COMPOUND INTEREST
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The time interval between the occasions at which interest is added to the account is called the compounding period . The chart below describes some of the common compounding periods:
| Compounding Period | Descriptive Adverb | Fraction of one year |
| 1 day | daily | 1/365 (ignoring leap years, which have 366 days) |
| 1 month | monthly | 1/12 |
| 3 months | quarterly | 1/4 |
| 6 months | semiannually | 1/2 |
| 1 year | annually | 1 |
The interest rate, together with the compounding period and the balance in the account, determines how much interest is added in each compounding period. The basic formula is this:
the interest to be added = (interest rate for one period)*(balance at the beginning of the period).Generally, regardless of the compounding period, the interest rate is given as an ANNUAL RATE (sometimes called the nominal rate) labeled with an r. Here is how the interest rate for one period is computed from the nominal rate and the compounding period:
| interest rate for one period | = | (nominal rate)*(compounding period as a fraction of a year) |
| = | (nominal rate)/(number of compounding periods in one year) |
| the interest to be added | = | (nominal rate)*(compounding period as a fraction of a year)*(balance at the beginning of the compounding period) |
| Compounded | Calculation | Interest Rate For One Period |
| Daily, each day, every 365th of a year | (.06)/365 | 0.000164384 |
| Monthly, each month, every 12th of a year | (.06)/12 | 0.005 |
| Quarterly, every 3 months, every 4th of a year | (.06)/4 | 0.015 |
| Semiannually, every 6 months, every half of a year | (.06)/2 | 0.03 |
| Annually, every year | .06 | .06 |
| 6% means 6 percent (from Medieval Latin for per centum, meaning "among 100"). 6% means 6 among 100, thus 6/100 as a fraction and .06 as a decimal. | ||
Here are some common units for this calculation:
| ||
| Nominal Interest Rate | Compounded | Interest Rate For One Period | Balance at the beginning of some period | Interest Added at the end of the same period |
| 15%/yr | Daily | 0.000410959=.15/365 | $10,000 | $4.11 |
| 5%/yr | Monthy | 0.004166667=.05/12 | $10,000 | $41.67 |
| 9%/yr | Quarterly | 0.0225=.09/4 | $10,000 | $225 |
| 5.5%/yr | Seminannually | 0.0275=.055/2 | $10,000 | $275 |
| 7.8%/yr86 | Annually | .078=.078/1 | $10,000 | $780 |
1. "Nominal" in ordinary English can indicate something formal, in name only, but not quite reality and perhaps something that needs further description. It fits well here, because the effect of compounding is a real rate of interest slightly higher than the nominal rate of interest. Click here to return to the first use of the word "nominal".
What Happens To An Account With Compounded Interest And No Withdrawals?
Consider now an account in which P0 is invested at the beginning of a compounding period, with a nominal interest rate r and compounding K times per year (so each compounding period is (1/K)th of one year). How much will be in the account after n compounding periods? Let P j denote the balance in the account after j compounding periods, including the interest earned in the last of these j periods. NOTE THAT WE HAVE JUST DEFINED A SEQUENCE OF REAL NUMBERS. To review what these sequences are, in general, see sequences of real numbers. Note that we have a recursive definition of this sequence:
| Pj+1=P j + the interest earned by Pj in one compounding period. |
| Pj+1 | = | Pj + the interest earned by Pj in one compounding period |
| = | Pj + (nominal rate)*(compounding period as a fraction of a year)*Pj | |
| = | Pj + r * (1/K) * Pj | |
| = | Pj + (r/K) * Pj | |
| = | Pj * (1 + r/K) |
| Values of "j" | Pj |
| j=0 | P1 = P0 * (1+r/K) |
| j=1 | P2 = P1 * (1+r/K) = P0 *(1+r/K) * (1+r/K) = P1 = P0 * (1+r/K)2 |
| j=2 | P3 = P2 * (1+r/K) = P0 *(1+r/K)2 * (1+r/K) = P0 * (1+r/K)3 |
| j=3 | P4 = P3 * (1+r/K) = P0 *(1+r/K)3 * (1+r/K) = P0 * (1+r/K)4 |
In general Pj = P0 * (1+r/K)j for non-negative whole numbers j. The rare person may wonder how we can leap to this conclusion about an infinite number of possible j's, given only four examples! This formula can be proved for all of the infinite number of possible j's by using the principle of mathematical induction. | |
For The Saver, There Is An Advantage To Compounding More Frequently. If One Fixes The Nominal Interest Rate And The Total Time The Account Collects Interest, More Frequent Compounding Produces More Interest. In the analysis below,we assume that the total time is a whole number multiple of compounding periods.
