Similar to most fixed income securities in Australia, Treasury Bonds are quoted and traded on a yield to maturity basis rather than on a price basis. This means the price is calculated by inputting the yield into the appropriate pricing formula.
The price per $100 face value is calculated using the following pricing formulae:
(1) Basic formula
\(\Large P = v^\frac{f}{d}\left(g \left( 1 + \require{enclose}a_{\enclose{actuarial}{n}} \right) + 100 v^{n} \right)\)
(2) Ex-interest formula
\(\Large P = v^\frac{f}{d}\left(g \require{enclose}a_{\enclose{actuarial}{n}} + 100 v^{n} \right)\)
The ex-interest period for Treasury Bonds is seven calendar days. With ex-interest Treasury Bonds the next coupon payment is not payable to a purchaser of the bonds. In this case, calculation of an ex-interest price is effected by the removal of the '\( 1 \)' from the term:
\( \large 1+\require{enclose}a_{\enclose{actuarial}{n}}\)
in formula \( (1) \), thereby adjusting for the fact that the purchaser will not receive a coupon payment at the next interest payment date.
(3) Near-maturing bonds formula (between the record date for the second last coupon and the record date for the final coupon)
\(\Large P = \LARGE \frac{100 + g}{1 + \left(\frac{f}{365}\right) i}\)
When a Treasury Bond goes ex-interest for the second last time it is treated as a special case. In this case formula (3) applies up until the record date for the final interest payment. There may be a slight discontinuity in the progress of the price of the bond around the time the bond goes ex-interest for the second last time but market participants can, if they wish, allow for this in their trading.
Where the maturity date coincides with a weekend or public holiday, the commonly accepted practice is to price near-maturing Treasury Bonds according to the actual date the principal and final interest are paid (and not the nominal maturity date).
(4) Near-maturing bonds formula (between the record date for the final coupon and maturity of the bond)
\(\Large P = \LARGE \frac{100}{1 + \left(\frac{f}{365}\right) i}\)
When a Treasury Bond goes ex-interest for the last time it is treated as a special case. In this case formula (4) applies from the time the bond goes ex-interest for the final time.
Where the maturity date coincides with a weekend or public holiday, the commonly accepted practice is to price near-maturing Treasury Bonds according to the actual date the principal and final interest are paid (and not the nominal maturity date).
In these formulae:
\(\large P=\) the price per $100 face value. \( P \) is rounded to three decimal places in formulae (1) and (2), and unrounded in formulae (3) and (4).
\(\large i=\) the annual percentage yield to maturity divided by 200 in formulae (1) and (2), or the annual percentage yield to maturity divided by 100 in formula (3) and (4).
\(\large v=\LARGE\frac{1}{1 + i} \)
\(\large f=\) the number of days from the date of settlement to the next interest payment date in formulae (1) and (2) or to the maturity date in formulae (3) and (4). In formulae (3) and (4), if the maturity date falls on a non-business day, the next good business day (as defined in the Information Memorandum) is used in the calculation of \(f\).
\(\large d=\) the number of days in the half year ending on the next interest payment date.
\(\large g=\) the half-yearly rate of coupon payment per $100 face value.
\(\large n=\) the term in half years from the next interest payment date to maturity.
\(\large \require{enclose}a_{\enclose{actuarial}{n}}=\large v + v^2 + ... + v^n = \LARGE \frac{1 - v^n}{i}\) \( . \mathrm{Except \, if\ \,} i = 0 \ \mathrm{\,then\,}\ \require{enclose}a_{\enclose{actuarial}{n}} = n \)
Worked Examples
(1) Basic formula
Consider the 2.75% 21 November 2029 Treasury Bond, with a yield to maturity of 1.10 per cent and settlement date of 12 September 2019.
