Quadratic Assignment Problem¶
Quadratic assignment problem (QAP) is the following problem.
Let \(N\) be a positive integer. Consider \(N\) factories to be built on \(N\) candidate sites. Each factory can be built on any of the candidate sites. Every two factories have trucks traveling to and from them, and their transportation volumes are known in advance. How can we minimize the sum of the amount transported x the distance traveled?
An application could be to determine the seating chart for a meeting so that people close to each other have seats closely.
Formulation¶
Let \(N\) potential factory locations be denoted by land \(0\), land \(1\), … , and \(N\) factories are denoted as factories \(0\), factories \(1\), …, factories \(N1\). Also let \(D_{i, j}\) denote the distance between land \(i\) and land \(j\), and \(F_{k, l}\) denote the transport volume between factory \(k\) and factory \(l\).
Variables¶
With \(N \times N\) binary variables \(q\), let \(q_{i, k}\) represent whether factory \(k\) is to be built on land \(i\).
For example, factory \(3\) will be built on land \(0\) if \(q\) has the following value.
factory 0 
factory 1 
factory 2 
factory 3 
factory 4 


land 0 
0 
0 
0 
1 
0 
land 1 
0 
1 
0 
0 
0 
land 2 
0 
0 
0 
0 
1 
land 3 
1 
0 
0 
0 
0 
land 4 
0 
0 
1 
0 
0 
Constraints¶
Each row and column of the binary variable table must have exactly one variable that is 1, so we place a onehot constraint on each row and column. Conversely, if these are satisfied, then there is only one way to determine which factory to build on which land.
Objective function¶
The objective function is the sum of transport volume x distance between factories. This can be expressed in the equation using \(q\) as follows.
Formulation¶
The above formulation, with \(N\times N\) binary variables \(q\), can be written as follows.
Problem setting¶
Before formulating with the Amplify SDK, we will create a problem. For simplicity, let the number of factories \(N=10\).
import numpy as np
N = 10
Next, we create a distance matrix \(D\) representing the distances between lands. The lands are randomly generated on the Euclidean plane. The distance matrix distance
is created as a twodimensional numpy.ndarray
.
rng = np.random.default_rng()
x = rng.integers(0, 100, size=(N,))
y = rng.integers(0, 100, size=(N,))
distance = (
(x[:, np.newaxis]  x[np.newaxis, :]) ** 2
+ (y[:, np.newaxis]  y[np.newaxis, :]) ** 2
) ** 0.5
print(distance)
[[ 0. 34.713 75.06 40.522 99.81 12.166 38.053 46.573 50.99 80.957]
[ 34.713 0. 87.664 64.351 98.25 46.271 6.403 69.426 41.195 105.646]
[ 75.06 87.664 0. 38.626 41.231 78.447 84.581 33.541 53.907 38.053]
[ 40.522 64.351 38.626 0. 73.539 41.012 63.891 6.083 49.254 41.617]
[ 99.81 98.25 41.231 73.539 0. 107.042 93.048 70.178 57.079 78.102]
[ 12.166 46.271 78.447 41.012 107.042 0. 50. 47. 61.612 78.447]
[ 38.053 6.403 84.581 63.891 93.048 50. 0. 68.622 36. 104.661]
[ 46.573 69.426 33.541 6.083 70.178 47. 68.622 0. 51.196 36.235]
[ 50.99 41.195 53.907 49.254 57.079 61.612 36. 51.196 0. 82.765]
[ 80.957 105.646 38.053 41.617 78.102 78.447 104.661 36.235 82.765 0. ]]
Also, we create a matrix \(F\) representing the amount of transport between factories, a random symmetric matrix of dimension 2, named flow
.
flow = np.zeros((N, N), dtype=int)
for i in range(N):
for j in range(i + 1, N):
flow[i, j] = flow[j, i] = rng.integers(0, 100)
print(flow)
[[ 0 10 45 35 77 50 5 81 87 4]
[10 0 99 70 37 96 72 61 24 41]
[45 99 0 68 88 74 24 3 65 67]
[35 70 68 0 49 10 69 31 14 43]
[77 37 88 49 0 21 10 22 25 93]
[50 96 74 10 21 0 65 5 51 43]
[ 5 72 24 69 10 65 0 40 30 62]
[81 61 3 31 22 5 40 0 48 31]
[87 24 65 14 25 51 30 48 0 18]
[ 4 41 67 43 93 43 62 31 18 0]]
Formulation with the Amplify SDK¶
In the formulation, we can use the Matrix
class for efficient formulation, since a quadratic term consisting of any two binary variables can appear in the objective function.
