Quote from: RetiredCoder on Today at 04:44:56 PM

So if we have 16.7M kangs at 3GH it means that each kang jumps 180 times per sec, it's 15.5M per day.
For #130 you have to use DP=30 at least, or better 32 (especially for #135).
So for DP=32 every kang finds a DP every 270 days (on average).
Correct?
Now the problem is obvious, isn't it?

Yeah, but still, what's your point? That JLP had it wrong? I already knew this!

Here's a benchmark on RTX 6000 Ada:

Code:

| DP | JP | GE | S_UV | S_RS | S_M | S_TRS | S_RS0 | REG | LM/SS/SL | Mo/s | ms/st | o/s/k |
| 32 | 64 | 1 | X | X | | | | | 0 | 1678 | | 46160 |
[/code[

So, a single kangaroo/thread results in a speed of 1678 Mk/s, which averages to a speed/kangaroo of [b]46160 jumps/s[/b]

So, at DP 32 every kangaroo yields a DP every 93000 seconds (around 1 DP / day / kangaroo).

But at the end of the day it doesn't really matter if you have a billion kangaroos producing a DP / day, or a single kangaroo producing a billion DPs per day, as long as the backend can handle both cases without hiccups.