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OllyDK

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07-03-2014
07:23 AM

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

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andre_rossouw

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07-03-2014
08:31 AM

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.

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andre_rossouw

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07-03-2014
08:31 AM

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.

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OllyDK

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07-04-2014
02:18 AM

Re: MirrorView/A bandwidth?

Thx a lot, Andre. That helped me out...

Have a nice weekend!

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