Chapter 6 - Integer Linear Programming : S-14

Chapter 6 - Integer Linear Programming : S-1 ———————————————————————————————————————————— Chapter 6 Integer Linear Programming

1. The optimal solutions to an LP problem occur at extreme points of the feasible region. There are a finite number of such extreme points for any given problem and they can be located and searched very efficiently. The optimal solution to an ILP does not necessarily occur at an extreme point of the relaxed feasible region, making it more difficult to locate these points.

2. M12 = MIN( 200/1, 1520/9, 2650/12) = 168.9 M22 = MIN( 200/1, 1520/6, 2650/16) = 165.7

3. Errata: The objective in this problem should be MIN rather than MAX. See file: Prb6_3.xlsm a. X1=6.25, X2=6.5 b. X1=8, X2=8 c. The optimal integer solution to an ILP cannot, in general, be obtained by rounding the continuous solution.

4. a. X1 + X2 + X4 + X6 = 2 b. X2 - X3 = 0 c. 2X5 - X3 - X4 £ 0 d. X2 + X4 - X5 £ 1

5. a. See file: Prb6_5.xlsm b. Select projects 2, 3 & 5. Total NPV = $573,000.

6. a. It would merely serve to reduce his net profit by $1,000. Since he must incur this cost regardless of what kind of hot tub he makes there is not reason to include this cost in the formulation of the model. b. MAX 350 X1 + 300 X2 - 900 Y1 - 800 Y2 ST 1 X1 + 1 X2 £ 200 9 X1 + 6 X2 £ 1,520 12 X1 + 16 X2 £ 2,650 X1 - M Y1 £ 0 X2 - M Y2 £ 0 X1, X2 must be nonnegative integers Y1,Y2 must be binary

7. a. MIN 700X1 + 600X2 + 900X3 + 1250X4 + 850X5 + 1000X6 + 1100X7

ST X1 + X4 + X7  1

X1 + X3 + X5  1

X2 + X3 + X4  1

X5 + X6  1

X1 + X6 + X7  1

X2 + X4 + X5  1

X2 + X3 + X7  1

X5 + X6 + X7  1

X3 + X7  1

X4 + X6  1 Xi binary b. See file Prb6_7.xlsm Chapter 6 - Integer Linear Programming : S-2 ————————————————————————————————————————————

c. X1 =0, X2 =0, X3 =1, X4 =1, X5 =0, X6 =1, X7 =0

8. a. MIN X1 + X2 + X3 + X4 + X5 + X6 + X7

ST X2 + X3 + X4 + X5 + X6  18

X3 + X4 + X5 + X6 + X7  17

X1 + X4 + X5 + X6 + X7  16

X1 + X2 + X5 + X6 + X7  16

X1 + X2 + X3 + X6 + X7  16

X1 + X2 + X3 + X4 + X7  14

X1 + X2 + X3 + X4 + X5  19

Xi  0 & integer

b. See file Prb6_8.xlsm

c. X1 =5, X2 =2, X3 =6, X4 =2, X5 =6, X6 =2, X7 =1 (alternate optimal solutions exist). d. 3

9. a. MIN: 32X1 + 80X2 + 32X3 + 80X4 + 32X5 + 80X6 + 32X7

ST X1 + X2  11

X1 + X2  11

X2 + X3 + X4  24

X2 + X3 + X4  16

X4 + X5 + X6  10

X4 + X5 + X6  22

X6 + X7  17

X6 + X7  6

X2 + X4  0.3*(X2+X3+X4)

X4 + X6  0.3*(X4+X5+X6)

X2  1

X6  1

Xi  0 & integer

b. See file Prb6_9.xlsm c. X1=9, X2=2, X3=16, X4=6, X5=15, X6=1, X7=16 Total cost = $2,512

10. a. Sum the sales potentials & divide by 5 = 210. This would give each sales rep an equal sales potential. b. Ui = amount by which rep i is under the target amount Oi = amount by which rep i is under the target amount Xij = 1 if rep i is assigned to account j; 0 otherwise MIN: U1 + O1 + U2 + O2 + U3 + O3+ U4 + O4+ U5 + O5 ST

