How to express the following Rate expression using rel_entr of CVX?

[Mark_L_Stone]

It looks like you already found most of my relevant posts.

Here's how to handle x*log(1+x/(x+y)) for x >= 0, y >= 0

x*log(1+x/(x+y)) for x >= 0, y >= 0 is jointly convex in x and y. Similarly to Xlog( 1+ Y/(X+Y) ): DCP rules, and build-in functions in cvx , we can derive that x*log(1+x/(x+y)) can be represented as rel_entr(x+y,2*x+y) + rel_entr(2*x+y,x+y)

refers to

Xlog( 1+ Y/(X+Y) ): DCP rules, and build-in functions in cvx

This can be entered into CVX as -2*rel_entr(x+y,x+2*y) - rel_entr(x+2*y,x+y) Derivation: x*log(1+y/(x+y)) = (x+y)*log(1+y/(x+y)) - y*log(1+y/(x+y)) The first term is -rel_entr(x+y,x+2*y). The second term can be handled with the result from the wizard of conic reformulation @Michal_Adamaszek in Writing x*log(1+x/y) , that x*log(1+x/y) = rel_entr(x+y,y) + rel_entr(y,x+y) , which in this case becomes -(rel_entr(x+2*y,x+y) + rel_entr(x+y,x+2*y)))

and then finished off with

Minimize log(1+1/x) where 0<x<inf

However, in Writing x*log(1+x/y) ,the wizard of conic reformulation,@Michal_Adamaszek produced the result x*log(1+x/y) = rel_entr(x+y,y) + rel_entr(y,x+y) Setting x = 1, then Interchanging the roles of x and y so as to match the problem in this thread, results in log(1 + 1/x) = rel_entr(x,x+1) + rel_entr(x+1,x) The RHS is accepted by CVX, and as a bonus, is already in a form needed by CVXQUAD. It executes without error. And you can see how to formulate the matrix analog using CVXQUAD’s q…

So essentially the same techniques can be used here.

x*log(1+y1/(x+y2)) = (x+y2)*log(1+y1/(x+y2)) - y2*log(1+y1/(x+y2)) = -rel_entr(x+y2,x+y2+y1) - y2*(rel_entr((x+y2)/y1,(x+y2)/y1+1)+rel_entr((x+y2)/y1+1,(x+y2)/y1))