ANNUITIES 587. When a person receives every year a certain sum of money, say $N, he is said to possess an annuity of $N. The right to receive this annuity may continue a certain number of years and then lapse, or it may be invested in him and his heirs forever; in the first case the annuity is terminable, in the second, perpetual. An example of a terminable annuity is a common arrangement in lending money where A lends B a certain sum, and B repays by a certain number of equal annual installments which are so adjusted as to cover both principal and interest. An example of a perpetual annuity is the case of a freehold estate which yields its owner a fixed income of $N per annum. In valuing annuities it is customary to speak of the whole sum which is paid annually, yet, in practice, the payment may be in semi-annual, quarterly, etc. installments; and this must be taken into account in calculating annuities. 588. Contingent Annuity. In some cases the annuity lasts only during the life of a certain named individual, called the nominee, who may or may not be the annuitant. In this and similar cases an estimate of the probable duration of human life enters into the calculation, and the annuity is said to be contingent. 589. Value of a Forborne Annuity. An annuitant B, who had the right to receive n successive payments in n annual installments, has for some reason or other not received these payments. The question is, what sum should he receive in compensation? Let P be the value of each payment and r the rate of interest. It is clear that the whole accumulated value of the annuity is the sum of the accumulated values of the n installments and that compound interest must in equity be allowed on each installment. The first installment will draw interest compounded annually for n 1 years and will acquire the value and so on until the nth installment, which will not draw interest, since the whole annuity is due at the time the nth installment is due. Hence the whole accumulated value of the annuity is A=P(1+r)-1+P(1+r)"-2+P(1+r)"¬3+ ... + P(1+r) +P; The quantity within the brackets is a G. P. in which a=1, r=1+r, nn; hence, according to formula (iii), 8540, 590. Since the four quantities A, P, r, and n are connected by equation (1), any one of these four quantities can be calculated if the other three are given. CALCULATION OF A, P, n, AND r 591. Formulae for calculating the values of A, P, n, and r by means of logarithms readily follow from equation (1), 589, thus: Given P, r, n; calculate the value of A. Applying logarithms to equation (1), 589, it follows that 592. I. (i) log A = log P+ log [(1 + r)” — 1] — log r. NOTE. The value of (1+r)" can be calculated by logarithms, and then substituted in the bracket. Application.-A father wishing to provide for a dowry for one of his children, invests $1250 each year for 20 years; what will the amount of the dowry be, including the last payment, if the interest is compounded annually at 4%? Use formula (i) and put P = 1250, r = .04, and n = 20. Thus, log A log 1250+ log [(1.04) 201] log .04 593. II. log P, thus (ii) = 3.096910+ log (1.19109) (8.602060 = 3.096910 + .075945 — 8.602060 + 10 = 4.570795. A $37221.58. 10) Given A, r, n; calculate P. Solve formula (i) for log Plog A+ log log [(1 + r)" — 1] r = log (Ar) log [1+r)" - 1]. Application.-How much money must a person invest annually for 15 years at 44% compound interest in order that a capital of $24000 may be accumulated with the fifteenth installment? Put in formula (ii) A = 24000, r = .045 and n = 15, then log Plog (24000 x .045) log [(1.045)151] 594. III. Given A, P, r; calculate the value of n. Solving equation (1), 589, for n, _log (1 +45) n = log (1+r) P Application. For how many years must $1400 be invested each year at 3% compound interest in order that a capital of $45260 including the last investment, may be accumulated? Substitute in formula (iii), A = $45260, P= $1400, and r = .035; then 595. IV. Given A, P, n; calculate the value of r. Equation (1), When (iv) is expanded, it involves r to the power n, and lower powers; the value of r can not be r can not be found immediately except when is 2, in which case the values of r are the roots of a quadratic equation; but the values of r can be derived by a method of successive tests to any desired degree of approximation. method is illustrated by the following problem: This On investing $1150 each year for 25 years at compound interest, the accumulated value of the total investment, including the last installment, is $50000; what was the rate of interest? when n= <4; therefore .05; for r = .05, and n = 25 it Since the result is greater than the number r = .05 is too large. Suppose, moreover, that r=.045; then for n = 25, it follows that Since the result is greater than 4 (2), the number r =.045 is too large. Suppose, further, that r = .044; then since n = 25, it follows that A = .044 is still too large. Suppose r = .043. Then it (4) the number follows that therefore r = .043 is too small. It follows from (4) and (5) that .043+.014 .043 < r < .044 and = .0435 is the value of r with an error less than .0005; the rate is therefore 4.35%, with a possible error of .05 of 1%. On continuing this process, as close an approximation as desired may be found. REFUNDING OF A DEBT BY ANNUITIES 596. Sinking Fund. To make the calculations for a sinking fund is to calculate the purchase price of a given annuity. Suppose that B desires, by paying down at once a sum of $E, to secure for himself and his heirs the right of receiving n annual payments of $P each, the first payment to be made m years hence. E is the sum of the present values of the n payments. The first payment is due in m years hence; its present value is, therefore, P The second is due in m +1 years hence; therefore the (1+r)m present value of $Pis P + (1+r)m (1 + r)m+1 + ... + (1+r)m+n−1; (1 1 COROLLARY. It the annuity is not "deferred," but begins at once, i. e., the first payment is due in one year, then m = 1, and 597. The Calculation of E or of P.-These two problems do not cause any difficulty. If the unknown is E, then from (3), 1596, (4) log E = log P + log [(1 + r)” — 1] - log r. n log (1+r) · Application. What is the value of a loan E which can be refunded by 34 annuities of $1500 each, the rate being 41% and the first annuity being paid at the end of one year? Put P 1500, r.045, and n = 34; using logarithms one finds (1+r)" 4. 46626; log Elog 1500+ log (3.46626)-n log (1.045)—log .045 = 3.176091 + 0.539862.649944-8.653213 × 10 598. If P is unknown it can be found by solving equation (4), 2597, for log P, etc. 599. To find n, solve equation (3) 596, for (1+r)", giving and determiner by successive approximations, employing the method of 595. 601. The Period of Time a Fraction of a Year.-Suppose that the interest is compounded every six months instead of every year (as was the case in 2597), and in this case suppose that P' is the sum paid each period of six months, the interest on $1 for six months, and n' the number of half-yearly payments; then it follows A = p, (1 + r′)n' — 1 that [2589, (i)] AP. A similar formula holds for payments made at the end of other fractional parts of a year. |