6619 CH08 UG 7/8/02 4:05 PM Page 109 8 C H A P T E R Linear Programming Modeling Applications: With Computer Analyses in Excel and QM for Windows TEACHING SUGGESTIONS Teaching Suggestion 8.1: Importance of Formulating Large LP Problems. Since computers are used to solve virtually all business LP problems, the most important thing a student can do is to get experience in formulating a wide variety of problems. This chapter provides such a variety. Teaching Suggestion 8.2: Note on Production Scheduling Problems. The Greenberg Motor example in this chapter is the largest problem in the book in terms of constraints, so it provides a good practice environment. An interesting feature to point out is that LP constraints are capable of tying one production period to the next. Teaching Suggestion 8.3: Solving Assignment Problems by LP. The example of the law firm of Ivan and Ivan in this chapter can clearly be solved more quickly using QM for Windows’ assignment program than by the LP program. Students should be asked why anyone would choose to use the LP approach. There are two answers: (1) many commercial LP programs do not contain assignment algorithms (which are more popular in academic software such as QM for Windows); and (2) the LP program can provide more sensitivity analysis and economic interpretation than is available in the assignment module. The assignment problem is treated in Chapter 10. Teaching Suggestion 8.4: Labor Planning Problem—Arlington Bank. This example is a good practice tool and lead-in for the Chase Manhattan Bank case at the end of the chapter. Without this example, the case would probably overpower most students. Teaching Suggestion 8.5: Ingredient Blending Applications. Three points can be made about the two blending examples in this chapter. First, both the diet and fuel blending problems presented here are tiny compared to huge real-world blending problems. But they do provide some sense of the issues to be faced. Second, diet problems that are missing the constraints that force variety into the diet can be terribly embarrassing. It has been said that a hospital in New Orleans ended up with an LP solution to feed each patient only castor oil for dinner because analysts neglected to add constraints forcing a well-rounded diet. ALTERNATIVE EXAMPLES Alternative Example 8.1: Natural Furniture Company manufactures three outdoor products, chairs, benches, and tables. Each product must pass through the following departments before it is shipped: sawing, sanding, assembly, and painting. The time requirements (in hours) are summarized in the tables below. The production time available in each department each week and the minimum weekly production requirement to fulfill contracts are as follows: Capacity (In Hours) Department Sawing Sanding Assembly Painting 450 400 625 550 Product Sawing Chairs Benches Tables 1.5 1.5 2.0 Product Minimum Production Level Chairs Benches Tables 100 50 50 Hours Required Sanding Assembly 1.0 1.5 2.0 2.0 2.0 2.5 Painting Unit Profit 1.5 2.0 2.0 $15 $10 $20 The production manager has the responsibility of specifying production levels for each product for the coming week. Let X1 Number of chairs produced X2 Number of benches produced X3 Number of tables produced The objective function is Maximize profit 15X1 10X2 20X3 Constraints 1.5X1 1.5X2 2.0X3 450 hours of sawing available 1.0X1 1.5X2 2.0X3 400 hours of sanding available 2.0X1 2.0X2 2.5X3 625 hours of assembly available 1.5X1 2.0X2 2.0X3 550 hours of painting available X1 2.0X2 2.0X3 100 chairs X2 2.0X3 50 benches X3 50 tables X1, X2, X3 0 Alternative Example 8.2: A phosphate manufacturer produces three grades, A, B, and C, which cost the firm $40, $50, and $60 per kilogram, respectively. The products require the labor and materials per batch that are shown on the following page. 109 6619 CH08 UG 7/8/02 4:05 PM 110 Page 110 CHAPTER 8 Labor hours Raw material #1 Raw material #2 LINEAR PROGRAMMING MODELING APPLICATIONS Grade A Grade B Grade C Available Resources 4 200 600 4 300 400 5 300 500 80 hr 6,000 kg 5,000 kg Problem 8-2 solved by computer: $50,000 invested in Los Angeles municipal bonds (X1) $0 invested in Thompson Electronics (X2) $0 invested in United Aerospace (X3) $175,000 invested in Palmer Drugs (X4) What mix of products would yield minimum cost? Objective function Minimize cost 40A 50B 60C Constraints $25,000 invested in Happy Days (X5) This produces an annual return on investment of $20,300. Minimize staff size X1 X2 X3 X4 X5 X6 8-3. where 4A 4B 5C 80 200A 300B 300C 6,000 600A 400B 500C 5,000 Labor: Raw material #1 Raw material #2 Xi number of workers reporting for