Mortgage calculations

A mortgage is taken out for 80000$. It is to be paid> by annual instalments > of 5000$ with the first payment being made at the> end of the first year that > the morgage was taken out. Interest of 4% is then> charged on any outstanding > debt. Find the total time taken to pay off the> morgage.

Answer: (It may help to followw this solution with a pen and paper)

[This first part is the Long Answer]

Let us denote the mortgage amount taken (principal) as P=80000

Let annual payments (in arrears) be given by C = 5000

Interest rate is given by r= 4% (annual rate) - or 0.04

let n be the number of years.

A step-by-step approach will be taken to address the problem.

At the end of the first year n=1:
Principal amount will have grown to P(1+r) e.g 80000(1+0.04) =83200

Payment made at the end of first year is C or 5000

Thus, amount left to pay is P(1+r)-C (or 83200-5000 = 78200)

we will not need numbers again until we are ready to solve the problem

End of second year n=2:
Interest is added to principal again
Thus new principal is (P(1+r)-C)*(1+r) = P(1+r)^2-C(1+r)

payment of C is again made.
Thus principal remaning is P(1+r)^2-C(1+r)-C


End of third year n=3:
Interest is again added to principal
Thus new principal is {P(1+r)^2-C(1+r)-C}*(1+r)=P(1+r)^3-C(1+r)^2-C(1+r)

payment of C is again made
Thus principal remaning is P(1+r)^3-C(1+r)^2-C(1+r)-C

You may see a pattern emerging - which will be illustrated next

We now extend this to year n

After the end of n years and the 'last' payment of C
The principal remaining is [P(1+r)^n]-[C(1+r)^(n-1)]-[C(1+r)^(n-2)]-...-[C(1+r)^2]-[C(1+r)]-[C]

notice the use of brackets to separate the terms.
Since this is the 'last' payment, the principal remaining must be zero.

Thus
P(1+r)^n = C(1+r)^n-1+...+C (note the terms in the equation above) ***

The terms on the right are as follows:

C, C(1+r),C(1+r)^2,...,C(1+r)^n-1 (note that there are n terms)

This is effectively a GEOMETRIC PROGRESSION (GP) with first term equal to C
and constant ratio equal to 1+r

Remembering that the sum of a GP is a(d^n-1)/(d-1)
where a is the first term, d is the constant ratio and n is the number of terms, this becomes for this series:
For our case, a=C and d=1+r

C([(1+r)^n]-1)/(1+r-1) = (C/r)([(1+r)^n]-1)

going back to equation *** above, we can rewrite it as
P(1+r)^n = (C/r)([(1+r)^n]-1)

By doing some algebra (1+r)^n= C/(C-Pr)

giving r= [C/(C-Pr)]^(1/n) - to be used to find the appropriate rate given everything else or
n= log [C/(C-Pr)]/log(1+r) ****
which is what we need for this problem.

for this example
n = log[5000/(5000-80000*0.04)]/log 1.04
n=26.049 years

[The short answer]

This problem is a constant payment mortgage.

The approach here is to recognise that a constant payment at regular time over a time interval is an annuity.

Then present value (PV) of an annuity of constant payment C starting in the next period is given by:

PV =(C/r)[1- (1/(1+r)^n)] note the brackets

For this mortgage, the present value would be equal to the mortgage amount, P

therefore P = PV above.

Solving for n would give you the same result as equation **** above
(The reader is expected to work the algebra to verify the result.