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Catalyst 2960X DRAM

sharlino
Level 1
Level 1

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

536870912415390896=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!

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