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Solve a small production problem with 2 products and 2 resource constraints Description An introductory examples that shows how to model a production problem with 2 products and 2 resource constraints Further explanation of this example:
This is an introductory example in the book 'Applications of optimization with Xpress'
Source Files By clicking on a file name, a preview is opened at the bottom of this page.
chess2.R ##################################### # This file is part of the # # Xpress-R interface examples # # # # (c) 2022-2024 Fair Isaac Corporation # ##################################### #' --- #' title: "Chess2" #' author: Y. Gu #' date: Jul.2021 #' --- #' #' ## ----setup, include=FALSE----------------------------------------------------- knitr::opts_chunk$set(echo = TRUE) knitr::opts_chunk$set(results = "hold") knitr::opts_chunk$set(warning = FALSE, message = FALSE) #' #' #' ## Brief Introduction To The Problem #' #' This is an introductory example in the book 'Applications of optimization with Xpress'. <https://www.fico.com/fico-xpress-optimization/docs/latest/examples/mosel/ApplBook/Intro/chess2.mos> #' #' This is a very small example, where the manufacture only produces two types of chess #' sets: a small set and large set. The time and materials per set required are known and #' within the limited lathe-hours and boxwood, we would like to maximize the profit by #' deciding how many small sets and large sets should be produced per week. #' #' The purpose of this example is to give readers experience in practical modeling and #' have a general sense of what modeling is. Instead of theoretical overview, a practical #' example is easier to get started with. In Chapter 1 of the book, full mathematical #' formulations of this example are provided, as well as further discussion of the results #' and explanations of the benefits of modeling. #' #' #' For this example, we need the package 'xpress'. #' ## ----Load The Package--------------------------------------------------------- library(xpress) #' #' #' Create a new empty problem, set the objective sense as maximization and give the problem #' a suitable name. #' ## ----Create The Problem------------------------------------------------------- # create a new problem prob = createprob() # change this problem to a maximization problem chgobjsense(prob, objsense = xpress:::OBJ_MAXIMIZE) # set the problem name setprobname(prob, "Chess2") #' #' #' Add the values we need for this example. #' ## ----Data--------------------------------------------------------------------- info.df <- data.frame( # time required on a lathe to make small and large sets lathe = c(3, 2), # amount of boxwood required for small and large sets boxwood = c(1, 3), # profit of small and large sets profit = c(5, 20) ) rownames(info.df) <- c("small", "large") # capacity of lathe-hours per week Lathe <- 160 # capacity of boxwood per week Boxwood <- 200 #' #' #' Create integer variables 'produce' for each product to represent the number of sets #' to make. #' ## ----Add Columns-------------------------------------------------------------- # create a vector 'produce' in 'info.df' to store the column indices. info.df$produce <- mapply( function(profit, name) xprs_newcol( prob, lb = 0, ub = Inf, coltype = "I", objcoef = profit, name = paste0("produce_", name) ), info.df$profit, rownames(info.df) ) #' #' #' Add the capacity constraints of this example. #' ## ----Add Rows, results='hide'------------------------------------------------- # cannot plan to use more of the lathe-hours than we have available xprs_addrow( prob, colind = info.df$produce, rowcoef = info.df$lathe, rowtype = "L", rhs = Lathe, name = "Lathe" ) # cannot plan to use more of the boxwood than we have available xprs_addrow( prob, colind = info.df$produce, rowcoef = info.df$boxwood, rowtype = "L", rhs = Boxwood, name = "Boxwood" ) #' #' #' Now we can solve the problem. #' ## ----Solve The Problem-------------------------------------------------------- setoutput(prob) summary(xprs_optimize(prob)) #' #' #' Display the solutions here. #' ## ----The Solutions------------------------------------------------------------ info.df$solution <- xprs_getsolution(prob) cat("The optimum profit is:", getdblattrib(prob, xpress:::MIPOBJVAL), "\n") for (i in 1:nrow(info.df)) { cat("Produce", info.df$solution[i], rownames(info.df[i, ]), "sets\n") } #' #' | |||||||||||

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