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Wireless Network Design: Optimization Models and Solution ...

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324 Nitin Salodkar <strong>and</strong> Abhay Kar<strong>and</strong>ikar<br />

In this chapter, we consider ergodic capacity (also termed as throughput capacity<br />

or expected capacity in the literature). This notion of capacity measures the rates<br />

achievable in the long run averaged over the channel variations. Here, the channel is<br />

assumed to vary sufficiently fast, yet the variations are slow enough such that ‘reasonably’<br />

long codes can be transmitted. We first derive an expression for the ergodic<br />

capacity of a single user point-to-point link under an average power constraint <strong>and</strong><br />

then extend the notion to multiuser scenario.<br />

14.2.1 Point-to-Point Capacity with Perfect Transmitter CSI<br />

Consider a single user wireless channel depicted in Figure 14.1. We assume transmitter<br />

has perfect knowledge of CSI. Let Pn denote the transmission power in slot n.<br />

Let X1 = x1,...,XM = xM be a given realization of channel states. We assume that<br />

the transmitter has an average power constraint of ¯P.<br />

Bits<br />

Fig. 14.1 Point-to-point transmission model<br />

X<br />

Feedback path with zero delay<br />

The restriction on the average power expenditure makes the problem of achieving<br />

capacity to be a power allocation problem. Specifically, the problem is to determine<br />

the transmission power in each slot that maximizes the information rate (<strong>and</strong><br />

hence achieves capacity), while keeping the average power expenditure below the<br />

prescribed limit. The problem can be stated as:<br />

max<br />

P 1,...,PM<br />

1<br />

M<br />

M �<br />

∑ log<br />

n=1<br />

1 + PnXn<br />

N0<br />

�<br />

, (14.3)<br />

subject to,<br />

M 1<br />

M ∑ Pn = ¯P. (14.4)<br />

n=1<br />

This is a constrained optimization problem <strong>and</strong> can be solved using st<strong>and</strong>ard Lagrangian<br />

relaxation [8]. Let x + denote max(0,x). A solution to the optimization<br />

problem stated in (14.3) <strong>and</strong> (14.4) is a policy that determines the optimal power in<br />

nth slot to be:

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