Consider the figure below in which a TCP sender and receiver communicate over a connection in which the sender->receiver segments may be lost. The TCP sender sends an initial window of 5 segments. Suppose the initial value of the sender->receiver sequence number is 362 and the first 5 segments each contain 638 bytes. The delay between the sender and receiver is 7 time units, and so the first segment arrives at the receiver at t=8. As shown in the figure below, 3 of the 5 segment(s) are lost between the segment and receiver.

Write a code in C that prints the following: Prints default TCP send and receive buffer sizes.Prints default UDP send and receive buffer sizes.Prints default SCTP send and receive buffer sizes.

TCP is a sliding window protocol. This means. а. there can be more than one segment "in flight" (sent but no ACK has been received) at a time. O b. the TCP retransmission timer slides up as ACKS are received. O c. there can only be one segment outstanding (sent but no ACK has been received) at at time.

Write a Java TCP MultipleClient-Server Program for following scenario.   NOTE Server Details are in the above figure Client-Server communicates continuously until the Client send Bye to the Server Use Multithreading for Multiple Clients NOTE : 1 cm =  0.39 inches 1 cm =  0.03 foot 1 cm =  0.01 meter   Sample Ouput: Client Server 1 – cm to inches 2 – cm to foot 3 – cm to meter Enter length in cms (int) : 10 Received from Server 3.9 inches   1 – cm to inches 2 – cm to foot 3 – cm to meter Enter length in cms (int) : BYE Received from Server BYE   Client DISCONNECTED Server Waiting for the Cline to be Connected… CONECTION ESTABLISHED   Received From Client 10 cms Sent to Client 3.9 inches   Received From Client BYE Sent to Client BYE     HINT : When client choses 1 for cm to inches and enter 10, then the client sends the string 110 to the Server. Where the first char ‘1’ represents the cm to inches and the other 2 chars represents the 10.