# Australian Treasury Bonds - Price Calculation with Accrual

In this document ASX Interest Rate Derivatives (on page 7) the Australian Commonwealth Treasury Bond (paying semi-anually) is valued as

$$P = v^{f/d} \cdot \left(\frac{c}{2} + \frac{c}{2}\cdot\sum_{k=1}^n v^k + 1\cdot v^n\right)\cdot 100$$ where $v=\frac{1}{1+y/2}$ (the "one-period" DF), $c=$ annual coupon, $f=$ number of days from settlement date of to next interest payment date (ranging from $0$ to ~$184$), $d=$ number of days in the half year ending on next interest payment date (usually ~$184$). So $f$ will decrease from $d$ to $0$ as we approach the next payment date.

The middle summand $\frac{c}{2}\cdot\sum_{k=1}^n v^k$ (PV of coupons) and last summand $1\cdot v^n$ (PV of nominal) are clear.

What confuses me is $v^{f/d}$ and the standalone $\frac{c}{2}$ at the front, which I assume account for accrued interest? But shouldn't accrued interest be calculated as $$\text{acrr} = \frac{c}{2}\cdot \frac{d-f}{d}$$ ?

Would anybody know how to interpret the first parts of this formula?

The factor $v^{f/d}$ multiplying the whole equation represents discounting the future bond flows over the period between the settlement date and the first coupon payment - it is not a whole period, so some fraction of the semi-annual DF factor is needed. With this convention, the fraction is obtained by raising the DF to the power of the ratio of number of days.
The first coupon is then the $\frac{c}2$ term inside the bracket. You can see that it can be incorporated into the sum over the other coupons simply by starting the summation at $k=0$ rather than $k=1$.