Ch09 - 1 CHAPTER 9 SOLUTIONS SECTION 9.2 1. Let xi = 1 if...

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Unformatted text preview: 1 CHAPTER 9 SOLUTIONS SECTION 9.2 1. Let xi = 1 if player i starts xi = 0 otherwise Then appropriate IP is max z = 3x 1 + 2x2 + 2x3 + x4 + 3x5 + 3x6 + x 7 s.t. x1 + x 3 + x5 + x 74 (guards) x3 + x 4 + x5 + x6 + x72 (forwards) x2 + x 4 + x61 (center) x 1 + x 2 + x3 + x 4 + x5 + x 6 + x7 = 5 3x1 + 2x2 + 2x3 + x4 + 3x5 + 3x6 + 3x710 (BH) 3x1 + x 2 + 3x3 + 3x4 + 3x5 + x 6 + 2x710 (SH) x1 + 3x 2 + 2x3 + 3x4 +3x5 + 2x6 + 2x710 (REB) x6 + x 31 -x4-x5 + 22y (If x 1>0 then x 4 + x52) x1 2(1-y) x2 + x31 x1,x 2,...x 7,y are all 0-1 variables 2. Let xi = Tons of water processed at site i yi = 1 if Pollution Control Station is built at Site i. yi = 0 otherwise Then the appropriate IP is min z = 100,000y1 + 60,000y2 + 40,000y 3 + 20x 1 + 30x 2 +40x 3 s.t. .40x 1 + .25x 2 + .20x380,000 (Pollutant 1) .30x1 + .20x2 + .25x 350,000 (Pollutant 2) x1 My1 x2 My2 x3 My3 All x ij0 and all yi = 0 or 1. M = (1/.20)(80,000) = 400,000 will do. This follows since at most 400,000 tons of water must be processed to remove the required pollutants. 3. Let x1 = Units of Product 1 produced x2 = Units of Product 2 produced yi = 1 if any Product i is produced yi = 0 otherwise Then the appropriate IP is max z = 2x1 + 5x 2 - 10y 1 - 20y 2 s.t. 3x 1 + 6x2120 x1 40y 1 x220y 2 x10,x 20, y1,y2 = 0 or 1 4. The statement (x 2 + x 3 = 2 implies x 4 = 1) is equivalent to x2 + x3-1>0 implies x 4 - 10. Thus by (28) and (29) we should add the constraints(M = 1 will do the job) 1 - x4y x2 + x 3 - 1(1 - y) y = 0 or 1 2 5a. This is equivalent to x2>0 implies x 11. Then (28) and (29) yield (M = 1 is ok): 1 - x1y x2(1 - y) y = 0 or 1 Instead you could have added the constraint y 1y2. 5b. We want x11 (1 - x10) or x 21 (1 - x20). From (26)' and (27)' we see that we must add the constraints 1 - x1y 1 - x2(1-y) y = 0 or 1. Instead we could have added the constraint y 1 + y31. 6. y1 = 1 if calculus is taken y1 = 0 otherwise y2 = 1 if OR is taken y2 = 0 otherwise y3 = 1 if Data Structures is taken y3 = 0 otherwise y4 = 1 if Bus Stat is taken y4 = 0 otherwise y5 = 1 if Comp. Sim. is taken y5 = 0 otherwise y6 = 1 if Intro. to Comp. is taken y6 = 0 otherwise y7 = 1 if Forecasting is taken y7 = 0 otherwise Then the appropriate IP is min z = y1 + y 2 + y3 + y 4 + y5 + y6 + y7 st y 1 + y2 + y 3 + y4 + y 72 (math) y2 + y4 + y 5 + y7 2 (operations research) y3 + y5 + y6 2 (computers) y4y1 y5y 6 y3y 6 y7y4 y1,y2,...y7 = 0 or 1 7. For x = 300, z1 = .4, z2 = .6, y1 = 1, all other variables equal 0. For x = 1200, z1 = z2 = y1 = y 2 =0, y3 = 1, z3 = .6, z4 = .4. 8. max z = 15,000z 2 + 27,000z3 + 77,000z4 + 8000z2' + 56,000z3' + 80,000z4' st 6z2 + 10z3 + 15z 4 + 4z 2' + 12z 3' + 15z 4'20 z1y 1, z2y1 + y2, z3y 2 + y3, z4y 3 z1'y1', z2'y1' + y 2', z3'y 2' + y3', z4'y3' y1 + y2 + y3 = 1 ,z1 + z2 + z3 + z4 = 1 y1' + y2' + y 3' = 1 ,z1' + z2' + z3' + z4' = 1 All y i and yi' = 0 or 1, all zi and zi'0 3 The number of IJ ads is given by 6z2 + 10z 3 + 15z 4 and the number of FS ads is given by 4z 2' + 12z 3' + 15z 4'. 9. Add the constraints x = y 1 + 2y2 + 3y3 + 4y 4 and y1 + y 2 + y3 + y 4 = 1 and yi = 0 or 1. Then x = i if and only if y i = 1 10. We need x + y-30 or 2x + 5y-120 or both. From (26)' and (27)' we see that we must add the constraints x + y-3Mz 2x + 5y-12M(1-z) z = 0 or 1 , M a large number 11. We have that x<3 (or 3-x>0) implies y3 (or 3-y0). From (28)' and (29)' we see that the following constraints are needed: y-3Mz, 3-x(1-z)M, z = 0 or 1, M a large number. 12. Let y 1 = 1 if NY is used, y 2 = 1 if LA is used, y 3 = 1 if Chicago is used, y 4 = 1 if Atlanta is used. Also y i = 0 if city i is not used. Define xij = units shipped from warehouse in city i to region j min z = 400y 1 + 500y 2 + 300y 3 + 150y 4 + 20x 11 + 40x 12 + 50x 13 + 48x 21 + 15x 22 + 26x 23 + 26x 31 + 35x 32 + 18x 33 + 24x 41 + 50x 42 +35x 43 st x 11 + x 21 + x31 + x 4180, x12 + x 22 + x32 + x 4270 x13 + x23 + x33 + x4340, x11 + x12 + x 13100y 1 x21 + x22 + x23100y 2, x 31 + x32 + x 33100y 3, x41 + x42 + x43100y 4, y1y2, y2 + y 41, y1 + y2 + y 3 + y42 , All x ij0, all yi = 0 or 1. 13. Let x i = Number of workers employed on Line i yi = 1 if Line i is used,y i=0 otherwise min z = 1000y 1 + 2000y 2 + 500x 1 + 900x2 s.t. 20x 1 + 50x 2120 30x1 + 35x 2150 40x 1 + 45x 2200 x17y1 x27y2 x10, x20, y1=0 or 1, y 2=0 or 1 14a. Let x i = 1 if disk i is used, xi = 0 otherwise min z = 3x1 + 5x 2 + x3 + 2x 4 + x5 + 4x 6 + 3x 7 + x8 + 2x 9 + 2x10 s.t. x1 + x 2 + x4 + x 5 + x8 + x91 (File 1) x1 + x31 (File 2) x2 + x5 + x7 + x 101 (File 3) x3 + x6 + x81 (File 4) x1 + x2 + x4 + x 6 + x7 + x 9 + x101 (File 5) xi = 0 or 1 (i=1,2,...10) 14b. If x3 + x 5>0, then x21 yields 1 - x22y x3 + x52(1 - y) y=0 or 1 (need M=2 because x 3 + x5 = 2 is possible) 15. Let P = number of Pear computers produced and A = number of apple computers produced. Let y 1 = 1 if any Pear computers are produced, y 1 = 0 otherwise, and y 2 = 1 if any Apple computers are produced and y 2 = 0 otherwise. Then the appropriate IP is max z = 400P + 900A - 5000y 1 - 7000y2 st P + 2A1200, 2P + 5A3000 P1200y 1, A600y2 A,P non-negative integers and y 1 and y2 are 0 or 1 4 16. Let H = homes built and A = apartments built. Also let y 1 = 1 if marina is built, y 1 = 0 otherwise and y 2 = 1 if tennis court is built, y 2 = 0, otherwise. Then a correct IP formulation is (objective function is in thousands of dollars of NPV) max z = 48A + 46H - 40A - 40H - 1200y 1 - 2800y 2 st A + H10,000, y1 + y2 = 1 3A - H30,000y 3, y 130,000(1 - y3) A, H non-negative integers and y i = 0 or 1 We chose M = 30,000 because 3A - H 30,000 will always hold. 17. Let Xi = number of units produced on machine i and Yi = 1 if machine i is used, Yi = 0 otherwise. See LINDO printout for optimal solution. Section 9.2 Problem 17 Printout MIN 1000 Y1 + 920 Y2 + 800 Y3 + 700 Y4 + 20 X1 + 24 X2 + 16 X3 + 28 X4 SUBJECT TO 2) - 900 Y1 + X1 <= 0 3) - 1000 Y2 + X2 <= 0 4) - 1200 Y3 + X3 <= 0 5) - 1600 Y4 + X4 <= 0 6) X1 + X2 + X3 + X4 = 2000 END INTE 4 OBJECTIVE FUNCTION VALUE 1) 37000.00 VARIABLE VALUE REDUCED COST Y1 1.000000 1000.000000 Y2 0.000000 920.000000 Y3 1.000000 -4000.000000 Y4 0.000000 