Actuarial Exam Course: Financial Mathematics Instructor: Dr. Chun Yu School of Statistics Jiangxi University of Finance and Economics Fall 2015 1 / 71
Module 3: Loan Repayment 1 The Amortization method of Loan Repayment 2 A Variable Payment Loan 3 Formulas for Level Payment Loan Amortization 4 Looking Forward and Looking Back 5 Monthly Payment Loans 6 An Installment Loan Example 7 Sinking Fund Repayment of a Loan 8 Capitalization of Interest and Negative Amortization 2 / 71
Section 3.1: The Amortization Method of Loan Repayment The amortization method is the most common method of loan repayment When a payment is made, it must be first applied to pay interest due and then any remaining part of the payment is applied to pay principal Example: Consider a loan for 30,000 with level payments to be made at the end of each year for 5 years at an annual rate of 8%. We already know how to find the annual payment: P = 7, 513.69. We will look at the first two payments to show how the method is applied. 3 / 71
The Amortization Method of Loan Repayment Payment 1: Beginning balance = 30, 000 Interest due = 30, 000(0.08) = 2, 400 Payment made = 7, 513.69 Interest paid = 2, 400 Principal paid = 7, 513.69 2, 400.00 = 5, 113.69 Balance after payment = 30, 000 5, 113.69 = 24, 886.31 Payment 2: Beginning balance = 24, 886.31 Interest due = 24, 886.31(0.08) = 1, 990.90 Payment made = 7, 513.69 Interest paid = 1, 990.90 Principal paid = 7, 513.69 1, 990.90 = 5, 522.79 Balance after payment = 24, 886.31 5, 522.79 = 19, 363.52 4 / 71
The Amortization Method of Loan Repayment The following table (3.1) shows the result of amortizing the loan over all 5 years Year Payment Interest Principal Balance 0 30,000.00 1 7,513.69 2,400.00 5,113.69 24,886.31 2 7,513.69 1,990.90 5,522.79 19,363.52 3 7,513.69 1,549.08 5,964.61 13,398.90 4 7,513.69 1,071.91 6,441.78 6,957.12 5 7,513.69 556.57 6,957.12 0.00 You should check the detail on a few more lines to verify your understanding of the method 5 / 71
The Amortization Method of Loan Repayment There are some key points here: The final balance is 0. The level payment pays off the loan as intended. As the balance declines over time the amount of interest due in each period decreases As the interest due goes down, the amount of principal paid in each period increases Note: all loan payments are assumed to be made at the end of the period under the amortization method. 6 / 71
Section 3.2: A Variable Payment Loan Many loans have payments that vary over time Example (3.3) A borrower would like to borrow 30,000 at 8% for 5 years, but would like to pay only 5,000 for the first two years and then catch up with a higher payment for the final three years. What is the payment for the final 3 years? 7 / 71
A Variable Payment Loan Solution: First we will find the loan balance in 2 years if the payment each year is 5000: the balance is 24,592. After the second payment of 5000, the borrower owes 24,592. He can pay this off in 3 years with a higher payment. The payment for the final 3 years is 9,542.52. 24, 592 = P 1 ( ) 1 3 1.08 0.08 P = 9, 542.52 8 / 71
A Variable Payment Loan Below we show the amortization table (3.4) for the loan. Year Payment Interest Principal Balance 0 30,000.00 1 5,000.00 2,400.00 2,600.00 27,400.00 2 5,000.00 2,192.00 2,808.00 24,592.00 3 9,542.52 1,967.36 7,575.16 17,016.84 4 9,542.52 1,361.35 8,181.17 8,835.67 5 9,542.52 706.85 8,835.67 0.00 9 / 71
A Variable Payment Loan Exercise (3.5) A borrower would like to borrow 30,000 at 8% for 5 years, but would like to pay only 4,000 for the first two years and then catch up with a higher payment for the final three years. What is the payment for the final 3 years? 10 / 71
A Variable Payment Loan Exercise (3.5) A borrower would like to borrow 30,000 at 8% for 5 years, but would like to pay only 4,000 for the first two years and then catch up with a higher payment for the final three years. What is the payment for the final 3 years? Answer: 10,349.63 11 / 71
Some Notation For a loan with periodic interest rate i: Loan Payment at time k: P mt k Loan Balance after P mt k is made: Bal k Principal paid in period k: P Rin k Note: the loan amount is Bal 0 For k 1 the amortization method is described by the recursive relations: Interest paid in P mt k+1 : Int k+1 = i(bal k ) Principal paid in P mt k : P Rin k+1 = P mt k+1 i(bal k ) Loan Balance after P mt k+1 : Bal k+1 = Bal k P Rin k+1 = (1 + i)bal k P mt k+1 12 / 71
