Thanks! Couldn't have been explained any clearer.
> Date: Tue, 8 Apr 2014 18:03:37 +0200 > From: [email protected] > To: [email protected] > Subject: Re: [OpenIndiana-discuss] Formula raidz storage space caclulation > > On 2014-04-08 17:05, Randy S wrote: > > Hi, > > > > yes even if you incorporate the kilo=1024 > > example: > > I filled in the example below on the site: > > RAID Mode: z1 > > Disk Size: 3 tb > > Quantity of Disks: 10 > > > > RAID-Z > > *Raw Storage: 30.0 TB / 30000.0 GB > > *Usable Storage: 24.6 TB / 25145.7 GB > > RAID-Z uses one disk for Parity much like RAID5 and requires at least three > > drives to be used. > > *Usable storage is the actual post-format amount where kilo = 1024, not 1000 > > > > If I use the (n-1)formula this would amount to: > > (10-1) * 3 TB = 27 TB (27000 GB) > > even if you use 1 TB=1024 GB: > > (10-1) * 3072 GB = 27.648 GB > > > > Somewhere theres about 2.5 TB being used for something in their > > calculations. > > Maybe someone can explain or show me where my calculation is going wrong? > > 1024 vs 1000 is another way around: a 3TB disk is > 3000000000000 / 1024 / 1024 / 1024 / 1024 = 2.72Tb or > 3000000000000 / 1024 / 1024 / 1024 = 2793.97Gb > > Times 9, and that amounts to 24.55Tb or 25145.7Gb > > (I am sometimes confused by "b" vs. "B" in such decimal vs. binary > notations, so maybe that should have been written another way around) > > HTH, > //Jim > > > _______________________________________________ > OpenIndiana-discuss mailing list > [email protected] > http://openindiana.org/mailman/listinfo/openindiana-discuss _______________________________________________ OpenIndiana-discuss mailing list [email protected] http://openindiana.org/mailman/listinfo/openindiana-discuss