start of work at period i (with i 1, 2, 3, 4, 5, or 6) X1 X2 12 SOLUTIONS TO PROBLEMS X2 X3 16 8-1. Since the decision centers about the production of the two different cabinet models, we let X4 X5 11 X3 X4 9 X5 X6 4 X1 number of French Provincial cabinets produced each day X2 number of Danish Modern cabinets produced each day X1 X6 3 All variables 0 The computer solution is to hire 30 workers: Objective: maximize revenue $28X1 $25X2 16 begin at 7 A.M. subject to 9 begin at 3 P.M. 3X1 2X2 360 hours (carpentry department) 112X1 1X2 200 hours (painting department) X1 X2 125 hours (finishing department) 3 4 X1 3 4 2 begin at 7 P.M. 3 begin at 11 P.M. An alternative optimum is 60 units (contract requirement) 3 begin at 3 A.M. X2 60 units (contract requirement) 9 begin at 7 A.M. X1, X2 0 7 begin at 11 A.M. Problem 8-1 solved by computer: 2 begin at 3 P.M. Produce 60 French Provincial cabinets (X1) per day Produce 90 Danish Modern cabinets (X2) per day Profit $3,930 (in four iterations) 9 begin at 7 P.M. 8-2. Let X1 dollars invested in Los Angeles municipal bonds X2 dollars invested in Thompson Electronics 0 begin at 11 P.M. 8-4. Let X1 number of pounds of oat product per horse each day X2 number of pounds of enriched grain per horse each day X3 dollars invested in United Aerospace X4 dollars invested in Palmer Drugs X3 number of pounds of mineral product per horse each day X5 dollars invested in Happy Days Nursing Homes Maximize return 0.053X1 0.068X2 0.049X3 0.084X4 0.118X5 subject to X1 X2 X3 X4 X5 $250,000 (funds to be invested) X1 .2 (X1 X2 X3 X4 X5) (municipal bonds) or .8X1 .2X2 .2X3 .2X4 .2X5 0 X2 X3 X4 .4 (X1 X2 X3 X4 X5) (combination of electronics, aerospace, and drugs) (X5 0.5X1) rewritten as 0.5X1 X5 0 (nursing home as percent of bonds) X1, X2, X3, X4, X5 0 Minimize cost 0.09X1 0.14X2 0.17X3 subject to 2X1 3X2 1X3 6 (ingredient A) 1 2 X1 1X2 12X3 2 (ingredient B) 3X1 5X2 6X3 9 (ingredient C) 1X1 112 X2 2X3 8 (ingredient D) 1 2 X1 21 X2 112 X3 5 (ingredient E) X1 X2 X3 6 (maximum feed/day) All variables 0 Solution: X1 113 X2 0 X3 313 cost 0.687 6619 CH08 UG 7/8/02 4:05 PM Page 111 CHAPTER 8 8-5. LINEAR PROGRAMMING MODELING APPLICATIONS Let Xij 1 if pitcher i is scheduled to go against opponent j, 0 otherwise X14 X15 X23 X24 X25 X33 X34 X35 X43 X44 X45 460 i 1, 2, 3, 4 stands for Jones, Baker, Parker, and Wilson, respectively, and X15 X24 X25 X33 X34 X35 X43 X44 X45 X53 X54 X55 470 j 1, 2, 3, 4 stands for Des Moines, Davenport, Omaha, and Peoria, respectively. X25 X34 X35 X43 X44 X45 X53 X54 X55 X63 X64 X65 440 where X15 X25 X35 X45 X55 X65 0.50(X13 X14 X15 X23 X24 X25 X33 X34 X35 X43 X44 X45 X53 X54 X55 X63 X64 X65) Objective: maximize overall probability of winning sum of probability of winning each game 0.6X11 0.8X12 0.5X13 0.4X14 0.7X21 0.4X22 0.8X23 0.3X24 0.9X31 0.8X32 0.7X33 0.8X34 0.5X41 0.3X42 0.4X43 0.2X44 All variables 0 Solving this on the computer results in the following solution: subject to X11 X12 X13 X14 1 (“Dead-Arm” Jones) X21 X22 X23 X24 1 (“Spitball” Baker) X31 X32 X33 X34 1 (“Ace” Parker) X41 X42 X43 X44 1 (“Gutter” Wilson) X11 X21 X31 X41 1 (Des Moines) X12 X22 X32 X42 1 (Davenport) X15 30 5-month leases in March X25 100 5-month leases in April X35 170 5-month leases in May X45 160 5-month leases in June X55 10 5-month leases in July All other variables equal 0. Total cost $677,100. X13 X23 X33 X43 1 (Omaha) As a result of this, there are 440 cars remaining at the end of August. X14 X24 X34 X44 1 (Peoria) 8-8. Solution: X12 1, X23 1, X34 1, X41 1, Total P 2.9 8-6. Let: X1 number of newspaper ads placed The linear program has the same constraints as in problem 8-7. The objective function changes and is now: Minimize cost Minimize cost $925X1 $2,000X2 subject to 0.04X1 0.05X2 0.40 (city exposure) 0.03X1 0.03X2 0.60 (exposure in northwest suburbs) 1260(X13 X23 X33 X43 X53 X63) 1600(X14 X24 X34 X44 X54 X64) X2 number of TV spots purchased 1850(X15 X25 X35 X45 X55 X65) Solving this on the computer results in the following solution: X1, X2 0 X15 30 5-month leases in March X25 100 5-month leases in April Note that the problem is not limited to unduplicated exposure (e.g., one person seeing the Sunday newspaper three weeks in a row counts for three exposures). X34 65 4-month leases in May X35 105 5-month leases in May X43 160 3-month leases in June Problem 8-6 solved by computer: X53 10 3-month leases in July Buy 20 Sunday newspaper ads (X1) Buy 0 TV ads (X2) This has a cost of $18,500. Perhaps the paint store should consider a blend of TV and newspaper, not just the latter. 8-7. 