Question 1. The Net Present Value

The net cash flow per month (inflow minus outflow) is:

mthly_net <- rep(75000-45000, times = 120)

The computation of NPV is standard. The only catch the candidate should pay attention to is that we are

working with effective instead of nominal interest rate:

NPV <- -2e06 +sum(mthly_net*(1 + 0.03)ˆ{-(1:120)/12})

For part b, we need to first adjust the operational cost to reflect inflation. Note that monthly revenue is not

adjusted.

inflationAdjustment <- rep(1.02ˆ{0:9},each = 12)

mthly_net <- 75000 - inflationAdjustment*45000

Above, I’ve used R’s vectorization to trim down the equation. You can equally well use the following code: mthly_net <- rep(75000, times = 120) - inflationAdjustment*rep(45000, times = 120)

The revised NPV is now:

NPV_revised <- -2e06 +sum(mthly_net*(1 + 0.03)ˆ{-(1:120)/12})

The equation of value is:

≠2 ◊ 106 +ÿ(120)Xj  ◊ (1 + i)≠j/12  = 0.

j=1

We can implement this in R and seach for its zero, as follows:

eqn_val <- function(i, ini, cashFlow){

-ini + sum(cashFlow*(1 + i)ˆ{-seq_along(cashFlow)/12})

}

uniroot(eqn_val, lower = 0, upper = 1, ini = 2e06, cashFlow = mthly_net)

## $root

## [1] 0.1055803

##

## $f.root

## [1] -2.669638

##

## $iter

## [1] 6

##

## $init.it

## [1] NA

##

## $estim.prec

## [1] 6.103516e-05

If you are nervous about writing a function of multiple arguments, this can work:

eqn_val <- function(j){

inflationAdjustment <- rep(1.02ˆ{0:9},each = 12)

mthly_net <- 75000 - inflationAdjustment*45000

-2e06 + mthly_net*(1 + i)ˆ{-seq_along(mthly_net)/12}

}

You can even rely on lexical scoping to do the job for you. You really should not, though.

eqn_val <- function(i){

-2e06 + mthly_net*(1 + i)ˆ{-seq_along(mthly_net)/12}

}

Question 2.

Note that the inflation adjustment vector has been defined above and has not changed since.                 mthly_net_B <- c(rep(0,times=2*12),rep(110000,times=8*12)) - inflationAdjustment*45000 NPV_B <- -2e06 +sum(mthly_net_B*(1 + 0.03)ˆ{-(1:120)/12})

The calculation of the rate of return is very similar to part a. This is because we have written eqn_val in a way that makes it reusable.

uniroot(eqn_val, lower = 0, upper = 1, ini = 2750000, cashFlow = mthly_net_B)

## $root

## [1] 0.07608493

##

## $f.root

## [1] -44.77227

##

## $iter

## [1] 7

##

## $init.it

## [1] NA

##

## $estim.prec

## [1] 6.103516e-05

We can recycle the equation of value in Question 1 for this.

i <- seq(from = 0, to = 0.2, by = 0.01)

NPV_sequence_A <- sapply(i, eqn_val, ini = 2e06, cashFlow = mthly_net)

NPV_sequence_B <- sapply(i, eqn_val, ini = 2750000, cashFlow = mthly_net_B)

plot(

x = i,

y = NPV_sequence_A,

xlab = "Effective Interest Rate",

ylab    = "Net Present Value",

col  = "red",

type = "l",

lty = 1

)

points(x = i, y = NPV_sequence_B, col = "blue", type = "l", lty = 2)

legend(

"topleft",

legend=c("Strat. A", "Strat. B"),

col=c("red", "blue"),

lty=1:2,

cex=0.8

)

 

0.00                   0.05                   0.10                   0.15                   0.20

Effective Interest Rate

Strategy B has heavier overhead and its profits are delayed. Income in the far future is punished more heavily

by discounting.

Question 3.

This is a standard amortisation scheme (output is omitted for brevity):

payment  <- mthly_net

outstanding <- intPart <- capPart <- balance <- retained <- rep(0,length(payment))

for(j in 1:length(payment)){

outstanding[j] <- ifelse(j==1, 2e06*0.70, balance[j-1])

intPart[j] <- ((1 + 0.025)ˆ{1/12} - 1)*outstanding[j]

capPart[j] <- payment[j] - intPart[j]

balance[j] <- outstanding[j]- capPart[j]

if(balance[j] < 0){

retained[j] <- abs(balance[j])

balance[j] <- 0

if(j < length(payment)){

retained[(j+1):length(payment)] <- payment[(j+1):length(payment)]

}

break

}