how to read dnet client

[danzigrules]

How do I tell how well I am doing?
1.94mkeys/sec that good? on the c2

1.33mkeys/sec on the 466

 

[Wolfie]

That sounds about right......

This site will tell you what the going rate is for what you have..... But remember..... this is if you don't have anything else running on your machines....

Wolf

[edit]I would also like to welcome you to our team. Hope you have a good time with us. Just remember. Don't take paulson too seriously. He doesn't know what to do with his...... wait.... I will not talk about that. 😛 hehehe[/edit]

 

[dennilfloss]

I get 1.32Mkeys/s on my Celeron 500 and my cpu% devoted to the client is ~96%.

According to that site, my herd should optimally crack 3.7 Mkeys/s. That will improve tomorrow, when WombatWoman's 2 computers join it. 🙂

 

[danzigrules]

1150 w/u day ok with two celey's ?
hope so cause that's all I got right now🙁
my c2 600 is the highest

celey 44 @ 525 is the other not bad then


need to figure out some memory tweaks for this c2

 

[dennilfloss]

That sounds right. I get about ~400wu/day but I find it quite variable since the packets are not all of equal size. I have seen some with 1*2^28 keys and some with 4*2^28 keys. That's why I think that the number of keys would be more reliable a statistics to rank teams. Maybe the DPC are crunching smaller blocks than us and that's why they're catching us. 😉

 

[danzigrules]

hell mine are like this:
[Oct 10 01:52:07 UTC] 2 RC5 packets (64 work units) remain in buff-in.rc5
[Oct 10 01:52:07 UTC] Projected ideal time to completion: 0.02:25:04.00
[Oct 10 01:52:07 UTC] 4 RC5 packets (69 work units) are in buff-out.rc5
.....10%.....20%.....30%.....40%.....50%.....60%.....70%.....80%.....90%....100
[Oct 10 03:06:04 UTC] Completed RC5 packet DBE4A356:E0000000 (32*2^28 keys)
0.01:13:57.26 - [1,935,878.47 keys/sec]
[Oct 10 03:06:04 UTC] Loaded RC5 32*2^28 packet DBE4A35C:E0000000
[Oct 10 03:06:04 UTC] Summary: 4 RC5 packets (100*2^28 keys)
0.02:53:16.11 - [1.94 Mkeys/s]
[Oct 10 03:06:04 UTC] 1 RC5 packet (32 work units) remains in buff-in.rc5
[Oct 10 03:06:04 UTC] Projected ideal time to completion: 0.01:12:32.00
[Oct 10 03:06:04 UTC] 5 RC5 packets (101 work units) are in buff-out.rc5
.....10%.....20%.....30%.....40%.....50%.....60%.....70%.....80%.....90%....100
[Oct 10 04:19:51 UTC] Completed RC5 packet DBE4A35C:E0000000 (32*2^28 keys)
0.01:13:47.21 - [1,940,273.02 keys/sec]
[Oct 10 04:19:51 UTC] Loaded RC5 32*2^28 packet DBE4A35E:E0000000
[Oct 10 04:19:51 UTC] Summary: 5 RC5 packets (132*2^28 keys)
0.04:07:03.32 - [1.94 Mkeys/s]
[Oct 10 04:19:51 UTC] 0 RC5 packets (0 work units) remain in buff-in.rc5
[Oct 10 04:19:51 UTC] 6 RC5 packets (133 work units) are in buff-out.rc5
[Oct 10 04:19:53 UTC] Connected to distributed.net 205.149.163.211:2064...
[Oct 10 04:19:53 UTC] The keyserver says: "Happy cracking. Hail Eris!"
[Oct 10 04:19:56 UTC] Retrieved RC5 work unit 159 of 159 (100.00% transferred)
[Oct 10 04:19:58 UTC] Sent RC5 work unit 133 of 133 (100.00% transferred)
.....10%.....20%.....30%.....40%.....50%.....60%.....70%....

 

[DanC]

Dennil - That's an interesting observation....

I wonder how that's calculated? - surely they must do a conversion.

I know... "Don't call me Shirley" 😛

 

[MikeyP]

I have seen some with 1*2^28 keys and some with 4*2^28 keys.

For the 4*2^28, you get credit for doing 4 work units. Each work unit is 2^28 keys.

 

[BurntKooshie]

PLEASE NOTE While smaller packets finish faster (WU's ie, 2 ^ 28 keys, takes the same amount of time), sending them takes up FAR more banwidth, and consume massive amounts of space compared to larger transfers in the log files. So if you have a machine, say, greater than a Penium, 2 ^ 33 is better! Sure it takes longer, and you won't get every last block out each day (they'll get out the next day, so they average out), but it GREATLY (by a factor of 32 in some cases) decrease the bandwidth used, and the size of the log files.

 

[LeBlatt]

...and the stress on Mika's box (our team proxy and stats generator, for new members)