A 3-year bond pays an annual coupon of 80 and repays its face value of 1,000 at maturity. It currently trades at 938. Find its yield to maturity y, that is, solve 938 = 80/(1+y) + 80/(1+y)^2 + 1080/(1+y)^3.

Step-by-step solution

1. 1

   Read the equation as "price = present value of the cash flows". Each term discounts one payment back to today at the same unknown rate $y$. The last term is $1080$, not $80$, because year 3 brings the final coupon and the $1{,}000$ face value: $80+1000=1080$. The yield to maturity is precisely the single discount rate that makes the present value equal the quoted price.

2. 2

   Substitute $x=1+y$ and clear the denominators. With $x=1+y$ the equation becomes

   $938=\frac{80}{x}+\frac{80}{x^2}+\frac{1080}{x^3}$

   and multiplying through by $x^3$ gives a cubic:

   $938x^3-80x^2-80x-1080=0$

   This shows why there is no tidy formula: for three or more periods the yield solves a polynomial of degree $\ge 3$, so it must be found numerically. The substitution is still worth doing, because it makes the function obviously decreasing in $x$ — a higher yield always means a lower price — so there is exactly one economically meaningful root.

3. 3

   Anchor the search with the par value. At $y=8\%$ (the coupon rate) the price would be

   $\frac{80}{1.08}+\frac{80}{1.08^2}+\frac{1080}{1.08^3}=74.07+68.59+857.34=1000.00$

   exactly par. The bond actually trades at $938<1000$, a discount, so the yield must be above $8\%$. This one check tells you which direction to search before doing any trial and error.

4. 4

   Bracket the root with two trials. Try $y=10\%$ ($x=1.10$):

   $\frac{80}{1.10}+\frac{80}{1.21}+\frac{1080}{1.331}=72.7273+66.1157+811.4200=950.2630$

   Still above $938$, so the yield is higher. Try $y=11\%$ ($x=1.11$):

   $\frac{80}{1.11}+\frac{80}{1.2321}+\frac{1080}{1.367631}=72.0721+64.9298+789.6867=926.6886$

   Now below $938$. The root is trapped between $10\%$ and $11\%$.

5. 5

   Interpolate, then refine. Linear interpolation between the two prices gives a first estimate

   $y\approx 0.10+0.01\cdot\frac{950.2630-938}{950.2630-926.6886}=0.10+0.01\cdot\frac{12.2630}{23.5744}=0.1052$

   One or two Newton steps on $f(x)=\frac{80}{x}+\frac{80}{x^2}+\frac{1080}{x^3}-938$ settle it at

   $x=1.1051578\quad\Longrightarrow\quad y=0.1051578$

6. 6

   Check the answer by pricing the bond again. With $y=10.51578\%$:

   $\frac{80}{1.1051578}+\frac{80}{1.1051578^2}+\frac{1080}{1.1051578^3}=72.3879+65.5000+800.1121=938.00$

   The price is reproduced to the cent, so $y\approx 10.52\%$. Note that a rounded guess such as $10.3\%$ prices the bond at about $943.1$ — over five dollars too high — which is why the bracketing step matters: do not stop at the first estimate that merely looks close.