View Full Version : How many operations per second can a standard PC perform
josephellengar
October 31st, 2008, 01:16 AM
I don't know- just a normal one that you'd buy today off the shelf. I just saw a news story that said that someone came up with a computer that can perform 50 trillion operations per second. Wow! So, what can mine do?
NovaAesa
October 31st, 2008, 01:20 AM
I guess it would all depend on the complexity of the particular operation you are talking about.
josephellengar
October 31st, 2008, 01:34 AM
I think that the 50 trillion operations was in FLOPS.
teqsun
October 31st, 2008, 01:47 AM
isnt a flop 32 bits of data?
ChanServ
October 31st, 2008, 02:29 AM
1
LaRoza
October 31st, 2008, 02:30 AM
I don't know- just a normal one that you'd buy today off the shelf. I just saw a news story that said that someone came up with a computer that can perform 50 trillion operations per second. Wow! So, what can mine do?
Vague question...
What is the hardward? I see no "standard PC" arch.
What is your CPU? Mine has a speed of 2.8 Ghz, therefore, it can do that many operations a second (1 hertz is an operation)
Keep in mind all the hardware works together, so you'd have to take into account the other hardware.
klange
October 31st, 2008, 02:48 AM
Technically, one hertz just means "1 / second". It can be one anything.
2.8GHz means 2,800,000,000Hz or 2.8 billion operations per second.
And then there's multi-core processors...
ModelM
October 31st, 2008, 02:54 AM
Perhaps there are some here who haven't heard of bogomips.
One way to measure the speed of a computer is in Millions of Instructions Per Second - or MIPS.
The kernel hackers put a quick routine into the kernel to measure your processor speed. It serves as only an estimate & shouldn't be measured against anything else so the numbers are Bogus and can't be compared to anything but themselves. Bogus MIPS - or BogoMIPS.
In a terminal type:
dmesg | grep Bogoand you should see a result that looks like this:
[ 15.522333] Calibrating delay using timer specific routine.. 5057.49 BogoMIPS (lpj=10114998)
[ 15.918108] Calibrating delay using timer specific routine.. 5053.97 BogoMIPS (lpj=10107947)
[ 15.918704] Total of 2 processors activated (10111.47 BogoMIPS).
That should give you something to play with for a while.
ad_267
October 31st, 2008, 02:57 AM
A flop is a floating point operation. http://en.wikipedia.org/wiki/FLOP
Most applications use double precision floats for operations (64 bits on a 32 bit PC)
LaRoza
October 31st, 2008, 03:44 AM
2.8GHz means 2,800,000,000Hz or 2.8 billion operations per second.
And then there's multi-core processors...
Yes, mine is dual core, I wasn't thinking about that.
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