Editing Interest (section)

==Calculation{{anchor|Calculation of interest|Types of interest}}==
===Simple interest===

'''Simple interest''' is calculated only on the principal amount, or on that portion of the principal amount that remains. It excludes the effect of [[compound interest|compounding]]. Simple interest can be applied over a time period other than a year, for example, every month.

Simple interest is calculated according to the following formula:

:<math>\frac {r \cdot B \cdot m}{n} </math>

where 
:''r'' is the simple annual [[interest rate]]
:''B'' is the initial balance
:''m'' is the number of time periods elapsed and
:''n'' is the frequency of applying interest.

For example, imagine that a credit card holder has an outstanding balance of $2500 and that the simple annual [[interest rate]] is 12.99% ''per annum'', applied monthly, so the frequency of applying interest is 12 per year. Over one month,

:<math>\frac{0.1299 \times \$2500}{12} = \$27.06</math>

interest is due (rounded to the nearest cent).

Simple interest applied over 3 months would be

:<math>\frac{0.1299 \times \$2500 \times 3}{12} = \$81.19</math>

If the card holder pays off only interest at the end of each of the 3 months, the total amount of interest paid would be

:<math>\frac{0.1299 \times \$2500}{12} \times 3 = \$27.06\text{ per month} \times 3\text{ months} =\$81.18</math>

which is the simple interest applied over 3 months, as calculated above. (The one cent difference arises due to rounding to the nearest cent.)

===Compound interest===
{{main|Compound interest}}
{{see also|Rate of return}}
Compound interest includes interest earned on the interest that was previously accumulated.

Compare, for example, a bond paying 6 percent semiannually (that is, coupons of 3 percent twice a year) with a certificate of deposit ([[Guaranteed Investment Certificate|GIC]]) that pays 6 percent interest once a year. The total interest payment is $6 per $100 par value in both cases, but the holder of the semiannual bond receives half the $6 per year after only 6 months ([[time preference]]), and so has the opportunity to reinvest the first $3 coupon payment after the first 6 months, and earn additional interest.

For example, suppose an investor buys $10,000 par value of a US dollar bond, which pays coupons twice a year, and that the bond's simple annual coupon rate is 6 percent per year. This means that every 6 months, the issuer pays the holder of the bond a coupon of 3 dollars per 100 dollars par value. At the end of 6 months, the issuer pays the holder:

:<math>\frac {r \cdot B \cdot m}{n} = \frac {6\% \times \$10\,000 \times 1}{2} = \$300</math>

Assuming the market price of the bond is 100, so it is trading at par value, suppose further that the holder immediately reinvests the coupon by spending it on another $300 par value of the bond. In total, the investor therefore now holds:

:<math>\$10\,000 + \$300 = \left(1 + \frac{r}{n}\right) \cdot B = \left(1 + \frac{6\%}{2}\right) \times \$10\,000</math>

and so earns a coupon at the end of the next 6 months of:

:<math>\begin{align}\frac {r \cdot B \cdot m}{n} 
&= \frac {6\% \times \left(\$10\,000 + \$300\right)}{2}\\
&= \frac {6\% \times \left(1 + \frac{6\%}{2}\right) \times \$10\,000}{2}\\
&=\$309\end{align}</math>

Assuming the bond remains priced at par, the investor accumulates at the end of a full 12 months a total value of:

:<math>\begin{align}\$10,000 + \$300 + \$309 
&= \$10\,000 + \frac {6\% \times \$10,000}{2} + \frac {6\% \times \left( 1 + \frac {6\%}{2}\right) \times \$10\,000}{2}\\
&= \$10\,000 \times \left(1 + \frac{6\%}{2}\right)^2\end{align}</math>

and the investor earned in total:

:<math>\begin{align}\$10\,000 \times \left(1 + \frac {6\%}{2}\right)^2 - \$10\,000\\
= \$10\,000 \times \left( \left( 1 + \frac {6\%}{2}\right)^2 - 1\right)\end{align}</math>

The formula for the '''annual equivalent compound interest rate''' is:

:<math>\left(1 + \frac{r}{n}\right)^n - 1</math>

where

:r is the simple annual rate of interest
:n is the frequency of applying interest

For example, in the case of a 6% simple annual rate, the annual equivalent compound rate is:

:<math>\left(1 + \frac{6\%}{2}\right)^2 - 1 = 1.03^2 - 1 = 6.09\%</math>

===Other formulations===
The outstanding [[balance (accounting)|balance]] ''B<sub>n</sub>'' of a loan after ''n'' regular payments increases each period by a growth factor according to the periodic interest, and then decreases by the amount paid ''p'' at the end of each period:

:<math>B_{n} = \big( 1 + r \big) B_{n - 1} - p,</math>

where

:''i'' = simple annual loan rate in decimal form (for example, 10% = 0.10. The loan rate is the rate used to compute payments and balances.)
:''r'' = period interest rate (for example, ''i''/12 for monthly payments) [http://www.fdic.gov/regulations/laws/rules/6500-1650.html#6500226.14]
:''B''<sub>0</sub> = initial balance, which equals the [[principal sum]]

By repeated substitution, one obtains expressions for ''B''<sub>''n''</sub>, which are linearly proportional to ''B''<sub>0</sub> and ''p'', and use of the formula for the partial sum of a [[geometric series]] results in

:<math>B_n = (1 + r)^n B_0 - \frac{(1+r)^n - 1}{r} p</math>

A solution of this expression for ''p'' in terms of ''B''<sub>0</sub> and ''B''<sub>''n''</sub> reduces to

:<math>p = r \left[ \frac{(1+r)^n B_0 - B_n}{(1+r)^n - 1} \right]</math>

To find the payment if the loan is to be finished in ''n'' payments, one sets&nbsp;''B''<sub>''n''</sub>&nbsp;=&nbsp;0.

The PMT function found in [[spreadsheet]] programs can be used to calculate the monthly payment of a loan:

:<math>p=\mathrm{PMT}(\text{rate},\text{num},\text{PV},\text{FV},) = \mathrm{PMT}(r,n,-B_0,B_n,)</math>

An interest-only payment on the current balance would be

:<math>p_I= r B. </math>

The total interest, ''I''<sub>''T''</sub>, paid on the loan is

:<math>I_{T} = np - B_0. </math>

The formulas for a regular savings program are similar, but the payments are added to the balances instead of being subtracted, and the formula for the payment is the negative of the one above. These formulas are only approximate since actual loan balances are affected by rounding. To avoid an underpayment at the end of the loan, the payment must be rounded up to the next cent.

Consider a similar loan but with a new period equal to ''k'' periods of the problem above. If ''r''<sub>''k''</sub> and ''p''<sub>''k''</sub> are the new rate and payment, we now have

:<math>B_k = B'_0 = (1 + r_k) B_0 - p_k. </math>

Comparing this with the expression for B<sub>k</sub> above, we note that

:<math>r_k = (1 + r)^k - 1</math>

and

:<math>p_k = \frac{p}{r} r_k. </math>

The last equation allows us to define a constant that is the same for both problems:

:<math>B^{*} = \frac{p}{r} = \frac{p_k}{r_k}</math>

and ''B''<sub>''k''</sub> can be written as

:<math>B_k = (1 + r_k) B_0 - r_k B^*.</math>

Solving for ''r''<sub>''k''</sub>, we find a formula for ''r''<sub>''k''</sub> involving known quantities and ''B''<sub>''k''</sub>, the balance after ''k'' periods:

:<math>r_k = \frac{B_0 - B_k}{B^{*} - B_0}</math>.

Since ''B''<sub>0</sub> could be any balance in the loan, the formula works for any two balances separate by ''k'' periods and can be used to compute a value for the annual interest rate.

''B''* is a [[scale invariant]], since it does not change with changes in the length of the period.

Rearranging the equation for ''B''<sup>*</sup>, one obtains a transformation coefficient ([[scale factor]]):

:<math>\lambda_k = \frac{p_k}{p} = \frac{r_k}{r} = \frac{(1 + r)^k - 1}{r} = k\left[1 + \frac{(k - 1)r}{2} + \cdots\right]</math> (see [[binomial theorem]])

and we see that ''r'' and ''p'' transform in the same manner:

:<math>r_k=\lambda_k r</math>
:<math>p_k=\lambda_k p</math>.

The change in the balance transforms likewise:

:<math>\Delta B_k=B'-B=(\lambda_k rB-\lambda_k p)=\lambda_k \, \Delta B </math>,

which gives an insight into the meaning of some of the coefficients found in the formulas above. The annual rate, ''r''<sub>12</sub>, assumes only one payment per year and is not an "effective" rate for monthly payments. With monthly payments, the monthly interest is paid out of each payment and so should not be compounded, and an annual rate of 12·''r'' would make more sense. If one just made interest-only payments, the amount paid for the year would be 12·''r''·''B''<sub>0</sub>.

Substituting ''p''<sub>''k''</sub> = ''r''<sub>''k''</sub> ''B''* into the equation for the ''B''<sub>''k''</sub>, we obtain

:<math>B_k=B_0-r_k(B^*-B_0)</math>.

Since ''B''<sub>''n''</sub> = 0, we can solve for ''B''*: 

:<math>B^{*} = B_0 \left(\frac{1}{r_n} + 1 \right).</math>

Substituting back into the formula for the ''B''<sub>''k''</sub> shows that they are a linear function of the ''r''<sub>''k''</sub> and therefore the ''λ''<sub>''k''</sub>:

:<math>B_k=B_0\left(1-\frac{r_k}{r_n}\right)=B_0\left(1-\frac{\lambda_k}{\lambda_n}\right)</math>.

This is the easiest way of estimating the balances if the ''λ''<sub>''k''</sub> are known. Substituting into the first formula for ''B''<sub>''k''</sub> above and solving for ''λ''<sub>''k''+1</sub>, we obtain

:<math>\lambda_{k+1}=1+(1+r)\lambda_k</math>.

''λ''<sub>0</sub> and ''λ''<sub>''n''</sub> can be found using the formula for ''λ''<sub>''k''</sub> above or computing the ''λ''<sub>''k''</sub> recursively from ''λ''<sub>0</sub> = 0 to ''λ''<sub>''n''</sub>.

Since ''p''&nbsp;=&nbsp;''rB''*, the formula for the payment reduces to

:<math>p=\left(r+\frac{1}{\lambda_n}\right)B_0</math>

and the average interest rate over the period of the loan is

:<math>r_\text{loan} = \frac{I_T}{nB_0} = r + \frac{1}{\lambda_n} - \frac{1}{n}, </math>

which is less than ''r'' if&nbsp;''n''&nbsp;>&nbsp;1.