700.000000 X1 800.000000 0.000000 X2 0.000000 4.000000 X3 1200.000000 0.000000 X4 0.000000 8.000000 ROW 2) 3) 4) 5) 6) SLACK OR SURPLUS DUAL PRICES 100.000000 0.000000 0.000000 0.000000 0.000000 4.000000 0.000000 0.000000 0.000000 -20.000000 18. Let Xi = number of book i produced and Yi = 1 if book i is produced at all and Yi = 0 otherwise. See LINDO printout for optimal solution. Section 9.2 Problem 18 MAX - 80000 Y1 - 50000 Y2 - 60000 Y3 - 30000 Y4 - 40000 Y5 + 25 X1 5 + 20 X2 + 23 X3 + 14 X4 + 18 X5 SUBJECT TO 2) - 5000 Y1 + X1 <= 0 3) - 4000 Y2 + X2 <= 0 4) - 3000 Y3 + X3 <= 0 5) - 4000 Y4 + X4 <= 0 6) - 3000 Y5 + X5 <= 0 7) X1 + X2 + X3 + X4 + X5 <= 10000 END SUB Y1 1.00000 INTE Y1 SUB Y2 1.00000 INTE Y2 SUB Y3 1.00000 INTE Y3 SUB Y4 1.00000 INTE Y4 SUB Y5 1.00000 INTE Y5 GIN X1 GIN X2 GIN X3 GIN X4 GIN X5 OBJECTIVE FUNCTION VALUE 1) 75000.00 VARIABLE VALUE REDUCED COST Y1 1.000000 80000.000000 Y2 1.000000 50000.000000 Y3 0.000000 60000.000000 Y4 0.000000 30000.000000 Y5 0.000000 40000.000000 X1 5000.000000 -25.000000 X2 4000.000000 -20.000000 X3 0.000000 -23.000000 X4 0.000000 -14.000000 X5 0.000000 -18.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 0.000000 3) 0.000000 0.000000 4) 0.000000 0.000000 5) 0.000000 0.000000 6) 0.000000 0.000000 7) 1000.000000 0.000000 19. Let Xi = number of computers produced (in thousands) at plant i and Yi = 1 if plant i is used and Yi = 0 otherwise. See LINDO printout for optimal solution. Section 9.2 Problem 19 MAX - 9 Y1 - 5 Y2 - 3 Y3 - Y4 + 2.5 X1 + 1.8 X2 + 1.2 X3 + 0.6 X4 SUBJECT TO 6 2) X1 + X2 + X3 + X4 <= 20 3) - 10 Y1 + X1 <= 0 4) - 8 Y2 + X2 <= 0 5) - 9 Y3 + X3 <= 0 6) - 6 Y4 + X4 <= 0 END INTE 4 1) 25.60000 VARIABLE VALUE Y1 1.000000 Y2 1.000000 Y3 0.000000 Y4 1.000000 X1 10.000000 X2 8.000000 X3 0.000000 X4 2.000000 ROW 2) 3) 4) 5) 6) REDUCED COST -10.000000 -4.600000 -2.400000 1.000000 0.000000 0.000000 0.000000 0.000000 SLACK OR SURPLUS DUAL PRICES 0.000000 0.600000 0.000000 1.900000 0.000000 1.200000 0.000000 0.600000 4.000000 0.000000 20. Let AB = 1 if rep is assigned to states A and B, etc. Then the correct LP is given on LINDO output. Section 9.2 Problem 20 Solution MAX 63 AB + 76 AC + 50 BD + 85 BE + 63 CD + 77 DE + 92 DG + 74 EF + 89 FG + 71 BC + 39 DF SUBJECT TO 2) AB + AC + BD + BE + CD + DE + DG + EF + FG + BC + DF = 2 3) AB + AC <= 1 4) AB + BD + BE + BC <= 1 5) AC + CD + BC <= 1 6) BD + CD + DE + DG + DF <= 1 7) BE + DE + EF <= 1 8) DG + FG <= 1 9) EF + FG + DF <= 1 END INTE 11 OBJECTIVE FUNCTION VALUE 1) 177.00000 VARIABLE VALUE AB .000000 AC .000000 BD .000000 BE 1.000000 CD .000000 REDUCED COST -63.000000 -76.000000 -50.000000 -85.000000 -63.000000 7 DE DG EF FG BC DF .000000 1.000000 .000000 .000000 .000000 .000000 -77.000000 -92.000000 -74.000000 -89.000000 -71.000000 -39.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) .000000 .000000 3) 1.000000 .000000 4) .000000 .000000 5) 1.000000 .000000 6) .000000 .000000 7) .000000 .000000 8) .000000 .000000 9) 1.000000 .000000 21. Let xij = number of air conditioners (in thousands) produced in city i for region j ( i = 1 is NY, j = 1 is East, etc.). Also let yj = 1 if factory is operated in city j, yj = 0 otherwise. Then the appropriate IP is(z is in thousands of dollars) min z = 6000y1 + 5500y2 + 5800y3 +6200y4 + 206x11 + 225x12 + 230x13 +290x14 + 225x21 + 206x22 + 221x23 + 270x24 + 230x31 + 221x32 + 208x33 + 262x34 + 290x41 + 270x42 + 262x43 + 215x44 st x11 + x21 + x31 + x41100 (East) x12 + x22 + x32 + x42150 (South) x13 + x23 + x33 + x43110 (Midwest) x14 + x24 + x34 + x4490 (West) x11 + x12 + x13 + x14150y1 (NY) x21 + x22 + x23 + x24150y2 (Atl) x31 + x32 + x33 + x34150y3 (Chic) x41 + x42 + x43 + x44150y4 (LA) (Either x1350 or x2350) 50 - x1350y 50 - x2350(1 - y) All xij integer; y, all yi = 0 or 1 22. Let Xi = 1 if word is used and Xi = 0 otherwise. The correct IP is on the LINDO printout. Section 9.2 Problem 22 MAX X1 + 7 X2 + 9 X3 + 4 X4 + 9 X5 + 4 X6 + 6 X7 + 9 X8 SUBJECT TO 2) X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 = 4 3) 2 X1 - X2 - 3 X3 - X5 - X6 + 2 X7 + X8 >= 0 END INTE 8 OBJECTIVE FUNCTION VALUE 1) 31.000000 VARIABLE VALUE X1 .000000 X2 1.000000 X3 .000000 X4 .000000 REDUCED COST -1.000000 -7.000000 -9.000000 -4.000000 8 X5 X6 X7 X8 1.000000 .000000 1.000000 1.000000 -9.000000 -4.000000 -6.000000 -9.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) .000000 .000000 3) 1.000000 .000000 NO. ITERATIONS= 13 BRANCHES= 1 DETERM.= 1.000E 0 23. Let Xij = 1 if machine i is used to do job j and Xij = 0 otherwise. Also let Yi = 1 if machine i is used at all and Yi = 0 otherwise. The correct IP is on the LINDO printout. Section 9.2 Problem 23 IN 42 X11 + 70 X12 + 85 X22 + 45 X23 + 58 X31 + 37 X34 + 58 X41 + 55 X43 + 38 X45 + 60 X52 + 54 X54 + 30 Y1 + 40 Y2 + 50 Y3 + 60 Y4 + 20 Y5 + 93 X13 SUBJECT TO 2) X11 + X31 + X41 = 1 3) X12 + X22 + X52 = 1 4) X23 + X43 + X13 = 1 5) X34 + X54 = 1 6) X45 = 1 7) X11 + X12 - 3 Y1 + X13 <= 0 8) X22 + X23 - 2 Y2 <= 0 9) X31 + X34 - 2 Y3 <= 0 10) X41 + X43 + X45 - 3 Y4 <= 0 11) X52 + X54 - 2 Y5 <= 0 END TITLE SECTION 9-2 PROBLEM 23 INTE 17 OBJECTIVE FUNCTION VALUE 1) 345.00000 VARIABLE VALUE X11 .000000 X12 .000000 X22 .000000 X23 .000000 X31 .000000 X34 .000000 X41 1.000000 X43 1.000000 X45 1.000000 X52 1.000000 X54 1.000000 Y1 .000000 Y2 .000000 Y3 .000000 REDUCED COST 42.000000 70.000000 85.000000 45.000000 58.000000 37.000000 58.000000 55.000000 38.000000 60.000000 54.000000 30.000000 40.000000 50.000000 9 Y4 Y5 X13 1.000000 1.000000 .000000 60.000000 20.000000 93.000000 ROW SLACK OR SURPLUS DUAL PRICES 2) .000000 .000000 3) .000000 .000000 4) .000000 .000000 5) .000000 .000000 6) .000000 .000000 7) .000000 .000000 8) .000000 .000000 9) .000000 .000000 10) .000000 .000000 11) .000000 .000000 24. Let Location 1 = Evansville, Location 2 = Indianapolis, Location 3 = South Bend. Let x ij = loaves per year (in hundreds of thousands) baked at location i for customer j. Let y i = 1 if a bakery is built in Location i, y i = 0, otherwise. Noting that a $1 cost each year is the equivalent of a $10 cost during the current year we obtain the following formulation: (objective function is in millions of dollars of NPV) min z = 5y 1 + 4y 2 + 4.5y 3 + .16x 11 + .34x 12 + .26x13 + .4x 21 +.3x 22 + .35x 23 + .45x 31 + .45x 32 + .23 x 33 st. x 11 + x 21 + x31 = 7, x12 + x 22 + x32 =4, x 13 + x23 + x33 = 3 x11 + x12 + x139y1, x 21 + x22 + x 239y2, x31 + x 32 +x 339y3 xij0, y i = 0 or 1 The optimal solution is to build bakeries in Evansville and Indianapolis and ship as follows: To Customer 1: 700,000 loaves from Evansville To Customer 2:400,000 loaves from Indianapolis To Customer 3: 200,000 loaves from Evansville and 100,000 loaves from Indianapolis. If Evansville or South Bend must produce at least 800,000 loaves per year add the following constraints: 800 - x11 - x12 -x13800y 4, 800 - x31 - x32 - x33800(1 - y4) y4 = 0 or 1 25. Let Region 1 = SE, Region 2 = NE, Region 3 = FW, Region 4 = MW, and let FF = Bank 1, R = Bank 2, PF = Bank 3 , and B = Bank 4. Define x ij = daily amount of checks for region i sent from bank j and let y j = 1 if bank j is used and y j = 0 if bank j is not used. Then the appropriate IP is max z = 1.05x 11 + .30x12 + .90x13 + .75x 14 + 1.2x 21 + .60x 22 + .75x 23 + .45x 24 + .60x31 + 1.2x 32 + .30x 33 + 1.65x 34 + .75x 41 +.60x42 + 1.05x 43 + .75x 44 - 50,000y1 - 40,000y 2 - 30,000y3 - 20,000y 4 st. x 11 + x 12 + x13 + x 14 = 40,000 x21 + x22 + x23 + x24 = 60,000 x31 + x32 + x33 + x34 = 30,000 x41 + x42 + x43 + x 44 = 50,000 x11 + x21 + x31 + x 4190,000y1 x12 + x22 + x32 + x4290,000y2 x13 + x23 + x33 + x4390,000y3 x14 + x24 + x34 + x 44 90,000y4 xij0, y j = 0 or 1 To understand the objective function observe that if x 11 = 1, then each day $7 remains in the bank and earns interest for the Clearinghouse. Thus on an annual basis x11 =1 "earns" Clearinghouse 7(.15) = $1.05. This explains the term 1.05x 11 in the objective function. 26. Let x ij = if city i is in district j, x ij = 0 otherwise. Also let y j = 1 if Republicans win district j, y j = 0 otherwise. Let r i = number of Republicans in city i and d i = number of democrats in city i. Then the appropriate IP is 10 j = 10 min z = j=1 st. yj i = 10 xij(ri + d i) 150,000 i=1 j = 1,2,3,4,5 i = 10 xij(ri + d i) 250,000 j = 1,2,3,4,5 i=1 j=5 x ij = 1 i = 1,2,...,10 j=1 i = 10 We must also ensure that if xij(ri - d i)>0, then i=1 yj = 1. To ensure this add constraints of the following form: 1-y j250,000zj j = 1,2,3,4,5 i = 10 x ij(ri - di) 250,000(1 - zj) j = 1,2,3,4,5 i=1 All variables must equal 0 or 1. 27. Let x i = 1 if service representative is located in city i. i = 1 is Boston, i = 2 is New York, i =3 is Philadelphia, i =4 is Washington. Let z j = 1 if there is a service representative within 150 miles of city j. j = 1 is Boston, j = 2 is New York, j = 3 is Philadelphia, j = 4 is Washington, j = 5 is Providence, j = 6 is Atlanta. Since profit per copier is $500 we wish to maximize z = 700(500)z 1 + 500(500)(1 - z1) + 1000(500)z 2 + 750(500)(1 - z2) +900(500)z 3 + 700(500)(1 - z3) +800(500)z 4 + 450(500)(1 - z4) +400(500)z 5 + 200(500)(1 - z5) +450(500)z 6 + 300(500)(1 - z6) - 80,000(x1 + x2 + x3 + x4) st. 1 - x1y1, z11-y1 1 - x2 - x3y2, z21 - y2 1 - x2 - x3 - x4y 3, z31 - y3 1 - x3 - x4y4, z41 - y4 1 - x1y 5, z51 - y5 1 - x2 - x3y6, z61 - y6 all xi, yj, zj are 0-1 variables. The constraints ensure that if zj = 1 and higher sales of copiers are "counted" by the objective function for city j , then there must be a service representative within 150 miles of city j. 28. Solving the appropriate transportation problem(using Table 20 costs and demands of 1000, 800 and 700 and supplies of 1000, 600 and 700) we find that 1000 men end up at Base 1, 600 at Base 2, and 700 at Base 3. Note:supply points = centers and demand points = bases and a dummy supply point with supply of 200 must be added. Using the variable definitions in the problem the appropriate IP is: min z = 5740S1 + 6030S2 + 6330S3 + 5920S4 + 6200S5 + 6280S6 +11,110L1 + 11545L2 + 11995L3 + 11380L4 + 11800L5 + 11920L6 + 12,160L7 st S1 + S2 + S3 + S4 + S5 + S67 L1 + L2 + L3 + L4 + L5 + L6 + L75 x11200S1, x21 + x22200S2, x32 + x33200S3, x42200S4, x53200S5, x61 + x63200S6, y11500L1, y21 + y22500L2, 11 y32 + y33500L3, y42500L4, y53500L5, y61 + y63500L6, y71 + y72 + y73500L7. x11 + x21 + x61 + y11 + y21 + y61 + y71 = 1000 x22 + x32 + x42 + y22 + y32 + y42 + y72 = 600 x33 + x53 + x63 + y33 + y53 + y63 + y73 = 700 All variables integer 29. Let xi1 = 1 if song i is on side 1, xi1 = 0 otherwise. xi2 = 1 if song i is on side 2, xi2 = 0 otherwise Any feasible solution to the following IP will do the job: max z = 0 st 144x11 + 5x21 + 3x31 + 2x41 + 4x51 + 3x61 + 5x71 + 4x8116 x11 + x12 = 1, x21 + x22 = 1, x31 + x32 = 1, x41 + x42 = 1 x51 + x52 = 1, x61 + x62 = 1, x71 + x72 = 1, x81 + x82 = 1 x11 + x31 + x51 + x81 = 2, x12 + x32 + x52 + x82 = 2 x21 + x41 + x61 + x813, x51 + x611 1 - x52y, x21 + x41 - 11 - y All variables = 0 or 1 30. Let xi,j = 1 if commercial of length i is assigned to block j and xi,j = 0 otherwise and yi = 1 if block i is used and yi = 0 otherwise. Then the appropriate formulation is min z = y1 + y2 + y3 + y4 + y5 + y6 + y7 + y8 st 15x15,1+16x16,1+20x20,1+25x25,1+30x30,1+35x35,1+40x40,1 +50x50,160y1 15x15,2+16x16,2+20x20,2+25x25,2+30x30,2+35x35,2+40x40,2 +50x50,260y2 15x15,3+16x16,3+20x20,3+25x25,3+30x30,3+35x35,3+40x40,3 +50x50,360y3 15x15,4+16x16,4+20x20,4+25x25,4+30x30,4+35x35,4+40x40,4 +50x50,460y4 15x15,5+16x16,5+20x20,5+25x25,5+30x30,5+35x35,5+40x40,5 +50x50,560y5 15x15,6+16x16,6+20x20,6+25x25,6+30x30,6+35x35,6+40x40,6 +50x50,660y6 15x15,7+16x16,7+20x20,7+25x25,7+30x30,7+35x35,7+40x40,7 +50x50,760y7 15x15,8+16x16,8+20x20,8+25x25,8+30x30,8+35x35,8+40x40,8 +50x50,860y8 All variables = 0 or 1 31. Let xic = 1 if product i is assigned to compartment c, xic = 0 otherwise. ei = shortage of product i (1 = super 2 = regular, 3 = unleaded), gic = gallons of product i in compartment c. min z = 10e1 + 8e2 + 6e3 st g11 + g21 + g312700 g12 + g22 + g322800 g13 + g23 + g331100 g14 + g24 + g341800 g15 + g25 + g353400 g11 + g12 + g13 + g14 + g15 + e1 - f1 = 2900 g21 + g22 + g23 + g24 + g25 + e2 - f2 = 4000 g31 + g32 + g33 + g34 + g35 + e3 - f3 = 4900 gi12700xi1, gi22800xi2, gi31100xi3, gi41800xi4 gi53400xi5 ( i = 1,2,3) x1j + x2j + x3j1 (j = 1, 2, 3, 4, 5) All xij = 0 or 1 All other variables 0 e1500,e2500,e3500 12 32. Let xik = 1 if k stores of type i in mall (type 1 = jewelry, etc.) xik = 0 otherwise k = 0,1,2,3. max z = .05(9x11 + 16x12 + 21x13 + 10x21 + 18x22 + 10x23 + 27x31 + 42x32 + 60x33 + 16x41 + 18x42 + 21x43 + 17x51 + 26x52 + 30x53 st 500x11 + 1000x12 + 1500x13 + 600x21 + 1200x22 + 1800x23 + 1500x31 + 3000x32 + 4500x33 + 700x41 + 1400x42 + 2100x43 + 900x51 + 1800x52 + 2700x5310,000 x11 + x12 + x13 = 1 x21 + x22 + x23 = 1 x31 + x32 + x33 = 1 x40 + x41 + x42 + x43 = 1 x51 + x52 + x53 = 1 all xij = 0 or 1. 