While I was at TechEd, I had a great time talking with our customers and hearing about their experiences as Exchange administrators. One of the areas that have come up in discussion a lot has to ...
Updated Jul 01, 2019
Version 2.0
Blog Post
Here is the basic info.
Server A
Time slice from 9 am to 6 pm.
Disk Transfers/Sec
Min/Avg/Max
34.400/339.052/1012.903
MsExchangeIS/Active User Count MsExchangeIS
Min/Avg/Max
519/920/1177
Disk Reads/sec
Min/Avg/Max
10.2/206.3/577.8
Disk Writes/sec
Min/Avg/Max
18.467/132.75/766.715
Total Mailboxes
563 on the server
Calculate your IOPS/user
I've seen a couple different options here.
Worst Case scenario
Max Disk Transfers/sec / Total Mailboxes on the server
1012/563
1.797
Max Disk Transfers/sec / Max Active user count from MSExchangeIS
1012/1177
.85
From the Exchange 2003 performance and scalability guide
IOPS/mailbox = (average disk transfer/sec) ÷ (number of mailboxes)
339/563
.602
and then
Avg Disk Transfers/sec / Avg Active user count from MSExchangeIS
339/920
.368
1. So which one is correct or should be used ?
The table Tables to lookup recommended maximum disk throughput per disk:
Table 1. Estimated maximum disk throughput for No Raid or Raid 0
2. Where do these numbers come from ? I understand the 180, but how do you reduce the achived throughput to 115 or 108 (using RAID 0+1)
3. The R:W ratio, is it as simple at looking at the averages on Disk Reads/sec and Disk Writes/Sec ?
So my ratio is roughly about 2:1 from the numbers listed above ?
Thanks this info is very good!