Easier way to calculate NPV with failure rates
I use this method, which thus far has yielded consistently correct answers. Can someone tell me if this is theoretically sound? (Obviously this won't work with multiple payoffs, but since all the numericals involve only initial CF and a final CF, I've been doing ok...)
I calculate prob of success by multiplying all the (1-p), where p=prob. of failure.
I then multiply this overall probability of success with the CF at the end. I get a modified CF.
I then calculate NPV using modified CF and initial CF, and get the right answer straightaway.
(The textbook method and the ones used in the explanations are really long and tedious, involving two stages of multiplication and subtraction, so I'm gonna use mine unless one of you brilliant folks can point out a hole in my theory) |