Stochastic Programming - Index of

236 STOCHASTIC PROGRAMMING
5. Show that for a convex function ϕ and any arbitrary z the subdifferential
∂ϕ(z) is a convex set. [Hint: For any subgradient (9.7) has to hold.]
6. Assume that you are faced with a large number of linear programs that
you need to solve. They represent all recourse problems in a two-stage
stochastic program. There is randomness in both the objective function
and the right-hand side, but the random variables affecting the objective
are different from, and independent of, the random variables affecting the
right-hand side.
(a) Argue why (or why not) it is a good idea to use some version of
bunching or trickling down to solve the linear programs.
(b) Given that you must use bunching or trickling down in some version,
how would you organize the computations
7. First consider the following integer programming problem:
min{cx | Ax ≤ h, x i ∈{0,...,b i }∀i}.
x
Next, consider the problem of finding Eφ(˜x), with
φ(x) =min{cy | Ay ≤ h, 0 ≤ y ≤ x}.
y
(a) Assume that you solve the integer program with branch-and-bound.
Your first step is then to solve the integer program above, but with
x i ∈{0,...,b i }∀i replaced by 0 ≤ x ≤ b. Assume that you get ˆx.
Explain why ˆx can be a good partitioning point if you wanted to find
Eφ(˜x) by repeatedly partitioning the support, and finding bounds on
each cell. [Hint: It may help to draw a little picture.]
(b) We have earlier referred to Figure 18, stating that it can be seen
as both the partitioning of the support for the stochastic program,
and partitioning the solution space for the integer program. Will the
number of cells be largest for the integer or the stochastic program
above Note that there is not necessarily a clear answer here, but you
should be able make arguments on the subject. Question (a) may be
of some help.
8. Look back at Figure 17. There we replaced one distribution by two others:
one yielding an upper bound, and one a lower bound. The possible values
for these two new distributions were not the same. How would you use the
ideas of Jensen and Edmundson–Madansky to achieve, as far as possible,
the same points You can assume that the distribution is bounded. [Hint:
The Edmundson–Madansky distribution will have two more points than
the Jensen distribution.]