cuopt-numerical-optimization-api-python

name	cuopt-numerical-optimization-api-python
version	26.08.00
description	Solve LP, MILP, QP (beta) with cuOpt Python API — linear/quadratic objectives, integer variables, scheduling, portfolio, least squares.
license	Apache-2.0
metadata	{"author":"NVIDIA cuOpt Team","tags":["cuopt","linear-programming","milp","qp","python"]}

cuOpt Numerical Optimization Skill (Python)

Model and solve LP, MILP, and QP problems using NVIDIA cuOpt's GPU-accelerated solver. The Python API surface (Problem, SolverSettings, solve) is shared across all three problem classes — only the objective form and a few rules change.

Before You Start

Use a formulation summary (parameters, constraints, decisions, objective) if available; otherwise ask for decision variables, objective, and constraints. Then confirm problem type (LP / MILP / QP — see below) and variable types.

Choosing LP vs MILP vs QP

Decide from the objective and variables:

If the objective is...	And variables are...	Use
Linear (sum of c_i * x_i)	All continuous	LP
Linear	Some integer or binary	MILP
Has squared (x*x) or cross (x*y) terms	Continuous (integer QP not supported)	QP (beta)

Prefer LP 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. Use QP only when the objective is genuinely quadratic (variance, squared error, kinetic energy).

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).
• Use QP when the objective minimizes variance, squared error, or any expression with x*x or x*y terms (portfolio optimization, least squares, regularized regression).

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()}")

QP Example (beta — MINIMIZE only)

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

# Portfolio variance minimization
problem = Problem("Portfolio")
x1 = problem.addVariable(lb=0, ub=1, vtype=CONTINUOUS, name="stock_a")
x2 = problem.addVariable(lb=0, ub=1, vtype=CONTINUOUS, name="stock_b")
x3 = problem.addVariable(lb=0, ub=1, vtype=CONTINUOUS, name="stock_c")

# Quadratic objective (variance) — MUST be MINIMIZE
problem.setObjective(
    0.04*x1*x1 + 0.02*x2*x2 + 0.01*x3*x3
    + 0.02*x1*x2 + 0.01*x1*x3 + 0.016*x2*x3,
    sense=MINIMIZE,
)

# Linear constraints
problem.addConstraint(x1 + x2 + x3 == 1, name="budget")
problem.addConstraint(0.12*x1 + 0.08*x2 + 0.05*x3 >= 0.08, name="min_return")

problem.solve(SolverSettings())
if problem.Status.name in ["Optimal", "PrimalFeasible"]:
    print(f"Variance: {problem.ObjValue}")

QP rules:

• MINIMIZE only — solver rejects MAXIMIZE for quadratic objectives. To maximize f(x), minimize -f(x).
• Continuous variables only — integer QP is not supported.
• Q should be PSD (positive semi-definite) for a convex problem; otherwise the solver may return a non-optimal stationary point.
• Beta — API may evolve; treat as production-capable for typical convex QP but expect occasional changes.

See references/qp_examples.md for least-squares, maximization-workaround, and matrix-form examples.

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

QP Status Values: Same set as LP. For QP debugging, print f"Actual status: '{problem.Status.name}'" and check that Q is PSD and variables are reasonably scaled.

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)
QP rejected with MAXIMIZE	QP only supports MINIMIZE	Negate the objective: minimize -f(x)
QP returns non-optimal	Q not PSD or variables badly scaled	Check Q is PSD; rescale variables to similar magnitudes

Getting Dual Values (LP / QP)

Duals and reduced costs are returned for LP and QP. They are not returned for a problem with quadratic constraints (every value comes back as NaN), so read them only when all constraints are linear. MILP returns no duals.

if problem.Status.name == "Optimal":
    constraint = problem.getConstraint("resource_a")   # linear constraint
    print(f"Dual value: {constraint.DualValue}")       # NaN if the model has quadratic constraints

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, QP)

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
portfolio	QP	Minimize portfolio variance; budget and min-return constraints
least_squares	QP	Minimize (x-3)² + (y-4)² (closest point)
maximization_workaround	QP	Maximize quadratic via minimize -f(x)

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