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#1
05-12-2010, 08:05 PM
 dubyajay Member Join Date: May 2009 College: UT in Austin - postgrad Posts: 60
Practice exam problem baffler

Ok, this may not baffle most of you but could anyone tell me where my thinking went wrong?

If a loan at 4% per annum is repaid by level annual installments that start one year after the loan is given, and the amount of principal in the first installment = \$100 and the principal in the last installment is \$237 - what is the amount of the original loan?

SO - I thought since P sub 1 (principal in the first payment) = R (v ^ n -t +1) and time = 1, so P sub 1 = R v^n = 100. And since P sub n = R v^n-n+1 would equal Rv which = 237... you could solve for R from the second equation and n from the first. I got n = 21 and R = 227.88 and solved for L.

The problem is, the answer given says for a level-payment loan, the principal repaid is in geometric progression w/ common ratio (1+i) (which I understand). So, 100(1+i)^(n-1) = 237 - this is the part I don't get. Why would the exponent be n-1 instead of n? Then they go on to say that n=23 and L = sum of principal repayments = 100 s angle 23 at 4% = 3667.79 (for all you people dying to know if you got the right answer).

Also, is my way completely wrong or could you also get to the right answer from my line of thinking but I did something incorrectly along the way?
#2
05-12-2010, 08:22 PM
 brandond Member Join Date: May 2008 Location: Kansas Studying for 3/MFE Posts: 249

Try it for a smaller number of payments, like 10 or something. If your first principal payment is at t=1 to get from t=1 to t=10 you multiply by (1+i)^9 or (1+i)^n-1 in general
#3
05-12-2010, 09:33 PM
 Laelsa Member Join Date: Nov 2007 Studying for C. College: Texas Favorite beer: Celebratory libations! Posts: 97

I think of it like this... The principal is found after the interest has accrued on the outstanding balance. P1, the first principal value, does not exist until the end of the 1st period. Pn does not exist until the end of the last period. If we are at t=n, then we must go back exactly n-1 periods to find the value of P1.

The key, to me, is the idea of allowing the interest to build before applying the payment and then, lastly, finding the principal. So there are n-1 periods between the 1st and nth principal.
#4
05-12-2010, 09:38 PM
 Laelsa Member Join Date: Nov 2007 Studying for C. College: Texas Favorite beer: Celebratory libations! Posts: 97

A side thought... If they asked, "What is the total interest paid on the loan?", would the answer be L*i(S angle n)? (the FV of the first period's interest as an annuity)
#5
05-12-2010, 10:12 PM
 Parasitic Ray Gun Member Join Date: Apr 2010 Studying for C College: University of State Posts: 167

Quote:
 Originally Posted by dubyajay SO - I thought since P sub 1 (principal in the first payment) = R (v ^ n -t +1) and time = 1, so P sub 1 = R v^n = 100. And since P sub n = R v^n-n+1 would equal Rv which = 237... you could solve for R from the second equation and n from the first. I got n = 21 and R = 227.88 and solved for L. ... Also, is my way completely wrong or could you also get to the right answer from my line of thinking but I did something incorrectly along the way?
Your method is correct up until Rv=237. You should find R=246.48, and then n=-ln(100/246.48)/ln(1.04)=23. Finally L=R*(1-1.04^(-23))/0.04=3661.91.

Quote:
 Originally Posted by Laelsa A side thought... If they asked, "What is the total interest paid on the loan?", would the answer be L*i(S angle n)? (the FV of the first period's interest as an annuity)
No. You could find the interest as the sum of annual payments less the sum of principal repaid, however: Rn - L = R*(n - a angle n).
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#6
05-13-2010, 03:22 PM
 dubyajay Member Join Date: May 2009 College: UT in Austin - postgrad Posts: 60

Thank you Ray Gun! I get stressed and make stupid mistakes. I see now what I did...
#7
05-13-2010, 10:10 PM
 Candidus Member Join Date: May 2010 Location: Greater Podunk, USA Studying for 2 College: Indiana University Alumnus Favorite beer: Glenlivet French Oak Posts: 58

It seems to me that there is a shortcut here. Since installments are annual, the last year of the loan is equivalent to a simple loan; the principal must have been 237 at the beginning of the last year, so that the loan will be paid off at the end. So you can reason that the total payment is 1.04 x 237 = 246.48. Knowing that, you also know that the interest paid at the end of the first year is 246.48 - 100 = 146.48. This is the annual interest on the original principal, since until the end of the first year, no principal payments have been made yet. Dividing that result by the annual interest should get you the principal, 146.48 / 0.04 = 3662.00 -- no logs, no powers, no counting periods, no rounding errors. You could probably do the math in your head if the lights went out and you solar calculator died. On the other hand, I don't know how many questions are really going to allow this kind of loop hole, so I probably won't spend a lot of time honing this line of reasoning.

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