7.1 Linear Optimization

The simplest optimization problem is a purely linear problem. A linear optimization problem (see also Sec. 12.1 (Linear Optimization)) is a problem of the following form:

Minimize or maximize the objective function

\[\sum_{j=0}^{n-1} c_j x_j + c^f\]

subject to the linear constraints

\[l_k^c \leq \sum_{j=0}^{n-1} a_{kj} x_j \leq u_k^c,\quad k=0,\ldots ,m-1,\]

and the bounds

\[l_j^x \leq x_j \leq u_j^x, \quad j=0,\ldots ,n-1.\]

The problem description consists of the following elements:

• \(m\) and \(n\) — the number of constraints and variables, respectively,

• \(x\) — the variable vector of length \(n\),

• \(c\) — the coefficient vector of length \(n\)

  \[\begin{split}c = \left[ \begin{array}{c} c_0 \\ \vdots \\ c_{n-1} \end{array} \right],\end{split}\]

• \(c^f\) — fixed term in the objective,

• \(A\) — an \(m\times n\) matrix of coefficients

  \[\begin{split}A = \left[ \begin{array}{ccc} a_{0,0} & \cdots & a_{0,(n-1)} \\ \vdots & \cdots & \vdots \\ a_{(m-1),0} & \cdots & a_{(m-1),(n-1)} \end{array} \right],\end{split}\]

• \(l^c\) and \(u^c\) — the lower and upper bounds on constraints,

• \(l^x\) and \(u^x\) — the lower and upper bounds on variables.

Please note that we are using \(0\) as the first index: \(x_0\) is the first element in variable vector \(x\).

The Fusion user does not need to specify all of the above elements explicitly — they will be assembled from the Fusion model.

7.1.1 Example LO1

The following is an example of a small linear optimization problem:

(7.1)\[\begin{split}\begin{array} {lccccccccl} \mbox{maximize} & 3 x_0 & + & 1 x_1 & + & 5 x_2 & + & 1 x_3 & & \\ \mbox{subject to} & 3 x_0 & + & 1 x_1 & + & 2 x_2 & & & = & 30, \\ & 2 x_0 & + & 1 x_1 & + & 3 x_2 & + & 1 x_3 & \geq & 15, \\ & & & 2 x_1 & & & + & 3 x_3 & \leq & 25, \end{array}\end{split}\]

under the bounds

\[\begin{split}\begin{array}{ccccc} 0 & \leq & x_0 & \leq & \infty , \\ 0 & \leq & x_1 & \leq & 10, \\ 0 & \leq & x_2 & \leq & \infty ,\\ 0 & \leq & x_3 & \leq & \infty . \end{array}\end{split}\]

We start our implementation in Fusion importing the relevant modules, i.e.

from mosek.fusion import *
import mosek.fusion.pythonic

Next we declare an optimization model creating an instance of the Model class:

    with Model("lo1") as M:

For this simple problem we are going to enter all the linear coefficients directly:

    A = [[3.0, 1.0, 2.0, 0.0],
         [2.0, 1.0, 3.0, 1.0],
         [0.0, 2.0, 0.0, 3.0]]
    c = [3.0, 1.0, 5.0, 1.0]

The variables appearing in problem (7.1) can be declared as one \(4\)-dimensional variable:

        x = M.variable("x", 4, Domain.greaterThan(0.0))

At this point we already have variables with bounds \(0\leq x_i\leq \infty\), because the domain is applied element-wise to the entries of the variable vector. Next, we impose the upper bound on \(x_1\):

        M.constraint(x[1] <= 10)

The linear constraints can now be entered one by one using the dot product of our variable with a coefficient vector:

        M.constraint("c1", x.T @ A[0] == 30.0)
        M.constraint("c2", x.T @ A[1] >= 15.0)
        M.constraint("c3", x.T @ A[2] <= 25.0)

We end the definition of our optimization model setting the objective function in the same way:

        M.objective("obj", ObjectiveSense.Maximize, x.T @ c)

Finally, we only need to call the Model.solve method:

        M.solve()

The solution values can be attained with the method Variable.level.

        sol = x.level()
        print('\n'.join(["x[%d] = %f" % (i, sol[i]) for i in range(4)]))

Listing 7.1 Fusion implementation of model (7.1). Click here to download.

from mosek.fusion import *
import mosek.fusion.pythonic

def main(args):
    A = [[3.0, 1.0, 2.0, 0.0],
         [2.0, 1.0, 3.0, 1.0],
         [0.0, 2.0, 0.0, 3.0]]
    c = [3.0, 1.0, 5.0, 1.0]

    # Create a model with the name 'lo1'
    with Model("lo1") as M:

        # Create variable 'x' of length 4
        x = M.variable("x", 4, Domain.greaterThan(0.0))

        # Create constraints
        M.constraint(x[1] <= 10)
        M.constraint("c1", x.T @ A[0] == 30.0)
        M.constraint("c2", x.T @ A[1] >= 15.0)
        M.constraint("c3", x.T @ A[2] <= 25.0)

        # Set the objective function to (c^t * x)
        M.objective("obj", ObjectiveSense.Maximize, x.T @ c)

        # Solve the problem
        M.solve()

        # Get the solution values
        sol = x.level()
        print('\n'.join(["x[%d] = %f" % (i, sol[i]) for i in range(4)]))