Scientific Calculations with Julia/Profit Maximization

Scientific Calculations with Julia Profit Maximization Personal Budget Optimization Cost Minimization Profit maximization is a classic example of linear programming, which in its standard form is expressed in three parts: A linear function to be maximized: Constraint problem of the following form: Non-negative variables The problem is usually expressed in matrix form as follows: Example calculation with Julia A company produces tables and cabinets that it intends to sell at prices of €400 and €200, respectively. Producing them requires 6 units of wood, 2 worker hours, and 100 minutes of machine time for the tables, and 10 units of wood, 4 worker hours, and 40 minutes of machine time for the cabinets. The company has 4,000 units of wood available daily, 250 workers working 8 hours a day, or 2,000 worker hours, and 50 machines operating 16 hours a day, or 48,000 machine hours. First, you need to install the JuMP and HiGHS packages. From the REPL, type: using Pkg Pkg.add("JuMP") Pkg.add("HiGHS") From Jupyter, in two code cells, type: ] add HiGHS ] add JuMP And then click Run. Then run the code to create the linear programming model: using JuMP using HiGHS model = Model(HiGHS.Optimizer) @variable(model, x >= 0) @variable(model, y >= 0) @objective(model, Max, 200y + 400x) @constraint(model, c1, 10y + 6x <= 4000) @constraint(model, c2, 4y + 2x <= 2000) @constraint(model, c3, 40y + 100x <= 48000) print(model) optimize!(model) max Subject to 200𝑦+400𝑥 6𝑥+10𝑦≤4000 2𝑥+4𝑦≤2000 100𝑥+40𝑦≤48000 𝑥≥0 𝑦≥0 Running HiGHS 1.5.3 [date: 1970-01-01, git hash: 45a127b78] Copyright (c) 2023 HiGHS under MIT licence terms Presolving model 3 rows, 2 cols, 6 nonzeros 3 rows, 2 cols, 6 nonzeros Presolve : Reductions: rows 3(-0); columns 2(-0); elements 6(-0) - Not reduced Problem not reduced by presolve: solving the LP Using EKK dual simplex solver - serial Iteration Objective Infeasibilities num(sum) 0 -3.7499962007e+01 Ph1: 3(5.6875); Du: 2(37.5) 0s 2 1.9789473684e+05 Pr: 0(0) 0s Model status : Optimal Simplex iterations: 2 Objective value : 1.9789473684e+05 HiGHS run time : 0.03 println("termination_status = ",termination_status(model)) println("primal_status = ",primal_status(model)) println("dual_status = ",dual_status(model)) println("objective_value = ",objective_value(model)) println("x = ",value(x)) println("y = ",value(y)) println("c1 = ",shadow_price(c1)) println("c2 = ",shadow_price(c2)) println("c3 = ",shadow_price(c3)) termination_status = OPTIMAL primal_status = FEASIBLE_POINT dual_status = FEASIBLE_POINT objective_value = 197894.73684210528 x = 421.05263157894734 y = 147.36842105263167 c1 = 5.2631578947368425 c2 = -0.0 c3 = 3.6842105263157894 That is, the company earns a maximum daily profit of €197,894 by producing 147 cabinets and 421 tables.