# Integer Linear Programming

## Introduction

In Integer Linear Programming (ILP), we seek to find a vector of integer numbers that maximizes (or minimizes) a linear cost function under a set of linear equality or inequality constraints [1]. In other words, it is an optimization problem where the cost function to be optimized and all the constraints are linear and the decision variables are integer.

## Mathematical Formulation

The ILP problem can be formulated as follows:

Given an $$n$$-dimensional vector $$\vec{c} = (c_1, c_2, \ldots, c_n)$$, an $$m \times n$$ matrix $$A = (a_{ij})$$ with $$i=1,\ldots,m$$ and $$j=1,\ldots,n$$, and an $$m$$-dimensional vector $$\vec{b} = (b_1, b_2, \ldots, b_m)$$, find an $$n$$-dimensional vector $$\vec{x} = (x_1, x_2, \ldots, x_n)$$ with integer entries that maximizes (or minimizes) the cost function:

\begin{align*} \vec{c} \cdot \vec{x} = c_1x_1 + c_2x_2 + \ldots + c_nx_n \end{align*}

subject to the constraints:

\begin{align*} A \vec{x} & \leq \vec{b} \\ x_j & \geq 0, \quad j = 1, 2, \ldots, n \\ x_j & \in \mathbb{Z}, \quad j = 1, 2, \ldots, n \end{align*}

# Solving with the Classiq platform

We go through the steps of solving the problem with the Classiq platform, using QAOA algorithm [2]. The solution is based on defining a pyomo model for the optimization problem we would like to solve.

from typing import cast

import networkx as nx
import numpy as np
import pyomo.core as pyo
from IPython.display import Markdown, display
from matplotlib import pyplot as plt


## Building the Pyomo model from a graph input

We proceed by defining the pyomo model that will be used on the Classiq platform, using the mathematical formulation defined above:

import numpy as np
import pyomo.core as pyo

def ilp(a: np.ndarray, b: np.ndarray, c: np.ndarray, bound: int) -> pyo.ConcreteModel:
# model constraint: a*x <= b
model = pyo.ConcreteModel()
assert b.ndim == c.ndim == 1

num_vars = len(c)
num_constraints = len(b)

assert a.shape == (num_constraints, num_vars)

model.x = pyo.Var(
# here we bound x to be from 0 to to a given bound
range(num_vars),
domain=pyo.NonNegativeIntegers,
bounds=(0, bound),
)

@model.Constraint(range(num_constraints))
def monotone_rule(model, idx):
return a[idx, :] @ list(model.x.values()) <= float(b[idx])

# model objective: max(c * x)
model.cost = pyo.Objective(expr=c @ list(model.x.values()), sense=pyo.maximize)

return model

A = np.array([[1, 1, 1], [2, 2, 2], [3, 3, 3]])
b = np.array([1, 2, 3])
c = np.array([1, 2, 3])

# Instantiate the model
ilp_model = ilp(A, b, c, 4)

ilp_model.pprint()

2 Set Declarations
monotone_rule_index : Size=1, Index=None, Ordered=Insertion
Key  : Dimen : Domain : Size : Members
None :     1 :    Any :    3 : {0, 1, 2}
x_index : Size=1, Index=None, Ordered=Insertion
Key  : Dimen : Domain : Size : Members
None :     1 :    Any :    3 : {0, 1, 2}

1 Var Declarations
x : Size=3, Index=x_index
Key : Lower : Value : Upper : Fixed : Stale : Domain
0 :     0 :  None :     4 : False :  True : NonNegativeIntegers
1 :     0 :  None :     4 : False :  True : NonNegativeIntegers
2 :     0 :  None :     4 : False :  True : NonNegativeIntegers

1 Objective Declarations
cost : Size=1, Index=None, Active=True
Key  : Active : Sense    : Expression
None :   True : maximize : x[0] + 2*x[1] + 3*x[2]

