Lean coding of simple linear optimization ortools models in Python

In a previous post on Google’s ortools module in Python I solved the linear optimization problem stated below:

The code written in my previous post can be reduced to fewer lines of code, resulting in lean code. In this post I will share a leaner version of my previous ortools-script.

Again, I start by importing pywraplp from ortools.linear_solver:

from ortools.linear_solver import pywraplp

Next, I create a solver instance (using GLOP solver) and store its reference to a reference handler:

solver = pywraplp.Solver.CreateSolver('linear_programming_examples', 'GLOP')

I now have to declare relevant optimization variables. In this example the optimization variables are x, y and z:

# declare variable x, with lower bound 0 and no upper bound
x = solver.NumVar(0, solver.infinity(), "x")
# declare variable y, with lower bound 0 and no upper bound
y = solver.NumVar(0, solver.infinity(), "y")
# declare variable z, with lower bound 0 and no upper bound
z = solver.NumVar(0, solver.infinity(), "z")

Now I can model the problem by specifying the objective functions and all relevant constraints. In the following lines of code this is realized in a very lean way:

# declare as max. problen and define objective function
solver.Maximize(x+2*y+3*z)
# add constraints
solver.Add(2*x+y+z <= 20)
solver.Add(x+y+z <= 15)
solver.Add(x-y-z >= 0)
<ortools.linear_solver.pywraplp.Constraint; proxy of <Swig Object of type 'operations_research::MPConstraint *' at 0x000001B9AEC956C0> >

The model is now complete and can be solved:

solver.Solve()

The optimal solution for x is outputed below:

print("x_opt: ", x.solution_value())
x_opt:  6.666666666666667

The optimal solution for y is outputed below:

print("y_opt: ", y.solution_value())
y_opt:  0.0

The optimal solution for z is outputed below:

print("z_opt: ", z.solution_value())
z_opt:  6.666666666666667

The optimal objective function value is outputed below:

print("optimal value: " + str(x.solution_value()+2*y.solution_value()+3*z.solution_value()))
optimal value: 26.666666666666668

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