Chapter 2 n Introduction to Linear Programming Learning Objectives 1. Obtain an overview of the kinds of problems linear programming has been used to solve. 2. Learn how to develop linear programming models for simple problems. 3. e able to identify the special features of a model that make it a linear programming model. 4. Learn how to solve two variable linear programming models by the graphical solution procedure. 5. Understand the importance of extreme points in obtaining the optimal solution. 6. Know the use and interpretation of slack and surplus variables. 7. e able to interpret the computer solution of a linear programming problem. 8. Understand how alternative optimal solutions, infeasibility and unboundedness can occur in linear programming problems. 9. Understand the following terms: problem formulation constraint function objective function solution optimal solution nonnegativity constraints mathematical model linear program linear functions feasible solution feasible region slack variable standard form redundant constraint extreme point surplus variable alternative optimal solutions infeasibility unbounded 2-1
Chapter 2 Solutions: 1. a, b, and e, are acceptable linear programming relationships. c is not acceptable because of 2 2 2. a. d is not acceptable because of 3 f is not acceptable because of 1 c, d, and f could not be found in a linear programming model because they have the above nonlinear terms. 8 4 4 8 b. 8 4 4 8 c. 8 4 Points on line are only feasible points 4 8 2-2
n Introduction to Linear Programming 3. a. (,9) (6,) b. (,6) (4,) c. (,2) Points on line are only feasible solutions 4. a. (4,) (2,) (,-15) 2-3
Chapter 2 b. (,12) (-1,) c. (1,25) Note: Point shown was used to locate position of the constraint line 5. a 3 c 2 1 b 1 2 3 2-4
n Introduction to Linear Programming 6. 7 + 1 = 42 is labeled (a) 6 + 4 = 42 is labeled (b) -4 + 7 = 42 is labeled (c) 1 8 (c) 6 4 (b) 2 (a) -1-8 -6-4 -2 2 4 6 8 1 7. 1 5 5 1 15 2 25 2-5
Chapter 2 8. 2 133 1 / 3 (1,2) -2-1 1 2 9. 2 (15,225) 1 (15,1) 1 2 3-1 -2 2-6
n Introduction to Linear Programming 1. 5 4 3 2 Optimal Solution = 12/7, = 15/7 Value of Objective Function = 2(12/7) + 3(15/7) = 69/7 1 1 2 3 4 5 6 + 2 = 6 (1) 5 + 3 = 15 (2) (1) 5 5 + 1 = 3 (3) (2) - (3) - 7 = -15 = 15/7 From (1), = 6-2(15/7) = 6-3/7 = 12/7 2-7
Chapter 2 11. = 1 Optimal Solution = 1, = 5 Value of Objective Function = 75 1 = 8 1 2 12. a. 6 5 4 3 2 Optimal Solution = 3, = 1.5 Value of Objective Function = 13.5 (3,1.5) 1 (,) 1 2 3 4 5 6 (4,) 2-8
n Introduction to Linear Programming b. 3 Optimal Solution =, = 3 Value of Objective Function = 18 2 1 (,) 1 2 3 4 5 6 7 8 9 1 13. a. c. There are four extreme points: (,), (4,), (3,1,5), and (,3). 8 6 4 Feasible Region consists of this line segment only 2 2 4 6 8 b. The extreme points are (5, 1) and (2, 4). 2-9
Chapter 2 c. 8 6 4 Optimal Solution = 2, = 4 2 2 4 6 8 14. a. Let F = number of tons of fuel additive S = number of tons of solvent base Max 4F + 3S 2/5F + 1 /2 S 2 Material 1 1 /5 S 5 Material 2 3 /5 F + 3 /1 S 21 Material 3 F, S 2-1
n Introduction to Linear Programming b. S F 15. a. c. Material 2: 4 tons are used, 1 ton is unused. d. No redundant constraints. 2-11
