Although it is conventional to quote interest rates on the basis of annual payments, it is not unusual to see a different length of time used to determine the compounding of interest earned or owed. For example, it is not unusual to have interest earned in a savings account accrue and be automatically reinvested (compounded) on a monthly basis. The result of doing this is that more money is earned than in the case of only a single instance of coumpounding at the end of the full year. This result is developed below through the use of an example.
Suppose that the annual rale of interest on money in a savings account is 12% (r = 0,12) and an individual places $1,000 into hej account. No withdrawals are made until the end of the year. If interest is compounded on an annual basis, the amount in her account at the end ofonc year will be SI.000X (1 +0.12) = $1,120. Alternatively, suppose that the bank computes its interest payments on a half-yearly basis and deposits the appropriate sum into their customers' savings accounts accordingly. This means that interest is compounded semiannually. Since an annual interest rate of 12% implies a semiannual interest rate of 6%. this means that the $1,000 on deposit earns $1,000 x 0.06 = $60 interest in the first half-year. When deposited into the account, this means that the principal for the second half-year is $1,060 which earns $1,060 x 0.06 = $63.60 and so the total value of the deposit at the end of the year is $1,000 -I- $60 + $63.60 = $1,123.60. A more mathematically convenient way to express this is to note that after the first half-year there is $ 1,000 x (1 -f 0.06) in the account and this amount earns 6% interest in the second half-year, implying that at the end of the year
[$1,000X (I +0.06)1 x (1 +0.06) = $1,000(1.06)- = $1,123.60
is the amount in the account.
Suppose that the bank instead decides to offer compounding every three months (i.e., each quarter of a year). The interest rate for a quarter of a yeai is 3% and compounding quarterly means the value of $ 1,000 at the end of a year will be
$1,000(1 +0.03)( 1 +0.03)1 I 4- 0.03)(] + 1.03) = $1,000(1.03 )4 = $1,125.51
If the bank offered compounding monthly, the relevant interest rate is 1% (per month) and (he value of $1,000 at the end of a year becomes
$1,000 (i + o.oi)(i + o.oi). ,.q + o.oi) = $1,0000 .oni: = $1,126.82
Not surprisingly, the more frequently interest payments are compounded, the greater is the value of the savings at the end of the year.
If we let r represent the annual interest rate and n the number of times per year that interest is compounded, then the relevant per period interest rate is r/n, and so in each period we need to apply the factor (1 4- r/n ) to determine the value of each dollar at the end of that period. It follows that the value of %P invested ;it an annual rate of interest r compounded n times per year is worth at the end of a year. (Try this formula for the example developed above where n = 1.4, and 12 for annual, quarterly, and monthly compounding respectively.)
By continuous compounding we mean that interest is compounded instantaneously or. in effect, n —* Thus* we can treat the term (I f r/n )" in the formuli
12 times n times n times above as a sequence with its limit limf,_^,<l -f r/n\", giving us the factor to be applied to the principal P in order to determine the value of $P invested lor a year with continuous compounding. It turns out that for the special case of r =■ I (100% interest rate), we get lim
Was this article helpful?