Jack1 one year ago 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?

1. campbell_st

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

2. campbell_st

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

3. ganeshie8

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

4. Jack1

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?

5. ganeshie8

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)

6. campbell_st

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

7. ganeshie8

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

8. campbell_st

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

9. Jack1

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

10. campbell_st

I don't think so as you only repay $38700 over the 3 years 11. Jack1 $$\Huge N = \frac{M \times (1 + r)^n -1}{r(1 + r)^n}$$ $$\Huge 80000 = \frac{M \times (1 + \frac{5.02}{1200})^n -1}{\frac{5.02}{1200}(1 + \frac{5.02}{1200})^n}$$ $$\Huge 80000 = \frac{M \times (1 + \frac{5.02}{1200})^{36} -1}{\frac{5.02}{1200}(1 + \frac{5.02}{1200})^{36}}$$ ==> M =$$248.64 / month? 12. Jack1 where am i going wrong? 13. campbell_st I think the amount need to closer to$700 per week

14. ganeshie8

I'm getting $3510 for monthly payment 15. ganeshie8 Presuming Tim will be paying it in 4 installements each month, dividing that by 4 gives the weekly payment of$877.5

16. Jack1

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

17. ganeshie8

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

18. campbell_st

well if you make M the subject its $M = \frac{80000 \times r(1+r)^{36}}{(1 + r)^{36} - 1}$

19. ganeshie8

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 ?

20. campbell_st

I'd suggest the repayment is about $2400 per month 21. Jack1 ok, i'm following so far 22. ganeshie8 what would be the Balance amount, $$B_2$$ after $$2$$ months ? 23. ganeshie8 at the end of 2nd month, the interest is calculated on $$B_1$$ and the monthly payment is subtracted : $\large B_2 = B_1(1+i) - M$ 24. ganeshie8 So it seems we're ending up with a recurrence relation $B_{n+1} = B_{n}(1+i) - M$ 25. ganeshie8 You need to find $$M$$ such that the loan gets paid off in 24 months, which is same as saying $$B_{24} = 0$$ 26. ganeshie8 good point! first notice, we have assumed that the payments are done at the end of each month 27. ganeshie8 So the balance amount at the end of 24th month has to be 0 for the loan to be closed in 24 months 28. ganeshie8 consequently $$B_{24}$$ has to be equal to $$0$$, and not $$M$$ 29. ganeshie8 you don't want any balance to be pending after 24 months right ? 30. Jack1 agreed 31. Jack1 so first month = B1 = 80334.7 - M 32. ganeshie8 Lets solve M for the general case, it would be simple, trust me 80334.7 and all looks messy 33. Jack1 cool, sounds good 34. ganeshie8 Let $$n+1=24$$, and $$B_{23+1} = 0$$ $B_{n+1} = B_{n}(1+i) - M$ becomes $B_{23+1} =B_{23}(1+i)-M = 0 \implies M = B_{23}(1+i)$ 35. ganeshie8 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 $M = \dfrac{B_0(1+i)^{24}}{(1+i)^{24}+(1+i)^{23} + \cdots + (1+i)+1}$ 36. ganeshie8 Not hard to see that the bottom is a geometric series, use the partial sum formula to simplify 37. Jack1 wow!... this is almost factorial looking 38. Jack1 ok... whats the partial sum formula? 39. ganeshie8 $x^n+x^{n-1} + \cdots + x+1 = \dfrac{x^n-1}{x-1}$ 40. Jack1 ok, so = $$\Large \frac {(1 + i)^{24} - 1}{(1 + i) - 1}$$ $$\Large \frac {(1 + i)^{24} - 1}{i}$$ ?? 41. Jack1 $$\large M = \dfrac{B_0(1+i)^{24}}{(1+i)^{24}+(1+i)^{23} + \cdots + (1+i)+1}$$ $$\Large M = \frac{i \times B_0(1+i)^{24}}{(1+i)^{24} + 1}$$ ?? 42. ganeshie8 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 43. Jack1 i think i did somethin wrong, as i tried to plug that in and got 333.66 recurring as my answer?? :( 44. ganeshie8 45. Jack1 yep, found my mistake, 3510.42 = monthly repayment for 2 yr period? 46. ganeshie8 Yes, for part b, simply change 24 to 36 47. Jack1 yes, match! 48. Jack1 awesome, thanks @ganeshie8 and @campbell_st !! 49. ganeshie8 don't forget the question is asking weekly payment, not monthly 50. Jack1 sáll good, can multiply it out to the total amount, then divide by 104 for 2 yrs, thanks tho 51. ganeshie8 that looks like a better idea ! xD 52. ganeshie8 you want to do : M*24 / 104 is it ? 53. Jack1 @mathstudent55 u been typing a novel in the background this whole time... what chu workin on man? 54. mathstudent55 $$\large A = \dfrac{Pr(1 + r)^n}{(1 + r)^n - 1}$$ $$\large A = \dfrac{80000\times \frac{0.0502}{12}(1 + \frac{0.0502}{12})^{24}}{(1 + \frac{0.0502}{12})^{24} - 1}$$ $$\large A = 3510.43/month$$ $$\large A = 810.10/week$$ -------------------------------------- $$\large A = \dfrac{Pr(1 + r)^n}{(1 + r)^n - 1}$$ $$\large A = \dfrac{80000\times \frac{0.0502}{12}(1 + \frac{0.0502}{12})^{36}}{(1 + \frac{0.0502}{12})^{36} - 1}$$ $$\large A = 2398.39/month$$ $$\large A = 532.98/week$$ 55. Jack1 =$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? 56. ganeshie8 57. mathstudent55 Sorry. i copied the number incorrectly.$2398.39/month = \$553.47/week

58. mathstudent55

You are correct.

59. Jack1

all good, thanks both!

60. Jack1

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

61. ganeshie8

It is a quick question if you had memorized the formula

62. ganeshie8

Too many complicated formulas to be memorized in financial math