From: Jason Cash (cash2001@swbell.net)
Date: Mon Jun 02 2003 - 10:35:26 GMT-3
One more then and I think this horse will be dead.
Based on that information, that being with the changes to 12.1 how should
queues be determined? Even in that link, the example is given with the
antiquated byte count determination.
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
Nguyen Hoang Long
Sent: Monday, June 02, 2003 2:24 AM
To: Hossam Mahmoud; Jason Cash; ccielab@groupstudy.com
Hossam,
check this:
http://127.0.0.1:8080/cc/td/doc/product/software/ios122/122cgcr/fqos_c/fqcpr
t
2/qcfconmg.htm#xtocid47
<quote>
CQ was modified in Cisco IOS Release 12.1. When the queue is depleted early,
or the last packet from the queue does not exactly match the configured byte
count, the amount of deficit is remembered and accounted for the next time
the
queue is serviced. Beginning with Cisco IOS Release 12.1, you need not be as
accurate in specifying byte counts as you did when using earlier Cisco IOS
releases that did not take deficit into account.
<quote>
----- Original Message -----
From: Hossam Mahmoud
To: Jason Cash ; 'Nguyen Hoang Long' ; ccielab@groupstudy.com
Sent: Monday, June 02, 2003 1:34 PM
Subject: RE: Custom Queue Confusion
I agree with you but do u know exactly which IOS version supports the new
borrowing technique.
As far as i know the LAB is still using 12.1. IOS ver 12.1 documentation
is
using the same calulation method used by Jason in his initial mail!!
So i think this is the calculation method expected at LABS.....is't it??
Thanks for advise
SAM
Jason Cash <cash2001@swbell.net> wrote:
That actually is the correct answer, but how did you come to get:
30:30:25:15 = 6:6:5:3 ?
How did you know to divide by 5 to get 6:6:5:3? Or did you use some
other
calculation?
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
Nguyen Hoang Long
Sent: Sunday, June 01, 2003 2:54 AM
To: Jason Cash; ccielab@groupstudy.com
Hi Jason,
I would do by this way for exact ratio:
IP:SNA:IPX:telnet = 30:30:25:15 = 6:6:5:3
therefore:
IP = 1500x6 = 9000 (30%)
SNA = 1500x6 = 9000 (30%)
IPX = 1500x5 = 7500 (25%)
Telnet = 1500x3 = 4500 (15%)
Since we're using new IOS, which permits the router to borrow from the
next
queue for a current queue, your calculation method is not necessary. I
prefer a more simple method.
Long
----- Original Message -----
From: "Jason Cash"
To:
Sent: Sunday, June 01, 2003 12:28 PM
Subject: Custom Queue Confusion
> I am getting really confused on the custom queuing! I have looked over
the
> previously provided links, but those examples don't address the
problem
I
am
> having. Whereas I am instructed:
>
>
>
> IP traffic 30% bw
>
> SNA traffic 30% bw
>
> IPX traffic 25% bw
>
> Telnet traffic 15% bw
>
>
>
> Here is what I do (average pkt 1500)
>
> 1) For each queue, divide the percentage of bandwidth you want to
allocate
> to the queue by the packet size, in bytes.
>
> 30/1500 = .02
>
> 30/1500 = .02
>
> 25/1500 = .0166
>
> 15/1500 = .01
>
>
>
> 2) Normalize the numbers by dividing by the lowest number:
>
> .02/.01 = 2
>
> .02/.01 = 2
>
> .0166/.01 = 1.66
>
> .01/.01 = 1
>
>
>
> 3) A fraction in any of the ratio values means that an additional
packet
> will be sent. Round up the numbers to the next whole number to obtain
the
> actual packet count.
>
> IP = 2
>
> SNA = 2
>
> IPX = 2
>
> Telnet = 1
>
>
>
> 4) Convert the packet number ratio into byte counts by multiplying
each
> packet count by the corresponding packet size.
>
> IP = 1500x2 = 3000
>
> SNA = 1500x2 = 3000
>
> IPX = 1500x2 = 3000
>
> Telnet = 1500x1 = 1500
>
>
>
> Now here is one place I get lost. IPX byte count is the same as IP and
SNA
> and I know that can't be right, but what do I do? 2 is the only
multiple
to
> go by.
>
>
>
> 5) To determine the bandwidth distribution this ratio represents,
first
> determine the total number of bytes sent after all three queues are
> serviced:
>
> Based on my info above I get: 3000+3000+3000+1500= 10500
>
>
>
> 6) Then determine the percentage of the total number of bytes sent
from
each
> queue:
>
> IP = 3000/10500 = .28 (28%)
>
> SNA = 3000/10500 = .28 (28%)
>
> IPX = 3000/10500 = .28 (28%)
>
> Telnet = 1500/10500= .14 (14%)
>
>
>
> This adds up to 98%. So I will wait for some advice before I proceed.
> Basically, when I a queue is rounded up and is equal to a greater
queue,
how
> is that queues byte count determined?
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