# Optimal Power and Bandwidth Allocation in a Gaussian Channel¶

by Robert Gowers, Roger Hill, Sami Al-Izzi, Timothy Pollington and Keith Briggs

from Boyd and Vandenberghe, Convex Optimization, exercise 4.62 page 210

Consider a system in which a central node transmits messages to \(n\) receivers. Each receiver channel \(i \in \{1,...,n\}\) has a transmit power \(P_i\) and bandwidth \(W_i\). A fraction of the total power and bandwidth is allocated to each channel, such that \(\sum_{i=1}^{n}P_i = P_{tot}\) and \(\sum_{i=1}^{n}W_i = W_{tot}\). Given some utility function of the bit rate of each channel, \(u_i(R_i)\), the objective is to maximise the total utility \(U = \sum_{i=1}^{n}u_i(R_i)\).

Assuming that each channel is corrupted by Gaussian white noise, the signal to noise ratio is given by \(\beta_i P_i/W_i\). This means that the bit rate is given by:

\(R_i = \alpha_i W_i \log_2(1+\beta_iP_i/W_i)\)

where \(\alpha_i\) and \(\beta_i\) are known positive constants.

One of the simplest utility functions is the data rate itself, which also gives a convex objective function.

The optimisation problem can be thus be formulated as:

minimise \(\sum_{i=1}^{n}-\alpha_i W_i \log_2(1+\beta_iP_i/W_i)\)

subject to \(\sum_{i=1}^{n}P_i = P_{tot} \quad \sum_{i=1}^{n}W_i = W_{tot} \quad P \succeq 0 \quad W \succeq 0\)

Although this is a convex optimisation problem, it must be rewritten in DCP form since \(P_i\) and \(W_i\) are variables and DCP prohibits dividing one variable by another directly. In order to rewrite the problem in DCP format, we utilise the \(\texttt{kl_div}\) function in CVXPY, which calculates the Kullback-Leibler divergence.

\(\text{kl_div}(x,y) = x\log(x/y)-x+y\)

\(-R_i = \text{kl_div}(\alpha_i W_i, \alpha_i(W_i+\beta_iP_i)) - \alpha_i\beta_iP_i\)

Now that the objective function is in DCP form, the problem can be solved using CVXPY.

```
#!/usr/bin/env python3
# @author: R. Gowers, S. Al-Izzi, T. Pollington, R. Hill & K. Briggs
import numpy as np
import cvxpy as cp
```

```
def optimal_power(n, a_val, b_val, P_tot=1.0, W_tot=1.0):
# Input parameters: α and β are constants from R_i equation
n = len(a_val)
if n != len(b_val):
print('alpha and beta vectors must have same length!')
return 'failed', np.nan, np.nan, np.nan
P = cp.Variable(shape=n)
W = cp.Variable(shape=n)
alpha = cp.Parameter(shape=n)
beta = cp.Parameter(shape=n)
alpha.value = np.array(a_val)
beta.value = np.array(b_val)
# This function will be used as the objective so must be DCP;
# i.e. elementwise multiplication must occur inside kl_div,
# not outside otherwise the solver does not know if it is DCP...
R = cp.kl_div(cp.multiply(alpha, W),
cp.multiply(alpha, W + cp.multiply(beta, P))) - \
cp.multiply(alpha, cp.multiply(beta, P))
objective = cp.Minimize(cp.sum(R))
constraints = [P>=0.0,
W>=0.0,
cp.sum(P)-P_tot==0.0,
cp.sum(W)-W_tot==0.0]
prob = cp.Problem(objective, constraints)
prob.solve()
return prob.status, -prob.value, P.value, W.value
```

## Example¶

Consider the case where there are 5 channels, \(n=5\), \(\alpha = \beta = (2.0,2.2,2.4,2.6,2.8)\), \(P_{\text{tot}} = 0.5\) and \(W_{\text{tot}}=1\).

```
np.set_printoptions(precision=3)
n = 5 # number of receivers in the system
a_val = np.arange(10,n+10)/(1.0*n) # α
b_val = np.arange(10,n+10)/(1.0*n) # β
P_tot = 0.5
W_tot = 1.0
status, utility, power, bandwidth = optimal_power(n, a_val, b_val, P_tot, W_tot)
print('Status: {}'.format(status))
print('Optimal utility value = {:.4g}'.format(utility))
print('Optimal power level:\n{}'.format(power))
print('Optimal bandwidth:\n{}'.format(bandwidth))
```

```
Status: optimal
Optimal utility value = 2.451
Optimal power level:
[1.151e-09 1.708e-09 2.756e-09 5.788e-09 5.000e-01]
Optimal bandwidth:
[3.091e-09 3.955e-09 5.908e-09 1.193e-08 1.000e+00]
```