back start next
[start] [1] [2] [3] [4] [5] [ 6 ] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44] [45] [46] [47] [48] [49] [50] [51] [52] [53] [54] [55] [56] [57]
6 a\t) (1.29) The definition of 6J is completely analogous to the definition of 5, except for the minus sign. The minus sign is necessary to make the force of discount a positive quantity. The denominator of formula (1.29) is positive but the numerator is negative, since a{t) is a decreasing function. Later in this section it will be seen that the force of discount bears a relationship to nominal and effective rates of discount similar to the relationship that the force of interest bears to nominal and effective rates of interest. However, it can be shown that 5, = 5,, so that we may dispense with 8, and use 6, . The proof is as follows: a\t) a\t) ±a{t) dt a\t) a\t) a{t)8t a\t) from formula (1.24) In theory, the force of interest may vary instantaneously. However, in practice it is often a constant. If the force of interest is constant over an interval of time, then the effective rate of interest will also be constant over that interval. This can be seen by using formula (1.26) over n measurement periods ( a positive integer) i,dt = e"" so that if 6, = 5 for 0 = a{n) = (1 + if < t <. n = I + i i=e I (1.30) (1.31a) (l + /yiog,(l +/) (1 + i) (1 + 0 which is constant for all t. Formulas (1.31a) and (1.32a) can also be written as series expansions: 8 . 8 . 8 1" 8 = log,(l +0 = + • • (1.3 (1.32fc) These two series will normally have a rapid rate of convergence, since in practice / and 8 are usually small positive numbers whose successive powers diminish quickly. Having related 8 and / immediately relates 8 and the other measures of interest described in this chapter. The following series of equalities is an expanded version of formula (1.22a) which summarizes much of the material contained in this chapter: = 1 + / = v" = (1 d) = 1 = e . (1.33) It is instructive to examine the behavior of the force of interest under simple interest and simple discount. For simple interest we have a(t) 1 +it 1 +it Similarly, for simple discount we have for 0 < t. (1.34) It is also possible to define a force of discount analogous to formula (1.24). For tliis pu ose, we use the discount function a~{t) instead of the accumulation function a(t). The definition of the force of discount at time t, denoted by Sl, is given by which expresses / as a function of 8. Taking the logarithm of formula (1.30) expresses 5 as a fiinction of / 8 = logd + 0. (1.32a) Formula (1.32a) may also be derived directly from the accumulation function for compound interest
a\t) \dt for 0 < r < \ld. (1.35) The upper restriction on values of t in formula (1.35) is necessary to have values of 8, that are finite and positive. As we would expect, 5, is a decreasing function of t for simple interest, but an increasing function of t for simple discount. The reader may be su rised to observe that although a constant force of interest 6 leads to a constant effective rate of interest /, the reverse is not necessarily true. To see this possibility, consider n measurement periods (n a positive integer). Again, we have f" ain) = (1 +0" = eJo . However, if we subdivide the nperiod integral into a sum of oneperiod integrals, we have .1 ,2 (1 + /)« = Cb,dt f b,dt + f b,dt + • • • + f" b,dt 8,dt 5,dt If we allow d, to vary within each of these measurement periods in such a way that all the oneperiod integrals are equal, then we would have the result that i is constant for each of the n periods, but 6, varies within each of these periods. A good practical example to illustrate this phenomenon is to invest 1 for n periods at compound interest for completed periods, but use simple interest over fractional periods. This produces a growth pattern of (1+0* for = I, 2, ... n, so that the effective rate of interest is constant over the n measurement periods. However, 6, varies within each of these periods, since 6, varies under simple interest. Other examples could be constructed to illustrate the same point, but they would generally involve more esoteric patterns for the variation in 8,. Another interesting insight into the nature of the force of interest can be obtained by analyzing 5 in terms of From formula (1.33) Now using a series expansion, we have = 6 + 1 + 1 m 2! + 1 3! and taking the limit as m approaches infinity, lim i"") = b. (1.36) m»oo This formula has intuitive appeal. Since f" is a nominal rate of interest convertible mthly, we can 1 1 1 as a nominal rate of interest convertible continuously. By an analogous argument, it is possible to show that lim d"") = 8. (1.37) m»oo