AIX: how to do a dump that contains the application related infos too

You can only dynamically increase/decrease the memory for an LPAR within the partition profiles' configured minimum/maximum. ie: Minimum 1GB, Maximum 10GB.

Assuming you want to work within this range, and the system has available memory, increasing the allocation dynamically is simple. From the Server view, select the LPAR you want to reconfigure, and in the menu either next to the name or in the bottom pane, navigate to: Dynamic Logical Partitioning -> Memory -> Add or Remove.

From here edit the values for the profile, it will be pre-populated with the current values, and hit apply. Note, if the allocation or more specifically de-allocation is large, you may want to increase the timeout value, as this operation can take a long time. If your HMC doesn't have a timeout option, you're probably running on some older code. If it times out, just retry the operation, and eventually it will reach your value.

If you do not available free memory on the system, you can take memory from another lpar using the same process. You can use the Dynamic Logical Partitioning -> Memory -> Move option to take from one LPAR and place it on another.

You can confirm the changes realtime on the host using topas or nmon.

Is there any chage sort of command on AIX? check /etc/shadow file thats where the expiry information is stored.

Update: It seems there is a passwdexpired subroutine that can be loaded and Checks the user's password to determine if it has expired. However, it seems to be used as root.

http://publib.boulder.ibm.com/infocenter/aix/v6r1/index.jsp?topic=%2Fcom.ibm.aix.basetechref%2Fdoc%2Fbasetrf1%2Fpasswdexpired.htm

This link has excellent documentation of what you would require

http://www.torontoaix.com/scripting/when_pwd_exp

As demonstrated earlier in the above article, the expiry of a password is governed by the maxage attribute.

For example:
maxage=0 means never to expire
maxage=2 means will expire in two weeks.

AIX stores the time in the epoch format in seconds, so first you must determine how many seconds in a week, as this is how maxage measures the time between password expiry, that is in week numbers. There are 86400 seconds in a day, so multiplying that by seven comes in at 604800. So there are 604800 seconds in a week. The next command you need to look at is the pwdadm, which in turn queries the file /etc/security/passwd. This file holds the values in seconds when a user last changed their password. Interrogating the file or using the pwdadm command will return the same result. For this demonstration, let us query the user spoll:

# grep -p "spoll:" /etc/security/passwd
spoll:
        password = EvqNjMMwJzXnc
        lastupdate = 1274003127
        flags =       ADMCHG

# pwdadm -q spoll
spoll:
        lastupdate = 1274003127
        flags = ADMCHG

You can see the lastupdate value in seconds from the above output. In other words, the last time the password was changed: 1274003127

Next, using the lsuser or interrogating the file with /etc/security/user, you can determine the number of weeks before the user spoll password will expire:

# grep -p "spoll:" /etc/security/user
spoll:
        admin = false
        maxage = 4

# lsuser -a maxage spoll
spoll maxage=4

You can see from the above output that the number of weeks before password expiry is 4. The next task is then to multiply the number of seconds in a week by the number of weeks before the user spoll password is due to expire. In this case, it is 4: 604800 * 4

# expr 604800 \* 4
2419200

Next, you need to add the maxage value in seconds (604800 * 4) to the last time the password was changed: 2419200 + 1274003127

# expr 2419200 + 1274003127
1276422327

You can now convert that number of seconds from UNIX epoch into a more meaningful current time presentation. You can use different tools, but for this demonstration you'll use gawk with the strftime function:

# gawk 'BEGIN {print strftime("%c",'1276422327')}'
Sun Jun 13 10:45:27 BST 2010

The above calculation gives the time of the next password expiry. So, you now know that user spoll's password was last changed on ( from the pwdadm command):

# gawk 'BEGIN {print strftime("%c",'1274003127')}'
Sun May 16 10:45:27 BST 2010

And that it will expire on:

Sun Jun 13 10:45:27 BST 2010

------------------Perl script-let--------

#!/bin/perl
use POSIX qw(strftime);
$maxage=4; 
$last_update = 1274003127
$max_week_seconds = 86400 * $maxage;
print strftime("%C ", localtime($max_week_seconds));