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8

(1 + 0* = \+ki +

Kk-\),2 { - -1)

2! 3!

A second example would be to evaluate e*as

= l-f)t5 + i + .+... 2! 3!

(2.1)

(2.2)

These formulas can be used for either positive or negative values of k, although convergence may be slow unless the absolute value of is small (e.g. for fractional periods of time). Other series expansions can be developed from the basic identities in Chapter 1. However, again it should be emphasized that using series expansions for calculation purposes is cumbersome and should be unnecessary except in unusual circumstances.

One method of crediting interest that is commonly encountered in practice is to use compound interest for integral periods of time, but to use simple interest for any fractional period. This approach is equivalent to using only the first two terms of the binomial expansion in formula (2.1) assuming 0 < it < 1.

It can be shown that the use of simple interest for a final fractional period is equivalent to performing a linear interpolation between (1 + i)" and

1000

1 4-

= $1648.72.

(J +1)"*, where is a non-negative integer. To see this we start with the linear inte olation

(1 + if+k (1 jt)(l + /)" + k{\ + 0"" (2.3)

= (1 + 0"[(1 - k) + {\ + i)] = (1 + Ol + ki) (2.4)

which is the formula if simple interest is used over the final fractional period. The use of simple interest introduces a bias, since it was shown in Section 1.5 that simple interest produces a larger accumulated value over fractional periods than does compound interest.

It is interesting to note that, in an analogous fashion, linear interpolation for v"** is equivalent to using simple discount over the final fractional period. Again, starting with linear interpolation

v+* = (1 - «+* = (1 - fc)(l - d)" + k{l - rf)"+ (2.5)

= (1 -"[(1 - A:) + A:(l - d)]

= Al-kd) (2.6)

which is the formula if simple discount is used over the final fractional period.

Example 2.1 Obtain numerical answers for Examples 1.7, 1.8,1.9, 1.10, and 1.12.

1. Example 1.7:

From the interest tables or by direct calculation

500(1.02)2° = 500(1.48595) = $742.97.

2. Example 1.8:

By direct calculation

1000(.97)2 = 1000(.69384) = $693.84.

3. Example 1.9:

By direct calculation

i(4) = 4[(.995)- - 1] = .0606, or 6.06%.

4. Example 1.10:

By direct calculation

lOOOe-* = $1648.72. If necessary, a series expansion could be used,

and discount were left in exponential form. This was done to facilitate the presentation of the basic principles without the necessity of the reader performing an undue amount of arithmetical computation. Naturally, in practical work, actual numerical answers are usually desired, and the purpose of this section is to discuss the various possible methods of obtaining such answers.

The advent of personal computers and inexpensive pocket calculators with exponential and logarithmic functions makes it feasible to obtain numerical answers to problems in interest by direct calculation. In fact, direct calculation is probably the easiest and most efficient method in most cases.

An alternative approach is to uie compound interest tables. Such tables appear in Appendix I and include values of v" and (1 + i)", as well as other functions which will be defined in succeeding chapters, for several rates of interest and for values of n from 1 to 50. Also, nonintegral values are given for v* and (1 + 0* for = 1/2, 1/4, and 1/12. Use of the compound interest tables is a convenient approach if required values appear in the tables.

As a last resort, if neither of the above techniques can be applied, direct calculation by hand can still be used. This may require the use of series expansions. One example would be to evaluate (1 + 0* using the binomial theorem



360 (2 - r,) + 30(M2 - M,) + (D2 - D,)

(2.7)

where /, = month of first date

D] = day of first date

y, = year of first date

M2 = month of second date

D2 = day of second date

Y2 = year of second date.

The third method is a hybrid and uses the exact number of days for the period of investment, but uses 360 days in a year. Simple interest on this basis is sometimes called the Bankers Rule and is often denoted by "actual/360." It will be shown in the exercises that the Bankers Rule is always more favorable to a lender than is exact simple interest. Also, it will be shown that the Bankers Rule is usually more favorable to a lender than is ordinary simple interest, although exceptions do exist.

In theory we could have a fourth method denoted by "30/actual" or "30/365." However, this method is almost never encountered in practice.

A further complication arises in a leap year. In most cases, February 29 is counted as a day and the year has 366 days. However, in some cases, February 29 is counted as a day, but the year is still counted as having 365 days. In other cases, February 29 is not counted as day, i.e. no interest is earned. The author has even encountered the assumption that all years have 365 1/4 days! A uniform approach to leap year does not seem to have emerged and different calculation bases are encountered in practice. It should be noted that under ordinary simple interest (30/360), leap year is irrelevant.

The above terms and discussion have been couched in terms of simple interest. However, the three commonly encountered calculation bases, i.e. actual/actual, 30/360, and actual/360, also are used for calculations on a compound interest basis.

It is assumed, unless stated otherwise, that in counting days interest is not credited for both the date of deposit and the date of withdrawal, but for only one of these two dates. If the difference in the two dates is calculated by normal procedures, then this will be the result. However, situations are occasionally encountered in practice in which interest is credited for both the date of deposit and the date of withdrawal, resulting in one extra days interest.

