Converting a maximizing problem into a minimizing program using linprog

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Kevin Wang 2020-9-5

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评论： Kevin Wang 2020-9-7

采纳的回答： Bruno Luong

Hi guys,

i have a question about the function linprog.

I want to use this function, and according to the matlab database the function linprog can be applied so that it solves for:

In my case i want to maximize the vector x. Can Anyone help me and tell me how to convert it? What effects will it have on the A matrix and the upper / lower bounds?

Thanks a lot!

Kev

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Bruno Luong 2020-9-5

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You just need to reverse the sign of f. Don't touch the rest.

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Bruno Luong 2020-9-7

编辑：Bruno Luong 2020-9-7

在 MATLAB Online 中打开

For a canonic form of LP

x = argmax f'*x
such that A*x <= b, x >= 0

(Method 3) It is also possible to solve the DUAL problem (2)

y = argmin b'*x
such that (-A'*y) <= -f, y >= 0

and finally x is retrieved as

x = Lagrange multiplier of (2)

f => b
A = -A'
b => -f
x => Lagrange multiplier of (2)

Example:

f = [6; 14; 13];
A = [0.5 2 1;
       1 2 4];
b = [24; 60];
% Primal argmax, method 1
x = linprog(-f, A, b, [], [], zeros(size(f)), [])
% Primal argmax, method 2
x = linprog(f, -A, b, [], [], [], zeros(size(f)));
x = -x
% Dual formulation, method 3
[y, ~, ~, ~, lambda] = linprog(b, -A', -f, [], [], zeros(size(b)), []);
x = lambda.ineqlin