# Network Modeling

Topics: Radeon R520, Flow network Pages: 2 (317 words) Published: March 27, 2013
5.a.

b.MIN5 X12 + 13 X13 + 45 X15 + 105 X17 + 27 X23 + 19 X24 + 50 X25 + 95 X27 +
14 X34 + 30 X35 + 32 X36 + 14 X43 + 35 X45 + 24 X46 + 35 X54 + 18 X56 + 25 X57 +
24 X64 + 18 X65 + 17 X67

ST-X12 - X13 - X15 - X17 = -1
+X12 - X23 - X24 - X25 - X27 = 0
+X13 + X23 + X43- X34 - X35 - X36 = 0
+X24 + X34 + X54 + X64 - X43 - X45 - X46 = 0
+ X15 + X25 + X35 + X45 + X65 - X54 - X56 - X57 = 0
+ X36 + X46 + X56 - X64 - X65 - X67 = 0
+ X17 + X27 + X57 + X67 = +1
Xij 0

The solution is: X13=X36=X67=1 with a minimum total cost of \$62.

17.a. One solution is:

b.310 feet.

21.This is a transportation problem.

25.Because there is a 10% loss of flow on all arcs going to node 4, a total of 702/0.9 = 780 units must flow into node 4. Thus, we can simply increase the demand at node 4 to 780 and assume no loss of flow occurs on arcs leading into this node. Similarly, only 608/1.05 = 579.05 units must flow into node 5. Thus, we can simply decrease the demand at node 5 to 579.05 and assume no gain of flow occurs on the arcs leading into this node.

30.MAXX71

ST+X71 -X12 - X13 - X14 = 0
+X12 -X23 - X25 = 0
+X13 +X23 +X43 -X35 -X36 - X37 = 0
+X14 -X43 -X46 = 0
+X25 +X35 -X57 = 0
+X36 +X46 -X67 = 0
+X37 +X57 +X67 = 0
0 X12 8
0 X13 9
0 X14 7
0 X23 7
0 X25 10
0 X35 8
0 X36 7
0 X37 9
0 X43 6
0 X46 9
0 X57 9
0 X67 11

The optimal solution is: X12 = 8, X13 = 9, X14 = 7, X25 = 8, X37 = 9, X46 = 7, X57 = 8, X67 = 7.
Maximal flow = 24 tons of sewage per hour.