CVX cannot perform the operation: {positive constant} ./ {real affine}

1

Hi there.

I am trying to solve a convex problem by CVX toolbox. When I was running my code, the toolbox told me that ‘cannot perform the operation: {positive constant} ./ {real affine}’.

The optimization problem is:

\min \limits _{\mathbf{X}_k} \quad \exp \left( \frac{1}{L \sum \limits _{k=1}^5 \mathbf{F}^H \mathbf{X}_k \mathbf{F}} \right) \text{Ei}\left( -\frac{1}{L \sum \limits _{k=1}^5 \mathbf{F}^H \mathbf{X}_k \mathbf{F}} \right)

where the variables \mathbf{X}_k, (k=1,2,3,4,5) are hermitian semidefinite matrices. L is a constant, \mathbf{F} is a known column vector. I am sure the objective function is a convex function.

Could you please help me? Thank you.

Brian

2

Are you doing exp* ei ？ Can you explain why this is convex？That might help you on expressing it with CVX.

3

Is Ei exponential integral? if so, I believe that is concave for denominator < 1 and convex for denominator > 1. Assuming L > 0, the denominator will always be >= 0, but not necessarily > 1.

And this is multiplied by the exp term and is supposedly convex? I’m not buying it, even for a single scalar X. Accordingly, I am marking this non-convex until you prove otherwise.

4

Okay, thank you for your kind reply.
\text{Ei} is exponential integral and L>0 yes.

Here is the proof of the the objective function:

Assume f(x) \triangleq -\exp \left( \frac{1}{x} \right) \text{Ei} \left( -\frac{1}{x} \right), second order derivative of f(x) is:

\frac{\partial^2 f(x)}{\partial x^2} = -\exp\left( \frac{1}{x} \right) x^{-4} \times \left[ - \text{Ei}\left( -\frac{1}{x} \right) -x\exp\left( -\frac{1}{x} \right)-2x \text{Ei}\left( -\frac{1}{x} \right) -x^2 \exp\left(- \frac{1}{x} \right) \right]

Assume h(x)=- \text{Ei}\left( -\frac{1}{x} \right) -x\exp\left( -\frac{1}{x} \right)-2x \text{Ei}\left( -\frac{1}{x} \right) -x^2 \exp\left(- \frac{1}{x} \right) , then

\frac{\partial h(x)}{\partial x} = -2 \text{Ei}\left( -\frac{1}{x} \right) -2x\exp \left( -\frac{1}{x} \right)

Hint: \frac{\partial \text{Ei} \left( -\frac{1}{x} \right)}{\partial x}=-\frac{\exp \left( -\frac{1}{x} \right)}{x}

Set g(x)\triangleq \text{Ei} \left( -\frac{1}{x} \right) + x \exp\left( -\frac{1}{x} \right)

\frac{\partial h(x)}{\partial x}=-2g(x)

g(x) \geq 0 for x \geq 0, \frac{\partial h(x)}{\partial x} \leq 0 for x \geq 0. Therefore, h(x) \leq h(0)=0. Natually, \frac{\partial ^2 f(x)}{\partial x^2} \leq 0 which means f(x) is concave. The objective function known as -f(x) is convex.

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Hi, thank you for your reply. You can see the reply for Mr Mark L Stone. Thank you again.

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Is f(x) \triangleq -\exp \left( \frac{1}{x} \right) \text{Ei} \left( -\frac{1}{x} \right) a monotone increasing function to x? If it is , you can just let x be the objective.

If it is not but still convex/concave, maybe you can use bisection to estimate the root of its derivative, then divide it into 2 monotone pieces.

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7

I now believe it is convex for the single scalar x case, and think it probably is for multiple scalar X_k. Do you have a proof of convexity for the multiple hermitian semidefinite case? Even if you do, I think your prospects for finding a CVX representation are rather poor.

8

Suppose g\left( \mathbf{X} \right) = L \sum \limits _{k=1}^5 \mathbf{F}^H \mathbf{X}_k \mathbf{F}, which is affine. For -f(x) is a convex function, -f\left( g\left( \mathbf{X} \right) \right) is convex.

9

Hello.

The first order derivative of the function g\left( \mathbf{X} \right) = \mathbf{F}^H \mathbf{X} \mathbf{F} is \mathbf{F}^H \mathbf{F}, the second order derivative is \mathbf{0}. Therefore, it is affine.

10

I don’t think any of the forum readers know how to represent your objective function in CVX.

Perhaps it would be a candidate for @Erling’s challenge, but I’m not sure exponential Integral qualifies, because i don’t think it can be written in AMPL or GAMS, unless we count AMPL Extended Function Library incorporation of GNU Scientific Library exponential integral.

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