X11 + X12 + X13 + X14 + X15 + X16 + X17 + X18 + X19 + X1,10 + U1 – O1 =210 X21 + X22 + X23 + X24 + X25 + X26 + X27 + X28 + X29 + X2,10 + U2 – O2 =210 X31 + X32 + X33 + X34 + X35 + X36 + X37 + X38 + X39 + X3,10 + U3 – O3 =210 X41 + X32 + X43 + X44 + X45 + X46 + X47 + X48 + X49 + X4,10 + U4 – O4 =210 Chapter 6 - Integer Linear Programming : S-3 ————————————————————————————————————————————

X51 + X52 + X53 + X54 + X55 + X56 + X57 + X58 + X59 + X5,10 + U5 – O5 =210 X11 + X21 + X31 + X41 + X51 = 1 X11 + X21 + X31 + X41 + X51 = 1 X12 + X22 + X32 + X42 + X52 = 1 X13 + X23 + X33 + X43 + X53 = 1 X14 + X24 + X34 + X44 + X54 = 1 X15 + X25 + X35 + X45 + X55 = 1 X16 + X26 + X36 + X46 + X56 = 1 X17 + X27 + X37 + X47 + X57 = 1 X18 + X28 + X38 + X48 + X58 = 1 X19 + X29 + X39 + X49 + X59 = 1 X1,10 + X2,10 + X3,10 + X4,10 + X5,10 = 1 Ui , Oi > 0 Xij binary

c. See file Prb6_13.xlsm d. Rep 1 gets customers 1 & 8 (potential = $204) Rep 2 gets customers 2 & 6 (potential = $209) Rep 3 gets customers 4 & 5 (potential = $207) Rep 4 gets customers 7 & 10 (potential = $218) Rep 5 gets customers 3 & 9 (potential = $212) Total deviation from theoretical optimal = 20 Teaching note: It should be noted that the sales reps are interchangeable here, so alternate optima exist. “Rep 4” could be used to represent the top sales person, “Rep 5” the #2 sales person, and so on.

11. Year 1 2 3 4 5 Minimum Capacity in MW 780 860 950 1060 1180

a. Let Xij = # of generators of size j purchased in year i.

MIN 300 X1,10 + 460 X1,25 + 670 X1,50 + 950 X1,100 + 250 X2,10 + 375 X2,25 + 558 X2,50 + 790 X2,100 + 200 X3,10 + 350 X3,25 + 465 X3,50 + 670 X3,100 + 170 X4,10 + 280 X4,25 + 380 X4,50 + 550 X4,100 + 145 X5,10 + 235 X5,25 + 320 X5,50 + 460 X5,100

ST 750 + 10 X1,10 + 25 X1,25 + 50 X1,50 + 100 X1,100  780

750 + 10 (X1,10 + X2,10 ) + 25 (X1,25 + X2,25) + 50 (X1,50 + X2,50 ) + 100 (X1,100 + X2,100)  860 750 + 10 (X1,10 + X2,10 + X3,10 ) + 25 (X1,25 + X2,25 + X3,25)

+ 50 (X1,50 + X2,50 + X3,50) + 100 (X1,100 + X2,100 + X3,100)  950 750 + 10 (X1,10 + X2,10 + X3,10 + X4,10) + 25 (X1,25 + X2,25 + X3,25 + X4,25)

+ 50 (X1,50 + X2,50 + X3,50 + X4,50) + 100 (X1,100 + X2,100 + X3,100 + X4,100)  1060 750 + 10 (X1,10 + X2,10 + X3,10 + X4,10 + X5,10) + 25 (X1,25 + X2,25 + X3,25 + X4,25 + X5,25)

+ 50(X1,50 + X2,50 + X3,50+ X4,50 + X5,50)+100(X1,100 + X2,100 + X3,100 + X4,100 + X5,100)1060

Xij  0 and integer b. See file: Prb6_11.xlsm c. X1,100 = X2,10 = X3,100 = X4,100 = X5,25 = X5,100 = 1, Total Cost = $3,115,000

