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14 persons life expectancy in the measurement of economic damages in personal injury and wrongful death lawsuits. If the injured party in such a lawsuit has a life expectancy of 15.7 years, then what is the correct value of the annuitycertain? This question does not have a clearcut answer and further interpretation is required. It is also possible to develop identities for annuities in which n is negative. However, such annuities are only of mathematical concern and are devoid of any practical significance. Such identities are developed in the exercises for sake of completeness. Consider next what happens if i < 0. The case in which = 0 is important in practice. If i = 0, then the present value or accumulated value of any annuity is merely the sum of the payments. Thus, in particular, we have = if j = 0. (3.23) If i < 0, then some interesting results occur. Present values become accumulated values, and vice versa, which has intuitive appeal. These results are developed in the exercises. Again, however, these results are more of theoretical than practical significance. 3.7 UNKNOWN TIME Thus far, in any problems involving annuities, we have assumed that n and / are both known. In Section 3.7 we will consider the case in which n is unknown, and in Section 3.8 we will consider the case in which i is unknown. In general, problems involving unknown time will not produce exact integral answers for n. These problems could be handled along the lines of Section 3.6 in which a smaller payment is made during the period following the last regular payment. However, in practice, this is seldom done because of the inconvenience and confusion of making a payment at a date which is not an integral number of periods from the dates all other payments are made. For example, making all regular payments on July 1 of each year for a period of years followed by a smaller payment on November 27 is not convenient for either party to the transaction. What is usually done in practice is either to make a smaller additional payment at the same time as the last regular payment, in effect making a payment larger than the regular payment, called a balloon payment, or to make a smaller payment one period after the last regular payment, called a drop payment. Naturally, the smaller payments in these two situations are not equal. nor would either be equal to the smaller payment made at an intermediate point as in Section 3.6. However, all these payments would be equivalent in value. Problems involving unknown time can best be illustrated by example. Example 3.6 An investment of $1000 is to be used to make payments of $100 at the end of every year for as long as possible. If the fund earns an annual effective rate of interest of 5%, find how many regular payments can be made and find the amount of the smaller payment: (1) to be paid on the date of the last regular payment, (2) to be paid one year after the last regular payment, and (3) to be paid during the year following the last regular payment, as described in Section 3.6. The equation of value is 100a = 1000 = 10. By inspection of the interest tables, we have 14 < n < 15. Thus, 14 regular payments can be made plus a smaller final payment. Figure 3.5 is a time diagram for this example. 0 1000 14 \4+ 15 h h h Figure 3.5 Time diagram for Example 3.6 In this figure X, X, and X are the smaller final payments for the above diree cases; arrows 1, 2, and 3 mark comparison dates for the above three cases; and derives its meaning from formula (3.22). 1. The equation of value at the end of the 14th year is I 100ij4 + Xy = 1000(1.05)". Thus, Xy = 1000(1.05)"  100ij4 = 1979.93  1959.86 = $20.07 . 2. The equation of value at the end of the 15th year is lOOi +2 = 1000(1.05)
Thus, x2 = 1000(1.05) ioo(T5i ) = 2078.93 2057.86 = $21.07 . It should be noted that 20.07(1.05) = 21.07, or that in general X,(l + i) = Xj. The reader should justify this result by general reasoning. 3. In this case the equation of value becomes lOOa = 1000 = 10 where 0 < * < 1. This can be written as 1 V 14 + t = 10 Thus, givuig (1.05)"+* = 2 14 + * = og,2 log, 1.05 .693147 .048790 14.2067 = .2067 . Thus, die exact final irregular payment from formula (3.22) is X3 = 100 (1.05) .2067 = $20.27 paid at time 14.2067. A common approximation to die exact amount would be 100* or $20.67. The exact answer obtained lies between die answers to 1 and 2, as we would expect. It is obvious that using formula (3.22) is not only inconvenient but that it is also more difficult to use if exact answers are required. Example 3.7 A fund of $25, is to be accumulated by means of deposits of $1000 made at the end of every year as long as necessary. If the fund earns an effective rate of interest of 8%, find how many regular deposits will be necessary and the size of a final deposit to be made one year after the last regular deposit. The equation of value is 10001 = 25,000 = 25 . By inspection of die interest tables, we have 14 < n < 15. Thus, it takes 14 regular deposits plus a smaller final deposit, X. The equation of value at die end of die 15di year is lOOOi. + = 25,000 Thus, X = 25,000  1000(551  1) = 25,000  26,152 = $1152 to the nearest dollar. What has happened here is that the last regular deposit brings the fund close enough to $25,000 that interest alone over the last year is sufficient to cause the fimd to exceed $25,000. The balance in die fund at die end of die 14di year is lOOOT = $24,215. The balance in the fund at the end of the 15th year with interest only over the last year is 24,215(1.08) = $26,152 which is in excess of the desired fund by $1152. The result agrees with the one above. This example should not be thought of as typical, since often a final deposit is necessary. However, it does illustrate diat pitfalls do exist and tiiat care must be used in obtaining reasonable answers. 