Post your S5R1 completion estimates here
Anonymous
10 Sep 2006 14:21:44 UTC
Topic 13450
(moderation:
Here'a a bit of raw info (Nov 5 2006). This is useful for seeing how much work available is short versus long workunits.
[root@einstein locality_scheduling]# ls work_available/*_ S5R1 | wc
2138 2138 64140
[root@einstein locality_scheduling]# ls working_set_removal/*_S5R1 | wc
3664 3664 128240
The sum 2138+3664 = 5802 is the number of data files [hl]_0050.0_S5R1 -> [hl]_1500.0_S5R1
[root@einstein locality_scheduling]# ls work_available/*0[0-3]*_S5R1 | wc
701 701 21030
[root@einstein locality_scheduling]# ls working_set_removal/*0[0-3]*_S5R1 | wc
1021 1021 35735.
So the number of data files with work remaining is 701 for short WU and 2138-701 = 1437 for long WU.
The original ratio of short/long was 350/1100 = 0.318. Current ratio is 701/1437 = 0.48
Conclusion is that we have proportionally more 'short' WU remaining than long one, compared with the beginning of the run. We are not running out of short WU.
Cheers,
Bruce
Post your S5R1 completion estimates here
)
You need to count more carefully! If I have objects numbered 0 and 1, that makes TWO objects, not one. The formula is Nhigh-Nlow+1.
For each IFO, the files are from 0050.0, 0050.5, ..., 1500.0. Count more carefully and you will see that this is 2901 files, not 2900.
Cheers,
Bruce