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10

, 100(.90703) + 200(.86384) + 500(. 7 84) .5236 800

which can be solved for /

log,.75236 .28454

= 5.832 years.

log,1.05 .04879

As expected, the true value of t is less than the value using the method of equated time.

2.7 UNKNOWN RATE OF INTEREST

Section 2.6 considered the case in which the length of the investment period is the unknown. In this section we consider the situation in which the rate of interest is the unknown. Problems involving the determination of an unknown rate of interest are widely encountered in practice, since it is often necessary to compute the rate of return involved in a particular transaction. Techniques to solve for an unknown rate of interest will be considered again in several later chapters for various important types of applications.

We consider four methods to use in determining an unknown rate of interest. The first of these is to solve the equation of value for / directly using a calculator with exponential and logarithmic functions. This method will work well if a single payment is involved and can occasionally be adapted to other situations as well.

The second method is to solve the equation of value for / by algebraic techniques. For example, an equation of value with integral exponents on all the terms can be written as an nib. degree polynomial in i. 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.

The third method is to use linear 1 1 11 in the interest tables. This technique was illustrated in Section 2.6 for determining unknown time and can be adapted for problems involving an unknown rate of interest as well.

The fourth method is successive approximation or iteration. In using iteration to solve for an unknown rate of interest, a fiinction involving i, denoted by , is determined using an equation of value, and iteration is used to find a value of i such fl at O = 0. The number of successive approximations will be greatly reduced if a starting value close to the true root is used. Good starting values can often be obtained by linear inte olation in the interest tables, i.e. by the third method described above. Iteration methods are discussed in more detail in Appendix V and these methods will be utilized in later chapters

for some commonly-encountered applications. An ad hoc iteration technique will be illustrated in Example 2.10.

Example 2.7 At what interest rate convertible quarterly would $1000 accumulate to $1600 in six years?

Let j = il so that the equation of value becomes

1000(1 + yf" = 1600 J = (1.6)"2" - 1 .

This equation can be solved direcUy, which gives

; = .019776.

The answer is

,<*) = 4; = .0791, or 7.91%.

Example 2.8 At what effective rate of interest will the present value of Tf2000 at the end of two years and $3000 at the end of four years be equal to $4000?

00 An equation of value is

4000 = 2000 2 + "

which can be rewritten as

" + 2 2 - 4 = 0 . This equation can be solved as a quadratic in v, which gives

v2 =

-2 ± v/4 + 4 • 3 • 4

2 • 3

Since 2 > 0, only the positive root is meaningful, and ve have

= .868517

9v2 = ±±J =

which gives

(1 + if = 1.151388

i = .0730, or 7.30% .

Example 2.9 At what interest rate convertibk semiannually would an investment of $1000 immediately and $2000 3 years Jrom now accumulate to $5000 10 years from now?



Let j = iV2 so that the equation of value becomes

1000(1 + + 2000(1 + y)" = 5000.

This problem cannot be solved analytically like the prior two examples, so we will use linear interpolation in the interest tables. Define

/) = 1000(1 + + 2000(1 + )" - 5000. We seek to find j, such that f{j) = Q. By trial and error

(.03 ) = 1000(1.80611) + 2000(1.51259) - 5000 = -168.71 .0350) = 1000(1.98979) + 2000(1.61869) - 5000 = 227.17

and performing a linear interpolation

.•nI -

which gives

; =.0300 . .0050 .0321

/<2> = 2(.0321) = .0642, or 6.42%.

Example 2.10 Obtain the answer to Example 2.9 to a higher level of accuracy using iteration.

We will use a calculator with exponential and logarithmic fiinctions and an ad hoc iteration method. From Example 2.9 we have

fif) = 1000(1 + )° + 2000(1 + )" - 5000 and a first approximation j = .0321. Note that fij) is an increasing function of for ; > 0. We have

.0321) = -6.114 .0322) = 1.759.

We used a higher value than .0321 in order to obtain a sign change for -Performing another linear interpolation,

0 + 6.114

; = .0321 + .0001

= .03218.

1.759 + 6.114 Cycling again with one more decimal place

.03218) = .18346 /.03217) = -.60420.

In this case we used a lower value than .03218 in order to obtain a sign change for -Performing another linear interpolation

j = .03217 + .00001 "tJi" n = -032178.

.18346 + .60420

Cycling again with one more decimal place

/(.032178) = .025919 /(.032177) = -.052851.

Thus, j = .032178, which is accurate to six decimal places, since

I .032178) I < I .032177)

The more accurate answer is

= 2(.032178) = .06436, or 6.436%.

The above procedure can be repeated as many times as necessary. One additional decimal place of accuracy is achieved on each cycle.

Although methods with faster rates of convergence do exist (see Appendix V), the above technique is simple and straightforward, and has a more rapid rate of convergence than most ad hoc methods. Moreover, a similar approach can be developed for a wide variety of problems in interest involving the determination of unknown rates of interest.

