Unfortunately, the original data are lost to posterity.
However, I can run you through another example to show you how it’s done. I’ll use annual pay bonds to make it easier to follow.
Suppose that the 1-year par rate is 2% and the 2-year par rate is 3%. As a binomial interest rate tree uses 1-period forward rates at each node, let’s compute the 1-year and 2-year spot rates, then the forward rates.
The 1-year spot rate – s1 – is easy: 2%.
To get the 2-year spot rate – s2 – we note that a 2-year bond paying a 3% coupon sells at par, so:
1,000 = 30 / 1.02 + 1,030 / (1 + s2)2
1,000 = 29.4118 + 1,030 / (1 + s2)2
970.5882 = 1,030 / (1 + s2)2
(1 + s2)2 = 1,030 / 970.5882 = 1.061212
1 + s2 = √1.061212 = 1.030152
s2 = 0.030152 = 3.0152%
The 1-year forward rate starting today – 1f0 – is easy: 2%.
To get the 1-year forward rate starting one year from today – 1f1 – we equate discount rates for 2 years:
(1 + s2)2 = (1 + 1f0)(1 + 1f1)
1.0301522 = 1.02(1 + 1f1)
1 + 1f1 = 1.0301522 / 1.02 = 1.061212 / 1.02 = 1.040404
1f1 = 0.040404 = 4.0404%
Now, to the tree. Let’s assume 10% volatility, so
rU = (e10%)2rD =e0.2rD ≈ 1.2214_r_D
The tree has to price the 2-year par bond correctly:
1,000 = 0.5[30 / 1.02 + 1,030 / (1.02)(1 + rU)] + 0.5[30 / 1.02 + 1,030 / (1.02)(1 + rD)]
2,000 = 30 / 1.02 + 1,030 / (1.02)(1 + rU) + 30 / 1.02 + 1,030 / (1.02)(1 + rD)
2,040 = 30 + 1,030 / (1 + rU) + 30 + 1,030 / (1 + rD)
1,980 = 1,030 / (1 + rU) + 1,030 / (1 + rD)
1,980(1 + rU)(1 + rD) = 1,030(1 + rD) + 1,030(1 + rU) = 2,060 + 1,030_r_D +1,030_r_U
1,980(1 + rD + rU + rDrU) = 2,060 + 1,030_r_D +1,030_r_U
1,980 + 1,980_r_D +1,980_r_U + 1,980 rDrU = 2,060 + 1,030_r_D +1,030_r_U
1,980 rDrU + 950_r_D + 950_r_U – 80 = 0
1,980_r_D(1.2214_r_D) + 950_r_D + 950(1.2214_r_D) – 80 = 0
2,418.3775_r_D2 + 2,110.3326_r_D – 80 = 0
We can solve this using the quadratic formula. I’ll spare you the details and go right to the solution(s):
rD = 3.6391%
rD = –90.9014%
The second one’s silly, so we’ll go with the first. That gives:
rU =1.2214 × 3.6391% = 4.4448%
Having done all of this, I’ll note that in practice nobody solves a quadratic equation to get these numbers. They use some numerical process (such as Excel’s Solver) to find the solutions.