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@xxxxxxxxxx> 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@xxxxxxxxxxxxxx [mailto:nobody@xxxxxxxxxxxxxx] On Behalf Of
Nguyen Hoang Long
Sent: Sunday, June 01, 2003 2:54 AM
To: Jason Cash; ccielab@xxxxxxxxxxxxxx
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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