33. Let xit = 1 if asset i is sold during year t and xit = 0 otherwise max z = 15x11 + 16x21 + 22x31 + 10x41 + 17x51 + 19x61 + 20x12 + 18x22 + 30x32 + 20x42 + 19x52 + 25x62 + 24x13 + 21x23 + 36x33 + 30x43 + 22x53 + 29x63 st 15x11 + 16x21 + 22x31 + 10x41 + 17x51 + 19x6120 20x12 + 18x22 + 30x32 + 20x42 + 19x52 + 25x6230 24x13 + 21x23 + 36x33 + 30x43 + 22x53 + 29x6335 x11 + x21 + x311 x21 + x22 + x231 x31 + x32 + x331 x41 + x42 + x431 x51 + x52 + x531 x61 + x62 + x631 All variables = 0 or 1 To implement a rolling planning horizon, we might solve a three year problem to determine which assets are sold during the current year. At the end of the current year we would update the information in Table 26 and then solve another problem with a three year horizon to determine which assets should be sold next year. 34a. Let yk = 1 if ladder k is replaced by a tower ladder and yk = 0 otherwise. max z = y1 + y2 + y3 +y4 + y5 + y6 + y7 st y2 + y31, y3+ y41, y1 + y51, y2 + y61, y3 + y61, All variables = 0 or 1 y4 + y71, y5 + y71 34b. max 7 - z1 - z2 - z3 - z4 - z5 - z6 - z7 or min z1 + z2 + z3 + z4 + z5 + z6 + z7 st z2 + z31, z3 + z41, z1 + z51, z2 + z61, z3 + z61, z4 + z71, z5 + z71 All variables = 0 or 1 This is indeed a set covering problem. The idea is that we are now minimizing the number of conventional ladders that are not replaced, and an alarm box is "covered" if at least one of the two closest ladders is not replaced. 35. Let xi = tons of steam produced on boiler i, yi = tons of steam processed on turbine j, ti = 1 if turbine i is used, ti = o otherwise, bi = 1 if boiler i is used, bi = 0 otherwise. min z = 10x1 + 8x2 + 6x3 + 2y1 + 3y2 + 4y3 st 4y1 + 5y2 + 6y38000 y1 + y2 + y3 = x1 + x2 + x3 x1500b1, x2300b2, x3400b3 x11000b1, x2900b2, x3800b3 y1300t1, y2500t2, y3600t3 y1600t1, y2800t2, y3900t3 13 All bi's and ti's = 0 or 1 All other variables 0 36. Let Ai = 1 if subsidiary i is grouped alone, Ai = 0 otherwise. Aij = 1 if subsidiaries i and j are grouped together, Aij = 0 otherwise. A123 = 1 if all three subsidiaries are grouped together, A123 = 0 otherwise min z = 35A1 + 80A2 + 100A3 + 75A12 + 180A23 + 170A13 + 250A123 st A1 + A12 + A13 + A123 = 1 A2 + A12 + A23 + A123 = 1 A3 + A23 + A13 + A123 = 1 All variables = 0 or 1 37. Let Xij = number of times type i room is used to meet type j requests. (i=1 is 50 seat room;i=2 is 100 seat room;i=3 is 150 seat room;i=4 is unfllfilled request. Then the correct IP is on LINDO output. Variables automatically came out integer so GIN command was omitted. Section 9.2 Problem 37 Printout MIN 2 X21 + 4 X31 + 300 X41 + 300 X42 + X33 + 100 X43 + 2 X24 + 4 X34 + 200 X44 SUBJECT TO 2) X14 <= 2 3) X14 + X11 <= 2 4) X11 <= 2 5) X11 <= 2 6) X24 <= 1 7) X21 + X24 <= 1 8) X21 <= 1 9) X23 <= 1 10) X34 + X32 <= 1 11) X31 + X34 + X32 <= 1 12) X31 + X32 <= 1 13) X31 <= 1 14) X33 <= 1 15) X21 + X31 + X41 + X11 = 3 16) X42 + X32 = 1 17) X33 + X43 + X23 = 1 18) X24 + X34 + X44 + X14 = 2 END LP OPTIMUM FOUND AT STEP 10 OBJECTIVE FUNCTION VALUE 1) 402.0000 VARIABLE VALUE X21 1.000000 X31 0.000000 X41 0.000000 X42 0.000000 X33 0.000000 X43 0.000000 X24 0.000000 X34 0.000000 X44 2.000000 X14 0.000000 X11 2.000000 X23 1.000000 X32 1.000000 REDUCED COST 0.000000 4.000000 0.000000 0.000000 1.000000 100.000000 100.000000 104.000000 0.000000 100.000000 0.000000 0.000000 0.000000 14 ROW 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17) 18) SLACK OR SURPLUS DUAL PRICES 2.000000 0.000000 0.000000 300.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 298.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 300.000000 0.000000 0.000000 1.000000 0.000000 1.000000 0.000000 0.000000 -300.000000 0.000000 -300.000000 0.000000 0.000000 0.000000 -200.000000 NO. ITERATIONS= 10 38. Let Xij = 1 if size i box is used to meet demand for type i,i+1,...j boxes. Let Yi = 1 if type i box is used at all and Yi = 0 if type i box is not used. Answer is on LINDO printout (also see file C9-2-38). Section 9.2 Problem 38 Printout MIN 13200 X11 + 9900 X12 + 16500 X13 + 23100 X14 + 6600 X15 15 + 13200 X16 + 6600 X17 + 9000 X22 + 15000 X23 + 21000 X24 + 6000 X25 + 12000 X26 + 6000 X27 + 13000 X33 + 18200 X34 + 5200 X35 + 10400 X36 + 5200 X37 + 16800 X44 + 4800 X45 + 9600 X46 + 4800 X47 + 3800 X55 + 7600 X56 + 3800 X57 + 7200 X66 + 3600 X67 + 3400 X77 + 1000 Y1 + 1000 Y2 + 1000 Y3 + 1000 Y4 + 1000 Y5 + 1000 Y6 + 1000 Y7 SUBJECT TO 2) X11 = 1 3) X12 + X22 = 1 4) X13 + X23 + X33 = 1 5) X14 + X24 + X34 + X44 = 1 6) X15 + X25 + X35 + X45 + X55 = 1 7) X16 + X26 + X36 + X46 + X56 + X66 = 1 8) X17 + X27 + X37 + X47 + X57 + X67 + X77 = 1 9) X11 - Y1 <= 0 10) X12 - Y1 <= 0 11) X13 - Y1 <= 0 12) X14 - Y1 <= 0 13) X15 - Y1 <= 0 14) X16 - Y1 <= 0 15) X17 - Y1 <= 0 16) X22 - Y2 <= 0 17) X23 - Y2 <= 0 18) X24 - Y2 <= 0 19) X25 - Y2 <= 0 20) X26 - Y2 <= 0 21) X27 - Y2 <= 0 22) X33 - Y3 <= 0 23) X34 - Y3 <= 0 24) X35 - Y3 <= 0 25) X36 - Y3 <= 0 26) X37 - Y3 <= 0 27) X44 - Y4 <= 0 28) X45 - Y4 <= 0 29) X46 - Y4 <= 0 30) X47 - Y4 <= 0 31) X55 - Y5 <= 0 32) X56 - Y5 <= 0 33) X57 - Y5 <= 0 34) X66 - Y6 <= 0 35) X67 - Y6 <= 0 36) X77 - Y7 <= 0 END INTEGER 35 OBJECTIVE FUNCTION VALUE 1) 72100.00 VARIABLE VALUE REDUCED COST X11 1.000000 13200.000000 X12 1.000000 9900.000000 X13 0.000000 16500.000000 X14 0.000000 23100.000000 16 X15 X16 X17 X22 X23 X24 X25 X26 X27 X33 X34 X35 X36 X37 X44 X45 X46 X47 X55 X56 X57 X66 X67 X77 Y1 Y2 Y3 Y4 Y5 Y6 Y7 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 1.000000 1.000000 1.000000 0.000000 0.000000 0.000000 1.000000 0.000000 1.000000 1.000000 1.000000 0.000000 0.000000 6600.000000 13200.000000 6600.000000 9000.000000 15000.000000 21000.000000 6000.000000 12000.000000 6000.000000 13000.000000 18200.000000 5200.000000 10400.000000 5200.000000 16800.000000 4800.000000 9600.000000 4800.000000 3800.000000 