How to solve custom convex functions

1

Hi,

I am new to CVX. So please excuse me if this question is very trivial.

I have a custom convex function L as defined below:

L(t) = 0        for t >= 1    
L(t) = (1-t)^2  for |t| < 1    
L(t) = -4t      for t <= -1

I am using this L(t) in my main objective function f(x) [minimize f(x) over x]:

f(x) = {other standard convex functions of x} + L(x)
subject to {some constraints}

My question is how do I declare L(t)?. I tried to define a new matlab function in L.m as follows:

function y = L(t)

   y = zeros(length(t), 1);

   for i = 1 : length(t)
       if abs(t(i)) < 1
           y(i,1) = (1 - t(i)) ^ 2;
       elseif t(i) <= -1
           y(i,1) = -4 * t(i);
       end
   end
end

But I get error while trying to solve the following problem (this is a hypothetical problem, where I have remove lot of other code to simplify):

xpos = [ randi([1, 20], 1, 30); randi([71, 90], 1, 30) ]';
ypos = ones(30, 1);

xneg = [ randi([61, 80], 1, 30); randi([11, 30], 1, 30) ]';
yneg = ones(30, 1) * -1;

x = [xpos;xneg];
y = [ypos;yneg];

cvx_begin quiet
    variable w(2)
    variable b

    minimize(0.5 * (w' * w) + sum(L(y.*(x*w + b))))
cvx_end

Am I violating any DCP rules of CVX in declaring L(t)? If I am not allowed to iterate over the elements of cvx variable (t in L, in this case), how else can I solve this problem?

Any help/pointer would be very helpful for me…

Thanks,

Ram.

2

Yes, you’re violating the rules in your definition of L(t). You can’t use variables in logical statements like if/then clauses. Functions must obey the same disciplined convex programming rules as any other constraint or objective.

Your function looks very similar to the huber_pos function, though. I cannot help but think you could make L out of an appropriate scaling and translation of that function.

3

As mcg suggests, you can express L(t) using huber_pos. Here’s the one-liner: huber_pos(1-t,2).

4

Wonderful, Bien! Thanks for contributing.