From: Ram Shummoogum (rshummoo@ca.ibm.com)
Date: Mon May 19 2003 - 09:09:07 GMT-3
If I assume that the packet size is 1500 Bytes then both you and the
answer is wrong.
The answer is wrong because queue 2 and queue 3 will be able to send a max
of 3 and 2 packets respectively.
dlsw 50%
ip 25%
ipx 15%
default 10%
Here is how it is calculated:
1.Divide the % by the packet size
2.rationalize the above
3. Multiply the results from step 2 to make them a whole number.
4.Multiply step 3 by the packet size.
Step1: 50/1500, 25/1500, 15/1500, 10/1500.
Step2: Divide everything by the smallest number which is (10/1500).
You get 5, 2.5, 1.5,1
Step3: Multiply by 2
You get 10, 5, 3, 2
Step 4: Multiply by 1500
You get 1500, 7500, 4500, 3000.
Verify your answer:
15000+7500+4500+3000= 30000
15000/30000=50%
7500/30000=25%
4500/30000=15%
3000/30000=10%
HTH
RAM
Tom Young <gitsyoung@yahoo.co.jp>@groupstudy.com on 05/19/2003 05:05:47 AM
Please respond to Tom Young <gitsyoung@yahoo.co.jp>
Sent by: nobody@groupstudy.com
To: ccielab@groupstudy.com
cc:
Subject: the queue-list
Hi, group
I really confused, I had asked this question for somebody
but hadn't got answer, it is that want to pay the half
bandwidth of serial port for dlsw, and 25% for ip , 15%
for IPX and 10% for the default. That is the answer as
below.
queue-list 1 protocol dlsw 0
queue-list 1 protocol ip 1
queue-list 1 protocol ipx 2
queue-list 1 default 3
queue-list 1 queue 0 byte-count 7500
queue-list 1 queue 1 byte-count 3750
queue-list 1 queue 2 byte-count 2250
But I think the bandwidth of serial port is 1.5M , that
1500000bps, is 150000Bps, so, 50% of serial port should be
75000 not the 7500. So I think the answer should be 75000,
37500 , 22500, right?
Hope someone could give me a correct answer, thanks alot .
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