Specifications

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Cisco AS5800 Operations, Administration, Maintenance, and Provisioning Guide
DOC-7810814=
Chapter 2 Commissioning
Task 1. Verifying Basic Setup
Table 2-3 describes the significant fields in the previous display:
Step 2 Convert bytes to megabytes (MB):
• Total processor memory = 9,4055,200 bytes = 89.7 MB
• Used processor memory = 42,346,480 bytes = 40.4 MB
• Free processor memory = 51,708,720 bytes = 49.3 MB
Total memory (89.7 MB) = Used memory (40.4 MB) + free memory (49.3 MB)
Step 3 Do some useful memory calculations:
Total Processor = Total RAM - Cisco IOS software (use the show version command to get the MB
assigned for all of Cisco IOS software + Processor)
cisco 7206 (NPE400) processor with 114688K/16384K bytes of memory.
114688 KB / (1024 KB / MB) = 112.0 MB
16384 KB = 16 MB
112 MB + 16 MB = 128 MB (what you purchased).
Note 112.0 MB - 89.7 MB = 22.3 MB. This means that 22.3 MB are not available for
processor memory.
Table 2-3 Show Memory Summary Output Field Descriptions
Field Description
Processor Processor memory. The Cisco IOS software image is initially read out of Flash
memory, decompressed, and placed in main memory. Routing tables and call control
blocks are also stored in main memory.
I/O Packets are temporarily stored in I/O memory.
Head Hexadecimal address of the head of the memory allocation chain.
Total(b) Summary of used bytes plus free bytes.
Used(b) Total number of bytes currently used for routing tables and call-processing
components.
Free(b) Total number of free bytes. The free memory size should be close to the largest block
available.
Lowest(b) Smallest amount of free memory since last boot.
Largest(b) Size of largest available free block. Whenever the largest available block is equal to
the free block, there is no fragmentation.