111 Let Xij number of new leases in month i for j-months, i 1, . . . , 6; j 3, 4, 5 Minimize cost subject to: All other variables equal 0. Total cost $752,950. 8-9. Let Xij number of students bused from sector i to school j Objective: minimize total travel miles 5XAB 8XAC 6XAE 0XBB 4XBC 12XBE 1260X13 1260X23 1260X33 1260X43 840X53 420X63 1600X14 1600X24 1600X34 1200X44 800X54 400X64 1850X15 1850X25 1480X35 1110X45 740X55 370X65 X13 X14 X15 420 390 4XCB 0XCC 7XCE 7XDB 2XDC 5XDE 12XEB 7XEC 0XEE subject to XAB XAC XAE 700 (number students in sector A) X13 X14 X15 X23 X24 X25 400 270 XBB XBC XBE 500 (number students in sector B) X13 X14 X15 X23 X24 X25 X33 X34 X35 430 130 XDB XDC XDE 800 (number students in sector D) XCB XCC XCE 100 (number students in sector C) 6619 CH08 UG 7/8/02 4:05 PM 112 Page 112 CHAPTER 8 LINEAR PROGRAMMING MODELING APPLICATIONS XEB XEC XEE 400 (number students in sector E) 16X1 81X2 74X3 83X4 7X5 XAB XBB XCB XDB XEB 900 (school B capacity) 14X6 8X7 26 XAC XBC XCC XDC XEC 900 (school C capacity) XAE XBE XCE XDE XEE 900 (school E capacity) All variables 0 22X1 28X5 19X6 63X7 Problem 8-11 solved by computer: Solution: XAB 400 The meal plan for the evening is XAE 300 No milk (X1 0) XBB 500 0.499 pound of ground meat (X2) XCC 100 0.173 pound of chicken (X3) XDC 800 No fish (X4 0) XEE 400 No beans (X5 0) Distance 5,400 “student miles” 8-10. Maximize number of rolls of Supertrex sold 20X1 6.8X2 12X3 65,000X4 X1 dollars spent on advertising where X2 dollars spent on store displays X3 dollars in inventory subject to X1 X2 X3 $17,000 (budgeted) $3,000 (advertising constraint) X2 0.105 pound of spinach (X6) 0.762 pound of white potatoes (X7) Each meal has a cost of $1.75 (in six iterations). The meal is fairly well-balanced (two meats, a green vegetable, and a potato). The weight of each item is realistic. This problem is very sensitive to changing food prices. Sensitivity analysis when prices change: X4 percent markup X1 50 All Xi 0 0.05X3 (or X2 0.05X3 0) (ratio of displays to inventory) X4 0.20 (markup ranges) X4 0.45 X1, X2, X3, X4 0 Problem 8-10 solved by computer: Spend $17,000 on advertising (X1). Spend nothing on in-store displays or on-hand inventory (X2 and X3). Take a 20% markup. The store will sell 327,000 rolls of Supertrex (in six iterations). This solution implies that no on-hand inventory or displays are needed to sell the product, probably due to an oversight on Mr. Kruger’s part. Perhaps a constraint indicating that X3 $3,000 of inventory should be held might be added. 8-11. Minimize total cost $0.60X1 2.35X2 1.15X3 2.25X4 0.58X5 1.17X6 0.33X7 subject to 295X1 1,216X2 394X3 358X4 128X5 118X6 279X7 1,500 295X1 1,216X2 394X3 358X4 128X5 118X6 279X7 900 .2X1 121.2X2 .4.3X3 3.2X4 3.2X5 14.1X6 2.2X7 4 16X1 1,296X2 .4.9X3 0.5X4 0.8X5 1.4X6 0.5X7 50 Milk increases 10 cents/lb: no change in price or diet Milk decreases 10 cents/lb: no change in price or diet Milk decreases 30 cents/lb (to 30 cents): potatoes drop out and milk enters, price $1.42/meal Ground meat increases from $2.35 to $2.75: price $1.93 and spinach leaves the optimal solution Ground meat increases to $5.25/lb: price $2.07 and meat leaves; milk, chicken, and potatoes in solution Fish decreases from $2.25 to $2.00/lb: no change Chicken increases to $3.00/lb: price $1.91 and meat, fish, spinach, and potatoes in solution If meat and fish are omitted from the problem, the solution is chicken 0.774 lb milk 1.891 lb potatoes 0.133 lb If chicken and meat are omitted; fish 0.679 lb spinach 0.0988 lb milk 2.188 lb 8-12. a. Let X1 no. of units of internal modems produced per week X2 no. of units of external modems produced per week X3 no. of units of circuit boards produced per week X4 no. of units of floppy disk drives produced per week X5 no. of units of hard drives produced per week X6 no. of units of memory boards produced per week 6619 CH08 UG 7/8/02 4:05 PM Page 113 CHAPTER 8 Objective function analysis: First find the time used on each test device: subject to 130X1 90Y1 40,000 (Aug. need, hours) hours on test device 1 7 X 3 X2 12 X3 6 X4 18 X5 17 X6 = 1 60 hours on test device 2 2 X 5 X2 3 X3 2 X4 15 X5 17 X6 1 60 hours on test device 3 130X2 90Y2 45,000 (Sept. need) 130X3 90Y3 35,000 (Oct. need) 130X4 90Y4 50,000 (Nov. need) 130X5 90Y5 45,000 (Dec. need) X1 350 (starting staff on Aug. 1) X2 X1 Y1 0.05X1 (staff on Sept. 1) X3 X2 Y2 0.05X2 (staff on Oct. 1) 5 X1 1X2 3 X3 2 X4 9 X5 2 X6 60 Thus, the objective function is maximize profit revenue material cost test cost X4 X3 Y3 0.05X3 (staff on Nov. 1) X5 X4 Y4 0.05X4 (staff on Dec. 1) All Xi, Yi 0 200X1 120X2 180X3 130X4 430X5 b. The computer-generated results are: 260X6 35X1 25X2 40X3 45X4 170X5 60X6 7 X 3 X2 12 X3 6 X4 18 X5 17 X6 15 1 60 2 X1 5 X2 3 X3 2 X4 15 X5 17 X6 12 60 5 X1 1X2 3 X3 2 X4 9 X5 2 X6 18 60 This can be rewritten as maximize profit $161.35X1 92.95X2 135.50X3 82.50X4 249.80X5 191.75X6 subject to 7 X1 3 X2 12 X3 6 X4 18 X5 17 X6 120 hours 60 2 X1 5 X2 3 X3 2 X4 15 X5 17 X6 120 hours 60 5 X1 1X2 3 X3 2 X4 9 X5 2 X6 100 hours 60 All variables 0 b. The solution is X1 496.55 internal modems X2 1,241.38 external modems X3 through X6 0 profit $195,504.80 c. The shadow prices, as explained in Chapters 7 and 9, for additional time on the three test devices are $21.41, $5.75, and $0, respectively, per minute. 