Wireless Network Design: Optimization Models and Solution ...

236 Dag Haugland and Di Yuan
not form a tree. Then the graph defined by the edges has at least two disconnected
components. Each of these graph components must have at least three edges, because
constraints (10.20)–(10.22) cannot be satisfied otherwise. In addition, at least
one of the components is a tree (i.e., contains no cycles), because there are |V | − 1
edges in total. Consider any such tree component, and any one of its leaves, say m.
Select arbitrarily a node n in another component. By (10.18), at node m, exactly one
outgoing q-variable to destination n equals one. Thus if we start at m and follow
this q-variable, we arrive at another node, say j. If j is not a leaf, another outgoing
q-variable will lead to a new node, which cannot be i because of (10.20). Repeating
the argument, we eventually arrive at a leaf node. At this point, (10.18) is violated
since the node has only one edge (used by the q-variable in the incoming direction).
We obtain a contradiction, and therefore RAP-MT is correct for ARAP.
We leave out the proof for SRAP, as the proof is quite lengthy and technical.
However, it follows from Section 5 in [45] that constraints (10.17)-(10.22) and
(10.24) define the set of feasible solutions to the Steiner tree problem. The additional
constraints (10.23) ensure that y is assigned values according to its definition
in Section 10.3.1, and hence RAP-MT is a correct model also for SRAP. Note that
Proposition 10.2 as well as the proof given for MET-F2 apply also to RAP-MT.
10.5.2 LP Strength
For the minimum Steiner tree problem, it was shown in [45] that a multi-tree formulation
is stronger than a multi-commodity flow formulation. We demonstrate that
this property is inherited by RAP.
Proposition 10.6. LP ∗ (RAP-F2) ≤ LP ∗ (RAP-MT).
Proof. Consider any feasible solution (q,y,z,w) to LP-RAP-MT. Define x t =
(q t − q s ) + ∀t ∈ D \ {s}. By (10.20), we get for all i ∈ V and t ∈ D \ {i,s} that
∑ j:(i, j)∈A x t i j − ∑ j:( j,i)∈A x t ji =
∑ j:(i, j)∈A,q t i j ≥qs i j
∑ j:(i, j)∈A,q t i j ≥qs i j
∑ j:(i, j)∈A
�
qt i j − qs �
i j − ∑ j:( j,i)∈A,qt ji≥qs �
ji
qt ji − qs �
ji =
�
qt i j − qs �
i j − ∑ j:( j,i)∈A,qt i j