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## E20-324 VideoILT Module 5 - Scenario 1 & 2 solutions missing

Hi,

I search the entire CD and can't find the solutions for scenario 1 - snapshot and scenario 2 - clone questions in Module 5.

Appreciate if any instructor can post the solutions.   Someone has posted their own solutions in the past but I am not sure if the answers are correct.

Thank you.

1 Solution

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## Re: E20-324 VideoILT Module 5 - Scenario 1 & 2 solutions missing

You have the method correct. Remember, though, that COFWs now perform 4 writes to the RL, so for LUN 1 you'd see 125 writes x 4 = 400 writes/s. For a R5 LUN, that's 400 x 4 = 1600 disk IOPs/150 = 10.7 disks. At this point you could decide that 10 disks would be OK [which is probably the case], or round up to the next available value for R5 4+1, which is 15 disks.

For LUN 2, you'd get 125/16 x 4 x 2 =  63 IOPs = 2 disks [you'd have to round up].

For LUN 3, you'd see 80 x 4 x 4 = 1280 disk IOPs/90 = 14.2 = 15 disks.

Yes, the host R/W ratio matters, because only host writes cause COFWs, so you need to first determine how many writes/s there are.

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## Re: E20-324 VideoILT Module 5 - Scenario 1 & 2 solutions missing

Post your answers to the questions and I'll let you know if they're correct.

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## Re: E20-324 VideoILT Module 5 - Scenario 1 & 2 solutions missing

Hi Andre,

I will work on Scenario 1- Snapshot first

LUN 1 : 500 random 4 kib IOPs , 3:1 R/W

( 500 / 4 ) * 3 ( 3 writes to the RLP ) = 375 IOPs

# of R5  10K SAS disks :

( read i/o + ( write i/o * wp ) / disk iops

( 0 + ( 375 * 4 ) / 150 ( for sas 10k)  = 10 drives

=======================================================

LUN 2 :  500 sequential 4 kib IOPs, 3:1 R/W

( 500 /4 ) = 125 ,  ( 125 / ( 64/4 kib) ) * 3 ( 3 writes to the RLP ) = 23.43 = 24 IOPs

# of R1/0 10K SAS

( 0 + ( 24 IOPs * 2 ) / 150  = 0.32 = 2 drives

============================================================

LUN 3 : 400 Random 8 kib IOPS , 4:1 R/W

( 400 / 8 ) * 3  = 150 IOPs

# of R5 7.2K  NL-SAS

(0 + ( 150 * 4) / 90 =  7 disks

==================================================================

One big question :

Does the R/W ratio change anything, I am not sure how to take that into consideration when doing the calculations ?

On LUN 3 it is 4:1 R/W

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## Re: E20-324 VideoILT Module 5 - Scenario 1 & 2 solutions missing

You have the method correct. Remember, though, that COFWs now perform 4 writes to the RL, so for LUN 1 you'd see 125 writes x 4 = 400 writes/s. For a R5 LUN, that's 400 x 4 = 1600 disk IOPs/150 = 10.7 disks. At this point you could decide that 10 disks would be OK [which is probably the case], or round up to the next available value for R5 4+1, which is 15 disks.

For LUN 2, you'd get 125/16 x 4 x 2 =  63 IOPs = 2 disks [you'd have to round up].

For LUN 3, you'd see 80 x 4 x 4 = 1280 disk IOPs/90 = 14.2 = 15 disks.

Yes, the host R/W ratio matters, because only host writes cause COFWs, so you need to first determine how many writes/s there are.

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## Re: E20-324 VideoILT Module 5 - Scenario 1 & 2 solutions missing

For LUN 1 , it is (500/4) * 4 writes = 400 writes/s.   4 writes is for the RLP right  ?

For the source LUN is it 1 read and 3 writes ?

For LUN 2,  I am confused with sequential.

So it is 125 / 16 x 4 x 2 , what is 16, 4 and 2 ?  ( if  2 is the WP ) , what is 16 and 4 then ?

For LUN 3,  80 x 4 x 4 - ( if  4 is the WP )  what is the 80 ? , what is the second number 4 ?

How do I factor in the R/W ratio into my calculations ? Given that LUN 1, 2 is 3:1 and LUN 3 is 4:1

Thank you.

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## Re: E20-324 VideoILT Module 5 - Scenario 1 & 2 solutions missing

I made a mistake - for LUN 1 you'd see 125 x 4 = 500 writes/s into the RL. The Source LUN would see 1 write from the host, plus a 64 kB chunk read for the COFW.

