CODEWITHSUNDEEP
rfunctionuniroot
onhalu
onhalu

Reputation: 745

Uniroot solution two equatations

I posted similar question a few days later, but now I have more complex problem. I have this dataset

product = data.frame(length = c(100, 200, 300, 400), qty = c(1, 2, 1, 3))

And price is defined by this equation (product[["length"]] * old_const * 2) + setup_price where old_const = 0.0158 and setup_price = 20.8

product[["old_price"]] = (product[["length"]] * old_const * 2) + setup_price

Now I would like to change constant setup_price to meet two requirements:

Total revenue with new prices is same as revenue with old prices AND Total revenue from first half of table is same as revenue from second half of table

y1 = sum((product[["length"]] * old_const * 2  + setup_price)* product[["qty"]])
y2 = sum((product[["length"]] * 2)* product[["qty"]])
y3 = sum((product[1:table_median, ][["length"]] * 2)* product[1:table_median, ][["qty"]])
y4 = sum((product[table_median:max(nrow(product)), ][["length"]] * 2) * product[table_median:max(nrow(product)), ][["qty"]])

where table_median = round(median(1:nrow(product)))

And my optimalization

const = seq(-2,2, by = 0.0001)
ev = NULL

for (i in 1:length(const)){
  ev[i] = (y1 - y2 * const[i]) + (y3*const[i] - y4)
}

plot(const, ev)

And I would like to get ev = 0 nad I dont know where exactly my 0 is.

There was recomandation to use uniroot function which I know how to use with one equation.

Upvotes: 0

Views: 36

Answers (1)

Nir Graham
Nir Graham

Reputation: 5157

I had to change belt_length to length to make sense of what you shared. I've gone ahead and simply refactored your for loop to involve a function, and then use that same function in uniroot

const = seq(-2,2, by = 0.0001)
ev = NULL

custfunc <- function(x){
  (y1 - y2 * x) + (y3*x - y4)
}
for (i in 1:length(const)){
  ev[i] = custfunc(const[i])
}

plot(const, ev)

uniroot(custfunc,interval = c(-2,2))

Upvotes: 1

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