NPV question

David Niven can purchase a new laptop computer today for $2500. The laptop provides an economic benifit of $900 per annum and has an expected life of 6 years. Laptops are expected to decline in price by 10% per annum and Davids discount rate is 10%. When is the best time for david to purcahse the laptop.?

Im stuck on this question...
I tried as shown in an example but i dont think it is correct...

some help would be great..

heres what i have so far:

Year : Cost : PV of Savings : NPV at purchase : NPV today
0 : -2500 : +900 : -1600 : -1600
1 : -2250 : +900 : -1350 : -1227.27
2 : -2025 : +900 : -1125 : -929.75
3 : -1822.50 : +900 : -922.25 : -692.90
4 : -1640.25 : +900 : -740.50 : -505.77
5 : -1475.975 : +900 : -575.975 : -357.64

They way i calculated NPV today was....

-1350/1.1^1
-1125/1.1^2.....so on..

am i correct?? where have i gone wrong if im wrong?

THANKS

You seem to be missing a couple of items:

1) $900 per annum and has an expected life of 6 years

This makes the NPV at Purchase

900v + 900v^2 + 900v^3 + 900v^4 + 900v^5 + 900v^6 - Cost, where v = 1/1.1. In summary, 3919.73 less Cost of Laptop.

2) "today"

This one is a little tricky. If we look ONLY at the time of purchase, we see that the cost continues to decline and the economic benefit is constant. This might suggest we NEVER buy the laptop. The better question would be, what is our choice TODAY of when we should buy?

Purchase Right now, we get 3919.73 - 2500 = 1419.74

Purchase Next year, but still looking at TODAY, we get (3919.73 - 2500/1.1)/1.1 = 1517.94

Two years from now, we get (3919.73 - 2500/(1.1^2))/(1.1^2) = 1565.90

'n' years from now, we get (3919.73 - 2500/(1.1^n))/(1.1^n).

This can be viewed as a maximization problem and you can bring all your favorite calculus to bear on the solution. On the other hand, if you insist on an integer number of years, you'll need to poke around a little in the neighborhood of the continuous solution.

Let's see what you get.