I had this problem on my homework for my investment science class. The name of the textbook and problem number will be kept secret to prevent people from using this solution to do their homework. The problem is to determine the worth of a roof that will last 5 more years with the conditions that the house will last forever, a new roof will last 20 years and costs $20k, the bank interest is currently 5% and the cost, interest and longevity of the roof will not change.

The value of the roof is the present value of having the roof lasting 5 years minus the present value of having to replace the roof immediately. This problem is solved by taking each senario and replacing the roofs forever.

PV = present value

PV_{roof} = PV_{5 years} – PV_{0 years}.

If we take the limiting case of each to infinity, we have

PV_{n years} = sum from k = 0 to infinity of -(replacement cost)/(1+r) ^{(new roof lifetime)*k+n}

So, how do we sum to infinity?

Let the sum, S = a + a*r + a*r^2 + …. + a*r^n

S*r = a*r + a*r^2 + a*r^3 + …. + a*r^{n+1}

S-S*r = S*(1-r) = a – a*r^(n+1) = a*(1 – r^{n+1})

S = a*(1-r^{n+1})/(1-r), for n = infinity and r < 1, S = a/(1-r)

PV_{roof} = $7k.

That’s a lot more than what I would think the roof would be worth. Who knew?

Advertisements

Like this:

LikeLoading...

Related

This entry was posted on Wednesday, July 11th, 2007 at 5:25 am and is filed under Miscellaneous. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

Read at your own risk. This site is intended to be a roundtable discussion. Make sure you thoroughly research things for yourself before relying on someone else’s opinion.

interesting… i like the clean design of the webpage as well.