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Sections 4.1, 4.2, 4.3, 4.4

Sections 4.1, 4.2, 4.3, 4.4. Suppose the payments for an annuity are level, but the payment period and the interest conversion period differ. One approach to computing the numerical value of the annuity is the following two-step procedure: (1) (2).

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Sections 4.1, 4.2, 4.3, 4.4

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  1. Sections 4.1, 4.2, 4.3, 4.4 Suppose the payments for an annuity are level, but the payment period and the interest conversion period differ. One approach to computing the numerical value of the annuity is the following two-step procedure: (1) (2) Find the rate of interest which is equivalent to the given rate of interest and convertible at the same frequency as payments are made. Using the rate of interest from in step 1, find the value of the annuity. Find the accumulated value at the end of six years of an investment fund in which $100 is deposited at the beginning of each quarter for the first three years and $50 is deposited at the beginning of each quarter for the second three years, if the fund earns 6% convertible monthly. The quarterly interest rate equivalent to 6% convertible monthly is (1.005)3– 1 = 0.015075 . The accumulated value at the end of six years is .. s–– 12 | 0.015075 .. s–– 12 | 0.015075 = 100 (1.015075)12 + 50 $2246.98

  2. A loan of $6000 is to be repaid with quarterly installments at the end of each quarter for four years. If the rate of interest charged on the loan is 8% convertible semiannually, find the amount of each quarterly payment. The quarterly interest rate equivalent to 8% convertible semiannually is (1.04)1/2– 1 = 0.019804 . If R denotes the quarterly payments, then the equation of value is = R $6000 a –– 16 | 0.019804 6000 –––––––––– = $441.21 R = a –– 16 | 0.019804

  3. Suppose we want to find the annual effective rate of interest i for which payments of $75 at the end of every quarter accumulate to $3000 at the end of six years. (a) (b) Letting j = i(4)/4, where i(4) is equivalent to the desired annual effective rate of interest, write the equation of value. 75 = 3000 s–– 24 | j 75[(1 + j)24– 1] – 3000j = 0 Solving the equation in part (a) for j = i(4)/4 is difficult. Use the TI-84 calculator to solve for j, and then use the Excel file Interest_Solver to find j. You should find j = i(4)/4 = 0.041803.

  4. Solve for j on the TI-84 calculator by doing the following: (Note: On the TI-83 calculator, the | 2nd || FINANCE | keys should be used in place of the | APPS | key and Finance option.) Press the | APPS | key, select the Finance option, and select the TVM_Solver option. Enter the following values for the variables displayed: N = 24 I% = 0 PV = 0 PMT = –75 FV = 3000 P/Y = 1 C/Y = 1 Select the BEGIN option for PMT , press the | APPS | key, and select the Finance option. Select the tvm_Pmt option, and after pressing the | ENTER | key, the desired result should be displayed. j = i(4)/4 = 0.041803

  5. Suppose we want to find the annual effective rate of interest i for which payments of $75 at the end of every quarter accumulate to $3000 at the end of six years. (a) (b) Letting j = i(4)/4, where i(4) is equivalent to the desired annual effective rate of interest, write the equation of value. 75 = 3000 s–– 24 | j 75[(1 + j)24– 1] – 3000j = 0 Solving the equation in part (a) for j = i(4)/4 is difficult. Use the TI-84 calculator to solve for j, and then use the Excel file Interest_Solver to find j. You should find j = i(4)/4 = 0.041803. Type the formula =75*((1+j)^24-1)-3000*j in cell A5. To avoid getting the solution j = 0 for the equation of value, type the restriction j >= 0.01 instead of j >= 0.