Chapter 2 b. Similar to part (a): the same feasible region with a different objective function. The optimal solution occurs at (78, ) with a profit of z = 2(78) + 9() = 14,16. c. The sewing constraint is redundant. Such a change would not change the optimal solution to the original problem. 16. a. variety of objective functions with a slope greater than -4/1 (slope of I & P line) will make extreme point (, 54) the optimal solution. For example, one possibility is 3S + 9D. b. Optimal Solution is S = and D = 54. c. Department Hours Used Max. vailable Slack Cutting and Dyeing 1(54) = 54 63 9 Sewing 5 / 6 (54) = 45 6 15 Finishing 2 / 3 (54) = 36 78 348 Inspection and Packaging 1 / 4 (54) = 135 135 17. 18. a. Max 5 + 2 + S 1 + S 2 + S 3 1-2 + 1S 1 = 42 2 + 3 + 1S 2 = 61 6-1 + 1S 3 = 125,, S 1, S 2, S 3 Max 4 + 1 + S 1 + S 2 + S 3 1 + 2 + 1S 1 = 3 3 + 2 + 1S 2 = 12 2 + 2 + 1S 3 = 1,, S 1, S 2, S 3 2-12
n Introduction to Linear Programming b. 14 12 1 8 6 4 Optimal Solution = 18/7, = 15/7, Value = 87/7 2 2 4 6 8 1 c. S 1 =, S 2 =, S 3 = 4/7 19. a. Max 3 + 4 + S 1 + S 2 + S 3-1 + 2 + 1S 1 = 8 (1) 1 + 2 + 1S 2 = 12 (2) 2 + 1 + 1S 3 = 16 (3),, S 1, S 2, S 3 2-13
Chapter 2 b. 14 (3) 12 1 (1) 8 6 4 Optimal Solution = 2/3, = 8/3 Value = 3 2/3 2 (2) 2 4 6 8 1 12 c. S 1 = 8 + 2 = 8 + 2/3-16/3 = 28/3 S 2 = 12-2 = 12-2/3-16/3 = S 3 = 16 2 - = 16-4/3-8/3 = 2. a. Max 3 + 2 + - S 1 = 4 3 + 4 + S 2 = 24 - S 3 = 2 - - S 4 =,, S 1, S 2, S 3, S 4 2-14
n Introduction to Linear Programming b. c. S 1 = (3.43 + 3.43) - 4 = 2.86 S 2 = 24 - [3(3.43) + 4(3.43)] = S 3 = 3.43-2 = 1.43 S 4 = - (3.43-3.43) = 2-15
Chapter 2 21. a. and b. 9 8 7 6 Constraint 2 5 Optimal Solution 4 Constraint 3 Constraint 1 3 2 Feasible Region 1 2 + 3 = 6 1 2 3 4 5 6 7 8 9 1 c. Optimal solution occurs at the intersection of constraints 1 and 2. For constraint 2, = 1 + Substituting for in constraint 1 we obtain 5 + 5(1 + ) = 4 5 + 5 + 5 = 4 1 = 35 = 35 = 1 + = 1 + 35 = 45 Optimal solution is = 35, = 45 d. ecause the optimal solution occurs at the intersection of constraints 1 and 2, these are binding constraints. 2-16
n Introduction to Linear Programming e. Constraint 3 is the nonbinding constraint. t the optimal solution 1 + 3 = 1(35) + 3(45) = 17. ecause 17 exceeds the right-hand side value of 9 by 8 units, there is a surplus of 8 associated with this constraint. 22. a. C 35 3 25 2 5 Inspection and Packaging Cutting and Dyeing 15 4 Feasible Region 1 Sewing 3 5 5 + 4C = 4 1 2 5 1 15 2 25 3 Number of ll-pro Footballs b. Extreme Point Coordinates Profit 1 (, ) 5() + 4() = 2 (17, ) 5(17) + 4() = 85 3 (14, 6) 5(14) + 4(6) = 94 4 (8, 12) 5(8) + 4(12) = 88 5 (, 168) 5() + 4(168) = 672 Extreme point 3 generates the highest profit. c. Optimal solution is = 14, C = 6 d. The optimal solution occurs at the intersection of the cutting and dyeing constraint and the inspection and packaging constraint. Therefore these two constraints are the binding constraints. e. New optimal solution is = 8, C = 12 Profit = 4(8) + 5(12) = 92 2-17
Chapter 2 23. a. Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced Max 24E + 18L 6E + 3L 21 Engine time L 28 Lady-Sport maximum 2E + 2.5L 1 ssembly and testing E, L b. L 7 6 Engine Manufacturing Time Number of EZ-Rider Produced 5 4 3 2 Frames for Lady-Sport Optimal Solution E = 25, L = 2 Profit = $96, 1 ssembly and Testing 1 2 3 4 5 Number of Lady-Sport Produced E c. The binding constraints are the manufacturing time and the assembly and testing time. 24. a. Let R = number of units of regular model. C = number of units of catcher s model. Max 5R + 8C 1R + 3 /2 C 9 Cutting and sewing R, C 1 /2 R + 1 /3 C 3 Finishing 1 /8 R + 1 /4 C 1 Packing and Shipping 2-18