Section 3.3: Formulas for Level Payment Loan Amortization It can be shown that for a level-payment loan with payment PMT: Interest paid in P mt t : Int t = P MT (1 v n t+1 ) (1) Principal paid in P mt t : P Rin t = P MT v n t+1 (2) 13 / 71
Formulas for Level Payment Loan Amortization Example (3.7) For the loan in table (3.1) we had n = 5, i = 0.08 and P MT = 7, 513.69. Then t = 4, n t + 1 = 2 and the principal in the 4th payment is P Rin 4 = 7513.69 1.08 2 = 6441.78 14 / 71
Formulas for Level Payment Loan Amortization Exercise (3.8) An annual loan for 10 years has interest rate 6% and level payment 1000. Find the amount of principal and interest in the 6th payment. 15 / 71
Exercise (3.8) An annual loan for 10 years has interest rate 6% and level payment 1000. Find the amount of principal and interest in the 6th payment. Answer: Principal 747.26; Interest 252.74 16 / 71
Formulas for Level Payment Loan Amortization Note that for a level payment loan one can say in general that Example (3.9) P Rin n+k = (1 + i) k P Rin n (3) For an 8% level payment loan in table (3.1), the amount of principal in the second payment is 5,522.79. Find the amount of principal in the 4th payment. Using formula (2) we have P Rin 2 = v (n 2+1) P MT = 5, 522.79 P Rin 4 = v (n 4+1) P MT = (1 + i) 2 v 4 P MT = 1.08 2 (5, 522.79) = 6441.78 17 / 71
Formulas for Level Payment Loan Amortization Exercise (3.10) For a 6% level payment loan, the amount of principal in the first payment is 5,321.89. Find the amount of principal in the 4th payment. 18 / 71
Exercise (3.10) For a 6% level payment loan, the amount of principal in the first payment is 5,321.89. Find the amount of principal in the 4th payment. Answer: 6,338.46 19 / 71
Section 3.4: Looking Forward and Looking Back Looking Forward: the Prospective Method Let us look again at the example loan from Table (3.1). The loan was for 30,000 at 8% annually for 5 years. The annual payment was 7,513.69 How to find the loan balance after the third payment? (More than one way!) New method: the loan balance after a payment is the present value of the remaining payments at the time of payment. So the loan balance after the third payment should be the present value of the remaining two payment at time 3 Check this! 20 / 71
Section 3.4: Looking Forward and Looking Back Method 1: Bal 3 = Bal 2 P Rin 3 = 19,363.52-5,964.61 = 13,398.91 Method 2: Bal 3 = (1 + i)bal 2 P mt 3 = (1.08)(19,363.52) - 7,513.69 = 13,398.91 New method: Bal 3 = P V = P a 2 0.08 = (7, 513.69) [ 1 ( 1.08) 1 2 0.08 ] = 13,398.90 21 / 71
Looking Forward: the Prospective Method Exercise (3.12) A loan made at an annual rate of 6.5% has 7 remaining payments of 950. What is the loan balance? 22 / 71
Looking Forward: the Prospective Method Exercise (3.12) A loan made at an annual rate of 6.5% has 7 remaining payments of 950. What is the loan balance? Answer: 5210.29 23 / 71
Looking Forward: the Prospective Method In actuarial notation, for a level payment loan with periodic payment P MT at a rate i for n periods, the balance after payment k is: Prospective balance = P MT a n k i 24 / 71
Looking Back: the Retrospective Method We just saw that you can find a loan balance by looking at the value of your remaining future obligations What about the past? Think about this by looking at what is owed on a loan and what has already been paid any point in time If the loan was for an amount P V and no payments were made by time k, you would owe P V (1 + i) k If you have actually made payments at times 1,..., k, then you have reduced the amount of that obligation by the future value at time k of those payments F V (Payments made at times 1,..., k) 25 / 71
Looking Back: the Retrospective Method If you have actually made payments at times 1,..., k, then you have reduced the amount of that obligation by the future value at time k of those payments F V (Payments made at times 1,..., k) The Retrospective balance is P V (1 + i) k F V (Payments made at times 1,..., k) For a level payment loan with payment P MT, the balance can be expressed in actuarial notation as P V (1 + i) k P MT s k (4) 26 / 71
