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Model a piecewise linear function using binary variables

Description

This example simulates the situation where we are buying a required number of items and we get discounts incrementally, and we want to minimize the total cost. This is a problem suitable to be modeled with SOS2 since we can express the cost as a piecewise linear function of the number of items bought. For SOS2 formulation, please refer to the file 'approx.R'. In this file we model the incremental price breaks using binary variables.

Further explanation of this example: This example is from the book 'Applications of optimization with Xpress'

Source Files
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price_breaks.R

#####################################
# This file is part of the          #
# Xpress-R interface examples       #
#                                   #
#   (c) 2021 Fair Isaac Corporation #
#####################################
#'
#' ---
#' title: "Price Breaks 2"
#' 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 example is from the book 'Applications of optimization with Xpress'](https://www.fico.com/fico-xpress-optimization/docs/latest/examples/mosel/ApplBook/Intro/pricebrinc2.mos). #' #' This example simulates the situation where we are buying a required number of items #' and we get discounts incrementally, and we want to minimize the total cost. #' #' This is a problem suitable to be modeled with SOS2 since we can express the cost as a #' piecewise linear function of the number of items bought. For SOS2 formulation, #' please refer to the file 'approx.R' and page 44 of the book, in this file we will #' model the incremental price breaks using binary variables. #' #' There are 4 break points:$0$,$B_1$,$B_2$and$B_3$, and 3 line segments, i.e. 3 #' different costs. We define binary variables$b_1$,$b_2$and$b_3$, where$b_i$is 1 #' if we have bought any item at a unit cost of$cost_i$. We also define decision #' variables$x_1$,$x_2$and$x_3$for the number of items bought at each cost. #' #' One important constraint here is that we cannot buy any item at the lower price without #' having bought the maximum number of items at the higher price, because otherwise the #' solution will tell us to buy all items at the lowest unit price. Besides, we need to #' set bounds for$x_1$,$x_2$and$x_3$to model the incremental discounts. For mathematical #' formulation of the constraints for these bounds, please refer to page 45 of the book #' 'Applications of optimization with Xpress'. #' #' #' For this example, we need the package 'xpress'. #' ## ----Load The Package--------------------------------------------------------- library(xpress) #' #' #' Create a new empty problem and give the problem a suitable name. #' ## ----Create The Problem------------------------------------------------------- # create a new problem prob <- createprob() # set the problem name setprobname(prob, "PriceBrinc2") #' #' Add the values we need for this example. #' ## ----Data--------------------------------------------------------------------- # demand DEM <- 150 # break points of cost function B <- c(0, 50, 120, 200) # three different unit costs cost.df <- data.frame(Break = 1:3, Cost = c(0.8, 0.5, 0.3)) #' #' #' Create variables$x_i$and binary variables$b_i$as mentioned in the introduction. #' ## ----Add Columns-------------------------------------------------------------- # 1. create a vector 'x' in 'cost.df' to store the indices of x: the number of # items bought at price Cost1, Cost2 and Cost3 cost.df$x <-
apply(cost.df, 1, function(x)
xprs_newcol(
prob,
lb = 0,
ub = Inf,
coltype = "C",
name = paste0("x_", x["Break"]),
objcoef = x["Cost"]
))

# 2. create a vector 'b' in 'cost.df' to store the indices of binary variables b1,
#    b2 and b3, where bi is 1 if we have bought any items at a unit cost of Cost_i.
cost.df$b <- apply(cost.df, 1, function(x) xprs_newcol( prob, lb = 0, ub = 1, coltype = "B", name = paste0("b_", x["Break"]), objcoef = NULL )) #' #' #' Add the four sets of constraints. #' ## ----Add Rows, results='hide'------------------------------------------------- # 1. satisfy the demand xprs_addrow( prob, colind = cost.df$x,
rowcoef = rep(1, nrow(cost.df)),
rowtype = "E",
rhs = DEM ,
name = "Meet the demand"
)

# 2. lower bounds on quantities
for (i in 1:(length(cost.df$Break) - 1)) { xprs_addrow( prob, colind = c(cost.df$x[i], cost.df$b[i + 1]), rowcoef = c(-1, (B[i + 1] - B[i])), rowtype = "L", rhs = 0, name = paste0("lb_", i) ) } # 3. upper bounds on quantities apply(cost.df, 1, function(x) xprs_addrow( prob, colind = c(x["x"], x["b"]), rowcoef = c(1, -(B[i + 1] - B[i])), rowtype = "L", rhs = 0, name = paste0("ub_", x["Break"]) )) # 4. sequence of price intervals, b1 >= b2 >= b3 for (i in 1:(length(cost.df$b) - 1)) {
prob,
colind = cost.df\$b[i:(i + 1)],
rowcoef = c(1, -1),
rowtype = "G",
rhs = 0,
name = paste0("sequence_", i)
)
}

#'
#'
#' Now we can solve the problem.
#'
## ----Solve The Problem--------------------------------------------------------
setoutput(prob)
summary(xprs_optimize(prob))

#'
#'
#' Display the solutions here.
#'
## ----The Solutions------------------------------------------------------------
solution <- xprs_getsolution(prob)

print(paste0(
"The minimum total cost is: ",
getdblattrib(prob, xpress:::MIPOBJVAL)
))
invisible(apply(cost.df, 1, function(x)
cat("The number of items bought at unit cost", x["Cost"], "is:", solution[x["x"] + 1], "\n")))

#'
#'