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An investor purchases two bonds with the following properties:

Bond 1: Has a face value $1000 and is redeemable at par. Pays coupons annually at a rate of 8.1% annual and was purchased for 1117.19.

Bond 2: Has a face value $1000 and is redeemable at par. Pays coupons annually at a rate of 6.5% annual and was purchased for 981.32.

If both bonds mature in the same number of years and the investor yields the same rate on both bonds, find the yield rate.

I have tried my best to work it out but there are too many variables and not sure how to go about it. Please help me find the yield rate

Alex Vincent
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2 Answers2

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I would just make a spreadsheet to compute the return on each bond for different terms. Most spreadsheets will compute rate of return for a cash flow. The first bond will have a yield of $8.1\%$ for very long terms because the effect of the premium will be reduced. At $1$ year the yield will be negative. The second bond will have a yield of $6.5\%$ per year for long terms for the same reason, but it will rise for short terms because of the discount. At $1$ year it will be something like $8.4\%$ because the discount will give an additional $1.9\%$ of yield. You can then try different terms to see where the yields match. I would guess around $8-9$ years as you have $13.6\%$ premium differential and $1.6\%$ yield differential, but that is a guess.

Ross Millikan
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It's like a CFA question.

Strictly speaking, the valuation of bonds depends on how you model interest rate term structure (ho-lee, HJM, or LIBOR etc.). But here I guess we assume the interest rate will be constant throughout the investment horizon. Plus, the convention is that bond pays semi-annual coupon at the rate of $(1 + C)^{0.5} - 1$ (or approx. $0.5C$), but here I just assume it pays once a year, at the rate of $C$ for simplicity.

So here is the answer (hopefully I got the numbers right, but even if not, you get how to do it): $$1117.19 = \frac {1000}{(1+r)^n} + \frac{81}{(1+r)} + \frac{81}{(1+r)^2} + ... + \frac{81}{(1+r)^n}=\frac{1000}{(1+r)^n} + 81 \times \frac{(1+r)^n - 1}{(1+r)^n \times r}$$ $$981.32 = \frac{1000}{(1+r)^n} + \frac{65}{(1+r)} + \frac{65}{(1+r)^2}+...+\frac{65}{(1+r)^n}=\frac{1000}{(1+r)^n} + 65 \times \frac{(1+r)^n - 1}{(1+r)^n \times r}$$ let $x = \frac {1000}{(1+r)^n}$, and $y=\frac{(1+r)^n - 1}{(1+r)^n\times r}$, we have $$1117.19=x + 81 \times y$$ $$981.32=x+65 \times y$$ Solving x and y, we have: $$\frac{1000}{(1+r)^n} =x= 429.348125$$ $$\frac{(1+r)^n - 1}{(1+r)^n \times r}=y=8.491875$$ Thus $$(1+r)^n = 2.329112303$$ $$r=0.06719975=6.72\%$$ $$n=13$$

Jay Zha
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