How to use a variable as a divisor in PuLP

Question

I was trying to resolve a LP problem with a constraint that is calculated by dividing variable A by variable B.

The simple version of the problem is as below:

Answer 1

I felt like zero shouldn't be allowed in my solution — and I included a variable that was the sum of x and y (think you're referring to it as A).

from pulp import LpProblem, LpStatus, LpVariable
from pulp import LpMinimize, LpMaximize, lpSum, value

# I feel like zero shouldn't be allowed for lower bound...
x = LpVariable("x", lowBound=1, cat="Integer")
y = LpVariable("y", lowBound=1, cat="Integer")
z = LpVariable("z", lowBound=1, cat="Integer")

model = LpProblem("Divison problem", LpMinimize)

model += x
model += z == x + y
model += z == 100

# Rather than division, we can use multiplication
model += x >= z * 0.5
model += y <= z * 0.4

model.solve()

print(LpStatus[model.status])

print(f"""
x = {int(x.varValue):>3}  #  60
y = {int(y.varValue):>3}  #  40
z = {int(z.varValue):>3}  # 100
""")

Answer 2

Linear Programming doesn't understand divisions, hence the error :) You have to reformulate it so that the division is formulated linearly. In this case:

prob += x / (x + y) > 0.5  
prob += y / (x + y) < 0.4

is equivalent to:

prob += x > 0.5 * (x + y)
prob += y < 0.4 * (x + y)

Which are linear constraints. Good luck!