CHAPTER 8 Linear Programming Applications 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 largest large problem in terms of the number 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: Labor Planning Problem—Hong Kong Bank of Commerce. 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. 4: 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: | | |Minimum | | |Capacity | |Production | |Department |(In Hours) |Product |Level | |Sawing |450 |Chairs |100 | |Sanding |400 |Benches |50 | |Assembly |625 |Tables |50 | |Painting |550 | | | |Hours Required |Unit | |Product |Sawing |Sanding |Assembly |Painting |Profit | |Chairs |1. 5 |1. 0 |2. 0 |1. 5 |$15 | |Benches |1. 5 |1. 5 |2. 0 |2. 0 |$10 | |Tables |2. 0 |2. 0 |2. 5 |2. 0 |$20 | The production manager has the responsibility of specifying production levels for each product for the coming week.

Formulate as a linear programming problem to maximize profit. 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

What mix of products would yield maximum profit? Solving with computer software we get: X1= 100 chairs; X2 = 50 benches; X3 = 112. 5 tables; profit = 4250. Alternative Example 8. 2: A phosphate manufacturer produces three grades of phosphate, A, B, and C, which yield profit of $40, $50, and $60 per kilogram, respectively. The products require the labor and materials per batch that are shown in the table. Each batch of Grade A phosphate yields 800kg of phosphate; each batch of Grade B phosphate yields 700kg of phosphate; and each batch of Grade C phosphate yields 800 kg. |Grade |Grade |Grade |Available | | |A |B |C |Resources | |Labor hours |4 |4 |5 |80 hr | |Raw material #1 |200 |300 |300 |6,000 kg | |Raw material #2 |600 |400 |500 |5,000 kg |

Formulate as an LP problem to maximize profit. Objective function Maximize profit = 40(800)A + 50(700)B + 60(800)C Constraints Labor:4A +4B +5C( 80 Raw material #1200A + 300B +300C( 6,000 Raw material #2600A + 400B +500C( 5,000 What mix of products would yield maximize profit? Solutions To Problems 8-1. Since the decision centers about the production of the two different cabinet models, we let X1= number of French Provincial cabinets produced each day X2= number of Danish Modern cabinets produced each day Objective: maximize revenue = $28X1 + $25X2 subject to 3X1 + 2X2( 360 hours (carpentry department) [pic] X1 + 1X2( 200 hours (painting department) [pic]X1 + [pic] X2( 125 hours (finishing department) X1( 60 units (contract requirement) X2( 60 units (contract requirement)

X1, X2( 0 Problem 8-1 solved by computer: Produce 60 French Provincial cabinets (X1) per day Produce 90 Danish Modern cabinets (X2) per day Revenue = $3,930 8-2. Let X1= dollars invested in Los Angeles municipal bonds X2= dollars invested in Thompson Electronics X3= dollars invested in United Aerospace X4= dollars invested in Palmer Drugs X5= dollars invested in Happy Days Nursing Homes Maximize return = 0. 53X1 + 0. 068X2 + 0. 049X3 + 0. 084X4 + 0. 118X5 subject to X1 + X2 + X3 + X4 + X5 ( $250,000 (funds) X1( . 2 (X1 + X2 + X3 + X4 + X5) (bonds) or ( . 8X1 – . 2X2 – . 2X3 – . 2X4 – . 2X5 ( 0 X2 + X3 + X4( . 4 (X1 + X2 + X3 + X4 + X5) (combination of electronics, aerospace, and drugs) or –0. 4X1 + 0. 6X2 + 0. 6X3 + 0. 6X4 – 0. 4X5 ( 0 (X5 ( 0. 5X1) rewritten as –0. 5X1 + X5 ( 0 (nursing home as percent of bonds) X1, X2, X3, X4, X5 ( 0 Problem 8-2 solved by computer: $50,000invested in Los Angeles municipal bonds (X1) $0invested in Thompson Electronics (X2) $0invested in United Aerospace (X3) 175,000invested in Palmer Drugs (X4) $25,000invested in Happy Days (X5) This produces an annual return on investment of $20,300. 8-3. Minimize staff size = X1 + X2 + X3 + X4 + X5 + X6 where Xi = number of workers reporting for start of work at period i (with i = 1, 2, 3, 4, 5, or 6) X1 + X2 ( 12 X2 + X3 ( 16 X3 + X4 ( 9 X4 + X5 ( 11 X5 + X6 ( 4 X1 + X6 (

