Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
11-29. a. Given the problem data, Grey can use the minimal-spanning tree model to determine
the least-cost approach to connect all houses to cable TV. As seen below, Grey should use
branches 1, 2, 3, 4, 6, 8, 9, and 11.
Beginning Data
Start Node
End Node
Cost
Branch 1
1
2
5
Branch 2
1
3
6
Branch 3
1
4
6
Results
Start
End
Node
Node
Cost
Include
Cost
Branch 1
1
2
5
Y
5
Branch 2
1
3
6
Y
6
Branch 3
1
4
6
Y
6
Branch 4
1
5
5
Y
5
Total
34
Solution steps
Starting
Ending
Cumulative
Branch
Node
Node
Cost
Cost
Branch 11
8
9
2
2
Branch 8
5
8
4
6
11-29. b. Grey can make the necessary calculations using the minimal-spanning tree model. The
results are below.
Results
Start
End
Node
Node
Cost
Include
Cost
Branch 1
1
2
5
Y
5
Branch 2
1
3
1
Y
1
Branch 3
1
4
1
Y
1
Branch 4
1
5
1
Y
1
Total
20
Solution steps
Starting
Ending
Cumulative
Branch
Node
Node
Cost
Cost
Branch 11
8
9
2
2
Branch 8
5
8
4
6
11-30. a. Using the shortest-route technique, George can determine the best way to go from
Quincy to Old Bainbridge. The data and results are below. As can be seen, the shortest route is to
take branches 2, 4, 7, 8 and 9 with a minimum distance of 1,200 miles.
Start Node
End Node
Distance
Branch 1
1
2
3
Branch 2
1
3
2
Branch 3
2
4
3
Branch 4
3
5
3
Start
End
Cumulative
Node
Node
Distance
Distance
Branch 2
1
3
2
2
Branch 4
3
5
3
5
Node
Node
Node
Node
1
2
3
4
Node 1
0
3
2
6
Node 2
3
0
5
3
Node 3
2
5
0
4
Node
Node
Node
Node
5
6
7
8
Node 1
5
9
7
12
Node 2
4
7
6
10
Node 3
3
7
5
10
11-30. b. George can use the shortest-route model to determine the impact of the changes. The
results are below. As you can see, the new shortest route is 1,000 miles (10 in the printout since
units are in 100s).
Start Node
End Node
Distance
Branch 1
1
2
3
Branch 2
1
3
2
Branch 3
2
4
3
Branch 4
3
5
1
Start
End
Cumulative
Node
Node
Distance
Distance
Branch 2
1
3
2
2
Node
Node
Node
Node
1
2
3
4
Node 1
0
3
2
4
Node 2
3
0
5
3
Node 3
2
5
0
2
Node
Node
Node
Node
5
6
7
8
Node 1
3
7
5
10
Node 2
4
7
6
10
Node 3
1
5
3
8
Node 4
1
4
3
7
11-31. a. South Side Oil and Gas can use the maximal-flow technique to determine the maxi-
mum flow through the network. As seen in the tables below, two paths are used with a total flow
rate of 1,500 gallons.
Total Flow 15
Start
End
Reverse
Node
Node
Capacity
Capacity
Flow
Branch 1
1
2
10
4
10
Branch 2
1
3
8
2
5
Branch 3
2
4
12
1
10
Branch 4
2
5
6
6
0
Iterations
Iteration
Path
Flow
Cumulative Flow
1
1→ 2→ 4→ 6→ 8
10
10
2
1→ 3→ 5→ 7→ 8
5
15
11-31. b. The results for South Side Oil and Gas are below. As you can see, the changes did not
have any impact on the maximal flow, which remains at 15 or 1,500 gallons. The calculations are
summarized below.
Total Flow 15
Start
End
Reverse
Node
Node
Capacity
Capacity
Flow
Branch 1
1
2
10
4
10
Branch 2
1
3
8
2
5
Branch 3
2
4
12
1
10
Branch 4
2
5
0
0
0
11-32. Given the problem data, the network module in QM for Windows gives the following
minimal-spanning tree results. The branches 1–3, 3–2, 3–5, 5–4 and 5–6 are used to connect the
nodes, and the total distance is 40.
Start node
End node
Cost
Include
Cost
1
2
12
1
3
8
Y
8
2
3
7
Y
7
2
4
10
Figure for Problem 11-32
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
End
Reverse
Node
Node
Capacity
Capacity
Flow
Branch 1
1
2
80
0
80
Branch 2
1
3
50
0
50
Branch 3
1
4
60
0
60
Branch 4
2
3
30
30
20
Branch 14
7
8
50
0
50
Iteration
Path
Flow
Cumulative Flow
1
1→ 2→ 5→ 8
60
60
2
1→ 3→ 6→ 8
50
110
11-34. QM for Windows indicates that total capacity is not affected. Other streets can be used to
still accommodate 190 cars.
Cumulative
Iteration
Path
Flow
Flow
1
1→ 2→ 5→ 8
60
60
2
1→ 4→ 7→ 8
50
110
3
30
140
11-35. Using the shortest-route technique in QM for Windows, we find the minimum total dis-
tance to be 16 as shown in the table.
Start
End
Cumulative
Node
Node
Distance
Distance
Branch 2
1
3
6
6
Branch 7
3
6
3
9
Branch 12
6
7
7
16
11-36. a. The solution is 4,900 feet. This is almost 1 mile. The solution along with the final
network is given below and on the next page.
