From: Motohiro Yamanaka (yamanaka@fsas.fujitsu.com)
Date: Tue Oct 29 2002 - 00:01:32 GMT-3
Hi,ALL
In this case, may I not think about DEFAULT?
Which is better?
1)
queue-list 2 protocol dlsw 1
queue-list 2 protocol ipx 2
queue-list 2 default 3
queue-list 2 queue 1 byte-count 1024
queue-list 2 queue 2 byte-count 512
queue-list 2 queue 3 byte-count 512
2)
queue-list 2 protocol dlsw 1
queue-list 2 protocol ipx 2
queue-list 2 protocol ip 3
queue-list 2 queue 1 byte-count 1024
queue-list 2 queue 2 byte-count 512
queue-list 2 queue 3 byte-count 512
thanks,
Motohiro
On Tue, 29 Oct 2002 01:41:55 +0000
"Sara Li" <saralilin@hotmail.com> wrote:
> 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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