3 All variables ( 0 The computer solution is to hire 30 workers: 16 begin at 7 a. m. 9 begin at 3 p. m. 2 begin at 7 p. m. 3 begin at 11 p. m. An alternative optimum is 3 begin at 3 a. m. 9 begin at 7 a. m. 7 begin at 11 a. m. begin at 3 p. m. 9 begin at 7 p. m. 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= number of pounds of mineral product per horse each day Minimize cost = 0. 09X1 + 0. 14X2 + 0. 17X3 subject to 2X1+ 3X2 + 1X3 ( 6 (ingredient A) [pic]X1+ 1X2 + [pic]X3 ( 2 (ingredient B) 3X1+ 5X2 + 6X3 ( 9 (ingredient C) 1X1+ 1[pic]X2 + 2X3 ( 8 (ingredient D) [pic]X1+ [pic]X2 + 1[pic]X3 ( 5 (ingredient E) X1+ X2 +X3 ( 6 (maximum feed/day) All variables ( 0 Solution: X1= 1[pic] X2= 0 X3= 3[pic] cost= 0. 87 8-5. Let E1, E2, and E3 represent the ending inventory for the three months respectively. Let RT1, RT2, and RT3 represent the reguar production for the three months and OT1, OT2, and OT3 represent the overtime production quantities during the three months respectively. Then the formulation is: Minimize cost: 300RT1 + 300RT2 + 300RT3 + 325OT1 + 325OT2 + 325OT3 + 20E1 + 20E2 + 20E3 subject to RT1 < 200 June regular production RT2 < 200 July regular production RT3 < 200 August regular production OT1 < 15 June overtime production OT2 < 15 July overtime production

OT3 < 15 August overtime production (E1 + RT1 + OT1= 195 Ending inventory from first month (E2 + E1 + RT2 + OT2= 215 Ending inventory from second month (E3 + E2 + RT3 + OT3= 205 Ending inventory from third month {All variables}? 0 Non-negativity constraints The optimal production schedule is to produce 200 each month during regular production and to use overtime to produce 10 units in July and 5 in August for a total cost of $184,975. 8-6. Let T = number of TV ads R = number of radio ads B = number of billboard ads N = number of newspaper ads Maximize total audience = 30,000T + 22,000R + 24,000B + 8,000N

Subject to 800T + 400R + 500B + 100N ( 15,000 ? ( 10 R (10 ? (10 ? (10 ? + R ( 6 500B + 100N ( 800T ?, R, ? , ? ( 0 Solution: T = 6. 875; R = 10; B = 9; N = 10; Audience reached = 722,250. If integer solutions are necessary, integer programming (see Chapter 11) could be used. 8-7. Let_X1= number of newspaper ads placed X2= number of TV spots purchased Minimize cost =$925X1 + $2,000X2 subject to0. 04X1 + 0. 05X2 ( 0. 40 (city exposure) 0. 03X1 + 0. 03X2 ( 0. 60 (exposure in northwest suburbs) X1, X2 ( 0 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). Problem 8-7 solved by computer: 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. 8-8. Let Xij = number of new leases in month i for j-months, i = 1, . . . , 6; j = 3, 4, 5 Minimize cost =1260X13 + 1260X23 + 1260X33 + 1260X43 + 840X53 + 420X63 + 1600X14 + 1600X24 + 1600X34 + 1200X44 + 800X54 + 400X64+ 1850X15 + 1850X25 + 1480X35 + 1110X45 + 740X55 + 370X65 subject to:X13 + X14 + X15 ( 420 – 390

X13 + X14 + X15 + X23 + X24 + X25 ( 400 – 270 X13 + X14 + X15 + X23 + X24 + X25 + X33 + X34 + X35 ( 430 – 130 X14 + X15 + X23 + X24 + X25 + X33 + X34 + X35 + X43 + X44 + X45 ( 460 X15 + X24 + X25 + X33 + X34 + X35 + X43 + X44 + X45 + X53 + X54 + X55 ( 470 X25 + X34 + X35 + X43 + X44 + X45 + X53 + X54 + X55 + X63 + X64 + X65 ( 440 X15 + X25 + X35 + X45 + X55 + X65 ( 0. 0(X13 + X14 + X15 + X23 + X24 + X25 + X33 + X34 + X35 + X43 + X44 + X45 + X53 + X54 + X55 + X63 + X64 + X65) All variables ( 0 Solving this on the computer results in the following solution: X15 = 305-month leases in March X25 = 1005-month leases in April X35 = 1705-month leases in May X45 = 1605-month leases in June X55 = 105-month leases in July All other variables equal 0. Total cost = $677,100. As a result of this, there are 440 cars remaining at the end of August. 8-9. The linear program has the same constraints as in problem 8-8.

