# Loan Payment

• Jan 28th 2010, 05:20 AM
Macleef
Loan Payment
I don't think I did the question right. Please check my solution and correct it! Is there a simple way of calculating this (it's very tedious...)?

Quote:

Consider a \$100,000 loan with the following 48-month repayment schedule. You are to repay an equal amount every month for the first 24 months, and every two months afterwards. There are 36 equal installments in total. The first repayment date is one month from now. The effective monthly interest rate is 0.50%. How much is each of the installments?
First 24 months (monthly):
PV = X(1/1.005)^1 + X(1/1.005)^2 + X(1/1.005)^3 +...+X(1/1.005)^24
= X(1/1.005^1 + ... + 1/1.005^24) = 22.5628662X

Last 24 months (bi-monthly):
PV = X(1/1.005)^2 + X(1/1.005)^4 + X(1/1.005)^6 + ... + X(1/1.005)^12
PV = X(1/1.005^2 + 1/1.005^4 + 1/1.005^6 + ... + 1/1.006^12
= 5.79497859X

(5.79497859X)/ (1.005^12) = 5.45832128X

(monthly and bi-monthly total):
PV = 5.45832128X + 22.5628662X = 28.0211875X

(Installments):
100,000 = 28.0211875X
X = 3568.73
• Jan 28th 2010, 07:04 PM
TKHunny
22.563 is good.

Whoops! 5.795 is no good. You have only six payments. Where's the other six?

Whoops! Even if 5.795 would have been right, 5.458 is no good. You came back only one year. Make that exponent 24.

You're definitely in the neighborhood. I get 3072.514
• Jan 29th 2010, 06:31 PM
Wilmer
Another way is use an equivalent rate during the last 2 years:
1.005^12 = (1 + x)^6 : x = 1.005^2 - 1

100000(1.005)^48 = p(1.005^24)[(1.005^24 - 1)/.005] + p[(1 + x)^12 -1]/x
Solve for p: 3072.51448....
• Jan 30th 2010, 07:16 AM
TKHunny
3) Or, do the whole thing monthly and subtract a slight variation of Wilmer's structure.

There are often many ways to proceed.