Hi everyone,
>From your helps of giving me several ideas, eventually I can solve the
posted problem. Here is the R code. It can be done by applying the
uniroot.all to the data frame together with the proper form of equation
(slightly modification of the original equation).
#Generate the sample data frame
customer.name = c("John","Mike","Peter")
product = c("Toothpaste","Toothpaste","Toothpaste")
cost = c(30,45,40)
mydata = data.frame(customer.name,product,cost)
#Original cost function - not used
#fcost = function(orders) 3.40 + (1.20 * orders^2)
#Slightly modification of the cost function to be a proper form for root
finding
#This is basically to set ==> 3.40 + (1.20 * orders^2) - fcost = 0
f.to.findroot = function(orders,fcost) 3.40 + (1.20 * orders^2) - fcost
#Using rootSolve package which contains uniroot.all function
library(rootSolve)
#Using plyr package which contains adply function
library(plyr)
#Use uniroot function to find the 'orders' variable (from the f.to.findroot
function) for each customer and put it into no.of.orders column in
mysolution data frame
#Replace 'fcost' with 'cost' column from mydata
#Interval of 0 to 1,000 is to make the f.to.findroot function have both
negative and positive sign, otherwise uniroot.all will give an error
mysolution = data.frame(adply(mydata, 1, summarize, no.of.orders =
uniroot.all(f.to.findroot,interval = c(0,1000),fcost=cost)))
mysolution
#Remove the redundant mydata as mysolution it is an extended version of
mydata
rm(mydata)
#Note uniroot.all can be used for both linear (e.g.orders^1) and non-linear
(e.g.orders^2) equations.
Thank you,
Prakasit Singkateera
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