Solution#1(10pts)

1. The given code follows the operation shown in figure 3.14(b). Here it's assumed that both A and B start at the same time. This results in each of the frame transmitted twice. This will create a problem if the code at A and B are not started simulataneously.
2. Remove the part of the code on B which transmits a frame before entering into the while loop.
3. With this modification B will never get a chance to send first and if A does not have anything to send then B will have to just wait till it gets something from A.

Solution#2(5+5pts)

Failure Condition for Protocol 2:
Stop and Wait protocol presented here fails if one of the frames transmitted is lost. For example, A sends frame to B and B sends back an ack which gets lost. Now A keeps on waiting for an ack as there is no timeout occuring at A. So A is now hung up. The same thing will happen if the frame sent by A gets lost. In this case B never receives any frame and A is expecting to receive an ack from B.
One set of parameters to simulate this is, sim 2 1000 40 20 20 3

Failure Condition for Protocol 3:
This protocol fails due to immature timeout at sender's end that is when the sender times out too early. The flow here is,

Suppose timeout interval = 3.

• Time 0: A send a packet to B seq = 0.
• Time 2: B receives a packet with seq=0 and sends an ack.
• Time 3: A times out and sends packet with seq=0.
• Time 4: B receives packet with seq=0 and transmits an ack.
• Time 5: A receives a first ack and sends packet with seq=1 which gets lost.
• Time 6: A receives an ack. (this is actually for second transmission of packet with seq=0). A now send next packet with seq=0.