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9 the effect of inflation which reduces the purchasing power of money over time. Inflation-adjusted calculations will be discussed in Section 9.4. As a consequence of the above principle, it is obvious that two or more amounts of money payable at different points in time cannot be compared until all the amounts are accumulated or discounted to a common date. This common date is called the comparison date, and the equation which accumulates or discounts each payment to the comparison date is called the equation of value. One device which is often helpful in the solution of equations of value is the time diagram. A time diagram is a one-dimensional diagram in which units of time are measured along the one dimension and payments are placed on the diagram at the appropriate points. Note that payments in one direction are placed on the top of the diagram and payments in the other direction are placed on the bottom of the diagram. The comparison date is denoted by an arrow. Figure 2.1 is an example of a time diagram used in the solution of Example 2.4. The time diagram is not necessary in the solution of equations of value; it IS merely an aid in visualizing the problem. With some practice the reader can usually dispense with a time diagram on simpler problems. However, time diagrams are usually helpful in the solution of more complex problems. One of the properties of compound interest is that the choice of the comparison date makes no difference in the answer obtained. Thus, there is a different equation of value for each comparison date, but they all produce the same answer. This important property of compound interest will be illustrated in the solution of Example 2.4. The reader is cautioned that under other patterns of interest, e.g. simple interest or simple discount, the choice of a comparison date does affect the answer obtained. This illustrates once again the inherent inconsistency in using simple interest or simple discount. t;. The reader should be aware that the problems involving accumulated values and present values already considered in the first two chapters are examples of equations of value. Example 2.4 illustrates a more general type of problem. Example 2.4 In return for a promise to receive $600 at the end of 8 fears, a person agrees to pay $100 at once, $200 at the end of 5 years, and to make a further payment at ffie end of 10 years. Find the payment at the end of 10 years if the nominal rate of interest is 8% convertible semiannually. We shall first work the problem with a comparison date of the present. The time diagram is shown in Figure 2.1. wording of a problem rrny be different depending upon the point of view. Examples and exercises phrased from both points of view appear, and the reader should not let the different phraseology be a source of confusion. As an example, recall the discussion in Section 1.7 involving the use of the words "paid" or "credited." To some readers the word "paid" may seem more normal from the vantage point of the borrower, while "credited" may seem more normal from the vantage point of the lender. Many other such examples could be cited - Complex financial transactions often involve more than two parties. For example, a business firm analyzing its rate of return on a major investment in a new plant is involved with a multiplicity of parties. However, the basic principles developed to analyze two-party transactions can readily be extended to analyze these more complex transactions. d) In practical applications involving interest the terminology can become confusing. Many terms have ambiguous meanings (e.g. see the discussion on the use of the word "discount" in Section 1.7). Furthermore, as we shall see in succeeding chapters some terms used unfortunately do not convey an intuitive description of the transactions involved (e.g. see the discussion on the terms "annuity-immediate" and "annuity-due" in Section 3.3). Finally, many parties involved in financial transactions simply do not always use terms with the precise meanings and definitions contained in this book. The reader is admonished in real-world applications to look beyond the stated terms and be certain to understand the exact nature of the financial transactions in question. There simply is not total consistency in terminology among the large and diverse number of parties involved in financial transactions involving interest. 2.5 EQUATIONS OF VALUE It is a fundamental principle in the theory of interest that the value of an amount of money at any given point in time depends upon the time elapsed since the money was paid in the past or upon the time which will elapse in the fiiture before it is paid. We have already seen this in many of the examples and exercises considered thus far in the first two chapters. This principle is often characterized as the recognition of the time value of money. This process would be in contrast to financial calculations not involving the effect of interest, in which case it would be said that such calculations do not recognize the time value of money. The reader is cautioned that "recognition of the time value of money" reflects the effect of interest, but not
