Hello there,

I am trying to run the code below. Yet it produces unexpected results. Under all constraints and the data given we would expect exactly one item of u(i,:,: ) to be set to 1 . Yet all u’s are always set to 0.

For information - one item u(i,h1,: ) can be set to 1 if and only if y(i,h1) has been set to 1, and only one index of u can be set to 1 in the chosen row.

Could you think of a possible reason? I am trying to find a bug in implementation yet I totally fail. All seems good but the results are bad and instead of getting some u’s=1 I am always getting zeros. What is more, I have tried to solve it with both Gurobi and Mosek, always receiving the same result no matter the solver.

------- My code: ----

```
clear all
n_hnon = 3;
n_href = 2;
n_h = n_hnon + n_href;
n_p = 15;
s_href = 0.8 + (0.93-0.8).* rand(n_href,1);
s_hnon = 0.1 + (0.3-0.1).*rand(n_hnon,1);
s_hs=[s_hnon;s_href];
H_max_non=100*ones(n_hnon,1);
H_max_ref=100*ones(n_href,1);
eta=0.2;
%% Optimization
cvx_begin
variable s_p2(n_p)
variable s_term1(n_p)
variable s_term2(n_p)
variable s_term3(n_p)
variable s_term4(n_p)
variable w(n_p,n_hnon)
variable b(n_p,n_href) binary
variable y(n_p,n_hnon) binary
variable u(n_hnon,n_href,n_p) binary
maximize sum(s_p2)
subject to
for p1=1:n_p
s_term1(p1) == eta*sum(s_hnon'.*y(p1,:))
end
for p2=1:n_p
s_term2(p2) == (1-eta)*sum(s_hnon'.*w(p2,:))
end
for p3=1:n_p
s_term3(p3) == sum(s_href'.*b(p3,:))
end
for p4=1:n_p
for h1 = 1:n_hnon
s_term4(p4) == sum((100.*squeeze(u(h1,:,p4))))
end
end
s_p2==s_term1+s_term2+s_term3+s_term4
%Product of binaries linearisation
w<=y
for p5=1:n_p
for h11=1:n_hnon
w(p5,h11)<=1-sum(u(h11,:,p5))
w(p5,h11)>=y(p5,h11)+(1-sum(u(h11,:,p5)))-1
end
end
for h12 = 1:n_hnon
sum(y(:,h12))<=H_max_non(h12)
end
for h2 = 1:n_href
sum(b(:,h2))+sum(sum(u(:,h2,:)))<=H_max_ref(h2)
end
%choose one
for p7=1:n_p
for h13=1:n_hnon
sum(u(h13,:,p7)) - y(p7,h13) <= 0
end
end
for p9=1:n_p
sum(y(p9,:))+sum(b(p9,:))==1
end
b==0
cvx_end
```