From: Krucker, Louis (louis.krucker@xxxxxxxxxxx)
Date: Thu Jun 06 2002 - 06:08:19 GMT-3
Hello
Here is my solution taken from cco
IP (1024/1024)*0.5= 0.5 / 0.5 = 1 byte-count=1024
DLSW (1024/512)*0.25= 0.5 / 0.5 = 1 byte-count=512
IPX (1024/256)*0.25= 1 / 0.5 = 2 byte-count=512
queue-list 1 protocol ip 1 tcp 2065
queue-list 1 protocol ipx 2
queue-list 1 protocol ip 3
queue-list 1 queue 1 byte-count 512
queue-list 1 queue 2 byte-count 512
queue-list 1 queue 3 byte-count 1024
interface s0
custom-queue-list 1
you can found all information here.
http://www.cisco.com/warp/public/cc/pd/ibsw/ibdlsw/tech/dls5_rg.htm
cheers
Louis
-----Original Message-----
From: Alex
To: ccielab@groupstudy.com
Sent: 06.06.2002 01:39
Subject: Challenge bite count for Custom Queuing
Please help to come up with the bite count for custom queue to
accommodate the following scenario
Configure the frame-relay between R3 and R2 to allocate 50% of the
bandwidth to IP 25% to DLSW and 25% to IPX DLSW packet size is 512 bytes
IP is 1024 bytes and IPX is 256 bytes. Use a maximum queue size of 2000.
Regards,
Alex
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