7600.000000 3800.000000 7200.000000 3600.000000 3400.000000 1000.000000 1000.000000 1000.000000 1000.000000 1000.000000 1000.000000 1000.000000 39a. W1 W2 W3 C1 C2 C3 C4 DUMMY -------------------------------------------------------P1 800 1000 1200 M M M M 0 300 -------------------------------------------------------P2 700 500 700 M M M M 0 200 --------------------------------------------------------P3 800 600 500 M M M M 0 300 17 ---------------------------------------------------------500 600 700 M M M M 0 200 P4 ------------------------------------------------------------700 600 500 M M M M 0 P5 400 -----------------------------------------------------------0 M M 40 80 90 50 0 1400 W1 -------------------------------------------------------------M 0 M 70 40 60 80 0 W2 --------------------------------------------------------------W3 M M 0 80 30 50 60 0 1400 --------------------------------------------------------------1400 1400 1400 200 300 150 250 500 Section 9.2 Problem 39 MIN 35Y1+45Y2+40Y3+42Y4+30Z1+40Z2+30Z3+40Y5+.8P11+P12+1.2P13 +.7P21+.5P22+.7P23+.8P31+.6P32+.5P33+.5P41+.6P42+.7P43+.7P51 +.6P52+.5P53+.04W11+.08W12+.09W13+.05W14+.07W21+.07W22+.06W23+.08W24 +.08W31+.03W32+.05W33+.06W34 SUBJECT TO -300Y1+P11+P12+P13<0 -200Y2+P21+P22+P23<0 -300Y3+P31+P32+P33<0 -200Y4+P41+P42+P43<0 -400Y5+P51+P52+P53<0 P11+P21+P31+P41+P51-W11-W12-W13-W14=0 P12+P22+P32+P42+P52-W21-W22-W23-W24=0 P13+P23+P33+P43+P53-W31-W32-W33-W34=0 -900Z1+W11+W12+W13+W14<0 -900Z2+W21+W22+W23+W24<0 -900Z3+W31+W32+W33+E34<0 W11+W21+W31>200 W12+W22+W32>300 W13+W23+W33>150 W14+W24+W34>250 END INTEGER 8 1) 671.5000 VARIABLE VALUE Y1 0.000000 Y2 0.000000 Y3 1.000000 Y4 1.000000 Z1 1.000000 Z2 0.000000 REDUCED COST 35.000000 45.000000 40.000000 40.000000 30.000000 40.000000 1400 18 Z3 Y5 P11 P12 P13 P21 P22 P23 P31 P32 P33 P41 P42 P43 P51 P52 P53 W11 W12 W13 W14 W21 W22 W23 W24 W31 W32 W33 W34 E34 1.000000 1.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 300.000000 200.000000 0.000000 0.000000 0.000000 0.000000 400.000000 200.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 300.000000 150.000000 250.000000 0.000000 30.000000 40.000000 0.290000 0.510000 0.700000 0.190000 0.010000 0.200000 0.290000 0.110000 0.000000 0.000000 0.120000 0.210000 0.190000 0.110000 0.000000 0.000000 0.060000 0.050000 0.000000 0.010000 0.030000 0.000000 0.010000 0.030000 0.000000 0.000000 0.000000 0.000000 39b. Let Pij = Number of tons produced at plant i and shipped to warehouse j, Wij = Number of tons sent from warehouse i to customer j, Yi = 1 if plant i is used, Yi = 0 otherwise, Zi = 1 if warehouse i is used, Zi = 0 otherwise. Then optimal solution is on LINDO printout. 40. Let T7 = Number of TA7 cables built, T8 = Number of TA8 cables built, SAT = Number of satellites sent up, FC = French circuits via cable, FS = French circuits via satellite, GC = German circuits via cable, GS = German circuits via satellite, SC = Swiss circuits via cable, SS = Swiss circuits via satellite, UC = UK circuits via cable, US = UK circuits via satellie. All circuit variables are in thousands, EF = Earth stations in France, EG = Earth stations in Germany, ES = Earth stations in Switzerland,, EU = Earth stations in UK, EST = total earth stations. The LINDO printout contains the optimal solution. Section 9.2 Problem 9.40 MIN 1600000 T7 + 2300000 T8 + 3000000 SAT + 6.2 GC + 5.8 SC + 120 EST SUBJECT TO 2) FC + FS = 20 3) SC + SS = 16 4) UC + US = 60 5) - 140 SAT + FS + SS + US + GS <= 0 6) - 8.5 T7 - 37.8 T8 + GC + SC + FC + UC <= 0 7) - EF - EG - ES - EU + EST = 0 19 8) - 0.19 EF + FS <= 0 9) - 0.19 EG + GS <= 0 10) - 0.19 ES + SS <= 0 11) - 0.19 EU + US <= 0 12) GC + GS = 60 END GIN T7 GIN T8 GIN EF GIN EG GIN ES GIN EU GIN SAT Optimal solution is z= 5374760 T8=1, EG=316, ES=84, EU=223, SAT=1, EST=623, FC=20, SS=15.96, UC=17.63, US=42.37, GS=60. All other variables = 0. 41. Let Ni = Number of reps assigned to district i, Xij = number of calls made to district j by reps assigned to district i, and Yi = 1 if any reps are assigned to district i, Yi=0 otherwise. See LINDO printout for solution. Section 9.2 Problem 41 TITLE SECT 9-2 PROBLEM 41 MIN 80 N1 + 80 N2 + 80 N3 + 80 N4 + 88 Y1 + 88 Y2 + 88 Y3 + 88 Y4 SUBJECT TO 2) - 160 N1 + X11 + 4 X12 + 5 X13 + 7 X14 <= 0 3) - 160 N2 + 4 X21 + X22 + 3 X23 + 5 X24 <= 0 4) - 160 N3 + 5 X31 + 3 X32 + X33 + 2 X34 <= 0 5) - 160 N4 + 7 X41 + 5 X42 + 2 X43 + X44 <= 0 6) X11 + X21 + X31 + X41 >= 50 7) X12 + X22 + X32 + X42 >= 80 8) X13 + X23 + X33 + X43 >= 100 9) X14 + X24 + X34 + X44 >= 60 10) N1 - 15 Y1 <= 0 11) N2 - 15 Y2 <= 0 12) N3 - 15 Y3 <= 0 13) N4 - 15 Y4 <= 0 END GIN N1 GIN N2 GIN N3 GIN N4 SUB Y1 1.00000 INTE Y1 SUB Y2 1.00000 INTE Y2 SUB Y3 1.00000 INTE Y3 SUB Y4 1.00000 INTE Y4 OBJECTIVE FUNCTION VALUE 1) 488.0000 20 VARIABLE VALUE N1 0.000000 N2 0.000000 N3 5.000000 N4 0.000000 Y1 0.000000 Y2 0.000000 Y3 1.000000 Y4 0.000000 X11 0.000000 X12 0.000000 X13 0.000000 X14 0.000000 X21 0.000000 X22 0.000000 X23 0.000000 X24 0.000000 X31 50.000000 X32 80.000000 X33 100.000000 X34 60.000000 X41 0.000000 X42 0.000000 X43 0.000000 X44 0.000000 REDUCED COST 80.000000 80.000000 80.000000 80.000000 88.000000 88.000000 88.000000 88.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 42. See EXCEL file S9_42.xls for duration calculations. For i=1,2,...6 let Xi = number of type i bonds purchased, X7 = number of t-bills purchased, Yi = 1 if any type i bonds are purchased, Yi = 0 otherwise. See LINDO printout for solution. Note: The constraint in row 9 ensures that the duration of the bond portfolio "matches" the duration of the payment stream. 