8-13. a. Let Xi no. of trained technicians available at start of month i Yi no. of trainees beginning in month i Minimize total salaries paid $2,000X1 2,000X2 2,000X3 2,000X4 2,000X5 900Y1 900Y2 900Y3 900Y4 900Y5 113 LINEAR PROGRAMMING MODELING APPLICATIONS Month Trained Technicians Available Trainees Beginning 350 346.2 328.8 384.6 365.4 13.7 (actually 14) 0 72.2 (actually 72) 0 0 Aug. Sept. Oct. Nov. Dec. Total salaries paid over the five-month period $3,627,279. 8-14. a. Let Xij acres of crop i planted on parcel j i 1 for wheat, 2 for alfalfa, 3 for barley where j 1 to 5 for SE, N, NW, W, and SW parcels Irrigation limits: 1.6X11 2.9X21 3.5X31 3,200 acre-feet in SE 1.6X12 2.9X22 3.5X32 3,400 acre-feet in N 1.6X13 2.9X23 3.5X33 800 acre-feet in NW 1.6X14 2.9X24 3.5X34 500 acre-feet in W 1.6X15 2.9X25 3.5X35 600 acre-feet in SW 5 ∑ 5 1.6 X1 j j 1 ∑ j 1 5 2.9 X2, j ∑ 3.5X 3, j 7, 400 j 1 water acre-feet total Sales limits: X11 X12 X13 X14 X15 2,200 wheat in acres ( 110,000 bushels) X21 X22 X23 X24 X25 1,200 alfalfa in acres ( 1,800 tons) X31 X32 X33 X34 X35 1,000 barley in acres ( 2,200 tons) Acreage availability: X11 X21 X31 2,000 acres in SE parcel X12 X22 X32 2,300 acres in N parcel X13 X23 X33 600 acres in NW parcel X14 X24 X34 1,100 acres in W parcel X15 X25 X35 500 acres in SW parcel 6619 CH08 UG 7/8/02 4:05 PM 114 Page 114 CHAPTER 8 LINEAR PROGRAMMING MODELING APPLICATIONS Objective function: The solution is infeasible. 5 maximize profit 8-16. This problem refers to Problem 8-15’s infeasibility. Some investigative work is needed to track down the issues. From a final simplex tableau, we find that constraints 5 and 11 still have artificial variables in the final solution. The two issues are: ∑ $2(50 bushels) X 1, j j 1 5 ∑ 5 $40(1.5 tons) X2, j j 1 ∑ ($50)(2.2 tons) X 3, j j 1 b. The solution is to plant X12 1,250 acres of wheat in N parcel X13 500 acres of wheat in NW parcel X14 31212 acres of wheat in W parcel X15 13712 acres of wheat in SW parcel X25 131 acres of alfalfa in SW parcel X31 600 acres of barley in SE parcel X32 400 acres of barley in N parcel Profit will be $337,862.10. Multiple optimal solutions exist. c. Yes, need only 500 more water-feet. 8-15. Amalgamated’s blending problem will have eight variables and 11 constraints. The eight variables correspond to the eight materials available (three alloys, two irons, three carbides) that can be selected for the blend. Six of the constraints deal with maximum and minimum quality limits, one deals with the 2,000 pound total weight restriction, and four deal with the weight availability limits for alloy 2 (300 lb), carbide 1 (50 lb), carbide 2 (200 lb), and carbide 3 (100 lb). Let X1 through X8 represent pounds of alloy 1 through pounds of carbide 3 to be used in the blend. Minimize cost 0.12X1 0.13X2 0.15X3 0.09X4 0.07X5 0.10X6 0.12X7 0.09X8 subject to manganese quality: 0.70X1 0.55X2 0.12X3 0.01X4 0.05X5 42 (2.1% of 2,000) 0.70X1 0.55X2 0.12X3 0.01X4 0.05X5 46 (2.3% of 2,000) silicon quality: 0.15X1 0.30X2 0.26X3 0.10X4 0.025X5 0.24X6 0.25X7 0.23X8 86 (4.3% of 2,000) 0.15X1 0.30X2 0.26X3 0.10X4 0.025X5 0.24X6 0.25X7 0.23X8 92 (4.6% of 2,000) carbon quality: 0.03X1 0.01X2 0.03X4 0.18X6 0.20X7 0.25X8 101 (5.05% of 2,000) 0.03X1 0.01X2 0.03X4 0.18X6 0.20X7 0.25X8 107 (5.35% of 2,000) Availability by weight: X2 300 X6 50 X7 200 X8 100 One-ton weight: X1 X2 X3 X4 X5 X6 X7 X8 2,000 1. 2. Requiring at least 5.05% carbon is not possible. Producing 1 ton from the materials is not possible. If constraints 5 and 11 are relaxed (or removed), one solution is X2 84 lb (alloy 2), X6 50 lb (carbide 1), X7 104 lb (carbide 2), and X8 100 lb (carbide 3). Cost $37.31. Each student may take a different approach and other recommendations may result. 