Note

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Optimization using bonmin

In this example we are going to explore mixed-integer non linear problems optimization using the bonmin interface. %%

import openturns as ot

ot.Log.Show(ot.Log.NONE)

List available algorithms

for algo in ot.Bonmin.GetAlgorithmNames():
    print(algo)

B-BB
B-OA
B-QG
B-Hyb
B-Ecp
B-iFP

Details and references on bonmin algorithms are available here .

Setting up and solving a simple problem

The following example will demonstrate the use of bonmin “BB” algorithm to solve the following problem:

\min - x_0 - x_1 - x_2

such that:

\begin{array}{l} (x_1 - \frac{1}{2})^2 + (x_2 - \frac{1}{2})^2 \leq \frac{1}{4} \\ x_0 - x_1 \leq 0 \\ x_0 + x_2 + x_3 \leq 2\\ x_0 \in \{0,1\}^n\\ (x_1, x_2) \in \mathbb{R}^2\\ x_3 \in \mathbb{N} \end{array}

The theoretical minimum is reached for x = [1,1,0.5,0]. At this point, the objective function value is -2.5

N.B.: equality and inequality constraints are required to be stated as g(x) = 0 and h(x) \geq 0, respectively. Thus the inequalities above will have to be restated to match this requirement:

\begin{array}{l} -(x_1 - \frac{1}{2})^2 - (x_2 - \frac{1}{2})^2 + \frac{1}{4} \geq 0\\ -x_0 + x_1 \geq 0 \\ -x_0 - x_2 - x_3 + 2 \geq 0\\ \end{array}

Definition of objective function

objectiveFunction = ot.SymbolicFunction(["x0", "x1", "x2", "x3"], ["-x0 -x1 -x2"])

# Definition of variables bounds
bounds = ot.Interval([0.0] * 4, [1, 1e6, 1e6, 5])

# Definition of constraints
# Constraints are defined as g(x) = 0 and h(x) >= 0
#    No equality constraint -> nothing to do
#    Inequality constraints:
h = ot.SymbolicFunction(
    ["x0", "x1", "x2", "x3"],
    ["-(x1-0.5)^2 - (x2-0.5)^2 + 0.25", "x1 - x0", "-x0 - x2 - x3 + 2"],
)

# Definition of variables types
CONTINUOUS = ot.OptimizationProblemImplementation.CONTINUOUS
BINARY = ot.OptimizationProblemImplementation.BINARY
INTEGER = ot.OptimizationProblemImplementation.INTEGER
variablesType = [BINARY, CONTINUOUS, CONTINUOUS, INTEGER]

# Setting up Bonmin problem
problem = ot.OptimizationProblem(objectiveFunction)
problem.setBounds(bounds)
problem.setVariablesType(variablesType)
problem.setInequalityConstraint(h)

bonminAlgorithm = ot.Bonmin(problem, "B-BB")
bonminAlgorithm.setMaximumCallsNumber(10000)
bonminAlgorithm.setMaximumIterationNumber(1000)
bonminAlgorithm.setStartingPoint([0.0] * 4)

Running the solver

bonminAlgorithm.run()

# Retrieving the results
result = bonminAlgorithm.getResult()
print(" -- Optimal point = " + str(result.getOptimalPoint()))
print(" -- Optimal value = " + str(result.getOptimalValue()))
print(" -- Evaluation number = " + str(result.getInputSample().getSize()))

-- Optimal point = [1,1,0.500256,0]
-- Optimal value = [-2.50026]
-- Evaluation number = 282