. Catcher's Model n Introduction to Linear Programming b. C 1 8 F 6 4 C & S P & S Optimal Solution (5,15) 2 2 4 6 8 1 Regular Model c. 5(5) + 8(15) = $3,7 R d. C & S 1(5) + 3 / 2 (15) = 725 F 1 / 2 (5) + 1 / 3 (15) = 3 P & S 1 / 8 (5) + 1 / 4 (15) = 1 e. Department Capacity Usage Slack C & S 9 725 175 hours F 3 3 hours P & S 1 1 hours 25. a. Let = percentage of funds invested in the bond fund S = percentage of funds invested in the stock fund Max.6 +.1 S.3 ond fund minimum.6 +.1 S.75 Minimum return + S = 1 Percentage requirement b. Optimal solution: =.3, S =.7 Value of optimal solution is.88 or 8.8% 2-19
Chapter 2 26. a. a. Let N = amount spent on newspaper advertising R = amount spent on radio advertising b. Max 5N + 8R N + R = 1 udget N 25 Newspaper min. R 25 Radio min. N -2R News 2 Radio N, R R 1 Radio Min udget Optimal Solution N = 666.67, R = 333.33 Value = 6, 5 N = 2 R Newspaper Min Feasible region is this line segment 5 1 N 27. Let I = Internet fund investment in thousands = lue Chip fund investment in thousands Max.12I +.9 1I + 1 5 vailable investment funds 1I 35 Maximum investment in the internet fund 6I + 4 24 Maximum risk for a moderate investor I, 2-2
n Introduction to Linear Programming 6 Risk Constraint lue Chip Fund (s) 5 4 3 Optimal Solution I = 2, = 3 $5,1 Maximum Internet Funds 2 1 Objective Function.12I +.9 vailable Funds $5, 1 2 3 4 5 6 I Internet Fund (s) Internet fund $2, lue Chip fund $3, nnual return $ 5,1 b. The third constraint for the aggressive investor becomes 6I + 4 32 This constraint is redundant; the available funds and the maximum Internet fund investment constraints define the feasible region. The optimal solution is: Internet fund $35, lue Chip fund $15, nnual return $ 5,55 The aggressive investor places as much funds as possible in the high return but high risk Internet fund. c. The third constraint for the conservative investor becomes 6I + 4 16 This constraint becomes a binding constraint. The optimal solution is Internet fund $ lue Chip fund $4, nnual return $ 3,6 2-21
Chapter 2 The slack for constraint 1 is $1,. This indicates that investing all $5, in the lue Chip fund is still too risky for the conservative investor. $4, can be invested in the lue Chip fund. The remaining $1, could be invested in low-risk bonds or certificates of deposit. 28. a. Let W = number of jars of Western Foods Salsa produced M = number of jars of Mexico City Salsa produced Max 1W + 1.25M 5W 7M 448 Whole tomatoes 3W + 1M 28 Tomato sauce 2W + 2M 16 Tomato paste W, M Note: units for constraints are ounces b. Optimal solution: W = 56, M = 24 Value of optimal solution is 86 29. a. Let = proportion of uffalo's time used to produce component 1 D = proportion of Dayton's time used to produce component 1 Maximum Daily Production Component 1 Component 2 uffalo 2 1 Dayton 6 14 Number of units of component 1 produced: 2 + 6D Number of units of component 2 produced: 1(1 - ) + 6(1 - D) For assembly of the ignition systems, the number of units of component 1 produced must equal the number of units of component 2 produced. Therefore, 2 + 6D = 1(1 - ) + 14(1 - D) 2 + 6D = 1-1 + 14-14D 3 + 2D = 24 Note: ecause every ignition system uses 1 unit of component 1 and 1 unit of component 2, we can maximize the number of electronic ignition systems produced by maximizing the number of units of subassembly 1 produced. Max 2 + 6D In addition, 1 and D 1. 2-22
