introductory of linear programming in using python or Pulp

solve.py

"""
This module is a demonstartion of using scipy to solve LP problem
following this tutorial: https://realpython.com/linear-programming-python

The problem os as following

maximize          z  = x + 2 y, same as minimizing -z = -x - 2y
subject to  2x +  y <= 20
           -4x + 5y <= 10
            -x + 2y >= -2, same as x - 2y <= 2 for scipy
            -x + 5y  = 15
                  x >= 0
                  y >= 0

the variable x and y are called decision variables

the output will be
     con: array([0.])          -> equality constraints residuals
     fun: -16.818181818181817  -> optimum cost
 message: 'Optimization terminated successfully.'
     nit: 3                    -> the number of iterations
   slack: array([ 0.        , 18.18181818,  3.36363636])
                               -> the differences between values of
                                  the left side and right side of the
                                  constrains
  status: 0                    -> 0 = found optimum solution
 success: True
       x: array([7.72727273, 4.54545455])  -> optimum values of the
                                              decision variables x & y
"""
from scipy.optimize import linprog

obj = [-1, -2]
lhs_ineq = [
    [2, 1],
    [-4, 5],
    [1, -2]
]
rhs_ineq = [ 20, 10, 2 ]

lhs_eq = [[-1, 5]]
rhs_eq = [15]

bounds = [
    (0, float("inf")),  # for x
    (0, float("inf")),  # for y
]

opt = linprog(c=obj,
              A_ub=lhs_ineq, b_ub=rhs_ineq,
              A_eq=lhs_eq, b_eq=rhs_eq,
              bounds=bounds,
              method="revised simplex")
print(opt)

solve_pulp.py

"""
This module is a demonstartion of using PuLP to solve LP problem
following this tutorial: https://realpython.com/linear-programming-python

The problem os as following

maximize          z  = x + 2 y
subject to  2x +  y <= 20
           -4x + 5y <= 10
            -x + 2y >= -2
            -x + 5y  = 15
                  x >= 0
                  y >= 0
where x are integers

the variable x and y are called decision variables

the output will be
     con: array([0.])          -> equality constraints residuals
     fun: -16.818181818181817  -> optimum cost
 message: 'Optimization terminated successfully.'
     nit: 3                    -> the number of iterations
   slack: array([ 0.        , 18.18181818,  3.36363636])
                               -> the differences between values of
                                  the left side and right side of the
                                  constrains
  status: 0                    -> 0 = found optimum solution
 success: True
       x: array([7.72727273, 4.54545455])  -> optimum values of the
                                              decision variables x & y
"""
from pulp import LpMaximize, LpProblem, LpStatus, lpSum, LpVariable
from pulp import GLPK

model = LpProblem(name="example", sense=LpMaximize)
x = LpVariable(name="x", lowBound=0, cat="Integer")
y = LpVariable(name="y", lowBound=0)

constraint_1 = 2 * x + 1 * y <= 20
constraint_2 = 4 * x - 5 * y >= -10
constraint_3 =    -x + 2 * y >= -2
constraint_4 =    -x + 5 * y == 15
cost_func    = x + 2 * y

model += constraint_1
model += constraint_2
model += constraint_3
model += constraint_4
model += cost_func

model.solve(solver=GLPK(msg=False))

for var in model.variables():
    print(f"{var.name}: {var.value()}")