12. a. MIN 450 X1 + 650 X2 + 550 X3 + 500 X4 + 525 X5 ST X1 + X3 ³ 1 X1 + X2 + X4 + X5 ³ 1 Chapter 6 - Integer Linear Programming : S-4 ————————————————————————————————————————————

X2 + X4 ³ 1 X3 + X5 ³ 1 X1 + X2 ³ 1 X3 + X5 ³ 1 X4 + X5 ³ 1 All Xi are binary where: X1 = 1 if Sanford is selected, 0 otherwise X2 = 1 if Altamonte is selected, 0 otherwise X3 = 1 if Apopka is selected, 0 otherwise X4 = 1 if Casselberry is selected, 0 otherwise X5 = 1 if Maitland is selected, 0 otherwise

b. See file: Prb6_12.xlsm c. X1= X4X5 =1 (Build at Sanford and Casselberry & Maitland) Minimum total cost = $1,475,000

13. a. X1 = batches of DVD players to produce X2 = batches of equalizers to produce X3 = batches of stereo tuners to produce

MAX (75*150) X1 + (50*150) X2 + (40*150) X3 ST (3*150) X1 + (2*150) X2 + (1*150) X3 £ 400,000 50,000/150 £ X1 £ 150,000/150 50,000/150 £ X2 £ 100,000/150 50,000/150 £ X3 £ 90,000/150

b. See file: Prb6_13.xlsm c. X1=334, X2=532, X3=600 Maximum profit = $11,347,500

14. a. MIN 21X1 + 23X2 + 25X3 + 24X4 + 20X5 + 26X6 + 1000Y1 + 950Y2 + 875Y3 + 850Y4 + 800Y5 + 700Y6 ST X1 + X2 + X3 + X4 + X5 + X6 = 1800 X1 - 500 Y1  0 X2 - 600 Y2  0 X3 - 750 Y3  0 X4 - 400 Y4  0 X5 - 600 Y5  0 X6 - 800 Y6  0 Xi  0 All Yi are binary

b. See file: Prb6_14.xlsm c. X1=500, X2=600, X4=100, X5=600 Total cost = $42,300 Chapter 6 - Integer Linear Programming : S-5 ————————————————————————————————————————————

15. a. Xij = 1 if item i is sold in year j, 0 otherwise

MAX CijXij

S.T. Xij = 1, for each item i

CijXij  30, for each year j

b. See file: Prb6_15.xlsm c. Total Cash Received $275,000

Year Item Sold 1 Car 2 Golf Clubs 3 Portrait 4 Desk 5 Piano 5 Humidor

16. a. Let Xi = 1 if project i is selected, 0 otherwise

MAX 650 X1 + 550 X2 + 600 X3 + 450 X4 + 375 X5 + 525 X6 + 750 X7 ST 7 X1 + 6 X2 + 9 X3 + 5 X4 + 6 X5 + 4 X6 + 8 X7 £ 20 250 X1 + 175 X2 + 300 X3 + 150 X4 + 145 X5 + 160 X6 + 325 X7 £ 950 X2 + X6£ 1 (implemented as X2 £ 1-X6 )

b. See file: Prb6_16.xlsm c. X1=X6=X7=1, Total NPV = $1,925,000

17. a. Let Xij = bushels (in 1000s) shipped from grove i to processing plant j Yij = 1 if Xij ³ 0, 0 otherwise

MIN $168 Y14 + $400 Y15 + $320 Y16 $280 Y24 + $240 Y25 + $176 Y26 $440 Y34 + $160 Y35 + $200 Y36 ST X14 + X15 + X16 = 275 X24 + X25 + X26 = 400 X34 + X35 + X36 = 300 X14 + X24 + X34 £ 200 X15 + X25 + X35 £ 600 X16 + X26 + X36 £ 225 Xij - MijYij £ 0 Xij ³ 0 Yij binary Note: Mij = MIN(supply of i, capacity of j)

b. See file: Prb6_17.xlsm c. The solution is: X15= 275, X24= 200, X26= 200, X35= 300, Y15= 1, Y24= 1, Y26= 1, Y35= 1. Chapter 6 - Integer Linear Programming : S-6 ————————————————————————————————————————————

Minimum trucking cost = $1,016.