8-17. X1 number of medical patients X2 number of surgical patients Maximize revenue $2,280X1 $1,515X2 subject to 8X1 2.5X2 32,850 (patient-days available 365 days 90 new beds) 3.1X1 2.6X2 15,000 (lab tests) 1X1 2.2X2 7,000 (x-rays) X2 2,800 (operations/surgeries) X1, X2 0 Problem 8-17 solved by computer: X1 2,791 medical patients X2 2,105 surgical patients revenue $9,551,659 per year To convert X1 and X2 to number of medical versus surgical beds, find the total number of hospital days for each type of patient: medical (2,791 patients)(8 days/patient) 22,328 days surgical (2,105 patients)(5 days/patient) 10,525 days total 32,853 days This represents 68% medical days and 32% surgical days, which yields 61 medical beds and 29 surgical beds. (Note that an alternative approach would be to formulate with X1, X2 as number of beds.) See the printout on the next page for the solution and sensitivity analysis. 8-18. This problem, suggested by Professor C. Vertullo, is an excellent exercise in report writing. Here is a chance for students to present management science results in a management format. Basically, the following issues need to be addressed in any report: (a) As seen in Problem 8-17, there should be 61 medical and 29 surgical beds, yielding $9,551,659 per year. (b) Referring to the QM for Windows printout, there are no empty beds. (c) There are 876 lab tests of unused capacity. (d) The x-ray is used to its maximum and has a $65.45 shadow price. (e) The operating room still has 695 operations available. 6619 CH08 UG 7/8/02 4:05 PM Page 115 CHAPTER 8 Printout for Problems 8-17 and 8-18 Sensitivity Analysis Printout for Problems 8-17 and 8-18 LINEAR PROGRAMMING MODELING APPLICATIONS 115 6619 CH08 UG 7/8/02 4:05 PM 116 8-19. Page 116 CHAPTER 8 a. LINEAR PROGRAMMING MODELING APPLICATIONS Let A1 tons of ore from mine A to plant 1 A2 tons of ore from mine A to plant 2 B1 tons of ore from mine B to plant 1 B2 tons of ore from mine B to plant 2 X1 tons shipped to Builder’s Home from plant 1 b. The solution is the same as problem 8-19 except the value of the objective function is $34,300. 8-21. Minimize time 12XA1 11XA2 8XA3 9XA4 6XA5 6XA6 6XG1 12XG2 7XG3 7XG4 5XG5 8XG6 8XS1 9XS2 6XS3 6XS4 7XS5 9XS6 subject to XA1 XA2 XA3 XA4 XA5 XA6 200 XG1 XG2 XG3 XG4 XG5 XG6 225 XS1 XS2 XS3 XS4 XS5 XS6 275 XA1 XG1 XS1 80 XA2 XG2 XS2 120 XA3 XG3 XS3 150 XA4 XG4 XS4 210 XA5 XG5 XS5 60 XA6 XG6 XS6 80 All variables 0 X2 tons shipped to Builder’s Home from plant 2 Y1 tons shipped to Homeowners’ Headquarters from plant 1 Y2 tons shipped to Homeowners’ Headquarters from plant 2 Z1 tons shipped to Hardware City from plant 1 Z2 tons shipped to Hardware City from plant 2 Minimize cost 6A1 8A2 7B1 10B2 13X1 19X2 17Y1 22Y2 20Z1 21Z2 subject to A1 A2 320 supply at A B1 B2 450 supply at B A1 B1 500 capacity at plant 1 A2 B2 500 capacity at plant 2 X1 X2 200 demand at Builder’s Home Y1 Y2 240 demand at Homeowners’ Headquarters Z1 Z2 330 demand at Hardware City A1 B1 X1 Y1 Z1 units shipped into plant 1 must equal units shipped out of plant 1 A2 B2 X2 Y2 Z2 units shipped into plant 2 must equal units shipped out of plant 2 Solution: Source (Station) Destination (Wing) Number of Trays 5A 5A 5A 3G 3G 3G 1S 1S 5 6 3 1 3 4 4 2 60 80 60 80 90 55 155 120 Optimal cost 4,825 minutes. Multiple optimal solutions exist. 8-22. Let Xi proportion of investment invested in stock i for i 1, 2, . . . , 5 All variables 0 Minimize beta 1.2X1 0.85X2 0.55X3 1.40X4 1.25X5 b. Solving this on the computer, we find the following solution: subject to A1 50 tons of ore from mine A to plant 1 A2 270 tons of ore from mine A to plant 2 B1 450 tons of ore from mine B to plant 1 X1 X2 X3 X4 X5 1 0.11X1 0.09X2 0.065X3 0.15X4 0.13X5 0.11 return should be at least 11% X1 200 tons shipped to Builder’s Home from plant 1 X1 0.35 Y1 240 tons shipped to Homeowners’ Headquarters from plant 1 X2 0.35 Z1 60 tons shipped to Hardware City from plant 1 Z2 270 X4 0.35 tons shipped to Hardware City from plant 2 X5 0.35 All other variables equal 0. Minimum total cost $19,160 8-20. a. The formulation is the same as the formulation in problem 8-19 except for a change in the objective function. We add the processing cost in the objective function, and the new objective function is: Minimize cost 28A1 30A2 25B1 28B2 13X1 19X2 17Y1 22Y2 20Z1 21Z2 All the constraints are the same as in the previous problem. total of the proportions must add to 1 no more than 35% in any single stock X3 0.35 Xi 0 for i 1, 2, . . . , 5 b. Solving this on the computer, we have X1 0 X2 0.10625 X3 0.35 X4 0.35 X5 0.19375 Minimum beta 1.015 6619 CH08 UG 7/8/02 4:05 PM Page 117 CHAPTER 8 LINEAR PROGRAMMING MODELING APPLICATIONS Return 0.11(0) 0.09(0.10625) 0.065(0.35) 0.15(0.35) 0.13(0.19375) 0.11 8-23. Let A 1,000 gallons of fuel to purchase in Atlanta L 1,000 gallons of fuel to purchase in Los Angeles N 0 (1,000 gallons of fuel to purchase in New Orleans) FN 11 (1,000 gallons of fuel remaining when plane lands in New Orleans) Total cost $31.2211 ( 1,000) H 1,000 gallons of fuel to purchase in Houston SOLUTIONS TO INTERNET HOMEWORK PROBLEMS N 1,000 gallons of fuel to purchase in New Orleans 8-24. To formulate this problem, we first add an activity G to represent the end of the project: FA fuel remaining when plane lands in Atlanta FL fuel remaining when plane lands in Los Angeles FH fuel remaining when plane lands in Houston Objective minimize XG subject to: XA 2 XB 3 FN fuel remaining when plane lands in New Orleans XC 1 Minimize cost 1.15A 1.25L 1.10H 1.18N XD XA 4 subject to XF XB 1 A FA 24 minimum amount of fuel board when leaving Atlanta A FA 36 maximum amount of fuel board when leaving Atlanta L FL 15 minimum amount of fuel board when leaving Los Angeles L FL 23 maximum amount of fuel board when leaving Los Angeles H FH 9 minimum amount of fuel board when leaving Houston H FH 17 maximum amount of fuel board when leaving Houston N FN 11 minimum amount of fuel board when leaving New Orleans N FN 20 XE XC 5 XE XD 5 XG XE 0 XG XF 0 All variables 0 Solution with QM for Windows: XA 2 XB 10 XC 6 XD 6 XE 11 XF 11 XG 11 maximum amount of fuel board when leaving New Orleans FL A FA (12 0.05(A FA 24)) This says that the fuel on board when the plane lands in Los Angeles will equal the amount on board at take-off minus the fuel consumed on that light. The fuel consumed is 12 (thousand gallons) plus 5% of the excess above 24 (thousand gallons). This simplifies to: 0.95A 0.95 FA FL 10.8 Similarly, Z 11 8-25. Let X1 number of Chaunceys mixed X2 number of Sweet Italians mixed X3 number of bourbon on the rocks mixed X4 number of Russian martinis mixed Maximize total drinks X1 X2 X3 X4 subject to FH L FL (7 0.05(L FL 15)) becomes 0.95L 0.95FL FH 6.25 1X1 4X3 52 oz (bourbon limit) FN H FH (3 0.05(H FH 9)) becomes 0.95H 0.95FH FN 2.55 1X1 223X4 64 oz (vodka limit) FL N FN (5 0.05(N FN 11)) becomes 0.95N 0.95FN FL 4.45 All variables 0 b. 117 The optimal solution is A 21.1579 (1,000 gallons of fuel to purchase in Atlanta) FA 6 (1,000 gallons of fuel remaining when plane lands in Atlanta) L 0 (1,000 gallons of fuel to purchase in Los Angeles) FL 15 (1,000 gallons of fuel remaining when plane lands in Los Angeles) H 6.2632 (1,000 gallons of fuel to purchase in Houston) FH 8 (1,000 gallons of fuel remaining when plane lands in Houston) 1X1 1X2 38 oz (brandy limit) 1X2 113X4 24 oz (dry vermouth limit) 1X1 2X2 36 oz (sweet vermouth limit) All variables 0 Because a Chauncey (X1) is 14 sweet vermouth, it requires 1 oz of that resource (each drink totals 4 oz). Problem 8-25 solved by computer: Mix 25.99 (or 26) Chaunceys (X1) Mix 5.00 (or 5) Sweet Italians (X2) Mix 6.50 (or 612) bourbon on the rocks (X3) Mix 14.25 (or 1414) Russian martinis (X4) This is a total of 51.75 drinks (in five iterations). 6619 CH08 UG 7/8/02 118 4:05 PM Page 118 CHAPTER 8 LINEAR PROGRAMMING MODELING APPLICATIONS 8-26. Minimize 6X11 8X12 10X13 7X21 11X22 11X23 4X31 5X32 12X33 subject to X11 X12 X13 150 X21 X22 X23 175 X31 X32 X33 275 All variables 0 8-27. Let Si 1 if Smith is assigned to Job i, 0 otherwise, for i 1, 2, 3, 4 Ji 1 if Jones is assigned to Job i, 0 otherwise, for i 1, 2, 3, 4 Di 1 if Davis is assigned to Job i, 0 otherwise, for i 1, 2, 3, 4 Ni 1 if Nguyen is assigned to Job i, 0 otherwise, for i 1, 2, 3, 4 Minimize days 4S1 5J1 4D1 5N1 10S2 14J2 13D2 11N2 8S3 8J3 9D3 7N3 9S4 10J4 12D4 11N4 subject to S1 J1 D1 N1 1 S2 J2 D2 N2 1 S3 J3 D3 N3 1 S4 J4 D4 N4 1 S1 S2 S3 S4 1 J1 J2 J3 J4 1 D1 D2 D3 D4 1 N1 N2 N3 N4 1 All variables 0 Solving this with QM for Windows, we have S2 1, J4 1, D1 1, and N3 1. So, Smith does Job 2, Jones does Job 4, Davis does Job 1, and Nguyen does Job 3. The total time is 31 days. 8-28. 2, 3. Let Xi number of BR54 produced in month i, for i 1, IXi number of BR54 units in inventory at end of month i, for i 0, 1, 2, 3. IYi number of BR49 units in inventory at end of month i, for i 0, 1, 2, 3. Minimize cost 80(X1 X2 X3) 95(Y1 Y2 Y3) 0.8(IX1 IX2 IX3) 0.95(IY1 IY2 IY3) Subject to: IX0 50 initial inventory of BR54 IY0 50 initial inventory of BR49 maximum production level in August maximum production level in September X3 Y3 1,100 maximum production level in October Y1 IY0 450 IY1 BR49 requirements for August Y2 IY1 420 IY2 BR49 requirements for September Y3 IY2 480 IY3 All variables 0 BR49 requirements for October A computer solution to this results in IX0 50, IX1 190, IX2 130, IX3 100, IY0 50, IY3 150, X1 460, X2 680, X3 470, Y1 400, Y2 420, Y3 630. All other variables 0. The total cost $267,028.50. SOLUTION TO RED BRAND CANNERS CASE 1. The main issue