Question

A researcher has funds to buy enough computing power to number-crunch a problem in 6 years. Computing power per dollar doubles every 15 months.
(a) When should he buy his computers to have finished the problem as soon as possible? Give your answer as a decimal in months.
(b) Suppose the problem would take months C on current computers. What is the largest value of C for which he should buy the computers immediately?

Thanks

Answer 1

This is an exponential function.

Answer 2

First you wanna find the exponential function that will determine how many years it will take to do the problem at a given time.

Answer 3

It's going to be in the form: \[ce^{kt}\]Solve for \(c\) using: \[6=ce^{k(0)}\]Then you can solve for \(k\) using:\[ce^{k(0)}=2ce^{k(15)}\]

Answer 4

The function for how much time it will take to solve the problem is: \[t+ce^{kt}\]So you want to use your calculus to find when this function is minimized.
Try taking the derivative and finding the critical points.
That is solve for \(t\) when the derivative equals \(0\):
\[0=\frac{d}{dt}(t+ce^{kt})\]

Answer 5

One revision I have to make is that:
\[ce^{k(0)}=2ce^{k(15)}\]Should be:\[ce^{k(0)}=2ce^{k(\frac{15}{12})}\]Since months need to be converted to years.

Answer 6

@T1ger5 Get all that?

Answer 7

Once you do part a, part b should be easy.

Answer 8

\[ce^{k(0)}=2ce^{k(15)}\]Should have been: \[2ce^{k(0)}=ce^{k(\frac{15}{12})}\]Definitely \(k\) should be negative. Hope you caught it!

Answer 9

Just one question when you are writing k(0) and k(15/12) what does that actually mean. Is k a constant or some function?

Answer 10

Well \(k\) is a constant. The parenthesis are meant to show that we are substituting a value into \(t\).

Answer 11

I understand part a), however, for part b) I keep coming to the conclusion that the largest possible value for c is 72. I know this is wrong but cannot really wrap my head around it.

Answer 12

I assume you found the time to wait (in months) before purchasing the computers, is given by
\[m= -\frac{15}{\ln2}\cdot \ln(\frac{15}{C\cdot \ln2}) \]
where C is the number of months needed to finish the job using current computer technology.

part (a) has C= 72 months
for part (b), we want the time to wait to be zero or less, which means buy the computers immediately.
Use zero as the maximum time we will wait, and solve for C, the number of months needed to do the job with current technology.
\[ -\frac{15}{\ln2}\cdot \ln(\frac{15}{C\cdot \ln2}) =0\]
\[ \ln(\frac{15}{C\cdot \ln2}) =0\]
\[ \frac{15}{C\cdot \ln2} =1\]
\[C= \frac{15}{ \ln2} \]
If C is longer than this, it makes sense to wait some positive number of months before purchasing the computers

Answer 13

Thank you you both for your answers and help. :)