Simple linear programming function in R, with an example.

linear-programming.R

lp <- function(givenName, givenQuantity, resources, needsChart) {
  if ( !all.equal(names(resources), colnames(needsChart)[2:length(colnames(needsChart))]) ) {
    warning("constraints and needsChart don't match.")
    return(NA)
  } else if ( dim(needsChart)[1] != 2 ) {
    warning("Can only handle two products.")
    return(NA)
  }
  
  sapply(givenQuantity, function(x) {
    resources.left <- resources - (x * c(needsChart[needsChart$name==givenName,2:dim(needsChart)[2]], recursive=T))
    poss <- min(as.integer(resources.left / c(t(needsChart[needsChart$name!=givenName,2:dim(needsChart)[2]]))))
    
    if ( poss < 0 ) return(NA)
    poss
  })
}

softballs.R

# Example for using the lp() function.
# Taken from _Spreadsheet modeling & decision analysis: 
# A practical introduction to business analytics_, 
# by Cliff Ragsdale

source('./linear-programming.R')

resources <- c(leather=6000, nylon=5400, core=3500, labor=2800, stitching=1500)

products = data.frame(
  name=c('softball', 'baseball'),
  leather=c(5, 4),
  nylon=c(6, 3),
  core=c(4,2),
  labor=c(2.5, 2),
  stitching=c(1,1)
)

softball.profit = 17-11
baseball.profit = 15-10.5

baseballs <- 1:1500
softballs <- lp('baseball', baseballs, resources, products)
ppf <- data.frame(baseballs, softballs, profit=(baseballs*baseball.profit+softballs*softball.profit))
optimum <- ppf[ppf$profit==max(ppf$profit, na.rm=T),][1,]

with(ppf, plot(baseballs, softballs, xlab="# Baseballs", ylab="# Softballs", type='l'))
with(optimum, {
  abline(h=softballs, col=gray(.8))
  abline(v=baseballs, col=gray(.8))
  points(baseballs, softballs, pch=20)
  text(baseballs, softballs, 
       sprintf('$%3$0.2f (%1$d, %2$d)', baseballs, softballs, profit),
       adj=c(-.03,-.5), ps=10)
})