Creating variables¶
To formulate using the Matrix
class, VariableGenerator
’s matrix()
method to issue variables.
from amplify import VariableGenerator
gen = VariableGenerator()
matrix = gen.matrix("Binary", N, N) # coefficient matrix
q = matrix.variable_array # variables
q
Creating the objective function¶
The matrix
created above is an instance of the class Matrix
, which has the following three properties.
quadratic
is numpy.ndarray
representing the coefficients of the second order terms, and its shape
is (N, N, N, N)
this time. quadratic[i, k, j, l]
corresponds to the coefficients of q[i, k] * q[j, l]
. That is, quadratic
must be set to a 4dimensional NumPy array such that quadratic[i, k, j, l] = distance[i, j] * flow[k, l]
linear
and constant
represent the coefficient and constant terms of the linear term, respectively, but since the objective function used in this problem contains only second order terms, we will not set them.
np.einsum("ij,kl>ikjl", distance, flow, out=matrix.quadratic)
Creating constraints¶
Impose a onehot constraint on each row and column of the variable array q
created in [](#Creating variables).
from amplify import one_hot
constraints = one_hot(q, axis=1) + one_hot(q, axis=0)
Creating a combinatorial optimization model¶
Let’s combine the objective function and constraints to create a model.
penalty_weight = np.max(distance) * np.max(flow) * (N  1)
model = matrix + penalty_weight * constraints
The penalty_weight
is applied to the constraints to give weight to the constraints. In Amplify AE, the solver used in this example, if you do not specify appropriate weights for the constraints, the solver will search in the direction of making the objective function smaller rather than trying to satisfy the constraints, and you will not be able to find a feasible solution. See Constraints and Penalty Functions for details.
Creating a solver client¶
Now, we will create a solver client to perform combinatorial optimization using Amplify AE. The solver client class corresponding to Amplify AE is FixstarsClient
class.
from amplify import FixstarsClient
client = FixstarsClient()
We also need to set the API token required to run Amplify AE.
Tip
After user registration, you can obtain a free API token that can be used for evaluation and validation purposes.
client.token = "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
We will set the solver’s timeout.
import datetime
client.parameters.timeout = datetime.timedelta(seconds=1)
Executing the solver¶
Finally, we will execute the solver using the created combinatorial optimization model and the solver client to find the solution to the quadratic programming problem.
from amplify import solve
result = solve(model, client)
The objective function value based on the best solution is shown below.
result.best.objective
206159.7553376566
The values of the variables in the optimal solution can be obtained in the form of a NumPy multidimensional array as follows.
q_values = q.evaluate(result.best.values)
print(q_values)
[[0. 1. 0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 1. 0. 0. 0.]
[1. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 1. 0. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 1. 0. 0.]
[0. 0. 0. 0. 0. 1. 0. 0. 0. 0.]
[0. 0. 0. 1. 0. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 1. 0. 0. 0. 0. 0.]
[0. 0. 0. 0. 0. 0. 0. 0. 0. 1.]
[0. 0. 0. 0. 0. 0. 0. 0. 1. 0.]]
Checking the results¶
We will visualize the results using matplotlib.
import matplotlib.pyplot as plt
import itertools
plt.scatter(x, y)
factory_indices = (q_values @ np.arange(N)).astype(int)
for i, j in itertools.combinations(range(N), 2):
plt.plot(
[x[i], x[j]],
[y[i], y[j]],
c="b",
alpha=flow[factory_indices[i], factory_indices[j]] / 100,
)