From: Igor M. (imanassypov@rogers.com)
Date: Mon Aug 25 2008 - 15:09:28 ART
Hey Lloyd,
thanks for reply
-why are you using the AR of 1536000?
serial lines are 1544000:
R4#sh int s1/0
Serial1/0 is up, line protocol is up
Hardware is M4T
Internet address is 154.1.0.4/24
MTU 1500 bytes, BW 1544 Kbit, DLY 20000 usec,
----------------------
I.M., M.Eng. P.Eng.
Network Architect
CI Investments
----------------------
--- On Mon, 8/25/08, Lloyd Ardoin <Lloyd@TheWizKid.biz> wrote:
From: Lloyd Ardoin <Lloyd@TheWizKid.biz>
Subject: RE: IE Lab16, QoS descrepancy
To: "Igor M." <imanassypov@rogers.com>
Received: Monday, August 25, 2008, 1:54 PM
Hi Igor,
I did this lab a while back but here is what I come up with.
bc = (AR - CIR) * tc /1000
bc= (1536000 - 512000) * 10/ 1000
bc= (1024000) * 10/1000
bc= 10240000/1000 = 10240
Lloyd V Ardoin
Network Engineer
Sagenet, LLC
918-270-7133
lardoin@sagenet.com
MCSE, CCDA, CCNP, CCSP, GSEC, GCFW, GCWN, CISSP
From: Igor M.
Sent: Mon 8/25/2008 12:41 PM
To: groupstudy
Subject: IE Lab16, QoS descrepancy
Hi,
If anyone has gone through lab 16 of the internetwork expert, I am having
troubles understanding their solutions for QoS question 7.1. To remind - R3,
R4, R5 where R3 is a hub, R3-R5 is provisioned at 512Kbps while R3-R4 is
provisioned at 1024Kbps. Both spokes (R4, & R5) are allowed to burst up to AR
if they have credit.
Now, if I calculate the pertaining values for lets say R3-R5 circuit, I come
up with
cir = 512000
bc = 5120 (to allow 10ms for Tc)
and be = Tc x (AR - CIR) = 10 x ( 1544 - 512 ) = 10320.
In their solution the same circuit R3-R5 be = 10240.
Where am I wrong in my calculations?
----------------------
I.M., M.Eng. P.Eng.
Network Architect
CI Investments
----------------------
Blogs and organic groups at http://www.ccie.net
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