3.3. The Operational
Policy Problem (Problem
HI)
101
Iteration 7
Arc
Ti
— ITJ
—bu
qu
/»·,·
State In kilter?
(1, 2)
0
-4
2
-2
2 =
C
Yes
(1, 3)
0
-5
5
0
0 = In Β
Yes
(2, 3)
4
-5
1
0
1 =
Β
Yes
(2, 4)
4
-7
3
0
1 > hi Β
Yes
(3, 2)
5
-4
1
2
0
In A
Yes
(3, 4)
5
-7
6
4
1 = hi A
Yes
(4, 1)
7
-0
0
7
2
la Αχ
No
1. Arc (2, 3) has changed from state C to state B; therefore flow can be
decreased in (2, 3) without driving it out of kilter.
2. State of arc (4, 1) is Αχ ; increase /« to hi.
3. Find path from node 1 to node 4 by labeling procedure.
Labeling procedure
Node
Label
1
(4+, 1)
3
(1
+ , 1)
2
(3-, 1)
4
(2+ 1)
Breakthrough has occurred.
In fact, arc (4, 1) is brought into kilter; therefore we have found the
optimum solution. The following tabulation gives the optimum flow values
for all the arcs.
Arc
—π,·
-6
QU
fu
State In kilter?
(1, 2)
0
-4
2
-2
2 = u 12
C
Yes
(1, 3)
0
-5
5
0 Ixt < 1 < ιΐχζ Β
Yes
(2, 3)
4
-5
1
0
0 = In
Β
Yes
(2, 4)
4
-7
3
0 hi < 2 < «24 Β
Yes
(3, 2)
5
-4
1
2
0 = ^32
A
Yes
(3, 4)
5
-7
6
4
1 = hi
A
Yes
(4, 1)
7
-4
0
7 3 - hx
A
Yes