# Price/Yield Conversion In Practice When working with Fixed Income trading systems we are often confronted with the requirement to implement a bond security Price/Yield conversion service. Many Finance textbooks describe the computation but the provided formulas are too academic and examples are provided only for the special case when the valuation date coincides with the coupon payment date. This post aims to provide a step-by-step explanation of the Price/Yield conversion used by market practitioners and valid for any valuation date.

We will begin with the process of computing the price, given the yield, of a Federal Republic of Germany debt security maturing on the 4th of January 2020 and paying an annual coupon of 3.25%.

## Present Value

Let’s lay down the pricing parameters first:

Name Value
Maturity 04/01/20
Coupon Rate 3.25%
Coupon frequency 1
Valuation date 04/10/16
Nominal value 100
Day count convention Actual/Actual
Previous coupon payment date 04/01/16
Next coupon payment date 04/01/17
Yield 0.2%

The next step is to determine the full price of the bond using the following equation:

$P^{full} = \frac{c_{1}}{(1 + y)^{T_{1}}} + \frac{c_{2}}{(1 + y)^{T_{2}}}+\dotsm+\frac{Nominal\ value + c_{n}}{(1 + y)^{T_{n}}}$

The full price of the bond is the price including accrued interest. It is also called “dirty price” or present value.

By convention, the traded price is the “clean price” which can be determined by subtracting the accrued interest from the “dirty price”.

It is calculated as follows:

$P^{clean} = P^{full} - Accrued\ Interest$

## Accrued Interest

Accrued interest equals the coupon times the fraction of the coupon period from the previous coupon payment date to the valuation date. This fraction is calculated by using the day count convention of the bond which is in our case Actual/Actual.

This gives us:

$Factor = \frac{Days(Previous\ coupon\ date,\ Valuation\ date) }{Days(Previous\ coupon\ date,\ Next\ coupon\ date)}$

Which by substituting our values equals:

$Factor = \frac{Days(04/01/16, 04/10/16) }{Days(04/01/16, 04/01/17)} = \frac{274}{366} = \textbf{0.7486338798}$

For regular coupon periods the coupon equation is:

$Coupon = \frac{Coupon\ rate \times Nominal }{Coupon\ frequency} = \frac{0.0325 \times 100}{1} = \textbf{3.25}$

At present, we have the values required for the computation of the Accrued Interest:

$Accrued\ Interest = Coupon \times Factor = 3.25 \times 0.7486338798 = \textbf{2.4330601093}$

## Time Periods

Let’s now go back to the equation of the “dirty price”:

$P^{full} = \frac{c_{1}}{(1 + y)^{T_{1}}} + \frac{c_{2}}{(1 + y)^{T_{2}}}+\dotsm+\frac{Nominal\ value + c_{n}}{(1 + y)^{T_{n}}}$

It is worth noting that yields $$(y)$$ are annually compounded as we use the German Fixed Rate Bonds method in which yields are compounded on the same frequency as the coupon frequency.

The T1…Tn time periods are thus expressed in years.

T1 equals to the remaining lifetime of the current coupon:

$T_{1} = \frac{Days(Valuation\ date,\ Next\ coupon\ date) }{Days(Previous\ coupon\ date,\ Next\ coupon\ date)} = \frac{92}{366} = \textbf{0.2513661202}$

The subsequent time periods are obtained by adding the year difference between the next coupon and the coupon ci to the remaining lifetime of the current coupon.

\begin{align} T_{2}& = T_{1} + 1\\ T_{3}& = T_{1} + 2\\ \vdots\\ T_{n}& = T_{1} + n - 1 \end{align}

## Clean Price

The Pfull equation thus becomes:

$P^{full} = \frac{3.25}{(1 + 0.002)^{0.2513661202}} + \frac{3.25}{(1 + 0.002)^{1.2513661202}}+\frac{3.25}{(1 + 0.002)^{2.2513661202}}+\frac{103.25}{(1 + 0.002)^{3.2513661202}}=\textbf{112.3071034523}$

And at the end we obtain the “clean price”:

$P^{clean} = 112.3071034523 - 2.4330601093 = \textbf{109.874043343}$

Which rounded to 5 decimal places is the exact value given by the Bloomberg YAS screen, the reference for market operators.

For convenience, we provide below the computation for each coupon period.

