# An application of the Xpress callbacks to a Benders decomposition algorithm.
#
# Solves the problem:
# min c1*x + c2*y
# st A1*x + A2*y <=b (m constraints)
# x binary n1-dimensional vector
# y >=0 continuous n2-dimensional vector
#
# NOTE: This example is for illustration/tutorial purposes only. There is no
# benefit in solving the generic problem defined below using Benders
# decomposition.
# Courtesy of Georgios Patsakis (UC Berkeley, Amazon) and Richard L.-Y. Chen (Amazon).
#
# (C) 2019-2025 Fair Isaac Corporation
# Import necessary packages.
import xpress as xp
import sys
# Initialize problem parameters.
c1 = [1,6,5,7] # n1 x 1
c2 = [9,3,0,2,3] # n2 x 1
b = [-3,-4,1,4,5] # m x 1
A1 = [[0, -2, 3, 2],
[-5, 0, -3, 1],
[1, 0, 4, -2],
[0, -3, 4, -1],
[-5, -4, 3, 0]]
A2 = [[3, 4, 2, 0, -5],
[0, 2, 3, -2, 1],
[2, 0, 1, -3, -5],
[-5, 3, -2, -3, 0],
[-2, 3, -1, 2, -4]]
m = len(b)
n1 = len(c1)
n2 = len(c2)
# Absolute accuracy for optimal objective.
ObjAbsAccuracy=0.00001
# SOLVE ORIGINAL PROBLEM WITHOUT DECOMPOSING
# Define Problem.
p = xp.problem()
# Define Variables.
x = [p.addVariable(vartype=xp.binary) for _ in range(n1)]
y = [p.addVariable() for _ in range(n2)] # positive real variable
# Define Constraints.
constr = [xp.Sum(A1[ii][jj]*x[jj] for jj in range(n1)) + \
xp.Sum(A2[ii][jj]*y[jj] for jj in range(n2)) \
<= b[ii] for ii in range(m)]
p.addConstraint(constr)
# Define Objective.
p.setObjective(xp.Sum(c1[jj]*x[jj] for jj in range(n1)) + \
xp.Sum(c2[jj]*y[jj] for jj in range(n2)) , \
sense=xp.minimize)
# Solve and retrieve solution.
p.optimize()
if p.attributes.solvestatus != xp.SolveStatus.COMPLETED or \
p.attributes.solstatus != xp.SolStatus.OPTIMAL:
raise RuntimeError('Problem could not be solved to MIP optimality')
yopt = p.getSolution(y)
print("The optimal objective is:", p.attributes.objval)
print("The first stage variables are:", p.getSolution(x))
print("The second stage variables are:", yopt)
print("The objective of the second stage is:",
sum(c2[jj]*yopt[jj] for jj in range(n2)))
# SOLVE DECOMPOSED PROBLEM
# We define the following functions:
# - subproblem: function that solves the subproblem for a given first stage
# variable xhat. If the subproblem has an optimal solution, it returns the
# the optimal dual multiplier associated with the non anticipativaty
# constraint, the optimal objective, and the label "Optimal". If the
# subproblem is infeasible, it returns the dual ray of unboundedness and
# the label "Infeasible".
#
# - integer_callback: An integer callback is triggered every time an integer
# solution is found. The function integer_callback is an input to the
# addPreIntsolCallback function and has a specified input - output format based
# on the Xpress documentation. It takes as input the main optimization
# problem name, a user data structure, a variable that indicates whether
# the integer solution comes from a heuristic or from a node relaxation,
# and the proposed update to the upper bound if this integer solution is
# accepted as feasible. The inputs are populated by the solver when the
# callback is triggered. The function returns whether the integer solution
# found should be accepted as feasible and the proposed update to the upper
# bound if the solution is feasible (which could be the same as the one
# proposed by the solver as part of the input).
# Subproblem Solver Function.
def subproblem(xhat):
# Input:
# - xhat: n1x1a array is the first stage variable,
# passed to the subproblem.
#
# Output:
# return (lamb: n1x1 array, beta: scalar, flag:string)
# - If the subproblem is Infeasible, the flag is 'Infeasible', and
# lamb (n1x1) and beta (1x1) are the components of the dual ray of
# unboundedness necessary to write a feasibility cut of the form:
# sum(x[i]*lamb[i] for all i) +beta >= 0
# - If the subproblem is Optimal, the flag is 'Optimal', beta (1x1) is the
# optimal objective, and lamb (n1x1) is the optimal dual multiplier
# associated with the non anticipativaty constraint, so that the
# optimality cut will have the form:
# theta >= sum(x[i]*lamb[i] for all i) + beta
# IMPORTANT NOTE: The subproblem needs to ensure tha upper bounds or non
# zero lower bounds should ONLY BE DEFINED THROUGH CONSTRAINTS, not as part
# of the variable definition. This is to ensure the feasibility cuts are
# correct.
# Side Note: It is a good practice to ensure that the subproblem is always
# feasible through modeling, by say adding slacks to all the constraints,
# hence avoiding feasibility cuts completely. This is because feasibility
# cuts tend to be extremely slow. In this code we implemented feasibility
# cuts for reasons of completeness.
# Define and Solve Subproblem.
# Initialize Problem.
r=xp.problem()
# Define Variables.
y=[r.addVariable() for _ in range(n2)] # Positive real variable - second stage.
