Financial mathematics increasing annuities

Question

I am having trouble solving this problem

Julie bought a house with a 100,000 mortgage for 30 years being repaid with payments at the end of each month at an interest rate of 8% compounded monthly. If Julie pays an extra 100 each month, what is the outstanding balance at the end of 10 years immediately after the 120th payment?

My attempt:

I first want to find the deposit per month.

I let $D$ be the deposit per month and since it increases by 100 each payment, I used an increasing annuity,

$D*100(Ia_{30|0.08}) = 100,000$

However, the $D$ I got was 8.12, which is clearly not right.

Can someone help?

Answer 1

At a nominal interest rate of $i^{(12)}=8\%$ compounded monthly, the effective interest rate per month is $i_m=\frac{i^{(12)}}{12}=0.67\%$. Let $L=100,000$. Without extra payment we have for $n=12\times 30=360$ months $$ L=P\,a_{\overline{n}|i_m}\quad\Longrightarrow P=\frac{L}{a_{\overline{n}|i_m}}=\frac{100,000}{a_{\overline{360}|0.67\%}}=\frac{100,000}{136.28}=733.76 $$ With an extra monthly payment $Q=100$ we have a periodic payment $P+Q=833.76$. Then we can find the outstanding loan balance $B_t$ $$ B_{t}=L(1+i_m)^{t}-(P+Q)\,s_{\overline{t}|i_m} $$ and at $t=120$ we have $$ B_{120}=\overbrace{100,000\times\underbrace{(1+0.67\%)^{120}}_{2.22}}^{221,964.02}\;-\,\overbrace{833.76\times\underbrace{\,s_{\overline{120}|0.67\%}}_{182.95}}^{152,533.92}=69,430.10 $$ Without the extra payment the OLB would be $$ B_{t}=L(1+i_m)^{t}-P\,s_{\overline{t}|i_m} $$ and at $t=120$ we would have $$ B_{120}=\overbrace{100,000\times\underbrace{(1+0.67\%)^{120}}_{2.22}}^{221,964.02}\;-\,\overbrace{733.76\times\underbrace{\,s_{\overline{120}|0.67\%}}_{182.95}}^{134,239.32}=87,724.70 $$ Using the prospective method we have $$ B_t=P\,a_{\overline{n-t}|i_m}\quad\Longrightarrow\; B_{120}=733.76\,a_{\overline{240}|0.67\%}=733.76\times 119.55=87,724.70 $$ that is obviously equal to the value founded with the retrospective method.

With an extra payment we will shorten the length $n$ of the loan repayment to a new $N$ $$ L=(P+Q)\,a_{\overline{N}|i_m}=(P+Q)\frac{1-(1+i_m)^{-N}}{i_m} $$ and then solving for $N$ $$ N=-\frac{\log\left(1-i_m\frac{L}{P+Q}\right)}{\log(1+i_m)}=241.9084704\approxeq 242 $$ that is the lenght of the loan $n=360$ has been shortened to $N=240$ if we make an extra payment.

Using the retrospective method we have $$ B_t=(P+Q)\,a_{\overline{N-t}|i_m}\quad\Longrightarrow B_{120}=833.76\,a_{\overline{122}|0.67\%}=833.76\times 83.27=69,430.10 $$

Answer 2

HINT:

I have no formulas, but good old spreadsheet works wonders: I calculated the monthly principal to be paid: $277.7777

Then the sum of the all the interest compounded monthly over the thirty years: $120,333.3333

TOTAL principal+interest=$220,333.3333

So amount to pay per month is $612.037 during 360 months.

If you decide to instead pay $712.037 during 120 months,

you will have outstanding balance at month 121 of $120,603.7037

(The decimals are there to show data is not rounded to cents).

Depending on the bank or country practices... I forgot to estimate the outstanding balance if those $100 go directly to principal: you will finish to pay in 317 months, thus reducing both the time and interest.

Outstanding Principal just after month 120 (no interest)=$62,145.11041