The proof is left as an exercise. This formula also has intuitive appeal. Since " is a nominal rate of discount convertible mthly, we can inte ret 5 as a nominal rate of discount convertible continuously. In essence, this is an alternate proof that the force of interest and the force of discount are equal. A third proof of this result is left as an exercise. The force of interest is a useful conceptual device, making the continuous growth of money at compound interest similar to growth functions encountered in the natural sciences. In theory, the most fundamental measure of interest is the force of interest. In practice, however, effective and nominal rates of interest and discount tend to be used more frequently because they are simpler for most people to comprehend and because most financial transactions involve discrete, not continuous, processes. This does not mean that the force of interest is devoid of practical significance. Besides being a useftil conceptual and analytical tool, it can be used in practice as an approximation to interest converted very frequently, such as daily. Also, some financial institutions actually have started to use continuous compounding in recent years. Example 1.10 Find the accumulated value of $1000 invested for ten years if the force of interest is 5%. The answer is
a(t) = e (1.26) If the form of is readily integrable, results may be obtained directly. If the form of 5 is not readily integrable, approximate methods of integration are necessary. The second type of variation considered involves changes in the effective rate of interest over a period of time. This type of variation is probably the one most commonly encountered in practice. As before, let i„ denote the effective rate of interest during the nth period from the date of investment. Then for integral f > 1, we have ait) = (1 + /,)(1 + /2) •• • iX + i,) = (1 + ik) (l) it = 1 If ij = 12 = • • • = V ~ familiar result a{t) = (1 + if is obtained. Present values with varying effective rates of interest can be handled similarly. Again, for integral t > 1 we have al(r) = (1 + Ji)i(l + ii) •••(!+ i,) = (1 + = v. (139) k=\ k=\ The approach followed in formulas (1.38) and (1.39) can be easily generalized to include nominal rates of interest or discount. All that is necessary is to count the number of interest conversion periods in each bracketed subinterval in which the rates are constant and then multiply the appropriate factors for each subinterval across the entire interval. This procedure is illustrated in the exercises. Frequently, in situations involving varying interest it is desired to find an equivalent level rate to the rates that vary. When this situation is encountered, the definition of equivalency given in Section 1.7 can be applied directly. However, it is important to note that the answers will depend on the period of time chosen for the comparison. Since interest is varying, the rate that would be equivalent over a period of one length would not be the same as that over a period of a different length. The measurement of interest 29 Example 1.11 Find the accumulated value of 1 at the end of n years if Using formula (1.26), the answer is Example 1.12 Find the accumulated value of $1000 at the end of 15 years if the effective rate of interest is 5% for the first 5 years, 4 1/2% for the second 5 years, and 4% for the third 5 years. Using formula (1.38), the answer is 1000(1.05)(1.045)(1.04). 1.11 SUMMARY OF RESULTS Table 1.1 summarizes much of the material of this chapter. Table 1.1 Summary of Relationships in Chapter 1 Rate of interest or discount 2 i Compound interest The accumulated value The present value of] ofl at time t = aft) at time t = a ~\t) wo;  (1 + 0   v = (l +J   ,(m)  1 +           (I  d)    (1  dy      1 d<>             Simple interest      /  1 +   (1 + it)   Simple discount      (I  dty   1 dt  
iCISES ,1.2 The accumulation and amount functions .Consider the amount function Aif) = t + 2t + . % Fmd the corresponding accumulation function a(f). b) Verify that a(r) satisfies the three properties of an accumulation function c) Find 1000(.05)(.0) ,ooor 1.10 VARYING INTEREST This section is concerned with situations involving varying interest. Two types of variation are considered. Other types of variation can be analyzed from basic principles. The first type of variation considered is a continuously varying force of interest. The basic formula for use in problems involving a varying force of interest is formula (1.26)
[start] [1] [2] [3] [4] [5] [ 6 ] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44] [45] [46] [47] [48] [49] [50] [51] [52] [53] [54] [55] [56] [57]
