Geometric series problem .

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January 30th 2009, 12:48 AM
mathaddict

Geometric series problem .

A house buyer borrows RM 50000 from a bank to buy a house which costs RM 70000 .The rate of interest charged by the bank is 9% per annum ,and is calculated based on the amount outstanding at the begining of each year . The house buyer is required to repay his loan in monthly installments for a period of 15 years . Assuming that the rate of interest is fixed for the entire duration of the loan , find the amount per month he has to repay the bank ?

My attempt :

It forms a sequence as follows :

$50000,50000(1.09),50000(1.09)^2,...,50000(1.09)^{n-1}$

The 15th term , $T_15=50000(1.09)^14=167086.35$
which is the amount he has to pay by the end of 15 years .

Thus , every month he has to pay $\frac{167086.35}{15\times12}=928.26$

My answer is obviously wrong . I wonder where my mistake is .. Thanks to the one who help me out ..
January 30th 2009, 02:18 AM
Grandad

Repayments

Hello mathaddict

Quote:

Originally Posted by mathaddict

A house buyer borrows RM 50000 from a bank to buy a house which costs RM 70000 .The rate of interest charged by the bank is 9% per annum ,and is calculated based on the amount outstanding at the begining of each year . The house buyer is required to repay his loan in monthly installments for a period of 15 years . Assuming that the rate of interest is fixed for the entire duration of the loan , find the amount per month he has to repay the bank ?

My attempt :

It forms a sequence as follows :

$50000,50000(1.09),50000(1.09)^2,...,50000(1.09)^{n-1}$

The 15th term , $T_15=50000(1.09)^14=167086.35$
which is the amount he has to pay by the end of 15 years .

Thus , every month he has to pay $\frac{167086.35}{15\times12}=928.26$

My answer is obviously wrong . I wonder where my mistake is .. Thanks to the one who help me out ..

Thanks for showing us your working here. What you're forgetting is that each year he pays not only the interest owing that year, but he also pays back some of the amount borrowed.

Are you allowed to use a spreadsheet to work out this answer? It's the easiest way.

If not, you'll have to let the monthly payment be x, and then say that, at the start of each new year, the amount owing

= amount owed at the start of the last year + interest payable during that year - 12x

Then work out how much interest he'll have to pay in the coming year to cover this new amount owing. Then repeat for the following year, and so on, up to the start of year 16. Then say that at the start of year 16 he owes nothing.

Grandad
January 30th 2009, 07:08 AM
Soroban

Hello, mathaddict!

This is an Amortization problem and requires a special formula.

I've seen this type of problem posted many times before.
It seem that the students are never given the formula,
. . nor are they prepared for the (very) long derivation.

Could it be that all those teachers are ignorant of the difficulties
. . involved with time-payment problems?
I'm beginning to suspect that this is so . . .

Quote:

A house buyer borrows $50,000 from a bank to buy a house.
The rate of interest 9% per annum.
The buyer is required to repay his loan in monthly installments for 15 years.
Assuming that the rate of interest is fixed for the entire duration of the loan,
find the amount per month he has to repay the bank?

The formula is: . $A \;=\;P\,\frac{i(1+i)^n}{(1+i)^n-1}$

. . where: . $\begin{Bmatrix}P &=& \text{principal} \\ i &=& \text{periodic interst rate} \\ n &=& \text{number of periods} \\ A &=& \text{periodic payment} \end{Bmatrix}$

We have: . $\begin{array}{ccc}P \:=\:50,\!000 \\<br /> i \:=\:\frac{9\%}{12} \:=\:0.0075 \\ n \:=\:15\!\cdot\!12 \:=\:180 \end{array}$

Hence: . $A \;=\;50,\!000\,\frac{0.0075(1.0075)^{180}}{1.0075^ {180}-1} \;=\;507.1332921$

Therefore, the monthly payment is: . $\boxed{\$507.13}$
January 30th 2009, 07:22 AM
mathaddict

Re :

Very big thank you to both of you !!! Soroban , just out of curioscity , how did the formula come or perhaps where did it come from ?
January 30th 2009, 08:47 AM
jonah

Quote:

Originally Posted by mathaddict

Very big thank you to both of you !!! Soroban , just out of curioscity , how did the formula come or perhaps where did it come from ?

Check this and especially this out. It is one way of deriving it. There are or course other ways.
Go through page 70 up to page 99.
Or you can just order this book if get tired of reading it online.
January 30th 2009, 01:05 PM
Grandad

Repayments

Hello mathaddict -

In fact, the interest is calculated at the start of each year for the whole year, as if the whole amount is owing for the whole year. So you have to use the values

P = 50000
i = 0.09
n = 15

to get the total payment each year, and then divide by 12, to get a monthly payment of 516.91.

Incidentally, I don't think it's really that hard to work out the formula for yourself. After $n$ years you'll find that the amount owing is

$P(1+i)^n - [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x$ , where $x$ is the annual payment

Inside the square brackets, it's a GP whose sum is $\frac{(1+i)^n - 1}{i}$

Set this sum equal to zero and solve for $x$ to obtain the formula.

Grandad
February 2nd 2009, 02:31 AM
mathaddict

Thanks all but i am having some problems with the series .

[tex] [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x[/math

$T_1=x$

$T_2=xi+x=x(1+i)$

$T_3=[x(1+i)]\cdot{i}+x$ but this is not equal to $x(1+i)^2$

Why is it so ??
February 2nd 2009, 03:21 AM
Grandad

GP

Hello mathaddict

Quote:

Originally Posted by mathaddict

Thanks all but i am having some problems with the series .

[tex] [1 + (1+i) + (1+i)^2 + ... + (1+i)^{n-1}]x[/math

$T_1=x$

$T_2=xi+x=x(1+i)$

$T_3=[x(1+i)]\cdot{i}+x$ but this is not equal to $x(1+i)^2$

Why is it so ??

Set your working out like this:

At the end of year 1, the amount owed is the original sum borrowed + the interest for the year - the total payments during the year

$= P + Pi - x$

$= P(1 + i) - x$

This, then is the amount owing at the start of year 2: the 'new $P$ ' if you like. So at the end of year 2, we'll replace $P$ by $P(1+i) - x$ , to get that the amount owing is:

$[P(1 +i) - x](1+i) - x$

$= P(1+i)^2 - x[1 + (1 + i)]$

Continue like this, multiplying by $(1+i)$ and subtracting $x$ each time. So at the end of the year 3, the amount owing is:

$P(1+i)^3 - x[1 + (1 + i) + (1 +i)^2]$

And, in a similar way, at the end of year $n$ , it is:

$P(1+i)^n - x[1 + (1 + i) + ...+ (1 +i)^{n-1}]$

So, inside the square brackets, [...], we have a GP whose first term is $1$ , common ratio $(1+i)$ and number of terms = $n$ , and whose sum is $\frac{(1+i)^n - 1}{(1+i) - 1}= \frac{(1+i)^n - 1}{i}$ .

Now if the whole sum is paid of after $n$ years, this total is zero. I.e.

$P(1+i)^n - x\frac{(1+i)^n - 1}{i} = 0$

Hence $x = \frac{iP(1+i)^n}{(1+i)^n - 1}$

OK now?

Grandad
February 2nd 2009, 03:43 AM
mathaddict

Re :

Clear enough , once again , thanks Grandad