18. a. X0 = number of efficiency apartments to build X1 = number of 1-bedroom apartments to build X2 = number of 2-bedroom apartments to build X3 = number of 3-bedroom apartments to build

MAX 350 X0 + 450 X1 + 550 X2 + 750 X3

ST X0 + X1 + X2 + X3 £ 40 500X0 + 700X1 + 800X2 + 1,000X3 £ 40,000 0 £ X0 £ 40 5 £ X1 £ 15 8 £ X2 £ 22 0 £ X3 £ 10 Xi integer

b. See file: Prb6_18.xlsm c. X0 = 0 , X1= 8 , X2 = 22 , X3 = 10 Maximum monthly rental income = $23,200 d. There are three binding constraints: the constraint restricting the total number of units to 40, the upper limit of 22 on the number of 2-bedroom apartments, and the upper limit of 10 on the number of 3-bedroom apartments.

19. a. The possible cutting patterns are summarized below:

Cutting Number of boards cut Pattern 7 ft 9 ft 11 ft 1 3 0 0 2 2 1 0 3 2 0 1 4 1 2 0 5 0 1 1 6 0 0 2

Xi = number of boards to which cutting pattern i is applied

MIN X1 + X2 + X3 + X4 + X5 + X6 ST 3X1 + 2X2 + 2X3 + 1X4 + 0X5 + 0X6 ³ 5,000 0X1 + 1X2 + 0X3 + 2X4 + 1X5 + 0X6 ³ 1,200 0X1 + 0X2 + 1X3 + 0X4 + 1X5 + 2X6 ³ 300 Xi ³ 0 & integer b. See file: Prb6_19.xlsm c. X1 = 1267, X3 = 300, X4 = 600, Total boards cut = 2167 (other solutions exist)

20. a. The possible cutting patterns are summarized below:

Cutting Pattern Roll Size 4 ft Strips 9 ft Strips 12 ft Strips Chapter 6 - Integer Linear Programming : S-7 ————————————————————————————————————————————

1 14 ft 3 0 0 2 14 ft 1 1 0 3 14 ft 0 0 1 4 18 ft 4 0 0 5 18 ft 2 1 0 6 18 ft 1 0 1 7 18 ft 0 2 0

Xi = number of rolls to which cutting pattern i is applied

MIN 1,000(X1 + X2 + X3 ) + 1,400( X4 + X5 + X6 + X7) ST 100(3X1 + 1X2 + 0X3 + 4X4 + 2X5 + 1X6 + 0X7) ³ 4,000 100(0X1 + 1X2 + 0X3 + 0X4 + 1X5 + 0X6 + 2X7) ³ 20,000 100(0X1 + 0X2 + 1X3 + 0X4 + 0X5 + 1X6 + 0X7) ³ 9,000 Xi ³ 0

b. See file: Prb6_20.xlsm c. X2 = 40, X3 = 90, X7 = 80 Minimum cost = $242,000

21. a. Let Pij = production at plant i allocated to meet demand for customer j Yi= 1 if any Xij ³ 0, 0 otherwise

MIN 35 P1X + 30 P1Y + 45 P1Z + 45 P2X + 40 P2Y + 50 P2Z + 70 P3X + 65 P3Y + 50 P3Z + 20 P4X + 45 P4Y + 25 P4Z + 65 P5X + 45 P5Y + 45 P5Z + 1000 (1,325 Y1 + 1,100 Y2 + 1,500 Y3 + 1,200 Y4 + 1,400 Y5) ST P1X + P1Y + P1Z £ 40,000 Y1 P2X + P2Y + P2Z £ 30,000 Y2 P3X + P3Y + P3Z £ 50,000 Y3 P4X + P4Y + P4Z £ 20,000 Y4 P5X + P5Y + P5Z £ 40,000 Y5 P1X + P2X + P3X + P4X + P5X ³ 40,000 P1Y + P2Y + P3Y + P4Y + P5Y ³ 25,000 P1Z + P2Z + P3Z + P4Z + P5Z ³ 35,000 Pij ³ 0 Yi binary

b. See file: Prb6_21.xlsm c The solution is: Build plants at locations 1, 4 and 5 (i.e., Y1=Y4=Y5 = 1). P1X=P1Y=P4X=20,000, P5Y=5,000, P5Z=35,000. Total cost = $7,425,000.

