Hi,

Some part of output:
================
$ hwmgr show scsi
SCSI DEVICE DEVICE DRIVER NUM DEVICE FIRST
HWID: DEVICEID HOSTNAME TYPE SUBTYPE OWNER PATH FILE VALID PATH
-------------------------------------------------------------------------
68: 0 mullins disk none 0 1 dsk0 [0/0/0]
69: 1 mullins disk none 0 1 dsk1 [0/1/0]
71: 4 mullins disk none 0 4 scp0 [2/0/0]
72: 5 mullins disk none 2 4 dsk2 [2/0/3]
73: 6 mullins disk none 2 4 dsk3 [2/0/4]
74: 7 mullins disk none 2 4 dsk4 [2/0/6]
75: 8 mullins disk none 2 4 dsk5 [2/0/7]
76: 9 mullins disk none 2 4 dsk6 [2/0/8]
77: 10 mullins disk none 0 4 dsk7 [2/0/9]
78: 11 mullins disk none 0 4 dsk8 [2/0/10]
79: 12 mullins disk none 0 4 dsk9 [2/0/11]
80: 13 mullins disk none 0 4 dsk10 [2/0/12]
================================================================
Main intention of doing this is to meet BID value=35 by using required number of disks.
a[x]=x (unknown value to programmer)varies from machine to machine.
And a[x] value can be found from below code ..
hwmgr show scsi > scsi.tmp
while read line; do
a[x]=`echo $line | grep "dsk" | awk '{ print $7 }'`
Coming to BID concept:
--------------------------------
BID value =(No of partitions * NUM PATH value).
In general each disk has 4 partitions (a,b,g,h) and and for each partition(a,b,g,h) their wll be NUM PATH value.Each partition NUM PATH value will be equal to NUM PATH value of that particular disk.
Example:- Suppose dsk1 has NUM PATH=4 then partition "a" has Total number of paths=4 partition "b" total number of paths=4 Partition "g" total number of paths=4 partition "h" total number of paths=4 Hence,if you use dsk1 then we can have BID value=(No of partitions * NUM PATH)=4*4=16.

So our requirement is to meet BID value=35.
After getting NUM PATH value by above code,we must use the required number of disks to meet BID=35.
Example:Let say dsk2 and dsk3 has NUMPATH=4 then we can get BID value=16+16=32) Still we didnt meet BID=35 and here we must note that we can't use the disk which has NUM PATH=4 now at this stage because 32+4=36 (not equal to 35) so we must use less than NUM PATH=4 , say if you find NUM PATH=1(less than NUM PATH=4)so we can proceede and here we must check how many partitions to be used to meet BID value=35.In this case only 3 partitions are required which has NUM PATH=1 so finally
32+3=35..
========================================================
I wrote the below code as per my above information..
#!/bin/ksh
main() {
set -x
int a[10],b[10];
# array a and b are used to store the paths and partitions for the disks until we get the
proper BID Value
int x = 0,y = 0, BID = 0;
int count=0,t=0;
while [[ $BID -le 35 ]] ; do
x=`expr $x + 1`;
hwmgr show scsi > scsi.tmp
while read line; do
a[x]=`echo $line | grep "dsk" | awk '{ print $7 }'`
done < scsi.tmp
b[x]=4;
BID+=a[x].b[x];
echo $BID;
if [[ $BID -gt 35 ]]; then
y = $BID - 35;
echo $y;
BID-= a[x].b[x];
if [[ count -eq 0 ]]; then
for i in b[x] ; do
t = a[x].i;
if [[ $y -eq $t ]] ; then
b[x] -= i;
BID += a[x].b[x];
count = 0;
break;
fi
count=1;
done
else
if [[ $count -eq 1 ]] ; then
a[x]=0;
b[x]=0;
fi
fi
else
if [[ $BID -eq 35 ]] ; then
break;
fi
fi
done
return;
}

Could you check it and correct me if you have better idea..
Request you to please provide inputs for this...