I would like a few different formulas or methods for doing a couple of conversions and calculations:
1) How can I convert petaFLOPS into a base $2$ number representing how many operations per second e.g. $2^{64}$? For example, if we assume a computer capable of $27$ petaFLOPS, how many operations per second can it perform in the format of $2^x$? I would like to be able to plug in different values into the formula to change the number of petaFLOPS e.g. $200, 1000$ and get different outputs.
2) If I get a result from the above formula, e.g. $2^{85}$ operations per second, I need a second formula to determine how long it would take in seconds to do $2^x$ operations e.g. $2^{100}$ operations.
My mathematical symbol understanding is not too good, so if you could write out the answer in simple, plain English without strange mathematical symbols and so on that would be very helpful.
Thank you!
1 petaFlop = $10^{15} \frac{\text{flop}}{\text{s}}$ and we want to find $n$ such that $2^n=k\cdot10^{15}$ where $k$ is the number of petaFlops
Take log base 2 of both sides:
$$\log_{2}{2^n}=n$$
$$\log_{2}{(k\cdot10^{15})}$$
So we know $n=\log_2{(k\cdot10^{15})}$
So 1 petaFlop is about $2^{49.82892}$ operations per second. But in general, $k$ petaFLOP is the same as $2^{\log_2{(k\cdot10^{15})}}$ FLOP.
Now, if you are running at $2^a$ operations per second and need to do $2^b$ operations, it will take
$$\frac{2^b \text{flop}}{2^a \frac{\text{flop}}{\text{s}}}=2^{b-a} \text{s}$$
Hope this helps. :)
In as non-math language as possible:
$$k\; \text{petaFLOP} = 2^{\log_2{(k\cdot10^{15})}} \;\text{FLOP}$$
And if you are running at $2^k \text{FLOP}$ and need to perform $2^x$ actions, it will take
$$2^{x-k}\; \text{s}$$