\(\Large P = v^\frac{f}{d}\left(g \left( 1 + \require{enclose}a_{\enclose{actuarial}{n}} \right) + 100 v^{n} \right)\)
where:
\( \large\ i=\Large\frac{1.10}{200}= \) \(\large 0.0055\)
\( \large\ v=\Large\frac{1}{1+i}=\frac{1}{1+0.0055}= \) \(\large 0.99453\)
\(\large \require{enclose}a_{\enclose{actuarial}{n}}=\Large \frac{1 - v^n}{i}=\frac{1-0.99453^{20}}{0.0055}=\)\(\large 18.8904\)
\( \large\ f = 70 \), the number of days from 12 September 2019 to 21 November 2019
\( \large\ d = 184 \), the number of days from 21 May 2019 to 21 November 2019
\( \large\ g =\Large\frac{2.75}{2}= \) \(\large 1.375\)
\( \large\ n = 20\), the number of half years from 21 November 2019 to 21 November 2029
\(\Large P = v^\frac{f}{d}\left(g \left( 1 + \require{enclose}a_{\enclose{actuarial}{n}} \right) + 100 v^{n} \right) = 0.99453^\frac{70}{184}\left(1.375 \left( 1 + 18.8904\right) + 100\times0.99453^{20} \right)\)
\(\Large P = 116.716\)
(2) Ex-Interest formula
Consider the 2.50% 21 May 2030 Treasury Bond, with a yield to maturity of 1.10 per cent and settlement date of 15 November 2019.
\(\Large P = v^\frac{f}{d}\left(g \left( \require{enclose}a_{\enclose{actuarial}{n}} \right) + 100 v^{n} \right)\)
where:
\( \large\ i=\Large\frac{1.10}{200}= \) \(\large 0.0055\)
\( \large\ v=\Large\frac{1}{1+i}=\frac{1}{1+0.0055}= \) \(\large 0.99453\)
\(\large \require{enclose}a_{\enclose{actuarial}{n}}=\Large \frac{1 - v^n}{i}=\frac{1-0.99453^{21}}{0.0055}=\)\(\large 19.78135\)
\( \large\ f = 6 \), the number of days from 15 November 2019 to 21 November 2019
\( \large\ d = 184 \), the number of days from 21 May 2019 to 21 November 2019
\( \large\ g =\Large\frac{2.50}{2}= \) \(\large 1.25\)
\( \large\ n = 21\), the number of half years from 21 November 2019 to 21 May 2030
\(\Large P = v^\frac{f}{d}\left(g \left( \require{enclose}a_{\enclose{actuarial}{n}} \right) + 100 v^{n} \right) = 0.99453^\frac{6}{184}\left(1.25 \left( 19.78135\right) + 100\times0.99453^{21} \right)\)
\(\Large P = 113.827\)
(3) Near-maturing bonds formula (between the record date for the second last coupon and the record date for the final coupon)
Consider the 2.75% 21 October 2019 Treasury Bond, with a yield to maturity of 1.00 per cent and settlement date of 26 September 2019.
\(\Large P = \LARGE \frac{100 + g}{1 + \left(\frac{f}{365}\right) i}\)
where:
\( \large\ i=\Large\frac{1.00}{100}= \) \(\large 0.01\)
\( \large\ f = 25 \), the number of days from 26 September 2019 to 21 October 2019
\( \large\ g =\Large\frac{2.75}{2}= \) \(\large 1.375\)
\(\Large P = \LARGE \frac{100 + g}{1 + \left(\frac{f}{365}\right) i}=\frac{100+1.375}{1+\left(\frac{25}{365}\right) \times0.01}\)
\(\Large P = 101.305613\)
(4) Near-maturing bonds formula (between the record date for the final coupon and maturity of the bond)
Consider the 2.75% 21 October 2019 Treasury Bond, with a yield to maturity of 1.00 per cent and settlement date of 16 October 2019.
\(\Large P = \LARGE \frac{100}{1 + \left(\frac{f}{365}\right) i}\)
where:
\( \large\ i=\Large\frac{1.00}{100}= \) \(\large 0.01\)
\( \large\ f = 5 \), the number of days from 16 October 2019 to 21 October 2019
\(\Large P = \LARGE \frac{100}{1 + \left(\frac{f}{365}\right) i}=\frac{100}{1+\left(\frac{5}{365}\right) \times0.01}\)
\(\Large P = 99.986303\)
