Regarding Challenge 1 of previous workshop, you have to find a peer and share your attempt. Improve your attempt with this feedback. If you solved the problem with an analytic formula, change the solution writing a loop.
Now modify your solution and write your code as a function. As any function, you need inputs, which are also called parameters of the function. Inside your function you need to process the inputs and then end up with an output or a set of outputs as results. It is said that each function also **returns* the values of the output.
For this challenge the inputs are:
Loan amount
Annual percentage rate
Frequency of annual compounding (we set as 12 a year, since we used months)
Fixed payment per period
The outputs have to be:
Number of periods needed to fully pay the loan
The amount of the last payment
Once you finish your function, test it with at least 2 different set of inputs, run your function and display your results.
Challenge 1 INPUT VECTORS
PV <- 6000000 MFP <- 50000 i <- 0.12/12 CALCULATE TIME. I calculated the time manually at first to see if my programming was correct
t <- -log(1-(PV*i/MFP))/log(1+i) t
First I rewrite the inputs to make it run only this chunk when needed to change it.PV <- 4000000 MFP <- 50000 i <- 0.12/12 EB <- PV T <- 0
Now I use while to create a loop as if I was making an amortization chart and get what I want. while (EB > 0) { First start counting the months T <- T+1 Then calculate interest I <- EB*i Then the payment(principal) of the month Pay <- MFP - I I designate my new ending balance EB <- EB - Pay I calculate the Principal paid of the month PPaid <- Pay + EB } print(paste( “It will take”, T, “months to pay and your last payment will be”, PPaid)) [1] “It will take 162 months to pay and your last payment will be 37035.3823180185” Challenge 2 COUPON
C <-5000000*.11/2 C [1] 275000 WITH ANNUAL INTEREST RATE = 8%
AR <- 0.08 exponentsb <- seq(1,40) PVbond <- C / (1 + AR)^exponentsb PVb<-sum(PVbond) cat(“The Price of the Bond is”, PVb) The Price of the Bond is 3279269 WITH ANNUAL INTEREST RATE = 11%
AR <- 0.11 exponentsb <- seq(1,40) PVbond <- C / (1 + AR)^exponentsb PVb<-sum(PVbond) cat(“The Price of the Bond is”, PVb) The Price of the Bond is 2461539 WITH ANNUAL INTEREST RATE = 13%
AR <- 0.13 exponentsb <- seq(1,40) PVbond <- C / (1 + AR)^exponentsb PVb<-sum(PVbond) cat(“The Price of the Bond is”, PVb) The Price of the Bond is 2099453
Challenges about Bond ValuationInclude - General approach (in CAPITAL LETTERS) - Sequence of steps needed
An investor purchasing a special Bonus from the UK Government is entitled to receive annual payments from the Government forever. What is the price of the bonus that pays $1,500 US annually if the next payment occurs one year from today? The market interest rate is 12%
To calculate the price we would need to use the present value formula, in which we divide the payment between the rate used, un this case it should be 1500/0.12
# To start we should debine the variables
payment<- 1500
rate<- 0.12
# Once we define , we need to compute the operation
present_value<- payment/rate
#Finally we add the output sentence
cat("The price of the special Bonus is","$", present_value)## The price of the special Bonus is $ 12500Find the bond value of the following 2 exercises:
The $1,000 face value if bond XYZ has a coupon rate of 6%, with interest paid semi-annually, and matures in 5 years. If the bond is priced to yield 8%, what is the bond’s value today?
The ABC bond has a 8% coupon rate (with interest paid semi-annually), a maturity value of $1,000, and matures in 5 years. If the bond is priced to yield 6%, what is the bond’s current price?
You have to write one FUNCTION to solve these 2 problems.
