How to reformulate this in a CVX way: Cube over Square

I meant someone can post why your problem is convex, if they have shown it is convex. Otherwise, I will assume it is not convex.

My prove process is like this.

So Dm is convex because sqrt(f) is convex and non-decreasing, and f is convex

Norm of an affine argument is convex, so Dm is convex. But the cube over square involving it, is not convex as far as I know.

So you mean that Dm^3/Tm^2 is not convex? How about Dm^2/Tm?

For Dm^2/Tm, you should be able to use quad_over_lin, in which case norm would never be explicitly used in the formulation.

For cube over square, @Erling 's approach, as linked in my earlier post, should work if one or both of the items being cubed or squared is concave (or affine), but would not work if either item is convex, which is your situation. That is because all argument of geo_mean mus be concave (or affine).

As the Dm is convex, the geo_mean couldn’t work. So, is there any way to express Dm^3/Tm^2 in cvx?

As far as I know, no. Please carefully read the "Why isn’t CVX accepting my model? READ THIS FIRST! " link provided above.

OK, thank you! I will check whether it is convex.

Hello~ Mr. Mark. Regard to the Dm^3/Tm^2, I have proved it is convex. As f=(Dm)^3 is convex,
According to the perspective operation of convex function,
the perspective function of f is g=Tm*(Dm/Tm)^3, which is also convex where Dm/Tm belongs to the f domain. (Dm and Tm are real and non-negative)

The reason of f=Dm^3 (sqrt((x1-x2)^2+(y1-y2)^2)^3) is convex is that
[x1,x2,y1,y2]H[x1,x2,y1,y2]’ (this equals 6*[x1-x2)^2+(y1-y2)^2)^3]) is always larger or equal to 0 when these 4 variables are real and non-negative, where H is the Hessian matrix of function f.

So, is there any way to express this in cvx? If still no, can I use the lower bound to replace this?

I will defer to @Michal_Adamaszek or someone else to address this.

Thank you for this! Mark~

First define variable Dm(M,1), then change

to Dm(i)>=norm(qm(i+1,:)-qm(i,:));. then expressing Dm^3/Tm^2 will do your job.

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Thank you for your help Jack. But Tm is a affine variable, could we do the division of the affine in CVX?

Please follow Mark’s advice above to express Dm^3/Tm^2.

Actually Mark’s advice cannot handle this as my expression in fact is ||qm+1-qm||^3/Tm^2. Thank you Jack, I am currently trying just dividing this expression to see whether could work or not.

Hello Mark. I recently used this “rotated_lorentz(1)” way to express ||qm+1 - qm||^3/Tm^2 like this:
variables qm(M+1,2) Tm(M,1) temp1(M,1) temp2(M,1)
for i=1:M
{qm(i+1,:)-qm(i,:),temp2(i),Tm(i)} == rotated_lorentz(1);
{temp2(i),temp1(i),qm(i+1,:)-qm(i,:)} == rotated_lorentz(1);
Then use temp1 in place of ||qm+1 - qm||^3/Tm^2. What do you think, is that right?

You have implicitly constrained qm(i+1,:)-qm(i,:) >= 0, and have nothing which does what norm does. it appears that you are trying to generalize @Michal_Adamaszek’s formulation for a scalar argument in a way which is not valid. Your qm(i+1,:)-qm(i,:) is not even a scalar, as it is in @Michal_Adamaszek’s formulation.

yes exactly, I am still stuck in this problem.

For the record, here is the formulation for norm(y)^3/x^2, as provided by @Michal_Adamaszek at How to reformulate this in a CVX way

Declare a variable t , add the constraint geo_mean([t,x,x])>=norm(y), and use t in place of norm(y)^3/x^2

Yes thank you Mark, I see. Thank you very much!