cuopt-lp-milp-api-python
cuOpt LP/MILP Skill
Model and solve linear and mixed-integer linear programs using NVIDIA cuOpt's GPU-accelerated solver.
Before You Start
Use a formulation summary (parameters, constraints, decisions, objective) if available; otherwise ask for decision variables, objective, and constraints. Then confirm variable types (see below) and interface (Python API recommended).
Choosing LP vs MILP
Prefer LP (all continuous variables) when the problem allows it. LP solves faster and has stronger optimality guarantees. Use MILP only when the problem logically requires whole numbers or yes/no decisions.
Problem types that need extra care: Multi-period planning and goal programming are easy to misinterpret. Double-check that rates and constraints apply to the right time period or priority level (AGENTS.md: verify understanding before code).
- Use LP when every quantity can meaningfully be fractional: flows, proportions, rates, dollars, hours, tonnes of material, etc.
- Use MILP when the problem mentions counts of discrete entities, yes/no choices, or either/or decisions (e.g. open a facility or not, assign a person to a shift, number of trucks).
Integer vs continuous from wording
Choose variable type from what the problem describes.
| Problem wording / concept | Variable type | Examples |
|---|---|---|
| Discrete entities (counts) | INTEGER | Workers, cars, trucks, machines, pilots, facilities, units to manufacture (when "units" means whole items), trainees, vehicles |
| Yes/no or on/off | INTEGER (binary, lb=0 ub=1) | Open a facility, run a machine, produce a product line, assign a person to a shift |
| Amounts that can be fractional | CONTINUOUS | Tonnes, litres, dollars, hours, kWh, proportion of capacity, flow volume, weight |
| Rates or fractions | CONTINUOUS | Utilization, percentage, share of budget |
| Unclear | Prefer INTEGER if the noun is a countable thing (a worker, a car); prefer CONTINUOUS if it's a measure (amount of steel, hours worked). If the problem says "whole" or "integer" or "number of", use INTEGER. |
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
| Problem | Likely Cause | Fix |
|---|---|---|
| Status never "OPTIMAL" | Using wrong case | Use "Optimal" not "OPTIMAL" |
| Integer var has fractional value | Defined as CONTINUOUS | Use vtype=INTEGER |
| Infeasible | Conflicting constraints | Check constraint logic |
| Unbounded | Missing bounds | Add variable bounds |
| Slow solve | Large problem | Set time limit, increase gap tolerance |
| Maximum recursion depth | Building big expr with chained + |
Use LinearExpression(vars_list, coeffs_list, constant) |
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)
| Model | Type | Description |
|---|---|---|
| lp_basic | LP | Minimal LP: variables, constraints, objective, solve |
| lp_duals | LP | Dual values and reduced costs |
| lp_warmstart | LP | PDLP warmstart for similar problems |
| milp_basic | MILP | Minimal MIP; includes incumbent callback example |
| milp_production_planning | MILP | Production planning with resource constraints |
Other reference
| Model | Type | Description |
|---|---|---|
| mps_solver | LP/MILP | Solve any problem from standard MPS file format |
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
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