From: Sara Li (saralilin@hotmail.com)
Date: Mon Oct 28 2002 - 22:41:55 GMT-3
i saw post earlier from this web site, here is the formula:
hope it helps:
> 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 - 2
>
> 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 -- 2* 256 = 512
So in the end we will come up with1 packet x 1024 = 1024
1 packet x 512 = 512
2 packets x 256 = 512
Total size = 2000 (adding all the 3 up >From: "Hamele Kassa" >Reply-To:
"Hamele Kassa" >To: "ccielab" >Subject: queuing >Date: Mon, 28 Oct 2002
19:41:00 -0500 > >group, > >If the question states as follows which
queueing mech I should use. > >bandwidth on interface 2048 and each
protocol should use exact bandwidth > >prot 1- 40% with packet size 512,
prot 2- 20% with packet size 1024 and prot 3 >- 40% with packet size 256
> >thanks, >hk
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