Production Planning (LP / MIP)

Problem Class	Solver	Key APIs	Links
Linear / Integer Programming (LP / MIP)	HiGHS (default)	m.integer(), m.constraint(), c, expr	Open in Playground Source Code

Production planning is a classic Linear Programming problem: determine the optimal quantity of each product to manufacture in order to maximize total profit while respecting resource constraints (raw materials, machine time, labor hours).

Mathematical Formulation

Let $x$ be the number of chairs and $y$ be the number of tables.

$$\begin{aligned} \text{Maximize} \quad & 45x + 80y & (\text{Profit}) \\ \text{Subject to} \quad & 5x + 20y \le 400 & (\text{Mahogany Wood limit}) \\ & 10x + 15y \le 450 & (\text{Carpentry Labor limit}) \\ & x, y \ge 0, \quad x, y \in \mathbb{Z} & (\text{Integer Quantities}) \end{aligned}$$

Implementation in WaSHI

import { Model, expr, c, maximize } from "@ohtaman/washi";

async function runProductionOptimization() {
  const m = new Model("FactoryProduction");

  // Decision variables: non-negative integer quantities
  const chairs = m.integer("chairs", { lb: 0 });
  const tables = m.integer("tables", { lb: 0 });

  // Objective: Maximize total profit ($45 per chair, $80 per table)
  m.objective(maximize(expr`45 * ${chairs} + 80 * ${tables}`));

  // Resource constraints: declarative formulas via c and fluent constraint()
  m.constraint(c`5 * ${chairs} + 20 * ${tables} <= 400`, "mahogany_limit")
   .constraint(c`10 * ${chairs} + 15 * ${tables} <= 450`, "labor_limit");

  const status = await m.solve();

  console.log(`Status: ${status}`);
  console.log(`Optimal Profit: $${m.objectiveValue}`);
  console.log(`Chairs: ${chairs.value}, Tables: ${tables.value}`);
}

runProductionOptimization();

Open in Playground →