From: Vijay S Jayaraman (vjayaram@in.ibm.com)
Date: Fri Feb 21 2003 - 08:20:22 GMT-3
yeah......Thats right...my mistake....
Regards,
Vijay.
soon ccie
<soonccie@yahoo.c To: Vijay S Jayaraman/India/IBM@IBMIN, CCIE FUN <ccieexam2002@yahoo.com>
om> cc: cannonr <cannonr@attbi.com>, ccielab@groupstudy.com, Emre Koyuncu
Sent by: <emrekoyuncu@hotmail.com>, nobody@groupstudy.com
nobody@groupstudy Subject: Re: Custom Queuing byte count calculation
.com
02/21/2003 02:39
PM
Please respond to
soon ccie
The doc says that byte is min quantity to be processed before moving to
tbe next queue; whereas queue len is how many packets allowed in the q
waiting to be processed. If there are more packets than the router can
process, i.e. more than 20 packets come in same time, they will be
dropped, but the router will NOT move to the next q before the min byte
count is reached.
Ex from doc:
(1) queue-list 9 queue 10 byte-count 1400: The byte count establishes
the MIN number of bytes the system allows to be delivered from a given
queue during a particular cycle.
(2) queue-list 3 queue 10 limit 40: The queue length limit is the MAX
number of packets that can be enqueued at any time, with the range
being from 0 to 32767 queue entries.
HTH
--- Vijay S Jayaraman <vjayaram@in.ibm.com> wrote:
> One thing that would be importent to note is the maximum queue size
> is just
> 20 packets (Hope the number is right)...
> so the byte count should fall within this queue size of 20
> packets.......
>
> If the queue limit of 20 packets is crossed, the next queue would be
> serviced so the large byte count would have no meaning in this
> case....
>
> of course the limit of 20 packets is pretty large in terms of byte
> counts
> considering average packet sizes.....
>
> Regards,
> Vijay.
>
>
>
>
>
>
> CCIE FUN
>
> <ccieexam2002@yah To: Emre Koyuncu
> <emrekoyuncu@hotmail.com>, cannonr <cannonr@attbi.com>,
> oo.com>
> ccielab@groupstudy.com
>
> Sent by: cc:
>
> nobody@groupstudy Subject: Re: Custom
> Queuing byte count calculation
> .com
>
>
>
>
>
> 02/20/2003 10:14
>
> AM
>
> Please respond to
>
> CCIE FUN
>
>
>
>
>
>
>
>
> Emre
> Thanks for the reply. This is exactly what i was
> looking for.
> So as long as i follow the rationale of percentage.
> my Total byte count for all the queues can be
> anything.
>
> I guess can it be 20000
>
> then
> www gets 50% or 10000
> FTP/Telnet get 15 % or 3000
> Fast0/0 gets 15% or 3000
> Remaining traffic get 20 % 4000
>
> Am i doing it correctly ???
> thanks
>
>
> --- Emre Koyuncu <emrekoyuncu@hotmail.com> wrote:
> > Hi ,
> >
> > Actually by using these calculations , you can get
> > the solution for packet
> > count not the byte count. The correct solution must
> > be the following way :
> > www traffic gets 50% =>> the value can be 5000 (it
> > can be anything as long
> > as it is rational with the other values)
> > FTP /Telnet get 15% =>> the value can be 1500
> > Traffic from e0/0 gets 15% =>> the value can be
> > 1500.
> > I assume the rest ,20% is going to default queue.
> > =>> the value can be 2000
> > By this way after transmitting 5000 bytes of www
> > traffic, you will begin
> > transmitting FTP/Telnet traffic for 1500 bytes after
> > that you will transmit
> > traffic from e0/0 for 1500 bytes after that the rest
> > of the traffic will be
> > transmitted for 2000 bytes the cycle goes on. In one
> > cycle which is 10000
> > bytes , www traffic transmits 5000 bytes which is
> > 50% of the total traffic,
> > FTP/telnet transmits 1500 bytes which is 15% of
> > total traffic, e0/0
> > transmits 1500 bytes , the rest will transmit 2000
> > bytes.
> > This way is easy for calculation. And you use the
> > queue-list byte-count
> > command.
> > But by using cannonr's calculation you have to
> > queue-list limit command. And
> > this will be the calculation with the packet count.
> >
> > Emre Koyuncu CCIE #10916
> >
> > ----- Original Message -----
> > From: "cannonr" <cannonr@attbi.com>
> > To: "CCIE FUN" <ccieexam2002@yahoo.com>;
> > <ccielab@groupstudy.com>
> > Sent: Tuesday, February 18, 2003 9:00 PM
> > Subject: Re: Custom Queuing byte count calculation
> >
> >
> > > You question would usually include a packet size
> > for each protocol. Here
> > is
> > > an example if you have a packet size for each.
> > >
> > > DLSW gets 25% and has a 512KB packet size
> > > Telnet gets 25% and has a 1500 KB packet size
> > > Citrix gets 50% and has a 1500 KB packet size
> > >
> > > To figure out byte count use the following
> > formula.
> > >
> > > Divide the percentage for each protocol by packet
> > size.... You get the
> > > following.
> > >
> > > DLSW 25/512 = .048848
> > > Telnet 25/1500 = .01666
> > > Citrix 50/1500 = .03333
> > >
> > > Divide the output of each by the smallest number.
> > >
> > > DLSW .048848/.01666 = 2.93
> > > Telnet .01666/.01666 = 1
> > > Citrix .03333/.01666 = 2.0006
> > >
> > > Round this number and multiply by byte count.
> > >
> > >
> > > DLSW 3X512 = 1536
> > > Telnet 1X1500 = 1500
> > > Citrix 2X1500 = 3000
> > >
> > >
> > > Now to verify that you are close to the proper
> > percentages by adding all
> > > byte counts and dividing by each one.
> > >
> > > 1536+1500+3000=6036
> > >
> > > DLSW 1536/6036=.25473
> > > Telnet 1500/6036=.2485
> > > Citrix 3000/6036=.4970
> > >
> > > If you do not round up in the earlier step, you
> > this formula will come a
> > > little closer to the exact numbers, but using DLSW
> > as an example, if you
> > > were to multiply 2.93 X 512, you actually get
> > 1500.16 as a byte counte.
> > If
> > > you use 1500 as your number, you will still send 3
> > packets before your
> > turn
> > > is up which equals 1536..... That's as close as
> > you can get!
> > >
> > >
> > > HTH
> > >
> > >
> > >
> > >
> > > ----- Original Message -----
> > > From: "CCIE FUN" <ccieexam2002@yahoo.com>
> > > To: <ccielab@groupstudy.com>
> > > Sent: Tuesday, February 18, 2003 9:37 PM
> > > Subject: Custom Queuing byte count calculation
> > >
> > >
> > > > Hello Group
> > > > Can anybody explain the best way to calculate
> > the
> > > > byte-count. i am working on a example
> > > > "it says that configure custome queueing on
> > frame
> > > > cloud so if congestion occurs then:
> > > >
> > > > www traffic gets 50% traffic
> > > > FTP /Telnet get 15%
> > > > Traffic from e0/0 gets 15%
> > > > rest of the traffic shares the remaining
> > bandwidth.
> > > >
> > > > Now how do i determine the byte-count in this
> > case.
> > > > I have worked on examples which provide the
> > > > predetermined byte-counts, the DOC CD has a nice
> > > > example on that.
> > > > But have always stumbled upon the kind of
> > examples i
> > > > explained above.
> > > >
> > > > any help would be appreciated.
> > > >
> > > > thanks
> > > >
> > > >
> > __________________________________________________
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>
>
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