RLP Size and Load

Well Andrea.. That did help me thanks.

I would like to discuss couple of questions:

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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?

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Since it's unmanned DR then we will assume the 30k iops all write iops. And mirrorview/A transfer size is 2k. So, the whole 64k chunk at source won't all be transfered to the DR.

My logic says:   30000 * 4 * 2 = 234 MB/s  is this correct?

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What if the same question requires to calculate the write throughput at the SRC LUN??

example:

Unmanned DR capable of performing 50000 IOPS. Using mirrorview/A. Primary image load is 8k IOs with R/W ratio:  5:1.

What is the maximum allowable total primary image throughput?

--> Here I am kinda clueless!!!

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Last question, is regarding Mirrorview/S.

SRC LUN 2TB. Current load 1000 4kB IOPS R/W  4:1

Mirror should resync within 24hr if fractured for 90min or less.

How much bandwidth is required?

--->  Mirrorview/s depends on concept of WIL and extents.

So, for 2TB SRC LUN, extent size is 1000KB. So in 90min of fracture:

amount to be sent to mirror in 90 min fracture = (1000/5) * 1000(extent size) * 90 * 60  =  1054 GB

BW = (amount in 90min) * 2(Round trip) / (24hrs)  = 1054*2/(24*60*60) = 24.4 MB/s = 195 Mb/s

Where did I go wrong?

Hi Saramir,

I'm current studying for this exam and it would be very helpfull if you could provide me the correct calculation - if you ever did find out the correct calculation for your secenario's?

thanks and kind regards

Tom

It's Andre ...

Your third question has been answered elsewhere on this board: see https://community.emc.com/message/680478#680478

For this 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 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.

For this question:

Unmanned DR capable of performing 50000 IOPS. Using mirrorview/A. Primary image load is 8k IOs with R/W ratio:  5:1.

What is the maximum allowable total primary image throughput?

Each update was caused by a write to the primary image. If the secondary performs a maximum of 5,000 (say) updates/s, then the primary can only perform 5,000 unique writes/s. That's the throughput you're looking for.

Finally I realize that COFW do 5 IO and of course there is original write IO, thats why LUN IOPS are devided by 6.... Thank you!!!

BUT on the forum, there are places where COFW is 3, than 4 and finally it is 5 :-)

Bear in mind that sometimes we may be talking about the effect of COFW on the Source LUN, sometimes on the Reserved LUN, and sometimes on both. In addition, we may be interested only in writes for some calculations, or reads and writes for others. Then there's the change in the internal operation of COFW to bear in mind as well ...

I'm glad it's a bit clearer now

andre.rossouw, can you please explain what the effect of the COFW is based on these different situations? I'm trying to understand how the COFW is determined and whether it has a direct relation to the read/write ratios.

Thank you!

Thank you andre.rossouw for the prompt reply!

So looking at the above example you state: "Each update to a MV/A secondary image causes a COFW, plus the write of the new data. That's 6 I/Os." Can you walk me through how you got 6?

Hi, Ernes!

We may need to look at the Source LUN, the Reserved LUN, or both, depending on what we're trying to achieve.

If, for example, you're asked about the performance effect of COFW activity on the Source LUN, you'll look only at that LUN. You would need to know the number of writes to the Source LUN [because reads don't cause COFW activity], the size of the writes [to determine if each causes only 1, or perhaps more, COFWs], and how random the pattern is. For exam purposes, 'random' means that each Source LUN write goes to a new area and therefore causes COFW activity, whereas 'sequential' means that the activity is purely sequential.

If you're asked about the effect on the Reserved LUN, you need the same information as before, plus you need to know if the secondary host is writing to the snapshot [once again, reads don't matter when we're only interested in COFW]. Once you have that information, you can determine the amount of write activity to the Reserved LUN, bearing in mind that there are data writes [the 64 kB chunk writes] and metadata [or map] writes, which are usually 8 kB in size.
The COFW impact on the Source LUN is caused by [1] the 64 kB chunk read, and [2] having to wait until COFW activity is completed before the host write can proceed. The COFW impact on the Reserved LUN is caused by the chunk write and the map write [which should all be cached writes].

Additional load on Source and RL can be caused by snapshot read activity. Our rule of thumb is an 80/20 mix, so we'd assume that 20% of the snapshot reads are satisfied by the RL, and the other 80% by the Source LUN.

Hope this helps ...

Yes, when the R/W ratio is 4:1 you need to divide the total I/O count by 5 to get the writes.

When a MV/A secondary image is updated, the update write is always a first write to the specific chunk, so will cause a COFW.

A COFW reads a chunk from the Source LUN [1 I/O], writes that chunk to the RL [1 I/O], and updates the RL map [3 I/Os], for a total of 5 I/Os. Add the host write [in this case, the MV/A update write] that caused the COFW, and you have 6 I/Os.

Remember also that those are 6 LUN I/Os, and not 6 disk I/Os, so to work out [if you have to] how many disk I/Os are involved you need to know the RAID type.

Thank you again! Appreciate the prompt reply.

Awesome, thank you!! Makes perfect sense.

One last question, I promise . I just found the following post: E20-324 VNX Certification - Scenarios. The RLP calculation by vpert for LUN 3 is off since the read/write is 4:1, correct?