Looking Back: the Retrospective Method Example (3.16) Use the retrospective method to check the balance after the third payment for the loan in Table (3.1) Solution The loan was for 30,000 at 8% annually for 5 years. The annual payment was 7,513.69. If no payments had been made you would owe 30, 000(1.08) 3 = 37, 791.36 You actually made 3 payments of 7,513.69. The future value of those payment at time 3 is 24,392.44. The total obligation (balance) is 37, 791.36 24, 392.44 = 13, 398.92 27 / 71
Looking Back: the Retrospective Method Exercise (3.17) A loan for 40,000 at a 7% annual rate has an annual payment of 9,755.63. Find the balance after the 4th payment. 28 / 71
Looking Back: the Retrospective Method Exercise (3.17) A loan for 40,000 at a 7% annual rate has an annual payment of 9,755.63. Find the balance after the 4th payment. Answer: 9117.40 29 / 71
Looking Forward and Looking Back More challenging questions Example (3.18) A loan at 10% annually has an initial payment of 100, and 9 further payments. The payment amount increases by 2% each year. Find the loan balance immediately after the fourth payment. Solution The payments are 100, 100(1.02), 100(1.02) 2,..., 100(1.02) 9. Immediately after the 4th payment the remaining payments are 100(1.02) 4,..., 100(1.02) 9. Using the prospective method, the balance immediately after the 4th payment is the present value of those remaining payments. 30 / 71
Example (3.18) Solution Bal 4 = 100(1.02)4 1.10 = 100(1.02)4 1.10 = 100(1.02)4 1.10 = 492.93 + + 100(1.02)9 [ 1.10 ( ) 6 1.02 1 + + 1.10 [ ( 1 1.02 ) 6 ] 1.10 1 ( 1.02 1.10 ) ( ) 1.02 2 + + 1.10 ( ) ] 1.02 5 1.10 31 / 71
Looking Forward and Looking Back Exercise (3.20) A loan at 8% annually has an initial payment of 1000, and 9 further payments. The payment amount decreases by 2% each year. Find the loan balance immediately after the third payment. 32 / 71
Looking Forward and Looking Back Exercise (3.20) A loan at 8% annually has an initial payment of 1000, and 9 further payments. The payment amount decreases by 2% each year. Find the loan balance immediately after the third payment. Answer: 4644.38 33 / 71
Looking Forward and Looking Back Example (3.19) A loan at 10% annually has an initial payment of 100, and 9 further payments. The payment amount increases by 10 each year. Find the loan balance immediately after the fourth payment. Solution The payments are 100, 110, 120,..., 190. Immediately after the 4th payment the remaining payments are 140, 150,..., 190. Using the prospective method, the balance immediately after the 4th payment is the present value of those remaining payments. 34 / 71
Example (3.19) Solution Using P = 140 and Q = 10 we have ( an nv n ) Bal 4 = P a n + Q i ( ) a6 6v 6 = 140a 6 + 10 0.1 ( ) 4.355 6(0.5645) = 140(4.355) + 10 0.1 = 706.57 35 / 71
Looking Forward and Looking Back Exercise (3.21) A loan at 8% annually has an initial payment of 100, and 9 further payments. The payment amount increases by 5 each year. Find the loan balance immediately after the sixth payment. 36 / 71
Looking Forward and Looking Back Exercise (3.21) A loan at 8% annually has an initial payment of 100, and 9 further payments. The payment amount increases by 5 each year. Find the loan balance immediately after the sixth payment. Answer: 208.60 37 / 71
Section 3.5: Monthly Payment Loans The exams have questions with scenarios like nominal rates convertible monthly payments Example (3.22) A thirty year monthly payment mortgage loan for 250,000 is offered at a nominal rate of 6% convertible monthly. Find the a) monthly payment b) the total principal and interest that would be paid on the loan over 30 years c) the balance in 5 years d) the principal and interest paid over the first 5 years 38 / 71
Solution for Example (3.22) a) The loan has monthly rate of 6% 12 = 0.5% Then we can find the monthly payment is 1498.88 b) The total principal paid is just the amount of the loan, 250,000. The total interest can be found as: Total payment-principal paid = 1498.88(360) 250, 000 = 289, 596.80 c) The balance in 5 years: Bal 60 = P V (1 + i) 60 P MT s 60 = (250, 000)(1.005) 60 (1498.88) ( 1.005 60 1 0.005 ) = 232, 635.89 d) First note that the amount of principal paid in 5 years is just the original amount of the loan less the balance in 5 years or 250, 000 232, 635.89 = 17, 364.11. The total interest paid in 5 years can be found as: Total payments-principal paid = 1498.88(60) 17364.11 = 72, 568.69 39 / 71
