How to express xlog(1+exp(-(y/x))) in CVX

1

Hi everyone,
I have -xlog(1+exp(-y/x)) in my optimization problem where x and y are variables(>=0). I don’t know how to express it in CVX. Could someone can help me to solve this problem?
Thanks in advance.

2

Assuming at the end you want to model something convex, then following should help you.
If

f(y) = log (1+exp(-y))

then the perspective function of that is

x * f(y/x) = x * log (1+exp(-y/x))

This is essentially the perspective function of log(sum exp) function, See

Recall

1+exp(-y)= exp(s)+exp(t)
s=0
t=-y.

1 Like
3

Thanks a lot for your help. The term -x(log(1+exp(-y/x))) is the objective function I want to maximize， can we just use the perspective function to replace?

4

If you understand section 7.1.1 in the previous link you should be able to write up the objective function an a CVX compliant way.

5

I read the previous link and related information. I’m sorry to bother you again. I just don’t know how to express (Kexp) in CVX. Is it coded like sum_i(ui)<=x; (u1, x,-q)==exponential(1)?
I would appreciate that if you can provide some examples or guidance.

6

{x1,x2,x3) in K_{exp} in the Mosek Modeling Cookbook is implemented in CVX as
{x3,x2,x1} == exponential(1)

See exp_cone in http://cvxr.com/cvx/doc/funcref.html#sets . But note that although not stated in the CVX Users’ Guide, this has to be implemented as
{x,y,z} == exponential(1)
to achieve what is shown in exp_cone.
The roles of what CVX Users’ Guide call x,y,z are the same as what is called x3,x2,x1 by Mosek Modeling Cookbook. I made this adjustment when showing the CVX code above corresponding to the Mosek Modeling Cookbook definition.