Stochastic Programming - Index of

NETWORK PROBLEMS 279
Figure 1
Network used to demonstrate definitions.
connected if for all Y ⊂N we have Q =[Y,N \Y ] ≠ ∅.
We shall also need the following sets:
F + (Y )={nodes j | k ∼ (i, j) fori ∈ Y }∪Y,
B + (Y )={nodes j | k ∼ (j, i) fori ∈ Y }∪Y.
The set F + (Y )containsY itself plus all nodes that can be reached directly
(i.e. in one step) from a node in Y . Similarly B + (Y )containsY and all nodes
from which Y can be reached directly.
Two other sets that are very similar to F + (Y )andB + (Y )are
F ∗ (Y )={nodes j |∃a directed path from some node i ∈ Y to node j}∪Y,
B ∗ (Y )={nodes j |∃a directed path from node j to some node i ∈ Y }∪Y.
Thus the sets F + and B + pick up immediate successors and predecessors,
whereas F ∗ and B ∗ pick up all successors and predecessors.
Example 6.1 Let us consider Figure 1 to briefly illustrate most of the
concepts we have introduced.
The node set N = {1, 2, 3, 4},andthearcsetE = {1, 2, 3, 4, 5}.Anexample
of an arc is 5 ∼ (2, 3), since arc 5 starts at node 2 and ends at node 3. Let
Y = {1, 3} and Y ′ = {2}. The network G(Y ) consists of nodes 1 and 3, and
arc 2, since that is the only arc connecting nodes in Y . Furthermore, for the
same Y and Y ′ ,wehave[Y,Y ′ ] + = {1}, since arc 1 is the only arc going from
nodes 1 or 3 to node 2. Similarly [Y,Y ′ ] − = {5}. If we define Q =[Y,N \Y ]
then Q + = {1, 4} and Q − = {5}. Therefore Q = {1, 4, 5} is a cut.
Again, with the same definition of Y ,wehave
Furthermore, we have
b(Y )=(1, 0, 1, 0) T , a(Q + )=(1, 0, 0, 1, 0) T .
F + ({1}) ={1, 2, 3}, F ∗ ({1}) ={1, 2, 3, 4},