Monthly Payment Loans Exercise (3.23) A fifteen year monthly payment mortgage loan for 250,000 is offered at a nominal rate of 6% convertible monthly. Find the a) monthly payment b) the total principal and interest that would be paid on the loan over 30 years c) the balance in 5 years d) the principal and interest paid over the first 5 years 40 / 71
Monthly Payment Loans Exercise (3.23) A fifteen year monthly payment mortgage loan for 250,000 is offered at a nominal rate of 6% convertible monthly. Find the a) monthly payment b) the total principal and interest that would be paid on the loan over 30 years c) the balance in 5 years d) the principal and interest paid over the first 5 years Answer: a) 2109.64; b) Principal 250,000, Interest 129,735.57; c) 190,022.75; d) Principal 59,977.25, Interest 66,601.27 41 / 71
Monthly Payment Loans Example (3.24) A thirty year monthly payment mortgage loan for 250,000 is offered at a rate of 6%. The borrower would like to have graduated payments where the first year s monthly payment is P, the second year s monthly payment is P + 100 and all subsequent monthly payments are P + 200. a) Find the initial payment P. b) Find the balance at the end of one year. 42 / 71
Monthly Payment Loans Example Solution a) Use the equation of value 250, 000 = P a 360 0.005 + v 12 (100)a 348 0.005 + v 24 (100)a 336 0.005 = P 166.7916 + 0.9419(100)(164.7434) + 0.8872(100)(162.5688) = P 166.7916 + 29, 940.28 This gives P = 1319.37 b) Using the prospective method, Bal 2 = (P + 100)a 348 0.005 + v 12 (100)a 336 0.005 = 1419.37(164.7434) + 0.9419(100)(162.5688) = 249, 144.19 43 / 71
Monthly Payment Loans Exercise (3.25) A fifteen year monthly payment mortgage loan for 250,000 is offered at a rate of 6%. The borrower would like to have graduated payments where the first year s monthly payment is P, the second year s monthly payment is P + 100 and all subsequent monthly payments are P + 200. a) Find the initial payment P. b) Find the balance at the end of one year. 44 / 71
Monthly Payment Loans Exercise (3.25) A fifteen year monthly payment mortgage loan for 250,000 is offered at a rate of 6%. The borrower would like to have graduated payments where the first year s monthly payment is P, the second year s monthly payment is P + 100 and all subsequent monthly payments are P + 200. a) Find the initial payment P. b) Find the balance at the end of one year. Answer: a) 1938.49; b) 241,507.13 45 / 71
Section 3.6: An Installment Loan Example With an installment loan, you pay regular payments of principal plus interest due at the end of each period Example (3.28) You have a 30,000 loan at 8% annually for 5 years. You agree to pay off the principal in installments of 6,000 per year, and to pay interest on the outstanding balance each year. The amortization table is below Year Payment Interest Principal Balance 0 30,000 1 8,400 2,400 6,000 24,000 2 7,920 1,920 6,000 18,000 3 7,440 1,440 6,000 12,000 4 6,960 960 6,000 6,000 5 6,480 480 6,000 0 46 / 71
An Installment Loan Example In order to find the interest at time 2, you would calculate 8% interest on the previous outstanding balance: 0.08(24, 000) = 1, 920. Then you would add that to the required principal of 6,000 for a total payment of 7,920. Note that now the principal is constant but the interest due and total payment decrease. Suppose that you wanted to find the interest due in the 4th payment. Note that Bal 3 = 30, 000 3(6000) = 12, 000 Int 4 = i(bal 3 ) = (0.08)(12, 000) = 960 The interest due in the 4th payment is 960, and the total payment is 6,960. 47 / 71
Exercise (3.29) You have a 30,000 loan at 8% annually for 30 years. You agree to pay off the principal in installments of 1,000 per year, and to pay interest on the outstanding balance each year. Find a) the interest due in the 11th payment b) the actual 11th payment 48 / 71
Exercise (3.29) You have a 30,000 loan at 8% annually for 30 years. You agree to pay off the principal in installments of 1,000 per year, and to pay interest on the outstanding balance each year. Find a) the interest due in the 11th payment b) the actual 11th payment Answer: a) 1600; b) 2600 49 / 71
