linprog

Syntax

linprog(f, [A], [b], [Aeq], [beq], [lb], [ub], [method=’simplex’])

Details

Solve the following optimization problem with a linear objective function and a set of linear constraints.

\(\min_xf^Tx\text{ such that}\begin{cases}A\cdot x\le b\\Aeq\cdot x=beq\\lb\le x\le ub\end{cases}\)

The result is a 2-element tuple. The first element is the minimum value of the objective function. The second element is the value of x where the value of the objective function is minimized.

Arguments

A and Aeq must be matrices with the same number of columns.

f, b and beq are vectors.

lb and ub are scalars or vectors with the same length as x indicating the lower bounds and upper bounds of x.

• If lb or ub is a scalar, all elements of x are subject to the same lower bound or upper bound constraint. If lb or ub is NULL, there is no lower bound or upper bound constraint for x.

• If lb or ub is a vector, an element of x is subject to the lower bound or upper bound constraint specified by the corresponding element of lb or ub.

method is a string indicating the optimization algorithm. It can be either ‘simplex’ (recommended) or ‘interior-point’.

Examples

Example 1. Find the minimum of x+2y subject to the constraints of

\(\begin{cases}x\le2\\y\le2\\x+y\ge2\end{cases}\)

$ f = [1, 2];
$ A = [-1, -1]$1:2;
$ b = [-2];
$ ub = 2;
$ re = linprog(f, A, b, , , , ub);

$ re[0];
2

$ re[1];
[2,0]

Example 2. Find the minimum of -3x1-2x2 subject to the constraints of

\(\begin{cases}2x_1+x_2\le10\\x_1+x_2\le8\\x_1\le4,x_2\in\Re\end{cases}\)

$ f = [-3, -2];
$ A = [2, 1, 1, 1]$2:2;
$ b = [10, 8];
$ ub = [4, NULL];
$ re = linprog(f, A, b, , , , ub);

$ re[0];
-18

$ re[1];
[2,6]