Date Payment T (Years) Discount Factor Present Value
04/01/17 3.25 0.2513661202 0.9994978959 3.2483681617
04/01/18 3.25 1.2513661202 0.9975028901 3.241884393
04/01/19 3.25 2.2513661202 0.9955118664 3.2354135658
04/01/20 103.25 3.2513661202 0.9935248168 102.5814373317
Sum 112.3071034523

# Computing the Yield

We’ll now take the opposite direction and compute the yield given the clean price of 109.874043.

As the above equation cannot be rearranged to give yield as an algebraic function of price the Newton-Raphson method is used to solve for y.

The Newton-Raphson iteration procedure will, starting from an initial estimate of the yield $$y_0$$, iterate using more and more precise estimates of the yield until the difference between the $$P^{clean}$$ derived from using the yield estimate and the given market price of the bond is less than 0.000001.

## First iteration

Using the initial seed value formula

$y_0 = \frac{Coupon}{P^{clean}}+\left[\frac{100}{P^{clean}}\right]^{TTM^{-1}} - 1$ where $$TTM$$ is the time to maturity of the bond, we obtain the initial value

$y_0 = \frac{3.25}{109.874043}+\left[\frac{100}{109.874043}\right]^{3.2513661202^{-1}} - 1 = \textbf{0.001033 }$

By substituting the initial yield into the price formula we obtain the price

$P^{clean} = P^{full} - AI = \frac{3.25}{(1 + 0.001033)^{0.2513661202}} + \frac{3.25}{(1 + 0.001033)^{1.2513661202}}+\frac{3.25}{(1 + 0.001033)^{2.2513661202}}+\frac{103.25}{(1 + 0.001033)^{3.2513661202}} -2.4330601093 =\textbf{110.208333}$

As this price is greater than the given price 109.874043 the above formula is recalculated with a higher yield.

## Second iteration

According to Newton-Raphson the $$y_{1}$$ is determined as:

$y_1 = 0.001033 - \frac{110.208333-109.874043}{-346.399}= \textbf{0.001998}$

where -346.399 is the derivative of the price equation with respect to yield.

It is determined by computing the derivative of the price equation and substituting $$y$$ with $$y_{0}$$.

$f^\prime(y)=\frac{dP}{dy}= \frac{-T_{1}\times c_{1}}{(1 + y)^{1+T_{1}}} + \frac{-T_{2}\times c_{2}}{(1 + y)^{1+T_{2}}}+\dotsm+ \frac{-T_{n}\times (Nominal\ value+c_{n})}{(1 + y)^{1+T_{n}}}$

Substituting $$y$$ with $$y_{0}$$ gives:

$f^\prime(y_{0})=\frac{-0.2513661202\times 3.25}{(1 + 0.001033)^{1.2513661202}} + \frac{-1.2513661202\times 3.25}{(1 + 0.001033)^{2.2513661202}}+\frac{-2.2513661202\times 3.25}{(1 + 0.001033)^{3.2513661202}}+ \frac{-3.2513661202 \times 103.25}{(1 + 0.001033)^{4.2513661202} }= \textbf{-346.399}$

Our new estimate of the yield is substituted into the price formula and gives us a new clean price of

$P^{clean} = P^{full} - AI = \frac{3.25}{(1 + 0.001998)^{0.2513661202}} + \frac{3.25}{(1 + 0.001998)^{1.2513661202}}+\frac{3.25}{(1 + 0.001998)^{2.2513661202}}+\frac{103.25}{(1 + 0.001998)^{3.2513661202}} -2.4330601093 =\textbf{109.874733}$

The difference between the obtained price and the market price is $$109.874733-109.874043=0.00069$$ which is greater than the tolerance 0.000001. A new iteration is thus required.

## Third iteration

As the price of the second iteration is greater than the market price we’ll add a small margin to $$y_{1}$$.

$y_2 = 0.001998-\frac{109.874733-109.874043}{-345}= \textbf{0.002}$

The derivative -345 is obtained by substituting $$y_{1}$$ in the derivative equation.

$f^\prime(y_{1})=\textbf{-345}$

Using the new yield in the price formula we get

$P^{clean} = P^{full} - AI = \frac{3.25}{(1 + 0.002)^{0.2513661202}} + \frac{3.25}{(1 + 0.002)^{1.2513661202}}+\frac{3.25}{(1 + 0.002)^{2.2513661202}}+\frac{103.25}{(1 + 0.002)^{3.2513661202}} -2.4330601093 =\textbf{109.874043}$

As the new price is equal to the market price no new iterations are required and the yield of the bond is set to 0.2%.