Question

Quick Question:

Compound Interest Calculation -
Tim borrows 80,000 from the bank at 5.02% p.a.

a) Assuming interest is calculated monthly, how much would Tim need to pay weekly to ensure the loan is paid off in 2 years?

b)Assuming interest is calculated monthly, how much would Tim need to pay weekly to ensure the loan is paid off in 3 years?

plz hlp?

Answer 1

well the annual interest needs to be changed to monthly

5.02/12 = % per month

the number of time period needs to be adjusted

t = 2 x 12 = 24

so the future value of the loan is

\[A = 80000(1 + \frac{\frac{5.02}{12}}{100})^{24}\]

after calculating the future value divide it by 24 to get the monthly repayment

Answer 2

for (b) the time periods increase to 36

so calculate the future value for 36 months of interest using the same formula

the divide it by 3 x 52 to get a weekly amount

Answer 3

I think, this should not be that easy as the Principal changes every month

Answer 4

that's what i was curious about... at a fixed amount, this is fine, but is there a formula that takes into account the repayments and the new Principal remaining?

Answer 5

We may proceed the same way.
To keep calculations simple, let the monthly payment be \(M\)
(we can divide by 4 to get the weekly payment later)

Answer 6

don't divide monthly by 4... as this will only result in 48 repayments instread of the necessary 52 per year

Answer 7

That is fine, the interest is being calculated monthly, so we need to anchor to months

Answer 8

well you can treat the same way as an annuity

\[N = \frac{M \times (1 + r)^n -1}{r(1 + r)^n}\]

N = the amount borrowed, M = montly repayments
r = interest rate per compounding period and n = number of time periods

Answer 9

x = ((y* ((1+(5.02/1200))^36))-1)/ (5.02/1200)*((1+(5.02/1200))^36)

so is this about right?
y = monthly repayments = 248.646

Answer 10

I don't think so as you only repay $38700 over the 3 years

Answer 11

\(\Huge N = \frac{M \times (1 + r)^n -1}{r(1 + r)^n}\)

\(\Huge 80000 = \frac{M \times (1 + \frac{5.02}{1200})^n -1}{\frac{5.02}{1200}(1 + \frac{5.02}{1200})^n}\)


\(\Huge 80000 = \frac{M \times (1 + \frac{5.02}{1200})^{36} -1}{\frac{5.02}{1200}(1 + \frac{5.02}{1200})^{36}}\)

==> M = $$248.64 / month?

Answer 12

where am i going wrong?

Answer 13

I think the amount need to closer to $700 per week

Answer 14

I'm getting $3510 for monthly payment

Answer 15

Presuming Tim will be paying it in 4 installements each month, dividing that by 4 gives the weekly payment of $877.5

Answer 16

ok, could u help me find where im messing up my sum though plz?
https://goo.gl/cfzl0r

Answer 17

computation will be easier if we don't use the formula as we don't understand it much

Answer 18

well if you make M the subject its

\[M = \frac{80000 \times r(1+r)^{36}}{(1 + r)^{36} - 1}\]

Answer 19

Let \(M\) be the required monthly payment so that the loan amount \(P\) to be paid off in 24 months
Let \(i\) = monthly interest rate = \(\frac{0.0502}{12}\)

Then, after 1 month the balance amount, \(B_1\) is given by
\[\large B_1 = P(1+i) - M\]

yes ?

Answer 20

I'd suggest the repayment is about $2400 per month

Answer 21

ok, i'm following so far

Answer 22

what would be the Balance amount, \(B_2\) after \(2\) months ?

Answer 23

at the end of 2nd month, the interest is calculated on \(B_1\) and the monthly payment is subtracted :

\[\large B_2 = B_1(1+i) - M\]

Answer 24

So it seems we're ending up with a recurrence relation
\[B_{n+1} = B_{n}(1+i) - M\]

Answer 25

You need to find \(M\) such that the loan gets paid off in 24 months,

which is same as saying \(B_{24} = 0\)

Answer 26

good point! first notice, we have assumed that the payments are done at the end of each month

Answer 27

So the balance amount at the end of 24th month has to be 0 for the loan to be closed in 24 months

Answer 28

consequently \(B_{24}\) has to be equal to \(0\), and not \(M\)

Answer 29

you don't want any balance to be pending after 24 months right ?

