Find i and find n

Question

The question is: One bond of face value 100 with semiannual coupons and r =0.025 costs 75.74. A similar bond with semiannual coupons and r=0.04 costs 112.13. Both are redeemable at par in n years and have the same yield rate i. (a) find i, (b) find n.

I know that the answers are supposed to be i=0.035 and n=27.5 years.

I got this but I think I am doing it wrong and I am not sure where to go next with it. Thank you.

i= i2 (yield)

j=i/2 rate per half year

Face value= 100

r=0.25

cost of first bond = =75.74

r=0.04

Cost of second bond = 112.13

Semi annual yield to maturity of 1st bond = $\text {(face value-price)}n - C \div \text {(Face value +Bond)}/2$ $= (100-75.74)/2 - 0.025 \div (100+75.74)/2$ $= 12.11/87.87 = 0.14$

Semi annual yield to maturity of 2st bond $= \text{(face value-price)}n - C \div \text{(Face value +Bond)}/2$ $= (100-112.13)/2-0.04) \div (100+112.13)/2$ $= -6.11/106.07 = -0.005$

Yield to maturity = 2* semi annual yield to maturity

1st bond = $2*0.14 = 0.28$

2nd Bond = $2*(-0.005) = -0.01$

Answer 1

Let be $P$ the price of the bond, $F$ the face value (par value) of the bond, $C$ the amount paid at maturity, $r$ the coupon rate, $m$ number of coupon payment periods.

The price of a bond is

$$P=Fr\,a_{\overline{m}|i}+Cv^{m}$$ where $a_{\overline{m}|i}=\frac{1-v^m}{i}$ and $v=\frac{1}{1+i}$.

In the problem $F=C=100$, $P_1=75.74$, $r_1=2.5\%$, $P_2=112.13$, $r_2=4\%$ and $m=2n$. So we have the system of equation $$\left\{ \begin{align*} P_2&=Fr_2\,a_{\overline{2n}|i}+Cv^{2n}\\ P_1&=Fr_1\,a_{\overline{2n}|i}+Cv^{2n} \end{align*}\right. $$ that is $$ \begin{align} 112.13&=4\,a_{\overline{2n}|i}+100v^{2n}\tag 1\\ 75.74&=2.5\,a_{\overline{2n}|i}+100v^{2n}\tag 2 \end{align} $$ From $(1)-(2)$ we find $a_{\overline{2n}|i}=\frac{112.13-75.74}{4-2.5}=24.26$ and substituting in $(1)$ or $(2)$ we find $v^{2n}=0.1509$ $$ \begin{align} v^{2n}&=0.1509\tag 3\\ 1-v^{2n}&=i\times 24.26 \tag 4 \end{align} $$ and then $$\boxed{i=\frac{1-0.1509}{24.26}=3.5\%}$$ and from $$ v^{2n}={(1+i)^{-2n}}={1.035^{-2n}}=0.1509 $$ we have $$\boxed{ n=-\frac{1}{2}\times \frac{\log 0.1509}{\log 1.035}\approx 27.5} $$

Answer 2

Let the yield be r and the number of years be n

First Bond Price relationship is

$75.74 = \frac{2.5}{1+\frac{r}{2}} +\frac{2.5}{(1+\frac{r}{2})^2}+\cdots + \frac{2.5}{(1+\frac{r}{2})^{2n}} +\frac{100}{(1+\frac{r}{2})^{2n}}\tag 1 $

$112.13 = \frac{4}{1+\frac{r}{2}} +\frac{4}{(1+\frac{r}{2})^2}+\cdots + \frac{4}{(1+\frac{r}{2})^{2n}} +\frac{100}{(1+\frac{r}{2})^{2n}} \tag 2$

Let us put $ i = \frac{1}{1+\frac{r}{2}} $

$75.74 = 2.5\left(i+i^2+\cdots+i^{2n}\right) +100.i^{2n}\tag 3 $

$112.13 = 4 \left(i+i^2+\cdots+i^{2n}\right) +100.i^{2n}\tag 4 $

Multiply $(3)$ by $4$ and $(4)$ by $2.5$ and subtract the second from first:

$4\times 75.74 - 2.5\times 112.13 = =1.5\times 100.i^{2n}$

$22.635 = 150i^{2n}$

$i^{2n} = .1509$

Simply subtract $(3)$ from $(4)$

$36.39 = 1.5i\left(1+i+\cdots+i^{2n-1}\right)$

$i\frac{(1-i^{2n})}{1-i} =\frac{36.39}{1.5} = 24.26$

$i(1-.1509) = (1-i)24.26$

$25.1091i = 24.26$

$ i = \frac{1}{1+\frac{r}{2}} = .96618$

$1+ \frac{r}{2} = \frac{1}{.96618}$

$1+ \frac{r}{2} = 1.035$

$ \frac{r}{2} = .035$

$ r = 2\times 0.035 = 0.07$

$i^{2n} = 0.1509$

$2n \log(0.96618) = \log(0.1509)$

$2n = 55$

Number of years $= 27.5$

Semi-Annual Yield $= 3.5$%

Yield to Maturity= $7.00$%