cuopt-numerical-optimization-api-python

What this skill does

# 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

```python
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)

```python
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)

```python
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:**

```python
# ✅ 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
```python
# 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
```python
# 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:
```python
# 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
prob