Hi There is a quesition on block layout in the raid10 far layout, that I would like to know more about. For 4 drives, and with 2 copies (-n 4 -p n2) I see several possible layouts, 3 of them are, showing the beginning of each raid0 section: Disks: a b c d Layout 1: 1 2 3 4 .............. 4 1 2 3 Layout 2: 1 2 3 4 .............. 3 4 1 2 Layout 3: 1 2 3 4 .............. 2 3 4 1 This gives 3! combinations for double faliure, or in all 6 possibilities. These are the combinations that will contain all blocks with only 2 working drives: Layout 1: a+c, b+d will work, total 2 combinations Layout 2: a+b, c+d, a+d b+c would work, total 4 combinations Layout 3: a+c, b+d would work, total 2 combinations So the best layout would be layout 2, as it provides a 4/6 chance = 67 % that it will survive the 2nd disk failure, while the 2 others only have 2/6 = 33 % chance of surviving the 2nd disk failure. How is the "-n 4 -p f2" layout actually done? I think one could do similar analyses for other numbers of drives. Is there any general pattern one (algorithm) one could use? I think one key is allocating copies in pairs. Maybe more could be gained by allocating the blocks in groups of powers of 2. Or maybe the only gain comes from the grouping in pairs as there are only 2 copies. best regards Keld -- To unsubscribe from this list: send the line "unsubscribe linux-raid" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