1 Constraint Declarations
monotone_rule : Size=3, Index=monotone_rule_index, Active=True
Key : Lower : Body                     : Upper : Active
0 :  -Inf :       x[0] + x[1] + x[2] :   1.0 :   True
1 :  -Inf : 2*x[0] + 2*x[1] + 2*x[2] :   2.0 :   True
2 :  -Inf : 3*x[0] + 3*x[1] + 3*x[2] :   3.0 :   True

5 Declarations: x_index x monotone_rule_index monotone_rule cost


## Setting Up the Classiq Problem Instance

In order to solve the Pyomo model defined above, we use the Classiq combinatorial optimization engine. For the quantum part of the QAOA algorithm (QAOAConfig) - define the number of repetitions (num_layers):

from classiq import construct_combinatorial_optimization_model
from classiq.applications.combinatorial_optimization import OptimizerConfig, QAOAConfig

qaoa_config = QAOAConfig(num_layers=3)


For the classical optimization part of the QAOA algorithm we define the maximum number of classical iterations (max_iteration) and the $$\alpha$$-parameter (alpha_cvar) for running CVaR-QAOA, an improved variation of the QAOA algorithm [3]:

optimizer_config = OptimizerConfig(max_iteration=90, alpha_cvar=0.7)


Lastly, we load the model, based on the problem and algorithm parameters, which we can use to solve the problem:

qmod = construct_combinatorial_optimization_model(
pyo_model=ilp_model,
qaoa_config=qaoa_config,
optimizer_config=optimizer_config,
)


We also set the quantum backend we want to execute on:

from classiq import set_execution_preferences
from classiq.execution import ClassiqBackendPreferences, ExecutionPreferences

backend_preferences = ExecutionPreferences(
backend_preferences=ClassiqBackendPreferences(backend_name="simulator")
)

qmod = set_execution_preferences(qmod, backend_preferences)

from classiq import write_qmod

write_qmod(qmod, "integer_linear_programming")


## Synthesizing the QAOA Circuit and Solving the Problem

We can now synthesize and view the QAOA circuit (ansatz) used to solve the optimization problem:

from classiq import show, synthesize

qprog = synthesize(qmod)
show(qprog)

Opening: https://platform.classiq.io/circuit/183bcd6c-83c8-4610-828f-d7cd40f9fb85?version=0.41.0.dev39%2B79c8fd0855


We now solve the problem by calling the execute function on the quantum program we have generated:

from classiq import execute

res = execute(qprog).result()


We can check the convergence of the run:

from classiq.execution import VQESolverResult

vqe_result = res[0].value
vqe_result.convergence_graph


# Optimization Results

We can also examine the statistics of the algorithm:

import pandas as pd

from classiq.applications.combinatorial_optimization import (
get_optimization_solution_from_pyo,
)

solution = get_optimization_solution_from_pyo(
ilp_model, vqe_result=vqe_result, penalty_energy=qaoa_config.penalty_energy
)

optimization_result = pd.DataFrame.from_records(solution)

probability cost solution count
6 0.019 3.0 [0, 0, 1] 19
15 0.015 3.0 [0, 0, 1] 15
24 0.011 2.0 [0, 1, 0] 11
106 0.002 2.0 [0, 0, 2] 2
3 0.025 2.0 [0, 1, 0] 25

And the histogram:

optimization_result.hist("cost", weights=optimization_result["probability"])

array([[<Axes: title={'center': 'cost'}>]], dtype=object)


Let us plot the solution:

best_solution = optimization_result.solution[optimization_result.cost.idxmax()]
best_solution

[0, 0, 1]


## Comparison to a classical solver

Lastly, we can compare to the classical solution of the problem:

from pyomo.opt import SolverFactory

solver = SolverFactory("couenne")
solver.solve(ilp_model)
classical_solution = [int(pyo.value(ilp_model.x[i])) for i in range(len(ilp_model.x))]
print("Classical solution:", classical_solution)

Classical solution: [0, 0, 1]