16.11.2012 Views

Wireless Network Design: Optimization Models and Solution ...

Wireless Network Design: Optimization Models and Solution ...

Wireless Network Design: Optimization Models and Solution ...

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

10 Integer Programming <strong>Models</strong> for Power-optimal Trees in <strong>Wireless</strong> <strong>Network</strong>s 231<br />

p (ik). But in any optimal integer solution, if there is any flow leaving node i <strong>and</strong><br />

arriving at destination t, then the flow amount is one unit, <strong>and</strong> it is carried by exactly<br />

one of the outgoing arcs of node i. Thus, ∑ N−1<br />

ℓ=k xt (iℓ) ≤ 1, <strong>and</strong> the same sum is zero if<br />

∑ N−1<br />

ℓ=k y (iℓ) = 0, which proves the validity of (10.8).<br />

In models MET-F1 <strong>and</strong> MET-F2, a node will use at most one of its c<strong>and</strong>idate<br />

power levels. This corresponds to the following inequalities:<br />

∑<br />

j∈V \{i}<br />

yi j ≤ 1,i ∈ V. (10.9)<br />

Although (10.9) is valid, it is redundant for defining the integer optimum. A<br />

simple proof based on contradiction is to assume that, at integer optimum, there are<br />

multiple y-variables being equal to one at some node. Setting these y-variables other<br />

than the one corresponding to the highest power to zero, it is easily verified that<br />

(10.5) <strong>and</strong> (10.8) remain satisfied, <strong>and</strong> the total power decreases. Hence constraints<br />

(10.9) are not needed for the correctness of the models. In fact, they have no impact<br />

on the LP optimum either, as stated in the proposition below.<br />

Proposition 10.2. The LP relaxations of models MET-F1 <strong>and</strong> MET-F2 have an<br />

optimal solution satisfying (10.9).<br />

Proof. Consider MET-F2. Assume (x,y) satisfies (10.7)-(10.8), <strong>and</strong> that y ∈ [0,1] |A| .<br />

We can assume that xt is acyclic for all t, which by (10.7) implies that ∑ N−1<br />

ℓ=k xt (iℓ) ≤ 1.<br />

Define ˆy ∈ [0,1] |A| by<br />

�<br />

y(ik), if ∑<br />

ˆy (ik) =<br />

N−1<br />

ℓ=k y (iℓ) ≤ 1,<br />

�<br />

N−1<br />

1 − ∑ℓ=k+1 y � +<br />

(iℓ) , otherwise.<br />

Then ∑ N−1<br />

ℓ=k ˆy (iℓ) = min � 1,∑ N−1<br />

ℓ=k y �<br />

(iℓ) ≤ 1. It follows that (x, ˆy) satisfies (10.8),<br />

<strong>and</strong> is hence feasible in the LP-relaxation of MET-F2. Since ˆy ≤ y, we also have<br />

∑(i, j)∈A pi j ˆyi j ≤ ∑(i, j)∈A pi jyi j, which completes the proof. For MET-F1 the proof<br />

is similar.<br />

The following proposition states the relation between the LP relaxations of MEP-<br />

F1 <strong>and</strong> MEP-F2. By this proposition <strong>and</strong> Proposition 10.1, the LP relaxations of<br />

the models in this section, following the order in which the models are presented,<br />

become increasingly stronger.<br />

Proposition 10.3. LP ∗ (MET-F1) ≤ LP ∗ (MET-F2).<br />

Proof. The inequality is obtained by observing that any feasible multi-commodity<br />

flow in MET-F2 can be transformed into a feasible single-commodity flow in MET-<br />

F1 by defining xi j = ∑t∈D\{s} xt i j ∀(i, j) ∈ A.<br />

In addition to the flow-based models, two cut-based models for MET were proposed<br />

in [10]. Their underlying idea is that for any node set S containing the source

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!