Not all practical problems involving interest require the counting of days. Many financial transactions are handled on a monthly, quarterly, semiannual, or annual basis. In these cases the counting methods described in this section are not required.

Terms to the sixth power must be used to achieve the same level of accuracy, which illustrates why series expansions should be used to obtain numerical answers only as a last resort.

5. Example 1.12:

From the interest tables, or by direct calculation

1000(1.05)(1.045)(1.04/ = 1000(1.27628)(1.24618)(1.21665) = $1935.05.

Example 2.2 Find the accumulated value of $5000 at the end of 30 years and 4 months at 6% per annum convertible semiannually: (1) assuming compound interest throughout, and (2) assuming simple interest during the final fractional period.

1. Assuming compound interest throughout, the answer is

5000(1.03)*° = $30,044.27 by direct calculation.

2. Assuming simple interest during the final fractional period, the answer is

5000(1.03)*<(1.02) = $30,047.18.

The answer to 2 is larger than 1, illustrating that simple interest produces larger accumulated values over fractional periods than compound interest does, although the difference is quite small.

2.3 DETERMINING TIME PERIODS

In practical problems involving interest it is necessary to determine the time period of an investment. Although there would appear to be no ambiguity in this process, different methods of counting the days in a period of investment have arisen in practice. Three methods are commonly encountered.

The first method is to use the exact number of days for the period of investment and to use 365 days in a year. Simple interest computed on this basis is sometimes called exact simple interest and is often denoted by "actual/actual." Appendix II contains a table numbering the days of the year, which facilitates counting the number of days between two given dates.

The second method assumes that each calendar month has 30 days and that the entire calendar year has 360 days. Simple interest computed on this basis is sometimes called ordinary simple interest and is often denoted by "30/360." Appendix II cannot be used for calculations on this basis. However, a formula for computing the number of days between two given dates is



2000(.08)

= $37.26

assuming that the year in question is not a leap year. Note that even if a table such as Appendix is unavailable, it is an easy matter to count the number of days and arrive at 85.

Using formula (2.7), the number of days is

360(0) + 30(9 - 6) + (10 - 17) = 83. Thus, the answer is

2000(.08)

= $36.89.

The counting has already been done above and the answer is

2000(.08)

= $37.78.

Not surprisingly, the answer using the Bankers Rule is greater than using either exact simple interest or ordinary simple interest.

2.4 THE BASIC PROBLEM

Broken down into its simplest terms, an interest problem involves four basic quantities:

1. The principal originally invested.

2. The length of the investment period.

3. The rate of interest.

4. The accumulated value of the principal at the end of the investment period.

If any three of these quantities are Icnown, then the fourth quantity can be determined. In the problems involving accumulated values considered so far. No. 4 is the unknown quantity; whereas, in the problems involving present values. No. 1 is the unknown quantity. Section 2.6 considers the case in which No. 2 is the unknown, while Section 2.7 considers the case in which No. 3 is the unknown.

The following observations may prove helpful in the solution of problems in interest:

a) The reader should assess the tools that will be available in performing the financial calculations. For example, the following considerations are relevant for the tools commonly utilized:

Interest tables

What interest rates are available?

Which functions are provided?

How many time periods have been calculated?

Are enough decimal places carried for sufficient accuracy?

Pocket calculators

Does the calculator have exponential and logarithmic functions?

Is it programmable?

Does it have built-in algorithms for financial calculations?

Personal computers

Is software already available to do the calculations involved?

Is the volume of calculations sufficiently large to warrant use of a personal computer?

Is the ability to quickly generate printouts of results important?

The approach to be used will depend on the tools available, the type and complexity of the calculations involved, and the volume of computation involved. For the bialance of this book we will assume the availability of a pocket calculator with exponential and logarithmic functions and we will utilize the interest tables in Appendix I. We will not assume that a personal computer is available; however, many applications can readily be handled by commonly available software or can readily be programmed.

b) The length of the investment period is measared in time units. It was mentioned in Chapter 1 that the fundamental measurement period is often assumed to be one year, and many problems are worked with this as the time unit, especially those involving effective rates of interest or discount.

However, if nominal rates of interest or discount are involved, often a time unit other than one year is most advantageous, The most convenient time unit to use is generally the interest conversion period. In problems involving continuous interest, however, some other time unit, such as a year, must typically be used.

c) An interest problem can be viewed from two ptrspectives, since it involves a financial transaction between two parties, 1 borrower and the lender. From either perspective, the problem is essentially the same; however, the

Example 2.3 Find the amount of interest that $2000 deposited on June 17 will earn, if the money is withdrawn on September 10 in the same year and if the rate of interest is 8%, on the following bases: (1) exact simple interest (actual/actual), (2) ordinary simple interest (30/360), and (3) the Bankers Rule (actual/360).

1. From Appendix , September 10 is day 253 and June 17 is day 168. The actual number of days in the period of invesUnent is 253 - 168 = 85. Thus, the answer is



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