From: Jaspreet Bhatia (jasbhati@xxxxxxxxx)
Date: Wed Jun 05 2002 - 21:32:27 GMT-3
Alex ,
There is a very simple formula for this kind of scenario .
let us write down the queues that you need to create first
1) Queue 1 - IP - packet size 1024 50 %
2) Queue 2 - DLSW - packet size 512 25 %
3) Queue 3 - IPX - packet size 256 25 %
STEP 1
Divide the percentile with the packet size
queue 1 - 50/1024 = 0.0488281
queue 2 - 25/512 = 0.0488281
queue 3 - 25/256 = 0.097656
STEP 2
now take the smallest value which is 0.0488281 and divide that by each value
queue 1 -- 1
queue 2 - 1
queue 3 - 0.5
STEP 3
multiply these values by the packet size to get the byte count for teh queue
queue1 -- 1*1024 = 1024
queue 2 -- 1* 512 = 512
queue 3 - 0.5* 256 = 128
Let me know if this works for you . This is the formula I use . Thanks
Jaspreet
At 06:39 PM 6/5/2002 -0500, Alex wrote:
>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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