Question about the VILT for Information Availability Design

 

VILT Information Availability  Design:

In Module 6, lesson 3, about 18:50 into the lesson, there is a review of Exercise 1: Full Volume Mirroring.

Is there a misleading statement or mistake in the calculations for LUN 2?

My question:

If the sequential reads for LUN2 replication are divided by 16, why aren't the initial sequential reads and/or writes for LUN2 divisible by 16?

I would appreciate it if you could find out if I am missing something, or if there is a problem with the explanation in the course.

Background:

The instructor reviews the calculations for LUN 1, random access.

In calculating the "additional I/O load on source due to replication", the instructor states that since the reads are sequential, only 125 reads from the source are needed for the replica.

LUN1 Calculations:

500 random 4KB IOPS, RAID5 3:1, 3:1 R/W => 375 reads, 125 writes.

Since the disks are RAID5, the write penalty is 4. The calculation for initial I/O's to the source is

375+ 4* 125 = 875 initial I/O's

The replica 'reads' the source. Since the reads are sequential, there are only 125 reads.

So the total I/O's are

875 initial I/O's + 125 reads for the replica = 1,000

LUN2 caculations:

500 sequential 4KB IOPS, RAID5 3:1, 3:1 R/W => 375 reads, 125 writes.

The instructor states that the initial I/O's are the same as for LUN1. (This is the sentenc I don't understand.)

The instructor then states that a different calculation is used for the replica 'reads', since the I/O is sequential. Each "chunk" is 64KB, and the blocksize is 4KB, which means that 16 blocks are read by each I/O.

Replica reads: 125 * 4KB/64KB = 7+ = 8 I/O's

Total I/O's are:

875 initial I/O's + 8 reads for the replica = 883

This led to my question above, repeated here:

If the sequential reads for LUN2 replication are divided by 16 (64KB/4KB= 16), why aren't the initial sequential reads and/or writes for LUN2 divisible by 16?