About xlog(1+y/x)

1

There is a function xlog(1+y/x) in my problem. Obviously it is convex when x,y>0. how can I use the cvx to solve it?
Thanks

2

It looks like you answered your own question in the other thread. Yes, this solution has been posted several times.

3

do you know how the feasibility problem works in cvx?
How dose the cvx knows the objective function if want to examine if the optimal value can reach some value?

4

I am not sure exactly what you are asking. You do not need to have an objective function in CVX. You can declare variable(s) and constraints, and let the solver called by CVX report on feasibility.

For example:

cvx_begin
variable x
1 <= x <= 0
cvx_end

Calling SeDuMi 1.34: 2 variables, 1 equality constraints

SeDuMi 1.34 (beta) by AdvOL, 2005-2008 and Jos F. Sturm, 1998-2003.
Alg = 2: xz-corrector, Adaptive Step-Differentiation, theta = 0.250, beta = 0.500
eqs m = 1, order n = 3, dim = 3, blocks = 1
nnz(A) = 2 + 0, nnz(ADA) = 1, nnz(L) = 1
it : by gap delta rate t/tP t/tD* feas cg cg prec
0 : 1.35E+01 0.000
1 : 5.23E+00 6.80E-01 0.000 0.0504 0.9900 0.9900 -0.11 1 1 1.7E+00
2 : 5.66E+04 2.23E-04 0.000 0.0003 0.9999 0.9999 -0.99 1 1
Primal infeasible, dual improving direction found.
iter seconds |Ax| [Ay]_+ |x| |y|
2 0.0 0.0e+00 0.0e+00 0.0e+00 1.0e+00

Detailed timing (sec)
Pre IPM Post
7.001E-03 1.500E-02 1.006E-03
Max-norms: ||b||=1, ||c|| = 0,
Cholesky |add|=0, |skip| = 0, ||L.L|| = 1.

Status: Infeasible
Optimal value (cvx_optval): +Inf

cvx_begin
variable x
1 <= x <= 2
cvx_end

Calling SeDuMi 1.34: 2 variables, 1 equality constraints

SeDuMi 1.34 (beta) by AdvOL, 2005-2008 and Jos F. Sturm, 1998-2003.
Alg = 2: xz-corrector, Adaptive Step-Differentiation, theta = 0.250, beta = 0.500
eqs m = 1, order n = 3, dim = 3, blocks = 1
nnz(A) = 2 + 0, nnz(ADA) = 1, nnz(L) = 1
it : by gap delta rate t/tP t/tD* feas cg cg prec
0 : 8.17E+00 0.000
1 : -1.90E-01 1.97E+00 0.000 0.2406 0.9000 0.9000 1.67 1 1 1.5E+00
2 : 1.08E-03 6.01E-02 0.000 0.0306 0.9900 0.9900 1.62 1 1 1.8E-02
3 : -1.00E-07 1.51E-06 0.000 0.0000 1.0000 1.0000 1.01 1 1
iter seconds digits cx by
3 0.0 Inf 0.0000000000e+00 0.0000000000e+00
|Ax-b| = 0.0e+00, [Ay-c]_+ = 0.0E+00, |x|= 7.1e-01, |y|= 0.0e+00

Detailed timing (sec)
Pre IPM Post
6.991E-03 1.501E-02 9.958E-04
Max-norms: ||b||=1, ||c|| = 0,
Cholesky |add|=0, |skip| = 0, ||L.L|| = 1.

Status: Solved
Optimal value (cvx_optval): +0

A feasible solution, x = 1.5000, was returned.

If you want to know whether a particular objective value can be reached, you can put that in as a constraint. Or just optimize with that objective and you’ll see what value of the objective can be reached.