Network Optimization Models CHAPTER 11
Network Models ALTERNATIVE EXAMPLES Alternative Example 11.1: Given the following network, perform the minimum-spanning tree technique to determine the best way to connect nodes on the network, while minimizing total distance.
We begin with node 1. Node 4 is the nearest node, and thus we connect node 1 to node 4. Given nodes 1 and 4, node 6 is the nearest, and we connect it to node 4. Now considering nodes 1, 4, and 6, we see that node 7 is the nearest to node 6 and we connect it. Node 5 is connected to node 7, and node 3 is connected to node 5 in the same way. Finally, node 2 is connected to node 1. Using the minimum-spanning tree technique, we can see that the total distance required to connect all nodes is 18. The following figure shows the results.
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Network Optimization Models
Alternative Example 11.2: Given the network in the figure on the next page, determine the maximum amount that can flow through the network.
We begin this problem by putting the maximum flow of 4 through nodes 1, 2, and 6. This is shown in the following figure. The flows have been adjusted along this path.
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Network Optimization Models Next, we will put the maximum flow of 1 through nodes 1, 4, 2, and 6. The adjusted flows are shown in the following figure.
Next, we put the maximum flow of 4 units through nodes 1, 4, and 6. The adjusted network is shown below.
This process continues. We put a maximum of 2 units through nodes 1, 3, 5, and 6. The maximum amount that can flow through the network is 11. The figure below shows the final results.
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Network Optimization Models
1–2–6
4 units
1–4–2–6
1 unit
1–4–6
4 units
1–3–5–6
2 units 11
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Network Optimization Models Alternative Example 11.3: Given the network in the following figure, determine the shortest route or path through the network.
The nearest node to node 1 is node 2. The distance is 50. Thus we put 50 in a box by node 2. The results of this step are shown in the following figure.
The next nearest node to node 1 is node 3. The distance is 100. Thus we put 100 in a box by node 3. The results of this step are shown in the following figure.
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Network Optimization Models We continue the process. The next-nearest node to node 1 is node 5. The distance between node 4 and 5 is 100 and the total distance between node 5 and node 1 is 200. Thus we put 200 in a box by node 5. The results of this step are shown in the following figure.
The next-nearest node to node 1 is node 4. Actually, there are two paths to node 4 with the same distance of 250. One path is nodes 1, 3, 5, and 4. The other path is nodes 1, 2, and 4. We put 250 in a box by node 4. The results of this step are shown in the following figure.
The final step is to consider node 6. We can get to node 6 through node 5 (distance of 200 to node 1) and node 4 (distance of 250 to node 1). Going through node 5 will minimize the total distance. We can see that the shortest route is 300 (200 from node 5 to node 1 and 100 from node 5 to node 6). The results are shown in the following figure.
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Network Optimization Models
11-9.
One optimal solution is shown. Connect 1–3, 1–4, 3–6, 6–7, 1–2, 4–5, 7–9, 9–8, 9–10, 10–11, 11–13, 13–14, and 14–12. Alternate solutions can be found by substituting 3–4 for 1–4 and substituting 9–12 for 13–14. Total distance = 45. 11-10. Flow Path
(Cars/Hour)
1–2–5–7–8
2
1–3–6–8
2
1–4–8
1
Total
5
11-11.
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Network Optimization Models
The shortest route is 1–3–5–7–10–13. The distance is 430 miles. 11-12. The minimal-spanning tree technique is needed to solve this problem. The minimum distance is 47 (4,700 feet). As you can see, the final solution has changed. 11-13. The maximal flow through the network is 7. This is higher by 2 cars from Problem 1110. The solution is given below. Flow Path
(Cars/Hour)
1–2–5–7–8
2
1–3–6–8
2
1–4–8
3
Total
7
11-14. This is the only optimum solution to this problem (177 units of length).
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Network Optimization Models 11-15. There are several possible solutions. One solution is presented below.
One solution: 1–4–6
40
1–2–5–6
55
1–3–5–6
45
1–4–5–6
27 167 widgets per day
Alternative solutions: Substitute 1–2–4–6 for 32 in lieu of 1–4–6 or 1–4–5–6 (or for some portion of the 32). 11-16. No, the changes do not have an impact on the final solution. With the changes, the optimal solution still has a shortest distance of 430 miles. The final network is given below. Note that we have increased the value for the paths 6–9 and 8–9 to a very high relative number (10,000) to ensure that these paths are forced out of the final solution. Figure for Problem 11-16
11-17. The solution to the minimal-spanning tree problem results in a minimum distance of 21 (2,100 yards). The final network follows. Figure for Problem 11-17
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Network Optimization Models
11-18. If the distance between nodes 6 and 7 becomes 5, the minimum distance changes to 23 (2,300 yards). The final network follows.
1–2–4–5–6
13
5–8–9
5
1–3–7
5 23
Another optimal solution exists.
