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15

= I,

1 •+

(l+g"-l-A:/,

The formula analogous to (3.29) is given by

. 2{k - n) • k{n-\y

(3.30)

(3.31)

The derivations of formulas (3.30) and (3.31) are left as exercises.

Appendix V contains other iteration algorithms not illustrated in this section which can also be applied to solve for unknown rates of interest.

Example 3.8 At what rate of interest, convertible quarterly, is \$16,000 the present value of \$1000 paid at the end of every quarter for five years?

Let J = tA, so that the equation of value becomes

\ 1 = 16,000

: = 16.

We will first illustrate interpolation in the interest tables. Defuie M = - - 16.

We seek to findy, such that ) = 0. By inspection of the interest tables .0200) = 16.3514 - 16 = .3514

.0250) = 15.5892 - 16 = -.4108 Now performing a linear interpolation

j = .0200 + .0050 „.* = .0223

.3514 + .4108

which gives

= 4(.0223) = .0892, or 8.92% .

We will next illustrate iteration. For a starting value we could use die value derived by linear interpolation above, i.e. Jq = .0223. However, we will instead apply formula (3.29) to obtain

(16) (21)

 0.0238 = .02283 = .02248 = .02234 Jl = 0.02345 = .02270 = .02243 = .02233 J2 = 0.02319 = .02261 = .02239 = .02231 0.02298 = .02253 = .02237 = .02230

We stop after 15 cycles, since die iteration is converging so slowly. We illustrate such a slow iteration, not because it is a good method to use, but ratiier to demonstrate the care which must be taken in practice to use iteration methods with a reasonable rate convergence.

We now try the Newton-Raphson iteration method using formula (3.28). We obtain die following results:

Jo = .0238 Jl = .0222459 2 = .0222623 = .0222623

We have achieved seven decimal places of accuracy after only two iterations! The slow ad hoc method given by formula (3.27) could not even achieve five decimal places of accuracy after 15 iterations. The power of the Newton-Raphson method is indeed evident.

Actually, a much faster ad hoc iteration dian diat exhibited by formula (3.27) can be achieved by using the technique illustrated in Example 2.10. Using this approach requires two steps in each cycle, but will achieve one additional decimal place of accuracy per cycle. The reader will be asked to perform such an iteration in the exercises.

The correct answer to six decimal places is thus

»W = 4(.0222623) = .089049, or 8.9049% .

Example 3.9 Derive the following alternative formula for a starting value to solve a-gi = and apply it to Example 3.8:

(3.32)

Consider n quantities: v, v , v ,. . ., v" . The arithmetic mean of these quantities

V + + +

n n

The geometric mean of these quantities is

1 + 2 + 3 + • • • + V "

n + 1

We first illustrate an iteration using formula (3.27). We obtain the following results:

(l + i)"" =

Formula (3.32) is now obtained immediately by substitution into formula (3.27). Applying formula (3.32) to Example 3.8 produces

Jo =

= .0225

which produces a starting value closer to the true value than formula (3.29).

It can be shown that an analogous formula for a starting value to solve i,- = k is given by

(3.33)

A third formula for starting values, which also produces excellent results, can be obtained as a by-product of material convered in Section 8.4 and will be given at that time.

3.9 VARYING INTEREST

Thus far we have assumed a level rate of interest throughout the term of the aimuity. In this section we will consider the situation in which the rate of interest can vary each period, but compound interest is still in effect. Other patterns of variation, not involving compound interest, are considered in Section 3.10.

As in Chapter 1, let i denote the rate of interest applicable for period it, i.e. for the interval from time : - 1 to time Jt. We consider first the present value of an n-period annuity-immediate.

Two patterns of variation could be involved. The first pattern would be for to be the applicable rate for period Jt regardless of when the payment is made. In this case the present value becomes

aj =(I+/,)-+(l+/,)-(l+i2)" + ---

+ (l+/,)-(l+J2)"-••(!+„)"= E (1+.)- 3-34)

t=l s=l

The second pattern would be to compute present values using rate if, for the payment made at fime over all A: periods. In this case the present value becomes

\-l L /-1 L ; \-2

= (1 + /i)- + (1 + /2)-+ • • • + a + i„y

= E (1 + it)--t-i

(3.35)

We now turn to accumulated values. We will consider values of j-, rather

than 5 so that all values of for A: = 1,2, . . . , will enter the formula. Again, the same two patterns of variation could be involved. If the applicable rate for period is ij regardless of when the payment is made, we have