z=[r.addVariable(lb=-xp.infinity) for _ in range(n1)] # Dummy copy variable,
# must have the exact shape as xhat.
epsilon=r.addVariable(lb=-xp.infinity) # Dummy variable to extract part of the
# infeasibility ray (if infeasible).
# Define Constraints.
# Dummy cosntraint for optimality/feasibility cuts.
# Note: make sure the index of the variable and of the position in the
# constraint array is the same (or has a known mapping).
dummy1= [z[i]==xhat[i] for i in range(n1)]
# Dummy constraint for feasibility cut.
dummy2= epsilon==1
# Second stage constraints, where RHS b has been multiplied by epsilon.
constr=[xp.Sum(A1[ii][jj]*z[jj] for jj in range(n1)) + \
xp.Sum(A2[ii][jj]*y[jj] for jj in range(n2)) \
- epsilon*b[ii]<=0 for ii in range(m)]
r.addConstraint(constr,dummy1,dummy2)
# Define Objective.
r.setObjective(xp.Sum(c2[jj]*y[jj] for jj in range(n2)),
sense=xp.minimize)
# Disable presolve. The only reason to do this is because in the case of
# an infeasible problem, the presolve might identify infeasibility of the
# problem without running simplex. Because of that, no dual infeasibility
# certificate will be found, hence a dual ray of unboundedness will not be
# returned, which means that a feasibility cut can not be formed. In
# practice, we should avoid infeasibility cuts altogether through modeling
# and have presolve on for a better performance in the subproblem. If an
# infeasibility cut can not be avoided, formulating the dual explicitly is
# the only option.
r.controls.presolve = 0
# Silence output.
r.controls.outputlog = 0
# Solve optimization.
r.optimize()
# Find the indices of constraint dummy1, which will be used to access the
# dual multipliers corresponding to the entries of xhat.
xind1=[dummy1[ii].index for ii in range(n1)]
if r.attributes.solvestatus != xp.SolveStatus.COMPLETED:
print("ERROR: Subproblem was not solved. Terminating.")
sys.exit()
# Take cases depending on subproblem status.
if r.attributes.solstatus == xp.SolStatus.OPTIMAL:
# Optimal subproblem.
print("Optimal Subproblem")
# Retrieve optimal dual multiplier and objective and return.
dualmult=r.getDuals()
lamb=[dualmult[ii] for ii in xind1]
beta=r.attributes.objval
return(lamb,beta,'Optimal')
elif r.attributes.solstatus == xp.SolStatus.INFEAS:
# Infeasible subproblem.
print("Infeasible Subproblem")
if not r.hasdualray ():
print ("Could not retrieve a dual ray, return no good cut instead:")
# This is just a cheat if the subproblem is found infeasible and
# the optimizer fails to find a dual ray. Since all the first stage
# variables are binary, we add a No-Good-Cut to exclude the point
# xhat instead of an infeasibility cut.
xhatones=set(ii for ii in range(n1) if xhat[ii]>=0.5)
lamb=[2*xhat[ii]-1 for ii in range(n1)]
beta=-sum(xhat)+1
else:
# Extract the dual ray.
dray = r.getDualRay()
print ("Dual Ray:", dray)
# Extract the part of the dual ray corresponding to the first stage
# variables xhat.
lamb=[dray[ii] for ii in xind1]
# Extract the constant from the dual ray entry of constraint dummy 2.
beta=dray[dummy2.index]
return(lamb,beta,'Infeasible')
else:
print("ERROR: Subproblem not optimal or infeasible. Terminating.")
sys.exit()
# Integer Callback Function
def integer_callback(p, data, soltype, cutoff):
# Input:
# NOTE: the input is populated by the solver when an integer solution is
# found.
# - p: an xp.problem() (the main problem from which the callbacks were
# triggered)
# - data: structure that is passed to the callback from the
# problem, and is specified in the addPreIntsolCallback function. No
# data is needed here, so the addPreIntsolCallback() call uses None.
# - soltype: 0 if the integer solution was found by a node relaxation,
# and not 0 if the integer solution was found by a heuristic
# - cutoff: the integer solution found by the solver is associated with an
# upper bound, if this integer solution is feasible to full problem. cutoff
# is the upper bound (slightly decreased) associated with that integer
# solution. If the user decides that the integer solution is indeed feasible,
# the user can modify that cutoff, if the default solver value misrepresents
# the upper bound, and return a new cutoff (newcutoff)
#
# Output:
# return (isreject, newcutoff)
# - isreject: True to reject integer solution, False to accept it
# - newcutoff: in case isreject is False, possible update to the cutoff.
print("Entering Integer Callback.")
if soltype ==0:
# In this case the integer solution was found by a node relaxation of
# the B&B tree, so we accept it as is. The node callback has taken
# care of ensuring the solution is feasible to the subproblem.
print("Integer solution from optimal node relaxation.")
print("Solution accepted. Cutoff:", cutoff)
return(False,None)
else:
print("Integer solution found from heuristic. Checking subproblem.")
