[Script] Breakdown of Oracle SGA into Solaris Locality Groups

Goal: for a given process, find out how the SGA was allocated in different locality groups on a system running Solaris operating system.

Download the shell script, sga_in_lgrp.sh. The script accepts any Oracle database process id as input, and prints out the memory allocated in each locality group.

Usage: ./sga_in_lgrp.sh <pid>

eg.,

# prstat -p 12820

   PID USERNAME  SIZE   RSS STATE   PRI NICE      TIME  CPU PROCESS/NLWP
 12820 oracle     32G   32G sleep    60  -20   0:00:16 0.0% oracle/2

# ./sga_in_lgrp.sh 12820

Number of Locality Groups (lgrp): 4
------------------------------------

lgroup 1 :   8.56 GB
lgroup 2 :   6.56 GB
lgroup 3 :   6.81 GB
lgroup 4 :  10.07 GB

Total allocated memory:  32.00 GB

For those who wants to have a quick look at the source code, here it is.

# cat sga_in_lgrp.sh

#!/bin/bash

# check the argument count
if [ $# -lt 1 ]
then
        echo "usage: ./sga_in_lgrp.sh <oracle pid>"
        exit 1
fi

# find the number of locality groups
lgrp_count=$(kstat -l lgrp | tail -1 | awk -F':' '{ print $2 }')
echo "\nNumber of Locality Groups (lgrp): $lgrp_count"
echo "------------------------------------\n"

# save the ism output using pmap
pmap -sL $1 | grep ism | sort -k5 > /tmp/tmp_pmap_$1

# calculate the total amount of memory allocated in each lgroup
for i in `seq 1 $lgrp_count`
do
        echo -n "lgroup $i : "
        grep "$i   \[" /tmp/tmp_pmap_$1 | awk '{ print $2 }' | sed 's/K//g' | awk '{ sum+=$1} END {printf ("%6.2f GB\n", sum/(1024*1024))}'
done

echo
echo -n "Total allocated memory: "
awk '{ print $2 }' /tmp/tmp_pmap_$1 | sed 's/K//g' | awk '{ sum+=$1} END {printf ("%6.2f GB\n\n", sum/(1024*1024))}'

rm /tmp/tmp_pmap_$1

Like many things in life, there will always be a better or simpler way to achieve this. If you find one, do not fret over this approach. Please share, if possible.