Section 3.7: Sinking Fund Repayment of a Loan When you use a sinking fund, you only pay the lender the interest at his stated rate i on the loan each period. You make level deposits to an account called a sinking fund that earns interest at a rate j We will use the notation SF D for the required sinking fund deposit 50 / 71
Sinking Fund Repayment of a Loan: Example (3.30) A 100,000 annual payment loan is made for a term of 10 years at 10% interest. The lender wants only payments of interest until the end of year 10 when the 100,000 must be repaid. The borrower will make level annual year-end payment to a sinking fund earning 8%. Find the sinking fund deposit and the balance in the sinking fund at times 3 and 4. Solution The first task is to find the required sinking fund deposit. Note that F V = 100, 000, interest rate for deposit j = 8%, it is easily to find P MT = 6, 902.95, i.e., make a deposit of 6,902.95 to the sinking fund each year and the borrower will also pay 100, 000(0.10) = 10, 000 to the lender resulting in a total loan payment of 16,902.95. 51 / 71
Sinking Fund Repayment of a Loan: Example (3.30) A 100,000 annual payment loan is made for a term of 10 years at 10% interest. The lender wants only payments of interest until the end of year 10 when the 100,000 must be repaid. The borrower will make level annual year-end payment to a sinking fund earning 8%. Find the sinking fund deposit and the balance in the sinking fund at times 3 and 4. Solution Next we will look at the balance in the sinking fund. The balance at time k is the future value of k payments of 6,920.95 to the fund. To get the balance at time 3, find F V = 22, 409.73. To get the balance at time 4, find F V = 31, 105.46. 52 / 71
Sinking Fund Repayment of a Loan Exercise (3.31) A 70,000 annual payment loan is made for a term of 10 years at 8% interest. The lender wants only payments of interest until the end of year 10 when the 70,000 must be repaid. The borrower will make level annual year-end payment to a sinking fund earning 6%. Find the sinking fund deposit and the balance in the sinking fund at times 5 and 6. 53 / 71
Sinking Fund Repayment of a Loan Exercise (3.31) A 70,000 annual payment loan is made for a term of 10 years at 8% interest. The lender wants only payments of interest until the end of year 10 when the 70,000 must be repaid. The borrower will make level annual year-end payment to a sinking fund earning 6%. Find the sinking fund deposit and the balance in the sinking fund at times 5 and 6. Answer: SF D = 5310.76; Balance: 29,937.23 and 37,044.22 54 / 71
Sinking Fund Repayment of a Loan We refer to the balance at time k as SF Bal k The general rule to find the amount of principal in a sinking fund payment is: Principal in kth payment = SF Bal k SF Bal k 1 In the previous example, the Principal in 4th payment is: SF Bal 4 SF Bal 3 = 31, 105.46 22, 409.73 = 8, 695.73 You can find the interest in that payment too Interest in 4th payment = Total P MT - Principal Paid = 16, 902.95 8695.73 = 8, 207.22 55 / 71
Sinking Fund Repayment of a Loan There is an alternative way to find the interest in a sinking fund payment In each period you pay interest to the lender but also earn interest on the sinking fund The difference between what you pay and what you earn is called your net interest At time 3, the balance in the sinking fund was 22,409.73 and the interest earned was 0.08(22, 409.73) = 1, 792.78 For the 4th payment we have: Net interest in 4th P MT = 10, 000 1, 792.78 = 8, 207.22 56 / 71
Sinking Fund Repayment of a Loan Exercise (3.32) A 70,000 annual payment loan is made for a term of 10 years at 8% interest. The lender wants only payments of interest until the end of year 10 when the 70,000 must be repaid. The borrower will make level annual year-end payment to a sinking fund earning 6%. Find the principal and interest paid in the 6th payment. 57 / 71
Sinking Fund Repayment of a Loan Exercise (3.32) A 70,000 annual payment loan is made for a term of 10 years at 8% interest. The lender wants only payments of interest until the end of year 10 when the 70,000 must be repaid. The borrower will make level annual year-end payment to a sinking fund earning 6%. Find the principal and interest paid in the 6th payment. Answer: Principal = 7106.99; Interest = 3803.77 58 / 71
