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Finding an LP subsystem with as many constraints as possible Description Given an infeasible LP, find the feasible subsystem of constraints of maximum cardinality. Further explanation of this example:
'Xpress Python Reference Manual'
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example_phase1.py # 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., 1983-2023 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.optimize() assert(p.attributes.solstatus == xp.SolStatus.INFEASIBLE) # 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.optimize() # 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) | |||||||||||

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