The objective function changes and is now: Minimize cost =1260(X13 + X23 + X33 + X43 + X53 + X63) + 1600(X14 + X24 + X34 + X44 + X54 + X64) + 1850(X15 + X25 + X35 + X45 + X55 + X65) Solving this on the computer results in the following solution: X15 = 305-month leases in March X25 = 1005-month leases in April X34 = 654-month leases in May X35 = 1055-month leases in May X43 = 1603-month leases in June X53 = 103-month leases in July All other variables equal 0. Total cost = $752,950. 8-10. Let Xij = number of students bused from sector i to school j Objective: minimize total travel miles = 5XAB+ 8XAC + 6XAE 0XBB+ 4XBC + 12XBE + 4XCB+ 0XCC + 7XCE + 7XDB+ 2XDC + 5XDE + 12XEB+ 7XEC + 0XEE subject to XAB + XAC + XAE= 700 (number of students in sector A) XBB + XBC + XBE= 500 (number of students in sector B) XCB + XCC + XCE= 100 (number of students in sector C) XDB + XDC + XDE= 800 (number of students in sector D) XEB + XEC + XEE= 400 (number students in sector E) XAB + XBB + XCB + XDB + XEB ( 900 (school B capacity) XAC + XBC + XCC + XDC + XEC ( 900 (school C capacity) XAE + XBE + XCE + XDE + XEE ( 900 (school E capacity) All variables ( 0 Solution: XAB= 400 XAE= 300 XBB= 500 XCC= 100 XDC= 800 XEE= 400

Distance = 5,400 “student miles” 8-11. Maximize number of rolls of Supertrex sold = 20X1 + 6. 8X2 + 12X3 – 65,000X4 whereX1= dollars spent on advertising X2= dollars spent on store displays X3= dollars in inventory X4= percent markup subject to X1 + X2 + X3 ( $17,000 (budgeted) X1( $3,000 (advertising constraint) X2( 0. 05X3 (or X2 – 0. 05X3 ( 0) (ratio of displays to inventory) [pic] (markup ranges) X1, X2, X3, X4 ( 0 Problem 8-11 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.

