Taylch07

83

PROBLEM SUMMARY

1. Shortest route

2. Shortest route

3. Shortest route

4. Shortest route

5. Shortest route

6. Shortest route

7. Shortest route

8. Shortest route

9. Shortest route

10. Shortest route

11. Shortest route

12. Shortest route

13. Shortest route

14. Shortest route

15. Shortest route

16. Minimal spanning tree











27. Maximal flow

28. Maximal flow

29. Maximal flow

30. Maximal flow

31 Maximal flow

32. Maximal flow

33. Maximal flow

34. Maximal flow

35. Maximal flow

36. Maximal flow

37. Maximal flow

38. Maximal flow

Chapter Seven: Network Flow Models

PROBLEM SOLUTIONS

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

10.

Step Permanent Set Branch Added Distance

1 {1} 1–3 73

2 {1,3} 1–2 89

3 {1,2,3} 1–4 96

4 {1,2,3,4} 3–7 154

5 {1,2,3,4,7} 2–5 164

6 {1,2,3,4,5,7} 3–6 167

7 {1,2,3,4,5,6,7} 4–8 177

8 {1,2,3,4,5,6,7,8} 7–9 208

9 {1,2,3,4,5,6,7,8,9} 7–10 239

10 {1,2,3,4,5,6,7,8,9,10} 9–12 263

11 {1,2,3,4,5,6,7,8,9,10,12} 8–11 283

12 {1,2,3,4,5,6,7,8,9,10,11,12} 10–13 323

11. 1 – 4 – 7 – 8 = 42 miles

12. 1 – 4 – 7 – 10 – 12 – 16 – 17 = 14 days

13. 1 – 3 – 11 – 14 = 13 days

14. 1 – 3 – 5 – 12 – 16 – 20 – 22 = 114

9.

99

15. This is an example of the application of the shortest route method to solve the scheduled replacement problem.The branch costs are determined using the formula,

cij = maintenance cost for year i, i + 1,…, + cost of purchasing new car at the beginning of year i-selling priceof a used car at the beginning of year j.

Solution: 1 - 4 - 7 = $64.5 = $64,500

A car should be sold at end of year 3 (beginning of year 4) and a new one purchased.

c

c

c

c

12

13

14

15

3 26 15 14

3 4 5 26 12 21 5

3 4 5 6 26 8 31 5

= + − == + + − == + + + − =

. .

. .

== + + + + − == + + + + + − == + +

3 4 5 6 8 26 4 43 5

3 4 5 6 8 11 26 2 56 5

3 4 5 616

17

. .

. .

.

c

c ++ + + + + − == + − == + + −

8 11 14 26 28 5 0 101

3 26 5 15 14 5

3 4 5 26 5 1223

24

.

. .

. .

c

c === + + + − == + + + + − == + +

22

3 4 5 6 26 5 8 32

3 4 5 6 8 26 5 4 44

3 4 5

25

26

27

c

c

c

. .

. .

. 66 8 11 26 5 2 57

3 27 15 15

3 4 5 27 12 22 5

3 4

34

35

36

+ + + − == + − == + + − == +

.

. .

c

c

c .. .

. .

. .

5 6 27 8 32 5

3 4 5 6 8 27 4 44 5

3 27 5 15 15 537

45

4

+ + − == + + + + − == + − =

c

c

c 66

47

56

57

3 4 5 27 5 12 23

3 4 5 6 27 5 8 33

3 28 15 16

= + + − == + + + − == + − ==

. .

. .c

c

c 33 4 5 28 12 23 5

3 28 5 15 16 567

+ + − == + − =

. .

. .c

1 2 3 4 5 6 7

98.5

56.557

44 44.543.5

31.5

21.5 22 22.5 23 23.5

32 32.5 33

14 14.5 15 15.5 16 16.5

Begining ofYear 1

End ofYear 6

17. 1 – 3, 4.11 – 4, 4.82 – 3, 3.64 – 8, 5.55 – 6, 2.16 – 7, 2.87 – 8, 2.77 – 9, 2.79 – 10, 4.6

32.9 = 32,900 feet

100

18.

