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Add indicator constraints Description Add a new row and an indicator constraint Further explanation of this example:
Xpress R Reference Manual
Source Files By clicking on a file name, a preview is opened at the bottom of this page.
Data Files indicator.R ##################################### # This file is part of the # # Xpress-R interface examples # # # # (c) 2021 Fair Isaac Corporation # ##################################### #' --- #' title: "Adding Indicator Constraints" #' --- #' #' #' An indicator constraint is made of a condition and a linear constraint. The #' condition is of the type `bin = value`, where `bin` is a binary variable and #' `value` is either 0 or 1. The linear constraint is any linear row. During global #' search, a row configured as an indicator constraint is enforced only when #' condition holds, that is only if the indicator variable `bin` has the specified #' value. #' #' We work with the facility location running #' example. Please familiarize yourself with this example first. #' #' We load the facility location example and optimize it first. ## ----Solving the Unmodified Facility Location Problem------------------------- library(magrittr) suppressMessages(library(xpress)) # quickly optimize the Facility location problem from a pipe p <- createprob() %>% readprob("flp.lp") %>% mipoptimize() print("MIP solution for the 5 facilities") xprs_getsolution(p)[1:5] #' #' # Adding a New Row #' #' We would like to express that if facility $F_5$ is open, then both $F_2$ and #' $F_4$ must be open, as well. Mathematically, we want to impose the implication #' #' $$ #' \begin{align} #' x_5 = 1 \Rightarrow x_2 + x_4 = 2 #' \end{align} #' $$ #' #' Since $x_5$ is currently open in the basic example problem, this additional #' restriction will be very expensive. Can we do without opening $x_5$? First, we #' now add the new row to the existing constraint matrix that expresses the #' cardinality restriction on variables $x_3$ and $x_5$ #' #' We use `addrows` to add the new row to the existing instance. #' The `addrows` function can be used to add only a single #' or even multiple new rows. #' For each row, we have to indicate its type and right hand side. #' Since we only add a single row, the rowtype and rhs can be passed #' as scalars. #' #' In addition, we pass the nonzero row coefficients in sparse row major format. #' When we pass column indices of these entries, we have to remember #' to use 0-based indexing, because these values are directly passed #' to Xpress. #' The 0-based column indices that correspond to variables $x_2$ and $x_4$ #' are 1 and 3. Here we use the syntax `c(2,4) - 1` which subtracts #' 1 from each 1-based index to obtain a 0-based index. #' ## ----Adding a New Row to the Problem------------------------------------------ print(addrows(p, rowtype = "E", rhs = 2, start=c(0,2), # 0-based indices colind = c(2,4) - 1,# 0-based indices rowcoef = c(1,1) ) ) #' #' We verify that the new linear constraint changes the #' solution to the problem (the linear constraint is now active) #' ## ----Verify row addition------------------------------------------------------ summary(mipoptimize(p)) print("MIP solution for the 5 facilities after adding the additional row") print(xprs_getsolution(p)[1:5]) #' # Declaration as Indicator Row #' #' Now we are ready to add the indicator implication between variable $x_5$ and the #' newly created row. #' Recall 0-based indexing: #' This newly created row has index `xpress:::ROWS - 1`. #' Similarly variable x_5 has column index 5 in R, but 4 in C #' #' The `print` statement assures us that we now have 1 indicator #' constraint in our formulation. #' ## ----Marking the New Row As Indicator----------------------------------------- print(setindicators(p, # 0-based. use the last added row rowind=getintattrib(p, xpress:::ROWS) - 1L, colind = 5L - 1L, complement = 1L # row is active when x_5 = 1 )) #' #' After resolving the indicator formulation, we immediately notice that $F_5$ is #' not opened, and therefore, the restriction on opening both $F_2$ and $F_4$ is #' lifted. #' ## ----Resolve with the Indicator Constraint------------------------------------ summary(mipoptimize(p)) print("MIP solution for the 5 facilities after adding the indicator constraint") print(xprs_getsolution(p)[1:5]) #' | |||||||||||

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