EBK COMPUTER NETWORKING
7th Edition
ISBN: 8220102955479
Author: Ross
Publisher: PEARSON
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Expert Solution & Answer
Chapter 7, Problem R20RQ
Program Description Answer
The data transfer between the corresponding node and the mobile host made through the home network but not surpassing the home network.
Therefore, the given statement is “False”.
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Students have asked these similar questions
TCP session sends 50 packets per second over an Ethernet Local Area Network (LAN). Each packet consists 2280B (excluding the preamble and cyclic redundancy check (CRC)). Calculate the size of the headers, and hence the TCP payload data. What therefore is the TCP throughput of the session?
An HTTP client opens a TCP connection using an initial sequence number (ISN) of 14,534 and the ephemeral port number of 59,100. The server opens the connection with an ISN of 21,732. Show the three TCP segments during the connection establishment if the client defines the rwnd of 4000 and the server defines the rwnd of 5000. Ignore the calculation of the checksum field.
A 3200 bit long TCP message is transmitted to the IP layer and becomes a datagram after adding a 160 bit header. The following Internet is connected by two LANs through routers. But the data part of the longest data frame that the second LAN can transmit is only 1200 bits. Therefore, datagrams must be segmented in the router. How many bits of data does the second LAN transmit to its upper layer?
Chapter 7 Solutions
EBK COMPUTER NETWORKING
Ch. 7 - Prob. R1RQCh. 7 - Prob. R2RQCh. 7 - Prob. R3RQCh. 7 - Prob. R4RQCh. 7 - Prob. R5RQCh. 7 - Prob. R6RQCh. 7 - Prob. R7RQCh. 7 - Prob. R8RQCh. 7 - Prob. R9RQCh. 7 - Prob. R10RQ
Ch. 7 - Prob. R11RQCh. 7 - Prob. R12RQCh. 7 - Prob. R13RQCh. 7 - Prob. R14RQCh. 7 - Prob. R15RQCh. 7 - Prob. R16RQCh. 7 - Prob. R17RQCh. 7 - Prob. R18RQCh. 7 - Prob. R19RQCh. 7 - Prob. R20RQCh. 7 - Prob. R21RQCh. 7 - Prob. R22RQCh. 7 - Prob. R23RQCh. 7 - Prob. P1PCh. 7 - Prob. P2PCh. 7 - Prob. P3PCh. 7 - Prob. P5PCh. 7 - Prob. P6PCh. 7 - Prob. P7PCh. 7 - Prob. P8PCh. 7 - Prob. P10PCh. 7 - Prob. P11PCh. 7 - Prob. P12PCh. 7 - Prob. P13PCh. 7 - Prob. P14PCh. 7 - Prob. P15PCh. 7 - Prob. P16P
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Similar questions
- 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?arrow_forwardTCP congestion control example. Consider the figure below, where a TCP sender sends 8 TCP segments at t = 1, 2, 3, 4, 5, 6, 7, 8. Suppose the initial value of the sequence number is 0 and every segment sent to the receiver each contains 100 bytes. The delay between the sender and receiver is 5 time units, and so the first segment arrives at the receiver at t = 6. The ACKs sent by the receiver at t = 6, 7, 8, 10, 11, 12 are shown. The TCP segments (if any) sent by the sender at t = 11, 13, 15, 16, 17, 18 are not shown. The segment sent at t=4 is lost, as is the ACK segment sent at t=7. TCP sender t=1 T t=2 t=3 t=4+ t=5- t=6+ t=11 t=12 t=13 t=14 t=15 t=16 t=17 t=18 I data segment data segment data segment data segment data segment data segment data segment data segment ACK ACK ACK ACK ACK ACK Ty A A V V htt TCP receiver t=6 t=7 t=8 t=9 t=10 t=11 t=12 t=13 What does the sender do at t=17? You can assume for this question that no timeouts have occurred.arrow_forwardTCP congestion control example. Consider the figure below, where a TCP sender sends 8 TCP segments at t = 1, 2, 3, 4, 5, 6, 7, 8. Suppose the initial value of the sequence number is 0 and every segment sent to the receiver each contains 100 bytes. The delay between the sender and receiver is 5 time units, and so the first segment arrives at the receiver at t = 6. The ACKS sent by the receiver at t = 6, 7, 8, 10, 11, 12 are shown. The TCP segments (if any) sent by the sender att = 11, 13, 15, 16, 17, 18 are not shown. The segment sent at t=4 is lost, as is the ACK segment sent at t=7. t=1 T data segment t=2+ data segment data segment-- t=3 TCP sender TCP receiver t=4+ t=5+ data segment - data segment t=6+ t36 data segment t=7 data segment t=8 data segment t=9 ACK + t=10 k -- ACK t=11 t=11 t=12 t=12 t=13 t=13 t=14 ACK -ACK ACK t=15 t=16 t=17 ACK t=18 What does the sender do at t=17? You can assume for this question that no timeouts have occurred.arrow_forward
