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Simple linear programming problem
2022-07-02 17:26:00 【__ Rain】


- For the parameters of the objective function c c c, Because the standard type is m i n min min, So when finding the maximum value, you need to add a minus sign to the function to find the minimum value , Write in function parameters − c -c −c that will do
- For the parameter of inequality A , b A,b A,b , The standard type is ≤ \leq ≤, So if the condition constraint is ≥ \geq ≥, First transform all inequalities into ≤ \leq ≤ Rewrite matrix A , b A,b A,b
c = [4000 3000];
A = [2 1; 1 1; 1 0];
b = [10 8 7];
lb = [0 0];
[x, fval] = linprog(-c, A, b, [], [], lb)
- Because there is no equation in the conditional constraint , therefore a e q , b e q aeq,beq aeq,beq use [ ] , [ ] [ \quad ],[ \quad ] [],[] Replace it
- next l b lb lb Can pass z e r o s zeros zeros Function quick creation , z e r o s ( 1 , 3 zeros(1,3 zeros(1,3 Create and return one 1 × 3 1\times3 1×3 Matrix
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