Stochastic Programming - Index of

DYNAMIC SYSTEMS 135
with z 0 given. Let α be a discount factor. What is often done in this case is
to solve for each s ∈ S the following problem
min
s.t.
T∑
α t r t (z t ,x t ,ξt s )+α T +1 Q(z T +1 )
t=0
z t+1 = G t (z t ,x t ,ξ s t )fort =0,...,T with z 0 given,
A t (z t ) ≤ x t ≤ B t (z t )fort =0,...,T,
⎫
⎪⎬
⎪⎭
(6.1)
where Q(z) represents the value of ending the problem in state z, yielding
an optimal solution x s =(x s 0 ,xs 1 ,...,xs T ). Now what We have a number of
different solutions—one for each s ∈ S. Shall we take the average and calculate
for each t
x t = ∑ p s x s t ,
s∈S
where p s is the probability that we end up on scenario s This is very often
done, either by explicit probabilities or by more subjective methods based on
“looking at the solutions”. However, several things can go wrong. First, if x is
chosen as our policy, there might be cases (values of s) forwhichitisnoteven
feasible. We should not like to suggest to our superiors a solution that might
be infeasible (infeasible probably means “going broke”, “breaking down” or
something like that). But even if feasibility is no problem, is using x a good
idea
In an attempt to answer this, let us again turn to event trees. In Figure 12 we
have T = 1. The top node represents “today”. Then one out of three things can
happen, or, in other words, we have a random variable with three outcomes.
The second row of nodes represents “tomorrow”, and after tomorrow a varying
number of things can happen, depending on what happens today. The bottom
row of nodes takes care of the rest of the time—the future.
This tree represents six scenarios, since the tree has six leaves. In the setting
of optimization that we have discussed, there will be two decisions to be made,
namely one “today” and one “tomorrow”. However, note that what we do
tomorrow will depend on what happens today, so there is not one decision for
tomorrow, but rather one for each of the three nodes in the second row. Hence
x 0 works as a suggested first decision, but x 1 isn’t very interesting. However, if
we are in the leftmost node representing tomorrow, we can talk about an x 1 for
the two scenarios going through that node. We can therefore calculate, for each
version of “tomorrow”, an average x 1 , where the expectation is conditional
upon being on one of the scenarios that goes through the node.
Hence we see that the nodes in the event tree are decision points and the arcs
are realizations of random variables. From our scenario solutions x s we can
therefore calculate decisions for each node in the tree, and these will all make