n Introduction to Linear Programming The linear programming model is: Max 2 + 6D 3 + 2D = 24 1 D 1, D The graphical solution is shown below. 1.2 D 1..8.6.4.2 3 + 2D = 24 2 + 6D = 3 Optimal Solution.2.4.6.8 1. 1.2 Optimal Solution: =.8, D = Optimal Production Plan uffalo - Component 1.8(2) = 16 uffalo - Component 2.2(1) = 2 Dayton - Component 1 (6) = Dayton - Component 2 1(14) = 14 Total units of electronic ignition system = 16 per day. 2-23
Chapter 2 3. a. Let E = number of shares of Eastern Cable C = number of shares of ComSwitch Max 15E + 18C 4E + 25C 5, Maximum Investment 4E 15, Eastern Cable Minimum 25C 1, ComSwitch Minimum 25C 25, ComSwitch Maximum E, C b. C 2 Minimum Eastern Cable Number of Shares of ComSwitch 15 1 5 Maximum Comswitch Maximum Investment Minimum Conswitch 5 1 15 Number of Shares of Eastern Cable E c. There are four extreme points: (375,4); (1,4);(625,1); (375,1) d. Optimal solution is E = 625, C = 1 Total return = $27,375 2-24
n Introduction to Linear Programming 31. 6 4 Feasible Region 2 Objective Function Value = 13 2 4 Optimal Solution = 3, = 1 6 3 + 4 = 13 32. 2-25
Chapter 2 33. a. Extreme Points Objective Function Value Surplus Demand Surplus Total Production Slack Processing Time ( = 25, = 1) 8 125 ( = 125, = 225) 925 125 ( = 125, = 35) 13 125 x 2 6 4 2 2 4 6 x 1 Optimal Solution: = 3, = 1, value = 5 b. (1) 3 + 4(1) = 7 Slack = 21-7 = 14 (2) 2(3) + 1 = 7 Surplus = 7-7 = (3) 3(3) + 1.5 = 1.5 Slack = 21-1.5 = 1.5 (4) -2(3) +6(1) = Surplus = - = 2-26
n Introduction to Linear Programming c. Optimal Solution: = 6, = 2, value = 34 34. a. 4 x 2 3 2 Feas ible Region (21/4, 9/4) 1 (4,1) 1 2 3 4 5 b. There are two extreme points: ( = 4, = 1) and ( = 21/4, = 9/4) c. The optimal solution is = 4, = 1 6 x 1 2-27
Chapter 2 35. a. Min 6 + 4 + S 1 + S 2 + S 3 2 + 1 - S 1 = 12 1 + 1 - S 2 = 1 1 + S 3 = 4,, S 1, S 2, S 3 b. The optimal solution is = 6, = 4. c. S 1 = 4, S 2 =, S 3 =. 36. a. Let T = number of training programs on teaming P = number of training programs on problem solving Max 1,T + 8,P T 8 Minimum Teaming P 1 Minimum Problem Solving T + P 25 Minimum Total 3 T + 2 P 84 Days vailable T, P 2-28
n Introduction to Linear Programming b. P 4 Minimum Teaming Number of Problem-Solving Programs 3 2 1 Minimum Total Days vailable Minimum Problem Solving 1 2 Number of Teaming Programs 3 T c. There are four extreme points: (15,1); (21.33,1); (8,3); (8,17) d. The minimum cost solution is T = 8, P = 17 Total cost = $216, 37. Regular Zesty Mild 8% 6% 81 Extra Sharp 2% 4% 3 Let R = number of containers of Regular Z = number of containers of Zesty Each container holds 12/16 or.75 pounds of cheese Pounds of mild cheese used =.8 (.75) R +.6 (.75) Z =.6 R +.45 Z Pounds of extra sharp cheese used =.2 (.75) R +.4 (.75) Z =.15 R +.3 Z 2-29
