Stochastic Programming - Index of

NETWORK PROBLEMS 283
then we do not need any of the inequalities is fairly obvious. The other part of
the proposition is much harder to prove, namely that if G(Y )andG(N \Y )
are both connected then the inequalities corresponding to Y and N\Y are
both needed. We shall not try to outline the proof here.
Proposition 6.2 might not seem very useful. A straightforward use could
still require the enumeration of all subsets of N , and for each such subset
achecktoseeifG(Y )andG(N \Y ) are both connected. However, we can
obtain more than that.
The first important observation is that the result refers to the connectedness
of two networks—both the one generated by Y and the one generated by N\Y .
Let Y 1 = N\Y . If both networks are connected, we have two inequalities that
we need, namely
b(Y ) T β ≤ a(Q + ) T γ
and
b(Y 1 ) T β = b(N \Y ) T β ≤ a(Q − ) T γ.
On the other hand, if at least one of the networks is disconnected, neither
inequality will be needed. Therefore, checking each subset of N means doing
twice as much work as needed. If we are considering Y and discover that both
G(Y )andG(Y 1 = N\Y ) are connected, we write down both inequalities at
the same time. An easy way to achieve this is to disregard some node (say
node n) from consideration in a full enumeration. This way, we will achieve
n ∈N\Y for all Y we investigate. Then for each cut where the connectedness
requirement is satisfied we write down two inequalities. This will halve the
number of subsets to be checked.
In some cases it is possible to reduce the complexity of a calculation by
collapsing nodes. By this, we understand the process of replacing a set of
nodes by one new node. Any other node that had an arc to or from one of
the collapsed nodes will afterwards have an arc to or from the new node: one
for each original arc. If Y is a set of nodes, we let A(Y )bethesetoforiginal
nodes represented by the present node set Y as a result of collapsing.
To simplify statements later on we shall also need a way to simply state
which inequalities we want to write down. An algorithm for the capacitated
case is given in Figure 4.
Note that we allow the procedure to be called with Y = ∅. This is a
technical devise to ensure consistent results, but you should not let that
confuse you at the present time. Based on Proposition 6.2, it is possible to
develop procedures that in some cases circumvent the exponential complexity
arising from checking all subsets of N . We shall use Figure 5 to illustrate some
of our points.
Proposition 6.3 If B + (i)∪F + (i) ={i, j} then nodes i and j can be collapsed
after the inequalities generated by CreateIneq({i}) have been created.