This post is more than 5 years old

Solved!

Go to Solution

8013

October 4th, 2012 01:00

# MirrorView /S bandwidth calculation

Hello,

After digging through all the VNX Unified Storage Solutions Desing Labs and their answers, I still have a problem.

It is regarding Mirrorview/S.

SRC LUN 2TB. Current load 1000 4kB IOPS R/W  4:1

Mirror should resync within 24hr if fractured for 90min or less.

How much bandwidth is required?

--->  Mirrorview/s depends on concept of WIL and extents.

So, for 2TB SRC LUN, extent size is 1000KB. So in 90min of fracture:

amount to be sent to mirror in 90 min fracture = (1000/5) * 1000(extent size) * 90 * 60  =  1054 GB

BW = (amount in 90min) * 2(Round trip) / (24hrs)  = 1054*2/(24*60*60) = 24.4 MB/s = 195 Mb/s

Where did I go wrong?

62 Posts

October 5th, 2012 10:00

Hi, Ron!

You are correct about the answer being related to extent size for the MirrorView/S mirror.

A 2 TB mirror (2,048 GiB) has an extent size of 1024 KiB. If it's fractured for 90 minutes, the amount of data marked dirty depends on the number of writes, and how random they are. 200 writes/s x 1024 KiB x 90 minutes x 60 seconds = 1,105,920,000 KiB. This data must be copied to the secondary in 24 hours, so required bandwidth - FOR THE RESYNCHRONIZATION - is 1,105,920,000 KiB / (24 x 60 x 60) seconds = 12,800 KiB/s.

Regular updates require 200 writes/s x 4 KiB = 800 KiB/s. The total bandwidth required is therefore (12,800 + 800) KiB/s = 13,600 KiB/s = 136,000 Kb/s = 136 Mb/s.

49 Posts

October 9th, 2012 02:00

Andre,

Thank you very much.

Ron

10 Posts

October 19th, 2012 15:00

Sorry .. regard the calculation:

A customer has a 2 TiB LUN which will be mirrored with MirrorView/S. Analyzer shows the current workload to be 500 8 KiB IOPs with a R/W ratio of 3:1. The customer requires that the mirror should resynchronize within 2 hours if fractured for 10 minutes or less. How much bandwidth is required to meet customer requirements?

500 / 4 * 1= 125(Write) * 1024(extent) * 10(min) * 60(sec)  = 76.800.000

76.800.000 KiB / (2*60*60) = 10.666 KiB/s

(125*8) = 1.000

1000+10.666 = 11.666 KiB/s = 116.000 Kb/s = 116 Mb/s

It's rigth ?

TKS

49 Posts

October 22nd, 2012 00:00

@vmauro,

Yes that seems right to me.
The only thing is that you should do the Kb/s : 1024 = Mb/s
so 116,666 Kb/s : 1024 = 113 Mb/s

10 Posts

October 22nd, 2012 02:00

TKS

about how many data in the fracture time ?

You have some formula?

19 Posts

December 26th, 2012 09:00

how is the extent size for the 2TB src LUN calculated? thanks

1 Message

January 9th, 2013 05:00

A 2 TB mirror (2,048 GiB) has an extent size of 1024 KiB. If it's fractured for 90 minutes, the amount of data marked dirty depends on the number of writes, and how random they are. 200 writes/s x 1024 KiB x 90 minutes x 60 seconds = 1,105,920,000 KiB. This data must be copied to the secondary in 24 hours, so required bandwidth - FOR THE RESYNCHRONIZATION - is 1,105,920,000 KiB / (24 x 60 x 60) seconds = 12,800 KiB/s.

Regular updates require 200 writes/s x 4 KiB = 800 KiB/s. The total bandwidth required is therefore (12,800 + 800) KiB/s = 13,600 KiB/s = 136,000 Kb/s = 136 Mb/s.

WHY: 200 writes/s x 1024 KiB x 90 minutes x 60 seconds ???

I don't undestand why 200 writes/s,

Tks

62 Posts

January 14th, 2013 12:00

The question stated 1000 IOPs with a 4:1 R/W ratio, hence 200 writes/s. Those writes continue while the secondary is fractured, and contribute to the data extents marked dirty. Once resynchronization starts, those writes are still occurring, in addition to resynchronization traffic.

The question is obviously simplified in that it assumes a constant data rate (and makes other assumptions as well).

4 Posts

April 11th, 2013 11:00

santechie

default extent size for 256GB is 128Kib

After that every 2Gib extent size is 1Kib

So for 2TB:

2048GiB-256GiB(128Kib known)=1792Gib

for 1792Gib Extent is 1792Gib/2=896Kib(Bcz for every 2Gib extent is 1Kib)

So total is 896+128=1024Kib is Extent size for 2TB

5 Practitioner

•

274.2K Posts

June 18th, 2013 10:00

Sorry about the late response but I just happened to find this when searching for something else.

The calculation as I understand is, if the mirror is fractured for 90min, then the amount of data marked dirty is

1024 kib (extent size) x 200 (writes) x 5400 sec (90 min) = 1105920000 kib or 1080000 mib or 1054.6875 gib of data changed on the source.

That data has to resync in 24 hrs, so
1105920000 kib / 86400 sec (24hrs) = 12800kibps or 12.5mibps.

62 Posts

June 18th, 2013 10:00

The easiest way to calculate extent size for a Clone or a MirrorView/S image is to take the LUN size in GB and use that as the block size for the extent, bearing in mind that theree's a 128 kB minimum. Some examples:

32 GB LUN: 128 kB extent

256 GB LUN: 128 kB extent (minimum size applies up to 256 GB LUN size)

500 GB LUN: 250 kB extent (500 blocks)

1 TB LUN: 512 kB extent (1,024 blocks)

2 TB LUN: 1,024 kB (1 MB) extent (2,048 blocks)

62 Posts

June 18th, 2013 10:00

That's correct for the resynchronization traffic alone. There's also the traffic generated by ongoing writes, and that's the additional 800 kB/s from the first answer.

5 Practitioner

•

274.2K Posts

June 19th, 2013 09:00

Thank you Andrew,

That one makes it clear.

View More

No Events found!

Top