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Fixed cost + variable cost

Open a production line only if it pays.

Chapter: V — Modelling tricks · lesson id fixed-cost · practice lesson

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A common pattern: opening a facility (or production line, or contract) costs a fixed amount, and then each unit produced has its own variable cost or revenue. You only want to pay the fixed cost if the line is actually used.

Model it with a binary y (line open?) and a continuous x (units produced). Tie them together with x ≤ M · y — if y = 0, x is forced to 0; if y = 1, x is free up to M.

Here we earn 2 per unit, pay a setup cost of 30 to open, and cap output at 50. Should the line open?

Predict before running

Setup cost 30, revenue 2/unit, max 50 units. Does the line open?

• No. Keep y = 0, profit = 0
• Yes. Open, profit 70
• Open partially (y = 0.5)

Starter code

model fixed_cost
var 0 <= x          # units produced
bin y               # 1 if production line is open
constr x <= 50 * y  # only produce if the line is open (M = 50)
max 2*x - 30*y
end

import elytica
def main():
  elytica.init_model("fixed_cost")
  elytica.run_model("fixed_cost")
  print("x =", elytica.get_variable_value("fixed_cost", "x"))
  print("y =", elytica.get_variable_value("fixed_cost", "y"))
  print("obj =", elytica.get_best_primal_bound("fixed_cost"))
  return 0

Hint

Profit with line open: 2·50 − 30 = 70. With line closed: 0. The solver will pick open.

What counts as correct

• no-error — Model solves
• numeric-match — Line opens (y = 1)
• numeric-match — Run at full capacity (x = 50)
• numeric-match — Profit = 2·50 − 30 = 70
• code-contains — Reads the objective from the solver (no hardcoding)
• code-contains — Reads variable values from the solver (no hardcoding)

Mastery variant

Higher setup cost but better unit margin. Worth opening the line?

model fixed_cost_v
var 0 <= x          # units produced
bin y               # 1 if production line is open
constr x <= 30 * y  # only produce if the line is open (M = 30)
max 5*x - 40*y
end

import elytica
def main():
  elytica.init_model("fixed_cost_v")
  elytica.run_model("fixed_cost_v")
  print("x =", elytica.get_variable_value("fixed_cost_v", "x"))
  print("y =", elytica.get_variable_value("fixed_cost_v", "y"))
  print("obj =", elytica.get_best_primal_bound("fixed_cost_v"))
  return 0

Read this through MCP: read with path: "lessons/fixed-cost", or resource hlpl://lessons/fixed-cost.