22. a. Additional Constraints: Y1+Y2 £ 1 and Y4+Y5 £ 1 (implement as Y1£ 1-Y2 and Y4£ 1-Y5) b. See file: Prb6_22.xlsm c. The solution is: Build plants at locations 1, 3 and 4 (i.e., Y1=Y3=Y4 = 1). P1X=15,000, P1Y=25,000, P3X=5,000, P3Z=35,000, P4X=20,000. Total cost = $7,800,000. Chapter 6 - Integer Linear Programming : S-8 ————————————————————————————————————————————

23. a. MIN 13 A1+ 9 B1+ 10 C1+ 11 A1+ 12 B2+ 8 C2 + 55 YA1+ 93 YB1+ 60 YC1+ 65 YA2+ 58 YB2+ 75 YC2 ST A1+ A2 = 3 B1+ B2 = 7 C1+ C2 = 4 0.4 A1+ 1.1 B1+ 0.9 C1£ 8 0.5 A2+ 1.2 B2+ 1.3 C2£ 6 A1- 3YA1 £ 0 B1- 7YB1 £ 0 C1- 4YC1 £ 0 A2- 3YA2 £ 0 B2- 7YB2 £ 0 C2- 4YC2 £ 0 Ai, Bi, Ci ³ and integer Yij binary

b. See file: Prb6_23.xlsm c. The optimal solution is: A2=3, B1=4, B2=3,C1=4, YA2=YB1=YB2=YC1=1 Total cost = $421

24. a. Let X1j = barrels of crude to buy from TX allocated to the production of product j X2j = barrels of crude to buy from OK allocated to the production of product j X3j = barrels of crude to buy from PA allocated to the production of product j X4j = barrels of crude to buy from AL allocated to the production of product j Yi = 1 if Xi > 0 and 0 otherwise

MIN 22 X1 + 21 X2 + 22 X3 + 23 X4 + 1500 Y1 + 1700 Y2 + 1500 Y3 + 1400 Y4 ST 2.00 X11 + 1.80 X21 + 2.30 X31 + 2.10 X41 ³ 750 2.80 X12 + 2.30 X22 + 2.20 X32 + 2.60 X42 ³ 800 1.70 X13 + 1.75 X23 + 1.60 X33 + 1.90 X43 ³ 1000 2.40 X14 + 1.90 X24 + 2.60 X34 + 2.40 X44 ³ 300 X1 = X11 + X12 + X13 + X14 X2 = X21 + X22 + X23 + X24 X3 = X31 + X32 + X33 + X34 X4 = X41 + X42 + X43 + X44 X1 - 1500 Y1 £ 0 X2 - 2000 Y2 £ 0 X3 - 1500 Y3 £ 0 X4 - 1800 Y4 £ 0 X1 - 500 Y1 ³ 0 X2 - 500 Y2 ³ 0 X3 - 500 Y3 ³ 0 X4 - 500 Y4 ³ 0 Chapter 6 - Integer Linear Programming : S-9 ————————————————————————————————————————————

b. See file: Prb6_24.xlsm c. Purchase 2,000 barrels from Oklahoma and 850 barrels from Pennsylvania. Total cost = $63,900.

25. a. MIN S1 + S2 + S3 S.T. X11 + X12 + X13 + X14 + X15 + S1 = 2700 X21 + X22 + X23 + X24 + X25 + S2 = 3500 X31 + X32 + X33 + X34 + X35 + S3 = 4200

