From: Nick Shah (nshah@xxxxxxxxxxxxxx)
Date: Wed Jun 05 2002 - 22:51:08 GMT-3
Jaspreet,
Correct me if I am wrong, but in step 2 of your logic, wont the Queue 3 be
2...
since we are dividing 0.097/0.0488 = 2
So in the end we will come up with
1 packet x 1024 = 1024
1 packet x 512 = 512
2 packets x 256 = 512
Total size = 2000 (adding all the 3 up)
rgds
Nick
----- Original Message -----
From: "Jaspreet Bhatia" <jasbhati@cisco.com>
To: "Alex" <afayn@yahoo.com>
Cc: <ccielab@groupstudy.com>
Sent: Thursday, June 06, 2002 10:32 AM
Subject: Re: Challenge bite count for Custom Queuing
> 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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