How much RAC does SETI@home need for pseudo-realtime analysis?

Greetings folks, I'm new to the whole BOINC scene so please forgive my ignorance. I have a technical question about the SETI@home project itself:

Assuming that SETI@home had sufficient server resources and bandwidth, how high of a sustained RAC does SETI@home need to process workunits as fast as it receives them from the data feeds it gets from the telescope? Disregarding the latency incurred by multiple validations and actually receiving the raw data in the first place. I am unclear as to whether this can be determined by looking at the result turn-around time on the Server status page.

For now I'm just curious about the standard multi-beam analysis and not Astropulse, although it would be fascinating to know that too if someone knows how to figure it out.

Thank you!

P.S. it would be awesome if someday this was figured out and calculated on the Server status page (perhaps as a ratio - Current RAC/Desired RAC?) It would give some of us something more "tangible" to focus on when determining what types of resources would be necessary to contribute a desired amount of processing to the project.

Greetings folks, I'm new to the whole BOINC scene so please forgive my ignorance. I have a technical question about the SETI@home project itself:

Assuming that SETI@home had sufficient server resources and bandwidth, how high of a sustained RAC does SETI@home need to process workunits as fast as it receives them from the data feeds it gets from the telescope? Disregarding the latency incurred by multiple validations and actually receiving the raw data in the first place. I am unclear as to whether this can be determined by looking at the result turn-around time on the Server status page.

For now I'm just curious about the standard multi-beam analysis and not Astropulse, although it would be fascinating to know that too if someone knows how to figure it out.

Thank you!

P.S. it would be awesome if someday this was figured out and calculated on the Server status page (perhaps as a ratio - Current RAC/Desired RAC?) It would give some of us something more "tangible" to focus on when determining what types of resources would be necessary to contribute a desired amount of processing to the project.

If what you mean is to process it as fast as the telescope can collect it (when it's actually collecting data which SAH can use), you're talking orders of magnitude more crunching horsepower than what we have currently available.

Joe Segur probably has a pretty good ballpark figure worked out from the last time the question came up.

Alinator

If what you mean is to process it as fast as the telescope can collect it (when it's actually collecting data which SAH can use), you're talking orders of magnitude more crunching horsepower than what we have currently available.

Joe Segur probably has a pretty good ballpark figure worked out from the last time the question came up.

Alinator

Thank you Alinator. From the mention of Joe's name I was able to track down a post:

http://setiathome.berkeley.edu/forum_thread.php?id=45805&nowrap=true#723956

From there I got some numbers to crunch...

If the receiver picks up 151,793 work units / hour (per Joe), and receives for an average of 9.5 hours on any given day (per Joe), and I assume that the average multi-beam workunit leads to 75 claimed credits, then we have 108,152,512.5 credits / day of work before replication.

If I then assume that a work unit needs to be crunched an average of 2.5 times, I come up with an average workload of 270,381,281.25 credits / day.

From http://boincstats.com/stats/project_graph.php?pr=sah:

SETI@home is currently calculating with a RAC of 45,648,330. Since each host that crunches a work unit also claims credit for it, I do not need to adjust for the average 2.5 crunches per work unit. Therefore, SETI@home is currently calculating at about 17% of the rate needed for "pseudo-realtime" analysis.

Where I could use more clarification at this point then, assuming my calculations aren't completely out in left field, is a better approximation for the average number of credits per work unit, and the average number of times a work unit is likely to be processed before being retired.

Of course this does not apply to Astropulse, which I guess is a more important figure to want to know at this point.

Hmmmm...

I'm not sure working straight from RAC is the best way to approach the question.

The telescope is recording 7, 2.5 MHz bandwidth beams, at 2 polarizations, or 14 channels of data, when the opportunity arises. From that, it gets divided into 256 10 KHz subbands.

After that, the splitter breaks the datastream into workunits of around 107 seconds in length, which then get further replicated into 2 tasks per WU for processing by the hosts.

In addition, we have to account for the fact that there is ~20% overlap in the time domain WU to WU, as well as some overlap (~10%, IIRC) from data drive to data drive.

So let's say we look at a 24 hour period of constant data collection, and forget about drive overlap for this period:

WU 'periods' per day = 86400 / (107 / .2) = 535

<edit> I seem to have a mistake here, I must recompute!

Hmmm... It looks like that should be: 86400 / (107 * .8) = 1,009.3458

Somebody want to double check that? Corrected values in parentheses.

Total Workunits per day = 535 * (7 * 2 * 256) = 1,917,440 (3,617,495.3271)

Therefore: Total Tasks Replicated per day = 3,834,880 (7,234,990.6542)

If we then take Winterknight's estimation of the failure rate, that makes it:

Total Workunits per day = 3,988,275 (7,524,390.2804)

Looking at the current # 1 Host, it seems to be running tasks in the 3000 seconds ballpark currently.

So we are talking on the order of 11,964,825,000 (22,573170841.2) seconds to run all the tasks in the 24 hours of telescope time collected. Since this host is an 8 banger, that works out to 1,495,603,125 (2,821646355.15) seconds for #1 host to do all the work collected in one day.

So divide that by 86400 to find how many #1 equivalents you need to do it all in one actual day and you get about 17310 (32,658) hosts.

Now take that number of #1 hosts and multiply by it's reported RAC (assuming it is at steady state currently) and you come up with a 'required' RAC of about 203,545,354 (384,014,312.076).

However, the host population isn't all # 1 hosts, so going to BOINCStats we find that the aggregate RAC for the project is:

Aggregate RAC per active host = 45,648,330 / 332,973 = 137.0932

So it would appear that as a group, we are about 6 orders of magnitude short of the horsepower we need to keep up if SAH was fortunate enough to have access to constant surveillance with a telescope equivalent to the gear currently at Arecibo!

Send that little tidbit to your Senators and House Representative to consider, before they start talking about giving Arecibo the financial ax, because it's old and 'obsolete', or not 'flexible' enough! :-D

Alinator

if the 33% is about accurate and I take a look at the Berkeley data density page.

Then I'dd say that of the last aprox. three years there were 190 'productive days', meaning that the Arecibo Observatory was 190/(356*3) was about 17% "SETI productive".

If this assumption is correct, then we are here a two times the "pseudo-realtime" analysis barrier!?

...
Average Credit Per Day Needed
--------------------------------------------------------------------------------
331,008 MB WUs / day
331,008 * 6.38 (AP WU factor) = 2,111,831.04 AP WUs / day
...

The 6.38 multiplier is for a group of MB WUs, data for each group of 256 MB Wus can also produce 6.38 AP WUs. So to calculate from individual MB WUs it's 6.38/256 ~= 0.025. Then 331,008 MB WUs / day gives about 8275 AP WUs / day.