If one fixes the initial balance (P 0), the nominal interest rate (r) and the duration of the deposit (T, in years) , you earn more interest with more compounding periods per year (K). The number of compounding periods that make up T will be KT. To avoid fractions of compounding periods, which were not analyzed above, assume that K is such that KT is a whole number. Then, by the formula above,P KT = P 0 * (1+r/K)KT.With T and r fixed (not changing) for this discussion, view the right-hand side above as a function of real variable K, say f(K). As long as 1+r/K is positive, this function will have a derivative:
(d/dK)[f(K)] = P 0 * (1+r/K)KT * [ T * ln(1+r/K) + K * T* (1/(1+r/K))*(-r/(K2 )) ].This simplifies somewhat: (d/dK)[f(K)] = P 0 * (1+r/K)KT * T* [ ln(1+r/K) - r/(K+r)) ] It well known that for x in the interval [0,1), we have ln(1+x) >= x - x2/2. If we substitute r/K for x and assume that r>0 and K>r, we find that ln(1+r/K) - r/(K+r)) >= (K-r)r2/(2 K2 (K+r)) > 0 ["ln" refers to the natural logarithm, the log to the base e.] Note that the derivative exists and is positive when P 0 , r, K, and T are all positive and K > r (which are natural assumptions about a savings account!). Since the derivative is positive, the original function f(K) is increasing. Thus, larger values of K make f(K) larger. If we make K larger and also make KT be an integer, then f(K) happens to coincide with P KT . Thus compounding more frequently produces more interest (subject to the assumption that T is a whole number multiple of the compounding period). If T is not a multiple of the compounding period, the conclusion depends strongly on the account's policies on withdrawals in the middle of a compounding period. For example, in some certificates of deposits the bank may charge a substantial penalty for "early" withdrawal.
What if we are utterly greedy, and insist that the bank compound our interest continuously?
What happens if we make the compounding period a millionth of a second, and ever smaller? Does the amount of interest increase forever without bounds, or do we reach a ceiling (a limit!) as we compound more and more frequently?
To answer these questions, consider g(K) = ln(f(K)):
g(K) = ln(P0) + (KT) * ln(1+r/K).As K approaches positive infinity, we have a race between two factors because KT is also approaching positive infinity (we assume that T is positive) while r/K approaches 0. As r/K approaches 0, 1+r/K approaches 1 and ln(1+r/K) approaches 0. Thus we seem to have infinity*0 in our limit as K approches positive infinity. Recall that L'Hôspital's rule applies to indeterminate forms 0/0 and infinity/infinity. Rewrite the difficult part of g(K) to take advantage of this rule:
g(K) = ln(P0) + ln(1+r/K) / [1/(KT)].Note that 1/(KT) is approaching 0, so that we have the indeterminate form of 0/0. By L'Hôspital's rule, examine the limit of a new ratio which is the ratio of the separate derivatives of the top and bottom of the indeterminate form:
{[1/(1+r/K)](-r/(K2)} / {-(KT)-2*T}After simplifying this new ratio, one has
[1/(1+r/K)] * (r/T) * [(KT)2] / (K2) = (rT) * [1/(1+r/K)].As K approaches positive infinity, this new ratio approaches (rT) * [1/(1+0)] = rT. Thus, g(K) has the limit ln(P0) + rT as K approaches positive infinity. Because ex is a continuous function, we can apply ex to the function g(K) to get f(K) back AND a limit for f(K) which is
e[ln(P0)+rT]=P 0*erT.Thus, compounding faster and faster does have a finite limit; this finite limit defines what economists (and bankers) mean by continuous compounding. If compounding is continuous at a nominal interest rate of r for a duration T (in years) with an beginning balance of P0, the balance at the end is
P 0*erT.Your comments and questions are welcome. Please use the email address at www.math.hawaii.edu Edited on September 6, 2006.
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