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## Introduction¶

In this tutorial, we briefly introduce how to build optimization problems using Qiskit’s optimization module. Qiskit introduces the QuadraticProgram class to make a model of an optimization problem. More precisely, it deals with quadratically constrained quadratic programs given as follows:

\begin{split}\begin{align} \text{minimize}\quad& x^\top Q_0 x + c^\top x\\ \text{subject to}\quad& A x \leq b\\ & x^\top Q_i x + a_i^\top x \leq r_i, \quad 1,\dots,i,\dots,q\\ & l_i \leq x_i \leq u_i, \quad 1,\dots,i,\dots,n, \end{align}\end{split}

where the $$Q_i$$ are $$n \times n$$ matrices, $$A$$ is a $$m \times n$$ matrix , $$x$$, and $$c$$ are $$n$$-dimensional vectors, $$b$$ is an $$m$$-dimensional vector, and where $$x$$ can be defined as binary, integer, or continuous variables. In addition to “$$\leq$$” constraints QuadraticProgram also supports “$$\geq$$” and “$$=$$”.

## Loading a QuadraticProgram from an LP file¶

As setup, you need to import the following module.

[1]:

from qiskit_optimization import QuadraticProgram
from qiskit_optimization.translators import from_docplex_mp


You start with an empty model. How to add variables and constraints to a model is explained in the section Directly constructing a QuadraticProgram.

Qiskit’s optimization module supports the conversion from Docplex model. You can easily make a model of an optimization problem with Docplex. You can find the documentation of Docplex at https://ibmdecisionoptimization.github.io/docplex-doc/mp/index.html

You can load a Docplex model to QuadraticProgram by using from_docplex_mp function.

## Loading a QuadraticProgram from a docplex model¶

[2]:

# Make a Docplex model
from docplex.mp.model import Model

mdl = Model("docplex model")
x = mdl.binary_var("x")
y = mdl.integer_var(lb=-1, ub=5, name="y")
mdl.minimize(x + 2 * y)
mdl.add_constraint((x + y) * (x - y) <= 1)
print(mdl.export_as_lp_string())

\ This file has been generated by DOcplex
\ ENCODING=ISO-8859-1
\Problem name: docplex model

Minimize
obj: x + 2 y
Subject To
c1: x - y = 3
qc1: [ x^2 - y^2 ] <= 1

Bounds
0 <= x <= 1
-1 <= y <= 5

Binaries
x

Generals
y
End



QuadraticProgram has a method prettyprint to generate a comprehensive string representation.

[3]:

# load from a Docplex model
mod = from_docplex_mp(mdl)
print(type(mod))
print()
print(mod.prettyprint())

<class 'qiskit_optimization.problems.quadratic_program.QuadraticProgram'>

Problem name: docplex model

Minimize
x + 2*y

Subject to
Linear constraints (1)
x - y == 3  'c0'

x^2 - y^2 <= 1  'q0'

Integer variables (1)
-1 <= y <= 5

Binary variables (1)
x



## Directly constructing a QuadraticProgram¶

We then explain how to make model of an optimization problem directly using QuadraticProgram. Let’s start from an empty model.

[4]:

# make an empty problem
print(mod.prettyprint())

Problem name: my problem

Minimize
0

Subject to
No constraints

No variables



The QuadraticProgram supports three types of variables:

• Binary variable

• Integer variable

• Continuous variable

When you add variables, you can specify names, types, lower bounds and upper bounds.

[5]:

# Add variables
mod.binary_var(name="x")
mod.integer_var(name="y", lowerbound=-1, upperbound=5)
mod.continuous_var(name="z", lowerbound=-1, upperbound=5)
print(mod.prettyprint())

Problem name: my problem

Minimize
0

Subject to
No constraints

Integer variables (1)
-1 <= y <= 5

Continuous variables (1)
-1 <= z <= 5

Binary variables (1)
x



You can set the objective function by invoking QuadraticProgram.minimize or QuadraticProgram.maximize. You can add a constant term as well as linear and quadratic objective function by specifying linear and quadratic terms with either list, matrix or dictionary.

Note that in the LP format the quadratic part has to be scaled by a factor $$1/2$$. Thus, when printing as LP format, the quadratic part is first multiplied by 2 and then divided by 2 again.

For quadratic programs, there are 3 pieces that have to be specified: a constant (offset), a linear term ($$c^{T}x$$), and a quadratic term ($$x^{T}Qx$$).

The cell below shows how to declare an objective function using a dictionary. For the linear term, keys in the dictionary correspond to variable names, and the corresponding values are the coefficients. For the quadratic term, keys in the dictionary correspond to the two variables being multiplied, and the values are again the coefficients.

