The MOSEK Python optimizer API manual Version 7.0 (Revision 141)

92 CHAPTER 7. ADVANCED API TUTORIAL
113 else:
114 print ("x%s = %s" % (basis[varsub[i]] - numcon, w1[varsub[i]]))
115
116 # Solve B^Tx = w2
117 nz = 1
118 varsub[0] = 0
119
120 nz = task.solvewithbasis(1, nz, varsub, w2)
121
122 print ("Solution to B^Tx = w2:")
123
124 for i in range(nz):
125 if basis[varsub[i]] < numcon:
126 print ("xc %s = %s" % (basis[varsub[i]], w2[varsub[i]]))
127 else:
128 print ("x %s = %s" % (basis[varsub[i]] - numcon, w2[varsub[i]]) )
129 except Exception as e:
130 print (e)
131
132 if name == ’ main ’:
133 main()
In the example above the linear system is solved using the optimal basis for (7.4) 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
and thus the basis is given by:
It can be verified that
[
B =
[ ] x
c
1
x 0
[ ] x
c
1
=
x 0
0 1
− 1 1
[ − 4
2
]
]