Optimization with equality and inequality constraints - Signal Processing and Optimization for Transceiver Systems - page 744

for otherwise we can define K to be the next smaller integer. So it follows that
A K − 1
λ x −

B K− 1 > 0 ,

and, since 1 /λ y ≥ 1 /λ x , we also have

A K − 1
λ y −

B K− 1 > 0 . (22 . 54)

This means that the solution (22.51) is such that the variable y K− 1 is forced to
be zero, even though the expression (22.54) would have yielded a positive value.
We will now show that such “forced zeros” violate optimality!

Forced zeros and optimality. The KKT condition (22.41) becomes, for k = K − 1,

A K− 1
( y K− 1 + B K− 1 ) 2

− + λ y − μ K− 1 =0 . (22 . 55)

If we force y K− 1 = 0, the KKT multiplier μ K− 1 has to be

A K − 1
B K− 1

μ K− 1 = λ y − < 0 , (22 . 56)

where the inequality follows from Eq. (22.54). But since the KKT multipliers are
required to satisfy μ k ≥ 0, Eq. (22.56) violates optimality.

Summarizing, whenever the expression
A k
λ y −

B k

is positive, we have to take y k to be equal to this expression instead of forcing it
to be zero. Otherwise KKT conditions, which are necessary for optimality, will
be violated. This shows that when there exists an optimal set

for some K
as described in Eq. (22.50), we cannot find a smaller optimal set {y k } as in Eq.
(22.51).

{ x k }

22.4.3 How does power in a channel depend on noise?

When we optimized the capacity of a set of independent channels by power
allocation we found that the optimal solution allocates more power to less noisy
channels. How about the MMSE solution discussed in this section? Does this
also allocate more power to less noisy channels? From Eq. (22.48) we see that
the power allocated to the k th channel (0 ≤ k ≤ K − 1) is

( k )= α k σ s = Aσ q k − σ q k =( A

− σ q k ) σ q k ,