21 Section 9.2 Problem 42 Solution MIN 500 Y1 + 500 Y2 + 500 Y3 + 500 Y4 + 500 Y5 + 500 Y6 + 5 X1 + 5 X2 + 5 X3 + 5 X4 + 5 X5 + 5 X6 SUBJECT TO 2) 872 X1 + 783 X2 + 1184 X3 + 759 X4 + 1000 X5 + 1020 X6 + 980 X7 = 251780 3) - 100 Y1 + X1 <= 0 4) - 100 Y2 + X2 <= 0 5) - 100 Y3 + X3 <= 0 6) - 100 Y4 + X4 <= 0 7) - 100 Y5 + X5 <= 0 8) - 100 Y6 + X6 <= 0 9) 6372 X1 + 5308 X2 + 7630 X3 + 6577 X4 + 6759 X5 + 6682 X6 + 245 X7 = 1125456.7 END TITLE SECTION 9-2 PROBLEM 42 INTE 6 OBJECTIVE FUNCTION VALUE 1) 1752.8130 VARIABLE VALUE REDUCED COST Y1 .000000 500.000000 Y2 .000000 500.000000 Y3 1.000000 436.626200 Y4 .000000 500.000000 Y5 1.000000 500.000000 Y6 .000000 500.000000 X1 .000000 .272699 X2 .000000 1.072937 X3 100.000000 .000000 X4 .000000 .093524 X5 50.562550 .000000 X6 .000000 .062990 X7 84.507610 .000000 43. Let xij = 1 if Plant i is used to produce type j cars (type 1 = Taurus, etc.) and let xij = 0 otherwise. Also let yj = 1 if plant j is used to produce any type of car, yj = 0 otherwise. Then expressing our objective function in billions of dollars we obtain min z = 7y1 + 6y2 + 4y3 + 2y4 + 6x11 + 8x12 + 4.5x13 + 7.5x21 + 9x22 + 5.5x23 + 8.5x31 + 9.5x32 + 6x33 + 9.5x41 + 11x42 + 7x43 st x11 + x21 + x31 + x41 = 1 x12 + x22 + x32 + x42 = 1 x13 + x23 + x33 + x43 = 1 x1jy1, j = 1,2,3 x2jy2 j = 1,2,3 x3jy3 j = 1,2,3 x4jy4 j = 1,2,3 22 x11+x12+x13 1 x21+x22+x231 x31+x32+x331 x41+x42+x431 1 - y1z y3 + y4 - 1 1 - z All variables = 0 or 1 44. Use the following LINGO model: MODEL: SETS: PROJECTS/1..10/:NPV,DO; YEARS/1..3/:AVAIL; YP(PROJECTS,YEARS):CASH; ENDSETS MAX=@SUM(PROJECTS:DO*NPV); @FOR(YEARS(J):@SUM(PROJECTS(I):DO(I)*CASH(I,J))<AVAIL(J)); @FOR(PROJECTS:@BIN(DO)); DATA: NPV=20, 30, 40, 50, 60, 70, 80, 90, 100, 130; AVAIL= 80, 60, 70; CASH = 6, 3, 5, 9, 5, 7, 12, 7, 9, 15, 9, 12, 18, 11, 12, 21, 13, 14, 24, 15, 16, 27, 17, 11, 30, 19, 20, 35, 21, 24; ENDDATA END Optimal solution is to do projects 1,2,9, and 10. 45. LINGO formulation is as follows: MODEL: SETS: PRODUCTS/1..4/:MADE,DEMAND,PRICE, FC,DO; RESOURCES/1..3/:AVAIL; RESPRO(RESOURCES,PRODUCTS):USAGE; ENDSETS MAX=@SUM(PRODUCTS(I):PRICE(I)*MADE(I)-FC(I)*DO(I)); @FOR(PRODUCTS(I):MADE(I)<=DEMAND(I)); @FOR(RESOURCES(I):@SUM(PRODUCTS(J):MADE(J)*USAGE(I,J))<=AVAIL(I)); @FOR(PRODUCTS(I):MADE(I)<=1000*DO(I)); @FOR(PRODUCTS(I):@BIN(DO(I))); DATA: FC= 30, 40, 50,60; PRICE= 2, 5, 6, 7; AVAIL=40, 60, 80; DEMAND= 40, 60,65, 70; USAGE= 1,2,3.5,4, 5,6,7,9, 3,4,5,6; ENDDATA 23 END Optimal solution is to make just 10 units of product 2. SECTION 9.3 1. 2. We wish to solve min z = 50x 1 + 100x 2 st 7x1 + 2x228 2x1 + 12x 224 x1, x20 SP 1 24 25 Note: In solving subproblems we have used the result discussed in Problem 8 to conclude that in each subproblem's optimal solution the "newest" constraint must be binding. Thus we know that some optimal solution to SP2 will have x 1 = 4. The two optimal solutions x 1 = 6, x2 = 1 and x 1 = 4 , x 2 = 2 have been found. 3. 26 4. 27 6. 7. 8. Choose c (0<c<1) so that NEWSOL = c(SOL0) + (1 - c)SOL1 has x 1 = i. Then NEWSOL satisfies all of SP 1's constraints and z-value for NEWSOL - z-value for SOL1 = c(z-value for SOL0) - z-value for SOL1)0. This follows because SOL0 is optimal for SP 0 and SOL1 may not be optimal for SP 0. Thus if SOL1 is optimal for SP 1 , then NEWSOL is also optimal for SP 1, and NEWSOL has x 1 = i, as desired. This result shows that when we branch on a constraint like x 12, some optimal solution to the new subproblem will have x1 = 2. This greatly simplifies the solution of the subproblems. 9a. Letting It = inventory at end of period t yields given LINDO printout (20 branches). Section 9.3 Problem 9a MIN 250 Y1 + 250 Y2 + 250 Y3 + 250 Y4 + 250 Y5 + 2 X1 + 2 X2 + 2 X3 + 2 X4 + 2 X5 + I1 + I2 + I3 + I4 + I5 28 SUBJECT TO 2) - 1270 Y1 + X1 <= 0 3) - 1270 Y2 + X2 <= 0 4) - 1270 Y3 + X3 <= 0 5) - 1270 Y4 + X4 <= 0 6) - 1270 Y5 + X5 <= 0 7) - X1 + I1 = - 220 8) - X2 - I1 + I2 = - 280 9) - X3 - I2 + I3 = - 360 10) - X4 - I3 + I4 = - 140 11) - X5 - I4 + I5 = - 270 END INTE 5 OBJECTIVE FUNCTION VALUE 1) 3680.000 VARIABLE VALUE REDUCED COST Y1 1.000000 250.000000 Y2 1.000000 250.000000 Y3 1.000000 250.000000 Y4 0.000000 -1020.000000 Y5 1.000000 250.000000 X1 220.000000 0.000000 X2 280.000000 0.000000 X3 500.000000 0.000000 X4 0.000000 0.000000 X5 270.000000 0.000000 I1 0.000000 1.000000 I2 0.000000 1.000000 I3 140.000000 0.000000 I4 0.000000 2.000000 I5 0.000000 3.000000 9b. See attached LINDO printout . (LP solution yields optimal solution!! Section 9.3 Problem 9b Printout MIN 250 Y1 + 250 Y2 + 250 Y3 + 250 Y4 + 250 Y5 + 2 X11 + 3 X12 + 4 X13 + 5 X14 + 6 X15 + 2 X22 + 3 X23 + 4 X24 + 5 X25 + 2 X33 + 3 X34 + 4 X35 + 2 X44 + 3 X45 + 2 X55 SUBJECT TO 2) X11 = 220 3) X12 + X22 = 280 4) X13 + X23 + X33 = 360 5) X14 + X24 + X34 + X44 = 140 6) X15 + X25 + X35 + X45 + X55 = 270 7) - 220 Y1 + X11 <= 0 8) - 280 Y1 + X12 <= 0 9) - 360 Y1 + X13 <= 0 10) - 140 Y1 + X14 <= 0 11) - 270 Y1 + X15 <= 0 12) - 280 Y2 + X22 <= 0 13) - 360 Y2 + X23 <= 0 14) - 140 Y2 + X24 <= 0 15) - 270 Y2 + X25 <= 0 16) - 360 Y3 + X33 <= 0 29 17) - 140 Y3 + X34 <= 18) - 270 Y3 + X35 <= 19) - 140 Y4 + X44 <= 20) - 270 Y4 + X45 <= 21) - 270 Y5 + X55 <= END INTE 5 0 0 0 0 0 OBJECTIVE FUNCTION VALUE 1) 3680.000 VARIABLE VALUE Y1 1.000000 Y2 1.000000 Y3 1.000000 Y4 0.000000 Y5 1.000000 X11 220.000000 X12 0.000000 X13 0.000000 X14 0.000000 X15 0.000000 X22 280.000000 X23 0.000000 X24 0.000000 X25 0.000000 X33 360.000000 X34 140.000000 X35 0.000000 X44 0.000000 X45 0.000000 X55 270.000000 REDUCED COST 250.000000 -30.000000 250.000000 110.000000 250.000000 0.000000 0.000000 2.000000 2.000000 4.000000 0.000000 1.000000 1.000000 3.000000 0.000000 0.000000 2.000000 0.000000 1.000000 0.000000 9c. Part (b) is faster. 9d. There are many fractional solutions for the Yi's that are feasible in LP relaxation for (9a) but not for (9b). For example, 1/1270<Y1<1/220 could yield a feasible solution with X11>0 for part (9a) but not for part (9b).Thus (9a) has a "bigger" feasible region to search and will take longer than (9b). Actually it has been shown that when solved as an LP the solution to (9b) will always yield integer values for the variables. SECTION 9.4 1. 2. Solving the LP relaxation yields z = 2, x1 = 0, x2 = 2. This solution has x 1 integer, so it is an optimal solution. 30 3. SECTION 9.5 1. We might choose up to eight "Item 1's" so we define items 1', 2', ... 8' all identical to Item 1. We might choose up to six Item 2's, so we define items 9', 10',... 14' all identical to Item 2. Finally, we might choose up to five Item 3's, so we define items 15', 16',... 19' all identical to Item 3. Letting xi = 1 if item i' is chosen and xi = 0 otherwise yields a 0-1 knapsack problem. 