# 9.1 Solving Linear Systems Involving the Basis Matrix¶

A linear optimization problem always has an optimal solution which is also a basic solution. In an optimal basic solution there are exactly $$m$$ basic variables where $$m$$ is the number of rows in the constraint matrix $$A$$. Define

$B \in \real^{m\times m}$

as a matrix consisting of the columns of $$A$$ corresponding to the basic variables. The basis matrix $$B$$ is always non-singular, i.e.

$\det (B) \neq 0$

or, equivalently, $$B^{-1}$$ exists. This implies that the linear systems

(9.1)$B \bar{x} = w$

and

(9.2)$B^T \bar{x} = w$

each have a unique solution for all $$w$$.

MOSEK provides functions for solving the linear systems (9.1) and (9.2) for an arbitrary $$w$$.

In the next sections we will show how to use MOSEK to

## 9.1.1 Basis identification¶

To use the solutions to (9.1) and (9.2) it is important to know how the basis matrix $$B$$ is constructed.

Internally MOSEK employs the linear optimization problem

(9.3)$\begin{split}\begin{array} {lccccl} \mbox{maximize} & & & c^T x & & \\ \mbox{subject to} & & & Ax - x^c & = & 0, \\ & l^x & \leq & x & \leq & u^x, \\ & l^c & \leq & x^c & \leq & u^c. \end{array}\end{split}$

where

$x^c \in \real^m \mbox{ and } x \in \real^n.$

The basis matrix is constructed of $$m$$ columns taken from

$\left[ \begin{array} {cc} A & -I \end{array} \right].$

If variable $$x_j$$ is a basis variable, then the $$j$$-th column of $$A$$, denoted $$a_{:,j}$$, will appear in $$B$$. Similarly, if $$x_i^c$$ is a basis variable, then the $$i$$-th column of $$-I$$ will appear in the basis. The ordering of the basis variables and therefore the ordering of the columns of $$B$$ is arbitrary. The ordering of the basis variables may be retrieved by calling the function Task.initbasissolve. This function initializes data structures for later use and returns the indexes of the basic variables in the array basis. The interpretation of the basis is as follows. If we have

$\mathtt{basis}[i] < \mathtt{numcon}$

then the $$i$$-th basis variable is

$x_{\mathtt{basis}[i]}^c.$

Moreover, the $$i$$-th column in $$B$$ will be the $$i$$-th column of $$-I$$. On the other hand if

$\mathtt{basis}[i] \geq \mathtt{numcon},$

then the $$i$$-th basis variable is the variable

$x_{\mathtt{basis}[i]-\mathtt{numcon}}$

and the $$i$$-th column of $$B$$ is the column

$A_{:,(\mathtt{basis}[i]-\mathtt{numcon})}.$

For instance if $$\mathtt{basis}[0] = 4$$ and $$\mathtt{numcon} = 5$$, then since $$\mathtt{basis}[0]< \mathtt{numcon}$$, the first basis variable is $$x_4^c$$. Therefore, the first column of $$B$$ is the fourth column of $$-I$$. Similarly, if $$\mathtt{basis}[1] = 7$$, then the second variable in the basis is $$x_{\mathtt{basis}[1]-\mathtt{numcon}} = x_2$$. Hence, the second column of $$B$$ is identical to $$a_{:,2}$$.

An example

Consider the linear optimization problem:

(9.4)$\begin{split}\begin{array} {lccl} \mbox{minimize} & x_0 + x_1 & & \\ \mbox{subject to} & x_0 + 2 x_1 & \leq & 2, \\ & x_0 + x_1 & \leq & 6, \\ & x_0, x_1 \geq 0. & & \end{array}\end{split}$

Suppose a call to Task.initbasissolve returns an array basis so that

basis[0] = 1,
basis[1] = 2.


Then the basis variables are $$x_1^c$$ and $$x_0$$ and the corresponding basis matrix $$B$$ is

$\begin{split}\left[ \begin{array} {cc} 0 & 1 \\ -1 & 1 \end{array} \right].\end{split}$

Please note the ordering of the columns in $$B$$ .

Listing 9.1 A program showing how to identify the basis.
import mosek
import sys

def streamprinter(text):
sys.stdout.write(text)
sys.stdout.flush()

def main():
numcon = 2
numvar = 2

# Since the value infinity is never used, we define
# 'infinity' symbolic purposes only
infinity = 0

c = [1.0, 1.0]
ptrb = [0, 2]
ptre = [2, 3]
asub = [0, 1,
0, 1]
aval = [1.0, 1.0,
2.0, 1.0]
bkc = [mosek.boundkey.up,
mosek.boundkey.up]

blc = [-infinity,
-infinity]
buc = [2.0,
6.0]

bkx = [mosek.boundkey.lo,
mosek.boundkey.lo]
blx = [0.0,
0.0]

bux = [+infinity,
+infinity]
w1 = [2.0, 6.0]
w2 = [1.0, 0.0]

try:
with mosek.Env() as env:
c,
0.0,
ptrb,
ptre,
asub,
aval,
bkc,
blc,
buc,
bkx,
blx,
bux)
if r != mosek.rescode.ok:
print("Mosek warning:", r)

#List basis variables corresponding to columns of B
varsub = [0, 1]

for i in range(numcon):
if basis[varsub[i]] < numcon:
print("Basis variable no %d is xc%d" % (i, basis[i]))
else:
print("Basis variable no %d is x%d" %
(i, basis[i] - numcon))

