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274 STOCHASTIC PROGRAMMING<br />

direction, is found to be in pos W , we can halve the number <strong>of</strong> points in A.<br />

In some cases we may therefore reduce the testing to one single problem.<br />

It is <strong>of</strong> importance to understand that the reduction in the size <strong>of</strong> A has<br />

two positive aspects. First, if we do not have (or do not know that we have)<br />

relatively complete recourse, the test for feasibility, and therefore generation <strong>of</strong><br />

feasibility cuts, becomes much easier. But equally important is the fact that<br />

it tells us something about our problem. If a column from H is in pos W ,<br />

we have found a direction in which we can move as far as we want without<br />

running into feasibility problems. This will, in a real setting, say something<br />

about the random effect we have modelled using that column.<br />

5.4 Bibliographical Notes<br />

Preprocessing and similar procedures have been used in contexts totally<br />

different from ours. This is natural, since questions <strong>of</strong> model formulations and<br />

infeasibilities are equally important in all areas <strong>of</strong> mathematical programming.<br />

For further reading, consult e.g. Roodman [7], Greenberg [3, 4, 5] or Chinneck<br />

and Dravnieks [1].<br />

An advanced algorithm for finding frames can be found in Wets and<br />

Witzgall [12]. Later developments include the work <strong>of</strong> Rosen et al. [8] and Dulá<br />

et al. [2]. The algorithm for finding a support was described by Tschernikov[9],<br />

and later also by Wets [11]. For computational tests using the procedure see<br />

Wallace and Wets [10]. Similar procedures for networks will be discussed in<br />

Chapter 6.<br />

For an overview <strong>of</strong> methods for extreme point enumeration see e.g.<br />

MattheissandRubin[6].<br />

Exercises<br />

1. Let W be the coefficient matrix for the following set <strong>of</strong> linear equations:<br />

x + 1 2 y − z + s 1 =0,<br />

2x + z + s 2 =0,<br />

x, y, z, s 1 , s 2 ≥ 0.<br />

(a) Find a frame <strong>of</strong> pos W .<br />

(b) Draw a picture <strong>of</strong> pos W , and find the generators <strong>of</strong> pol pos W by<br />

simple geometric arguments.<br />

(c) Find the generators <strong>of</strong> pol pos W by using procedure support in<br />

Figure 5. Make sure you draw the cones pos W and pol pos W after<br />

each iteration <strong>of</strong> the algorithm, so that you see how it proceeds.

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