From: Chris Home (clarson52@comcast.net)
Date: Thu Apr 17 2003 - 22:46:09 GMT-3
So is it correct that if the credits be uses are made from unused bc in the
previous tc interval, that if there is more data then bc in the current
interval that those credits from the previous interval can be used as be to
transmit the excess (in the current tc interval)? And that this happens
every interval, not really be in the first interval of the first second?
This would make sense to me as I have never understood the be in first
interval or when that first proverbial 1 second period started. (With the
first packet? Some static clocking mechanism?) So it sounds as if it is a
process that happens every tc interval. Correct?
----- Original Message -----
From: "Brian Dennis" <brian@labforge.com>
To: "'OhioHondo'" <ohiohondo@columbus.rr.com>; <ANDF@nnpi.com>;
<ccielab@groupstudy.com>
Sent: Thursday, April 17, 2003 1:06 PM
Subject: RE: FRTS Revealed (was RE: MINCIR = CIR?)
> It is from the previous time interval (Tc) not the previous second. That
> is why that Tech Tip from CCO about Frame-relay traffic shaping creates
> so much confusion.
>
> Brian Dennis, CCIE #2210 (R&S/ISP-Dial/Security)
> Director of CCIE Training and Development - IPexpert, Inc.
> Mailto: brian@ipexpert.net
> Toll Free: 866.225.8064
> Outside U.S. & Canada: 312.321.6924
>
> -----Original Message-----
> From: OhioHondo [mailto:ohiohondo@columbus.rr.com]
> Sent: Thursday, April 17, 2003 9:28 AM
> To: Brian Dennis; ANDF@nnpi.com; ccielab@groupstudy.com
> Subject: RE: FRTS Revealed (was RE: MINCIR = CIR?)
>
> Brian
>
> Just for a clarification -- are the credits that Be uses in the 1st time
> interval built up over the 1 second period (using all of the unused Bc
> bits
> in a 1 second period) OR over a single Tc period (Using the unused Bc
> bits
> of let's say a 125 millisecond period).
>
> If the first is true, then Be could be a much larger value than Bc.
>
> -----Original Message-----
> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com]On Behalf Of
> Brian Dennis
> Sent: Thursday, April 17, 2003 11:09 AM
> To: ANDF@nnpi.com; ccielab@groupstudy.com
> Subject: RE: FRTS Revealed (was RE: MINCIR = CIR?)
>
>
> Since Be is made up of unused Bc bytes from the previous interval (Tc)
> Be can not contain more bytes than Bc. So if Bc is 4000 how can you ever
> have a Be value more than 4000?
>
> Brian Dennis, CCIE #2210 (R&S/ISP-Dial/Security)
> Director of CCIE Training and Development - IPexpert, Inc.
> Mailto: brian@ipexpert.net
> Toll Free: 866.225.8064
> Outside U.S. & Canada: 312.321.6924
>
> -----Original Message-----
> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
> ANDF@nnpi.com
> Sent: Thursday, April 17, 2003 6:46 AM
> To: brian@cyscoexpert.com; ccielab@groupstudy.com
> Subject: RE: FRTS Revealed (was RE: MINCIR = CIR?)
>
> Brian,
>
> I appreciate your input on this. On page 66 of the 2002 CCIE Power
> Session
> slides they show Be = 32000 when CIR = 32000, port speed = 64000, Bc =
> 4000
> and Tc = 125ms. Are they incorrect? Shouldn't Be have been 4000?
>
> Thanks,
>
> Andy Fernandes
>
> -----Original Message-----
> From: Brian McGahan [mailto:brian@cyscoexpert.com]
> Sent: Thursday, April 17, 2003 12:40 AM
> To: 'OhioHondo'; 'Jonathan V Hays'; ccielab@groupstudy.com
> Cc: 'Mike Williams'
> Subject: FRTS Revealed (was RE: MINCIR = CIR?)
>
>
> Due to incorrect documentation, FRTS is one of the most
> misunderstood topics within the CCIE domain. Since this is my favorite
> lecture which I usually end up giving 5 or 6 times per week, I'll just
> go
> for the complete explanation to start with :)
>
> To calculate FRTS values, we will use the following formulas:
>
> Bc = (CIR * Tc) / 1000
> Be = ((AR - CIR) * Tc)/1000
>
>
> In the CCO example you are referencing, the hub has an
> access-rate
> of 192000bps, while the remote only has an access-rate of 64000bps.