8 9 10 600 Figure 2.1 Time Diagram for Example 2.4 Since interest is convertible semiannually, we will count time periods in half-years. The equation of value is 100 + 200v "> + Xvo = 600v 1 at 4% 600v*- 100 - 200v< 600(.53391) - 100 - 200(.67556) .45639 = $186.76. We could also have chosen a different comparison date and obtained a different equation of value. For example, if the comparison date were chosen to be the end of the 10th year, then the arrow in the time diagram would be under 10 and the equation of value would be 100(1.04)20 + 200(1.04)° + X = 600(1.04)" X = 600(1.04)" - 100(1.04)2° - 200(1.04)° = 600(1.16986)- 100(2.19112)-200(1.48024) = $186.76. Thus, the same answer is obtained. The two equations of value are equivalent. If both sides of the first one are multiplied by (1.04)°, the second one is obtained. The reader can verify that if other comparison dates are chosen, the same answer is obtained. 2.6 UNKNOWN TIME As discussed in Section 2.4, if any three of the four basic quantifies entering into an interest problem are given, then the fourth can be determined. In this section we consider the situation in which the length of the investment period is the unknown. The best method of solving for unknown time involving a single payment is to use logarithms. This technique will be illustrated in Example 2.5. As indicated above in Section 2.4, we are assuming the availability of a pocket calculator with exponential and logarithmic functions. An alternative approach with less accuracy that can be used if such a calculator is unavailable is Unear inte olation in the interest tables. This technique is also illustrated in Example 2.5. A situation sometimes arises in which several payments made at various points in time are to be replaced by one payment numerically equal to the sum This approximation is denoted by t and is often called using the method of equated time. It is possible to prove that the value of t is always greater than the true value of t, or, alternatively, that the present value using the method of equated time is smaller than the true present value. Consider quantities each equal to v 2 quantities each equal to v, and so forth until there are s„ quantities each equal to v". The arithmetic mean of these quantities is Sjv + S2V + 5j + 2 + The geometric mean of these quantities is 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 ijv> + S2v + • • • + s/" 5j + 2 + + 5„ of the other payments. The problem is to find the point in time at which the one payment should be made such that it is equivalent in value to the payments made separately. Let amounts s,S2, . . ,s„ be paid at times fj, fj. • • • . respectively. The problem is to find time t, such tiiat + $2 + • • • + paid at time t is equivalent to tiie payments of , • • . *n made separately. The flindamental equation of value is (S, + :?2 + • • • + n> = + 2 + • • • + V" - which is one equation in one unknown t. As an exercise, the reader will be Risked to find an exact expression for t. As a first approximation, t is often calculated as a weighted average of the various times of payment, where the weights are the various amounts paid, i.e. llflfL = k=X (2.9)
The left-hand side is the true present value which exceeds the present value given by the method of equated time on the right-hand side. Thus, the value of t from formula (2.9) is always greater than the true value of t from formula (2.8). The method of equated time is useful in analyzing the average length of financial transactions. This application will be discussed further in Section 9.8. Another interesting question often asked is how long it takes money to double at a given rate of interest. We can analyze this problem as follows: giving (1 + 0" = 2 n loggd + 0 = logg2 It is possible to derive an approximation to the exact result given in formula (2.10) as follows: log, 2 loged + 0 .6931 log,(l + 0 The second factor evaluated for i = 8% is 1.0395. Thus we have (2.11) This is frequently called the rule of 72, since it can be applied immediately by dividing 72 by the rate of interest expressed as a percentage (i.e. as 100/). The rule of 72 produces surprisingly accurate results over a wide range of interest rates. Illustrative values are provided in Table 2.1.  Table 2.1 Length of Time It Takes Money to Double | Rate of interest | Rule of 72 | Exact value | | | 17.67 | | | 11.90 | | | 9.01 | | | 7.27 | | | 6.12 | | | 4.19 |
Example 2.5 Find the length of time necessary for $1000 to accumulate to $1500 if invested at 6% per annum compounded semiannually: (1) by use of logarithms, and (2) by interpolating in the interest tables. Let n be the number of half-years. The equation of value is 1000(1.03)" = 1500 (1.03)" =1.5. 1. Using logarithms n log J.03 = logl.5 log e 1-5 .405465 log, 1.03 Thus, the number of years is 6.859. .029559 = 13.717. 2. From the interest tables, (1.03)" = 1.46853 and (1.03)" = 1.51259, so that 14 13 < n < 14. Performing a linear interpolation 1.50000 - 1.46853 n = 13 -I- 1.51259 - 1.46853 = 13.714. Thus, the number of years is 6.857. This method is equivalent to the assumption of simple interest during the final fraction of an interest conversion period. This point was discussed in more detail in Section 2.2. Example 2.6 Payments of $100, $200, and $500 are due at the ends of years 2, 3, and 8, respectively. Assuming an effective rate of interest of 5% per annum, find the point in time at which a payment of $800 would be equivalent: (1) by the method of equated time, and (2) by an exact method. 1. By the method of equated time using formula (2.9) 100 • 2 200 • 3 -I- 500 100 -h 200 -I- 500 2. The exact equation of value is 800v = lOOv + 200v + 500v = 6 years.
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