Can formulation (sum(affine/concave) < constant) be transformed to a convex expression?

I am noob in convex optimization and I have some questions :sweat:
The expression image seems nonconvex, where a_n and b_n are variables, c is a constant. is it possible to transform it into a convex expression??
(We know that image can be convex since it can be transformed to image .)

Anyone can give me some suggestions? thanks !!!

when you transform it to an<=c*log(1+bn) ,in order to be accepted by cvx,the left side must be convex
and the right must be concave,log(1+bn) are concave when (1+bn) are concave and >0; which have menthioned in Convex optimization in part3,不说了,急着出门。 :rofl:

I have deleted my previous reply, because it is incorrect.

I think so. however, thanks for your kind attitude.

求和之后问题就变得比较麻烦了 :sweat:,脑阔疼

This is an English language forum. Thank you.

Yes, i know. As for the problem itself, I try to use English as much as i can.

In your previous reply, if the a_n is a binary variable, is it possible to use a slack variable to replace d_n*b_n ?

Section 2.8 of shows how to linearize the product of binary variable with continuous variable.

ok, i saw this pdf in your previous answer in other posts, maybe it can help, thanks.

You can do variable substitution. Firstly, you should define x_n =\frac{a_n}{\log(1+b_n)}, \frac{a_n}{\log(1+b_{n}^{max})} \leq x_n \leq \frac{a_n}{\log(1+b_{n}^{min})}. Secondly,
you can replace \frac{a_n}{\log(1+b_n)} in the expression with x_n. Then, you have a new expression \sum_{n=1}^{N}x_n \leq c, \\ \textbf{s.t.} \frac{a_n}{\log(1+b_{n}^{max})} \leq x_n \leq \frac{a_n}{\log(1+b_{n}^{min})}. Finally, the above expression is a convex set.

that is a good idea :grinning:

@Yenfy I presume you are assuming a_n \ge 0.

Perhaps your formulation assumes a_n and b_n do not appear elsewhere in the model?

How are a_n and b_n recovered from x_n?

I,m sorry, it’s my fault. I thought a_n was a parameter. If a_n is a variable, my above solution is wrong.

@Yenfy If a_n were a parameter and b_n is constrained to be >= 0, the easiest thing to do in CVX would be
sum(a.*inv_pos(log(1+b))) <= c

When I use inv_pos() in CVX (not this problem.), the result usually become infeasible. :upside_down_face:,but when I comment out the constraint contains inv_pos() expression, it can be solved and the result can satify the commented consraints. :expressionless:, that’s very confusing.

@TomK Please provide a reproducible example, clearly showing what you say.

I have some new questions about function quad_over_lin()

It seems that y cannot be a vector?
if x, y are all 1×N vectors, how to formulate sum(x(n)^2/y(n)) in CVX, N is a large constant, so I cannot formulate it as ((x(1)^2/y(1)) +(x(2)^2/y(2)) +…+(x(n)^2/y(n))) in CVX because it is too long.

y can be a vector.

help quad_over_lin

quad_over_lin Sum of squares over linear.
Z=quad_over_lin(X,Y), where X is a vector and Y is a scalar, is equal to
SUM(ABS(X).^2)./Y if Y is positive, and +Inf otherwise. Y must be real.

If X is a matrix, quad_over_lin(X,Y) is a row vector containing the values
of quad_over_lin applied to each column. If X is an N-D array, the operation
is applied to the first non-singleton dimension of X.

quad_over_lin(X,Y,DIM) takes the sum along the dimension DIM of X.
A special value of DIM == 0 is accepted here, which is automatically
replaced with DIM == NDIMS(X) + 1. This has the effect of eliminating
the sum; thus quad_over_lin( X, Y, NDIMS(X) + 1 ) = ABS( X ).^2 ./ Y.

In all cases, Y must be compatible in the same sense as ./ with the squared
sum; that is, Y must be a scalar or the same size as SUM(ABS(X).^2,DIM).

Disciplined convex programming information:
    quad_over_lin is convex, nonmontonic in X, and nonincreasing in Y.
    Thus when used with CVX expressions, X must be convex (or affine)
    and Y must be concave (or affine).

I want to minimize image
This is my code:
% cvx_solver SeDuMi
cvx_solver SDPT3
cvx_precision low

variable a(5)
variable V_slack(5)

minimize( sum( quad_over_lin(a,V_slack) ) )

The error becomes
image :upside_down_face: