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## E20-324 - Mirrorview/A

I have been working on a sample question.  A secondary MV/A system is capable of performing 90,000 LUN IOPs .

If the total primary image workload consists of 4 kib random I/Os with a read write ratio of 4:1.

What is the maximum bandwidth required for the link between the systems ?

I worked out the following solution but it looks like my answer is wrong.

The number of writes is 18,000, a COFW plus original LUN generates 5 I/Os.

Therefore the system can cope with 90,000/5 = 18,000 COFWs.

Hence the bandwidth required is 18,000 updates/s x 4 kib = 72,000 or 70.3 Mb/s.

However, the answer given is 600 Mb/s.

Any help appreciated. Thank you.

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## Re: E20-324 - Mirrorview/A

A COFW generates 5 I/Os, plus the original write, for a total of 6. 90,000 / 6 = 15,000 COFWs/s. Since each host I/O is 4 kB, that's 15,000 x 4 kB = 60,000 kB/s = 600,000 kb/s = 600 Mb/s.

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## Re: E20-324 - Mirrorview/A

A COFW generates 5 I/Os, plus the original write, for a total of 6. 90,000 / 6 = 15,000 COFWs/s. Since each host I/O is 4 kB, that's 15,000 x 4 kB = 60,000 kB/s = 600,000 kb/s = 600 Mb/s.

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## Re: E20-324 - Mirrorview/A

Thanks for all the help, much appreciated.

I am a little confused because Lab 6 on page 31 uses only 5 I/Os:

Q .Two VNX systems which can each perform 24,000 LUN IOPs will be used in a

MirrorView/A environment.

One of them is used exclusively as a secondary for MV/A, with no other activity. What is

the maximum allowable number of purely random 4 KiB writes that can be supported

on the primary MV/A images? Assume that the update takes exactly as long as the

configured update cycle

A COFW plus original LUN write generates 5 I/Os, therefore the

secondary system can cope with 24,000 / 5 = 4,800 COFWs [4,800 updates] per second.

This is the number of unique writes on the primary side.

Q1.  Why does the Lab 6 solution use 5 I/Os instead of 6  , are the 2 questions different ?

Thank you.

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## Re: E20-324 - Mirrorview/A

As noted elsewhere, a COFW performs 5 I/Os with recent VNX software [plus the original write]. Be careful of that.

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## Re: E20-324 - Mirrorview/A

Thanks so much for the explanation.

One last question on MV Write Response Time on Module 5 page 107:

Q1.  Protocol converter delay = 4ms    What is the 4 in ( 4 x 2 x 0.5ms delay ) ?

Is 4 the latency ? confused with Module 5 page 102 , latency times 4 for FC

What is the 2 ?  is it the roundtrips  ?

Q2.  Signal Propagation Delay = 1.6 (4 x 04.ms)

Again , what is the 4 ?

Thank you once again.

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