Using linear interpolation practice problem

Cathy Moran, CFA, is estimating a value for an infrequently traded bond with six years to maturity, an annual coupon of 7%, and a single-B credit rating. Moran obtains yields-to-maturity for more liquid bonds with the same credit rating:

5% coupon, eight years to maturity, yielding 7.20%.
6.5% coupon, five years to maturity, yielding 6.40%.
The infrequently traded bond is most likely trading at: ?

Answer:
Using linear interpolation, the yield on a bond with six years to maturity should be 6.40% + (1 / 3)(7.20% – 6.40%) = 6.67%. A bond with a 7% coupon and a yield of 6.67% is at a premium to par value.

My question: Please explain why the “1/3” is used.

6.40% + (1 / 3)(7.20% – 6.40%) = (2/3) 6.4% + (1/3) 7.20%

You can also think of it as 6 yrs is 1/3 of the distance from 5 yrs to 8 yrs. :nerd_face:

\frac{6\ years - 5\ years}{8\ years - 5\ years} = \frac{1\ year}{3\ years} = \frac13

I like your explanation. Now how can I think of the 2/3 that you used?

\frac{6\ years - 8\ years}{5\ years - 8\ years} = \frac{-2\ years}{-3\ years} = \frac23

6 years is ⅔ of the way from 8 years to 5 years.

Excellent.

(2/3) 6.4% + (1/3) 7.20%

Now why did he multiply the (2/3) with 6.4% and conversely, (1/3) with 7.20%?

This is one where your intuition can lead you astray. You might think that the ⅓ weight goes with 5 years and the ⅔ weight with 8 years (because 6 is ⅓ of the way from 5 to 8, and ⅔ of the way from 8 to 5).

However, if you draw a time line with 5 years, 6 years, and 8 years in anything close to proper scale, you’ll see that 6 is closer to 5 than to 8, so 5 gets the bigger weight and 8 gets the smaller weight. It would be a lot easier to see if instead of a 6-year bond you had a 5.1-year bond: 5.1 years is right on top of 5 years, so it should get a weight of 29/30, and 8 years get a weight of 1/30.

Excellent! That explanation has now led to all of the pieces being understood. Thanks a million!

My pleasure.