Question

A researcher has funds to buy enough computing power to number-crunch a problem in 5 years. Computing power per dollar doubles every 23 months.
When should he buy his computers to have finished the problem as soon as possible? Give your answer as a decimal in months, accurate within 0.1 months
Suppose the problem would take c months on current computers. What is the largest value of c for which he should buy the computers immediately? Give an answer as a decimal accurate to 0.1 months.

Answer 1

If he buys today it will take 5*12 = 60 months.
If he waits 23 months it will take 23 + 30 months.
30 months = 60*e^(r*23)
30/60 = 1/2 = e^23r
ln(1/2) = 23r; r = ln(1/2)/23
If he waits w months, the total time will be
\[w+60e^{w\ln(1/2)/23}\]
Taking the derivative of this w.r.t. w and setting = to 0:
\[0 = 1+\frac{60}{23}ln(1/2)e^{w\ln(1/2)/23}\]
\[ -\frac{23}{60} = ln(1/2))e^{w\ln(1/2)/23}\]
\[ -\frac{23}{60ln(1/2)} = e^{w\ln(1/2)/23}\]
\[ ln(-\frac{23}{60ln(1/2)}) = w\ln(1/2)/23\]
\[\frac{23}{ln(1/2)} ln(-\frac{23}{60ln(1/2)}) = w\]
I get 19.7 months

Answer 2

thank you so much i have been stuck on this for ages