Stochastic Programming - Index of

NETWORK PROBLEMS 299
1
Figure 16
2.
1
(0,2)
(-1,0)
2
4
2
(0,1) 3
2
1
4
3
3
(0,1)
8
6
(-3,1)
1
(0,0)
4 7 5
5
+/- 1
2
1
(-2,2)
2
(0,0)
-/+ 1
Arc capacities after the update based on φ(E ˜ξ, 0) and nodes 1 and
units, and we have so far solved for a demand of 1. Therefore we must now
look at a demand of 1 and a supply of 1 in node 4, based on the arc capacities
in Figure 16. In that figure we have updated the capacities from Figure 15
based on the solutions for node 2.
A supply in node 4 gives us the solution
y 4+ =(0, 0, 0, 0, 0, 0, 1, 0) T ,
with a cost of 2. One unit demand, on the other hand, gives us
y 4− =(0, 0, 0, 0, 0, −1, 0) T ,
with a cost of −2. The parameters are therefore d + 4
arc capacities in Figure 17.
What we have found so far is as follows:
=2=d− 4 . This leaves the
φ(ξ,η) =22+H(η)
{
5(ξ1 − 2) if ξ
+
1 ≥ 2,
3(ξ 1 − 2) if ξ 1 < 2,
{
3ξ2 if ξ
+
2 ≥ 0,
ξ 2 if ξ 2 < 0,
{
2(ξ4 +1) ifξ
+
4 ≥−1,
2(ξ 4 +1) ifξ 4 < −1.
If, for simplicity, we assume that all distributions are uniform, we easily