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We saw earlier how to deduce the interest rate from a bond price. Suppose you are offered a 1 year $1000 bond for $970.87. What's the interest rate? Generally, you would look at a good bond table like the one at Yahoo and the interest rate will be listed. If you find you have to compute it yourself, here's the procedure. If the interest rate is i, then the math is:

970.87 * (1+i) = 1000.

1+i = 1000 / 970.87 = 1.03.

i = 3%.

Let's do a more complicated example now. Suppose it's a 2 year bond for $10,000 with interest payments of $150 at 6 months, 12 months, and 18 months. We're offered this bond for $9,862.88. What's the interest rate?

We have to work out the six month interest rate first. The time period we can use is the least common denominator of the payment schedule. This will most often be either a month, six months, or a year. In this case it's six months. We'll call the six month rate x. We already know how to bring all these payments back to present value:

PV = $9,862.88 = $150 / 1+x + $150 / (1+x)^{2} + $150 / (1+x)^{3} + $10,000 / (1+x)^{4}

PV (1+x)^{4} = $9,862.88 (1+x)^{4} = $150 (1+x)^{3} + $150 (1+x)^{2} + $150 (1+x) + $10,000

$9,862.88 (1+x)^{4} - $150 (1+x)^{3} - $150 (1+x)^{2} - $150 (1+x) - $10,000 = 0.

As you can see, this leads to a 4th order polynomial, which cannot be solved with a formula. If this were a 20 year bond with payments every six months, it would lead to a 40th order polynomial. What can we do? What you do is ask your friendly neighborhood engineer to help out. Although the polynomial cannot be solved with a formula, we already (almost) know the answer: it's about 4%, more or less. All you need to do is write down the polynomial P and its derivative D.

Our first guess for the answer is x_{1} = 4% = .04.

Our second guess will be x_{2} = x_{1} - P( x_{1} ) / D( x_{1} ).

Our third guess will be x_{3} = x_{2} - P( x_{2} ) / D( x_{2} ).

Our third guess will be good enough. Sir Isaac Newton worked out this method of finding these solutions. He was the director of the England Mint and responsible for all of England's money.

Let's work the example above.

P(x) = $9,862.88 (1+x)^{4} - $150 (1+x)^{3} - $150 (1+x)^{2} - $150 (1+x) - $10,000 = 0

D(x) = 4 * $9,862.88 (1+x)^{3} - 3 * $150 (1+x)^{2} - 2 * $150 (1+x) - $150

P(.04) = $11,538.19 - $168.73 - $162.24 - $156 - $10,000 = $1051.205

D(.04) = $44,377.45 - $486.72 - $312 - $150 = $43428.87

x_{2} = .04 - P(.04) / D(.04) = .04 - 1051.21 / 43428.87 = .01579

P(.016) = $10,509.39 - $157.32 - $154.84 - $152.40 - $10,000 = $36.56

D(.016) = $41,375.57 - $464.52 - $304.81 - $150 = $40,431.52

x_{3} = .016 - P(.016) / D(.016) = .016 - 36.56 / 40,431.52 = .016 - .0011 = .01489

Remember, this is a six month rate so the yearly rate is (1+x_{3})^{2} = 1.03000 = 3%.

We can check how good this guess is by running the polynomial one more time.

P(.01489) = $10,464.02 - $156.81 - $154.50 - $152.24 - $10,000 = $0.03

Newton's method is very powerful. Our error in our original guess was a bit over $1000. Our second guess gave us an error of just over $35, and our third guess has an error of about 3 cents.