> Therefore, the fastest speed that can be sustained over the second is
> 64000bps. This is what we want to average per second, which is Cisco's
> term
> CIR.
>
> Hub:
> AR = 192000
> CIR = 64000
>
> Remote:
> AR = 64000
> CIR = 64000
>
> Since CIR <= 640Kbps, the default Tc is 125ms. (I'll explain
> this
> later) Now we have the following:
>
> Hub:
> AR = 192000
> CIR = 64000
> Tc = 125
>
> Remote:
> AR = 64000
> CIR = 64000
> Tc = 125
>
> Now that we have both CIR and Tc, we can solve for Bc. Since
> both
> the CIR and Tc are the same for the Hub and the Remote, Bc will be the
> same:
>
> Bc = (CIR * Tc)/1000
> Bc = (64000 * 125)/1000
> Bc = (8000000)/1000
> Bc = 8000 bits
>
> Now we have:
>
> Hub:
> AR = 192000
> CIR = 64000
> Tc = 125
> Bc = 8000
>
> Remote:
> AR = 64000
> CIR = 64000
> Tc = 125ms
> Bc = 8000
>
> Now we must solve for Be, starting with Hub:
>
> Be = ((AR - CIR) * Tc)/1000
> Be = ((192000 - 64000) * 125) / 1000
> Be = ((128000) * 125) / 1000
> Be = (16000000) / 1000
> Be = 16000
>
> Now the Remote:
>
> Be = ((AR - CIR) * Tc)/1000
> Be = ((64000 - 64000) * 125) / 1000
> Be = ((0) * 125) / 1000
> Be = (0) / 1000
> Be = 0
>
>
> So far, our calculations have nothing to do with how our line is
> actually provisioned. Instead, we have simply calculated how our token
> bucket will shape the traffic. The example then states "In case of
> congestion, it can drop down to 32Kbps at the minimum." This statement
> implies that the provider has provisioned 32000bps for us on this VC. In
> order to dynamically react to congestion from the cloud, we must enable
> BECN
> adapt, which is where the following two statements come
> from:
>
> frame-relay mincir 32000
> frame-relay adaptive-shaping becn
>
>
>
> The following statement in the document is wrong:
>
> "Ideally for data PVCs Bc = CIR/8 so that Tc = 125msec."
>
> There is a maximum Tc value per CIR. The greater the CIR the
> smaller the Tc value must be. Unfortunately, this behavior is not
> documented. In the following example, the frame-relay map-class X is
> applied to an interface with one VC. We can test this behavior based on
> the
> following:
>
> Suppose we have a CIR of 640000. If Bc = CIR/8, then Bc =
> 640000/8
> = 80000.
>
> map-class frame-relay X
> frame-relay cir 640000
> frame-relay bc 80000
> no frame-relay adaptive-shaping
> end
>
> Rack2R6#show traffic-shape
>
> Interface Se0/0
> Access Target Byte Sustain Excess Interval Increment
> Adapt
> VC List Rate Limit bits/int bits/int (ms) (bytes)
> Active
> 604 640000 10000 80000 0 125 10000 -
>
>
>
> This indicates that we have 8 intervals per second (1000ms per
> second, 125ms per interval), and are sending 80000 bits per interval,
> which
> works out to an average of 640000bps. So far so good.
>
> Now let's change our CIR to 800000bps:
>
> Current configuration:
> !
> map-class frame-relay X
> frame-relay cir 800000
> frame-relay bc 100000
> no frame-relay adaptive-shaping
> end
>
> Rack2R6#sh traffic
>
> Interface Se0/0
> Access Target Byte Sustain Excess Interval Increment
> Adapt
> VC List Rate Limit bits/int bits/int (ms) (bytes)
> Active
> 604 800000 6300 100000 0 63 6300 -
>
>
> Notice that our interval is not 125ms now, but our Bc still
> conforms
> to the formula Bc = CIR/8. In actuality, we have (bear with
> me) 1000/63 = 15.873015873015873015873015873016 intervals per second. At
> 100000 bits per interval, we are averaging about 1587301bps, which is
> clearly beyond 800Kbps. This is due to the fact that for a CIR of
> 800Kbps,
> the maximum Tc value is 63ms.
>
> Instead, let us compute Bc using 63 as the Tc:
>
> Bc = (CIR * Tc) / 1000
> Bc = (800000 * 63) / 1000
> Bc = 50400
>
> Rack2R6#sh run map-class
> Building configuration...
