How to calculate the bandwidth with MirrowView /S

You’d probably want to ask that question in the current “Ask the Expert” discussion, because that’s all about replication!

Hi Ron,

We've move your question to the Ask the Expert forum, where it can be promptly addressed.

Thanks,

Bob.

Good Aftenoon Ron!

We have Ask the Expert Discussion going on about Replication considerations on the Clariion VNX system, where our community expert will be answering questions on Replication Considerations. I have notified the forum moderators to include this question as part of ATE here: https://community.emc.com/thread/149825

RRR : If history repeats, I will ROFL!

Thanks, I will follow it there.

Ron

Did you get a sufficient answer? Or at least a way on how to calculate?

Yes, Andre Rossouw helped me.
This is the answer from Andre:

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.

Andre is almost right, since he accidently switched the TB and TiB. 2 TB = 2,000 GB. 2 TiB = 2.048 TB, not the other way around.

But the way he approaches the question is correct. How much data will your primary system be behind in that 90 minutes. The 1000 IOps you mentioned, is that host IOps or storage IOps? If storage then 1000 IOps in R:w of 4:1 means 200 IOps are writes. If those IOps are host IOps, you have to take the write penalty into the equasion as well. And the randomness he's talking about depends on how often an extend of 1 MiB will be written to. The more the less extends will be changed, but worst case scenario means that every write will actually change a unique 1 MiB extend, which will then have to be replicated.

The 1000 IOPs are indeed storage IOPs.

Thanks

Hi there,

I have moved this discussion to the VNX Support Forum so that it is available to that subject matter audience as an answered question.

Thanks and regards,

Mark

you're welcome (a bit late, but better late than never )

How did you get 200w?

Well, it says that the host is doing 1000 IOps in a 4:1 ratio, so that means for each 4 reads there's 1 write. So projected on the number 1000 that means 800 reads and 200 writes.

Thank you.