This post is more than 5 years old
33 Posts
0
4606
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.
andre_rossouw
62 Posts
1
July 8th, 2013 11:00
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.
paul881
33 Posts
1
July 9th, 2013 01:00
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.
andre_rossouw
62 Posts
2
July 9th, 2013 12:00
As noted elsewhere, a COFW performs 5 I/Os with recent VNX software [plus the original write]. Be careful of that.
paul881
33 Posts
0
July 10th, 2013 04:00
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.
andre_rossouw
62 Posts
0
July 10th, 2013 05:00
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.
Alp Arslan
34 Posts
0
December 4th, 2016 00:00
So it means Now we need to calculate for 5 I/Os for CoFW.