Understand our server system

Hi all

I'm sorry for the long question, that it is. I wonder if someone can help me understand how our system of hard drives and servers are configured. Our main server is a Poweredge 2950 with five physical hard disks. Three in a Raid 5 configuration and the other two on Raid 1. The Raid 1 virtual disk size is 1.86 Tb. Which is what Managing Open Server Administrator said that the physical disks are two discs with each to 1.86.

The size of Raid 5 is Go 930 and OMSA says it's three discs each 465 GB. Do I understand correctly that, in a Raid 5 configuration, three disks are required, but the size is only made up of two discs? If not, where did the another 465 GB?

Thank you

Magnus

Nitman,

That's right, Raid 5 requires at least 3 drives. With the reduction of the size, it's because of the parity. Raid 5 striped blocks and distributed parity. This means that parity is going through and is stored on all readers. Unlike Raid 3, who had completely separate players who stored parity. So, essentially when you set up the table, the parity will be an equal percentage % of the installed readers. If player 3, a 1/3 of the space is designated for parity. 4 discs 1/4 of each drive is used and so on. The reason for this is so that if any single drive has failed, and then the remaining disks have complete information on the remaining disks. What makes the redundant table for a single drive failure. Raid 1 is basically a mirror, where both discs have the same data, so if one fails, all the data are kept.

Let me know if it helps.

Tags: Dell Servers

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