I have the price range price <- c(2.5,2.6,2.7,2.8)
and my dataset have several time t. For each time t, I have a corresponding cost c and demand quantity d.
I need to find the optimal price for each time t to maximise my required profit function (p-c)*d.
How can I achieve that?
The sample of mydata looks like this, I have 74 observations in total:
t | c | d |
---|---|---|
1 | 0.8 | 20 |
2 | 0.44 | 34 |
3 | 0.54 | 56 |
4 | 0.67 | 78 |
5 | 0.65 | 35 |
Here is my code but it reports error, can anybody help me to fix it? Much thanks!
max <-data.frame()
for (i in mydata$t) {
for (p in price) {
profit <- ((p-mydata$c)*mydata$d)
max <- max %>% bind_rows(data.frame(time=mydata$t,
price=p,
cost=mydata$c,
profit = profit
))
}
}
maxvalue <- max %>% group_by(time) %>% max(profit)
CodePudding user response:
Since you did not provide a piece of your data which I could use, this is a bit of a guess, but the idea would be:
dat <- as.data.table(mydata)
# Iterate through each value of t and get the price for which (p-c)*d is the highest
result <- dat[, p[which.max((p-c)*d))], t]
CodePudding user response:
Ok! I did not realize you kept the price outside your table. Then try adding all possibilities to the table first this:
dat <- data.table(t= 1:5,
c= c(0.8,0.44,0.54,0.67,0.65),
d= c(20,34,56,78,35))
# Add all possible prices as an extra column (named p)
# Note that all lines will be repeated accordingly
dat <- dat[, .(p= c(2.5,2.6,2.7,2.8)), (dat)]
# Iterate through each value of t and get the price for which (p-c)*d is the highest
result <- dat[, .(best_price= p[which.max((p-c)*d)]), t]