Optimization with equality and inequality constraints - Signal Processing and Optimization for Transceiver Systems - page 735

i

R s

+

R l

v

−

Figure 22.1 . An electrical circuit with a voltage source v , source resistance R s , and
load resistance R l .

If μ = 0 we have to have R l = 0 (because of Eq. (22.24)), which means μ< 0

from the preceding equation.

Since this contradicts the first KKT condition

(22.22), we cannot have μ = 0. Setting μ = 0 in Eq. (22.25) yields the solution

(22.21) indeed. Thus, from a formal view point, the KKT condition in this
example yields μ = 0, which is equivalent to ignoring the constraint R l ≥ 0.

Maximixing power delivered to R s . An interesting modification of the problem is
this: what is the optimal load resistance R l such that the power delivered to the
fixed source resistance R s (assumed nonzero) is maximized? The power in R s is
given by

s = v
R s + R l

2 R s .

P

Proceeding as before we now find that the stationary equation of the KKT con-
ditions yields

2 v 2 R s
( R s + R l ) 3 −

μ =0 .

This shows that μ cannot be zero (since R s = 0). Since μR l = 0 (third KKT

condition), it follows that R l has to be zero. Thus, maximum power is delivered
to R s when the load R l is zero, as one would expect! We say that the solution is
at the boundary of the constraint set, because R l cannot be any smaller.

22.3 Maximizing channel capacity

In many communication systems we are faced with maximizing the quantity

N− 1

x k
Q k ) .

C =0 . 5 log 2 (1 +

i =0

This quantity is proportional to the capacity of a parallel set of N independent
channels [Cover and Thomas, 1991] with input powers x k and zero-mean additive
Gaussian noise components with powers Q k , as shown in Fig. 22.2.