Answer 30

agreed

Answer 31

so first month = B1 = 80334.7 - M

Answer 32

Lets solve M for the general case, it would be simple, trust me
80334.7 and all looks messy

Answer 33

cool, sounds good

Answer 34

Let \(n+1=24\), and \(B_{23+1} = 0\)

\[B_{n+1} = B_{n}(1+i) - M\]

becomes
\[B_{23+1} =B_{23}(1+i)-M = 0 \implies M = B_{23}(1+i)\]

Answer 35

Maybe next substitute \(B_{23} = B_{22}(1+i) -M\) and get
\[M = (B_{22}(1+i)-M)(1+i) = B_{22}(1+i)^2 - M(1+i)\]
recursively substituting we end up with
\[M = (B_{22}(1+i)-M)(1+i) = B_{0}(1+i)^{24} - M(1+i)^{23} - M(1+i)^{22} - \cdots -M(1+i)\]

isolating \(M\) we get
\[M = \dfrac{B_0(1+i)^{24}}{(1+i)^{24}+(1+i)^{23} + \cdots + (1+i)+1}\]

Answer 36

Not hard to see that the bottom is a geometric series, use the partial sum formula to simplify

Answer 37

wow!... this is almost factorial looking

Answer 38

ok... whats the partial sum formula?

Answer 39

\[x^n+x^{n-1} + \cdots + x+1 = \dfrac{x^n-1}{x-1}\]

Answer 40

ok, so
=
\(\Large \frac {(1 + i)^{24} - 1}{(1 + i) - 1}\)

\(\Large \frac {(1 + i)^{24} - 1}{i}\) ??

Answer 41

\(\large M = \dfrac{B_0(1+i)^{24}}{(1+i)^{24}+(1+i)^{23} + \cdots + (1+i)+1}\)


\(\Large M = \frac{i \times B_0(1+i)^{24}}{(1+i)^{24} + 1} \)

??

Answer 42

looks good ! plugging that in and rearranging we end up with a nice looking formula for monthly payment :

\[M = B_0*i*\dfrac{(1+i)^n}{(1+i)^n-1}\]

where \(B_0\) = loan amount (starting principal)
\(i\) = periodic interest rate
\(n\) = total number of periods

Answer 43

i think i did somethin wrong, as i tried to plug that in and got 333.66 recurring as my answer?? :(

Answer 44

http://www.wolframalpha.com/input/?i=80000*0.0502%2F12*%281%2B0.0502%2F12%29%5E24%2F%28%281%2B0.0502%2F12%29%5E24-1%29

Answer 45

yep, found my mistake, 3510.42 = monthly repayment for 2 yr period?

Answer 46

Yes, for part b, simply change 24 to 36

Answer 47

yes, match!

Answer 48

awesome, thanks @ganeshie8 and @campbell_st !!

Answer 49

don't forget the question is asking weekly payment, not monthly

Answer 50

sáll good, can multiply it out to the total amount, then divide by 104 for 2 yrs, thanks tho

Answer 51

that looks like a better idea ! xD

Answer 52

you want to do : M*24 / 104 is it ?

Answer 53

@mathstudent55 u been typing a novel in the background this whole time... what chu workin on man?

Answer 54

\(\large A = \dfrac{Pr(1 + r)^n}{(1 + r)^n - 1} \)

\(\large A = \dfrac{80000\times \frac{0.0502}{12}(1 + \frac{0.0502}{12})^{24}}{(1 + \frac{0.0502}{12})^{24} - 1} \)

\(\large A = $3510.43/month \)

\(\large A = $810.10/week\)

--------------------------------------

\(\large A = \dfrac{Pr(1 + r)^n}{(1 + r)^n - 1} \)

\(\large A = \dfrac{80000\times \frac{0.0502}{12}(1 + \frac{0.0502}{12})^{36}}{(1 + \frac{0.0502}{12})^{36} - 1} \)

\(\large A = $2398.39/month \)

\(\large A = $532.98/week\)

Answer 55

= $810.10 per week for a 2 yr loan,
and
= $553.47 per week for a 3 yr loan... i think?
wait how did u get 532 per week?

Answer 56

$553.47 looks correct

http://www.wolframalpha.com/input/?i=%2880000*0.0502%2F12*%281%2B0.0502%2F12%29%5E36%2F%28%281%2B0.0502%2F12%29%5E36-1%29%29*36%2F%2852*3%29

Answer 57

Sorry. i copied the number incorrectly.
$2398.39/month = $553.47/week

Answer 58

You are correct.

Answer 59

all good, thanks both!

Answer 60

and sorry it didnt turn out to be a quick question (looks like i lied in the OP) :-|

Answer 61

It is a quick question if you had memorized the formula

Answer 62

Too many complicated formulas to be memorized in financial math