# In any other case, we need to solve the subproblem.
# Retrieve xhat and thetahat from the heuristic solution.
xhat = p.getCallbackSolution([x[ii] for ii in range(n1)])
thetahat = p.getCallbackSolution(theta)
print("Solution tested x=",xhat, "and theta=",thetahat)
print("Cutoff is:",cutoff)
# Solve the subproblem.
(dual_mult,opt,status)=subproblem(xhat)
if status=='Infeasible':
# In the case of an infeasible subproblem, reject the solution found.
return (True,None)
elif status=='Optimal':
if thetahat>=opt-ObjAbsAccuracy:
# In this case the vector (xhat,thetahat) is feasible, so accept
# it and accept the cutoff.
print("Accepting pair x=",xhat,",theta=",thetahat )
print("Accept new cutoff:",cutoff)
return(False,None)
else:
# In this case the solution (xhat,thetahat) is infeasible so
# reject it.
return (True,None)
else:
print("We shouldn't reach this point.")
print("Rejecting pair x=",xhat,",theta=",thetahat )
return(True,None)
# Node Callback Function
def node_callback(p, data):
# Input:
# Note: the input is populated by the solver when the relaxation of a node
# is solved
# - p: the xp.problem corresponding to the main problem, from which the
# callback is triggered.
# - data: structure that is passed to the callback from the
# problem, and is specified in the addOptnodeCallback function. No
# data is needed here, so the addOptnodeCallback() call uses None.
#
# Output:
# return 0
print("We entered a node callback")
# Obtain node relaxation solution.
xind = [x[ii].index for ii in range(n1)]
thetaind = [theta.index]
xhat = p.getCallbackSolution([x[ii] for ii in range(n1)])
thetahat = p.getCallbackSolution(theta)
print("Solution tested x=",xhat, "and theta=",thetahat)
# Now let's see if the solution satisfies the subproblem induced constraints.
# Solve the subproblem.
(dual_mult,opt,status)=subproblem(xhat)
if status=='Infeasible':
# Create feasibility cut to add to node.
coefficients=dual_mult
rhs=-opt
print("Add cut. Cut stats:")
print("coefficients:",coefficients)
print("<=rhs:",rhs)
print("column indices:",xind)
# Add the cut (this will reject the point).
p.addCuts([1],['L'],[rhs],[0,len(xhat)],xind,coefficients)
print("Rejecting pair x=",xhat,",theta=",thetahat, "with a cut." )
return 0
elif status=='Optimal':
if thetahat>=opt-ObjAbsAccuracy:
# In this case the vector (xhat,thetahat) is feasible, so accept.
print("Accepting pair x=",xhat,",theta=",thetahat )
print("An integer callback will be triggered for that pair later.")
return 0
else:
# In this case the solution (xhat,thetahat) is infeasible so reject
# it and add an optimality cut.
coefficients=[1]+[-dual_mult[ii] for ii in range(len(dual_mult))]
rhs=opt-sum(dual_mult[ii]*xhat[ii] for ii in range(len(dual_mult)))
print("Store cut. Cut stats:")
print("coefficients:",coefficients)
print(">=rhs:",rhs)
print("column indices:",thetaind+xind)
# Add the cut (this will reject the point).
p.addCuts([1],['G'],[rhs],[0,len(xhat)+1],thetaind+xind,coefficients)
print("Rejecting pair x=",xhat,",theta=",thetahat,"with a cut." )
return 0
else:
print("ERROR: Subproblem not optimal or infeasible. Terminating.")
sys.exit()
# Main Problem.
# Define main problem.
p=xp.problem()
# Define first stage variables.
x=[p.addVariable(vartype=xp.binary) for _ in range(n1)]
theta=p.addVariable() # (positive) second stage cost - change the default lower
# bound if second stage cost may not be positive.
# Define objective.
p.setObjective(xp.Sum(c1[jj]*x[jj] for jj in range(n1)) + theta,
sense=xp.minimize)
# Add Integer callback. i.e., when an integer solution is found, the function
# integer_callback will be called by the solver.
p.addPreIntsolCallback(integer_callback, None, 0)
# Add node_callback. i.e., when a node relaxation is solved, the function
# node_callback will be called by the solver.
p.addOptnodeCallback(node_callback, None, 0)
# Deactivate presolve. This is to ensure that the solver will not eliminate
# variables, so the indices we retrieve for the variables and the cuts we add
# correspond to/contain existing optimization variables. Presolve can be
# activated, but additional steps are required for that (presolving the cuts
# the same way as the rows of the matrix).
p.setControl({"presolve":0,"mippresolve":0,"symmetry":0})
# Solve the main problem.
p.optimize()
if p.attributes.solvestatus == xp.SolveStatus.COMPLETED and \
p.attributes.solstatus == xp.SolStatus.OPTIMAL:
# Print results.
print("The optimal objective is:",p.attributes.objval)
print("The first stage variables are:",p.getSolution(x))
else:
print("Could not solve the problem")
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