To calculate the bond value of this 2 bonds, we need to use 2 formulas, the Annuity:
A=(c/i)*[1-(1/((1+i)^n)]and the present value of a future value
PV= FV/((1+i)^n)Once we get the result of this 2, we need to sum the result, and that would provide us the bond value
#We start by providing our function, in which we need to use the future value (1000,1000), time (5,5), yield (8%, 6%), coupon rate (6%, 8%), and the number of periods( in this case is semi annually, this represents 2 capitalizations per year)
bond_value <- function(FV,t,Y,i,n){
# We start with the annuity formula to get the present value of the cash flows
# A=(c/i)*[1-(1/((1+i)^n)]
# We calculate first of all the fixed payment
FP<- i/(n*FV)
# with the fixed payment, we calculate the present value of the cash flows
PV_cashF <- FP/(Y/n)*(1-(1/((1+(Y/n))^(t*n))))
# Now we use the second formula to calculate the present value of a future value
# PV= FV/((1+i)^n)
PV_FutVal <- FV/((1+(Y/n))^(t*n))
# Finally, as I mentioned, I need to sum both outputs
return(PV_cashF+PV_FutVal)
}# For the XYZ bond:
#The $1,000 face value if bond XYZ has a coupon rate of 6%, with interest paid semi-annually, and matures in 5 years. If the bond is priced to yield 8%, what is the bond’s value today?
cat("The value of the XYZ bond is", bond_value(FV=1000, t=5, Y=0.08, i=0.06, n=2))## The value of the XYZ bond is 675.5644#For the ABC bond:
#The ABC bond has a 8% coupon rate (with interest paid semi-annually), a maturity value of $1,000, and matures in 5 years. If the bond is priced to yield 6%, what is the bond’s current price?
cat("The value of the ABC bond is", bond_value(FV=1000, t=5, Y=0.06, i=0.08, n=2))## The value of the ABC bond is 744.0943The HIJ bond has a current price of $800, a maturity value of $1,000, and matures in 5 years. If interest is paid semi-annually and the bond is priced to yield 8%, what is the bond’s annual coupon rate?
You have to write a function to solve this problem.
We need to get the coupon rate from the bond price formula
BP= (iFV/(Y/n))[1-(1/(1+(Y/n)^(t*n))]+ (FV/(1+(Y/n)^(t*n)))
This takes us to
i= (BP- (FV/(1+(Y/n)^(tn))))(Y/n)/[1-(1/(1+(Y/n)^(t*n))]*FV
#Now let's start by generating our function
coupon_rate <- function(BP,FV,t,Y,n) {
# We use the formula we get to obtain the coupon rate
cr<- (BP- (FV/(1+(Y/n)^(t*n))))*(Y/n)/(FV*((1-(1/(1+(Y/n)^(t*n))))))
# Finally we multiply the coupon rate times the periodicity which is 2 cause it's semiannually
return(cr*n)
}# Now we put the output as we want it
cat("The coupon rate of the HIJ bond is", coupon_rate(BP =800,FV=1000,t=5,Y=0.08,n=2),"%")## The coupon rate of the HIJ bond is -1.53314e+12 %# watch out, the result is not correctYou are planning to save for retirement over the next 30 years. To do this, you will invest $6,000 a month in a stock account and $3,000 a month in a bond account. The return of the stock account is expected to be 12% compounded monthly, and the bond account will pay 5% annual interest rate also compounded monthly. When you retire, you will combine your money into an account with a 9% annual return. How much can you withdraw each month from your account assuming a 25-year withdrawal period? At the end of 25 years your balance must be zero.
Hint: check if you can use the functions you wrote in the previous challenges to solve this challenge.
To get our output we may need to calculate the present value
A=(c/i)*[1-(1/((1+i)^n)]and later on get the future value from this formula
PV= FV/((1+i)^n)Meaning,
FV=PV*((1+i)^n)#We should start with the present value
# We formulate our function
pv_retirement <- function(c,t,i,n) {
pv <- c/(i/n) * (1 - 1/(1+i/n)^(t*n))
return(pv)
}# I change according to the values I got for the investments
stock <- pv_retirement(c=6000,t=30,i=0.12,n=12)
bond <- pv_retirement(c=3000,t=30,i=0.05,n=12)
# We get the future value according to the formula previously mentioned
# FV=PV*((1+i)^n)
stock_2 <- stock*(1+0.12/12)^(30*12)
bond_2 <- bond*(1+0.05/12)^(30*12)
# Sum both
balance <- stock_2 + bond_2
cat("The balance of 30 years would be", "$",balance)## The balance of 30 years would be $ 23466561# Now I get the monthly withdrawal
i_account<- 0.09
wtdwl<- balance*(i_account/12)/(1-(1/(1+i_account/12)^(25*12)))
cat("the monthly withdrawal shoul be",wtdwl)## the monthly withdrawal shoul be 196930.5