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 = The accumulated value at the end of six years is 100 ( ) s–– 12 | s–– 12 | = $

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 = If R denotes the quarterly payments, then the equation of value is R = $6000 a –– 16 | R = 6000 –––––––––– = a –– 16 | $441.21

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. s–– 24 | j 75 = 3000 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 = [(1 + j) 24 – 1] – 3000j = 0

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 = 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:

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. s–– 24 | j 75 = 3000 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 = [(1 + j) 24 – 1] – 3000j = 0 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.

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

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 per conversion 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 v k + v 2k + … + v n =v k [1 + v k + (v k ) 2 + … + (v k ) n/k – 1 ] = 1 – v n v k —— = 1 – v k 1 – v n ———— = (1 + i) k – 1 (1 – v n ) / i —————— = [(1 + i) k – 1] / i a – n| s – k|

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

… ………… Conversion Periods  Payment Periods  nknk 2 k  1k  1 12 k  nknk  1 1 k 2k2k3k3k k = n nknk Accumulated Value Present Value Annuity Immediate Annuity Due 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.

… ………… Conversion Periods  Payment Periods  nknk 2 k  1k  1 12 k  nknk  1 1 k 2k2k3k3k k = n nknk Accumulated Value Present Value Annuity Immediate Annuity Due 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. a – n| s – k| a – n| a – k | s – n| s – k| s – n| a – k |

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

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| a – 3| (1.005) 36 += s –– 36| a – 3| (1.005) 36 += $

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

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 + 5v 1/4 + 5(v 1/4 ) 2 + 5(v 1/4 ) 3 + … 15 = 1 + v 1/4 + (v 1/4 ) 2 + (v 1/4 ) 3 + … 15 = 1 —— 1 – v 1/4 v = 1 14 —— = — 1 + i 15 4 i = 15 — –1 = or %

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 per interest 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 [v 1/m + v 2/m + … + v n ] = [1 + v 1/m + (v 1/m ) 2 + … + (v 1/m ) mn – 1 ] = 1 – v n ——— = 1 – v 1/m 1 – v n ————— = (1 + i) 1/m – 1 1 – v n —— i (m) (m) a – n| = 1 — m v 1/m — m v 1/m — m 1 — m = i—i(m) i—i(m) a – n|

The accumulated value of this annuity immediately after the last payment is (1 + i) 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 (m) s – n| 1 – v n —— i (m) = (1 + i) n – 1 ———— i (m) = i—i(m) i—i(m) s – n|..(m) a – n| = 1 – v n —— d (m) = i—d(m) i—d(m) a – n|..(m) s – n| = (1 + i) n – 1 ———— d (m) = i—d(m) i—d(m) s – n|

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

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) = [(1.04) 1/2 – 1] = Observe that when applying formulas derived for payments of 1/m, we must multiply by Pm when each actual payment is P. If P denotes the quarterly payments, then the equation of value is

We have m = and n = from which we have i = and i (m) = [(1.04) 1/2 – 1] = If P denotes the quarterly payments, then the equation of value is (P)(2) = $6000 P = 3000 –––––––––– = $ (2) a – 8| 0.04 i — i (2) a – 8| ––––––––––––––– = (1.0099)( ) 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 161/28 If P denotes the quarterly payments, then the equation of value is a –– 8 | 0.04 s –– 1/2| 0.04 P = $6000 P =$ /2  1 ––––––––– = 1  1.04  8 $

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 (4)(5) 20 —— = 75 d (4) i = 14 — –1 = 15  4 or %..(4) a –  | = 4 —— = d (4) 15