Question

A researcher has funds to buy enough computing power to number-crunch a problem in 10 years. Computing power per dollar doubles every 16 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, accurate within 0.1 months. b)Suppose the problem would take cmonths on current computers. What is the largest value of cfor which he should buy the computers immediately? Give an answer as a decimal accurate to 0.1 months.

Answer 1

what class is this for?

Answer 2

calculus

Answer 3

http://openstudy.com/study#/updates/4e814d840b8bc11dd53e2b2f

Answer 4

i dont understand where the 30 comes from at 23+30 months.

Answer 5

i have solved for part a using the link above. i need to find b now

Answer 6

\[
f(t)=\begin{cases}
0&,0\leq c\\
2^{\frac{c}{16}}t&,c<t
\end{cases}
f(t)=120\\
2^\frac{c}{16}t=120\\
t=\frac{120}{2^\frac{c}{16}}\\
T=c+120\left(2^{-\frac{c}{16}}\right)\\
\frac{dT}{dc}=1+120\left(2^{-\frac{c}{16}}\right)\left(-\frac{\ln(2)}{16}\right)=0\\
2^{-\frac{c}{16}}=\frac{16}{120\ln(2)}\\
-\frac{c}{16}=4-3\log_2(15)-\log_2\left(\ln(2)\right)\\
c=48\log_2(15)+16\log_2\left(\ln(2)\right)-64
\]
Hopefully I didn't get it wrong.

Answer 7

thanks for your work. i get 115 from the last c equation but it is not the answer

Answer 8

The derivative is correct but I didn't solve dT/dc correctly.

Answer 9

\[
-\frac{c}{16}=4-3-\log_2(15)-\log_2\left(\ln(2)\right)\\
c=16\log_2(15)+16\log_2\left(\ln(2)\right)-16=38.0500\\
T=c+120\left(2^{-\frac{c}{16}}\right)=61.1331\]

Answer 10

I think part b is asking for asking forthe problem's computation time on current computer such that solving the problem now will be the fastest instead of waiting the computational power to improve.

Answer 11

Or:\[
T=c+n2^{-\frac{c}{16}}\\
\frac{dT}{dc}(0)=0\text{ for }n\in \mathbb{R}
\]