Stochastic Programming - Index of

RECOURSE PROBLEMS 165
for feasibility. We should like to check for feasibility in such a way that if the
given problem is not feasible, we automatically come up with a generator of
pol pos W . For the discussion, we shall use Figure 5.
We should like to find a σ such that
σ T t ≤ 0 for all t ∈ pos W.
This is equivalent to requiring that σ T W ≤ 0. In other words, σ should be
in the cone pol pos W . But, assuming that the right-hand side h(ξ) − T (ξ)ˆx
produces an infeasible problem, we should at the same time require that
σ T [h(ξ) − T (ξ)ˆx] > 0,
because if we later add the constraint σ T [h(ξ) − T (ξ)x] ≤ 0 to our problem,
we shall exclude the infeasible right-hand side h(ξ) − T (ξ)ˆx without leaving
out any feasible solutions. Hence we should like to solve
max
σ
{σT (h(ξ) − T (ξ)ˆx) | σ T W ≤ 0, ‖σ‖ ≤1},
where the last constraint has been added to bound σ. We can do that, because
otherwise the maximal value will be +∞, and that does not interest us since
we are looking for the direction defined by σ. If we had chosen the l 2 norm, the
maximization would have made sure that σ came as close to h(ξ) − T (ξ)ˆx as
possible (see Figure 5). Computationally, however, we should not like to work
with quadratic constraints. Let us therefore see what happens if we choose
the l 1 norm. Let us write our problem differently to see the details better. To
do that, we need to let the unconstrained σ be replaced by σ 1 − σ 2 ,where
σ 1 ,σ 2 ≥ 0. We then get the following:
max{(σ 1 −σ 2 ) T (h(ξ)−T (ξ)ˆx) | (σ 1 −σ 2 ) T W ≤ 0, e T (σ 1 +σ 2 ) ≤ 1, σ 1 ,σ 2 ≥ 0},
where e is a vector of ones. To more easily find the dual of this problem, let
us write it down in a more standard format:
max(σ 1 − σ 2 ) T (h(ξ) − T (ξ)ˆx) dual variables
W T σ 1 − W T σ 2 ≤ 0
y
e T σ 1 + e T σ 2 ≤ 1 t
σ 1 ,σ 2 ≥ 0
From this, we find the dual linear program to be
min{t | Wy+ et ≥ (h(ξ) − T (ξ)ˆx), −Wy+ et ≥−(h(ξ) − T (ξ)ˆx), y,t ≥ 0}.
Note that if the optimal value in this problem is zero, we have Wy =
h(ξ) − T (ξ)ˆx, so that we do indeed have h(ξ) − T (ξ)ˆx ∈ pos W , contrary