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RAID 101
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This is not a RAID in the strict sense, because JBOD does not provide
any redundancy. If any one drive in the JBOD-type array fails, the whole array fails
and all data on it is lost. Typical usage of JBOD layout is just to create a
disk of larger capacity by merging two or more smaller disks. This is only
practical if the disks have different capacity. For the disks of equal capacity,
RAID 0 is better because it provides the same capacity increase, the same
non-redundant layout with no disk space overhead and features faster read/write speed
in typical applications. JBOD can provide speed increase if two operations are
requested simultaneously on data blocks that are stored on different drives, but
this is relatively rare situation (let's say for example that read of blocks D1,
D2, D3 and D11, D12 is requested simultaneously; in such a case two requests can
be performed in parallel, increasing overall I/O speed). Minimum of two disks is required for a
hardware JBOD. Minimum of one disk is
required for a software JBOD (Windows NT/2000/2003 "Spanned volume", which
allows volume to occupy nonadjacent regions of the same physical disk). There is no disk
space overhead. The following exception may or may not apply in your case:
hardware RAID controller may support a single-disk JBOD configuration - this is
just a trick to allow a single drive to be attached to the controller, without
actually RAIDing anything. The same applies to RAID0 consisting of 1 member
disk.

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RAID 5 utilizes a parity function to provide redundancy and data
reconstruction. Typically, an "exclusive OR" ("XOR") binary function is used to
compute parity for a given "row" of the array. Anyway, the parity is computed as
a function of several data blocks P=P(D1, D2, ... DN-1) for N disk layout. In
case of a single drive failure, the inverse function is used to compute data
from the remaining data blocks and parity block.
Let's say for example that the Disk 3 fails in configuration illustrated
below.
- Data blocks D1 and D2 will be read directly from their corresponding
disks (which are operational).
- Parity block P1,2 is really not needed (does not contain user data) so
it will be just discarded.
- Data block D3 will be read from its corresponding disk (Disk 2).
- Data block D4, which is missing because its drive is offline will be
reconstructed using D3 and P3,4 like this: D4=Pinverse(D3,P3,4)
During normal operation, read speed gain is (N-1) times, because requests
will be evenly routed to N-1 disks (parity read is not needed during normal
operations). Write procedure is more complicated, and actually imposes some
speed penalty. Let's say we need to write block D1. We also need to update its
corresponding parity block P1,2. There are two ways to accomplish this:
- Read D2; compute P1,2=P(D1,D2); write D1 and P1,2;
- Read D1;old and P1,2;old; compute P1,2
from these data; write D1 and P1,2.
Both of these ways require at least one read operation as the overhead. This
read operation can not be parallelized in any way with its corresponding write
operation, so write speed should decrease (by the factor of two, assuming equal
read and write speed). Most current implementations mitigate this effect by
maintaining the entire "row" (D1, D2 and D3) in
the cache.
Minimum of three disks is required to implement RAID5. Storage space
overhead equals the capacity of a single
member disk and does not depend on the number of disks.

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RAID type (level) reference table |
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RAID Type |
Number of disks required |
Fault-tolerance |
Speed increase with N
disks in the array* |
Disk space overhead** |
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JBOD (Span) |
2+ |
None |
Uncertain (heavily depends on
the volume layout). No significant increase in typical applications. |
None |
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RAID 0 (Stripe) |
2+ |
None |
N times increase on both
reads and writes |
None |
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RAID 1 (Mirror) |
Exactly 2 |
Single disk failure |
Double speed on reads;
No gain on writes |
100% |
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RAID 5 (Stripe with parity) |
3+ |
Single disk failure (this
will cause read speed to degrade) |
(N-1) times increase on
reads;
up to 2x loss of speed on writes |
A capacity equivalent to that
of one member disk is used to hold checksums |
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RAID 0+1 (Mirrored stripe
set) |
4+ |
Single disk failure; half of
dual disk failures, depending on the
location/assignment of failed drives. |
N times increase on reads;
N/2 times increase on writes |
100% |
*Note: speed increase is a very rough
estimate based on the assumption that disk traffic consists mostly of linear
(sequential) reads of large data chunks. This estimation also assumes
controller(s) are capable of overlapped operations and there are no problems
with bus throughput (since several RAIDed high-performance drives can easily
saturate the bus like PCI).
**Note: disk space overhead
calculation is based on the assumption that all member disks are of equal
capacity. If the disks are not equal in size, the smallest of all disks will be
used as a "column" size.
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Copyright © 2001-2008 Zero Assumption Recovery [Data recovery forum]
There are currently 98 visitors browsing the site.
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Weekend discount in effect 40 hours 54 min left
Sounds good, but it didn't identify my camera as a drive.
I use an Olympus C740UZ with an XD picture card, and Windows XP.
Windows reads the camera as removeable disk E: but ZA failed to pick it up.
I tried all the different processes suggested by ZA help, without success.
Shame, because all the other reviews suggest it's a nice little bit of software.
Try it anyway, you've got nothing to lose.
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