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paul881

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07-03-2013
05:17 AM

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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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.

andre_rossouw

2 Iron

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07-08-2013
11:32 AM

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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.

andre_rossouw

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07-08-2013
11:32 AM

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paul881

2 Bronze

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07-09-2013
01:19 AM

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

andre_rossouw

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07-09-2013
12:37 PM

Re: E20-324 - Mirrorview/A

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paul881

2 Bronze

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07-10-2013
04:50 AM

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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A FC write makes 2 round trips, therefore traverses the link 4 times - that's where the 4 comes from. The 2, when using protocol converters, reflects the number of converters in the link - FC->IP, then IP->FC.

andre_rossouw

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07-10-2013
05:44 AM

Re: E20-324 - Mirrorview/A

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Alp Arslan

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12-04-2016
12:41 AM

Re: E20-324 - Mirrorview/A

So it means Now we need to calculate for 5 I/Os for CoFW.

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