Y11 + Y21 + Y31  1

Y12 + Y22 + Y32  1

Y13 + Y23 + Y33  1

Y14 + Y24 + Y34  1

Y15 + Y25 + Y35  1

Xi1 - 2500Yi1  0, i=1 to 3

Xi2 - 2000Yi2  0, i=1 to 3

Xi3 - 1500Yi3  0, i=1 to 3

Xi4 - 1800Yi4  0, i=1 to 3

Xi5 - 2300Yi5  0, i=1 to 3 Yj binary

Xi  0

b. See file: Prb6_25.xlsm c. Compartment 1, 2500 gallons of diesel Compartment 2, 2000 gallons of regular unleaded Compartment 3, 1500 gallons of regular unleaded Compartment 4, 1800 gallons of premium unleaded Compartment 5, 2300 gallons of premium unleaded Minimum shortage = 300 gallons

26. a. A = Amount to invest in investment A B = Amount to invest in investment B C = Amount to invest in investment C D = Amount to invest in investment D E = Amount to invest in investment E M1 = Amount to invest in the money market in 2001 M2 = Amount to invest in the money market in 2002 M3 = Amount to invest in the money market in 2003

MAX 1.3 B + 1.65 C + 1.3 D + 1.05 M3 ST A + C + E + M1 = 1,000,000 0.45 A + 1.05 M1 - B – M2 = 0 1.05 A + 1.25 E + 1.05 M2 – D – M3 = 0 A - M YA £ 0 B - M YB £ 0 C - M YC £ 0 D - M YD £ 0 E - M YE £ 0 A - 50000 YA ³ 0 B - 50000 YB ³ 0 C - 50000 YC ³ 0 D - 50000 YD ³ 0 E - 50000 YE ³ 0 Chapter 6 - Integer Linear Programming : S-10 ————————————————————————————————————————————

A, B, C, D, E, M1, M2, M3 ³ 0 Yi binary

b. See file: Prb6_26.xlsm c. A=100,000, B= 0, C=0, D=152,250, E=0, M1=0, M2=45,000, M3=0 Maximum amount of money at the beginning of 2004 = $197,925

27. a. Let Bi = beginning inventory in week i Ni = the number of batches ordered in week i Ei = inventory held at the end of week i Di = demand in week i Yi = 1 if Bi > 0, 0 otherwise

MIN 125*(N1+ N2+ N3+ N4) + 50*(Y1+ Y2+ Y3+ Y4) + 15*(E1+ E2+ E3+ E4)/100 ST Bi=Ei-1, i=2,3,4 Ei = Bi + 100*Ni - Di , i=1,2,3,4 Ni - M Yi£ 0 Ei ³ 0 Yi binary Ni ³ 0 and integer

b. See file: Prb6_27.xlsm c. The optimal solution is: N1=6, N3=5, Y1=1, Y3=1. Minimum total cost = $1,565.

28. a. Let Xi = 1 if design change i is selected, 0 otherwise

MIN 150 X1 + 350 X2 + 50 X3 + 450 X4 + 90 X5 + 35 X6 + 650 X7 + 75 X8 + 110 X9 + 30 X10 ST 50X1 + 75 X2 + 25 X3 + 150 X4 + 60 X5 + 95 X6 + 200 X7 + 40 X8 + 80 X9 + 30 X10 ³ 400 X4 £ 1 - X7 Xi binary

b. See file: Prb6_28.xlsm c. The optimal solution is: X4 = X5 = X6 = X9 = X10 = 1. Minimum total cost = $715,000.

29. a. See file Prb6_29.xlsm b. Build Red Snappers at sites 4, 8 & 9. Build Olive Groves at sites 1, 6 & 10. NPV=$105.8. c. See file Prb6_29.xlsm d. Build Red Snappers at sites 4, 7 & 9. Build Olive Groves at sites 1, 6 & 10. NPV=$100.3.

30. a. MAX 70(X11+X21+X31)+50(X12+X22+X32)+60(X13+X23+X33)+80(X14+X24+X34)

ST X11+X21+X31  4800

X12+X22+X32  2500

X13+X23+X33  1200

X14+X24+X34  1700

X11+X12+X13+X14  3000

X21+X22+X23+X24  6000

X31+X32+X33+X34  4000

40X11+25X12+60X13+55X14  145000 Chapter 6 - Integer Linear Programming : S-11 ————————————————————————————————————————————

40X21+25X22+60X23+55X24  180000

40X31+25X32+60X33+55X34  155000

0.9(X31+X32+X33+X34)  X11+X12+X13+X14  1.1(X31+X32+X33+X34)

0.4Xij  X21+X22+X23+X24  0.6Xij

Xij  MjYij,  ij

Yij  1,  i

Xij  0 Yij binary

b. See file Prb6_30.xlsm

c. Load 4500 tons of commodity 1 in center hold (X21=4500), 1888.89 tons of commodity 2 in rear hold (X32=1888.89), 1700 tons of commodity 4 in forward hold (X14=1700). Profit = $545,444.

