<< Chapter < Page | Chapter >> Page > |
With some additional assumptions, it can be shown that the KKT conditions can find a global minimizer.
Definition 3 A function $f$ is said to be affine over $\Omega $ if $f\left({\sum}_{i}^{n}{a}_{i}{x}_{i}\right)={\sum}_{1}^{n}{a}_{i}f\left({x}_{i}\right)$ for all ${x}_{1},...,{x}_{n}\in \Omega $ and all weights $\left\{{a}_{i}\right\}$ obeying ${\sum}_{i}^{n}{a}_{i}=1$ .
Theorem 2 (Karush-Kuhn-Tucker Sufficient Conditions) If $f$ and ${h}_{j}$ , $j=1,...,m$ are convex functions and ${g}_{i}$ , $i=1,...,n$ are affine functions, and if the KKT condition are satisfied at a feasible point ${x}_{0}\in \Omega $ then ${x}_{0}$ is a global minimizer of $f$ over $\Omega $ .
Fix ${x}_{1}\in \Omega $ let $d={x}_{1}-{x}_{0}$ . Define a functional $x\left(t\right)=t{x}_{1}+(1-t){x}_{0}={x}_{0}+td$ over $t\in [0,1]$ . Then, define the constraints limited over the set of points $x\left(t\right)$ :
Therefore, all points $x\left(t\right)\in \Omega $ are feasible. Furthermore, note that ${H}_{j}\left(0\right)={h}_{j}\left({x}_{0}\right)=0\ge {H}_{j}\left(t\right)={h}_{j}\left({x}_{t}\right)$ if $j\in J\left({x}_{0}\right)$ . Now, we compute the derivatives of these two functions with respect to $t$ :
and for $j\in J\left({x}_{0}\right)$ ,
Now consider the function $F\left(t\right)=f\left(x\right(t\left)\right)$ : its derivative is given by
where we use the third KKT condition. Since $f\left(x\right)$ is convex and $x\left(t\right)$ is affine, then $F\left(t\right)=f\left(x\right(t\left)\right)$ is convex in $t\in [0,1]$ . Thus $\frac{\partial F}{\partial t}$ is nondecreasing and $\frac{\partial F\left(t\right)}{\partial t}\ge \frac{\partial F\left(0\right)}{\partial t}\ge 0$ for $t\in [0,1]$ . Thus, $F\left(1\right)\ge F\left(0\right)$ or $f\left({x}_{1}\right)\ge f\left({x}_{0}\right)$ . Since ${x}_{1}$ was arbitrary, ${x}_{0}$ is a global minimum of $f$ on $\Omega $ .
Example 2 (Channel Capacity) The Shannon capacity of an additive white Gaussian noise channel is given by $C=\frac{1}{2}{log}_{2}(1+\frac{P}{N})$ , where $P$ is the transmitted signal power and $N$ is the noise variance. Assume that $n$ channels are available with a total transmission power ${P}_{T}={\sum}_{i=1}^{n}{P}_{i}$ available among the channels, where ${P}_{i}$ denotes the power in the ${i}^{th}$ channel. We wish to assign a power profile $P={[{P}_{1},...,{P}_{n}]}^{T}$ that maximizes the total capacity for the set of channels
where ${N}_{i}$ represents the variance of the noise in the ${i}^{th}$ channel.
To solve the problem, we set up an objective function to be minimized
and also set up the constraints
as the values of the powers must be nonnegative. We start by computing the gradients of these functions: for $f$ , we must compute the directional derivative
where the gradient has entries ${\left(\nabla f\left(p\right)\right)}_{i}=-{\left(2(ln2)({N}_{i}+{P}_{i})\right)}^{-1}$ .
For the constraints, it is straightforward to see that $\nabla g\left(P\right)=\mathbf{1}$ and $\nabla {h}_{i}\left(P\right)=-{e}_{i}$ , $i=1,...,n$ .
We begin by assuming that the solution ${P}^{*}$ is a regular point. Then the KKT conditions give that for some $\lambda $ and nonnegative ${\mu}_{1},...,{\mu}_{m}$ we must have
The second set of constraints can be written as
Consider each inequality constraint ${h}_{i}$ .
To simplify, write $r=\frac{1}{2\lambda (ln2)}$ ; then, we have two possibilities for each channel $i$ from above:
Thus the power is allocated among the channels using the formula ${P}_{i}^{*}=max(0,r-{N}_{i})$ , and the value of $r$ is chosen so that the total power constraints is met:
This is the famous water-filling solution to the multiple channel capacity problem, illustrated in [link] .
Notification Switch
Would you like to follow the 'Signal theory' conversation and receive update notifications?