\ = (1 + n> + (1 + + n-l) +•••+(! + i„)a + i„-i). . . (1 + /,) n t

= E (1 + wi)- (3.36)

/=1 s=l

Alternatively, if the payment made at time earns at rate over the rest of the accumulation period, we have

=(l+g + (l+/„.i) +

= E a + wi)-/=1

+ (1 + r,)"

(3.37)

Accumulated values of the annuity-immediate can be obtained from accumulated values of the annuity-due by using formula (3.16), i.e. sfY] ~

The reader should distinguish carefully between the two different patterns of computing interest described above, since both occur in practice. In essence, formulas (3.34) and (3.36) assume that the interest rate for any given period is the same for all payments whose value is affected by interest during that period. On the other hand, formulas (3.35) and (3.37) assume that each payment has an associated interest rate which remains level over the entire period for which present values or accumulated values are being computed. The former case will be seen to have an important practical application in Section 10.2, while the latter case will be seen to have an important practical application in Section 9.6.

In practice, changes in interest rates often do not occur each period, but only once every several periods. In such cases, values can be obtained direcUy

However, we know that the arithmetic mean of n positive numbers, not all of which are equal, is greater than the geometric mean, and thus we have

+ 1

We replace the right-hand side by a slightly larger number and assume that

n V

We turn next to finding a value for 5. If we assume that 1 invested at

at time n, then we

time t, where r = 1, 2, . . . n - 1 will accumulate to would have

ait)

(3.39)

This would seem to be an appropriate procedure in certain cases, e.g. a problem involving a varying force of interest over the n periods. However, it does not produce correct results in all cases. For example, suppose we wish to find the accumulated value of an n-period annuity-immediate in which each payment is invested at simple interest from the date of payment until the end of the n periods. The accumulated value of such an annuity would be

1 + (1 + 0 + (1 + 20 + • . . + [1 + (n - 1) n . This leads to a generalized version of formula (3.3)

= E « (3.40)

which will produce the correct answer for the above simple interest example.

However, it can easily be shown that formulas (3.39) and (3.40) do not produce the same result under simple interest. In fact, it can be shown that formulas (3.39) and (3.40) do not produce the same results in general for any pattern of interest other than compound interest.

For those readers who like symmetry and completeness, we could use formula (3.40) to find another expression for

n-l . n-l

, (3.41)

fto (n) a(«),to

It is not su rising that formulas (3.38) and (3.41) produce different answers unless compound interest is involved.

ambiguous, but serious distortions will result if a significant period of time is involved in the calculation.

Also, the subject of annuity values not involving compound interest has appeared in various forms in the literature. Thus, the reader should be exposed to some of the results which have appeared and the ambiguities involved.

It would seem logical to return to the accumulation function as the basis to compute annuity values. The present value of an n-period annuity-immediate is equal to the sum of the present values of the individual payments. Thus, a generalized version of formula (3.1) is

from annuity values given earlier in this chapter. This approach is generally easier to use tnan to directly apply the formulas given above.

Example 3.10 Find the accumulated value of a 10-year annuity-immediate of \$100 per year if the effective rate of interest is 5% for the first 6 years and 4% for the last 4 years.

We could apply formula (3.36); however, the following approach is simpler. The accumulated value of the first six payments after six years is

10051.05-

This value is accumulated to the end of the 10 years at 4%, giving

100 .05 (1-04). The accumulated value of the last four payments is

1005.04-

100[i51.05(-04) + .04

= 100[(6.8019)(1.16986) + 4.2465 = \$1220.38

Example 3.11 Rework Example 3.10 if the first 6 payments are invested at an effective rate of interest of 5% and if the final 4 payments are invested at 4%.

We could apply formula (3.37), but again a simpler approach is available. The setup is similar to Example 3.10 and the answer is

100 [ijq 05(1-05)+ 04 = 100[(6.8019)(1.21551) + 4.2465 = \$1251.43 .

The reader should justify die relative magnitude of die answers to Examples 3.10 and 3.11 by general reasoning.

3.10 ANNUITIES NOT INVOLVING COMPOUND INTEREST

The valuation of annuities not involving compound interest is fiill of pitfalls and requires careful analysis and interpretation to obtain reasonable results. Multiple values for annuities can result, which in turn requires qualification as to the basis on which the values are calculated. In fact, it is best to avoid dealing with annuities not involving compound interest, if possible.

Unfortunately, it is not always possible to avoid the subject. For example, on occasion some courts have required the use of simple interest in computing the value of an annuity involving lost income in personal injury and wrongful death lawsuits. The "correct" procedure to compute such values is not only

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