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[802.3ae_Serial] FW: Re: Questions on additional allowable insertion losses

I didn't see this message come back to me, and I got no responses. So, I am sending it again.
 
Tom
 
-----Original Message-----
From: Lindsay, Tom
Sent: Tuesday, February 05, 2002 9:24 PM
To: Serial reflector 802.3ae (E-mail)
Cc: T11.2 reflector (E-mail)
Subject: [T11.2] Re: Questions on additional allowable insertion losses

Piers -
 
To avoid the issues raised by my earlier email including my last sentence, if we calculate everything at the max wavelength and min width, then the rows will still add up to the power budget. However, quite a few numbers change:
 
  Power budget
  Channel insertion loss (slight)
  Unallocated margin (some cable plants, not documented)
  Allocation for penalties and margin
  Additional allowable insertion loss
  VECP
  Stressed OMA
 
I think that is it. Most of the changes are small, but nonzero. Obviously the triple tradeoff curves stay the same.
 
Is this what we want to do?
 
Tom
-----Original Message-----
From: Lindsay, Tom
Sent: Monday, February 04, 2002 12:52 PM
To: Serial reflector 802.3ae (E-mail)
Cc: T11.2 reflector (E-mail)
Subject: Questions on additional allowable insertion losses

A question - my understanding of the values in "additional insertion loss allowed" (for example in Table 52-10) is that they are determined by the difference between unallocated margins from their cable plant type and the one with the least unallocated margin. Is this correct?
 
That is, if 50 micron 2000 MHz-km has 0.8 dB unallocated margin (and it is the least of all -S variants), and if 50 micron 500 MHz-km has 1.3 dB unallocated margin, then 50 micron 500 MHz-km can allow up to 0.5 dB (the difference) of addtional insertion loss.
 
If this is true, then I find that the unallocated margin differences change considerably across the triple-tradeoff curves. For example, the differences become quite small at longer wavelengths and narrower spectral widths.
 
Three simple proposals:
1. Eliminate the row entirely, as the values become quite small in some cases.
2. Use the values determined by max center wavelength and most narrow spectral width.
3. Explain to me how I am confused by this (preferred option).
 
Note that unless all rows are determined by the same spectral conditions, some of these options will affect how the rows add up to equal the power budget.
 
Comments?

Tom