Value
1–3
9
3–7
6
7–12
8
Figure for Problem 11-36a
b. Eliminating the paths 6–11, 7–12, and 17–20 has changed the shortest route to 5,500 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 dis-
tance 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.
Value
1–4
10
4–8
10
8–13
8
Figure for Problem 11-36b
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 short-
est path is 6,400 feet (64). The solution along with the final network follows.
Flow
1–2
10
2–5
15
5–10
8
Figure for Problem 11-36c
SOLUTIONS TO INTERNET HOMEWORK PROBLEMS
11-37. The shortest distance from farm 1 to farm 6 is found using QM for Windows. The results
are: Total distance = 17
Start
End
Cumulative
Node
Node
Cost
Distance
Branch 2
1
3
10
10
11-38. Using the minimal-spanning tree in QM for Windows, the minimum amount of cable is
22 miles.
Start
End
Cumulative
Node
Node
Cost
Cost
Branch 1
1
2
8
8
Branch 3
2
3
4
12
11-39. Using the minimal-spanning tree in QM for Windows, the minimum amount of cable is
27 miles.
Start
End
Node
Node
Cost
Include
Cost
Branch 1
1
2
8
Y
8
Branch 2
1
3
10
Branch 6
3
4
6
Branch 7
3
5
2
Y
2
Branch 8
4
5
3
Y
3
Branch 9
4
6
6
11-40. Using QM for Windows, the maximal flow is 7 hundred. The paths are
Iteration
Path
Flow
Cumulative Flow
1
1→ 3→ 4→ 6
3
3
11-41. The increased capacity from node 1 to node 3 does not increase the maximal flow. It is
still 7 hundred. There is already sufficient capacity from node 1.
SOLUTION TO BINDER’S BEVERAGE CASE
This is a shortest-route problem. With the data given in the problem, the shortest-route model
can be used to determine the minimum time in minutes required to go from the plant to the ware-
house in east Denver. The results are on the next page. As you can see, the best route is to take
North Street to I-70. At Exit 137, South Street is taken to the warehouse. This route takes one
hour (60 minutes).
Data
Start Node
End Node
Distance
North Street
1
2
20
I 70—A
2
4
5
I 70—B
4
8
10
High Street—A
1
3
20
High Street—B
3
4
20
6 Ave—A
4
5
15
6 Ave—B
5
6
25
6 Ave—C
6
10
40
Rose Street—A
6
7
20
Shortest Path
Total distance = 60
Start
End
Cumulative
Node
Node
Distance
Distance
North Street
1
2
20
20
Minimum distance matrix
Node
Node
Node
Node
Node
1
2
3
4
5
Node 1
0
20
20
25
30
Node 2
20
0
25
5
20
Node 3
20
25
0
20
15
Node 10
60
40
55
35
50
Node
Node
Node
Node
Node
6
7
8
9
10
Node 1
55
50
35
45
60
Node 2
45
35
15
25
40
Node 3
40
35
30
40
55
Node 4
40
30
10
20
35
SOLUTION TO SOUTHWESTERN UNIVERSITY TRAFFIC PROBLEMS
This is a maximal-flow problem. Using QM for Windows we have the following results:
Iteration
Path
Flow
Cumulative Flow
1
1→ 2→ 5→ 8
12
12
1. The capacity without any expansion is 28 (thousand) cars per hour. This would indicate that a
serious problem will exist if there are 33,000 cars per hour leaving the stadium. The problem is
2. To get the capacity to 33, we must add an additional 5 units. You could add 3 units of capacity
from node 1 to node 4. This matches the inflow to the outflow at node 4. Also, expanding the
SOLUTIONS TO INTERNET CASE
SOLUTION TO RANCH DEVELOPMENT PROJECT CASE
1. The minimum distance that will connect all houses to the water and sewer lines is 10,000 feet
(100). The solution along with the final network follows:
Start
End
Branch
Node
Node
Cost
Include
Cost
Branch 1
1
2
3
Y
3
Branch 2
1
5
2
Y
2
Branch 13
8
9
2
Y
2
Branch 14
9
13
7
Y
7
Branch 15
10
11
8
Branch 16
10
15
11
Branch 17
11
12
2
Y
2
Branch 28
19
24
5
Y
5
Branch 29
20
21
4
Y
4
Branch 30
21
22
1
Y
1
Branch 31
21
25
4
Y
4
Branch 32
22
23
6
Y
6
2. Moving footprint number 16 to accommodate the expansion of the pond area has increased
the minimum total distance to 10,100 feet (101). A decision now has to be made about whether
the increased distance and cost for the water and sewer system is worth the additional expected
property prices. The solution along with the final network follows.
Start
End
Branch
Node
Node
Cost
Include
Cost
Branch 1
1
2
3
Y
3
Branch 2
1
5
2
Y
2
Branch 3
2
3
1
Y
1
Branch 9
5
10
5
Y
5
Branch 10
6
7
2
Y
2
Branch 11
6
11
4
Y
4
Branch 12
7
12
4
Branch 13
8
9
2
Y
2
Branch 14
9
13
7
Y
7
Branch 25
17
23
8
Y
8
Branch 26
18
19
2
Y
2
Branch 27
19
20
2
Y
2
Branch 28
19
24
5
Y
5
Branch 29
20
21
4
Y
4
