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Solve a simple MIP using Benders decomposition

Description
Solve a simple MIP using Benders decomposition. Courtesy of Georgios Patsakis (UC Berkeley, Amazon) and Richard L.-Y. Chen (Amazon).

Further explanation of this example: 'Xpress Python Reference Manual'

Source Files
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benders_decomp.py

#
# An application of the Xpress callbacks to a Benders decomposition algorithm
#
# Courtesy of Georgios Patsakis (UC Berkeley, Amazon) and Richard L.-Y. Chen (Amazon)
#

################################################################################
##################   BENDERS DECOMPOSITION USING CALLBACKS   ###################
################################################################################
#### NOTE: This example is for illustration/tutorial purposes only. There is no
#### benefit in solving the generic problem defined below using Benders
#### decomposition.

#### 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

################################################################################
#########################  PROBLEM INITIALIZATIONS   ###########################
################################################################################

#### 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        ######################
################################################################################
#### Solves the problem without decomposing.

#### Define Problem
p = xp.problem()

#### Define Variables
x = [xp.var(vartype=xp.binary) for i in range(n1)]
y = [xp.var() for i 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)]

#### 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.solve()

if p.getProbStatus() != xp.mip_optimal:
raise RuntimeError('Problem could not be solved to MIP optimality')

yopt = p.getSolution(y)
print("The optimal objective is:", p.getObjVal())
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
####   addcbpreintsol function and has a specified input - output format based
####   on the Xpress documentation. It takes as input the master 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=[xp.var() for i in range(n2)] # positive real variable - second stage
z=[xp.var(lb=-xp.infinity) for i in range(n1)] # dummy copy variable
# must have the exact shape as xhat
epsilon=xp.var(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)]

# 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 alltogether 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.setControl({"presolve":0})

# Silence output
r.setControl ('outputlog', 0)
# Solve optimization
r.solve()

# Find the indices of constraint dummy1, which will be used to access the
# dual multipliers corresponding to the entries of xhat
xind1=[r.getIndex(dummy1[ii]) for ii in range(n1)]

# Take cases depending on subproblem status
if r.getProbStatus()==xp.lp_optimal:
# Optimal subproblem
print("Optimal Subproblem")
# Retrieve optimal dual multiplier and objective and return
dualmult=r.getDual()
lamb=[dualmult[ii] for ii in xind1]
beta=r.getObjVal()
return(lamb,beta,'Optimal')

elif r.getProbStatus()==xp.lp_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 (dray)
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[r.getIndex(dummy2)]
return(lamb,beta,'Infeasible')
else:
print("ERROR: Subproblem not optimal or infeasible. Terminating.")
sys.exit()

################################################################################
#### Integer Callback Function
def integer_callback(p, data, isheuristic, cutoff):
# Input
# NOTE: the input is populated by the solver when an integer solution is
# found.
# - p: an xp.problem() (the master problem from which the callbacks were
# trigger)
# - data: structure that is passed to the callback from the
#   problem, and is specified in the addcbpreintsol function. No
#   data is needed here, so the addcbpreintnode() call uses None.
# - isheuristic: 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 isheuristic ==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 heuristic solution in vectorized form
s=[]
p.getlpsol(s,None,None,None) # This will load the solution to array s

# Obtain indices of x and theta in the original problem
xind=[p.getIndex(x[ii]) for ii in range(n1)]
thetaind=[p.getIndex(theta)]

# Construct xhat and thetahat based on the vector s where the full
# solution was stored and the variable indices
xhat=[s[ii] for ii in xind]
thetahat=s[thetaind[0]]
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, simply 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 master problem, from which the
# callback is triggered.
# - data: structure that is passed to the callback from the
#   problem, and is specified in the addcboptnode function. No
#   data is needed here, so the addcboptnode() call uses None.
# Output
# return 0

print("We entered a node callback")

# Obtain node relaxation solution
s=[]
p.getlpsol(s,None,None,None)
xind=[p.getIndex(x[ii]) for ii in range(n1)]
thetaind=[p.getIndex(theta)]
xhat=[s[ii] for ii in xind]
thetahat=s[thetaind[0]]
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("coefficients:",coefficients)
print("<=rhs:",rhs)
print("column indices:",xind)
# Add the cut (this will reject the point)
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)
print("Rejecting pair x=",xhat,",theta=",thetahat,"with a cut." )
return 0
else:
print("ERROR: Subproblem not optimal or infeasible. Terminating.")
sys.exit()

################################################################################
#### Master Problem

# Define master problem
p=xp.problem()

# Define first stage variables
x=[xp.var(vartype=xp.binary) for i in range(n1)]
theta=xp.var() # (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.

# Add node_callback. i.e., when a node relaxation is solved, the function
# node_callback will be called by the solver.

# 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 master problem
p.solve()

if p.getProbStatus() == xp.lp_optimal:

# Print results
print("The optimal objective is:",p.getObjVal())
print("The first stage variables are:",p.getSolution(x))
else:
print("Could not solve the problem")