31. a. See file: Prb6_31.xlsm b. Build towers in areas 8, 11, 19 & 22; Profit = $377 thousand. c. Build towers in areas 3, 5, 6, 14, 17, 22, 25; Profit = $219 thousand.

32. a. Let Xi = 1 if an ambulance is located in region i Let Yi = 1 if region i is being serviced within 4 minutes

MAX 21 Y1 + 35 Y2 + 15 Y3 + 60 Y4 + 20 Y5 + 37 Y6 ST X1 + X2 + X3 + X4 + X5 + X6 =2 Y1 < 1X1 + 1X2 + 1X3 + 0X4 + 0X5 + 0X6 Y2 < 1X1 + 1X2 + 0X3 + 0X4 + 0X5 + 0X6 Y3 < 1X1 + 0X2 + 1X3 + 1X4 + 1X5 + 0X6 Y4 < 0X1 + 0X2 + 1X3 + 1X4 + 0X5 + 0X6 Y5 < 0X1 + 0X2 + 1X3 + 0X4 + 1X5 + 1X6 Y6 < 0X1 + 0X2 + 0X3 + 0X4 + 1X5 + 1X6

Xi binary, Yi binary

b. See file: Prb6_32.xlsm c. The optimal solution is: X3 = X5 = 1. A total of 153,000 people can be reached within 4 minutes using these locations. d. 3 ambulance are required, in areas 1, 3, and 5 e. You get the same solution as in part d.

33. a. See file Prb6_33.xlsm b. State Week AZ 6 CA 6 CT 2 GA 3 IL 5 MA 4 ME 6 MN 4 MT 5 Chapter 6 - Integer Linear Programming : S-12 ————————————————————————————————————————————

NC 1 NJ 4 NV 1 OH 2 OR 4 TX 3 VA 1 Maximum pieces processed in any week: 353,856

34. a. See file Prb6_34.xlsm b. Operate plants 2, 3, 5 & 6, warehouses 3 & 4. Total cost = $855,000 c. See file Prb6_34.xlsm

35. a. See file: Prb6_35.xlsm b.

c. Chapter 6 - Integer Linear Programming : S-13 ————————————————————————————————————————————

36. a. See file Prb6_30.xlsm b. Build the lines from Fairbanks to Tok, Skagway to Whitehorse, Watson Lake to New Hazelton, Ft Nelson to Chetwynd, Tok to Whitehorse, Whitehorse to Watson Lake, and Watson Lake to Ft. Nelson. Total cost = $119 million.

37. a. See file Prb6_37.xlsm. Place hubs in Andersen, Berekley, Dillon, Edgefield, Hampton,. Kershaw, Orangeburg & Union. Total hubs = 8, Total coverage = 55. b. See file Prb6_37.xlsm. Place hubs in Bamberg, Colleton, Florence, Greenville, Greenwood, Hampton, Kershaw, Marion, Orangeburg & Union. Total hubs = 10, Total coverage = 77.

X 1 = 2.25 X 2 = 1.50 X <= 2 Obj = 25.5 1 X 1 >= 3

X 1 = 2.0 X 1 = 3.0 X 2 = 1.67 X 2 = 0.0 Obj = 25.3 Obj = 18 X 2 >= 2 X 2 <= 1

X 1 = 2.0 X 1 = 1.5 X 2 = 1.0 X 2 = 2.0 Obj = 20 Obj = 25.0

X 1 <= 1 X 1 >= 2

X 1 = 1.0 X 2 = 2.33 Infeasible Obj = 24.6

X 2 <= 2 X 2 >= 3

X = 1.0 1 X 1 = 0.0 X = 2.0 2 X 2 = 3.0 Obj = 22.0 Obj = 24.0

optimal solution Chapter 6 - Integer Linear Programming : S-14 ————————————————————————————————————————————

