JAWAHAR NARAYANANCIS 617 EXAM QUESTIONSQuestion .1. Imagine that a two way handshake rather than a three handshake were used to set up connections. In other words, a third messagewas not required. Are deadlocks possible? Give an example or show that none exist?Deadlocks are possible. For example, a packet arrives at A out of the blue, and A acknowledges it. The acknowledgement gets lost, but A is now open while B knows nothing at all about what has happened. Now the same thing happens to B, and both are open, but expecting different sequence numbers. Timeouts have to be introduced to avoid the deadlocks.Show a deadlock that cannot be resolved with timeouts.Question .2. Datagram fragmentation and reassembly are handled by IP and are invisible to TCP. Does this mean that TCP does not have to worry about data arriving in the wrong order?TCP has to take care of the data arriving in the wrong order. Even though each Datagram arrives intact, it is possible that datagrams arrive in the wrong order, so TCP has to be prepared to reassemble the parts of a message properly.Why the datagrams can arrive in the wrong order?Question .3. Give one advantage of RPC on UDP over transactional TCP. Give one advantage of T/TCP over RPC.RPC has an advantage over UDP in that it takes only two packets instead of three. However, RPC has a problem if the reply does not fit in one packet.Question .4. The Protocol field in the IPv4 header is not present in the fixed IPv6 header. Why not?The Protocol field tells the destination host which protocol handler to give the IP packet to. Intermediate routers do not need this information,so it is not needed in the main header. Actually, it is present there, but it is disguised. The Next header field of the last (extension) header is used for this purpose.Question .5. What is the total size of the minimum TCP MTU, including TCPand IP overhead but not including the data link layer overhead?The default segment size is 536 bytes. TCP adds 20 bytes and IP adds 20 packets, making the default 576 bytes in total.Question .6. The maximum payload of a TCP segment is 65, 495 bytes. Why is this number chosen?The entire TCP segment must fit in the 65, 515-byte payload field of an IP packet. Since the TCP header is a minimum of 20 bytes, only 65,495 bytes are left for TCP data.This is a wrong answer.Question .7. Can a machine with a single DNS name have multiple IP addresses? How could this occur?Yes it is possible for a machine with a single DNS name to have multiple IP addresses. An IP address consists of a network number and a host number. If a machine has two Ethernet cards, it can be on two separate networks, and if so, it needs two IP addresses.Question .8. DNS uses UDP instead of TCP. If a DNS packet is lost, there is no automatic recovery. Is this a problem, and if so, what is the solution? DNS is idempotent. Operations can be repeated without harm. When a process makes a DNS request, it starts a timer. If the timer expires, itjust makes the request again. No harm is done. Question .9. In addition to being subject to loss, UDP packets have a maximum length, potentially as low as 576 bytes. What happens when a DNS name to be looked up exceeds this length? Can it be sent in two packets?The problem does not occur. DNS names must be shorter than 256bytes. The standard requires this. Thus, all DNS names fit in a single minimum length