This post is more than 5 years old
16 Posts
0
2205
August 12th, 2013 00:00
E20-324 MirrorView/A unmanned DR badwidth
Hi I am studying for E20-324 and I'm struggling with this question as I can not match my answer with the provided answers.
Here is the question:
Your customer wants to use a VNX system at an unmanned DR site. The system will be used exclusively as a secondary MirrorView/A system, and it is capable of performing 60,000 LUN IOPs in this role. The primary system is also a VNX.
If the primary image workload consists of 1 KiB random IOs with a R/W ratio of 3:1, what is the maximum bandwidth required for the link between the VNX systems?
240 Mb/s
120 Mb/s
360 Mb/s
600 Mb/s
I tried to answer the question this way:
Each update to secondary image cause 5 + 1 = 6 I/O
60,000 / 6 = 10,000
10,000 * 1 Kib = 10,000 Kib
Which is not in the answers , can you please help me to sort this out?
Thanks,
Golnaz
andre_rossouw
62 Posts
0
August 19th, 2013 06:00
Hi, golnaz!
I did take the 2 kB MV/A granularity into account; I didn't have to use it in the calculation, though, because the host write size was 4 kB, obviously larger than 2 kB, and an integer multiple of 2 kB.
In the second example you mentioned, where the secondary system can perform 60,000 LUN IOPs and the host write size is 1 kB, the MV/A granularity is important. Even though host writes are 1 kB, MV/A will transfer 2 kB across the wire, so the bandwidth required is:
10,000 updates/s x 2 kB/update = 20,000 kB/s = 200,000 kb/s = 200 Mb/s
Bear in mind that for serial communications, like network links, bandwidth is measured in decimal units rather than binary units, meaning that there are 1,000,000 bits in a Mb/s and not 1,048,576 (1024 x 1024) bits.
christopher_ime
2K Posts
0
August 17th, 2013 21:00
Each time I calculate it, I get 240 Mb/s. Is that the answer?
Why are you saying 6 I/O per update on the secondary? CoFW should be just 5 I/O if I'm not mistaken.
MATH:
=====
1) 60000 LUN IOPs / 5 IO for CoFW per update = 15000 updates per sec
2) The key with MirrorView/A is that the transfer granularity is 2KB and so even though there are 1KB (random) updates it is a 2KB transfer :
2KB * 15000 updates/s = 30000 KB/s is the maximum bandwidth required
But the choices are in Mb/s (bits). So...
3) Convert to Kb/s (bits)
30000 KB/s * 8 b/B (bits per Byte) = 240000 Kb/s
4) Convert to Mb/s ("networking/serial" multiplier of 1000 and not 1024)
240000 Kb/s / 1000 = 240 Mb/s
christopher_ime
2K Posts
0
August 17th, 2013 21:00
Can you explain the "+1"?
christopher_ime
2K Posts
0
August 17th, 2013 22:00
Okay, so I do see it documented as 6 LUN I/O's per CoFW. Sorry about that.
"1 chunk read, 1 chunk write, 1 host I/O, 3x map updates."
Plugging that consideration into my calculations, then I see what you mean. I don't get any of the answers. Hopefully someone else can tell us where the discrepancy is.
MATH
====
1) 60000 LUN IOPs / 6 LUN IO for CoFW per update = 10000 updates per sec
2) The key with MirrorView/A is that the transfer granularity is 2KB and so even though there are 1KB (random) updates it is a 2KB transfer :
2KB * 10000 updates/s = 20000 KB/s is the maximum bandwidth required
But the choices are in Mb/s (bits). So...
3) Convert to Kb/s (bits)
20000 KB/s * 8 b/B (bits per Byte) = 160000 Kb/s
4) Convert to Mb/s ("networking/serial" multiplier of 1000 and not 1024)
160000 Kb/s / 1000 = 160 Mb/s
christopher_ime
2K Posts
0
August 17th, 2013 22:00
Doh... I can't do math apparently. Ignore my calculations above.
Hmmm... good question. I think it comes down to what each update on the secondary per 2KB transfer in LUN IOPs are.
golnaz1
16 Posts
0
August 18th, 2013 18:00
Hi Christopher Imes ,
Please have a look at this post which has been answered by andre.rossouw
https://community.emc.com/thread/144071?start=0&tstart=0
I completely confused and I have my exam in 2 days
He did not take into consideration 2 KiB granuality of MirrorView/A ...
He answered :
-----------------------------------------------
Your customer wants to use a VNX system at an unmanned DR site. The system will be used exclusively as a secondary MirrorView/A system, and it is capable of performing 30,000 LUN IOPs in this role. The primary system is also a VNX. If the primary image workload consists of 4 KiB random IOs with a R/W ratio of 3:1, what is the maximum bandwidth required for the link between the VNX systems?
Each update to a MV/A secondary image causes a COFW, plus the write of the new data. That's 6 I/Os. If the VNX can do 30,000 IOPs, it can do 30,000/6 = 5,000 updates/s. The host write is 4 KiB, so if the data is aligned (and we'll assume it is, since the question doesn't say that it isn't), that's 5,000 updates/s x 4 Kib/update = 20,000 KiB/s. That's how much bandwidth we need.
------------------------------------------------
So if I follow him:
60000 LUN IOPs / 6 LUN IO for CoFW per update = 10000 updates per sec
10,000 updates/s x 1 KiB/update = 10,000 KiB/s
if
10,000 KiBs * 8 = 80,000 kib/s
80,000 / 1024 = 78 Mib
I am totally lost !!
golnaz1
16 Posts
0
August 19th, 2013 18:00
Thank you very much Andre!
It became clear for me now. The correct answer will be 200 Mb/s which is not included in the answers. Your point about serial links was very helpful .