39. Candidate problems could be selected on a depth first, or breadth first basis. Other strategies might include selecting the candidate with the best relaxed objective function value, or the lowest sum of integer infeasibilities. All of these strategies are heuristic. Case 6-1: Optimizing A Timber Harvest 1. See file: Case6_1.xlsm (This solution models the problem as a fixed-charge network flow problem. Other formulations may be possible.) 2. Harvest areas 2, 3, 6, 8, 9, and 12; Total profit = $16,000 3. Harvest areas 1, 2, 3, 4, 5, 6, 8, 9, and 12; Total profit = $17,000

Case 6-2: Power Dispatching At Old Dominion

1. Xij = megawatts generated at plant i on day j Rij =1, if generator at plant i is running on day j; 0, otherwise. Sij = 1, if generator at plant i is started-up on day j; 0, otherwise.

MIN 5 (X11 + X12 + … + X17) + 4 (X21 + X22 + … + X27) + 7 (X31 + X32 + … + X37) + 200 (R11 + R12 + … + R17) + 300 (R21 + R22 + … + R27) + 250 (R31 + R32 + … + R37) + 800 (S11 + S12 + … + S17) + 1000 (S21 + S22 + … + S27) + 700 (S31 + S32 + … + S37)

ST X11 + X21 + X31  4300

X12 + X22 + X32  3700

X13 + X23 + X33  3900

X14 + X24 + X34  4000

X15 + X25 + X35  4700

X16 + X26 + X36  4800

X17 + X27 + X37  3600

X1i  2100 R1i  0 , i = 1, 2, …, 7

X2i  1900 R2i  0 , i = 1, 2, …, 7

X3i  3000 R3i  0 , i = 1, 2, …, 7

R1,i-1 + R1i  S1i  0 , i = 2, 3, …, 7

R2,i-1 + R2i  S2i  0 , i = 2, 3, …, 7

R3,i-1 + R3i  S3i  0 , i = 2, 3, …, 7

Xij  0 Rij, Sij binary

2. See file: Case6_2.xlsm 3. Total Cost = $142,750 Plant 1 2 3 4 5 6 7 New River 2100 1800 2000 2100 2100 2100 1700 Galax 1900 1900 1900 1900 1900 1900 1900 James River 300 0 0 0 700 800 0

4. Old Dominion's marginal cost of producing the last 300 MW of the 4300 MW demanded on day 1 is $3,050. (This result is obtained by reducing the demand on day 1to 4000 MW, re-optimizing the problem, and subtracting the objective value ($139,700) from the value obtained in question 2.) If they can buy 300 MW for less than $3050, they should do so and not operate the James River plant on day 1. 5. This plan has the Galax and New River plants running at or near capacity each day. If demand is a little higher than projected, it would be wise to keep the James River plant online continuously as well to avoid several start ups. Case 6-3: The MasterDebt Lockbox Problem See file: Case6_3.xlsm Chapter 6 - Integer Linear Programming : S-15 ————————————————————————————————————————————

1. Use Dallas and New York, Total cost = $224,000 2. Use Sacramento and Atlanta, Total cost = $225,750 Case 6-4: Removing More Snow in Montreal

1. See file: Case6_4.xlsm 2. Disposal Site Sector 1 2 3 4 5 1 0.0 1.0 0.0 0.0 0.0 2 1.0 0.0 0.0 0.0 0.0 3 0.0 0.0 1.0 0.0 0.0 4 0.0 0.0 1.0 0.0 0.0 5 0.0 0.0 1.0 0.0 0.0 6 0.0 0.0 0.0 1.0 0.0 7 0.0 0.0 0.0 1.0 0.0 8 0.0 0.0 0.0 0.0 1.0 9 1.0 0.0 0.0 0.0 0.0 10 0.0 0.0 0.0 1.0 0.0

3. $54,700 4. 250 = lots! 5. Increase capacity at site 2. Cost savings = $54,700-$48,968 = $5,732