Cuopt Lp Milp Api Python

Integer vs continuous from wording

Choose variable type from what the problem describes.

Rule of thumb: If the quantity is "how many things" (people, vehicles, items, sites), use INTEGER. If it's "how much" (mass, volume, money, time) or a rate, use CONTINUOUS unless the problem explicitly requires whole numbers.

Quick Reference: Python API

LP Example

from cuopt.linear_programming.problem import Problem, CONTINUOUS, MAXIMIZE
from cuopt.linear_programming.solver_settings import SolverSettings

# Create problem
problem = Problem("MyLP")

# Decision variables
x = problem.addVariable(lb=0, vtype=CONTINUOUS, name="x")
y = problem.addVariable(lb=0, vtype=CONTINUOUS, name="y")

# Constraints
problem.addConstraint(2*x + 3*y <= 120, name="resource_a")
problem.addConstraint(4*x + 2*y <= 100, name="resource_b")

# Objective
problem.setObjective(40*x + 30*y, sense=MAXIMIZE)

# Solve
settings = SolverSettings()
settings.set_parameter("time_limit", 60)
problem.solve(settings)

# Check status (CRITICAL: use PascalCase!)
if problem.Status.name in ["Optimal", "PrimalFeasible"]:
    print(f"Objective: {problem.ObjValue}")
    print(f"x = {x.getValue()}")
    print(f"y = {y.getValue()}")

MILP Example (with integer variables)

from cuopt.linear_programming.problem import Problem, CONTINUOUS, INTEGER, MINIMIZE

problem = Problem("FacilityLocation")

# Binary variable (integer with bounds 0-1)
open_facility = problem.addVariable(lb=0, ub=1, vtype=INTEGER, name="open")

# Continuous variable
production = problem.addVariable(lb=0, vtype=CONTINUOUS, name="production")

# Linking constraint: can only produce if facility is open
problem.addConstraint(production <= 1000 * open_facility, name="link")

# Objective: fixed cost + variable cost
problem.setObjective(500*open_facility + 2*production, sense=MINIMIZE)

# MILP-specific settings
settings = SolverSettings()
settings.set_parameter("time_limit", 120)
settings.set_parameter("mip_relative_gap", 0.01)  # 1% optimality gap

problem.solve(settings)

# Check status
if problem.Status.name in ["Optimal", "FeasibleFound"]:
    print(f"Open facility: {open_facility.getValue() > 0.5}")
    print(f"Production: {production.getValue()}")

CRITICAL: Status Checking

Status values use PascalCase, NOT ALL_CAPS:

# ✅ CORRECT
if problem.Status.name in ["Optimal", "FeasibleFound"]:
    print(problem.ObjValue)

# ❌ WRONG - will silently fail!
if problem.Status.name == "OPTIMAL":  # Never matches!
    print(problem.ObjValue)

LP Status Values: Optimal, NoTermination, NumericalError, PrimalInfeasible, DualInfeasible, IterationLimit, TimeLimit, PrimalFeasible

MILP Status Values: Optimal, FeasibleFound, Infeasible, Unbounded, TimeLimit, NoTermination

Common Modeling Patterns

Binary Selection

# Select exactly k items from n
items = [problem.addVariable(lb=0, ub=1, vtype=INTEGER) for _ in range(n)]
problem.addConstraint(sum(items) == k)

Big-M Linking

# If y=1, then x <= 100; if y=0, x can be anything up to M
M = 10000
problem.addConstraint(x <= 100 + M*(1 - y))

If-then "must also produce"

When the problem says if we do X then we must also do Y, enforce both (i) the binary link and (ii) that Y is actually produced:

# y_X <= y_Y (if we do X, we must "do" Y)
problem.addConstraint(y_X <= y_Y)
# Production of Y when Y is chosen: produce at least 1 (or a minimum) when y_Y=1
problem.addConstraint(production_Y >= 1 * y_Y)  # or min_amount * y_Y

Otherwise the solver can set y_Y=1 but production_Y=0, satisfying the binary link but not the intent.

Building large expressions

Chained + over many terms can hit recursion limits in the API. Prefer building objectives and constraints with LinearExpression:

from cuopt.linear_programming.problem import LinearExpression

# Build as list of (vars, coeffs) instead of v1*c1 + v2*c2 + ...
vars_list = [x, y, z]
coeffs_list = [1.0, 2.0, 3.0]
expr = LinearExpression(vars_list, coeffs_list, constant=0.0)
problem.addConstraint(expr <= 100)

See reference models in this skill's assets/ for examples.

Piecewise Linear (SOS2)

# Approximate nonlinear function with breakpoints
# Use lambda variables that sum to 1, at most 2 adjacent non-zero

Solver Settings

settings = SolverSettings()

# Time limit
settings.set_parameter("time_limit", 60)

# MILP gap tolerance (stop when within X% of optimal)
settings.set_parameter("mip_relative_gap", 0.01)

# Logging
settings.set_parameter("log_to_console", 1)

Common Issues

Getting Dual Values (LP only)

if problem.Status.name == "Optimal":
    constraint = problem.getConstraint("resource_a")
    shadow_price = constraint.DualValue
    print(f"Shadow price: {shadow_price}")

Reference Models

All reference models live in this skill's assets/ directory. Use them as reference when building new applications; do not edit them in place.

Minimal / canonical examples (LP & MILP)

Other reference

Quick command to list models: ls assets/ (from this skill's directory).

When to Escalate

Use troubleshooting and diagnostic guidance if:

• Infeasible and you can't determine why
• Numerical issues