With circuit switching, total transmission time T = i+x/d+k*d

Witch packet switching, total transmission time T=x/d+k*d+(k-1)*p/d

Think 10% as the probability of 0 packet transmission attempted in one slot, so by Poisson Distribution, P(k)=(G^k*e^-G)/k! we have:

P(0)=e^-G=0.1 è G=2.3

S=G*e^{-G è} S=0.23

Since G>1, so the channel is overloaded

Probability of a collision on around i is 2^-i, the probability that the first k-1 attempts fail, followed by a success on round k is :

P_k=( Õ (1/2ⁱ))(1-1/2^k) (where i = 0,1…. k-1)

The expected number of rounds is then just S kP_k

In order to detect collision on the channel, the transmission time T=pkt/transmission rate has to be greater than 2*RTT. So for the faster Ethernet, since the transmission rate is increased by 10, we will have to decrease the propagation delay by 10 so that it can still detect the collision. So the maximum wire length in Fast Ethernet is 1/10 as long as in Ethernet.

Case 1: If spanning tree protocol is used, the new bridge will announces itself and the spanning tree algorithm computes a spanning tree for the new configuration, which will not cause any problems because no matter how many bridges you connect, you always end up with a spanning tree.

Case 2: If no spanning tree protocol is used, there will be multiple paths between LAN segments. Then frames will cycle and multiply within the interconnected LAN, thereby crashing the entire network. See Pg. 371-372 on Kurose Book.