2.8 PRACTICAL EXAMPLES

Virtually everyone is exposed to interest calculations regularly in their financial affairs. However, in practice interest calculations do not always follow the exact procedures outlined in this book. The purpose of this section is to familiarize the reader with some of the variations encountered in practice. The examples in this section are by no means exhaustive, but they are illustrative of some "real-world" applications of interest. Some terms for types of investments with which the reader may be unfamiliar are used in this section, e.g. "certificate of deposit" and "money market fund." Definitions of these terms are given in Section 8.8.

One type of advertisement frequently seen in the newspapers quotes two different rates on deposits. As this is being written, the author is looking at two such advertisements. In the first, a bank is quoting a one-year certificate of deposit as having "7.91% rate/8.15% yield." In the second, a savings and loan is quoting a money market fimd as having "8.00% interest rate/8.30% annual yield." What is the meaning attached to these two numbers? In the above illustrations the former number is a nominal rate, while the latter is an annual effective rate. The reader should verify that i = .0791 is equivalent to / = .0815 and that ) = .0800 is equivalent to / = .0830. In the former case the frequency of compounding was stated in the advertisement, but in the latter case it was not. The monthly frequency was ascertained by trial and error.

Section 2.3 deals with some of the variations in counting days. One advertisement the author has seen indicates that a savings bank credits 6% compounded daily which produces a yield of 6.27%. Using either a 360 or 365-day year produces an answer of 6.18%. How can an answer of 6.27% be obtained? After some trial and error, it was discovered that the savings bank was using a mix of 360-day and 365-day years as follows:



.06 360

- 1 = .0627.

When interest is not compounded daily, some interesting variations appear. Consider an investment account which credits interest monthly and on which deposits and withdrawals are occurring. One common variation credits interest on the average daily balance. Another common variation credits interest on the minimum daily balance. Yet a third variation credits interest on the beginning balance, reduced by any withdrawals, but not incremented by any deposits. In this latter case deposits do not start earning interest until the beginning of the following month.

Even the frequency of compounding can be tricky. Usually compounding occurs at some regular, predetermined interval. However, the author has encountered one money market fund which compounded interest whenever the interest rate being credited was changed. Thus, in a month with unchanged rates interest would be compounded only once, while in another more volatile month interest might be compounded as many as six or more times.

It is also important to distinguish between rates of interest and rates of discount. For example, the United States Treasury issues Treasury bills for periods of 13, 26, and 52 weeks. The rates on "T-bills," as they are often called, are computed as rates of discount. On the other hand, rates on longer-term Treasury securities are computed as rates of interest. Thus, rates on Treasury bills cannot be directly compared with rates on longer-term Treasury securities unless converted to equivalent rates. Treasury bills are discussed further in Section 7.2.

Rates of discount are also encountered in short-term commercial transactions. For example, if $10,000 is borrowed on a discount basis at 12% for one month, then the borrower receives $9900 immediately and repays $10,000 at the end of the month. Such short-term transactions are often computed on a simple discount basis, as just illustrated. The net effect is equivalent to using a rate of discount convertible at the same frequency as the term of the loan. Thus, if a stated rate, such as the 12% used above, is used for loans of varying lengths, then the equivalent effective rates will differ depending upon the length of the loans.

Credit cards have an interesting way of charging interest. The interest generally is computed on the ending balance of the prior month. Thus, no interest is assessed on new charges during a month for that month, i.e. interest does not begin to be charged until the following month. In essence, this is an interest-free loan from the date of charge until the end of the month. Persons who always pay their credit card charges in full each month are, in essence,

EXERCISES

2.2 Obtaining numerical results

1. $10,000 is invested for four months at 12.6%, where interest is computed using a quadratic to approximate an exact calculation. Find the accumulated value.

Find the present value of $5000 to be paid at the end of 25 months at a rate of discount of 8% convertible quarterly:

a) Assuming compound discount throughout.

b) Assuming simple discount during the final fractional period.

Find an expression for the fraction of a period at which the excess of present values computed at simple discount over compound discount is a maximum.

Method A assumes simple interest over final fractional periods, while Method assumes simple discount over final fractional periods. The annual effective rate of interest is 20%. Find the ratio of the present value of a payment to be made in 1.5 years computed under Method A to that computed under Method B.

2.3 Determining time periods

If an invesUnent was made on the day the United States entered World War , i.e. December 7, 1941, and was terminated at the end of the war on August 8, 1945, for how many days was the money invested:

borrowing money for short periods of time without paying any interest at all. On the other hand, the interest rate charged on those accounts which do carry outstanding balances from month-to-month tends to be very high.

Another feature that investors need to be careful to consider is a penalty for early withdrawal. For example, such penalties apply to many certificates of deposit. If a two-year certificate of deposit credits an annual rate of 9%, the purchaser is likely to have a rude awakening upon surrendering after one year and learning that less than 9% will be paid. The penalties for early withdrawal often involve either a reduction in the credited interest rate, or not crediting interest for a certain period of time, or some combination of the two. Other types of early withdrawal penalties are also encountered.

These illustrations should give the reader a flavor of some of the types of applications encountered. The fact that so many variations exist and that consistent terminology is not used from situation to situation can make seemingly simple financial analysis difficult. In real-world applications the reader is cautioned to look below the surface and ascertain exactly how Calculations involving interest actually will be done. Otherwise, comparisons among various options for borrowing and lending may well be flawed.



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