Stochastic Programming - Index of

RECOURSE PROBLEMS 205
is best achieved by looking ahead using (5.2). Our general advice is therefore
that in the setting of two (or more) stages one should seek a strategy that
minimizes the final number of cells, and that it is worthwhile to pay quite a
lot per iteration to achieve this goal.
3.6 Simple Recourse
Let us consider the particular simple recourse problem
min{c T x + E˜ξQ(x, ˜ξ) | Ax = b, x ≥ 0}, (6.1)
where
Q(x, ξ) =min{q +T y + + q −T y − | y + − y − = ξ − Tx, y + ≥ 0,y − ≥ 0}.
Hence we assume
and in addition
W =(I,−I),
T (ξ) ≡ T (constant),
h(ξ) ≡ ξ,
q = q + + q − ≥ 0.
In other words, we consider the case where only the right-hand side is
random, and we shall see that in this case, using our former presentation
h(ξ) =h 0 + ∑ i hi ξ i , we only need to know the marginal distributions of the
components h j (ξ) ofh(ξ). However, stochastic dependence or independence
of these components does not matter at all. This justifies the above setting
h(ξ) ≡ ξ.
By linear programming duality, we have for the recourse function
Q(x, ξ)
=min{q +T y + + q −T y − | y + − y − = ξ − Tx, y + ≥ 0,y − ≥ 0}
=max{(ξ − Tx) T π |−q − ≤ π ≤ q + }. (6.2)
Observe that our assumption q ≥ 0 is equivalent to solvability of the secondstage
problem. Defining
χ := Tx,
the dual solution π ⋆ of (6.2) is obvious:
π ⋆ i = {
q
+
i if ξ i − χ i > 0,
−q − i if ξ i − χ i ≤ 0.