How to put 'or' into contraints in Pulp in python

Question

I have a problem, where I have to find the optimal cost of 3 given motor.

Motor 1 has a range of 100 - 300 Motor 2 has a range of 400 - 1000 Motor 3 has a range of 50 - 250

They have a target value of 600

Motor 1 price is 5000 Motor 2 price is 5500 Motor 3 price is 5250

The equation looks like this:

Cost = Motor1 * 5000 + Motor2 * 5500 + Motor3 * 5250.

And a very important part, NOT every motor needs to run.

I have a python code, that can calculate it, but I can give it to it that not every motors needs to be inclued. Here is the code:

from pulp import LpProblem, LpVariable, LpMinimize

def find_lowest_cost():
    # Define the problem
    problem = LpProblem("Motor Optimization", LpMinimize)

    # Define the decision variables
    x = LpVariable("Motor1", lowBound=100, cat='Integer')  # Power of motor 1
    y = LpVariable("Motor2", lowBound=0, cat='Integer')  # Power of motor 2
    z = LpVariable("Motor3", lowBound=50, cat='Integer')  # Power of motor 3

    # Define the objective function (cost)
    problem += x * 5000 + y * 5500 + z * 5250

    # Define the constraints
    problem += x >= 100  # Motor 1 lower bound
    problem += x <= 300   # Motor 1 upper bound
    problem += y >= 350  # Motor 2 lower bound
    problem += y <= 1000  # Motor 2 upper bound
    problem += z >= 50  # Motor 3 lower bound
    problem += z <= 250  # Motor 3 upper bound
    problem += x + y + z == 500  # Total power constraint

    # Solve the problem
    problem.solve()

    # Retrieve the optimal solution
    lowest_cost = problem.objective.value()
    best_combination = (x.value(), y.value(), z.value())
    return lowest_cost, best_combination

cost, combination = find_lowest_cost()
print("Lowest cost:", cost)
print("Motor combination:", combination)

I tried to add 'or' to the "Define the Constraints' part, but it did not help

    problem += x >= 100 or x ==0 # Motor 1 lower bound
    problem += x <= 300  # Motor 1 upper bound
    problem += y >= 350 or y == 0 # Motor 2 lower bound
    problem += y <= 1000  # Motor 2 upper bound
    problem += z >= 50 or z == 0 # Motor 3 lower bound
    problem += z <= 250  # Motor 3 upper bound
    problem += x + y + z == 500  # Total power constraint

So my Questions is, how to implement that 'OR' into my code.

Thank you in advance

Is motor 2's lower bound 350 or 0? Those are inconsistent, and the constraints defining bounds should be moved to bounds on the variable declaration. — Reinderien, Jun 01 '23 at 12:54
See: https://stackoverflow.com/questions/76372321/how-can-i-both-lower-limit-an-optimization-variable-and-allow-it-to-be-zero/76380393#76380393 — Erwin Kalvelagen, Jun 01 '23 at 12:56
Are those motor speeds really integers? It seems more realistic for them to be continuous — Reinderien, Jun 01 '23 at 13:03

score 0 · Accepted Answer · answered Jun 01 '23 at 13:15

I make some assumptions:

Continuous power of motors, not integral
Observe the minima in your variable bounds, not the redundant and inconsistent constraints added later
Use 500 as a target, not 600

You need binary selection variables, like this:

from pulp import LpProblem, LpVariable, LpMinimize, LpContinuous, lpDot, LpBinary, lpSum

powers = (
    LpVariable('Motor1', cat=LpContinuous, upBound=300),
    LpVariable('Motor2', cat=LpContinuous, upBound=1000),
    LpVariable('Motor3', cat=LpContinuous, upBound=250),
)
used = LpVariable.matrix(name='MotorUsed', cat=LpBinary, indices=range(len(powers)))

problem = LpProblem(name='Motor_Optimization', sense=LpMinimize)
problem.objective = lpDot(powers, (5000, 5500, 5250))

problem.addConstraint(name='target', constraint=lpSum(powers) == 500)

for power, power_min, use in zip(
    powers,
    (100, 0, 50),
    used,
):
    problem.addConstraint(power >= power_min*used)
    problem.addConstraint(power <= 1000*used)

problem.solve()
combination = [p.value() for p in powers]
print('Lowest cost:', problem.objective.value())
print('Motor combination:', combination)

Result - Optimal solution found

Objective value:                2550000.00000000
Enumerated nodes:               0
Total iterations:               0
Time (CPU seconds):             0.01
Time (Wallclock seconds):       0.01

Option for printingOptions changed from normal to all
Total time (CPU seconds):       0.01   (Wallclock seconds):       0.01

Lowest cost: 2550000.0
Motor combination: [300.0, 0.0, 200.0]

How to put 'or' into contraints in Pulp in python

1 Answers1