700 x sc + x sw + x ss = 300 x ec + x ew + x es = 900 x + x + x = 600 wc ww ws xcc + xcw + xsc = 500 x nc + x sc + x + x wc + x cc < 1,200 x nw + x sw + x ew + x ww + x cw < 1,200 x ns + x ss + x es + x ws + x cs < 1,200 x j > cc Z = 14, a) Add the following 3 constraints to the original formulation: xss < 150 xww < 300 xcc < 250 x nc = 700 xnw = 0 x Sw = 150 x ss = 150 x es = 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 x Sw = 150 x ss = 150 xec = 50 xes = 850 xwc = 300 xww = 300 xcc =

8 x a1 > 30,000 x 1 > 40,000 x p1 > 50,000 x a1 + x m1 + x p1 + S 1 = 500,000 x a2 + x m2 + -*>2 + S 2 = S ' 2x a1 x a3 + x m3 + x s3 = S x a x m1 x a4 + x s4 = S x a3 + 13x ml + x j, S j > 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 sj 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) x a1 = 410,000 x m1 = 40,000 x a2 = 492,000 x p1 = 50,000 x a3 = 642,400 Z = $1,015,056 x a4 = 845, a) x 1 = no. of homeowner s policies x 2 = no. of auto policies x 3 = no. of life policies maximize Z = 35x x x 3 14x x x 3 < 35,000 6x 1 + 3x x 3 < 20,000 x 1, x 2, b) x 1 = 2,500 Z = $87,500 x 3 > a) x 1 = no. of issues of Daily Life x2 = no. of issues of Agriculture Today x 3 = no. of issues of Surf s Up maximize Z = 2.25x x x 3 x 1 + x 2 + x 3 > 5,000.01x x x 3 < 120.2x 1 +.5x 2 +.3x 3 < 3,000 x 1 < 3,000 x 2 < 2,000 x 3 < 6,000 x 1, x 2, x 3 > 0 4-8

9 b) x os = 60 Sp ecial: x s + x c s + x m s = 200 lbs. x cs = 80 Dark: x bd + x md = 72 lbs. x ms = 60 R e gu la r : x b [ + x 0r + x mr = 138 lbs. x,d = xmd = 7.2 xbr = 45.2 xor = 10 x m = 82.8 Z = $1, a) x 1 = $ amount borrowed for six months in July y. t = $ amount borrowed in month i (i = 1, 2,..., 6) for one month c i = $ amount carried over from month i to i + 1 minimize Z =.11x ^ y t 6 i = 1 July: x 1 + y ,000 - c 1 = 60,000 August: c 1 + y 2 +30,000 - c 2 = 60,000 + y 1 September: c 2 +y 3+ 40,000 - c 3 = 80,000 + y 2 October: c 3+y ,000 - c 4 = 30,000 + y 3 November: c 4 + y ,000 - c 5 = 30,000 + y 4 December: c 5 + y ,000 - c 6= 20,000 + y 5 End: x 1 + y 6 < c 6 b) Solution x 1, y, c. t > 0 x 1 = 70,000 y3 = 40,000 y4 = 20,000 y1 = y2 = y5 = y6 = 0 c1 = 30,000 c5 = 30,000 c 6 = 110,000 Z = $10,700 c) 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 c 6 = 130,000 Z = $9,

10 4-10

Chapter 4. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall 4-1

Chapter 4. Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall 4-1 Linear Programming: Modeling Examples Chapter 4 4-1 Chapter Topics A Product Mix Example A Diet Example An Investment Example A Marketing Example A Transportation Example A Blend Example A Multiperiod