in this case is how to allocate 3 million pounds of tomatoes. The overall objective is to maximize total sales less variable costs. These costs include production and selling expenses. Twenty percent of the crop was grade A and the rest was grade B. In setting up the constraints, the amount of grade A tomatoes cannot exceed 20% of 3 million pounds. Thus not more than 600,000 pounds of grade A tomatoes can be used. Similarly, not more than 2,400,000 pounds of grade B tomatoes can be used. Furthermore, the demand for 50,000 cases of tomato juice and 80,000 cases of tomato paste should be met. The demand for whole tomatoes is not a constraint in this problem. Finally, minimum quality requirements should be met. This includes an average of 8 points per pound for whole tomatoes and 6 points per pound for tomato juice. There is no constraint for tomato paste. Another issue is whether or not to buy 80,000 additional pounds of grade A tomatoes. This would increase the amount of available grade A tomatoes from 600,000 pounds to 680,000 pounds. To answer this question, a new formulation can be made using the new 680,000-pound constraint and a price of 8.5 cents per pound for the 80,000 additional pounds of grade A tomatoes in the objective function. A faster way to resolve this issue is to use postoptimality analysis, or shadow prices. Using this approach, you compare the value of the 80,000 additional tomatoes with the cost, which is 8.5 cents per pound. 2. Yi number of BR49 produced in month i, for i 1, 2, 3. X1 Y1 1,100 X2 Y2 1,100 X3 IX2 500 IX3 BR54 requirements for October X13 X23 X33 300 Cost $4,525. ending inventory of BR49 X2 IX1 740 IX2 BR54 requirements for September X12 X22 X32 100 X11 25, X13 125, X23 175, X31 175, X32 100 ending inventory of BR54 IY3 150 X1 IX0 320 IX1 BR54 requirements for August X11 X21 X31 200 The solution is: IX3 100 The problem can be formulated using LP as follows: X1 pounds of whole A tomatoes X2 pounds of whole B tomatoes X3 pounds of juice A tomatoes X4 pounds of juice B tomatoes X5 pounds of paste A tomatoes X6 pounds of paste B tomatoes Maximize: 0.0822X1 0.0822X2 0.066X3 0.066X4 0.074X5 0.074X6 6619 CH08 UG 7/8/02 4:05 PM Page 119 CHAPTER 8 LINEAR PROGRAMMING MODELING APPLICATIONS subject to 1X1 1X2 14,400,000 1X3 1X4 1,000,000 1X5 1X6 2,000,000 1X1 1X3 1X2 1X5 1X4 1X1 3X2 600,000 1X6 2,400,000 0 3X3 1X4 0 All variables 0 The first constraint refers to the 14 million pounds of whole tomatoes—800,000 cases at 18 pounds per case—that constitutes maximum demand. Similarly, the maximum demand for tomato juice is 50,000 cases at 20 pounds per case or 1 million pounds, and the maximum demand for tomato paste is 80,000 cases at 25 pounds per case or 2 million pounds, and these are constraints 2 and 3. Constraints 4 and 5 reflect the availability of grade A and grade B tomatoes, respectively, and the last two constraints are the quality constraints. The requirements that canned tomatoes must average at least 8 points means that at least three-fourths of the tomatoes must be grade A: X1 0.75(X1 X2) X1 3X2 0 Similarly, the requirements that tomato juice must average at least 6 points means that at least one-fourth of the tomato juice must be grade A, and that is the last constraint. The coefficients in the objective function are the unit profits. A case of whole tomatoes (grade A and grade B) sells for $4. The variable cost (less the tomatoes) is $2.52. Since the tomatoes are already on hand (and no salvage appears to be possible), they represent a sunk cost and are not part of the decision process. Since there are 18 pounds per case, the unit profit is (4.00 2.52)/18 0.0822. Similar analyses hold for the other terms in the objective function. The solution of the linear programming problem is X1 525,000 X3 75,000 X5 0 X2 175,000 X4 225,000 X6 2,000,000 The maximum profit is $225,340. All of the grade A tomatoes are used. The shadow price for the slack variable in constraint 4 is 0.0903. Each additional pound of grade A tomatoes costing 8.5 cents will increase profits by 0.093 0.0850 0.0053. A sensitivity analysis indicates that up to an additional 600,000 pounds of grade A tomatoes could be purchased without affecting the solution basis. SOLUTION TO CHASE MANHATTAN BANK CASE This very advanced and challenging scheduling problem can be solved most expeditiously using linear programming, preferably integer programming. Let F denote the number of full-time employees. Some number, F1, of them will work 1 hour of overtime between 5 P.M. and 6 P.M. each day and some number, F2, of the full-time employees will work overtime between 6 P.M. and 7 P.M. There