CODEWITHSUNDEEP
rmathequation
Matt
Matt

Reputation: 53

Iterative equations in R

I have the equation C.sul.f = C.sul.i + (R * C.sil.i) / (1 + R/D), where C.sul.i = 200000, C.sil.i = 100, R = 100, and D = 2130.

The calculation requires that the for every iteration from 1:100000, the C.sul.f from the previous iteration is used for the C.sul.i of the current iteration. Essentially,

Iteration 1 --> C.sul.f1 = C.sul.i + (R * C.sil.i) / (1 + R/D)

Iteration 2 --> C.sul.f2 = C.sul.f1 + (R * C.sil.i) / (1 + R/D)

Iteration 3 --> C.sul.f3 = C.sul.f2 + (R * C.sil.i) / (1 + R/D)

.

.

Iteration n --> C.sul.fn = C.sul.fn-1 + (R * C.sil.i) / (1 + R/D)

Could I please get some help how I would structure this script in R

I appreciate any help.

Thank you in advance.

Upvotes: 1

Views: 156

Answers (3)

ThomasIsCoding
ThomasIsCoding

Reputation: 101618

If you like to have a recursion answer, you can try defining a user function f like below

f <- function(n) {
  if (n==1) return(c.sul.i + (R * c.sil.i) / (1 + R/D))
  f(n-1) + (R * c.sil.i) / (1 + R/D)
}

For efficiency considerations, answer by @SirTain using a math solution is the best.

Upvotes: 0

SirTain
SirTain

Reputation: 369

niko has an excellent solution using Reduce, but for your example (and for all series) I recommend first applying a bit of series reduction to see if there's a simple general solution.

More Math, Faster Code

In your series, the only recursive element is effectively a constant value added to the previous one. If you expand it out, it becomes fairly obvious:

C.sul.f1 = C.sul.i + (R * C.sil.i) / (1 + R/D)

C.sul.f2 = C.sul.i + (R * C.sil.i) / (1 + R/D) + (R * C.sil.i) / (1 + R/D)

C.sul.f3 = C.sul.i + (R * C.sil.i) / (1 + R/D) + (R * C.sil.i) / (1 + R/D) + (R * C.sil.i) / (1 + R/D)

in general:

C.sul.fn = C.sul.i + n * ((R * C.sil.i) / (1 + R/D))

So for the most efficient function, I recommend:

c_sul_i <- 200000
c_sil_i <- 100
R <- 100
D <- 2130

C.sul.fn <- function(n) {
     return(C.sul.i + n * (R * C.sil.i) / (1 + R/D) )
}

sapply(1:100000,C.sul.fn)

Or, if you prefer single line implementation:

sapply(1:100000,function(x) {C.sul.i + x * (R * C.sil.i) / (1 + R/D)})

Upvotes: 2

niko
niko

Reputation: 5281

This is the archetype of problem Reduce solves:

c_sul_i <- 200000
c_sil_i <- 100
R <- 100
D <- 2130

Reduce(
  f = function(x, y) x + (R * c_sil_i) / (1 + R/D),
  x = 1:5,
  init = c_sul_i + (R * c_sil_i) / (1 + R/D),
  accumulate = TRUE
  )
# [1] 209551.6 219103.1 228654.7 238206.3 247757.8 257309.4

PS: Note that for example purposes I only computed the first 5 values. For your case just swap 1:5 with 1:100000.

Upvotes: 3

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