Specifications

3-22
Cisco AS5x00 Case Study for Basic IP Modem Services
11/24/1999
Section 3 Commissioning the Cisco AS5800 Hardware
Verifying Basic Setup
The following exercise explains how to inspect and calculate memory usage:
Step 1
Display the memory status report. Note that the largest-memory block is close to the free-memory
block, which is good. There is no fragmentation.
5800-NAS#show memory summmary
Head Total(b) Used(b) Free(b) Lowest(b) Largest(b)
Processor 6164D4E0 94055200 42346480 51708720 50435436 51592056
I/O 7000000 16777216 6433400 10343816 10343816 10343772
PCI 4B000000 4194304 618584 3575720 3575720 3575676
Caution
Do not enter the
show memory summary
command with the
terminal length 0
command
enabled. Otherwise, you will produce many screens of output which might interrupt your
session.
Table 3-3 describes the significant fields in the previous display:
Step 2
Convert bytes to megabytes:

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)
Tip: MB = bytes / (1024 X 1024)
Table 3-3 Show Memory Summary Output Field Descriptions
Field Description
Processor
Processor memory. The Cisco IOS image is initially read out of Flash memory,
decompressed, and placed into 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.
In the example, there is a small amount of fragmentation.