Chapter 2 Cost of Cheese = Cost of mild + Cost of extra sharp = 1.2 (.6 R +.45 Z) + 1.4 (.15 R +.3 Z) =.72 R +.54 Z +.21 R +.42 Z =.93 R +.96 Z Packaging Cost =.2 R +.2 Z Total Cost = (.93 R +.96 Z) + (.2 R +.2 Z) = 1.13 R + 1.16 Z Revenue = 1.95 R + 2.2 Z Profit Contribution = Revenue - Total Cost = (1.95 R + 2.2 Z) - (1.13 R + 1.16 Z) =.82 R + 1.4 Z Max.82 R + 1.4 Z.6 R +.45 Z 81 Mild.15 R +.3 Z 3 Extra Sharp R, Z Optimal Solution: R = 96, Z = 52, profit =.82(96) + 1.4(52) = $13,28 38. a. Let S = yards of the standard grade material per frame P = yards of the professional grade material per frame Min 7.5S + 9.P.1S +.3P 6 carbon fiber (at least 2% of 3 yards).6s +.12P 3 kevlar (no more than 1% of 3 yards) S + P = 3 total (3 yards) S, P 2-3
n Introduction to Linear Programming b. 5 P Professional Grade (yards) 4 3 2 1 total carbon fiber Extreme Point S = 15 P = 15 Extreme Point S = 1 P = 2 Feasible region is the line segment kevlar 1 2 3 4 5 6 Standard Grade (yards) S c. Extreme Point Cost (15, 15) 7.5(15) + 9.(15) = 247.5 (1, 2) 7.5(1) + 9.(2) = 255. The optimal solution is S = 15, P = 15 d. Optimal solution does not change: S = 15 and P = 15. However, the value of the optimal solution is reduced to 7.5(15) + 8(15) = $232.5. e. t $7.4 per yard, the optimal solution is S = 1, P = 2. The value of the optimal solution is reduced to 7.5(1) + 7.4(2) = $223.. lower price for the professional grade will not change the S = 1, P = 2 solution because of the requirement for the maximum percentage of kevlar (1%). 39. a. Let S = number of units purchased in the stock fund M = number of units purchased in the money market fund Min 8S + 3M 5S + 1M 1,2, Funds available 5S + 4M 6, nnual income M 3, Minimum units in money market S, M, 2-31
Units of Money Market Fund Chapter 2 M x 2 2 8x8S 1 + 3x 3M = 2 62, 15 Optimal Solution 1. 5 5 1 15 2 x 1 S Units of Stock Fund Optimal Solution: S = 4, M = 1, value = 62 b. nnual income = 5(4) + 4(1) = 6, c. Invest everything in the stock fund. 4. Let P 1 = gallons of product 1 P 2 = gallons of product 2 Min 1P 1 + 1P 2 1P 1 + 3 Product 1 minimum 1P 2 2 Product 2 minimum 1P 1 + 2P 2 8 Raw material P 1, P 2 2-32
n Introduction to Linear Programming 8 P 2 Number of Gallons of Product 2 6 4 2 1P 1 +1P 2 = 55 (3,25) Feasible Region Use 8 gals. 2 4 6 8 Number of Gallons of Product 1 P 1 Optimal Solution: P 1 = 3, P 2 = 25 Cost = $55 41. a. Let R = number of gallons of regular gasoline produced P = number of gallons of premium gasoline produced Max.3R +.5P.3R +.6P 18, Grade crude oil available 1R + 1P 5, Production capacity 1P 2, Demand for premium R, P 2-33
Chapter 2 b. P 6, Gallons of Premium Gasoline 5, 4, 3, 2, 1, Production Capacity Maximum Premium Optimal Solution R = 4,, P = 1, $17, Grade Crude Oil R 1, 2, 3, 4, 5, 6, Optimal Solution: 4, gallons of regular gasoline 1, gallons of premium gasoline Total profit contribution = $17, Gallons of Regular Gasoline c. Constraint Value of Slack Variable Interpretation 1 ll available grade crude oil is used 2 Total production capacity is used 3 1, Premium gasoline production is 1, gallons less than the maximum demand d. Grade crude oil and production capacity are the binding constraints. 2-34
n Introduction to Linear Programming 42. x 2 14 12 Satis fies Constraint # 2 1 8 6 Infeas ibility 4 2 Satis fies Constraint # 1 2 4 6 8 1 x 1 12 43. 4 x 2 3 Unbounded 2 1 1 2 3 x 1 44. a. 4 2 x 2 Objective Function Optimal Solution (3/16, 3/16) Value = 6/16 2 4 x 1 b. New optimal solution is =, = 3, value = 6. 2-35