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 needed. 8-12. 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 6X1 + 81X2 + 74X3 + 83X4 + 7X5+ 14X6 + 8X7 ( 26 22X1 + 28X5 + 19X6 + 63X7( 50 All Xi ( 0 Problem 8-12 solved by computer: The meal plan for the evening is No milk (X1 = 0) 0. 499 pound of ground meat (X2) 0. 173 pound of chicken (X3) No fish (X4 = 0) No beans (X5 = 0) 0. 105 pound of spinach (X6) 0. 762 pound of white potatoes (X7) Each meal has a cost of $1. 75. 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: 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. 33 lb If chicken and meat are omitted; fish= 0. 679 lb spinach= 0. 0988 lb milk= 2. 188 lb 8-13. 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 Objective function analysis: First find the time used on each test device: hours on test device 1 [pic] hours on test device 2 [pic] hours on test device 3 [pic] Thus, the objective function is aximize profit = (revenue) – (material cost) – )test cost) = (200X1 + 120X2 + 180X3 + 130X4 + 430X5 + 260X6 – 35X1 – 25X2 – 40X3 – 45X4 – 170X5 – 60X6)[pic] [pic] [pic] This can be rewritten as maximize profit =$161. 35X1 + 92. 95X2 + 135. 50X3 + 82. 50X4 + 249. 80X5 + 191. 75X6 subject to 7X1 + 3X2 + 12X3 + 6X4 + 18X5 + 17X6 < 120(60) Minutes on test device 1 2X1 + 5X2 + 3X3 + 2X4 + 15X5 + 17X6 < 120(60) Minutes on test device 2 5X1 + 1X2 + 3X3 + 2X4 + 9X5 + 2X6 < 100(60) Minutes on test device 3 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 Chapter 7 and Module 7, for additional time on the three test devices are $21. 41, $5. 75, and $0, respectively, per minute. 8-14. 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 subject to 130X1 – 90Y1( 40,000 (Aug. need, hours) 130X2 – 90Y2( 45,000 (Sept. need) 130X3 – 90Y3( 35,000 (Oct. need) 130X4 – 90Y4( 50,000 (Nov. need) 130X5 – 90Y5( 45,000 (Dec. eed) 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) X4= X3 + Y3 – 0. 05X3 (staff on Nov. 1) X5= X4 + Y4 – 0. 05X4 (staff on Dec. 1) All Xi, Yi ( 0 b. The computer-generated results are: | |Trained | | | |Technicians |Trainees | |Month |Available |Beginning | |Aug. 350 |13. 7 (actually 14) | |Sept. |346. 2 |0 | |Oct. |328. 8 |72. 2 (actually 72) | |Nov. |384. 6 |0 | |Dec. |365. 4 |0 | Total salaries paid over the five-month period = $3,627,279. 8-15. a. Let Xij = acres of crop i planted on parcel j wherei = 1 for wheat, 2 for alfalfa, 3 for barley = 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 [pic] 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 Objective function: maximize profit [pic] b. The solution is to plant X12= 1,250 acres of wheat in N parcel X13= 500 acres of wheat in NW parcel X14= 312[pic] acres of wheat in W parcel X15= 137[pic] 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. 8-16. 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: 1 0. 70X1 + 0. 55X2 + 0. 12X3 + 0. 01X4 + 0. 05X5 ( 42 (2. 1% of 2,000) 2 0. 70X1 + 0. 55X2 + 0. 12X3 + 0. 01X4 + 0. 05X5 ( 46 (2. 3% of 2,000) silicon quality: 3 0. 15X1 + 0. 30X2 + 0. 26X3 + 0. 10X4 + 0. 025X5 + 0. 24X6 + 0. 25X7 + 0. 23X8 ( 86 (4. 3% of 2,000) 4 0. 15X1 + 0. 30X2 + 0. 26X3 + 0. 10X4 + 0. 025X5 + 0. 24X6 + 0. 25X7 + 0. 23X8 ( 92 (4. 6% of 2,000) carbon quality: 5 0. 03X1 + 0. 01X2 + 0. 03X4 + 0. 18X6 + 0. 20X7 + 0. 25X8 ( 101 (5. 5% of 2,000) 6 0. 03X1 + 0. 01X2 + 0. 03X4 + 0. 18X6 + 0. 20X7 + 0. 25X8 ( 107 (5. 35% of 2,000) Availability by weight: 7 X2 ( 300 8 X6 ( 50 9 X7 ( 200 10 X8 ( 100 One-ton weight: 11 X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 = 2,000 The solution is infeasible. 8-17. This problem refers to Problem 8-16’s infeasibility. Some investigative work is needed to track down the issues. The two issues are: 1. Requiring at least 5. 05% carbon is not possible. 2. Producing 1 ton from the materials is not possible. If constraints 5 and 11 are relaxed (or removed), one solution is X2 = 83. lb (alloy 2), X6 = 50 lb (carbide 1), X7 = 83. 6 lb (carbide 2), and X8 = 100 lb (carbide 3). Cost = $34. 91. Each student may take a different approach and other recommendations may result. 8-18. X1= number of medical patients X2= number of surgical patients Maximize revenue = $2,280X1 + $1,515X2 subject to 8X1 + 5X2( 32,850 (patient-days available = 365 days ( 90 new beds) 3. 1X1 + 2. 6X2 ( 15,000 (lab tests) 1X1 + 2X2( 7,000 (x-rays) X2( 2,800 (operations/surgeries) X1, X2( 0 Problem 8-18 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. 8-19. 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 8-18, 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 because the slack for constrain 1 has a value of 0.. (c) There are 876 (the slack for constraint 2) lab tests of unused capacity. (d) The x-ray is used to its maximum (slack for constraint 3 is 0) and has a $65. 5 dual price. The revenue would increase by this amount for each additional x-ray. (e) The operating room still has 695 operations available (the slack for constraint 4). [pic] [pic] 8-20. For the Low Knock Oil Company example it was originally assumed that a one to one ratio of raw materials (crude oil) to finished goods (gasoline). In reality, that ratio is closer to 46%. Hence, the example problem needs to be modified with 0. 46 as the coefficient throughout the first two constraints as follows: Minimize 30X1 + 30X2 + 34. 80X3 + 34. 80X4 subject to: 0. 46X1 + 0. 46X3( 25000 0. 46X2+ 0. 46X4( 32000 0. 10 X1 + 0. 15X3 ( 0 0. 05X2 – 0. 25X4 ( 0 The rounded solution is X1 = 32609; X2 = 57971; X3 = 21739; X4 = 11594; Cost = 3877391 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 Solution: |Source |Destination |Number of |(Station) |(Wing) |Trays | |5A |5 |60 | |5A |6 |80 | |5A |3 |60 | |3G |1 |80 | |3G |3 |90 | |3G |4 |55 | |1S |4 |155 | |1S |2 |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 Minimize beta = 1. 2X1 + 0. 85X2 + 0. 55X3 + 1. 40X4 + 1. 25X5 subject to X1 + X2 + X3 + X4 + X5 = 1total of the proportions must add to 1 0. 11X1 + 0. 09X2 + 0. 065X3 + 0. 15X4 + 0. 13X5 ( 0. 11return should be at least 11% X1 ( 0. 5no more than 35% in any single stock X2 ( 0. 35 X3 ( 0. 35 X4 ( 0. 35 X5 ( 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 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 H = 1,000 gallons of fuel to purchase in Houston N = 1,000 gallons of fuel to purchase in New Orleans 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

