financial break even point given a stream of cashflows in R

Suppose I have every month an income of 1200
The interest rate is 1% - so after 1 year the price will increase 1%

I would like to find out how many years it will take until each investment will break even

Suppose an investment costs 200,000 with a momthly income of 1200

My first year and subsequent years annual income will be:

firstYear = 1200 * 12
additionalYears = (1200*12)*(1+0.01)^c(1:5)
c(firstYear, additionalYears)

14400.00 14544.00 14689.44 14836.33 14984.70 15134.54

I would like to make the "5" in the above example dynamic until it find the breakeven point.

In this example I have:

sum(c(firstYear, additionalYears))
198854.3

So the investment did not breakeven yet. Adjusting it to "12" gives me the breakeven point:

firstYear = 1200 * 12
additionalYears = (1200*12)*(1+0.01)^c(1:12)
sum(c(firstYear, additionalYears))

If possible I would like to determine the month of that year it will break even (so given this example it breakseven in month 12 of year 12, but others might break even in month 8 of year 6 etc.

Instead of using a loop, you can use vectors. Set the max years to 100, create a vector of 100 incomes to grow and a vector of 100 growth factors. Multiply the two and get a cumulative sum of the cost (negative) and the incomes. Count the number of times the sum is negative, that is your break even.

cost = -200000 # negative cost
income = 1200*12 # annual income
i = 0.01 # interest rate to grow income after year 0.

# repeat 14400 101 times, multiply it by (1+r)^n - R is vectorised
income100 = rep(income, 101) * ((1+i) ^ seq(0,100)) 

# subtract the cost from the cumulative sum of income 
cumincome = cost + cumsum(income100) 

# how many are negative?
yrs = sum(cumincome < 0)

# how much left to recover in next year
yrs + (-cumincome[yrs] / income100[yrs+1])

[1] 13.06991

Putting this into a function

break_even_years <- function(cost, income, interest=0, period = "monthly"){
  
  if(cost >= 0) cost = -cost
  
  if(period == "monthly") income = income * 12
  
  income100 = rep(income, 101) * ((1+interest) ^ seq(0,100))
  cumincome = cost + cumsum(income100)
  # how many are negative?
  yrs = sum(cumincome < 0)
  
  # how much left to recover in next year
  yrs + (-cumincome[yrs] / income100[yrs+1])
  
}

Using the function

purrr::map2(
  .x = cost,
  .y = investment,
  ~ break_even_years(.x, .y, interest = 0.03, "annual"))

[[1]]
[1] 28.90435

[[2]]
[1] 23.75858

[[3]]
[1] 6.391264

[[4]]
[1] 3.505453

Benchmarking

library(microbenchmark)

microbenchmark(break_even_years(200000,1000,0.01), 
               find_break_even_year(1000, 200000, 0.01, 100), times = 1000)

Unit: microseconds
                                          expr   min      lq      mean median      uq
          break_even_years(200000, 1000, 0.01)  50.9   87.10  257.4185  119.0  159.05
 find_break_even_year(1000, 200000, 0.01, 100) 853.5 1247.05 3432.5157 1556.2 2391.35
      max neval
  36938.0  1000
 145980.6  1000

I think this answers my question. If anybody can help with not using the forloop function that would be very helpful.

library(tidyverse)
investment = c(1000, 2000, 5000, 27000)
interest_rate = 0.03
cost = c(45000, 67900, 34678, 98367)
max_years = 100
future_value = list()

find_break_even_year <- function(CF, investment, interest_rate, max_years){
  for (year in 1:max_years){
    #print(year)
    future_value[[year]] <- CF * (1 + interest_rate)^year
    future_value_sums = sum(unlist(future_value))
    if(future_value_sums >= investment)
      return(year)
  }
}

purrr::map2(
  .x = investment,
  .y = cost,
  ~ find_break_even_year(.x, .y, interest_rate = 0.03, max_years = 100)
)