Carl, There's a public report for a similar issue in now: http://now.netapp.com/NOW/cgi-bin/bol?Type=Detail http://now.netapp.com/NOW/cgi-bin/bol?Type=Detail&Display=80268 &Display=80268

The equation to combine these is: if (Low >= 0) x = High * 2^32 + Low if (Low < 0) x = (High + 1) * 2^32 + Low

In this instance: High = 2 Low = -1834563852 since Low < 0, use the second equation: x = (2+1) * 2^32 + (-1834563852) = 12884901888 - 1834563852 = 11050338036

Good luck, -Joshua

-----Original Message----- From: Carl Howell [mailto:chowell@uwf.edu] Sent: Thursday, January 25, 2007 1:20 PM To: toasters@mathworks.com Subject: SNMP Question

What would be the correct way to calculate the total size of an aggregate with the following SNMP information:

dfHighTotalKBytes = 2

dfLowTotalKBytes = -1834563852

I thought it would be 2*4294967295 + 1834563852 = 10,424,498,442

But this comes out 625,839,594 short of the df -rk output:

Aggregate total = 11,050,338,036

The MIB is at version 1.17.3.

Thanks,

--Carl