3.8 UNKNOWN RATE OF INTEREST In this section we consider the situation in which the rate of interest is the unknown. As noted in Section 2.7, problems involving the determination of an tJnknown rate of interest are widely encountered in practice. We consider three methods to use in determining an unknown rate of interest. The first of these is to solve for i by algebraic techniques. For example, our fundamental defmition for an nyear annuityimmediate  V I I (3.1) is an nth degree polynomial in v. If the roots of this polynomial can be determined algebraically, then i is immediately determined. This method is generally practical only for small values of n. Alternatively, we can express or l/a in terms of i and solve by algebraic techniques. As series expansions, we have n{n + 1) n{n + l)(n + 2) 2 «1 2! 3! (3.24)
2 12 (3.25) The derivations of formulas (3.24) and (3.25) are left as exercises. Formula (3.25) is generally preferred to formula (3.24), since the rate of convergence is faster. The use of formula (3.24) or (3.25) is not a particularly attractive method of solving numerically for an unknown rate of interest. However, these formulas do have value in connection with the derivation of other more useful approximation formulas later in this section and in Chapter 7. The second method is to use linear interpolation in the interest tables. This method is quite similar to that used in Section 2.7. It should be noted that the accuracy of linear inte olation will depend on how closely together the interest rates appearing in the tables are tabulated. The third, and by far the best, method is successive approximation or iteration. This method can be applied to all problems of this type and can produce whatever level of accuracy is required by carrying out enough iterations. It does require the use of a calculator with exponential and logarithmic fiinctions, whereas the first two methods above do not. The reader is referred to Appendix V for more background on iteration methods. Iteration can easily be applied when an equation of the form i = 8ii) (3.26) exists and converges to the true value of i, which satisfies the formula exactly. Assume some starting value, labeled Iq. Then generate a value by In general, /j ? /g < so that another trial will be necessary. Then generate a value «2 by If the iteration is converging, then the successive values Jq, i, ... will converge to the true value i. In practice, iterations will be carried out until iji = i to the required degree of accuracy. The reader should note that the symbol was used in Chapter 1 to represent the rate of interest during the sth period from the date of investment. Consider the problem in which the value of is given as some constant and it is desired to find the rate of interest i which will produce this value. One simple, straightforward iteration method can be obtained directly from formula (3.2) l(l + 0" (3.27) Unfortunately, as we will see in Example 3.8, the rate of convergence for this iteration formula is quite slow. A powerful method defined in Appendix V, which converges very rapidly, js the NewtonRaphson iteration method. The iteration formula for solving aji = by this method is = I, 1(1 i(i + g"i{i + i,(n + i)} (3.28) The deviation of formula (3.28) is left as an exercise. Formula (3.28) is a bit complicated, and may not be worth the trouble for isolated calculations. However it is an excellent method if a large number of calculations need to be performed, since it does have a very rapid rate of convergence. Applying an iteration method requires the use of a starting value. The number of iterations will be greatly reduced if a starting value close to the true root is used. Good starting values can be obtained by linear inte olation in the interest tables, i.e. by the second method described above. However, a more convenient method in determining starting values than interpolation in the interest tables is to apply directly an approximation formula based only on n and k. Such a formula can be derived from the first two terms of formula (3.25) which gives n + 1 2n nk nk 2ink) (3.29) kin + 1) Formula (3.29) is analogous to a commonlyused formula for computing approximate bond yields given in Section 7.6. A verbal inte retation of formula (3.29) will be given at that time. Analogous results to formulas (3.28) and (3.29) can be derived for accumulated values. The NewtonRaphson iteration formula for solving n]t = * is
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