19.

20.

21.

1–21–32–43–65–66–77–8

1–32–33–44–65–65–86–7

1–22–33–64–85–66–77–97–89–10

1–42–43–64–65–76–77–8

16.

1–32–43–44–64–75–7

minimum distance = 70

101

1

2

3

4

5

6

7

8

9

10

11

12

13

14

112

107 76

63

67

92 88

84 61

8273

95

86

22.

23.

25.

24.

Total sidewalk = 1,086 ft.Length of ductwork = 790 ft 1–4

2–33–43–55–65–75–8

1–22–42–53–44–75–66–88–9

26. 1 – 2 = 481 – 4 = 524 – 7 = 353 – 5 = 395 – 6 = 295 – 8 = 565 – 9 = 486 – 7 = 809 – 10 = 719 – 12 = 7110 – 11 = 3811 – 14 = 5712 – 13 = 105Total = 729

102

27.

103

28.

104

29.

105

30.

106

107

31.

108

Maximal flow network:

32.

109

33.

34.

35. This problem is solved using Excel with theaddition of a cost constraint to the normal linearprogramming formulation for a maximum flowproblem, as follows,

Maximize Z = x10, 1

subject to:

solution

Z = == =

16

684 684 000

16 000 units

Total cost

,

$ ,

x x

x x

x x

x x

x x

x x

12 45

13 57

14 58

25 59

26 68

35 6

6 8

2 10

8 4

0 1

6 2

0

= == == == == == 9

36 7 10

8 10

9 10

6

2 3

6

7

== =

==

x x

x

x

,

,

,

3 5 7

2212 13 14 25 26 35 36

57 58 59

( ) ( ) ( )

(

x x x x x x x

x x x

+ + + + + + 4 45( )x++ + + )) ( )

,

+ ++ ≤

19

16 70068 69

9 10

x x , ,+ +12 147 10 8 10x x

x

x x

x

x x

x x

x x

x x

12 57

13

14 59

25 68

26 69

35 7

7 3

10

8 1

9 6

6 8

7

≤ ≤≤ x58 4≤≤ ≤≤ ≤≤ ≤≤ ,,

,

,

,

10

36 8 10

45 9 10

10 1

8

5 7

10 7

100

≤

≤ ≤≤ ≤

≤

x x

x x

x

x x x x

x x x

x x x

x x x

10 1 12 13 14

12 25 26

13 35 36

14 45 46

0

0

0

0

, − − − =

− − =− − =− − =

xx x x x

x x

x

x

25 57 58 59

26 68 69

57

58

0

0

0

+ + − − − =+ − − =− =+

x x

x x

x

35 45

36

7,10

xx

x

x

68

69

8,10

− =

+ − =+ + − =

x

x x

x x x

8 10

59 9 10

7 10 9 10 10 1

0

0

0

,

,

, , ,

110

36.

111

37. 1–2–6–10–12–13–15 = 25

1–2–6–12–13–15 = 35

1–2–9–12–15 = 10

1–3–6–10–12–15 = 30

1–3–7–10–13–15 = 20

1–4–7–10–13–15 = 20

1–4–7–6–10–13–15 = 10

1–4–7–6–12–15 = 5

1–4–8–13–15 = 25

1–5–8–13–15 = 35

1–5–8–14–15 = 5

1–5–11–14–15 = 30

maximum flow = 250

Branch Allocation Branch Allocation

1–2 70 7–6 151–3 50 8–13 601–4 60 8–14 51–5 70 9–12 452–6 25 10–12 552–9 45 10–13 503–6 30 11–14 303–7 20 12–13 604–7 35 12–15 454–8 25 13–15 1705–8 40 14–15 355–11 306–10 656–12 57–10 40

38. 1–2 = 16 3–4 = 162–5 = 12 4–10 = 45–7 = 12 9–10 = 42–6 = 12 10–13 = 86–7 = 4 13–14 = 47–12 = 16 4–8 = 1212–15 = 22 8–14 = 121–3 = 22 14–15 = 163–9 = 69–11 = 211–12 = 213–12 = 4

Total traffic = 38,000 cars.