[6]:

# Add objective function using dictionaries
mod.minimize(constant=3, linear={"x": 1}, quadratic={("x", "y"): 2, ("z", "z"): -1})
print(mod.prettyprint())

Problem name: my problem

Minimize
2*x*y - z^2 + x + 3

Subject to
No constraints

Integer variables (1)
-1 <= y <= 5

Continuous variables (1)
-1 <= z <= 5

Binary variables (1)
x



Another way to specify the quadratic program is using arrays. For the linear term, the array corresponds to the vector $$c$$ in the mathematical formulation. For the quadratic term, the array corresponds to the matrix $$Q$$. Note that the ordering of the variables ($$x$$ in the mathematical formulation) is the order in which the variables were originally declared in the QuadraticProgram object.

[7]:

# Add objective function using lists/arrays
mod.minimize(constant=3, linear=[1, 0, 0], quadratic=[[0, 1, 0], [1, 0, 0], [0, 0, -1]])
print(mod.prettyprint())

Problem name: my problem

Minimize
2*x*y - z^2 + x + 3

Subject to
No constraints

Integer variables (1)
-1 <= y <= 5

Continuous variables (1)
-1 <= z <= 5

Binary variables (1)
x



You can access the constant, the linear term, and the quadratic term by looking at Quadratic.objective.{constant, linear, quadratic}, respectively. As for linear and quadratic terms, you can get a dense matrix (to_array), a sparse matrix (coefficients), and a dictionary (to_dict). For dictionaries, you can specify whether to use variable indices or names as keys. Note that the quadratic terms are stored in a compressed way, e.g., {('x', 'y'): 1, ('y', 'x'): 2} is stored as {('x', 'y'): 3}. You can get the quadratic term as a symmetric matrix by calling to_array(symmetric=True) or to_dict(symmetric=True). If you call to_dict(name=True), you can get a dictionary whose keys are pairs of variable names.

[8]:

print("constant:\t\t\t", mod.objective.constant)
print("linear dict:\t\t\t", mod.objective.linear.to_dict())
print("linear array:\t\t\t", mod.objective.linear.to_array())
print("linear array as sparse matrix:\n", mod.objective.linear.coefficients, "\n")
print(
)

constant:                        3
linear dict:                     {0: 1}
linear array:                    [1 0 0]
linear array as sparse matrix:
(0, 0)       1

quadratic dict w/ index:         {(0, 1): 2, (2, 2): -1}
quadratic dict w/ name:          {('x', 'y'): 2, ('z', 'z'): -1}
symmetric quadratic dict w/ name:        {('y', 'x'): 1, ('x', 'y'): 1, ('z', 'z'): -1}
[[ 0  2  0]
[ 0  0  0]
[ 0  0 -1]]

[[ 0  1  0]
[ 1  0  0]
[ 0  0 -1]]

(0, 1)       2
(2, 2)        -1


You can add linear constraints by setting name, linear expression, sense and right-hand-side value (rhs). You can use senses ‘EQ’, ‘LE’, and ‘GE’ as Docplex supports.

[9]:

# Add linear constraints
mod.linear_constraint(linear={"x": 1, "y": 2}, sense="==", rhs=3, name="lin_eq")
mod.linear_constraint(linear={"x": 1, "y": 2}, sense="<=", rhs=3, name="lin_leq")
mod.linear_constraint(linear={"x": 1, "y": 2}, sense=">=", rhs=3, name="lin_geq")
print(mod.prettyprint())

Problem name: my problem

Minimize
2*x*y - z^2 + x + 3

Subject to
Linear constraints (3)
x + 2*y == 3  'lin_eq'
x + 2*y <= 3  'lin_leq'
x + 2*y >= 3  'lin_geq'

Integer variables (1)
-1 <= y <= 5

Continuous variables (1)
-1 <= z <= 5

Binary variables (1)
x



You can add quadratic constraints as well as objective function and linear constraints.

[10]:

# Add quadratic constraints
linear={"x": 1, "y": 1},
quadratic={("x", "x"): 1, ("y", "z"): -1},
sense="==",
rhs=1,
)
linear={"x": 1, "y": 1},
quadratic={("x", "x"): 1, ("y", "z"): -1},
sense="<=",
rhs=1,
)
linear={"x": 1, "y": 1},
quadratic={("x", "x"): 1, ("y", "z"): -1},
sense=">=",
rhs=1,
)
print(mod.prettyprint())

Problem name: my problem

Minimize
2*x*y - z^2 + x + 3

Subject to
Linear constraints (3)
x + 2*y == 3  'lin_eq'
x + 2*y <= 3  'lin_leq'
x + 2*y >= 3  'lin_geq'

x^2 - y*z + x + y == 1  'quad_eq'
x^2 - y*z + x + y <= 1  'quad_leq'
x^2 - y*z + x + y >= 1  'quad_geq'

Integer variables (1)
-1 <= y <= 5

Continuous variables (1)
-1 <= z <= 5

Binary variables (1)
x



You can access linear and quadratic terms of linear and quadratic constraints as in the same way as the objective function.

