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Two variables, weighted objective

Constraints + objective in one go.

Chapter: I — Foundations · lesson id two-vars-objective · practice lesson

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Real models have multiple variables and constraints. Here we have two — x and y, both non-negative — and a single structural constraint x + y <= 5.

The objective coefficients tell the solver which variables it should care about. With max 1*x + 3*y, each unit of y is worth three times each unit of x. Given the budget x + y ≤ 5, the solver puts everything on y.

Try flipping max to min — the optimum jumps to the origin.

Predict before running

With max 1*x + 3*y and budget x + y <= 5, where does the solver put the budget?

• All on x (x = 5, y = 0)
• All on y (x = 0, y = 5)
• Split evenly (x = y = 2.5)

Starter code

model weighted
var 0 <= x
var 0 <= y
constr x + y <= 5
max 1*x + 3*y
end

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

Hint

Predict the optimum before running. Whichever variable has the larger coefficient wins.

What counts as correct

• no-error — Model solves
• numeric-match — Solver leaves x at 0 (smaller coefficient)
• numeric-match — Solver pushes y to the budget (5)
• code-contains — Reads variable values from the solver (no hardcoding)

Mastery variant

Different budget, different coefficients. Predict which variable wins this time.

model weighted_v
var 0 <= a
var 0 <= b
constr 2*a + b <= 12
max 5*a + 2*b
end

import elytica
def main():
  elytica.init_model("weighted_v")
  elytica.run_model("weighted_v")
  print("a =", elytica.get_variable_value("weighted_v", "a"))
  print("b =", elytica.get_variable_value("weighted_v", "b"))
  return 0

Read this through MCP: read with path: "lessons/two-vars-objective", or resource hlpl://lessons/two-vars-objective.