How to run please this example perfectly to get optimal solution .. Thank you

f = [0 ; - 10 ; 0 ; - 6; - 20] ; 
A = [1 0 0 0 0 ;
     0 1 0 0 0 ;
     0 0 1 0 0 ;
     0 0 0 1 0 ;
     0 0 0 0 1] ;
B = [4 ; 3 ; 3 ; 2 ; 2] ; 
Aeq = [1 - 1 - 1 0 0 ; 0 0 1 - 1 - 1] ;
beq = [0; 0; 0; 0; 0]; 
lb = zeros (5, 1) ;
ub = zeros (5, 1) ;
x = linprog(f,A,B,Aeq,beq,lb,ub)
[x,fval] = linprog(f,A,Aeq,beq,lb,ub);

Answer 1

[Steven Lord]

lb = zeros (5, 1) ;
ub = zeros (5, 1) ;

So based on the way you've constructed the bounds, all the elements of your solution have to be greater than or equal to 0 and less than or equal to 0. The only possible solution that satisfies both sets of bounds is zeros(5, 1).

Are you sure that ub shouldn't be something like ones(5, 1)?

[Torsten]

Good catch.

[Dalia ElNakib]

Thank You so much

It's related to this Example solution

[Torsten]

f = -[0 ; 10 ; 0 ; 6; 20] ; 
Aeq = [1 -1 -1 0 0 ; 0 0 1 -1 -1] ;
beq = [0; 0]; 
lb = zeros(5,1) ;
ub = [4; 3; 3; 2; 2] ;
[I,fval] = linprog(f,[],[],Aeq,beq,lb,ub)

Optimal solution found.

I = 5×1

4 2 2 0 2

fval = -60

[Steven Lord]

That layout is potentially misleading. It only lists the first three of the "Physical constraints [that] limit the currents". I wonder if the other two were listed after the circuit diagram on the next page. That separate would make them easy to overlook.

[Torsten]

I wonder if the other two were listed after the circuit diagram on the next page.

No, they are mentionned in the text part.

[Steven Lord]

Yes, I saw that. I would have expected them also to appear in the "Therefore, the problem can be posed as the following linear program:" section as inequalities like the last three bound constraints. But due to a lack of space on the page I wonder if students solving the problem could overlook the last two equations (which I assumed appeared after the diagram on page 262.)

[Dalia ElNakib]

Many Thanks for your clarifications.

Really appreciate your efforts.

Answer 2

[Torsten]

f = [0 ; -10 ; 0 ; -6; -20] ; 
A = [1 0 0 0 0 ;
     0 1 0 0 0 ;
     0 0 1 0 0 ;
     0 0 0 1 0 ;
     0 0 0 0 1] ;
B = [4 ; 3 ; 3 ; 2 ; 2] ; 
Aeq = [1 -1 -1 0 0 ; 0 0 1 -1 -1] ;
beq = [0; 0]; 
lb = zeros (5, 1) ;
ub = zeros (5, 1) ;
[x,fval] = linprog(f,A,B,Aeq,beq,lb,ub)

Optimal solution found.

x = 5×1

0 0 0 0 0

fval = 0

[Dalia ElNakib]

f = [0 ; -10 ; 0 ; -6; -20] ; 
A = [1 0 0 0 0 ;
     0 1 0 0 0 ;
     0 0 1 0 0 ;
     0 0 0 1 0 ;
     0 0 0 0 1] ;
B = [4 ; 3 ; 3 ; 2 ; 2] ; 
Aeq = [1 -1 -1 0 0 ; 0 0 1 -1 -1] ;
beq = [0; 0]; 
lb = zeros (5, 1) ;
ub = ones(5, 1);
[x,fval] = linprog(f,A,B,Aeq,beq,lb,ub);

Optimal solution found.

[Dalia ElNakib]

@Torsten@Steven Lord

Thank you so much

Both work to ge the optimal solution

Either:

lb = zeros (5, 1) ;

ub = zeros (5, 1) ;

Or

lb = zeros (5, 1) ;

ub = ones (5, 1) ;

f = [0 ; -10 ; 0 ; -6; -20] ; 
A = [1 0 0 0 0 ;
     0 1 0 0 0 ;
     0 0 1 0 0 ;
     0 0 0 1 0 ;
     0 0 0 0 1] ;
B = [4 ; 3 ; 3 ; 2 ; 2] ; 
Aeq = [1 -1 -1 0 0 ; 0 0 1 -1 -1] ;
beq = [0; 0]; 
lb = zeros (5, 1) ;
ub = zeros(5, 1);
[x,fval] = linprog(f,A,B,Aeq,beq,lb,ub);

Optimal solution found.