hck
December 27, 2019, 5:01pm
1
cvx_begin
maximize(t)
subject to
R1u = log(1+Pu*h0./(a1.*(H^2+(x-s1x).^2+(y-s1y).^2)))./log(2).*a1;
R2u = log(1+Pu*h0./(a2.*(H^2+(x-s2x).^2+(y-s2y).^2)))./log(2).*a2;
R3u = log(1+Pu*h0./(a3.*(H^2+(x-s3x).^2+(y-s3y).^2)))./log(2).*a3;
R1v = log(1+P1.*h0./(b1.*(H^2+(x-d1x).^2+(y-d1y).^2)))./log(2).*b1;
R2v = log(1+P2.*h0./(b2.*(H^2+(x-d2x).^2+(y-d2y).^2)))./log(2).*b2;
R3v = log(1+P3.*h0./(b3.*(H^2+(x-d3x).^2+(y-d3y).^2)))./log(2).*b3;
B0/(101*Rt)*sum(R1u) >= t;
B0/(101*Rt)*sum(R2u) >= t;
B0/(101*Rt)*sum(R3u) >= t;
B0/(101*Rt)*sum(R1v) >= t;
B0/(101*Rt)*sum(R2v) >= t;
B0/(101*Rt)*sum(R3v) >= t;
Pall=P1+P2+P3;
Ball=a1+a2+a3+b1+b2+b3;
for i=1:101
Pall(i) <= Pv;
P1(i) >= 0;
P2(i) >= 0;
P3(i) >= 0;
Ball(i) <= 1;
a1(i) >= 0;
a2(i) >= 0;
a3(i) >= 0;
b1(i) >= 0;
b2(i) >= 0;
b3(i) >= 0;
end
cvx_end
the errors are:
出错 ./ (line 19)
出错 exam01 (line 41)h0./(a1. (H^2+(x-s1x).^2+(y-s1y).^2)))./log(2).*a1;
If you are willing to have a1 implicitly constrained to be positive, you can use inv_pos(a1).. Otherwise it is non-convex. Similarly for the other expressions.
Also, use of the for loop is not incorrect, but it is faster and simpler to use
Pall <= Pv
P1 >= 0
etc.
hck
December 28, 2019, 8:21am
4
i change my code to this, but still have some problems.
cvx_begin
variables t a1(101) a2(101) a3(101) b1(101) b2(101) b3(101) P1(101) P2(101) P3(101)
maximize(t)
subject to
R1u = log(1+inv_pos(a1).*Pu*h0./(H^2+(x-s1x).^2+(y-s1y).^2))./log(2).*a1;
R2u = log(1+inv_pos(a2).*Pu*h0./(H^2+(x-s2x).^2+(y-s2y).^2))./log(2).*a2;
R3u = log(1+inv_pos(a3).*Pu*h0./(H^2+(x-s3x).^2+(y-s3y).^2))./log(2).*a3;
R1v = log(1+inv_pos(b1).*P1.*h0./(H^2+(x-d1x).^2+(y-d1y).^2))./log(2).*b1;
R2v = log(1+inv_pos(b2).*P2.*h0./(H^2+(x-d2x).^2+(y-d2y).^2))./log(2).*b2;
R3v = log(1+inv_pos(b3).*P3.*h0./(H^2+(x-d3x).^2+(y-d3y).^2))./log(2).*b3;
B0/(101*Rt)*sum(R1u) >= t;
B0/(101*Rt)*sum(R2u) >= t;
B0/(101*Rt)*sum(R3u) >= t;
B0/(101*Rt)*sum(R1v) >= t;
B0/(101*Rt)*sum(R2v) >= t;
B0/(101*Rt)*sum(R3v) >= t;
Pall=P1+P2+P3;
Ball=a1+a2+a3+b1+b2+b3;
Pall <= Pv;
P1 >= 0;
P2 >= 0;
P3 >= 0;
Ball <= 1;
cvx_end
the errors are:
出错 exam01 (line 41)Pu h0./(H^2+(x-s1x).^2+(y-s1y).^2))./log(2).*a1;
it seems like the function in log() is a convex function, but it’s illegal
hck
December 28, 2019, 9:40am
5
it seems that the function in log() must be concave. so i rewrite the function in log(), this time the error that Illegal operation: log( {convex} ) is eliminated. but there is another error. it seems that R1u={convex}.a1 is violating the DCP ruleset where a1 is a variable. it seems that a variable can not be a part of .
cvx_begin
variables t a1(101) a2(101) a3(101) b1(101) b2(101) b3(101) P1(101) P2(101) P3(101)
maximize(t)
subject to
R1u = -log(1-inv_pos(1+a1.*((H^2+(x-s1x).^2+(y-s1y).^2)./Pu*h0)))./log(2).*a1;
R2u = -log(1-inv_pos(1+a2.*((H^2+(x-s2x).^2+(y-s2y).^2)./Pu*h0)))./log(2).*a2;
R3u = -log(1-inv_pos(1+a3.*((H^2+(x-s3x).^2+(y-s3y).^2)./Pu*h0)))./log(2).*a3;
R1v = -log(1-inv_pos(1+b1.*((H^2+(x-d1x).^2+(y-d1y).^2)./P1*h0)))./log(2).*b1;
R2v = -log(1-inv_pos(1+b2.*((H^2+(x-d2x).^2+(y-d2y).^2)./P2*h0)))./log(2).*b2;
R3v = -log(1-inv_pos(1+b3.*((H^2+(x-d3x).^2+(y-d3y).^2)./P3*h0)))./log(2).*b3;
B0/(101*Rt)*sum(R1u) >= t;
B0/(101*Rt)*sum(R2u) >= t;
B0/(101*Rt)*sum(R3u) >= t;
B0/(101*Rt)*sum(R1v) >= t;
B0/(101*Rt)*sum(R2v) >= t;
B0/(101*Rt)*sum(R3v) >= t;
Pall=P1+P2+P3;
Ball=a1+a2+a3+b1+b2+b3;
Pall <= Pv;
P1 >= 0;
P2 >= 0;
P3 >= 0;
Ball <= 1;
cvx_end
the errors are:
出错 exam01 (line 41)((H^2+(x-s1x).^2+(y-s1y).^2)./Pu h0)))./log(2).*a1;
Whoops, I misread the last a1as being in the denominator and not seeing the a1 inside the log. Hopefully my answer to your subsequent question How to write x*log(1+A/(B*x))? solves your problem. And maybe you’ve now implemented that, so see my answer to your question Why is my program failed? .
hck
December 28, 2019, 2:25pm
8
Thank you Mark! I used rel_entr() to rewrite the R1u expression, and the errors are fixed! But I still can’t get the result as shown in Why is my program failed? . I will try to solve this problem according to your suggestion. Thanks again!