Solution#4.1
With pure ALOHA the usable bandwidth is 0.184 x 56 kbps = 10.3 kbps. Each station requires 10 bps, so N = 10300/10 = 1030 stations.
Solution#4.3
Each terminal makes one request every 200 sec, for a total load of 50 requests/sec. Hence G = 50/8000 = 1/160.
Solution#4.5
(a) From the Poisson law, P0 = 0.1 = e-G. This gives G = 2.3.
(b) Using S = G e-G gives S = 0.23.
(c) Whenever G > 1 the channel is overheaded, so it is overheaded.
Solution#4.19
The round trip propagation time of the cable is 10 µsec. A complete transmission has four phases:
transmitter siezes cable (10 µsec)
transmit data (25.6 µsec)
receiver seizes cable (10 µsec)
acknowledgement sent (3.2 µsec)
The sum of these is 48.8 µsec. In this period 224 data bits are sent for a rate of 4.7 Mbps.
Solution#4.25
The situation is same as a station being unable to pass a token. In both cases, after a station has passed a token, it stays alert waiting to see if the next station either passes the token or sends a frame. If neither happens, it issues who_follows frame to begin removing the failed station from the logical ring. It does not matter if the token recepient failed just before accepting the token or just after. In both cases, th etoken passer keeps paying attention until it sees some activity from the token recipient.
Solution#4.28
Measured from the time of token capture, it takes 25.6 µsec to transmit a packet. Additionally, a token must be transmitted, taking 0.8 µsec, and the token must propagate 20 meters, tyaking another 0.1 µsec. Thus we have sent 224 bits in 26.5 µsec for a data rate of about 8.5 Mbps. With these parameters the ring beats the Ethernet in terms of effective bandwidth.
Solution#4.28
The new bridge announces itself and spanning tree algorithm computes a spanning tree for a new configuration. The new topology will put one of the bridges into standby mode, so it will be available as a spare in case one of the others fails. This type of configuration provides extra reliability at extra cost and is not unusual. It does not cause any problems because no matter how many bridges you connect, you always end up with a spanning tree.