# Reinvestment yield rate with bonds

A 1000 par value 10-year bond is purchased with 9% semiannual coupons for 950. The coupon payments can be reinvested at a nominal rate of 8% convertible semiannually. Calculate the nominal annual yield rate convertible semiannually over the ten-year period.

There is an equation for reinvestment with bonds

$P(1+i')^n=Fr*\frac{(1+j)^n-1}{j}+C$

where P is the purchase price, coupons are Fr, number of periods is n, the redemption rate is C, j is the reinvestment rate, and the yield rate is $i'$.

For this problem, I'm looking for $i'$ and I already know $n=20$, $Fr=950$, $j=.04$, and $C=1000$.

So I can get $P(1+i')^{20}=Fr*\frac{(1.045)^{20}-1}{.045}+10000$, but how do I get the purchase price?

• Didn't you say that the purchase price was 950 in the first line? Commented Nov 14, 2017 at 3:50
• You are right and I am dumb. Commented Nov 14, 2017 at 3:55
• I wouldn't say dumb, but I could say polite if you gave my comment a positive vote. Commented Nov 14, 2017 at 4:20

$Fr=1000*.045=45$

$P(1+i′)n=Fr∗\frac{(1+j)^n−1}{j}+C$

$950(1+i')^{20}=45*45\frac{1.04^{20}-1}{.04}+1000$

$i'\approx .046103$

Multiply by 2 because of semiannual rates and then $i'\approx .092212$

Face value $F=1000$, coupon rate $r=\frac{r^{(2)}}{2}=\frac{9\%}{2}=4.5\%$, number of coupon payment periods $n=2\times 10=20$, amount paid at maturity $C=F=1000$ (par value bond), reivestment rate $i=\frac{i^{(2)}}{2}=\frac{8\%}{2}=4\%$, coupon price $P=950$.

At the end of 10 years, we will receive the principal of $1,000$, plus the accumulated value of reinvested coupons equal to $$Fr\times s_{\overline{n}|i}=Fr\times\frac{(1+i)^n-1}{i}=45 \times s_{\overline{20}|4\%}\approx 1,340$$ for a total of $$Fr\times s_{\overline{n}|i}+C\approx 2,340.$$

Thus over $10$ years, we turned our initial investment of $950$ into $2,340$, and, if we write $j^{(2)}$ for the nominal annual yield rate convertible semiannually, then we must have $$P\left(1+\frac{j^{(2)}}{2}\right)^n=Fr\times s_{\overline{n}|i}+C$$ that is $$950\left(1+\frac{j^{(2)}}{2}\right)^{20}=2,340$$ and then

$$j^{(2)}=2\left[\left(\frac{2,340}{950}\right)^{1/20}-1\right]\approx 9.22\%$$