112

CASE SOLUTION: AROUND THEWORLD IN 80 DAYS

Using QSB + the following optimal route for PhileasFogg was determined (which is approximately the sameroute he travelled in the book).

1(London) – 6(Paris) – 7(Barcelona) – 12(Naples)– 17(Athens) – 22(Cairo) – 23(Aden) –28(Bombay) – 31(Calcutta) – 33(Singapore) –35(Hong Kong) – 37(Shanghai)– 39(Yokohama) – 41(San Francisco) –47(Denver) – 50(Chicago) – 53(New York)– 1(London) = 81 days

Note that 3 additional “end” nodes were added to thenetwork for computer solution – 55, 56 and 57. Node 55replaced node 3 (Casablanca); node 56 replaced node 2(Lisbon), and node 57 replaced node 1 (London) at the

end of the network. Additional branches were added toconnect Casablanca with Lisbon (56–57) and Lisbonwith London (56–57).

Although it appears that Phileas Fogg lost his wager,recall that, as in the novel, he travelled toward the eastand eventually crossed the international date line. Thissaved him one day and allowed him to win his wageronce he realized (just in time) the error in hiscalculations.

CASE SOLUTION: BATTLE OF THEBULGE

(1) Verdun – (2) Stenay = 8(1) Verdun – (3) Montmedy = 23(1) Verdun – (5) Etain = 26(2) Stenay – (11) Bouillon = 8(3) Montmedy – (6) Virton = 10

113

(3) Montmedy – (11) Bouillon = 15(4) Longuyon – (3) Montmedy = 2(4) Longuyon – (7) Longwy = 5(5) Etain – (4) Longuyon = 7(5) Etain – (8) Briey = 10(5) Etain – (9) Havange = 9(6) Virton – (13) Tintigny = 10(7) Longwy – (14) Arlon = 9(8) Briey – (10) Thionville = 10(9) Havange – (15) Luxembourg = 8(10) Thionville – (15) Luxembourg = 7(10) Thionville – (9) Havange = 3(11) Bouillon – (12) Florenville = 7(11) Bouillon – (16) Paliseul = 16(12) Florenville – (13) Tintigny = 7(13) Tintigny – (17) Neufchateau = 12(14) Arlon – (18) Martelange = 8(15) Luxembourg – (19) Diekirch = 21(16) Paliseul – (20) Recogne = 16(17) Neufchateau – (18) Martelange = 12(18) Martelange – (21) Bastogne = 23(19) Diekirch – (21) Martelange = 3(19) Diekirch – (21) Bastogne = 18(19) Diekirch – (18) Martelange = 3(20) Recogne – (21) Bastogne = 16

Total flow = 57,000 troops

CASE SOLUTION: NUCLEAR WASTEDISPOSAL AT PAWV POWER ANDLIGHT

This is a “modified” shortest route problem. Instead ofthe minimum time as the objective function thepopulation traveled through should be minimized. Thetime (which would normally be the objective function)should be a constraint ≤ 42 hours.

Solution:

(1) Pittsburgh - (4) Akron

(4) Akron - (6) Toledo

(6) Toledo - (10) Chicago

(10) Chicago - (16) Davenport/Moline/Rock Island

(16) Davenport/Moline/Rock Island - (19) Des Moines

(19) Des Moines - (23) Omaha

(23) Omaha - (28) Cheyenne

(28) Cheyenne - (31) Salt Lake City

(31) Salt Lake City - (33) Nevada Site

Total time = 39.95 hours

Total population (Z) = 15.83 million

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Taylch07