Hey Guys

I have disks stripped at raid5. We have 10 I/O channels
Does it mean i need to have 10 different file systems for the Database containers for the optimal performance.

Can we have optimal performance when we use single filesystem and disks are stipped on raid5 with 10 I/O channels

thx ahead

Tony

I am not going to comment on the "10 I/O channels" because I not sure I understand what that exactly means.

But keep in mind that unless you have a data warehouse application with a lot of table scans, spreading out a tablespace to multiple containers will not make much difference in performance. If you application is mainly OLTP with well written queries that use indexes, and you have fairly good bufferpool hit ratio, I would not worry too much about multiple containers.

Hi,
if I understand correctly the hard disk for optimal performance should be configured something like this:
disk1 --> apply to disk 1 array --> create one RAID5 logical disk on 1 array
disk2 --> apply to disk 1 array --> create one RAID5 logical disk on 1 array
disk3 --> apply to disk 1 array --> create one RAID5 logical disk on 1 array
disk4 --> apply to disk 2 array --> create one RAID5 logical disk on 2 array
disk5 --> apply to disk 2 array --> create one RAID5 logical disk on 2 array
disk6 --> apply to disk 2 array --> create one RAID5 logical disk on 2 array
etc.
If you have more disks then you can create more arrays. If using RAID5 at least 3 disks must be associated to one array.
If you have less then 6 disks then only one option is possible: all disk in one RAID array.

If you will install Windows on system then you will have c: to array 1 and d: for array 2. On Unix/Linux there will need to mount both disk to file system

Then create database (if you are using v8 and above) with "automatic" syntax and define c:, d:.

This will by default create two containers spread to two disk arrays.

Bear in mind each array has its own RAID5 field.

Hope this helps,
Grofaty

My system is OLTP system on UNIX servers.
Can all these disk arrays point to the same file system or each array should have its own file system.

atl_geek,
for each disk array create file system and then mount one file system to another one. The main idea is DB2 will have two physical paths to store data to disk (in case of two disk arrays). In version 8.2.2 (v8.1 fixpack 9) and above you can create database using the following syntax:
The above DB2 command will tell DB2 you have two physical storage paths that can be used in parallel. DB2 creates one tablespace and two containers, each containers points to one disk array.

Can you please write your version of database, execute 'db2level' command and tell as more about your database type, execute: 'db2limc -l' command.

Hope this helps,
Grofaty

Can we say each disk array represents one file system

A disk array could be one or more file systems.

If you use multiple containers for a tablespace, then each container should be a different physical array (not just a different file system).

What is the command to find out the disks in each array
Also what is the command to find out what disk arrays are allocated to each file system.

Ps: For AIX systems

Thx alot for your help

each hardware raid5 system is a self contained unit which appears as a hdisk to AIX. One should not care for the individual disks in the raid5 array; you have no control over it. You create a VG(volume group) using 1 or more hdisks; then you create LV(logical volumes); lvs can be used as raw disk; also a JFS or JFS2 filesystem can be created on lv. Make the segment size and prefetchsize of any DMS tablespace an integral multiple of Raid Stripesize.