QUIZ 3 MAT 340 ANNUITIES Part II LOANS Part I Please do your work on a separate sheet of paper and circle your final answers. 1. Calculate the present value of an annuity immediate that has a sequence of quarterly payments of 5,12,19,26,...,103. Assume an annual effective rate of 4%. First let's define the quarterly effective interest rate: 1 j=1.04 4 1 Then, recognize that we have an arithmetic progression of payments with the first payment, P = 5, and the common difference, Q = 7. Also, the last payment can be defined as 103 = 5 + 14(7). If you have an annuity with terms: P, P+Q, P+2Q, P+3Q,...P+(n-1)Q. The present value can be found with the following formula: Pa n i Q a n i nv n =5a15 i j7 a 15 j 15 v 15 =731.78 j 2. You are given an annuity-immediate paying 10 annually for twenty years. After the twenty years, the payments decrease by one per year until it reaches a payment of 1. The payments of one continue forever. The annual effective rate of interest is 6%. Calculate the present value of this annuity. PV =10a 20 0.06v 20 Da 9 0.06v 29 a 0.06 =10 a20 0.06v 20 9 a 9 0.06 v 29 1 0.06 0.06 =129.1991
3. Jane receives a 10-year increasing annuity-immediate paying 100 the first year and increasing by 100 each year thereafter. Mary receives a 10-year decreasing annuityimmediate paying X the first year and decreasing by X/10 each year thereafter. At an annual interest rate of 5%, both annuities have the same present value. Calculate X. Jane has an annuity of 100, 200, 300,..., 1000. So, the present value of Jane is: 100 Ia 10 0.05=3937.33 Mary has an annuity of X, X-X/10, X-2X/10, X-3X/10,... X-9X/10 So, using the formula that was used in problem 1 and setting it equal to Jane's Present Value: 3937.3349=Pa n i Q a n i nv n i = Xa10 0.05 X 10 a 3937.3349= X [a10 0.05 1 10 a 10 0.05 10 v 10 0.05 X =864.1015 10 0.05 10 v 10 ] 0.05 4. Calculate a 6 0.05 a n i = 1 vn = 1 v6 ln1.05 =5.2016 5. Annual payments of 500 are made at the beginning of each year for 30 years to an annuity earning an annual effective rate of 7%. The interest is immediately reinvested into another fund earning 4.5% annual effective interest. At the end of the 30 years, what is the accumulated value of the 30 payments and the reinvested interest? The first fund just has a future value of the thirty payments of 500 = 15,000. The second fund is receiving payments of 35,70,105,...,1050 at the end of each year for 30 years.
So, the accumulated total value is : 15,00035 Is 30 0.045=15,00035 s 30 0.045 30 0.045 =15,00026,251.86=41,251.5872 6. A loan of 10,000 is amortized by equal semiannual payments for 4 years at an effective annual rate of 5%. Assume the first payment is made 6 months after the receiving the loan. Construct a complete amortization table, including interest/principal portion of each payment, and outstanding balance after each payment. First, we need to calculate the payment amount: 10,000=K a 8 j, where j=1.05 1 So, the payment amount = 1,392.86 T Pmt Interest portion Principal portion Outstanding Balance 0 10000 1 1392.86 246.95 1145.91 8854.09 2 1392.86 218.65 1174.21 7679.88 3 1392.86 189.66 1203.2 6476.68 4 1392.86 159.94 1232.92 5243.76 5 1392.86 129.5 1263.36 3980.4 6 1392.86 98.3 1294.56 2685.83 7 1392.86 66.33 1326.53 1359.3 8 1392.86 33.57 1359.29 0.01 7. A 300,000 home loan is amortized by equal monthly payments for 25 years, starting one month from the time of the loan at a nominal rate of 7% compounded monthly. How much total interest is paid during the last 10 years of the loan? First, find the outstanding balance after 15 years of payments. The payment amount will be: 300,000=K a 300 0.07. K =2,120.34. 12 Then, the outstanding balance after the 180 th payment, using the prospective method is:
OB 180 =2120.34 a OB t =Ka n t i 120 0.07 120 =182,616.9463 If 120 payments of 2,120.34 where made during the last ten years, there was an out of pocket contribution of 254,440.80 to pay off an outstanding balance of 182,616.95. Therefore, the total amount of interest paid would be the difference = 71,823.85. 8. A 35 year loan is to be repaid in equal annual installments. The amount of interest paid in the 8 th installment is 135. The amount of interest paid in the 22 nd installment is 108. Calculate the amount of interest paid in the 29 th installment. I t =K 1 v n t1 I 8 =135=K 1 v 35 81 =K 1 v 28 I 22 =108=K 1 v 35 221 =K 1 v 14 Solve both of the above equations for K and then you can set them equal to each other: 135 108 135 1 v 28= 1 v 14 108 =1 v28 1 v 14 1.25= 1 v14 1v 14 1 v 14 1.25=1v 14 0.25=v 14 Substituting this into equations above, we see that the payment is 144. We are trying to find the amount of interest in the 29 th payment: I 29 =K 1 v 35 291 =K 1 v 7 =1441 v 14 =1441 0.25=72 9. Seth borrows X for four years at an annual effective interest rate of 8%, to be repaid with equal payments at the end of each year. The outstanding loan balance immediately after the third payment is 559.12. Calculate the principal repaid in the second payment.
OB t =K a n t i OB 3 =K a 1 0.08=559.12 K = 559.12 =603.8496 a 1 0.08 PR t =K v n t1 PR 2 =603.85 v 4 21 = 603.85 1.08 3 =479.36