3.5 Constrained optimization with inequality constraints  (Page 2/2)

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 $x\left(t\right)=t{x}_{1}+\left(1-t\right){x}_{0}={x}_{0}+td$ over $t\in \left[0,1\right]$ . Then, define the constraints limited over the set of points $x\left(t\right)$ :

$\begin{array}{cc}\hfill {G}_{i}\left(t\right)& ={g}_{i}\left(x\left(t\right)\right)={g}_{i}\left(t{x}_{1}+\left(1-t\right){x}_{0}\right)=t\phantom{\rule{0.277778em}{0ex}}g\left({x}_{i}\right)+\left(1-t\right)g\left({x}_{0}\right)=0,\hfill \\ \hfill {H}_{j}\left(t\right)& ={h}_{j}\left(x\left(t\right)\right)={h}_{j}\left(t{x}_{1}+\left(1-t\right){x}_{0}\right)\le t{h}_{j}\left({x}_{1}\right)+\left(1-t\right){h}_{j}\left({x}_{0}\right)\le 0;\hfill \end{array}$

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 $t$ :

$\frac{\partial G}{\partial t}=0=\partial {g}_{i}\left({x}_{0},d\right)=⟨\nabla {g}_{i}\left({x}_{0}\right),d⟩,$

and for $j\in J\left({x}_{0}\right)$ ,

$0\phantom{\rule{0.277778em}{0ex}}\ge \phantom{\rule{0.277778em}{0ex}}\frac{\partial {H}_{j}}{\partial t}=\partial {h}_{j}\left({x}_{0},d\right)=⟨\nabla {h}_{j}\left({x}_{0}\right),d⟩.$

Now consider the function $F\left(t\right)=f\left(x\left(t\right)\right)$ : its derivative is given by

$\frac{\partial F}{\partial t}=\partial f\left({x}_{0},d\right)=⟨\nabla f\left({x}_{0}\right),d⟩=-\sum _{i=1}^{n}⟨\nabla {g}_{i},d⟩{x}_{i}-\sum _{i=1}^{n}{\mu }_{j}⟨\nabla {h}_{j},d⟩\ge 0,$

where we use the third KKT condition. Since $f\left(x\right)$ is convex and $x\left(t\right)$ is affine, then $F\left(t\right)=f\left(x\left(t\right)\right)$ is convex in $t\in \left[0,1\right]$ . 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 $t\in \left[0,1\right]$ . 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 $C=\frac{1}{2}{log}_{2}\left(1+\frac{P}{N}\right)$ , 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 $P={\left[{P}_{1},...,{P}_{n}\right]}^{T}$ that maximizes the total capacity for the set of channels

$C\left(P\right)=\sum _{i=1}^{n}C\left({P}_{i}\right)=\sum _{i=1}^{n}\frac{1}{2}{log}_{2}\left(1,+,\frac{{P}_{i}}{{N}_{i}}\right),$

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

$f\left(P\right)=-C\left(P\right)=-\sum _{i=1}^{n}C\left({P}_{i}\right)=-\sum _{i=1}^{n}\frac{1}{2}{log}_{2}\left(1+\frac{{P}_{i}}{{N}_{i}}\right)$

and also set up the constraints

$\begin{array}{cc}\hfill g\left(P\right)& =\sum _{i=1}^{n}{P}_{i}-{P}_{T}={\mathbf{1}}^{T}P-{P}_{T},\hfill \\ \hfill {h}_{i}\left(P\right)& =-{P}_{i}=-{e}_{i}^{T}P,\phantom{\rule{3.33333pt}{0ex}}i=1,...,n,\hfill \end{array}$

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

$\begin{array}{cc}\hfill \delta f\left(p;h\right)& =\frac{\partial }{\partial \alpha }{\left(f\left(p+\alpha h\right)\right)|}_{\alpha =0}={\left(\left(-,\sum _{i=1}^{n},\frac{{h}_{i}/{N}_{i}}{2\left(ln2\right)\left(1,+,\frac{{P}_{i}}{{N}_{i}},+,\alpha ,\frac{{h}_{i}}{{N}_{i}}\right)}\right)|}_{\alpha =0},\hfill \\ & =-\sum _{i=1}^{n}\frac{{h}_{i}}{2{N}_{i}\left(ln2\right)\left(1,+,\frac{{P}_{i}}{{N}_{i}}\right)}=-\sum _{i=1}^{n}\frac{{h}_{i}}{2\left(ln2\right)\left({N}_{i}+{P}_{i}\right)}=⟨\nabla f\left(p\right),h⟩,\hfill \end{array}$

where the gradient has entries ${\left(\nabla f\left(p\right)\right)}_{i}=-{\left(2\left(ln2\right)\left({N}_{i}+{P}_{i}\right)\right)}^{-1}$ .

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

$\begin{array}{cc}\hfill \sum _{i=1}^{n}{\mu }_{i}{P}_{i}^{*}& =0,\hfill \\ \hfill -\frac{1}{2\left(ln2\right)\left({N}_{i}+{P}_{i}^{*}\right)}+\lambda -{\mu }_{i}& =0,\phantom{\rule{3.33333pt}{0ex}}i=1,...,n.\hfill \end{array}$

The second set of constraints can be written as

$\begin{array}{cc}\hfill \frac{1}{2\left(ln2\right)\left({N}_{i}+{P}_{i}^{*}\right)}& =\lambda -{\mu }_{i},\hfill \\ \hfill {N}_{i}+{P}_{i}^{*}& =\frac{1}{2\left(ln2\right)\left(\lambda -{\mu }_{i}\right)}.\hfill \end{array}$

Consider each inequality constraint ${h}_{i}$ .

• If ${h}_{i}$ is inactive, then ${P}_{i}^{*}>0$ and ${\mu }_{i}=0$ . Then,
$\begin{array}{cc}\hfill {N}_{i}+{P}_{i}^{*}& =\frac{1}{2\lambda \left(ln2\right)},\hfill \\ \hfill {P}_{i}^{*}& =\frac{1}{2\lambda \left(ln2\right)}-{N}_{i}>0.\hfill \end{array}$
• If ${h}_{i}$ is active, then ${P}_{i}^{*}=0$ and so
$\begin{array}{cc}\hfill {N}_{i}& =\frac{1}{2\left(ln2\right)\left(\lambda -{\mu }_{i}\right)},\hfill \\ \hfill {\mu }_{i}& =\lambda -\frac{1}{2{N}_{i}\left(ln2\right)}\ge 0,\hfill \\ \hfill \frac{1}{2{N}_{i}\left(ln2\right)}& \le \lambda ,\hfill \\ \hfill \frac{1}{2\lambda \left(ln2\right)}& \le {N}_{i}.\hfill \end{array}$

To simplify, write $r=\frac{1}{2\lambda \left(ln2\right)}$ ; then, we have two possibilities for each channel $i$ from above:

• If $r-{N}_{i}>0$ (i.e., if ${N}_{i} ), then ${P}_{i}^{*}=r-{N}_{i}$ .
• If $r-{N}_{i}\le 0$ (i.e., if $r\le {N}_{i}$ ) then ${P}_{i}^{*}=0$ .

Thus the power is allocated among the channels using the formula ${P}_{i}^{*}=max\left(0,r-{N}_{i}\right)$ , and the value of $r$ is chosen so that the total power constraints is met:

$\sum _{i=1}^{n}max\left(0,r-{N}_{i}\right)={P}_{T}.$

This is the famous water-filling solution to the multiple channel capacity problem, illustrated in [link] .

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