# Example: given an infeasible LP, find an (infeasible) solution that minimize
# the total distance from the constraints.
#
# Then solve the obtained MaxFS problem.
#
# (C) Fair Isaac Corp., 1983-2020
import xpress as xp
x = xp.var()
y = xp.var()
# build a very simple problem with pairs of incompatible constraints
p = xp.problem()
p.addVariable(x, y)
lhs1 = 2*x + 3*y
lhs2 = 3*x + 2*y
lhs3 = 4*x + 5*y
p.addConstraint(lhs1 >= 6, lhs1 <= 5)
p.addConstraint(lhs2 >= 5, lhs2 <= 4)
p.addConstraint(lhs3 >= 8, lhs3 <= 7)
p.solve()
assert(p.getProbStatus() == xp.lp_infeas)
# We verified the problem is infeasible. Add one binary for each
# constraint to selectively relax them.
m = p.attributes.rows
# get the signs of all constraints: 'E', 'L', or 'G'. Note that this example
# only works with inequality constraints only
sign = []
p.getrowtype(sign, 0, m - 1)
# big-M, large-enough constant to relax all constraints (quite conservative
# here)
M = 1e3
matval = [M]*m
for i in range(m):
if sign[i] == 'L':
matval[i] = -M
# Add m new binary columns
p.addcols([1]*m, # obj. coefficients (as many 1s as there are constraints)
range(m + 1), # cumulative number of terms in each column:
# 0,1,2,...,m as there is one term per column
range(m), matval, # pairs (row_index, coefficient) for each column
[0]*m, [1]*m, # lower, upper bound (binary variables, so {0,1})
['b_{}'.format(i) for i in range(m)], # names are b_i, with i is the
# constraint index
['B']*m) # type: binary
p.solve()
# Print constraints constituting a Maximum Feasible Subsystem
b = p.getSolution(range(p.attributes.cols - m, p.attributes.cols))
maxfs = [i for i in range(m) if b[i] > 0.5]
print('MaxFS has ', len(maxfs), 'constraints:', maxfs)