This is Subject Network 1 if you dont know all the answers dont answer it R9. In

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This is Subject Network 1 if you dont know all the answers dont answer it

R9. In our rdt protocols, why did we need to introduce sequence numbers?

R16. Consider the Telnet example discussed in Section 3.5. A few seconds after the user types the letter ‘C,’ the user types the letter ‘R.’ After typing the letter ‘R,’ how many segments are sent, and what is put in the sequence number and acknowledgment fields of the segments?

P2. Consider below Figure. What are the source and destination port values in the segments flowing from the server back to the clients’ processes? What are the IPaddresses in the network-layer datagrams carrying the transport-layer segments?

P5. Suppose that the UDP receiver computes the Internet checksum for the received UDP segment and finds that it matches the value carried in the checksum field. Can the receiver be absolutely certain that no bit errors have occurred? Explain.

P9. Give a trace of the operation of protocol rdt3.0 when data packets and acknowledgment packets are garbled. Your trace should be similar to that used in the textbook (using timing diagram).

P25. We have said that an application may choose UDP for a transport protocol because UDP offers finer application control (than TCP) of what data is sent in a segment and when. a. Why does an application have more control of what data is sent in a segment? b. Why does an application have more control on when the segment is sent?

P26. Consider transferring an enormous file of L bytes from Host A to Host B. Assume an MSS of 536 bytes.

a. What is the maximum value of L such that TCP sequence numbers are not exhausted? Recall that the TCP sequence number field has 4 bytes.

b. For the L you obtain in (a), find how long it takes to transmit the file. Assume that a total of 66 bytes of transport, network, and data-link header are added to each segment before the resulting packet is sent out over a 155 Mbps link. Ignore flow control and congestion control so A can pump out the segments back to back and continuously.

Web client host C Web server B Per-connection HTTP processes Transport- layer demultiplexing source port: 7532 dest. port 80 source port: dest. port 80 dest. IP: 261 source IPdest. IP source IP: Web client host A source port: dest. port 26145 80 source IP dest. IP-

Explanation / Answer

R9 :

Sequence numbers are required for a receiver to find out whether an arriving packet contains new data or is a retransmission.This is provide basically acknowledgement data so for network confirmation sequence number is required.

R16 : PLease provide image or circuit

R2: the Host A, which sending segment who have port 26145 and destination port 80. same doing by host B. B sending response back to Host A using source port 80 and destination port 26145.These are provide basically acknowledgement data For the Host C , server B will send the response back with source port 80 and destination port 26145 and 7532. The IP addresses in the datagrams that are heading back to A or
C will have the source IP set to the servers IP address, and the destination IP set to the appropriate clients address, which it specified in its own request it had originally sent to the server.

R5 : Consider sending an application message over a transport protocol. With TCP, the application writes data to the connection and TCP will grab bytes without necessarily putting a single message in the TCP segment. UDP encapsulates in a segment whatever the application gives it. So, if the application gives UDP an application message, this message will be payload of the UDP segment Thus, with UDP, an application has more control of what data is sent in a segment.so Whenever a digit is changed to a new digit, there is a bit error that occurs. Recall that a checksum as defined on UDP Checksum.

R9 : The sender is in state “Wait” and the receiver is in state “Wait for 0 ”. The scenarios for corrupted data and ackwedgement ACK.


P 25 :

a) UDP encapsulates in a segment whatever the application gives it. So, if the application gives UDP an application message, this message will be payload of the UDP segment. Thus, with UDP, an application has more control of what data is sent in a segment. the application do es not directly assign data to the sending segment. Instead, it puts the data to sending buer and the TCP proto cols will decide whether putting all the data into one segment or dividing them into several parts and sending them with several segments.

B)it has control and congestion control, as well as the retransmission
mechanisms. With them,though   the data can be sent correctly and reliably, the arriving time could be much later the sending time, and time inteveral could b e unexpected.

P 26

a) There are 2^32 = 4,294,967,296 possibe sequence numbers.

b) First we need to caculate segment.
(2^32)/536 = 8,012,999

The number of segments is. 66 bytes of header
8,012,999 * 66 = 528,857,934 byte

The number of segments is. 66 bytes of header get added to each segment giving a total of 528,857,934 bytes of header.
The total number of bytes transmitted is 2^32 + 528,857,934 = 4.824 × 10^9 bytes

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