[11]:

lin_geq = mod.get_linear_constraint("lin_geq")
print("lin_geq:", lin_geq.linear.to_dict(use_name=True), lin_geq.sense, lin_geq.rhs)
print(
lin_geq.rhs,
)

lin_geq: {'x': 1.0, 'y': 2.0} ConstraintSense.GE 3
quad_geq: {'x': 1.0, 'y': 1.0} {('x', 'x'): 1.0, ('y', 'z'): -1.0} ConstraintSense.GE 3


You can also remove linear/quadratic constraints by remove_linear_constraint and remove_quadratic_constraint.

[12]:

# Remove constraints
mod.remove_linear_constraint("lin_eq")
print(mod.prettyprint())

Problem name: my problem

Minimize
2*x*y - z^2 + x + 3

Subject to
Linear constraints (2)
x + 2*y <= 3  'lin_leq'
x + 2*y >= 3  'lin_geq'

x^2 - y*z + x + y == 1  'quad_eq'
x^2 - y*z + x + y >= 1  'quad_geq'

Integer variables (1)
-1 <= y <= 5

Continuous variables (1)
-1 <= z <= 5

Binary variables (1)
x



You can substitute some of variables with constants or other variables. More precisely, QuadraticProgram has a method substitute_variables(constants=..., variables=...) to deal with the following two cases.

• $$x \leftarrow c$$: when constants have a dictionary {x: c}.

• $$x \leftarrow c y$$: when variables have a dictionary {x: (y, c)}.

## Substituting Variables¶

[13]:

sub = mod.substitute_variables(constants={"x": 0}, variables={"y": ("z", -1)})
print(sub.prettyprint())

Problem name: my problem

Minimize
-z^2 + 3

Subject to
Linear constraints (2)
-2*z <= 3  'lin_leq'
-2*z >= 3  'lin_geq'

z^2 - z == 1  'quad_eq'
z^2 - z >= 1  'quad_geq'

Continuous variables (1)
-1 <= z <= 1



If the resulting problem is infeasible due to lower bounds or upper bounds, the methods returns the status Status.INFEASIBLE. We try to replace variable x with -1, but -1 is out of range of x (0 <= x <= 1). So, it returns Status.INFEASIBLE.

[14]:

sub = mod.substitute_variables(constants={"x": -1})
print(sub.status)

Infeasible substitution for variable: x

QuadraticProgramStatus.INFEASIBLE


You cannot substitute variables multiple times. The method raises an error in such a case.

[15]:

from qiskit_optimization import QiskitOptimizationError

try:
sub = mod.substitute_variables(constants={"x": -1}, variables={"y": ("x", 1)})
except QiskitOptimizationError as e:
print("Error: {}".format(e))

Error: 'Cannot substitute by variable that gets substituted itself: y <- x 1'


Note: When you display your problem as LP format using export_as_lp_string, Binaries denotes binary variables and Generals denotes integer variables. If variables are not included in either Binaries or Generals, such variables are continuous ones with default lower bound = 0 and upper bound = infinity. Note that you cannot use ‘e’ or ‘E’ as the first character of names due to the specification of LP format.

[16]:

mod = QuadraticProgram()
mod.binary_var(name="e")
mod.binary_var(name="f")
mod.continuous_var(name="g")
mod.minimize(linear=[1, 2, 3])
print(mod.export_as_lp_string())

\ This file has been generated by DOcplex
\ ENCODING=ISO-8859-1
\Problem name: CPLEX

Minimize
obj: _e + 2 f + 3 g
Subject To

Bounds
0 <= _e <= 1
0 <= f <= 1

Binaries
_e f
End


[17]:

import qiskit.tools.jupyter

%qiskit_version_table


### Version Information

SoftwareVersion
qiskitNone
qiskit-terra0.25.1
qiskit_optimization0.5.0
System information
Python version3.8.17
Python compilerGCC 11.3.0
Python builddefault, Jun 7 2023 12:29:56
OSLinux
CPUs2
Memory (Gb)6.769481658935547
Fri Sep 01 13:49:07 2023 UTC

### This code is a part of Qiskit

[ ]: