###### Interbank Offered Rates

Floating Rate Bonds A floating rate bond is expressed as a reference rate... **Read More**

An annualized and compounded yield on a fixed rate bond depends on the number of periods in a year, called the periodicity of the annual rate, which typically matches the frequency of coupon payments.

For instance, the annual yield-to-maturity on a 3-year zero-coupon bond priced at 85 per 100 of par value could be calculated as follows:

- for annual compounding, annual yield-to-maturity is 5.567%. Periodicity for this annual yield is 1 (This is also known as an
**effective annual rate**since there is 1 compounding period).

$$\begin{align}\frac { 100 }{ (1+r)^{ 1 } } &=85.00\\ \Rightarrow r &=(\frac{100}{85})^{\frac{1}{3}}-1\\ \therefore r &=5.567\% \ (\text{per year})\end{align}$$

- For semi-annual compounding, annual yield-to-maturity is 5.491%. Periodicity is 2.

$$\begin{align}\frac { 100 }{ (1+r)^{ 1 } } &=85.00\\ \Rightarrow r &=(\frac{100}{85})^{\frac{1}{6}}-1\\ \therefore r &=2.746\% \ (\text{per 6 months})\end{align}$$

This is equivalent to 2.746×2=5.491% (per year)

The annual rate with the periodicity of two is termed as **semi-annual bond basis yield** or **semi-annual bond equivalent yield**. Note that “semi-annual bond basis yield” is different from “yield per semi-annual period.” For instance, if a bond yield is 3% per semiannual period, then the annual yield is 6% stated on a semi-annual bond basis.

- For quarterly compounding, annual yield-to-maturity is 5.454%. The periodicity for this annual yield-to-maturity is 4.

$$\begin{align}\frac { 100 }{ (1+r)^{ 1 } } &=85.00\\ \Rightarrow r &=(\frac{100}{85})^{\frac{1}{12}}-1\\ \therefore r &=1.364\% \ (\text{per 6 months})\end{align}$$

This is equivalent to 4×1.3664%=5.4656% annually.

- For monthly compounding, annual yield-to-maturity is 5.430\%. Periodicity is 12.

$$\begin{align}\frac { 100 }{ (1+r)^{ 1 } } &=85.00\\ \Rightarrow r &=(\frac{100}{85})^{\frac{1}{36}}-1\\ \therefore r &=0.452\% \ (\text{per quarter})\end{align}$$

This is equivalent to 12×0.452%=5.424% annually.

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The general formula for converting annual percentage rate from *m *periods per year (\(APR_m \)) to an annual percentage rate for *n * periods per year (\(APR_n \)) is given below.

$$( 1+\frac{APR_m}{m})^m= (1+\frac{APR_n}{n})^n $$

Using algebra, we can convert between different periodicities. For example, let’s convert 6% from a periodicity of two to a periodicity of four.

Using the above algebra, we can convert between different periodicities. For example, let’s convert 6% from a periodicity of two to a periodicity of four.

$$\begin{align} (1+\frac{0.06}{2})^2 &= (1+\frac{ APR_4 } {4} )^4 \\ (1+\frac{0.06}{2})^ {\frac{2}{4}} &= 1+\frac{ APR_4 } {4}1.0149 &= 1+\frac{ APR_4 } {4}\\ 4 × (1.0149 -1) &= APR_4\\ \Rightarrow APR_4 &= 0.0596\end{align}$$

The periodicity being higher means more compounding is done each year. As such, it makes sense that the rate is lower than 6%.

The general rule of thumb is that compounding more frequently at a lower annual rate is equivalent to compounding less regularly at a higher annual rate.

The 2-year, 6% semi-annual coupon payment bond is priced at 105 per 100 of par value. Calculate the yield-to-maturity and its equivalent annual rate compounded monthly.

**Solution**

Using the general bond pricing formula:

$$PV=\frac{PMT}{(1+r)^1}+ \frac{PMT}{(1+r)^2}+…+\frac{PMT}{(1+r)^N}$$

We can either find the yield-to-maturity using the financial calculator or just trial and error, and then we use linear interpolation for approximation. Let us handle the trial and error method.

Note that this bond is trading at a premium since the price is higher than the par value, and therefore we expect the yield-to-maturity to be less than the coupon rate. Now, using the above equation we have:

$$p=\frac{3}{(1+r)^1}+ \frac{3}{( 1+r )^2}+…+\frac{3}{(1+r)^4}$$

Let us start with \(r_1=1.5 \%\), so:

$$p_1=\frac{3}{(1.015)^1}+ \frac{3}{( 1.015 )^2}+…+\frac{3}{(1.015)^4}=105.7816$$

Now, at \(r_1=2 \%\)

$$p_1=\frac{3}{(1.02)^1}+ \frac{3}{( 1.02 )^2}+…+\frac{3}{(1.02)^4}=103.8077$$

Now we use the linear interpolation:

$$\begin{align}\frac{p-p_1}{p_2-p_1 }&≈\frac{r-r_1}{r_2-r_1}\\ \frac{105-105.7816}{103.8077-105.7816 } &≈\frac{r-0.015}{0.02-0.015}\\ \Rightarrow r &=0.015+ (\frac{105-105.7816}{103.8077-105.7816 })(0.02-0.015)=0.01698 \end{align}$$

So that the annual rate is:

$$0.01698×2=0.03396=3.396 \%$$

Using the following formula, we can calculate the annualized monthly rate as:

$$\begin{align}(1+\frac{0.03396}{2})^2&=(1+(\frac{APR_12}{12})^{12}\\ \Rightarrow APR_{12}&=12[(1+\frac{0.03396}{2})^{\frac{2}{12}}-1]=0.03372=3.372 \%\end{align}$$

QuestionA hypothetical bond has an annual percentage rate of 8% with a periodicity of 12. This bond’s semi-annual rate is

closestto:

- 8.01%
- 8.10%
- 8.13%

SolutionThe correct answer is

C.\( (1+\frac{0.08}{12})^{12} = (1+\frac{ APR_2 } {2} )^2 \)

\( (1+\frac{0.08}{12})^ { \frac{12}{2} } = 1+\frac{ APR_2} {2} \)

\( 2 × (1+\frac{0.08}{12})^ {\frac{12}{2} }) – 1 = APR_2 \)

\( APR_2 = 0.0813 \)