Comment Re:I suspect a mistake in units... (Score 1) 453
Date Posted: Monday, July 09, 2001 3:08 PM
Some info you will need if you wanna do "quick math" :
The dnet client processes and transmits packets of data. Those packets contain blocks. One block contains 2^28 keys to test. The user gets credited for blocks he has processed.
One packet can contain any number of blocks between 1 and 32. But the size in bytes of a packet is always the same, whatever number of blocks it contains, because a packet determines a range of blocks, with a starting point and a length, not just a series of blocks.
I just took the log from my dnet proxy for today (since the last time it had cleared its outbound buffer), and found out I had processed 91 packets, containing 1497 blocks. The buffer was 15.296 bytes on disk. That is 168 bytes per packet, plus 8 bytes for the control header. Let's ignore those measly 8 bytes.
Assuming he was producing 60K blocks per day, that would cause an outbound internet traffic of 60.000 x 168 /1024 = 9,844.75 kilobytes per day, assuming using 1 block packets (worst case). If using only largest packets (best case), you divide this by 32, and get 307.62 KB per day.
Double this to account for inbound traffic used to fetch new data to process, and you get a whooping 615.23K per day, best case, or 19,687.5, worst case.
Knowing that the default client config at that time was to use 2 block packets, the amount should get no higher than 8.5 MB a day.
I don't know what line speed they use at that school, but that gives an idea of how little bandwidth he may have used.
Now, David's lawyer should have access to all information concerning the case, including how many packets he submitted to dnet from the school to know exactly how much BW he used and how many seconds that took.
Some info you will need if you wanna do "quick math" :
The dnet client processes and transmits packets of data. Those packets contain blocks. One block contains 2^28 keys to test. The user gets credited for blocks he has processed.
One packet can contain any number of blocks between 1 and 32. But the size in bytes of a packet is always the same, whatever number of blocks it contains, because a packet determines a range of blocks, with a starting point and a length, not just a series of blocks.
I just took the log from my dnet proxy for today (since the last time it had cleared its outbound buffer), and found out I had processed 91 packets, containing 1497 blocks. The buffer was 15.296 bytes on disk. That is 168 bytes per packet, plus 8 bytes for the control header. Let's ignore those measly 8 bytes.
Assuming he was producing 60K blocks per day, that would cause an outbound internet traffic of 60.000 x 168
Double this to account for inbound traffic used to fetch new data to process, and you get a whooping 615.23K per day, best case, or 19,687.5, worst case.
Knowing that the default client config at that time was to use 2 block packets, the amount should get no higher than 8.5 MB a day.
I don't know what line speed they use at that school, but that gives an idea of how little bandwidth he may have used.
Now, David's lawyer should have access to all information concerning the case, including how many packets he submitted to dnet from the school to know exactly how much BW he used and how many seconds that took.
