CODEWITHSUNDEEP
rfinanceirr
dhhs91
dhhs91

Reputation: 11

Computing IRR using optim

I want to find the internal rate of return (IRR), basically the 'rate' that makes my NPV function go to zero, using the optim function.

My current code for the NPV function (which works) is:

npv <- function(rate, cf){
    r_v <- rep (rate,length (cf))
    t_v <- as.numeric (seq(1:length (cf)))
    pv <- cf * exp (-t_v*r_v)
    sum (pv)
}

I tried using the following optim function:

InternalRateReturn <- optim(c(0,1), npv, cf = testcf2, gr = NULL, method = "L-BFGS-B", lower = -Inf, upper = Inf,control=list(), hessian = FALSE)

but it is not coming back with the correct answer for InternalRateReturn$par as opposed to using the uniroot method below.

May I ask how to modify this code (to reiterate, I just want to optimize the rate in the npv function such that the npv function equals zero)?

The IRR function using uniroot is as per below:

irr1 <- function(cf) {
    uniroot(npv, c(0, 1), cf=cf)$root
}

Upvotes: 1

Views: 1347

Answers (2)

dmi3kno
dmi3kno

Reputation: 3055

There's pretty cool example of using optim to calculate IRR by Matt Brigida

### IRR Function:  Takes a vector of payments and returns a list which includes the internal rate of return ($IRR) and possible word of warning ($beware) ----

irr <- function(x, period = 1, starting.value = .1){

### This should detect the number of sign changes.  Should correctly not warn if there are many negative cash flows (so long as there is only 1 change in sign).

    irr.func <- function(r){ ( sum(x / (1 + r)^{0:(length(x)-1)}) )^2 }
    result <- optim(par = starting.value, fn = irr.func, method = "Brent", lower = -1000000, upper = 1000000)

    ## detecting number of sign changes
    x.ge.0 <- 1 * (x >= 0)
    changes <- diff(x.ge.0)
    changes <- changes * changes
    num.changes <- sum(changes)

    if( num.changes > 1) {

        statement <- "Your cash flows change more than once -- so you may have multiple IRRs. This function will only return the first IRR it finds. To find the others, you can try different starting values.  However, note the IRR does not make sense if the signs change more than once (try Modified IRR or NPV)."
        value <- period * result$par
        return(list(beware = statement, IRR = value))

    } else {

        return(list(IRR = period * result$par))

    }
}

Matt also has a pretty useful function for more realistic modified IRR (which does not make unreasonable assumptions about the reinvestment rate)

### Modified IRR (MIRR) Function:  Takes a vector of payments and returns the MIRR by ----

mirr <- function(x, period = 1, starting.value = .1, discount.rate = 0.1, investment.rate = 0.05){

    ## move cash flows
    ## negative
    cf.neg <- (x < 0) * x
    ## discounted
    pv.cf.neg <- cf.neg / (1 + discount.rate)^{0:(length(x)-1)}
    pv <- sum(pv.cf.neg)

    ## positive
    cf.pos <- (x > 0) * x
    fv.cf.pos <- cf.pos * (1 + investment.rate)^{0:(length(x)-1)}
    fv <- sum(fv.cf.pos)

    mirr.per.period <- ( fv / abs(pv) )^{1 / (length(x))} - 1

    return( period * mirr.per.period )
}

Upvotes: 2

landroni
landroni

Reputation: 2988

If you simply need to compute IRR or NPV (or MIRR), and since it is not clear why you would absolutely need to use optim, you may simply consider packages financial or FinCal instead of hacking your own function. Like this:

> require(financial)
> cf(c(-123400, 36200, 54800, 48100), i = 2.5)

Cash Flow Model

Flows:
      1       2       3       4 
-123400   36200   54800   48100 

 IRR%: 5.96 
 NPV Extremes at I%:  

   I%     NPV     NFV     NUS
1 2.5 8742.13 9414.32 3060.95

> require(FinCal)
> npv(c(-123400, 36200, 54800, 48100), r = 0.025)
[1] 8742.134
> irr(c(-123400, 36200, 54800, 48100))
[1] 0.05959787

Upvotes: 0

Related Questions