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## LUN/SCSI ID on 11iv3

Hey folks,
I was wondering how to determine the decimal LUN/SCSI ID for a disk. on 11iv2 there was a formula that would garner that information. But when I try to use that same formula on 11iv3 it gave the wrong ID's.

9 REPLIES 9
Acclaimed Contributor

## Re: LUN/SCSI ID on 11iv3

Example?

Hope this helps!
Regards
Torsten.

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Exalted Contributor

## Re: LUN/SCSI ID on 11iv3

Shalom,

How about you share the formula. Then maybe with a frame of reference and a few minutes I can give you a new formula.

11.31 uses built in multi pathing, which may be interfering with your formula's ability to do good calculations.

SEP
Steven E Protter
Owner of ISN Corporation
http://isnamerica.com
http://hpuxconsulting.com
Founder http://newdatacloud.com

## Re: LUN/SCSI ID on 11iv3

This is a post I found on the 11iv2:

will explain as per the follwoing example..

disk 1623 0/0/1/1/0/4/0.2.4.0.0.15.7 sdisk CLAIMED DEVICE HP OPEN-E
/dev/dsk/c2t15d7 /dev/rdsk/c2t15d7
disk 239 0/0/1/1/0/4/0.2.4.0.1.0.0 sdisk CLAIMED DEVICE HP OPEN-E
/dev/dsk/c29t0d0 /dev/rdsk/c29t0d0

If you notice, the Calculation of the LUN number is done with the last 3 decimal digits in the H/W path. You have indeed used the correct formula to compute the LUN number. Ex: 0/0/1/1/0/4/0.2.4.0.0.15.7 whould yield me..

0*128 + 15*8 + 7 = 127

That is, LUN number 127.

Honored Contributor

## Re: LUN/SCSI ID on 11iv3

It should not be any different on 11iv3..

What are you seeing ?

11iv3 uses the agile view in order to multipathing to work.

e.g. /dev/disk/disk1, disk2, disk3 etc..
ioscan -m dsf to see the relationship of agile name to device paths.

## Re: LUN/SCSI ID on 11iv3

disk 848 0/4/0/0/0/1.186.16.0.0.7.7 sdisk CLAIMED DEVICE FALCON IPSTOR DISK
/dev/dsk/c64t7d7 /dev/rdsk/c64t7d7

this disk should be lun 63

using the calculation of before it comes up as lun 112
Honored Contributor
Solution

## Re: LUN/SCSI ID on 11iv3

0.7.7

8 * 7 = 56
56 + 7 = 63

??
Honored Contributor

## Re: LUN/SCSI ID on 11iv3

I think you formula is now being evaluated differently.

What it should be doing (with the 0.7.7) is:

(0*128)+(7*8)+7=
0+56+7=
63

What I think it is doing is:

8*(7+7)=
8*14=
112

If you have an automated process that is doing these calculation, you may need to add parenthesis to the formula to make sure things are evaluated as they should be.