Chapter 2 45. a. b. Feasible region is unbounded. c. Optimal Solution: = 3, =, z = 3. d. n unbounded feasible region does not imply the problem is unbounded. This will only be the case when it is unbounded in the direction of improvement for the objective function. 46. Let N = number of sq. ft. for national brands G = number of sq. ft. for generic brands Problem Constraints: N + G 2 Space available N 12 National brands G 2 Generic 2-36
n Introduction to Linear Programming Extreme Point N G 1 12 2 2 18 2 3 12 8 a. Optimal solution is extreme point 2; 18 sq. ft. for the national brand and 2 sq. ft. for the generic brand. b. lternative optimal solutions. ny point on the line segment joining extreme point 2 and extreme point 3 is optimal. c. Optimal solution is extreme point 3; 12 sq. ft. for the national brand and 8 sq. ft. for the generic brand. 2-37
Chapter 2 47. 6 x 2 5 Processing Time 4 3 2 (125,225) lternate optima 1 (25,1) 1 2 3 4 x 1 lternative optimal solutions exist at extreme points ( = 125, = 225) and ( = 25, = 1). or Cost = 3(125) + 3(225) = 15 Cost = 3(25) + 3(1) = 15 The solution ( = 25, = 1) uses all available processing time. However, the solution. ( = 125, = 225) uses only 2(125) + 1(225) = 475 hours. Thus, ( = 125, = 225) provides 6-475 = 125 hours of slack processing time which may be used for other products. 2-38
n Introduction to Linear Programming 48. Possible ctions: i. Reduce total production to = 125, = 35 on 475 gallons. ii. Make solution = 125, = 375 which would require 2(125) + 1(375) = 625 hours of processing time. This would involve 25 hours of overtime or extra processing time. iii. Reduce minimum production to 1, making = 1, = 4 the desired solution. 49. a. Let P = number of full-time equivalent pharmacists T = number of full-time equivalent physicians The model and the optimal solution are shown below: MIN 4P+1T S.T. 1) P+T >=25 2) 2P-T>= 3) P>=9 Optimal Objective Value 52. Variable Value Reduced Cost P 9.. T 16.. 2-39
Chapter 2 Constraint Slack/Surplus Dual Value 1. 1. 2 2.. 3. 3. The optimal solution requires 9 full-time equivalent pharmacists and 16 full-time equivalent technicians. The total cost is $52 per hour. b. Current Levels ttrition Optimal Values New Hires Required Pharmacists 85 1 9 15 Technicians 175 3 16 15 The payroll cost using the current levels of 85 pharmacists and 175 technicians is 4(85) + 1(175) = $515 per hour. The payroll cost using the optimal solution in part (a) is $52 per hour. Thus, the payroll cost will go up by $5 5. Let M = number of Mount Everest Parkas R = number of Rocky Mountain Parkas Max 1M + 15R 3M + 2R 72 Cutting time 45M + 15R 72 Sewing time.8m -.2R % requirement Note: Students often have difficulty formulating constraints such as the % requirement constraint. We encourage our students to proceed in a systematic step-by-step fashion when formulating these types of constraints. For example: M must be at least 2% of total production M.2 (total production) M.2 (M + R) M.2M +.2R.8M -.2R 2-4