>
> Current configuration:
> !
> map-class frame-relay X
> frame-relay cir 800000
> frame-relay bc 50400
> no frame-relay adaptive-shaping
> end
>
> Rack2R6#sh traffic-shape
>
> Interface Se0/0
> Access Target Byte Sustain Excess Interval Increment
> Adapt
> VC List Rate Limit bits/int bits/int (ms) (bytes)
> Active
> 604 800000 6300 50400 0 63 6300 -
>
>
> We still have 1000/63 = 15.873015873015873015873015873016
> intervals
> per second. If we are sending 50400bits every
> 15.873015873015873015873015873016ms, we will be averaging 800000bps.
>
> I'm not sure exactly where the cutoff values are for the max Tc,
> but
> the router will tell you the appropriate Tc value for a CIR if you enter
> it
> in the map-class, apply it, then show traffic-shape.
>
>
> Hope I didn't confuse you all further :)
>
>
> Brian McGahan, CCIE #8593
> Director of Design and Implementation
> brian@cyscoexpert.com
>
> CyscoExpert Corporation
> Internetwork Consulting & Training
> Toll Free: 866-CyscoXP
> Outside US: 847.674.3392
> Fax: 847.674.2625
>
>
> > -----Original Message-----
> > From: OhioHondo [mailto:ohiohondo@columbus.rr.com]
> > Sent: Wednesday, April 16, 2003 10:08 PM
> > To: Jonathan V Hays; 'Brian McGahan'; ccielab@groupstudy.com
> > Cc: 'Mike Williams'
> > Subject: RE: MINCIR = CIR?
> >
> > This new equation/explanation for calculating FRTS parameter -- I've
> never
> > seen it before. It doesn't make sense with anything I've read on the
> > subject. How does it relate to the CCO iformation found at:
> >
> >
> http://www.cisco.com/en/US/tech/tk713/tk237/technologies_configuration_e
> xa
> > mp
> > le09186a00800942f8.shtml
> >
> > where:
> > Bc=CIR*Tc
> > Be= (AR*Tc) - Bc
> >
> > Where is the source of your information?
> >
> >
> > -----Original Message-----
> > From: nobody@groupstudy.com [mailto:nobody@groupstudy.com]On Behalf Of
>
> > Jonathan V Hays
> > Sent: Wednesday, April 16, 2003 9:00 PM
> > To: 'Brian McGahan'; ccielab@groupstudy.com
> > Cc: 'Mike Williams'
> > Subject: RE: MINCIR = CIR?
> >
> >
> > > -----Original Message-----
> > > From: Brian McGahan [mailto:brian@cyscoexpert.com]
> > > Sent: Wednesday, April 16, 2003 1:33 PM
> > > To: 'Jonathan V Hays'; ccielab@groupstudy.com
> > > Cc: 'Mike Williams'
> > > Subject: RE: MINCIR = CIR?
> > >
> > [snip]
> > > The following formulas hold true for Cisco's implementation of
> > > Frame-Relay Traffic Shaping:
> > >
> > > Bc = (CIR * Tc)/1000
> > > Be = ((AR - CIR) * Tc)/1000
> > >
> >
> > Brian,
> >
> > What units are you using to require division by 1000? Most Cisco
> > documents seem to give Bc= CIR * Tc.
> >
> >
> http://www.cisco.com/en/US/tech/tk713/tk237/technologies_configuration_e
> > xample09186a00800942f8.shtml
> >
> > See the document I cited in my earlier post (included above) under
> > "Nonconfigurable parameters - interval (Tc)" -
> > ---
> > "The time interval during which you send the Bc bits in order to
> > maintain the average rate of the CIR in seconds.
> >
> > Tc = Bc/CIR in seconds.
> >
> > The range for Tc is between 10 ms and 125 ms."
> > ---
> > In the example given in the Cisco document, Bc is 8000 bits, CIR is
> > 64000 bps, then Tc = 8000 bits / 64000 bps = 1/8 second.
> >
> > Or if you use Bc = CIR * Tc, then
> > Bc = 64000 bps * 1/8 second = 8000 bits.
> >
> > What does dividing by 1000 get me?
> >
> > Thanks.
This archive was generated by hypermail 2.1.4 : Thu May 01 2003 - 13:35:57 GMT-3