# solve Bx = w1
# varsub contains index of non-zeros in b.
#  On return b contains the solution x and
# varsub the index of the non-zeros in x.
nz = 2

nz = task.solvewithbasis(0, nz, varsub, w1)
print("nz = %s" % nz)
print("Solution to Bx = w1:")

for i in range(nz):
if basis[varsub[i]] < numcon:
print("xc %s = %s" % (basis[varsub[i]], w1[varsub[i]]))
else:
print("x%s = %s" %
(basis[varsub[i]] - numcon, w1[varsub[i]]))

# Solve B^Tx = w2
nz = 1
varsub[0] = 0

nz = task.solvewithbasis(1, nz, varsub, w2)

print("Solution to B^Tx = w2:")

for i in range(nz):
if basis[varsub[i]] < numcon:
print("xc %s = %s" % (basis[varsub[i]], w2[varsub[i]]))
else:
print("x %s = %s" %
(basis[varsub[i]] - numcon, w2[varsub[i]]))
except Exception as e:
print(e)

if __name__ == '__main__':
main()


In the example above the linear system is solved using the optimal basis for (9.4) and the original right-hand side of the problem. Thus the solution to the linear system is the optimal solution to the problem. When running the example program the following output is produced.

basis[0] = 1
Basis variable no 0 is xc1.
basis[1] = 2
Basis variable no 1 is x0.

Solution to Bx = b:

x0 = 2.000000e+00
xc1 = -4.000000e+00

Solution to B^Tx = c:

x1 = -1.000000e+00
x0 = 1.000000e+00


Please note that the ordering of the basis variables is

$\begin{split}\left[ \begin{array} {c} x^c_1 \\ x_0 \end{array} \right]\end{split}$

and thus the basis is given by:

$\begin{split}B = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 1 \end{array} \right]\end{split}$

It can be verified that

$\begin{split}\left[ \begin{array}{c} x^c_1 \\ x_0 \end{array} \right] = \left[ \begin{array} {c} -4 \\ 2 \end{array} \right]\end{split}$

is a solution to

$\begin{split}\left[ \begin{array}{cc} 0 & 1 \\ -1 & 1 \end{array} \right] \left[ \begin{array} {c} x^c_1 \\ x_0 \end{array} \right] = \left[ \begin{array} {c} 2 \\ 6 \end{array} \right].\end{split}$

## 9.1.2 Solving arbitrary linear systems¶

MOSEK can be used to solve an arbitrary (rectangular) linear system

$Ax = b$

using the Task.solvewithbasis function without optimizing the problem as in the previous example. This is done by setting up an $$A$$ matrix in the task, setting all variables to basic and calling the Task.solvewithbasis function with the $$b$$ vector as input. The solution is returned by the function.

An example

Below we demonstrate how to solve the linear system

(9.5)$\begin{split}\left[ \begin{array} {cc} 0 & 1 \\ -1 & 1 \end{array} \right] \left[ \begin{array} {c} x_0 \\ x_1 \end{array} \right] = \left[ \begin{array}{c} b_1 \\ b_2 \end{array} \right]\end{split}$

with two inputs $$b=(1,-2)$$ and $$b=(7,0)$$ .

import mosek

aval,
asub,
ptrb,
ptre,
numvar,
basis):
# Since the value infinity is never used, we define
# 'infinity' symbolic purposes only
infinity = 0

skx = [mosek.stakey.bas] * numvar
skc = [mosek.stakey.fix] * numvar

for i in range(len(asub)):

for i in range(numvar):

for i in range(numvar):
mosek.boundkey.fr,
-infinity,
infinity)

# Define a basic solution by specifying
# status keys for variables & constraints.

def main():
numcon = 2
numvar = 2

aval = [[-1.0],
[1.0, 1.0]]
asub = [[1],
[0, 1]]

ptrb = [0, 1]
ptre = [1, 3]

# Directs the log task stream to the user specified
lambda msg: sys.stdout.write(msg))
# Put A matrix and factor A.
# Call this function only once for a given task.

basis = [0] * numvar
b = [0.0, -2.0]
bsub = [0, 1]

aval,
asub,
ptrb,
ptre,
numvar,
basis)

# now solve rhs
b = [1, -2]
bsub = [0, 1]
nz = task.solvewithbasis(0, 2, bsub, b)
print("\nSolution to Bx = b:\n")

# Print solution and show correspondents
# to original variables in the problem
for i in range(nz):
if basis[bsub[i]] < numcon:
print("This should never happen")
else:
print("x%d = %d" % (basis[bsub[i]] - numcon, b[bsub[i]]))

b[0] = 7
bsub[0] = 0

nz = task.solvewithbasis(0, 1, bsub, b)

print("\nSolution to Bx = b:\n")
# Print solution and show correspondents
# to original variables in the problem
for i in range(nz):
if basis[bsub[i]] < numcon:
print("This should never happen")
else:
print("x%d = %d" % (basis[bsub[i]] - numcon, b[bsub[i]]))

if __name__ == "__main__":
try:
main()
except:
import traceback
traceback.print_exc()


The most important step in the above example is the definition of the basic solution, where we define the status key for each variable. The actual values of the variables are not important and can be selected arbitrarily, so we set them to zero. All variables corresponding to columns in the linear system we want to solve are set to basic and the slack variables for the constraints, which are all non-basic, are set to their bound.

The program produces the output:

Solution to Bx = b:

x1 = 1
x0 = 3

Solution to Bx = b:

x1 = 7
x0 = 7