Recurrence relations help.

Iancar29

Any help with how to find the closed formula is much appreciated guys.


Q. Your house mortgage grows at a rate of 5% per month. You originally borrow 100,000 euros for the house, and every month you make a payment of 1000 euros. How many months will it take before you owe nothing on the house?


I know for say population growth the eqn would be as follows for this example
Increase of 1.14% and original number 2000. Then...

An = (1.14) ^ n ( 2000 ) ...

Is the mortgage question similar ?

Really appreciate any help

Thanks

MathsManiac

I think you may have made an error in giving us the question. That's a completely unrealistic interest rate. 5% per month, compounded monthly, is equivalent to an annual rate (APR) of about 80%! Also, your interest the very first month would be €5000, and, since you're only paying back €1000, your mortgage would be growing rapidly for ever.

Let's suppose the question was meant to say: 0.5%. (This is a much more reasonable rate - about 6.2% per annum APR.)

Now to the question.

Have you heard of the idea of the "present value" of a future payment?

On the day you took out the mortgage, the sum of the present values of all the future payments has to equal the amount you borrowed.

The present value of the payment you make after 1month is €1000/(1.005).
The present value of the payment you make after 2 months is €1000/(1.005^2).
The present value of the payment you make after 3 months is €1000/(1.005^3).
...
The present value of the payment you make after n months is €1000/(1.005^n).

These form a geometric series, whose first term is 1000/1.005 and whose common ratio 1/1.005, and whose sum must equal €100,000.

That should be enough info for you to solve it. Post back if you're still stuck.

Iancar29

Hey MathsManiac, thanks for your reply!
Yes sorry, that should of been 0.5% .


I kinda understand the present value. So that future payment
no.1 is €1000/(1.005)
no.2 is €1000/(1.005^2).
... and so on. That i get. But how that joins the sequence i dunno.

Im still struggling to form the eqn.

Atm i have : An = r( An-1 ) ; where A is the original amount and r is that common ratio?

But this then goes to 100,000(n) = 1/1.005^n (100,000 n-1 ) ?? ... that definitely aint right from the looks of it .

Sorry for the hassle


I know the answer is n = 137 months but i dont care about that , i just want to know how to get the equation to get to there.

Yakuza

The answer is that payback occurs in month 139 (assuming the first monthly payment is at the end of month one). It is only 137 if you pay back the first €1000 immediately (in effect, borrowing €99000) (this is the difference between an "annuity immediate"and an "annuity due" (the timing of the payments))
As MM says, you need to think about the present value of your series of 1000.

Plugging in your data into the formula in the link above shows (treating the time period as months) yields:

100000 = 1000/.005 * (1 - (1/(1.005)^n).

Solve for n.

Hint : Using logs : n = ln(0.5)/ln((1/1.005)) = 138.9757

Iancar29

Yakuza wrote: »

The answer is that payback occurs in month 139 (assuming the first monthly payment is at the end of month one). It is only 137 if you pay back the first €1000 immediately (in effect, borrowing €99000) (this is the difference between an "annuity immediate"and an "annuity due" (the timing of the payments))
As MM says, you need to think about the present value of your series of 1000.

Plugging in your data into the formula in the link above shows (treating the time period as months) yields:

100000 = 1000/.005 * (1 - (1/(1.005)^n).

Solve for n.

Hint : Using logs : n = ln(0.5)/ln((1/1.005)) = 138.9757

Thanks Yakuza!

It was the (1 - (1/(1.005)^n) part that i couldn't figure out.

Solving for n was no problem then.


I'll remember the formula then to be as follows

Loan = Payment/rate * {1-(1/rate)^n}

Thats interesting about the n=139 , dunno why the loan would be one you pay back a payment straight away! Very odd.

Thanks again Guys!