RE: Exposure with unprotected disk units

Scott,

Yes, all drives in a RAID parity set must be the same capacity.

Yes, you can ride out the loss of a user ASP without negative consequences
to the data on the other ASP's.

You're mixing the terms ASP and parity set in your note; they are not
necessarily equivalent.  But if ASP three is constituted entirely of those
two unprotected drives, then you can lose a drive in that ASP without losing
data anywhere else.  Your system should stay up.

Regards,
Andy Nolen-Parkhouse

> In reviewing one of the old systems I am responsible for, I noticed that
> some 17.54gb disk units were added onto the system (as ASP 3) and are in
> an
> unprotected state.  I think if we had added 4gb disk units in these
> locations, we simply could have added them to the existing parity set, but
> since they are 17.54gb they can't be added.  Is this correct?
> 
>                                   Size     %    I/O   Request   Read
> Write
> Read  Write     %       --Protection--
> Unit  Type    (M)  Used    Rqs  Size (K)    Rqs    Rqs    (K)    (K)  Busy
> ASP  Type  Status
>    1  6607   4194  84.8     .5      10.3     .3     .2   12.5    6.7     0
> 1  DPY   ACTIVE
>    2  6607   3670  84.7     .5      15.3     .3     .2   18.6   11.4     0
> 1  DPY   ACTIVE
>    3  6607   3670  84.7     .5      18.2     .2     .2   24.0   10.9     0
> 1  DPY   ACTIVE
>    4  6607   3670  85.0     .5      11.7     .3     .2   17.2    4.9     0
> 1  DPY   ACTIVE
>   17  6607   3145  23.8     .0       5.4     .0     .0    4.0    5.9     0
> 2  DPY   ACTIVE
>    <snip>
>   23  6714  17548  18.6     .1      56.9     .1     .0   66.8   37.4     0
> 3
>   24  6714  17548  18.6     .2      67.6     .1     .0   76.6   50.2     0
> 3
> 
> 
> What could I expect if we lost either drive 23 or 24?  Can the system ride
> out the loss of a user ASP?  If not, and it crashes, when the failed disk
> is replaced and the system is brought back up, is there any data loss in
> the ASPs that *are* RAID protected? (This is a V4R4 system).
> 
> Scott Lindstrom