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Topic # 7361 9-Apr-2006 12:48
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does anybody know:
1: The new contention ratios...are they based on a physical connection to the DSLAMS? For example my current 2m/128 is connected to the same physical rack, or block of 50 users contending for the 2mbs of backhaul bandwidth?
2: On the old or new ratios...are all the DSL connections that are contending for the bandwidth on the same plan?  e.g. 256/128 3.5/128 3.5/512
3: So Most importantly - Does this mean that if I "upgrade" my connection to a plan with 512k upload...I'll likely be contending with all high maintenance (heavy internet users) (147 of them!!) instead of the adverage joe searching for p0rn?


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Juha
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  Reply # 32887 12-Apr-2006 19:19
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Telecom provisions 24kbit/s per user and month for UBS on the ATM backhaul; I thought it was 40kbit/s for Xtra, but apparently the dreaded retail equivalence has been introduced there too.

So, if an ISP has 100 users, it'll get a committed information rate of 24*100=2,400kbit/s. If the ATM circuit isn't loaded to the hilt, users can burst up to their "target speed" but not all the time. Telecom has configured the ATM circuit so that it allows bursting up to the top speed for a certain number of cells at the time, and then drops users down to 24kbit/s. I'm not sure what that cell number is, but have heard it's 96 at the time - so, 96 * 53 bytes only, but remember that each cell has only a 40 byte IP payload.






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Wannabe Geek


  Reply # 32945 13-Apr-2006 14:01
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Thanks for the reply Juha. I did know about the 24kbps CDR (Commited Data Rate) per user, that is the new minimum speed per 148 contending users - (148 * 24kbps = 3.5Mbps).  40kbps was the CDR with the old contention ratio, @ 50:1.  As far as I know once Telecom completes its "upgrade" all users will be on 148:1 - can anybody confirm this? 

 
 
 
 


Juha
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  Reply # 32949 13-Apr-2006 14:46
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esky: Thanks for the reply Juha. I did know about the 24kbps CDR (Commited Data Rate) per user, that is the new minimum speed per 148 contending users - (148 * 24kbps = 3.5Mbps). 40kbps was the CDR with the old contention ratio, @ 50:1. As far as I know once Telecom completes its "upgrade" all users will be on 148:1 - can anybody confirm this?


Telecom would be able to, but... Wink




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  Reply # 34218 28-Apr-2006 17:08
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I'm not sure what that cell number is, but have heard it's 96 at the time - so, 96 * 53 bytes only, but remember that each cell has only a 40 byte IP payload.


The encapsulation is as follows SDH->ATM->IP->UDP->L2TP->PPP->IP

So per ATM cell there is 8 bytes of overhead since CBR is employed. Then you will have 20 bytes for IP overhead, 8 for UDP, assuming 12 for L2TP, 10 for PPPoA and 20 for IP. Now you can start adding on your TCP/UDP of whatever transport protocol your using overhead into the mix at the end before you actually transport some meaningful data.

Of course not all that is added per cell (IP onwards) but it does add up to a decent percentage of your remaining data unit in the ATM cell.

Juha
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  Reply # 34236 28-Apr-2006 20:17
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Fraktul:
The encapsulation is as follows SDH->ATM->IP->UDP->L2TP->PPP->IP

So per ATM cell there is 8 bytes of overhead since CBR is employed. Then you will have 20 bytes for IP overhead, 8 for UDP, assuming 12 for L2TP, 10 for PPPoA and 20 for IP. Now you can start adding on your TCP/UDP of whatever transport protocol your using overhead into the mix at the end before you actually transport some meaningful data.

Of course not all that is added per cell (IP onwards) but it does add up to a decent percentage of your remaining data unit in the ATM cell.


8+20+8+12+10+20 = 78 bytes

???




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  Reply # 34260 29-Apr-2006 02:24
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Of course not all that is added per cell (IP onwards) but it does add up to a decent percentage of your remaining data unit in the ATM cell.


Each cell does not have this overhead, only the ATM overhead. The end user layer 3 data units are fragmented down through the encapsulation layers as needed into the ATM cells. Each layers overhead still it adds to the total overhead before usable end user layer 4 data unit however.

Perhaps not explaining it very well, but say you took an Ethernet frame and encapsulated it in ATM cells (this is just an example)... You have 8 bytes overhead for the ATM cell with CBR and you have 14 bytes for the Ethernet frame overhead which is fragmented over the ATM cells so the 8 bytes is a given per cell but the Ethernet overhead averages over the cells is a measurable percentage. The overhead itself fits in the first cell but im just talking about averages.

As you point out its 70 odd bytes of overhead for each layer 3 data unit so this actually spans two cells.

If you think this is incorrect please point out, this is only my understanding over the overhead calculations. I know the actual list of encapsulation protocols involved is correct however :)

Juha
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  Reply # 34264 29-Apr-2006 11:14
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No, I'm not saying it's incorrect - I really can't remember how ATM works anymore. Years ago in Singapore I had an IBM made (from memory) ATM card with LAN Emulation, plus one made by a Danish manufacturer, in my desktop computer and made an effort to understand how it all worked then.

However, Ethernet killed off ATM at that level and I never really had a need to look at it until Telecom NZ rolled out DSL...




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