This function uses the per-specified data frame cars Cars93. I'm trying to figure out how i can also return the number of iterations by returning the variable 'passes' at the end as well as the number of resources aka 'needed'.
factory.function <- function (cars.output=1, trucks.output=1) {
factory <- matrix(c(40,1,60,3),nrow=2,
dimnames=list(c("labor","steel"),c("cars","trucks")))
available <- c(1600,70); names(available) <- rownames(factory)
slack <- c(8,1); names(slack) <- rownames(factory)
output <- c(cars.output, trucks.output); names(output) <- colnames(factory)
passes <- 0 # How many times have we been around the loop?
repeat {
passes <- passes + 1
needed <- factory %*% output # What do we need for that output level?
# If we're not using too much, and are within the slack, we're done
if (all(needed <= available) &&
all((available - needed) <= slack)) {
break()
}
# If we're using too much of everything, cut back by 10%
if (all(needed > available)) {
output <- output * 0.9
next()
}
# If we're using too little of everything, increase by 10%
if (all(needed < available)) {
output <- output * 1.1
next()
}
# If we're using too much of some resources but not others, randomly
# tweak the plan by up to 10%
# runif == Random number, UNIFormly distributed, not "run if"
output <- output * (1+runif(length(output),min=-0.1,max=0.1))
}
return(output)
}
factory.function()