  6. Select options Tools > Solver to solve the equation of value as indicated below (and if necessary, first use options Tools > Add-Ins) (c) Using the solution j = i(4)/4 = 0.041803 found in part (b), find the desired annual effective rate of interest. 4 i(4) i = 1 + — – 1 = (1 + j)4 – 1 = 4 0.17799 or 0.178

  7. A second approach with annuities where the payment period and the interest conversion period differ involves algebraic analysis. First, consider annuities payable less frequently than interest is convertible. We let k = n = i = the number of interest conversion periods in one payment period, the term of the annuity measured in interest conversion periods, rate of interest perconversion period. It follows that n / k = the number of annuity payments made. (Note that n / k must be an integer, but k need not be an integer.) The present value of an annuity which pays 1 at the end of each interval of k interest conversion periods is vk + v2k + … + vn = vk[1 + vk + (vk)2 + … + (vk)n/k– 1] = 1 – vn vk—— = 1 – vk 1 – vn ———— = (1 + i)k– 1 (1 – vn) / i —————— = [(1 + i)k– 1] / i a – n| s – k|

  8. The accumulated value of this annuity immediately after the last payment is a – n| s – n| (1 + i)n = s – k| s – k| The present value of an annuity which pays 1 at the beginning of each interval of k interest conversion periods is 1 + vk + v2k + … + vn – k = 1 + vk + (vk)2 + … + (vk)n/k– 1 = 1 – vn —— = 1 – vk (1 – vn) / i ———— = (1 – vk) / i a – n| a – k| The accumulated value of this annuity k interest conversion periods after the last payment is a – n| s – n| (1 + i)n = a – k| a – k|

  9. … … … … … 1 2 k  1 1 2 k  1 1 2 k  1 1 2 k  1 Conversion Periods  k 2k 3k n k k = n n k Payment Periods  n k  1 1 2 3 Note that in general n / k must be an integer, but k need not be an integer. Also, note that for any of the four formulas, putting double dots (..) above the symbol in the numerator and above the symbol in the denominator results in exactly the same formula. Present Value Accumulated Value Annuity Immediate Annuity Due

  10. … … … … … 1 2 k  1 1 2 k  1 1 2 k  1 1 2 k  1 Conversion Periods  k 2k 3k n k k = n n k Payment Periods  n k  1 1 2 3 Note that in general n / k must be an integer, but k need not be an integer. Also, note that for any of the four formulas, putting double dots (..) above the symbol in the numerator and above the symbol in the denominator results in exactly the same formula. Present Value Accumulated Value s – n| a – n| Annuity Immediate s – k| s – k| Annuity Due a – n| s – n| a – k | a – k |

  11. The present value of a perpetuity-immediate which pays 1 at the end of each interval of k interest conversion periods is vk + v2k + v3k + … = vk[1 + vk + (vk)2 + (vk)3 + … ] = 1 vk—— = 1 – vk 1 ———— = (1 + i)k– 1 1 / i —————— = [(1 + i)k– 1] / i 1 i s – k| The present value of a perpetuity-due which pays 1 at the beginning of each interval of k interest conversion periods is 1 —— = 1 – vk 1 1 + vk + v2k + v3k + … = 1 + vk + (vk)2 + (vk)3 + … = i a – k|

  12. Find the accumulated value at the end of six years of an investment fund in which $100 is deposited at the beginning of each quarter for the first three years and $50 is deposited at the beginning of each quarter for the second three years, if the fund earns 6% convertible monthly. (Note that this was done earlier using the other approach.) s –– 36| 0.005 s –– 36| 0.005 100 (1.005)36+ = 50 a – 3| 0.005 a – 3| 0.005 39.3361 39.3361 $2247.01 100 (1.005)36+ = 50 2.9702 2.9702

  13. An investment of $3000 is used to make payments of $500 at the end of every year for as long as possible with a smaller final payment made at the same time as the last regular payment. If interest is 8% convertible semiannually, find the number of payments and the amount of the final payment. If the smaller final payment were equal to 0, then the equation of value would be a – n| 0.04 500 = 3000 s – 2| 0.04 = 6 = 6(2.0400) = 12.24 a – n| 0.04 s – 2| 0.04 From the TI-84 calculator, we find that < n < , which implies that 17 18 8 regular payments and a smaller 9th payment denoted as R are made. The equation of value at the end of 8 years is s –– 16| 0.04 R + 500 = 3000(1.04)16 R = $269.79 s – 2| 0.04 The amount of the final payment is $500 + $269.79 = $769.79

  14. A series of payments of $5 are made every 3 months forever, with the first payment made immediately. At what annual effective rate of interest is the present value of these payments equal to $75? The equation of value is 75 = 5 + 5v1/4 + 5(v1/4)2 + 5(v1/4)3 + … 15 = 1 + v1/4 + (v1/4)2 + (v1/4)3 + … 1 —— 1 – v1/4 15 = 4 1 14 —— = — 1 + i 15 v = 4 15 — – 1 = 14 0.31781 or 31.781% i =

