Conference netcad::hub_mgnt

>   How do the M ports affect ring length? 
>   
>   I understand the A and B ports are on the ring so the cable 
>   comming in and out of the conc. must be counted. How is the cable
>   from the M ports accounted for ?   
    
        Active M ports are part of the ring.  (See figure 3-6 on page 3-12 of
    the FDDI System Level Description to see an example of how the ring is
    laid out.)
    
>   How is the token rotation timer handled on M ports ? 
>   
>   Again the conc. itself must be part of the calculation.  
>   Does the conc. arbitrate for the token in behalf of the M ports?
    
        Each station on the ring participates on their own during ring
    initialization.
    
>    Lets extend this line of questioning to cover the Gigaswitch/FDDI.
>    
>    I would presume that it being such a fast switch, it does not 
>    contend for tokens or participate in the TRT calcualation.  
    
        Each GIGAswitch/FDDI port is a station on a ring and follows the
    same rules as everyone else is supposed to follow; I.E. it waits for
    the token, and participates in TRT negociation.  (With the exception of
    the full duplex case, which is not special to GIGAswitch/FDDI).
    
        You might find additional information in the UPSAR::FDDI notes
    conference, since these are general FDDI questions...

    
    The token flow in our concentrators (and any other concentrator for
    that matter) is as follows: 
    
    1) token on the primary ring enters the A port of the concentrator
    2) It exits via the first active M port, travels through the station 
       attached to the M port, and back in the M port
    3) It repeats (2) for each active M port
    4) Finally, the token leaves the concentrator via the B port 
    
    The fiber connecting stations and concetrator M ports is included in
    the allowable extent of the net. 
    
    Think of the concentrator as a smart repeater (i.e. it connects media 
    segments together, but doesn't do more than that from the point of view
    of the token protocol). 

    Thank you for taking the time to reply to such basic 
    questions.  
    
    
    re .1 
    
    Thanks again, the diagram on 3-12 of the system level description makes 
    more sense now. 
    
    I have added the FDDI conf. to my notebook  
                
    re .2 
    
    >> 
    >> M ports themselves don't affect length 
    
    >> The fiber connecting stations and concetrator M ports is included in
    >> the allowable extent of the net.
     
    Are you saying the fibers traveling to AND from the M ports are counted 
    in the ring length (extent) but the ports themselves are do not add any
    length?  
    
    
    Jim
        
    

    re -.1 
    
    The specs allow 2 rings, with 100 km of fiber in each ring. The 200 km
    number is the extent of the ring when the ring is wrapped. There are
    delays through each station, and each concentrator port. But, they are
    factored into the calculations for length, so a network designer only
    needs to worry about the fiber length, not the actual timing budget. 
    
    

>                     <<< Note 1665.4 by NETCAD::B_CRONIN >>>
>                                 -< Yes, but >-

>    The specs allow 2 rings, with 100 km of fiber in each ring. The 200 km
>    number is the extent of the ring when the ring is wrapped. There are
>    delays through each station, and each concentrator port. But, they are
>    factored into the calculations for length, so a network designer only
>    needs to worry about the fiber length, not the actual timing budget. 

Let me see if I understand the explanation.

Raj Jain's FDDI Handbook has an example of ring latency caluclation.(page 385)
It says:
	A ring with 16 stations and a total length of 20km. Using two-fiber
	cable, this would correspond to a cable lenght of 10km.
	Light waves travel along the fiber at a speed of 5.085 us/km.
	The station delay, the delay between receiving a bit and repeating
	it on the transmitter side, is of the order of 1us per station.

	Hence, 
	Ring latency D = (20 km) x (5.085 us/km) + (16 stations) x (1 us)
	               = 0.12 ms

Question --
	1. Why the length of 20 km is chosen?
	   -- Is it for a worst case sinario? Under a normal (operating) 
	      condition, "10 km" of the primary ring length should be adequate.

	2. Should a concentrator be considered as a station in the caluculation 
	   -- they are actually "repeating" bits.
 
The reason for asking is for me to understand elements around caluculating the
ring utilization with the fomula:

                                Ring latency
			U = 1 - ------------
                                Average TRT

Thanks,
_Tak