11-19. The maximum number of cars that can flow from the hotel complex to Disney World is 13 (1,300 cars per hour). Solution to Problem 11-19
1–2 1–3 1–4 2–6 3–7 4–8 6–9 10–11 7–10 8–10 9–11
Flow 3 8 2 3 8 2 3 10 8 2 3
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Network Optimization Models Total maximum flow: 13. 11-20. The impact of the construction project to increase the road capacity around the outside roads from International Drive to Disney World would increase the number of cars per hour to 1,700 per hour (17). The increase is 400 cars per hour as would be expected. The solution follows. Solution to Problem 11-20
Flow 1–2 5 1–3 8 1–4 2 1–5 2 2–6 5 3–7 8 4–8 2 5–8 2 6–9 5 10–11 12 7–10 8 8–10 4 9–11 5 Total maximum flow: 17. 11-24. The impact of the emergency repair is that nodes 6 and 7 cannot be used. All flow in and out of these nodes is 0. As a result, the flow from the origin to the final network node has been reduced to 2,000 gallons per hour (2). The solution is shown in the following table. Solution to Problem 11-24
1–2 1–4 2–5 4–8 5–9 8–11 9–12 11–13 12–14 13–14 Total maximum flow: 2.
Flow 1 1 1 1 1 1 1 1 1 1
11-25. The shortest route from node 1 to node 16 is 74 kilometers. The solution along with the final network is shown in the following table and figure.
1–3 3–7
Value 15 11
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Network Optimization Models 7–11 18 11–14 16 14–16 14 Shortest path: 1–3–7–11–14–16 Total shortest distance: 74. Figure for Problem 11-25
11-26. The impact of closing two nodes (nodes 7 and 8) is to increase the shortest route from 74 to 76 kilometers. Note that all paths into and from nodes 7 and 8 have their values changed to a very high relative number (10,000) to force these paths out of the final solution. The solution along with the final network is given below.
Value 20 10 12 16 18
1–2 2–6 6–9 9–13 13–16 Shortest path: 1–2–6–9–13–16 Total shortest distance: 76.
Figure for Problem 11-26
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Network Optimization Models
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Network Optimization Models 11-33. Using the maximal-flow technique in the network module of QM for Windows we have a maximum flow of 190 as shown in the table.
Maximal Network Flow 190 Start Node Branch 1 1 Branch 2 1 Branch 3 1 Branch 4 2 Branch 5 2 Branch 6 3 Branch 7 3 Branch 8 3 Branch 9 4 Branch 10 5 Branch 11 5 Branch 12 6 Branch 13 6 Branch 14 7 Iteration 1 2 3 4 5
End Node 2 3 4 3 5 4 5 6 7 6 8 7 8 8
Path 1 2 5 8 1 3 6 8 1 4 7 8 1 2 3 4 7 6 5 8 1 4 3 5 6 8
Capacity 80 50 60 30 60 40 70 60 80 20 90 30 70 50
Reverse Capacity 0 0 0 30 0 40 0 0 0 20 0 30 0 0
Flow 80 50 60 20 60 –10 10 50 70 –20 80 –20 60 50
Flow 60 50 50 20
Cumulative Flow 60 110 160 180
10
190
11-34. QM for Windows indicates that total capacity is not affected. Other streets can be used to still accommodate 190 cars.
Iteration 1 2 3 4 5 6
Path 1 2 5 8 1 4 7 8 1 3 4 7 6 8 1 2 3 5 6 8 1 3 5 8 1 4 3 5 8
Flow 60 50 30 20 20 10
Cumulative Flow 60 110 140 160 180 190
11-35. Using the shortest-route technique in QM for Windows, we find the minimum total distance to be 16 as shown in the table.
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Network Optimization Models
Branch 2 Branch 7 Branch 12
Start Node 1 3 6
End Node N ode 3 6 7
Distance 6 3 7
Cumulative Distance 6 9 16
along with the final 11-36. a. The solution is 4,900 feet. This is almost 1 mile. The solution along network is given below and on the next page. 1–3 3–7 7–12 12–16 16–20 20–23 23–25 Shortest path: 1–3–7–12–16–20–23–25 Total shortest distance: 49.
Value 9 6 8 5 6 7 8
Figure for Problem 11-36a
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Network Optimization Models b. Eliminating the paths 6–11, 7–12, and 17–20 has changed changed the shortest route to 5,500 feet feet (55). This is higher than the solution in part a, as you would expect. The solution (below) along with the final network (on the next page) are given. When using the software, the distance for paths 6–11, 7–12, and 17–20 should be increased to a very high relative value (10,000) to force the paths out of the solution. 1–4 4–8 8–13 13–16 16–20 20–23 23–25 Shortest path: 1–4–8–13–16–20–23–25 Total shortest distance: 55.
Value 10 10 8 6 6 7 8
Figure for Problem 11-36b
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Network Optimization Models c. In addition to eliminating paths 6–11, 7–12, and 17–20 from the network, the paths used in the solution presented in part b are also eliminated. Thus we eliminate the path 1–4–8–13– 16–20–23–25. Again, this is done in the software by increasing the distances along these paths to a very high relative value (10,000) to force them out of the solution. The new shortest path is 6,400 feet (64). The solution along with the final network follows. 1–2 2–5 5–10 10–14 14–18 18–22 22–25 Shortest path: 1–2–5–10–14–18–22–25 Total shortest distance: 64.
Flow 10 15 8 5 5 6 15
Figure for Problem 11-36c
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