will be seven sets of part-time employees; Pj will be the number of part-time employees who begin their workday at hour j, j 1, 2, . . . , 7, with P1 being the number of workers beginning at 9 A.M., P2 at 10 A.M., . . . , P7 at 3 P.M. Note that because part-time 119 employees must work a minimum of 4 hours, none can start after 3 since the entire operation ends at 7 P.M. Similarly, some number of part-time employees, Qj, leave at the end of hour j, j 4, 5, . . . , 9. The workforce requirements for the first two hours, 9 A.M. and 10 A.M., are: P.M. F P1 14 F P1 P2 25 At 11 A.M. half of the full-time employees go to lunch; the remaining half go at noon. For those hours: 0.5F P1 P2 P3 26 0.5F P1 P2 P3 P4 38 Starting at 1 P.M., some of the part-time employees begin to leave. For the remainder of the straight-time day: F P1 P2 P3 P4 P5 Q4 55 F P1 P2 P3 P4 F P1 P2 P5 P6 Q4 Q5 60 F P1 P2 P3 P4 P5 F P1 P6 P7 Q4 Q5 Q6 51 F P1 P2 P3 P4 P5 P6 F P1 P7 Q4 Q5 Q6 Q7 29 For the two overtime hours: F1 P1 P2 P3 P4 P5 P6 F1 P1 P2 P7 Q4 Q5 Q6 Q7 Q8 14 F2 P1 P2 P3 P4 P5 P6 P7 F1 P1 P2 Q4 Q5 Q6 Q7 Q8 Q9 9 If the left-hand sides of these 10 constraints are added, one finds that 7F hours of full-time labor are used in straight time (although 8F are paid for), F1 F2 full-time labor hours are used and paid for at overtime rates, and the total number of part-time hours is 10P1 9P2 8P3 7P4 6P5 5P6 4P7 6Q4 5Q5 4Q6 3Q7 2Q8 Q9 128.4 which is 40% of the day’s total requirement of 321 person-hours. This also leads to the objective function. The total daily labor cost which must be minimized is Z 8(10.11)F 8.08(F1 F2) 7.82(10P1 9P2 8P3 7P4 6P5 5P6 4P7 6Q4 5Q5 4Q6 3Q7 2Q8 Q9) Total overtime for a full-time employee is restricted to 5 hours or less, an average of 1 hour or less per day per employee. Thus the number of overtime hours worked per day cannot exceed the number of full-time employees: F1 F2 F Since part-time employees must work at least 4 hours per day, Q4 P1 for those leaving at the end of the fourth hour. At the end of the fifth hour, those leaving must be drawn from the P1 Q4 remaining plus the P2 that arrived at the start of the second hour: Q5 P1 P2 Q4 Similarly, for the remainder of the day, Q6 P1 P2 P3 Q4 Q5 Q7 P1 P2 P3 P4 Q4 Q5 Q6 6619 CH08 UG 7/8/02 120 4:05 PM Page 120 CHAPTER 8 LINEAR PROGRAMMING MODELING APPLICATIONS Q8 P1 P2 P3 P4 P5 Q4 Q5 Q6 Q7 Q9 P1 P2 P3 P4 P5 P6 Q4 Q5 Q6 Q7 Q8 To ensure that all part-timers who began at 9 A.M. do not work more than 7 hours: Q4 Q5 Q6 Q7 P1 Similarly, Q4 Q5 Q6 Q7 Q8 P1 P2 Q4 Q5 Q Q7 Q8 Q9 P1 P2 P3 Finally, to ensure that all part-time employees leave at some time: P1 P2 P3 P4 P5 P6 P7 Q4 Q5 Q6 Q7 Q8 Q9 The resulting problem has 16 integer variables and 22 constraints. If integer programming software is not available, the linear programming problem can be solved and the solution rounded, making certain that none of the constraints have been violated. Note that the integer programming solution might also need to be adjusted—if F is an odd integer, 0.5F will not be an integer and the requirement that “half” of the full-time employees go to lunch at 11 A.M. and the other half at noon will have to be altered by assigning the extra employee to the appropriate hour. 1. The least-cost solution requires 29 full-time employees, 9 of whom work two hours of overtime per day. In actuality, 18 of the full-time employees would work overtime on two different days and 9 would work overtime on one day. Fourteen of the full-time workers would take lunch at 11 A.M. and the other 15 would take it at noon. Eleven part-timers would begin at 11 A.M., with 9 of them leaving at 3 P.M. and the other 2 at 4 P.M. Fifteen part-time employees would work from noon until 4 P.M., and 5 would work from 2 P.M. until 6 P.M. The resulting cost of 232 hours of straight time, 18 hours of overtime, and 126 hours of part-time work is $3,476.28 per day. This solution is not unique—other work assignments can be found that result in this same cost. 2. The same staffing would be used every day. In fact, one would expect different patterns to present themselves on different days; for example, Fridays are usually much busier bank days than the others. In addition, the person-hours required for each hour of the day are assumed to be deterministic. In a real situation, wide fluctuations will be experienced in a stochastic manner. The optimal solution results in a considerable amount of idle time, partly caused by the restriction that employees can start at the beginning of an hour and leave at the end. Eliminating this restriction might yield better results at the risk of increasing the problem size.