  15. Now, consider annuities payable more frequently than interest is convertible. We let m = n = i = the number of payment periods in one interest conversion period, the term of the annuity measured in interest conversion periods, rate of interest perinterest conversion period. It follows that mn = the number of annuity payments made. The present value of an annuity which pays 1/m at the end of each mth of an interest conversion period is 1 — m (m) a – n| = [v1/m + v2/m + … + vn] = 1 – vn ——— = 1 – v1/m 1 – vn ————— = (1 + i)1/m– 1 v1/m — m v1/m — m 1 — m [1 + v1/m + (v1/m)2 + … + (v1/m)mn– 1] = 1 – vn —— i(m) i — i(m) = a – n|

  16. The accumulated value of this annuity immediately after the last payment is 1 – vn —— i(m) (1 + i)n– 1 ———— i(m) i — i(m) (m) s – n| = (1 + i)n = = s – n| The present value of an annuity which pays 1/m at the beginning of each mth of an interest conversion period (similar to the notation and derivation for an annuity-due) is The accumulated value of this annuity 1/mth of an interest conversion period after the last payment is 1 – vn —— d(m) i — d(m) ..(m) a – n| = = a – n| (1 + i)n– 1 ———— d(m) i — d(m) ..(m) s – n| = = s – n|

  17. Observe that i — i(m) ii — + — i(m)m ..(m) a – n| (1 + i)1/m i(m) 1 + — m = (m) a – n| = = a – n| a – n| and that i — i(m) ii — + — i(m)m ..(m) s – n| (1 + i)1/m i(m) 1 + — m = (m) s – n| = = s – n| s – n| Appendix 4 (at the end of Chapter 4) in the textbook displays several of these types of formulas. The present value of a perpetuity-immediate which pays 1/m at the end of each mth of an interest conversion period is 1 — i(m) (m) a – | = The present value of a perpetuity-due which pays 1/m at the beginning of each mth of an interest conversion period is 1 — d(m) ..(m) a – | =

  18. Observe that when applying formulas derived for payments of 1/m, we must multiply by Pm when each actual payment is P. A loan of $6000 is to be repaid with quarterly installments at the end of each quarter for four years. If the rate of interest charged on the loan is 8% convertible semiannually, find the amount of each quarterly payment. We have m = and n = from which we have i = and i(m) = . 2 8 0.04 2[(1.04)1/2– 1] = 0.039608 If P denotes the quarterly payments, then the equation of value is

  19. We have m = and n = from which we have i = and i(m) = . 2 8 0.04 2[(1.04)1/2– 1] = 0.039608 If P denotes the quarterly payments, then the equation of value is (2) a – 8| 0.04 = (P)(2) $6000 3000 ––––––––––––––– = (1.0099)(6.732745) 3000 –––––––––– = $441.21 P = i — i(2) a – 8| 0.04 Observe that we may apply the formulas derived for annuities payable less frequently than interest is convertible to annuities payable more frequently than interest is convertible by setting the number of interest conversion periods in one payment period k equal to 1/m .

  20. Observe that we may apply the formulas derived for annuities payable less frequently than interest is convertible to annuities payable more frequently than interest is convertible by setting the number of interest conversion periods in one payment period k equal to 1/m . A loan of $6000 is to be repaid with quarterly installments at the end of each quarter for four years. If the rate of interest charged on the loan is 8% convertible semiannually, find the amount of each quarterly payment. We have n / k = and k = ; therefore, n = . We also have that the effective interest rate per interest conversion period is 16 1/2 8 0.04 . If P denotes the quarterly payments, then the equation of value is a –– 8 | 0.04 = P $6000 s –– 1/2| 0.04 1.041/2 1 ––––––––– = 1  1.048 P = $6000 $441.21

  21. A series of payments of $5 are made every 3 months forever, with the first payment made immediately. At what annual effective rate of interest is the present value of these payments equal to $75? The equation of value is (4)(5) ..(4) a – | 75 = 20 —— = 75 d (4) 4 —— = d (4) 15  4 14 — – 1 = 15 0.31781 or 31.781% i =

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