2. Let item 1 = bedroom set,... item 5 = TV set. Letting x i = 1 if item i is chosen and x i = 0 otherwise yields the following knapsack problem: max z = 60x 1 + 48x 2 + 14x 3 + 31x 4 + 10x 5 st 800x 1 + 600x 2 + 300x3 + 400x 4 + 200x51100 xi = 0 or 1 From the following tree we find the optimal solution to be z = 79, x 2 = x4 =1, x 1 = x3 = x 5 = 0. 31 3. Letting x i = 1 if item i is chosen and x i = 0 otherwise yields the following knapsack problem: max z = 5x 1 + 8x2 + 3x3 + 7x4 32 st 3x1 + 5x 2 + 2x3 + 4x 46 xi = 0 or 1 We obtain the following tree (for each subproblem any omitted variable equals 0): Note that since optimal objective function value for any candidate solution associated with a branch must be an integer, SP 3 can at best yield a z-value of 10, so we need not branch on SP 3. Thus the optimal solution is z = 10, x 1 = x2 = 0, x3 = x4 = 1. SECTION 9.6 1. In tree we jumptrack by branching on best bound. Thus x 34 = 1 branch is "followed" first. This yields us to a feasible solution having (D = total delay) D = 22 minutes. Branching on the x 44 = 1 branch yields a feasible solution having D = 20. All other branches must have D20, so we have found an optimal sequence. Job 1, then job 2, then job 3 and then job 4 is optimal (it has a total delay of 20 minutes. Job 2 then job 1, then job 3, and then job 4 is also optimal. 33 2. Let LFR = City 1, LFP = City 2, LR = City 3, and LP = City 4. Then we obtain the following branch and bound tree: 1-2-4-3-1 with total setup time of 330 minutes is optimal. Thus we should produce gasolines in order LFR-LFP-LP-LR-LFR. 3. Add a city 1' and arcs (2,1'), (3,1'), and (4,1'). Cost of arc from (i,1') = cost of arc from (i,1). Now find shortest Hamiltonian path from 1 to 1'. 4a. This is a minimum spanning tree problem. The solution would be to connect the following pairs of pins: 1 and 2, 1 and 3, and 1 and 4. 34 4b. The solution in problem 4a has three wires touching pin 1, so it can no longer be optimal. We may find the correct solution by solving the following TSP: 0 1 2 3 4 0 0 0 0 0 0 1 0 0 1 2 2 2 0 1 0 3 2.9 3 0 2 3 0 3 4 0 2 2.9 3 0 Observe that we will never have more than two wires touching any pin. 5a. Let's start at LFR = 1. Then we go to 2, then to 4, and finally to 3. Again we obtain 1-2- 4-3-1 which is the minimum cost path. 5b. Begin with (1,2)-(2,1). Arc Replaced Replacement Arcs Added Cost (1,2) (1,3)-(3,2) 120+130-50 = 200 (1,2) (1,4)-(4,2) 140+120-50 = 210 (2,1) (2,3)-(3,1) 140+90-60 = 170* (2,1) (2,4)-(4,1) 110+130-60 = 180 We now have the subtour (1,2)-(2,3)-(3,1). Arc Replaced Replacement Arcs Added Cost (1,2) (1,4)-(4,2) 140+120-50 = 210 (2,3) (2,4)-(4,3) 110+80-140 = 50* (3,1) (3,4)-(4,1) 60+130-90 = 100 We now have obtained the optimal tour 1-2-4-3-1 with length 330 minutes. 6. Let location 6 = packaging area and d ij = distance between locations i and j. Then we have a 6 city TSP with a distance matrix having d ij as its i-j'th element. To find out the order in which clothes should be picked up start at location 6. 7. Let x ij = 1 if there is a queen in row i and column j, and x ij = 0 otherwise. The entry "OK" indicates that for this node, the queens we have already placed cannot capture each other. "x" indicates a node where the queens already placed can capture each other. Following the "OK" entries yields the following solution: X X X X 35 8. Factory 1, W1 node means that Warehouse 1 is used to serve Factory 1, etc. The W1-Factory 1 and W4-Factory 2 path yields an optimal assignment: Warehouse 1 to Factory 1, Warehouse 4 to Factory 2, Warehouse 2 to Factory 3, Warehouse 3 to Factory 4, and Warehouse 5 to Factory 5. A total cost of $35,000 is incurred. 9. The waste associated with cutting Sheet i and then Sheet j is the i-j'th entry in the following matrix: 36 Sheet 1 Sheet 2 Sheet 3 Sheet 1 M 0.7 0.5 Sheet 2 0.5 M 0.4 Sheet 3 0.8 0.9 M Sheet 4 0.4 0.5 0.3 Beginning 0.3 0.4 0.2 Sheet 4 Beginning 0.0 0.3 0.9 0.2 0.2 0.5 M 0.1 0.7 M For example, if we cut Sheet 2 and then Sheet 3, Sheet 2 ends 0.8 yds. from beginning of roll, so we must go ( 1 - 0.8) + (0.2) = 0.4 yds. from end of Sheet 2 to beginning of Sheet 3. Solving the above problem as an assignment problem we find an optimal solution z = 1.7, x13 = x 34 = x25 = x42 = x51 = 1. Since this solution has no subtours, it is optimal. Thus cutting Sheet 1, then Sheet 3, then Sheet 4, then Sheet 2, and then making a cut to bring us back to the beginning of the roll will waste 1.7 yards. 10. The distance matrix is given by 1 2 3 4 5 1 0 51/2 171/2 6 501/2 2 51/2 0 2(2)1/2 17 1/2 291/2 1/2 1/2 3 17 2(2) 0 51/2 3 1/2 1/2 4 6 17 5 0 21/2 1/2 1/2 5 50 29 3 21/2 0 Using LINGO we found optimal drilling order to be 1-2-4-5-3-1 with total length 14.89. 11. Utilizing branch and bound we find Job 1 should be done first, Job 2 2nd, Job 3 3rd, and Job 4 last. Total penalty cost = 47. SECTION 9.7 1. From the tree we find that the optimal solution to the IP is z = 4, x1 = x2 = x4 = x5 = 1, x 3 = 0. 37 2. The best completion of node 1 ( x 1 = x3 = 1, x2 = 0, z =3) is feasible, and is therefore optimal. 3. Let xi = 1 if project i is chosen and xi = 0 otherwise. Then the appropriate IP is: max z = 5x 1 + 9x2 + 6x3 + 3x4 + 2x5 st 4x 1 + 6x2 + 5x 3 + 4x 4 + 3x 510 x1 + x 21, x3 + x41, x2x5 All x i = 0 or 1 From the tree we find an optimal solution is z = 11, x1 = x3 = 1, x2 = x4 = x 5 =0. 4. From the tree we find an optimal solution to be z = 2, x 2 = x4 = 1, x 1 = x3 = x 5 = x6 = 0. 38 5. The x 1 = 0 branch has no feasible completion. All nodes descending from the x1 = 1 branch are eventually found to be infeasible. An easy way to see that the problem is infeasible is to note that all four guards must play. Their rebounding total is 6. No matter who fills out the starting lineup, the highest rebounding total that can be attained is 9, which is less than 5(2) = 10. 6. Suppose two different representations exist. Then un'2n + un-1'2n-1 +...+ 2u1' + u0' = un2n + un-12n-1 +...