Hello!
I have a WS-C2960RX-48LPS-L, 15.2(4)E1, RELEASE SOFTWARE (fc2).
sh ver | i memory
cisco WS-C2960X-48LPS-L (APM86XXX) processor (revision A0) with 524288K bytes of memory.
sh memory summary
Head Total(b) Used(b) Free(b) Lowest(b) Largest(b)
Processor 74EEA80 397565104 54963616 342601488 330226084 314572656
I/O 1E000000 16777216 12533192 4244024 4053660 4242920
Driver te 3B00000 1048576 44 1048532 1048532 1048532
I thought that sum of Processor, I/O and Driver text total bytes columns gives the exact number of total DRAM size installed in a box. A simple math shows that is not.
Processor (397565104)+I/O(16777216)+Driver text(1048576) = 415390896 bytes
524288KB=536870912 bytes
536870912−415390896=121480016
My question is: what purpose is 121480016 bytes of RAM for ?
I did find some old topics with similar questions, but I have not understand those explanations at 100 %. Please, shed some light.
Thanks!