For sequential I/O, a number of writes will occur before the next chunk is touched. The number of writes depends on the size of the chunk - fixed at 64 kB - and the size of the I/O - 4 kB in this case. The first write touches a virgin chunk, so causes a COFW. When the next 4 kB write arrives, it's no longer the first write, because the chunk has been touched already. This continues until write #17, which will again be the first write to a virgin chunk. So, we see that 64 kB / 4 kB = 16, which means that only one write out of 16 causes a COFW. If the I/O size was 8 kB, every 8th write would cause a COFW, so we'd divide the number of writes by 8.

For LUN 2, then, we have 125 writes/s / 16 [because it's sequential, and 64 kB/4 kB] x 4 [4 writes to the RL] x 2 [the write penalty for R1 or R1/0] = 63 disk IOPs.

LUN 3 uses R5, so we have 80 writes/s x 4 writes/COFW x 4 [write penalty for R5] = 1280 disk IOPs.

You need the R/W ratio if you're given the number of IOPs and the R/W ratio. As examples, 500 IOPs with a R/W ratio of 3:1 gives you 125 writes/s, whereas 500 IOPs with a R/W ratio of 4:1 gives you 100 writes/s. You need the ratio of writes to total IOPs, so it's (writes / (reads + writes)).

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## Re: E20-324 VideoILT Module 5 - Scenario 1 & 2 solutions missing

Thanks for the explanation.

I just wanted to confirm that I understood you correctly.   Please verify that this is correct

So, in the case of LUN 2 :  500 sequential 4 kib IOPs, 3:1 R/W,

We have 3:1 R/W which is 25% of 500  = 125 writes

In the case of LUN 3 : 400 Random 8 kib IOPS , 4:1 R/W,

We have 4:1 R/W which is 20% of 400 = 80 writes

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That's correct.

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## Re: E20-324 VideoILT Module 5 - Scenario 1 & 2 solutions missing

Hi Andre,

I just read your paper on Choosing the Right CLARiiON®  Data Replication Method: A Performance-Based Approach and it is a great paper !

I am finishing my e20-324 videoILT course however I can't figure out the solutions on lab 9: Consolidation case study.

I suppose I am overwhelmed with the information provided by the question hence I am not really sure which are the pertinent pieces of information that I have to use.

Appreciate if you can help me with this question.

The solution is the following:

Proposed configuration

Primary array

VNX5300 DPE with 8 x 3.5 inch 600 GB 10 k rpm SAS drives

additional drives - (6) 3.5 inch 600 GB 10 k rpm SAS drives

- (12) 3.5 inch 2 TB 7.2 k rpm NL-SAS drives

hot spares - (1) 3.5 inch 600 GB 10 k rpm SAS

(1) 3.5 inch 2 TB 7.2 k rpm NL-SAS

dual control stations

Local and remote protection suites

Note 1 : amounts below does NOT include the systems drives (the first four drives in enclosures 0_0)

est. usable capacity = 17.01 TB

est. checkpoint/snapshot capacity of 5% = 970 GB

est. rule of thumb performance = 2760 backend IOPS

Note 2 : amounts below DO include the systems drives (the first four drives in enclosures 0_0)

est. usable capacity = 18.58 TB

est. checkpoint/snapshot capacity of 5% = 980 GB

est. rule of thumb performance = 3320 backend IOPS

Note 3 : amounts below does NOT include the systems drives (the first four drives in enclosures 0_0)

est. usable capacity = 11.78 TB

est. rule of thumb performance = 1200 backend IOPS

Note 4 : amounts below DO include the systems drives (the first four drives in enclosures 0_0)

est. usable capacity = 13.35 TB

est. rule of thumb performance = 1760 backend IOPS

Q1.  How did the solution arrived at the number of drives in the primary array :

Primary array

VNX5300 DPE with 8 x 3.5 inch 600 GB 10 k rpm SAS drives

additional drives - (6) 3.5 inch 600 GB 10 k rpm SAS drives................... and the rest of it.

Q2.  Note 1 :  how is 17.01 TB and the 2760 IOPs obtained ?  I added up all the disks capacities in the primary array and is no where near this figure.  How did they get the IOPs ?

Q3.  Note 2 :  18.58 TB, 3320 IOPs ?   I use the VNX capacity calculator and this is the capacity I obtained :

But I am not sure if I am doing this right.

Q4.  Is the checkpoint / snapshot capacity usually 5% ?

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