Scientific Calculations with Julia/Cost Minimization

Scientific Calculations with Julia Cost Minimization Profit Maximization Shortest Path Cost minimization is a classic example of linear programming that in its standard form is expressed in three parts: A linear function to be minimized: 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 needs three types of wood units x1, x2, x3 to produce chairs, which cost €10, €20, and €5, respectively. Type 1 and 2 units must be more than 300, type 2 units more than 100, and type 1 and 3 units more than 300. Therefore, the model to analyze is as follows: First, 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, x1 >= 0) @variable(model, x2 >= 0) @variable(model, x3 >= 0) @objective(model, Min, 10*x1 +20*x2+ 5*x3) @constraint(model, c1, x1+x2>=300) @constraint(model, c2, x2>=100) @constraint(model, c3, x1+x3>=300) print(model) optimize!(model) min 10𝑥1+20𝑥2+5𝑥3 Subject to 𝑥1+𝑥2≥300 𝑥2≥100 𝑥1+𝑥3≥300 𝑥1≥0 𝑥2≥0 𝑥3≥0 Running HiGHS 1.5.3 [date: 1970-01-01, git hash: 45a127b78] Copyright (c) 2023 HiGHS under MIT licence terms Presolving model 2 rows, 2 cols, 3 nonzeros 0 rows, 0 cols, 0 nonzeros Presolve : Reductions: rows 0(-3); columns 0(-3); elements 0(-5) - Reduced to empty Solving the original LP from the solution after postsolve Model status : Optimal Objective value : 4.5000000000e+03 HiGHS run time : 0.00 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("x1 = ",value(x1)) println("x2 = ",value(x2)) println("x3 = ",value(x3)) 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 = 4500.0 x1 = 200.0 x2 = 100.0 x3 = 100.0 c1 = -5.0 c2 = -15.0 c3 = -5.0 That is, the company has a minimum cost of €4500, purchasing 200 units of type 1 wood, 100 units of type 2 wood, and 100 units of type 3 wood to produce the chairs, satisfying the indicated constraints.