Greetings of the day ,
Is the below understanding correct ?
In a half rack machine there are 7 cells and all are mirror copies of other , so the effective storage space = disk space in one cell .
Further if we have normal redundancy in each cell , then the effective storage space of the whole exadata will be half of the disk space in the cell.
Is there any rule that all the cells must have equal disk space?
In ASM perspective ,
Is there any rule like all the all the fail groups ( 2 or 3 depends on the redundancy level) must have equal size with in a disk group?
Each cell is not a full copy of the other cells. ASM diskgroups on Exadata utilize each cell as a failgroup, meaning that no 2 copies of an allocation unit are placed on the same cell. A half rack of high capacity storage (3TB drives) has 252TB of raw space (12 disks * 7 cells) ~126TB of usable space with normal redundancy, and ~84TB of usable space in high redundancy. As far as disk sizes, on ASM you generally want all disks to be the same size to keep everything balanced evenly. I have a short writeup on Exadata diskgroup storage at http://blog.oracle-ninja.com/2011/12/exadata-diskgroup-planning
no 2 copies of an allocation unit are placed on the same cell.
How many copies will be stored ?
Normal redundancy - 2copies
High redundancy - 3copies
Can i say like there is 1 to 1 relationship between cell and fail group?
How the striping happens in 12 cell disks of each cell?
How many cell failures can be tolerated?
Hi udayjampani -
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