Stochastic Programming - Index of

292 STOCHASTIC PROGRAMMING
It is not quite clear if capacity increases can take on any values, or just some
predefined ones. Also, the cost structure of the possible investments are not
yet clear. Even so, we are asked to analyse the problem, and create a better
basis for decisions.
The first thing we must do, to use the procedures of this chapter, is to
make sure that, technically speaking, we have a network (as defined at the
start of the chapter). A close look will reveal that a network must have equality
constraints at the node, i.e. flow in must equal flow out. That is not the case
in our little network. If City 3 has spare capacity, we do not have to send
extra sewage to the city, we simply leave the capacity unused if we do not
need it. The simplest way to take care of this is to introduce some new arcs
in the network. They are shown with dotted lines in Figure 10. Finally, to
have supply equal to demand in the network (remember from Proposition 6.1
that this is needed for feasibility), we let the external flow in node 4 be the
negative of the sum of external flows in the other three nodes.
You may wonder if this rewriting makes sense. What does it mean when
“sewage” is sent along a dotted line in the figure The simple answer is that
the amount exactly equals the unused capacity in the city to which the arc
goes. (Of course, with the given numbers, we realize that no arc will be needed
from node 4 to node 1, but we have chosen to add it for completeness.)
Now, to learn something about our problem, let us apply Proposition 6.2
to arrive at a number of inequalities. You may find it useful to try to write
them down. We shall write down only some of them. The reason for leaving
out some is the following observation: any node set Y that is such that Q +
contains a dotted arc from Figure 10 will be uninteresting, because
a(Q + ) T γ = ∞,
so that the inequality says nothing interesting. The remaining inequalities are
as follows (where we have used that all existing pipes have a capacity of 5 per
unit time).
⎫
β 1 ≤ 10 + x 1 + x 3 + x 4 ,
β 2 ≤ 10 + x 1 + x 2 ,
β 3 ≤ 5 + x 3 , ⎪⎬
β 1 + β 2 + β 3 ≤ 10 + x 1 + x 3 + x 4 ,
(4.1)
β 1 + β 2 ≤ 15 + x 1 + x 2 + x 3 + x 4 ,
β 1 + β 3 ≤ 10 + x 1 + x 3 + x 4 , ⎪⎭
β 2 + β 3 ≤ 10 + x 1 + x 3 .
Letusfirstnotethatifwesetallx i = 0 in (4.1), we end up with a number
of constraints that are not satisfied for all possible values of β. Hence, as we
already know, there is presently a chance that sewage will be dumped.
However, our interest is mainly to find out about which investments to
make. Let us therefore rewrite (4.1) in terms of x i rather than β i :