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тАО11-10-2003 10:02 PM
тАО11-10-2003 10:02 PM
disk busy I/O
how can i determine if the the total number of disk transfers for systemconsumed the most disk transfers by scale , in other words "when it considered as a high "
3 REPLIES 3
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тАО11-11-2003 12:49 AM
тАО11-11-2003 12:49 AM
Re: disk busy I/O
If you're using HPUX:
You can use SAM.
You can use IOSTAT too.
But your best friend is to use SAR ("sar -d 5 100)
If your pockets are deep, invest on installing GlancePlus software.
If using Windows, then use the perf tools relating to disk available to you.
You can use SAM.
You can use IOSTAT too.
But your best friend is to use SAR ("sar -d 5 100)
If your pockets are deep, invest on installing GlancePlus software.
If using Windows, then use the perf tools relating to disk available to you.
Hakuna Matata.
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тАО11-11-2003 01:14 AM
тАО11-11-2003 01:14 AM
Re: disk busy I/O
Things to look for in sar -d are:
a. % busy greater than 50%
b. avque greater than 3
c. avwait greater than avserv.
- This means the i/os are waiting longer than the
the time it takes to process them. Bad...
d. However:
- %busy high and queue length low, is okay, because
it means the disk is working, but is keeping up.
e. avserv < 6 ms is good!
If a disk rotates at 10,000 revolutions per minutes, then
how long does one revolution take:
10,000 revolutions
10,000 RPM = ------------------
60 seconds
60 seconds
1 revolution = --------------
10,000
= .006 seconds
= 6 milliseconds
f. Average seek time would be 1/2 rotation time. Also
called "latency time".
g. Average transfer rate for one block of data on one
cylinder would then be 6 milliseconds.
a. % busy greater than 50%
b. avque greater than 3
c. avwait greater than avserv.
- This means the i/os are waiting longer than the
the time it takes to process them. Bad...
d. However:
- %busy high and queue length low, is okay, because
it means the disk is working, but is keeping up.
e. avserv < 6 ms is good!
If a disk rotates at 10,000 revolutions per minutes, then
how long does one revolution take:
10,000 revolutions
10,000 RPM = ------------------
60 seconds
60 seconds
1 revolution = --------------
10,000
= .006 seconds
= 6 milliseconds
f. Average seek time would be 1/2 rotation time. Also
called "latency time".
g. Average transfer rate for one block of data on one
cylinder would then be 6 milliseconds.
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тАО11-11-2003 05:37 PM
тАО11-11-2003 05:37 PM
Re: disk busy I/O
thanx capparroso & abramson ,
if the average disk I/O is about 30K ,
does it requires any upgrade or change ?
attached a graph from OVO/OVR to know your opinion .
regards
if the average disk I/O is about 30K ,
does it requires any upgrade or change ?
attached a graph from OVO/OVR to know your opinion .
regards
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