cvx_begin quiet
variable cap1
[Plosskw,Qlosskw]=LoadFlow (cap1);
minimize(sum(Qlosskw));
subject to
0<=cap1<=Q_max
cvx_end
function [Plosskw,Qlosskw]=LoadFlow (cap1);
m(loc1,3)=m_a(loc1,3)-cap1;
…
…
…
The following error occurred converting from cvx to double:
Error using double
Conversion to double from cvx is not possible.
Error in LoadFlow (line 5)
m(loc1,3)=m_a(loc1,3)-cap1;
I’m guessing that my answer Using double incorrectly. Cannot convert from cvx to double applies.
And don’t use quiet until you are sure everything is working well.
I used this expression but it didn’t work, actually I tried expression m(loc1,3) after variable cap1
and
expression m(loc1,3) in function but none of them worked. Could it be because of the function definition in cvx?
Show s complete reproducible code, including all the input data, and use the Preformatted text icon.applied to the entire code. Don’t have function calls unless you show us the function.
cvx_begin
variable cap1
[Qlosskw]=LoadFlowAnalysiss (m1,line,cap1);
minimize(sum(Qlosskw));
subject to
0<=cap1<=40;
cvx_end
function [Qlosskw]=LoadFlowAnalysiss (m,line,cap1);
m_a=m;
loc1=14;
m(loc1,3)=m_a(loc1,3)-cap1;
....
end
m1 is matrix 33* 3 and line is 32* 5
Where is the expression statement?
I tried expression m(loc1,3) once right after the variable cap1 and again in the function But none of them worked
Show a complete program. with the expression statement included. If you have two versions, then show them both.
When I put expression m(loc1,3)=m_a(loc1,3)-cap1; inside the function
No CVX model exists in this scope.
Error in LoadFlowAnalysiss (line 5)
expression m(loc1,3)=m_a(loc1,3)-cap1;
and right after the variable cap1
cvx_begin quiet
variable cap1
expression m(14,3)
…
The following error occurred converting from cvx to double:
Error using double
Conversion to double from cvx is not possible.
Error in LoadFlowAnalysiss (line 5)
m(loc1,3)=m_a(loc1,3)-cap1;
You’re showing us bits and pieces, show us the complete program, at least up to where the CVX error message occurs. And you have not provided the part where we see what the value of loc1 is.
And don’t use the quiet option.
I don’t think you can be declaring variables or expressions inside a function called from somewhere where the cvx_begiin … cvx_ebnd is. There should be another way to do what you want. Why don’t you try writing a program in which all the CVX expressions are in the one function (or script or command line).
global m1
global line
mm=[ 1 0 0
2 100 60
3 90 40
4 120 80
5 60 30
6 60 20
7 200 100
8 200 100
9 60 20
10 60 20
11 45 30
12 60 35
13 60 35
14 120 80
15 60 10
16 60 20
17 60 20
18 90 40
19 90 40
20 90 40
21 90 40
22 90 40
23 90 50
24 420 200
25 420 200
26 60 25
27 60 25
28 60 20
29 120 70
30 200 600
31 150 70
32 210 100
33 60 40 ];
% line No. sending recieving resistance(ohm) reactance(ohm)
% node node
line=[ 1 1 2 0.0922 0.0470
2 2 3 0.4930 0.2511
3 3 4 0.3660 0.1864
4 4 5 0.3811 0.1941
4 5 6 0.8190 0.7070
6 6 7 0.1872 0.6188
7 7 8 0.7114 0.2351
8 8 9 1.0300 0.7400
9 9 10 1.0440 0.7400
10 10 11 0.1966 0.0650
11 11 12 0.3744 0.1238
12 12 13 1.4680 1.1550
13 13 14 0.5416 0.7129
14 14 15 0.5910 0.5260
15 15 16 0.7463 0.5450
16 16 17 1.2890 1.7210
17 17 18 0.7320 0.5740
18 2 19 0.1640 0.1565
19 19 20 1.5042 1.3554
20 20 21 0.4095 0.4784
21 21 22 0.7089 0.9373
22 3 23 0.4512 0.3083
23 23 24 0.8980 0.7091
24 24 25 0.8960 0.7011
25 6 26 0.2030 0.1034
26 26 27 0.2842 0.1447
27 27 28 1.0590 0.9337
28 28 29 0.8042 0.7006
29 29 30 0.5075 0.2585
30 30 31 0.9744 0.9630
31 31 32 0.3105 0.3619
32 32 33 0.3410 0.5302];
% m_initrial=mm;
% l_initial=line;
[r1,c1]=size(mm);
totalm=sum(mm(:,2));
m1=mm;
totalload=sum(pppower)*10;
for ii=12
m1(:,2)=(mm(:,2)/totalm).*totalload(ii);
m1(:,3)=(mm(:,3)/totalm).*totalload(ii);
end
cvx_begin
variable cap1
expression m(14,3)
[Plosskw,Qlosskw,Voltage_Mag,Voltage_Angle,Ibrp]=LoadFlowAnalysiss (m1,line,cap1);
minimize(sum(Qlosskw));
subject to
0<=cap1<=40;
cvx_end
function [Plosskw,Qlosskw,Voltage_Mag,Voltage_Angle,Ibrp]=LoadFlowAnalysiss (m,l,cap1);
m_a=m;
loc1=14;
m(loc1,3)=m_a(loc1,3)-cap1;
....
....
....
Ibrp=[abs(Ibr) angle(Ibr)*180/pi];
PL(1,1)=0;
QL(1,1)=0;
% losses
for f=1:br
Pl(f,1)=(Ibrp(f,1)^2)*R(f,1);
Ql(f,1)=X(f,1)*(Ibrp(f,1)^2);
PL(1,1)=PL(1,1)+Pl(f,1);
QL(1,1)=QL(1,1)+Ql(f,1);
end
Plosskw=(Pl)*100000;
Qlosskw=(Ql)*100000;
PL=(PL)*100000;
QL=(QL)*100000;
Voltage_Mag = vbp(:,1);
Voltage_Angle = vbp(:,2)*(pi/180);
m in LoadFlowAnalysiss has the value m1from from the calling program, and m1 is a double precision array. Then you are assigning a CvX expression into an element of it. It is a CVX expression because it includes cap, which is the variable cap from the calling program. So you did exactly what the CVX User’s Guide warned you not to do.
Try delclaring the expression you need in the calling program and pass that expression as an argument to the function. Then assign the individual element of the expression inside the function.
like that?
cvx_begin
variable cap1
expression m(14,3)
[Plosskw,Qlosskw,Voltage_Mag,Voltage_Angle,Ibrp]=LoadFlowAnalysiss (m1,line,cap1,m(14,3));
minimize(sum(Qlosskw));
subject to
0<=cap1<=40;
cvx_end
function [Plosskw,Qlosskw,Voltage_Mag,Voltage_Angle,Ibrp]=LoadFlowAnalysiss (m,l,cap1,mnew);
m_a=m;
loc1=14;
m(loc1,3)=mnew;Try that. You can put whos n the function that is called and see what it says.
Also include m by itself on a line with no semicolon or anything else. CVX should telll you what that is (if it is any kind of CVX thing).