n Introduction to Linear Programming The optimal solution is M = 65.45 and R = 261.82; the value of this solution is z = 1(65.45) + 15(261.82) = $45,818. If we think of this situation as an on-going continuous production process, the fractional values simply represent partially completed products. If this is not the case, we can approximate the optimal solution by rounding down; this yields the solution M = 65 and R = 261 with a corresponding profit of $45,65. 51. Let C = number sent to current customers N = number sent to new customers Note: Number of current customers that test drive =.25 C Number of new customers that test drive =.2 N Number sold =.12 (.25 C ) +.2 (.2 N ) =.3 C +.4 N Max.3C +.4N.25 C 3, Current Min.2 N 1, New Min.25 C -.4 N Current vs. New 4 C + 6 N 1,2, udget C, N, 2-41
Chapter 2 2, N udget Current Min. Current 2 New 1,.3C +.4N = 6 Optimal Solution C = 225,, N = 5, Value = 8,75 New Min. 1, 2, 3, C 52. Let S = number of standard size rackets O = number of oversize size rackets Max 1S + 15O.8S -.2O % standard 1S + 12O 48 Time.125S +.4O 8 lloy S, O, 2-42
n Introduction to Linear Programming 53. a. Let R = time allocated to regular customer service N = time allocated to new customer service Max 1.2R + N R + N 8 25R + 8N 8 -.6R + N R, N, b. Optimal Objective Value 9. Variable Value Reduced Cost R 5.. N 3.. Constraint Slack/Surplus Dual Value 1. 1.125 2 69.. 3. -.125 Optimal solution: R = 5, N = 3, value = 9 HTS should allocate 5 hours to service for regular customers and 3 hours to calling on new customers. 54. a. Let M 1 = number of hours spent on the M-1 machine M 2 = number of hours spent on the M-2 machine Total Cost 6(4)M 1 + 6(5)M 2 + 5M 1 + 75M 2 = 29M 1 + 375M 2 Total Revenue 25(18)M 1 + 4(18)M 2 = 45M 1 + 72M 2 Profit Contribution (45-29)M 1 + (72-375)M 2 = 16M 1 + 345M 2 2-43
Chapter 2 Max 16 M 1 + 345M 2 M 1 15 M-1 maximum M 2 1 M-2 maximum M 1 5 M-1 minimum M 2 5 M-2 minimum 4 M 1 + 5 M 2 1 Raw material available M 1, M 2 b. Optimal Objective Value 545. Variable Value Reduced Cost M1 12.5. M2 1. 145. Constraint Slack/Surplus Dual Value 1 2.5. 2. 145. 3 7.5. 4 5.. 5. 4. The optimal decision is to schedule 12.5 hours on the M-1 and 1 hours on the M-2. 55. Mr. Krtick s solution cannot be optimal. Every department has unused hours, so there are no binding constraints. With unused hours in every department, clearly some more product can be made. 56. No, it is not possible that the problem is now infeasible. Note that the original problem was feasible (it had an optimal solution). Every solution that was feasible is still feasible when we change the constraint to lessthan-or-equal-to, since the new constraint is satisfied at equality (as well as inequality). In summary, we have relaxed the constraint so that the previous solutions are feasible (and possibly more satisfying the constraint as strict inequality). 57. Yes, it is possible that the modified problem is infeasible. To see this, consider a redundant greater-thanor-equal to constraint as shown below. Constraints 2,3, and 4 form the feasible region and constraint 1 is redundant. Change constraint 1 to less-than-or-equal-to and the modified problem is infeasible. 2-44
n Introduction to Linear Programming Original Problem: Modified Problem: 58. It makes no sense to add this constraint. The objective of the problem is to minimize the number of products needed so that everyone s top three choices are included. There are only two possible outcomes relative to the boss new constraint. First, suppose the minimum number of products is <= 15, then there was no need for the new constraint. Second, suppose the minimum number is > 15. Then the new constraint makes the problem infeasible. 2-45