- A packet switch receives a packet and determines the outbound link to which the packet should be forwarded. When the packet arrives, one other packet is halfway done being transmitted on this outbound link and four other packets are waiting to be transmitted. Packets are transmitted in order of arrival. Suppose all packets are 1,500 bytes and the link rate is 2 Mbps. What is the queuing delay for the packet? More generally, what is the queuing delay when all packets have length L, the transmission rate is R, x bits of the currently-being-transmitted packet have been transmitted, and n packets are already in the queue?arrow_forwardIP datagrams on a specific network can carry a maximum of only 440 bytes in the data portion. A node on this network, running an application using TCP generates a TCP segment with 1,900 bytes in the data portion. How many IP packets are transmitted to carry this TCP segment, and what are their sizes (including the header of 20 bytes)?arrow_forwardWe 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. Why does an application have more control of what data is sent in a segment and when the segment is sent?arrow_forward
- Q: Describe in detail how TCP packets flow in the case of TCP handoff, along with the information on source and destination addresses in the various head- ers.arrow_forwardSuppose Host A sends two TCP segments back to back to Host B over a TCP connection.The first segment has sequence number 90; the second has sequence number 110. Suppose that the first segment is lost but the second segment arrives at B. In the acknowledgment that Host B sends to Host A, what will be the acknowledgment number?arrow_forwardHost A and B are communicating over a TCP connection. Host B has already received from Host A all bytes up through byte 23. Suppose Host A then sends two segments to Host B back-to-back. The first and the second segments contain 30 and 50 bytes of data, respectively. In the first segment, the sequence number is 24, the source port number is 3000, and the destination port number is 80. Host B sends an acknowledgment whenever it receives a segment from Host A. A. In the second segment sent from Host A to B, what arethe sequence number_________,source port number __________,and destination port number__________?B. If the second segment arrives after the first segment, in the acknowledgment of the second segment, what arethe acknowledgment number___________,the source port number__________,and the destination port number __________?C. If the second segment arrives before the first segment, in the acknowledgment of the first arriving segment,what is the acknowledgment number ___________?D.…arrow_forward
- Consider a TCP connection between Host A and Host B. Suppose that the TCP segments traveling from Host A to Host B have source port number x and destination port number y. What are the source and destination port numbers for the segments traveling from Host B to Host A?arrow_forwardConsider the datagrams of each layer in the TCP/IP model, e.g., Frame at the Link Layer, Packet at the Internet Layer. Explain the functioning of the following protocols based on their datagram header fields only. Note: You do not need to describe the header field but based on the header field, explain how the protocol works, e.g., Acknowledgement field in a TCP Header is responsible for providing reliable data transfer and in case of data loss or ACK not received, segment is re-sent. You do not need to consider every field of the headers but wherever there are more than 4 fields, then based on 5 fields, describe its functioning, if less than or equal to 4 fields then consider all the fields. Ethernet Frame 11 Frame IPv4 Packet IPv6 Packet TCP Segment UDP Segment HTTParrow_forwardConsider TCP client with ISN = 20120 communicating with a TCP server with ISN = 30130 -client has 4500 data bytes to send to the server -client MSS = 1500 bytes -server has 6000 data bytes to send back -server MSS = 3000 bytes -show all the client-server transactions to transfer all data in both directions (sequence numbers, ack numbers for all arrows).arrow_forward
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