From: John Conzone (jkconzone@xxxxxxxx)
Date: Mon Jul 10 2000 - 21:01:20 GMT-3
I need some help with understanding 2500 memory. Here's a show ver
from a 2500:
cisco 2500 (68030) processor (revision A) with 16384K/2048K bytes
of memory.
Processor board ID 01101479, with hardware revision 00000000
Bridging software.
SuperLAT software copyright 1990 by Meridian Technology Corp).
X.25 software, Version 2.0, NET2, BFE and GOSIP compliant.
TN3270 Emulation software.
Basic Rate ISDN software, Version 1.0.
Cisco-ET Extended Temperature platform.
1 Ethernet/IEEE 802.3 interface(s)
2 Serial network interface(s)
1 ISDN Basic Rate interface(s)
32K bytes of non-volatile configuration memory.
16384K bytes of processor board System flash (Read ONLY)
Okay, this means I have 16 M of flash and 2M of DRAM, correct?
Total of 18M. Is the 16 M of flash reflected on the last line? Is this
memory on simms that can be changed? Okay, here's another Sho Ver from
another 2500(actually a CPA 2513):
cisco 2500 (68030) processor (revision D) with 2048K/2048K bytes of
memory.
Processor board ID 02386057, with hardware revision 00000000
Bridging software.
SuperLAT software copyright 1990 by Meridian Technology Corp).
X.25 software, Version 2.0, NET2, BFE and GOSIP compliant.
TN3270 Emulation software (copyright 1994 by TGV Inc).
1 Ethernet/IEEE 802.3 interface.
1 Token Ring/IEEE 802.5 interface.
2 Serial network interfaces.
32K bytes of non-volatile configuration memory.
16384K bytes of processor board System flash (Read ONLY)
This one says 2048/2048, which means a total of 4M, right? Which is
Flash and which is DRAM? Why do I still have the "16384K bytes of
processor board System flash (Read ONLY)" on the last line. I thought
that was my flash? I'm buying some 2500's and want to make sure I get
the right memory upgrades and know how to install them.
Thanks!
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