NPV?

August 1, 2007
By

If you aren’t very good at math, check out some of my other posts. This won’t interest you. If you are very good at math, I need your assistance, because I’m not. If you’re worried about “academic dishonesty,” don’t be. I don’t have any classes this summer.

I understand that the NPV of a consol bond or perpetuity would be C/(1+r) where C is the stated coupon rate, and r is the discount rate. Let’s assume that C is expected to grow by g% each year, and that this instrument is also a perpetuity. What would be the mathematical expression for such an instrument? For practicality’s sake, assume g < r

This is as far as I can get:

NPV = Sum of [C/(1+r)] + [(C*(1+g))/(1+r)^2] + … + [(C*(1+g)^(n-1))/(1+r)^(n)]

Obviously, the sum increases as n increases, but at an ever decreasing rate, the velocity of which is determined by the spread between g and r – distant future growth (in terms of NPV) approaches irrelevance (zero) as n increases, and holding g constant, it approaches zero faster as r increases.

How can I simplify this so that I can plug it into my calculator? If I had taken more calculus, I’d be able to figure this out. I have taken the liberty of putting the formula into an xls spreadsheet and cascading it, which allows me to change the values of C, g, r; with a discount rate of 15% and a growth rate of 1%, the NPV appears to be growing about a penny per year, past 100 years, so it’s still quite a long term deal.

but if you know of a better way, please advise.

8 Responses to NPV?

  1. dwz on August 1, 2007 at 7:45 pm

    WTF?

  2. David Z on August 1, 2007 at 8:48 pm

    LOL.

  3. Geek on August 2, 2007 at 11:07 am

    If the NPV increases by a penny after 100 years, what is the breakpoint where the NPV increases by less than a dollar?

    I suppose I could fire up Excel and figure it out for myself but I’m lazy, what can I say :)

  4. David Z on August 2, 2007 at 11:10 am

    Yeah, I did it in excel, but I was looking for a formula along the lines of C/(1+r), although I know it would be a little more complicated than that. Oh well, I tried.

  5. Zach on August 3, 2007 at 12:36 am

    You can factor out the C from the sum.

    Then, because 1/x^N is the same as 1/((x^n)*x), you can factor out a (1+r) from the denominator:

    ((1+g)^(n-1))/((1+r)^(n-1) * (1+r) )

    so that your terms now have the same exponent, (n-1).

    Then you factor out the extra (1+r) from the denominator and put it outside with the C from step 1.

    Then, once you choose a value for g, you have an ordinary geometric series.

  6. David Z on August 3, 2007 at 6:58 am

    That’s what I was looking for!

  7. Dan Z on August 7, 2007 at 8:45 pm

    You might be a nerd. Youre treading a dangerous line here.

  8. David Z on August 8, 2007 at 11:13 am

    Trust me – it’ll be worth it.

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