FN = fuel remaining when plane lands in New Orleans Minimize cost = 4. 15A + 4. 25L + 4. 10H + 4. 18N subject to A + FA ( 24minimum amount of fuel on board when leaving Atlanta A + FA ( 36maximum amount of fuel on board when leaving Atlanta L + FL ( 15minimum amount of fuel on board when leaving Los Angeles L + FL ( 23maximum amount of fuel on board when leaving Los Angeles H + FH ( 9minimum amount of fuel on board when leaving Houston H + FH ( 17maximum amount of fuel on board when leaving Houston N + FN ( 11minimum amount of fuel on board when leaving New Orleans N + FN ( 20maximum amount of fuel on 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 flight. 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, FH = L + FL – (7 + 0. 05(L + FL – 15)) becomes 0. 95L + 0. 95FL – FH = 6. 25 FN = H + FH – (3 + 0. 05(H + FH – 9)) becomes 0. 95H + 0. 95FH – FN = 2. 55 FA = N + FN – (5 + 0. 05(N + FN – 11)) becomes 0. 95N + 0. 95FN – FA = 4. 45 All variables ( 0 The optimal solution is A=18 (1,000 gallons of fuel to purchase in Atlanta) FA=6 (1,000 gallons of fuel remaining when plane lands in Atlanta) L=3 (1,000 gallons of fuel to purchase in Los Angeles)

FL=12 (1,000 gallons of fuel remaining when plane lands in Los Angeles) H=1 (1,000 gallons of fuel to purchase in Houston) FH=8 (1,000 gallons of fuel remaining when plane lands in Houston) N=5 (1,000 gallons of fuel to purchase in New Orleans) FN=6 (1,000 gallons of fuel remaining when plane lands in New Orleans) Total cost = 112. 45 (( 1,000) Solutions to Internet Homework Problems 8-24. 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 1X1 +4X3 ( 52 oz (bourbon limit) 1X1 +1X2 ( 38 oz (brandy limit) 1X1 +2[pic]X4 ( 64 oz (vodka limit) X2 +1[pic]X4 ( 24 oz (dry vermouth limit) 1X1 +2X2 ( 36 oz (sweet vermouth limit) All variables ( 0 Because a Chauncey (X1) is [pic] sweet vermouth, it requires 1 oz of that resource (each drink totals 4 oz). Problem 8-27 solved by computer: Mix 25. 99 (or 26) Chaunceys (X1) Mix 5. 00 (or 5) Sweet Italians (X2) Mix 6. 50 (or 6[pic]) bourbon on the rocks (X3) Mix 14. 25 (or 14[pic]) Russian martinis (X4) This is a total of 51. 75 drinks (in five iterations). 8-25. Minimize 6X11 + 8X12 + 10X13 + 7X21 + 11X22 + 11X23 + 4X31 + 5X32 + 12X33 subject to X11 + X12 + X13( 150 X21 + X22 + X23( 175 X31 + X32 + X33( 275 X11 + X21 + X31= 200 X12 + X22 + X32= 100

X13 + X23 + X33= 300 All variables ( 0 The solution is: X11 = 25, X13 = 125, X23 = 175, X31 = 175, X32 = 100 Cost = $4,525. 8-26. Let Xi = number of BR54 produced in month i, for i = 1, 2, 3. Yi = number of BR49 produced in month i, for i = 1, 2, 3. 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 = 50initial inventory of BR54 IY0 = 50initial inventory of BR49 IX3 = 100ending inventory of BR54 IY3 = 150ending inventory of BR49

X1 + Y1 ( 1,100maximum production level in August X2 + Y2 ( 1,100maximum production level in September X3 + Y3 ( 1,100maximum production level in October X1 + IX0 = 320 + IX1 BR54 requirements for August X2 + IX1 = 740 + IX2 BR54 requirements for September X3 + IX2 = 500 + IX3 BR54 requirements for October Y1 + IY0 = 450 + IY1 BR49 requirements for August Y2 + IY1 = 420 + IY2 BR49 requirements for September Y3 + IY2 = 480 + IY3 BR49 requirements for October All variables ( 0 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 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. . , . . . , P7 at 3 p. m. Note that because part-time employees must work a minimum of 4 hours, none can start after 3 p. m. 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: 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 0P1 + 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 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.

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