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3 Posts
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2672
MirrorView/A bandwidth?
Hi all!
Somehow I miss a point in calculating required bandwidth for MV/A. I hope someone can clear things up.
Let´s say we got two VNX systems, the secondary is able to do 80,000 LUN IOPS. The load on the primary image is 1kb random 4:1. What is the max required bandwidth between the two VNX systems?
Thanks a lot and best regards,
Olly
andre_rossouw
62 Posts
2
July 3rd, 2014 08:00
Hi, Olly!
There are a few parts to this question:
a. In a MV/A environment, the secondary system limits [though it may not be the only limit] how many updates/s can be performed based on its performance. In this case we know that the secondary can do 80,000 LUN IOPs. We also know that a MV/A update causes 6 I/Os [primary system write, read of secondary chunk, write of secondary chunk, 3x map updates], so that means we can perform at most 80,000/6 updates/s = 13,333 updates/s.
b. The transfer granularity of MV/A is 2 kB, so even though our host write size is 1 kB, each host write causes 2 kB to cross the link as an update
c. The R/W ratio is irrelevant in this question - we're calculating maximum required bandwidth, not the maximum number of I/Os at the primary
So, maximum required bandwidth is 13,333 updates/s x 2 kB/update = 26,666 kB/s. You could divide that by 1,024 to give MB/s, in which case the answer is ~26 MB/s. Translated into serial speed [multiplying by 10 to change bytes to bits, and also account for protocol overhead], that's ~260 Mb/s.
OllyDK
3 Posts
0
July 4th, 2014 02:00
Thx a lot, Andre. That helped me out...![](https://prod-care-community-cdn.sprinklr.com/community/687062f5-603c-4f5f-ab9d-31aa7cacb376/ReMirrorViewAbandwidth-56628bce-f312-440d-a09b-d3b9df7704ca-512917881)
Have a nice weekend!