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Unformatted text preview: 2010 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1802 Financial Mathematics LN4: Amortization Schedules and Sinking Funds (Chapter 5) Amortization Method The borrower repays the lender by a series of payments at regular intervals. The payment is applied first to interest due on the unpaid balance with whatever is left going as principal repayment to reduce the loan balance. Let L = amount borrowed, n = number of payments R = level payment, i = effective rate of interest per payment period. R R R     1 2 ··· n L Clearly, L = Ra n  i . Let B k be the outstanding balance just after the k th payment. B k = Ra n k  i (prospective) OR B k = L (1 + i ) k Rs k  i (retrospective). Note that B k + t = (1 + i ) t B k , < t < 1 and B k +1 = (1 + i ) B k R . Example. A loan is being repaid by 15 annual payments at the end of each year. The first 5 installments are 4,000 each, the next 5 are 3,000 each, and the final 5 are 2,000 each. Find expressions for the outstanding loan balance immediately after the second 3,000 installment. Soln: 1. Prospectively The second 3,000 installment occurs at time 7. OB P 7 = 3 , 000 a 3  + 2 , 000 v 3 a 5  = 2 , 000 a 8  + 1 , 000 a 3  = 1 , 000( a 3  + 2 a 8  ) . 1 2. Retrospectively OB R 7 = L (1 + i ) 7 4 , 000 s 5  (1 + i ) 2 3 , 000 s 2  where L = 2 , 000 a 15  + 1 , 000 a 10  + 1 , 000 a 5  OR L = 4 , 000 a 5  + 3 , 000 v 5 a 5  + 2 , 000 v 10 a 5  = a 5  (4 , 000 + 3 , 000 v 5 + 2 , 000 v 10 ) . 2 Splitting a Payment into Principal and Interest Let P k = principal repaid in the k th installment and I k = amount of interest paid in the k th installment Assuming level installments of R , R = P k + I k . P k = R I k = R i · OB k 1 = R iRa n k +1  = R ( 1 i ( 1 v n k +1 i ) ) = Rv n k +1 and I k = i · OB k 1 = i · R ( 1 v n k +1 i ) = R (1 v n k +1 ) = R P k . Amortization Schedule for a Loan of a n  Repaid Over n Periods at Rate i 2 Payment Interest Principal Outstanding a n  1 1 ia n  = 1 v n v n a n  v n = a n 1  2 1 ia n 1  = 1 v n 1 v n 1 a n 1  v n 1 = a n 2  ....
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