Using indicators

# Construct a problem from scratch with variables of various
# types. Adds indicator constraints and shows how to retrieve
# such data once it has been added to the problem using the
# API functions.
# - Example discussed in mipformref whitepaper -
#
# (C) 1983-2026 Fair Isaac Corporation

import xpress as xp

m = xp.problem()

m.controls.miprelstop = 0

# All variables used in this example.
v1 = m.addVariable(lb=0, ub=10, vartype=xp.continuous)
v2 = m.addVariable(lb=1, ub=7, vartype=xp.continuous)
v3 = m.addVariable(lb=5, ub=10, threshold=7, vartype=xp.semicontinuous)
vb = m.addVariable(vartype=xp.binary)

# Indicator constraints consist of a tuple with a condition on a
# binary variable and a constraint.
ind1 = (vb == 0, v1 + v2 >= 6)
ind2 = (vb == 1, v1 + v3 >= 4)

# Adds the first indicator constraint.
m.addIndicator(ind1)
# Adds another indicator constraint and the second one defined above.
m.addIndicator((vb == 1, v2 + v3 >= 5), ind2)

# Returns the indicator constraint condition (indicator variable and complement flag)
# associated to the rows in a given range.
ind_inds, ind_flags = m.getIndicators(0, 2)

print("Indicator variables and flags: ", ind_inds, ind_flags)

m.setObjective(xp.Sum(v1,v2,v3))

m.optimize()

print("v1: ", m.getSolution(v1),
      ", v2: ", m.getSolution(v2),
      "; sol vector: ", m.getSolution(),
      "; obj: ", m.attributes.objval,
      sep="")  # Default separator between strings is " ".

# Indicator constraint definition with nonlinear constraints.
# - Example discussed in mipformref whitepaper -
#
#     min   b1 + b2
#     s.t.  b1 = 1  ->  x1*x2 = 12
#           b2 = 0  ->  sin(x2) >= 0.5
#           b1, b2 in {0,1}
#
# (C) 2026 Fair Isaac Corporation

import xpress as xp
from xpress.enums import SolStatus

p = xp.problem(name="NL indicator constraints")

x = p.addVariables(2, name="x")
# Binary variables driving the indicator constraints
b = p.addVariables(2, vartype=xp.binary, name="b")

# Indicator constraint: condition (b[i] == 0 or 1) paired with the
# constraint that must hold when the condition is true.
ind1 = (b[0] == 1, x[0] * x[1] == 12)  # b[0]=1 -> x[0]*x[1]=12
ind2 = (b[1] == 0, xp.sin(x[1]) >= 0.5)  # b[1]=0 -> sin(x[1])>=0.5

# Both constraints are nonlinear (quadratic / trigonometric)
p.addIndicator(ind1, ind2)

p.setObjective(b[0] + b[1], sense=xp.ObjSense.MINIMIZE)

p.optimize()

print("Problem status:", p.attributes.solstatus.name)

if p.attributes.solstatus in [SolStatus.FEASIBLE, SolStatus.OPTIMAL]:
    for i in range(2):
        print(f"x[{i}] = {p.getSolution(x[i])}")

    # getIndicators returns the indicator variable index and the sense flag
    # for each row.
    ind_vars, ind_flags = p.getIndicators(0, 1)
    for i in range(2):
        print(f"indicator {i}: b[{i}] = {p.getSolution(b[i])}, flag = {ind_flags[i]}")