4.1 Amortization Amortization method: repay a loan by means of installment payments at periodic intervals This is an example of annuity We already know how to calculate the amount of each payment Our goal: find the outstanding principal Two methods to compute it: –prospective –retrospective
Two Methods Prospective method: outstanding principal at any point in time is equal to the present value at that date of all remaining payments Retrospective method: outstanding principal is equal to the original principal accumulated to that point in time minus the accumulated value of all payments previously made Note: of course, this two methods are equivalent. However, sometimes one is more convenient than the other
Examples (p. 75-76) (prospective) A loan is being paid off with payments of 500 at the end of each year for the next 10 years. If i =.14, find the outstanding principal, P, immediately after the payment at the end of year 6. (retrospective) A 7000 loan is being paid of with payments of 1000 at the end of each year for as long as necessary, plus a smaller payment one year after the last regular payment. If i = 0.11 and the first payment is due one year after the loan is taken out, find the outstanding principal, P, immediately after the 9 th payment.
One more example… (p. 77) (Different frequency) John takes out 50,000 mortgage at 12.5 % convertible semi-annually. He pays off the mortgage with monthly payments for 20 years, the first one is due one month after the mortgage is taken out. Immediately after his 60 th payment, John renegotiates the loan. He agrees to repay the remainder of the mortgage by making an immediate cash payment of 10,000 and repaying the balance by means of monthly payments for ten years at 11% convertible semi-annually. Find the amount of his new payment.
4.2 Amortization Schedule Goal: divide each payment (of annuity) into two parts – interest and principal Amortization schedule – table, containing the following columns: –payments –interest part of a payment –principal part of a payment –outstanding principal DurationPaymentInterestPrincipal Repaid Outstanding Principal 05000.00 113875.05600.00787.054212.95 213875.05505.55881.503331.45 313875.05399.77987.282344.17 413875.05281.301105.751238.42 513875.05148.611238.440 Example: 5000 at 12 % per year repaid by 5 annual payments Amortization schedule:
Example t - 1 t Payment X Outstanding principal P Interest earned during interval (t-1,t) is i P Therefore interest portion of payment X is i P and principal portion is X - i P A 1000 loan is repaid by annual payments of 150, plus a smaller final payment. If i =.11, and the first payment is made one year after the time of the loan, find the amount of principal and interest contained in the third payment Recall: in practical problems, the outstanding principal P can be found by prospective or retrospective methods
General method 01t2 n 1111 ….. a n| present value = outst. principal at 0 ….. a n-t| outstanding principal at t interest portion of (t+1)-st payment = i a n-t| = 1 – v n-t t+1 1 principal portion of (t+1)-st payment = 1 – (1 – v n-t ) = v n-t If each payment is X then interest part of k th payment = X (1 – v n-k+1 ) principal part of k th payment = X∙v n-k+1
Example (p. 79) A loan of 5000 at 12% per year is to be repaid by 5 annual payments, the first due one year hence. Construct an amortization schedule
General rules to obtain an amortization schedule I.Take the entry from “Outs. Principal” of the previous row, multiply it by i, and enter the result in “Interest” II.“Payment” – “Interest” = “Principal Repaid” III.“Outs. Principal” of prev. row - “Principal Repaid” = “Outs. Principal” IV.Continue DurationPaymentInterestPrincipal Repaid Outstanding Principal 05000.00 113875.05600.00787.054212.95 213875.05505.55881.503331.45 313875.05399.77987.282344.17 413875.05281.301105.751238.42 513875.05148.611238.440 i = 12 %
Example (p. 80) A 1000 loan is repaid by annual payments of 150, plus a smaller final payment. The first payment is made one year after the time of the loan and i =.11. Construct an amortization schedule
4.3 Sinking Funds Alternative way to repay a loan – sinking fund method: –Pay interest as it comes due keeping the amount of the loan (i.e. outstanding principal) constant –Repay the principal by a single lump-sum payment at some point in the future
012 n interest i L iLiL iLiL ….. Loan L lump-sum payment L Lump-sum payment L is accumulated by periodic deposits into a separate fund, called the sinking fund Sinking fund has rate of interest j usually different from (and usually smaller than) i If (and only if) j is greater than i then sinking fund method is better (for borrower) than amortization method
Examples (p. 82) John borrows 15,000 at 17% effective annually. He agrees to pay the interest annually, and to build up a sinking fund which will repay the loan at the end of 15 years. If the sinking fund accumulates at 12% annually, find – the annual interest payment –the annual sinking fund payment –his total annual outlay –the annual amortization payment which would pay off this loan in 15 years Helen wishes to borrow 7000. One lender offers a loan in which the principal is to be repaid at the end of 5 years. In the mean-time, interest at 11% effective is to be paid on the loan, and the borrower is to accumulate her principal by means of annual payments into a sinking fund earning 8% effective. Another lender offers a loan for 5 years in which the amortization method will be used to repay the loan, with the first of the annual payments due in one year. Find the rate of interest, i, that this second lender can charge in order that Helen finds the two offers equally attractive.
4.4 Yield Rates Investor: –makes a number of payments at various points in time –receives other payments in return There is (at least) one rate of interest for which the value of his expenditures will equal the value of the payments he received (at the same point in time) This rate is called the yield rate he earns on his investment In other words, yield rate is the rate of interest which makes two sequences of payments equivalent Note: to determine yield rate of a certain investor, we should consider only payments made directly to, or directly by, this investor
Examples (p. 83 – p. 85) Herman borrows 5000 from George and agrees to repay it in 10 equal annual instalments at 11%, with first payment due in one year. After 4 years, George sells his right to future payments to Ruth, at a price which will yield Ruth 12% effective –Find the price Ruth pays. –Find George’s overall yield rate.
At what yield rate are payments of 500 now and 600 at the end of 2 years equivalent to a payment of 1098 at the end of 1 year? Henri buys a 15-year annuity with a present value of 5000 at 9% at a price which will allow him to accumulate a 15- year sinking fund to replace his capital at 7%, and will produce an overall yield rate of 10%. Find the purchase price of the annuity.