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

5.10. PROBLEM MODIFICATION AND REOPTIMIZATION 65
and
maximize 1.5x 0 + 2.5x 1 + 3.0x 2
subject to 3x 0 + 4x 1 + 3x 2 ≤ 100000,
3x 0 + 2x 1 + 3x 2 ≤ 50000,
2x 0 + 3x 1 + 2x 2 ≤ 60000,
(5.11)
x 0 , x 1 , x 2 ≥ 0.
After changing the A matrix we can find the new optimal solution by calling
Task.optimize again.
5.10.3 Appending variables
We now want to add a new product with the following data:
Product no. Assembly (minutes) Polishing (minutes) Packing (minutes) Profit ($)
3 4 0 1 1.00
This corresponds to creating a new variable x 3 , appending a new column to the A matrix and setting
a new value in the objective. We do this in the following code.
[ production.py ]
122 # Append a new varaible x 3 to the problem */
123 task.appendvars(1)
124
125 # Set bounds on new varaible
126 task.putbound(mosek.accmode.var,
127 task.getnumvar()-1,
128 mosek.boundkey.lo,
129 0,
130 +inf)
131
132 # Change objective
133 task.putcj(task.getnumvar()-1,1.0)
134
135 # Put new values in the A matrix
136 acolsub = array([0, 2])
137 acolval = array([4.0, 1.0])
138
139 task.putacol(task.getnumvar()-1, # column index
140 acolsub,
141 acolval)
After this operation the problem looks this way:
maximize 1.5x 0 + 2.5x 1 + 3.0x 2 + 1.0x 3
subject to 3x 0 + 4x 1 + 3x 2 + 4x 3 ≤ 100000,
3x 0 + 2x 1 + 3x 2 ≤ 50000,
2x 0 + 3x 1 + 2x 2 + 1x 3 ≤ 60000,
(5.12)