If x1 and x2 are two variables, how do I get the product of x1 and x2 in CVX? If I represent it as x1.* x2, the error “Disciplined convex programming error: Cannot perform the operation: {convex} .* {convex}” will be appear. How to solve this problem? Please help.

It may or may not be possible depending on in what context you want to use it. Please write the full constraint/problem you want to model.

The code I have written is as follows:

cvx_begin

variable t(para.T+1, 2)

variable u1(1, para.T)

variable u2(1, para.T)

variable u3(1, para.T)

variable v1(1, para.T)

variable v2(1, para.T)

variable temp(1, para.T)

expression lower_bound(1, para.T)

for ii=1:para.T

lower_bound(ii)=ave_k1(ii)*u1(ii)+ ave_k2(ii)*u2(ii)+ave_k3(ii)*u3(ii)+(ave_g1k1(ii)+ave_g1k2(ii)+ave_g1k3(ii))*v1(ii)+(ave_g2k1(ii)+ave_g2k2(ii)+ave_g2k3(ii))*v2(ii);

A(ii)=norm(W(:,1)).^2.* (B_k(:,1).* pow_pos(inv_pos(u1(ii)),2)+2*C_g1k(:,1).*inv_pos(v1(ii)).*inv_pos(u1(ii)));

```
temp(ii)= log(A+sigma2)/log(2);
```

end

maximize(1/para.T*(sum(lower_bound-temp)))

subject to

80 <= u1;

80 <= u2;

80 <= u3;

20 <= v1;

20 <= v2;

for jj=1:para.T

pow_pos(norm(t(jj, :)-para.place_user1(1:2)),2)+6400+u10(jj)^2-2* u10(jj)* u1(jj)<=0;

pow_pos(norm(t(jj, :)-para.place_user2(1:2)),2)+6400+u20(jj)^2-2* u20(jj)* u2(jj)<=0;

pow_pos(norm(t(jj, :)-para.place_user3(1:2)),2)+6400+u30(jj)^2-2* u30(jj)* u3(jj)<=0;

pow_pos(norm(t(jj, :)-para.place_S1(1:2)),2)+400+v10(jj)^2-2* v10(jj)* v1(jj)<=0;

pow_pos(norm(t(jj, :)-para.place_S2(1:2)),2)+400+v20(jj)^2-2* v20(jj)* v2(jj)<=0;

end

for ii=1:para.T

pow_pos(norm(t(ii+1,:)-t(ii, :)),2)<=Vmax^2;

end

cvx_end

As shown in the above, the term " inv_pos(v1(ii)).* inv_pos(u1(ii))" in the expression “A(ii)” is not allowed. There is an error "Disciplined convex programming error: Cannot perform the operation: {convex} .* {convex} ". So how should I express the product of two CVX variables?

Have you proven this optimization problem is convex? I haven’t figured out exactly what you have, but it doesn’t look likely to be convex to me.