Stochastic Programming - Index of

RECOURSE PROBLEMS 181
Looking back at our example in (4.1), we find the Jensen lower bound by
calculating φ(E ˜ξ) =φ(0). That has been solved alreadyinSection1.3,where
we found that φ(0) = 126.
3.4.2 Edmundson–Madansky Upper Bound
Again let ˜ξ be a random variable. Let the support Ξ = [a, b], and assume that
q(ξ) ≡ q 0 . As in the previous section, we define φ(ξ) =Q(ˆx, ξ). (Remember
that x is fixed at ˆx.) Consider Figure 14, where we have drawn a linear function
U(ξ) between the two points (a, φ(a)) and (b, φ(b)). The line is clearly above
φ(ξ) for all ξ ∈ Ξ. Also this straight line has the formula cξ + d, and since we
know two points, we can calculate
φ(b) − φ(a)
c = , d = b
b − a
b − a φ(a) − a
b − a φ(b).
We can now integrate, and find (using the linearity of U(ξ))
EU(˜ξ) =
φ(b) − φ(a)
E
b − a
˜ξ +
b
b − a φ(a) −
= φ(a) b − E ˜ξ
b − a + φ(b)E ˜ξ − a
b − a .
a
b − a φ(b)
In other words, if we have a function that is convex in ξ over a bounded
support Ξ = [a, b], it is possible to replace an arbitrary distribution by a
two point distribution, such that we obtain an upper bound. The important
parameter is
p = E ˜ξ − a
b − a ,
so that we can replace the original distribution with
P {˜ξ = a} =1− p, P {˜ξ = b} = p. (4.2)
As for the Jensen lower bound, we have now shown that the Edmundson–
Madansky upper bound can be seen as either changing the distribution and
keeping the problem, or changing the problem and keeping the distribution.
Looking back at our example in (4.1), we have two independent random
variables. Hence we have 2 2 = 4 LPs to solve to find the Edmundson–
Madansky upper bound. Since both distributions are symmetric, the
probabilities attached to these four points will all be 0.25. Calculating this
we find an upper bound of
1(106.6825 + 129.8625 + 122.1375 + 145.3175) = 126.
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