How to solve b≤Cx≤z with linear programming?
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Hello,
I'd like to calculate the following. For example:
There is a Matrix C m*n(5x4), vector b (b1,b2,b3,b4,b5) and vector z (z1,z2,z3,z4,z5), their values are given. I need to find such vector x (x1,x2,x3,x4,x5) that:
*my calculation:
b1<= Cm1n1*x1 + Cm2n1*x2 + Cm3n1*x3 + Cm4n1*x4 + Cm5n1*x5 < z1
.
.
.
b5<= Cm1n5*x1 + Cm2n5*x2 + Cm3n5*x3 + Cm4n5*x4 + Cm5n5*x5 < z5
Also, the vector x must have positive values.
I was adviced to resolve it by linear programmning. So I have had a look at LINPROG. As I was not sure what to put instead of "f" (Linear objective function vector) I set it to a vector of 0's as I also did that for vector of lower bounds.
x = linprog(f,C,b,[],[],lb)
The vector x which was calculated gave me positive values. Then I have used these x-values for *my calculation (see above please), which gave me results which are far bellow the vector b values.
Does anybody know what I am doing wrong? How to get x above d but below z?
Many thanks. LZ
6 Kommentare
Sean de Wolski
am 9 Apr. 2012
Your question isn't clear to me. If you look at the doc for linsolve, you'll see that you're trying to find an optimal x that minimizes f'x. If f is all zeros, you can pick any x you want that meets your above constraints and it will lie on the hyperplane defining the minimum of f'x, i.e. zeros in all n-dimensions.
Antworten (2)
Teja Muppirala
am 10 Apr. 2012
You can solve this with LINPROG, but I think you need help in understanding how to formulate your constraints properly.
Constraint 1: C*x < z
This is fine as it is.
Constraint 2: C*x > b
Multiply this equation by -1 and rewrite it as
-C*x < -b
Now, we will need to combine both of these constraints into a single matrix.
[C; -C]*x < [z; -b]
This inequality is saying the exact same thing as the two inequalities above. It is just using matrix multiplication. Now LINPROG can accept the [C; -C] and [z; -b] as its inputs to define the constraint.
As a concrete example:
C =[ -1.3617 1.0655 0.0909 -0.5503
-0.6411 -0.2132 -2.1390 0.9111
-0.0097 -0.2970 0.2654 -0.7814
0.5465 0.4400 -0.2322 -0.7191
0.3432 -0.4943 0.4786 -0.3528];
b = [-1; -2; -3; -4; -5];
z = [2; 3; 4; 5; 6];
lowerbound_on_x = zeros(size(C,2),1);
f = zeros(size(C,2),1);
x = linprog(f,[-C;C],[-b;z],[],[],lowerbound_on_x)
This gives me:
x =
1.9958
4.2650
0.3293
2.3048
You can then confirm C*x
ans =
0.5883
-0.7932
-2.9996
1.2334
-2.0788
This satisfies b < C*x < z and x > 0
Sean de Wolski
am 9 Apr. 2012
%Cx<z
%Cx>b
C = rand(5);
b = (1:5)';
z = (11:15)';
x = [C;C]\[z; b]
constraintsMet = all(C*x>=b)&&all(C*x<z)
Maybe?
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