All of the examples can be found in Jupyter notebook form here.

Optimal advertising

This example is taken from https://web.stanford.edu/~boyd/papers/pdf/cvx_applications.pdf.

Setup:

We have $m$ adverts and $n$ timeslots
The total traffic in time slot $t$ is $T_t$
The number of ad $i$ displayed in period $t$ is $D_{it} \geq 0$
We require $\sum_{i=1}^m D_{it} \leq T_t$ since we cannot show more than $T_t$ ads during time slot $t$.
We require $\sum_{t=1}^n D_{it} \geq c_i$ to fulfill a contract to show advertisement $i$ at least $c_i$ times.

Goal: Choose $D_{it}$.

For some empirical $P_{it}$ with $0 \leq P_{it} \leq 1$, we obtain $C_{it} = P_{it}D_{it}$ clicks for ad $i$, which pays us some number $R_i > 0$ up to a budget $B_i$. The ad revenue for ad $i$ is $S_i = \min( R_i \sum_t C_{it}, B_i )$ which is concave in $D$. We aim to maximize $\sum_i S_i$.

using Random
using Distributions: LogNormal
Random.seed!(1);


m = 5; # number of adverts
n = 24; # number of timeslots
SCALE = 10000;
B = rand(LogNormal(8), m) .+ 10000;
B = round.(B, digits=3); # Budget

P_ad = rand(m);
P_time = rand(1,n);
P = P_ad * P_time;

T = sin.(range(-2*pi/2, stop=2*pi-2*pi/2, length=n)) * SCALE;
T .+= -minimum(T) + SCALE; # traffic
c = rand(m); # contractual minimum
c *= 0.6*sum(T)/sum(c);
c = round.(c, digits=3);
R = [rand(LogNormal(minimum(c)/c[i]), 1) for i=1:m]; # revenue

# Form and solve the optimal advertising problem.
using Convex, SCS;
D = Variable(m, n);
Si = [min(R[i]*dot(P[i,:], D[i,:]'), B[i]) for i=1:m];
problem = maximize(sum(Si),
               [D >= 0, sum(D, dims=1)' <= T, sum(D, dims=2) >= c]);
solve!(problem, () -> SCS.Optimizer(verbose=0));

Plot traffic.

using Plots
plot(1:length(T), T, xlabel="hour", ylabel="Traffic")

Plot P.

heatmap(P)

Plot optimal D.

heatmap(evaluate(D))

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