Sinking Fund Repayment of a Loan We denote the loan amount by L and the term by n Recall that the loan interest rate is i and the sinking fund rate is j The sinking fund deposit satisfies the equation SF D(s n j ) = L. Thus SF D = L s n j (5) The interest payable to the lender each period is Li, so the total loan payment each period is given by ( ) L 1 + Li = L + i (6) s n j s n j 59 / 71
Sinking Fund Repayment of a Loan The balance in the sinking fund at time k is given by: SF Bal k = SF Ds k j = L s k j s n j (7) The principal paid in payment k is SF Bal k SF Bal k 1 = SF Ds k j SF Ds k 1 j = SF D(s k j s k 1 j ) = SF D(1 + j) k 1 60 / 71
Sinking Fund Repayment of a Loan Thus we have a formula that is useful for exam problems The principal paid in sinking fund payment k = SF D(1 + j) k 1 (8) The interest paid in payment k can be given in two ways Net interest = Total payment - Principal paid = (SF D + Li) SF D(1 + j) k 1 (9) Interest to lender - Interest on sinking fund = Li SF Bal k 1 (j) (10) 61 / 71
Sinking Fund Repayment of a Loan Example (3.40) For the sinking fund loan in Example (3.30) we found that SF D = 6, 902.95 and the principal paid in the 4th payment was 8695.73. We can check the principal paid amount using formula (8). SF D(1 + j) k 1 = 6, 902.95(1.08) 3 = 8, 695.73 62 / 71
Sinking Fund Repayment of a Loan Exercise (3.41) Use formula (8) to verify the principal paid amount in Exercise (3.32). 63 / 71
Sinking Fund Repayment of a Loan Example (3.42) For a sinking fund loan SF D = 5310.76 and the amount of principal in the third payment is 5967.17. What is the interest rate? Solution 1 + j = P rincipal = SF D(1 + j) k 1 = 5310.76(1 + j) 2 = 5967.17 5967.17 5310.76 = 1.06 j = 0.06 64 / 71
Sinking Fund Repayment of a Loan Exercise (3.43) For a sinking fund loan SF D = 5066.43 and the amount of principal in the 4th payment is 6206.59. What is the interest rate? 65 / 71
Sinking Fund Repayment of a Loan Exercise (3.43) For a sinking fund loan SF D = 5066.43 and the amount of principal in the 4th payment is 6206.59. What is the interest rate? Answer: 7% 66 / 71
Section 3.8: Capitalization of Interest and Negative Amortization Capitalization of interest and negative amortization occur when the payment made is less than the interest on the loan Example (3.44) A borrower would like to borrow 30,000 at 8% for 5 years, but would like to pay only 2,000 for the first two years and then catch up with a higher payment for the final three years. What is the payment for the final 3 years? Solution First we will find the loan balance in 2 years if the payment each year is 2000. The balance is 30,832. After the second payment of 2000, the borrower owes 30,832. He can pay this off in 3 years with a higher payment. Set P V = 30, 832 and interest rate i = 8%, we can find the payment for the final 3 years is 11,963.85 67 / 71
Example (3.44) Below we show the amortization table for the loan. Year Payment Interest Principal Balance 0 30,000.00 1 2,000.00 2,400.00-400.00 30,400.00 2 2,000.00 2,432.00-432.00 30,832.00 3 11,963.85 2,466.56 9,497.29 21,334.71 4 11,963.85 1,706.78 10,257.07 11,077.64 5 11,963.85 886.21 11,077.64 0.00 Note what happened in yeras 1 and 2. The total payment was less than the interest required so the principal paid amount was negative. When the negative principal paid was subtracted form the prior balance, the effect was to add the shortfall to the balance of the loan. In period 1 the borrower has a shortfall of 400, and this means that he now owes 30, 000 + 400 = 30, 400. 68 / 71
Capitalization of Interest and Negative Amortization Such an increase in balance is called negative amortization because the amount of principal amortized is negative. It is also referred to as capitalization of interest since the unpaid interest is turned into capital when it is added to the balance of the loan. 69 / 71
Capitalization of Interest and Negative Amortization Exercise (3.45) Find the principal paid, interest paid and balance in year 1 if the initial payment were 1500 instead of 2000. 70 / 71
Capitalization of Interest and Negative Amortization Exercise (3.45) Find the principal paid, interest paid and balance in year 1 if the initial payment were 1500 instead of 2000. Answer: Interest = 2400; Principal = -900; Balance = 30,900 71 / 71