# 6.1 Linear Optimization¶

The simplest optimization problem is a purely linear problem. A linear optimization problem 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 $$x$$.

## 6.1.1 Example LO1¶

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

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

under the bounds

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

Solving the problem

To solve the problem above we go through the following steps:

1. (Optionally) Create an environment.

4. Optimization.

5. Extracting the solution.

Below we explain each of these steps.

Create an environment.

The user can start by creating a MOSEK environment, but it is not necessary if the user does not need access to other functionalities, license management, additional routines, etc. Therefore in this tutorial we don’t create an explicit environment.

We create an empty task object. A task object represents all the data (inputs, outputs, parameters, information items etc.) associated with one optimization problem.

    # Create a task object
# Attach a log stream printer to the task


We also connect a call-back function to the task log stream. Messages related to the task are passed to the call-back function. In this case the stream call-back function writes its messages to the standard output stream. See Sec. 7.4 (Input/Output).

Before any problem data can be set, variables and constraints must be added to the problem via calls to the functions Task.appendcons and Task.appendvars.

        # Append 'numcon' empty constraints.
# The constraints will initially have no bounds.

# Append 'numvar' variables.
# The variables will initially be fixed at zero (x=0).


New variables can now be referenced from other functions with indexes in $$0, \ldots, \mathtt{numvar}-1$$ and new constraints can be referenced with indexes in $$0, \ldots , \mathtt{numcon}-1$$. More variables and/or constraints can be appended later as needed, these will be assigned indexes from $$\mathtt{numvar}$$/$$\mathtt{numcon}$$ and up.

Next step is to set the problem data. We loop over each variable index $$j=0, \ldots, \mathtt{numvar}-1$$ calling functions to set problem data. We first set the objective coefficient $$c_j = \mathtt{c[j]}$$ by calling the function Task.putcj.

            task.putcj(j, c[j])


Setting bounds on variables

The bounds on variables are stored in the arrays

        # Bound keys for variables
bkx = [mosek.boundkey.lo,
mosek.boundkey.ra,
mosek.boundkey.lo,
mosek.boundkey.lo]

# Bound values for variables
blx = [0.0, 0.0, 0.0, 0.0]
bux = [+inf, 10.0, +inf, +inf]


and are set with calls to Task.putvarbound.

            # Set the bounds on variable j
# blx[j] <= x_j <= bux[j]


The Bound key stored in bkx specifies the type of the bound according to Table 6.1.

Table 6.1 Bound keys as defined in the enum boundkey.

Bound key

Type of bound

Lower bound

Upper bound

boundkey.fx

$$u_j = l_j$$

Finite

Identical to the lower bound

boundkey.fr

Free

$$-\infty$$

$$+\infty$$

boundkey.lo

$$l_j \leq \cdots$$

Finite

$$+\infty$$

boundkey.ra

$$l_j \leq \cdots \leq u_j$$

Finite

Finite

boundkey.up

$$\cdots \leq u_j$$

$$-\infty$$

Finite

For instance bkx[0]= boundkey.lo means that $$x_0 \geq l_0^x$$. Finally, the numerical values of the bounds on variables are given by

$l_j^x = \mathtt{blx[j]}$

and

$u_j^x = \mathtt{bux[j]}.$

Defining the linear constraint matrix.

Recall that in our example the $$A$$ matrix is given by

$\begin{split}A = \left[ \begin{array}{cccc} 3 & 1 & 2 & 0 \\ 2 & 1 & 3 & 1 \\ 0 & 2 & 0 & 3 \end{array} \right].\end{split}$

This matrix is stored in sparse format in the arrays:

        asub = [[0, 1],
[0, 1, 2],
[0, 1],
[1, 2]]
aval = [[3.0, 2.0],
[1.0, 1.0, 2.0],
[2.0, 3.0],
[1.0, 3.0]]


The array aval[j] contains the non-zero values of column $$j$$ and asub[j] contains the row indices of these non-zeros.

Using the function Task.putacol we set column $$j$$ of $$A$$

            task.putacol(j,                  # Variable (column) index.
asub[j],            # Row index of non-zeros in column j.
aval[j])            # Non-zero Values of column j.