+ 2u1 + u0 and for some k uk/=uk' but um = um' holds for m>k. For simplicity, suppose that uk = 1 and uk' = 0 (if uk = 0 and uk' = 1 similar reasoning applies). Then we must have (1) 2k + uk-12k-1 + ... + 2u1 + u0 = uk-1'2k-1 + ... + 2u1' + u0'. Note, however, that uk-1'2k-1 + ... + 2u1' + u0' 2k-1 + 2k-2 +...1 = 2k - 1<2 k1ssssssss. This last inequality shows that the right hand side of (1) must be less than the left -hand side of (1), so (1) cannot hold. This 39 contradiction proves that with binary notation there cannot be two different representations of an integer. Section 9.8 1. Since both constraints have a fractional part of 1/2 in the optimal tableau, we arbitrarily choose to use the first constraint to yield the cut: x2 + 7s1/22 + s2/22 = 3 + 1/2 or x2 - 3 = 1/2 - 7s1/22 - s2/22 or 1/2 - 7s1/22 - s2/220. Adding this constraint with a slack variable s 3 yields the following tableau: z x 1 x 2 s1 s2 s3 RHS ____________________________________________ 1 0 0 56/11 30/11 0 126 ____________________________________________ 0 0 1 7/22 1/22 0 7/2 ____________________________________________ 0 1 0 -1/22 3/22 0 9/2 ____________________________________________ 0 0 0 -7/22 -1/22 1 -1/2 ____________________________________________ The dual simplex pivoting rule indicates that s1 should enter in row 3 yielding the following tableau: z x 1 x 2 s1 s2 s3 RHS ____________________________________________ 1 0 0 0 2 16 118 ____________________________________________ 0 0 1 0 0 1 3 ____________________________________________ 0 1 0 0 1/7 -1/7 32/7 ____________________________________________ 0 0 0 1 1/7 -22/7 11/7 ____________________________________________ We now arbitrarily choose row 2 to generate the next cut: x1 + s2/7 - s3 + 6s3/7 = 4/7 + 4 or x1 - s3 -4 = 4/7 - s2/7 -6s3/7 yielding the cut - s2/7 - 6s3/7-4/7. Adding a slack variable s4 to the cut yields the following tableau: z x 1 x 2 s1 s2 s3 s4 RHS ____________________________________________________ 1 0 0 0 2 16 0 118 ____________________________________________________ 0 0 1 0 0 1 0 3 ____________________________________________________ 0 1 0 0 1/7 -1/7 0 32/7 _____________________________________________________ 0 0 0 1 1/7 -22/7 0 11/7 ______________________________________________________ 0 0 0 0 -1/7 -6/7 1 -4/7 _____________________________________________________ Entering s2 in the last constraint yields the following (optimal) tableau. z x 1 x 2 s1 s2 s3 s4 RHS 40 ____________________________________________________ 1 0 0 0 0 4 14 110 ____________________________________________________ 0 0 1 0 0 1 0 3 ____________________________________________________ 0 1 0 0 0 -1 1 4 _____________________________________________________ 0 0 0 1 0 -4 1 1 ______________________________________________________ 0 0 0 0 1 6 -7 4 _____________________________________________________ This tableau yields the optimal solution z = 110 x1 = 4 x 2 = 3. 2. Since the second constraint has the fractional part closest to 1/2, we use it to generate the cut: x 2 + e1/5 - e2 + 2e2/5 = 1 + 3/5 or x2 - e2 - 1 = 3/5 - 2e2/5 - e1/5. This yields the cut 3/5 - 2e2/5 - e1/50. Adding a slack variable s3 yields the following tableau: z x1 x2 e1 e2 s3 RHS _______________________________________________________________ 1 0 0 -4/5 -18/5 0 88/5 _______________________________________________________________ 0 1 0 -2/5 1/5 0 4/5 _______________________________________________________________ 0 0 1 1/5 -3/5 0 8/5 _______________________________________________________________ 0 0 0 -1/5 -2/5 1 -3/5 _______________________________________________________________ We now enter e1 into the basis yielding the following optimal tableau: z x1 x2 e1 e2 s3 RHS _______________________________________________________________ 1 0 0 0 -2 -4 20 _______________________________________________________________ 0 1 0 0 1 -2 2 _______________________________________________________________ 0 0 1 0 -1 1 1 _______________________________________________________________ 0 0 0 1 2 -5 3 _______________________________________________________________ This tableau yields the optimal solution z = 20, x 1 = 2, x 2 = 1. 3. From row 2 we obtain the following cut: x2 + s1/3 + s2/6 = 2 + 1/2 or x2 - 2 = 1/2 - s1/3 - s2/6 or adding a slack s3 yields the cut 1/2 - s1/3 - s2/6 + s30. We now obtain the following tableau: z x 1 x 2 s1 s 2 s 3 RHS _____________________________________________ 1 0 0 -2/3 -5/6 0 -15/2 _____________________________________________ 0 1 0 1/3 -1/12 0 5/4 _____________________________________________ 41 0 0 1 1/3 1/6 0 5/2 _____________________________________________ 0 0 0 -1/3 -1/6 1 -1/2 _____________________________________________ Entering s1 into the basis yields the following tableau z x 1 x 2 s1 s 2 s 3 RHS _____________________________________________ 1 0 0 0 -1/2 -2 -13/2 _____________________________________________ 0 1 0 0 -1/4 1 3/4 _____________________________________________ 0 0 1 0 0 1 2 _____________________________________________ 0 0 0 1 1/2 -3 3/2 _____________________________________________ From the last constraint we obtain a cut as follows: s1 + s2/2 - 3s3 = 1 + 1/2 or s1 - 3s3 - 1 = 1/2 - s2/2 or 1/2 - s2/20 or adding a slack variable s4, 1/2 - s2/2 + s4 = 0. Adding this constraint we obtain the following tableau: z x 1 x 2 s1 s2 s3 s4 RHS _________________________________________________________ 1 0 0 0 -1/2 -2 0 -13/2 _________________________________________________________ 0 1 0 0 -1/4 1 0 3/4 _________________________________________________________ 0 0 1 0 0 1 0 2 _________________________________________________________ 0 0 0 1 1/2 -3 0 3/2 _________________________________________________________ 0 0 0 0 -1/2 0 1 -1/2 __________________________________________________________ Entering s2 we obtain the following optimal tableau: z x 1 x 2 s1 s2 s3 s4 RHS _________________________________________________________ 1 0 0 0 0 -2 -1 -6 _________________________________________________________ 0 1 0 0 0 1 -1/2 1 _________________________________________________________ 0 0 1 0 6 1 0 2 _________________________________________________________ 0 0 0 1 0 -3 1 1 _________________________________________________________ 0 0 0 0 1 0 -2 1 __________________________________________________________ This tableau yields the optimal solution z = -6, x1 = 1, x 2 = 2. ...
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This homework help was uploaded on 04/02/2008 for the course IEOR 160 taught by Professor Hochbaum during the Fall '07 term at University of California, Berkeley.

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