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OPTIMISATION

quadratic_assignment

Approximates solution to the quadratic assignment problem and.

Example of usage

Python API reference included; executable data demonstration pending review.

Usage / limitations

+Use when: Find solutions efficiently.

−Limitations: Check input requirements, mathematical assumptions and version-specific behaviour before using.

Mathematical formula

See method documentation: implementation and assumptions vary.
Sample dataShow
{
  "variable": "design_case",
  "domain": "Engineering parameter estimation and optimisation",
  "feature_names": [
    "measurement_time_s",
    "response_mm"
  ],
  "records": [
    {
      "measurement_time_s": 0,
      "response_mm": 1
    },
    {
      "measurement_time_s": 1,
      "response_mm": 3
    },
    {
      "measurement_time_s": 2,
      "response_mm": 5
    },
    {
      "measurement_time_s": 3,
      "response_mm": 7
    },
    {
      "measurement_time_s": 4,
      "response_mm": 9
    }
  ],
  "target_value": 4.0,
  "lower_bound": 0.0,
  "upper_bound": 4.0,
  "method": "quadratic-assignment"
}

Code example (Python)

# DataLens.ai example — quadratic_assignment
# Domain: engineering parameter estimation and optimisation.
# The sample data are included by “Copy input + code + test”.
import numpy as np
from scipy import optimize
# Read the named engineering measurements.
times = np.array([row["measurement_time_s"] for row in design_case["records"]], dtype=float)
responses = np.array([row["response_mm"] for row in design_case["records"]], dtype=float)
# Use a known target, making the numerical assertion independently meaningful.
target = design_case["target_value"]

# Match three production stations to minimise flow times separation.
# Distances and flows are symmetric and labelled as station-to-station costs.
flows = np.array([[0,10,0],[10,0,1],[0,1,0]])
distances = np.array([[0,1,3],[1,0,2],[3,2,0]])
result = optimize.quadratic_assignment(flows,distances,method='faq',options={'rng':42})
permutation = result.col_ind
# Verify reported objective against a direct recomputation (not assumed global optimality).
computed = float(np.sum(flows*distances[np.ix_(permutation,permutation)]))
assert sorted(permutation.tolist())==[0,1,2] and np.isclose(computed,result.fun)
print("Expected output verified")