There are many alternative formats for entering the $$A$$ matrix. See functions such as Task.putarow, Task.putarowlist, Task.putaijlist and similar.

Finally, the bounds on each constraint are set by looping over each constraint index $$i= 0, \ldots,\mathtt{numcon}-1$$

        # Set the bounds on constraints.
# blc[i] <= constraint_i <= buc[i]
for i in range(numcon):


Optimization

After the problem is set-up the task can be optimized by calling the function Task.optimize.

        task.optimize()


Extracting the solution.

After optimizing the status of the solution is examined with a call to Task.getsolsta. If the solution status is reported as solsta.optimal the solution is extracted in the lines below:

            xx = task.getxx(mosek.soltype.bas)


The Task.getxx function obtains the solution. MOSEK may compute several solutions depending on the optimizer employed. In this example the basic solution is requested by setting the first argument to soltype.bas.

Catching exceptions

We catch any exceptions thrown by MOSEK in the lines:

except mosek.Error as e:
print("ERROR: %s" % str(e.errno))
if e.msg is not None:
print("\t%s" % e.msg)
sys.exit(1)


The types of exceptions that MOSEK can throw can be seen in Sec. 15.5 (Exceptions). See also Sec. 7.3 (Errors and exceptions).

Source code

The complete source code lo1.py of this example appears below. See also lo2.py for a version where the $$A$$ matrix is entered row-wise.

Listing 6.1 Linear optimization example. Click here to download.
import sys
import mosek

# Since the value of infinity is ignored, we define it solely
# for symbolic purposes
inf = 0.0

# Define a stream printer to grab output from MOSEK
def streamprinter(text):
sys.stdout.write(text)
sys.stdout.flush()

def main():
# Attach a log stream printer to the task

# Bound keys for constraints
bkc = [mosek.boundkey.fx,
mosek.boundkey.lo,
mosek.boundkey.up]

# Bound values for constraints
blc = [30.0, 15.0, -inf]
buc = [30.0, +inf, 25.0]

# Bound keys for variables
bkx = [mosek.boundkey.lo,
mosek.boundkey.ra,
mosek.boundkey.lo,
mosek.boundkey.lo]

# Bound values for variables
blx = [0.0, 0.0, 0.0, 0.0]
bux = [+inf, 10.0, +inf, +inf]

# Objective coefficients
c = [3.0, 1.0, 5.0, 1.0]

# Below is the sparse representation of the A
# matrix stored by column.
asub = [[0, 1],
[0, 1, 2],
[0, 1],
[1, 2]]
aval = [[3.0, 2.0],
[1.0, 1.0, 2.0],
[2.0, 3.0],
[1.0, 3.0]]

numvar = len(bkx)
numcon = len(bkc)

# Append 'numcon' empty constraints.
# The constraints will initially have no bounds.

# Append 'numvar' variables.
# The variables will initially be fixed at zero (x=0).

for j in range(numvar):
# Set the linear term c_j in the objective.

# Set the bounds on variable j
# blx[j] <= x_j <= bux[j]

# Input column j of A
asub[j],            # Row index of non-zeros in column j.
aval[j])            # Non-zero Values of column j.

# Set the bounds on constraints.
# blc[i] <= constraint_i <= buc[i]
for i in range(numcon):

# Input the objective sense (minimize/maximize)

# Solve the problem
# Print a summary containing information
# about the solution for debugging purposes

# Get status information about the solution

if (solsta == mosek.solsta.optimal):

print("Optimal solution: ")
for i in range(numvar):
print("x[" + str(i) + "]=" + str(xx[i]))
elif (solsta == mosek.solsta.dual_infeas_cer or
solsta == mosek.solsta.prim_infeas_cer):
print("Primal or dual infeasibility certificate found.\n")
elif solsta == mosek.solsta.unknown:
print("Unknown solution status")
else:
print("Other solution status")

# call the main function
try:
main()
except mosek.Error as e:
print("ERROR: %s" % str(e.errno))
if e.msg is not None:
print("\t%s" % e.msg)
sys.exit(1)
except:
import traceback
traceback.print_exc()
sys.exit(1)