Ok, did some more reading, so I hope I now have it right now LOL.
This was from the link posted back a few posts, basically pages
FTTP - Brown Field Capture and
FTTP GEA Network Architecture
The downstream bandwidth per strand is 2.5Gbps, so the 288 strand cable has a total bandwidth of 720Gbps downstream, and the upstream per strand is 1.25Gbps with a total per 288 strand cable of 360Gbps.
- There are multi 288 strand fibre cable (COF 200 Cable) that leaves the exchange.
- The 288 strands go into a Aggregation Node that takes 48 strands and passes the rest through to go onto the next Aggregation Node.
- These 48 strands gets put into 12 cables with 4 strands that goes into a 32 way splitter resulting in 128 strands.
- These 128 strands then goes into the FibreDP Hardware that takes < 20 strands and passes the rest through to the next FibreDP Hardware daisy chained to it.
- Now a 4 strand fibre is used where only 1 strand of it is connected to one of the <20 strands and goes up the pole to the address etc.
Hopefully that is more right.
So each strand is shared with 32 connections (Downstream: 2.5Gbps / 32 = 78.125Mbps, Upstream: 1.25Gbps / 32 = 39.0625Mbps when hammered)
Granted we lose a fair amount of downstream when everyone hammers the downstream at the same time, this is unlikely but could happen, where as the upstream no matter how much you hammer it you will always be able to get 30Mbps which makes sense.
So in theory a 288 strand cable (COF 200 Cable) can handle around 9,216 connections, so only 2 of these cables would be required to handle all our FTTP connections which is about 18,176 give or take and 1 cable for all the FTTC connections which is about 5,377 give or take.
So if all above is correct our exchange would only require 3 lots of the 288 strand cables to support it and also leaving loads free for future use.
Paul