4. Consider following figure. TCP sender wants to send a total of 20 segments to the receiver and sendsan initial window of 5 segments at t = 1, 2, 3, 4, & 5, respectively. Suppose the initial value of thesequence number is 50 and every segment sent to the receiver each contains 512 bytes. The delaybetween the sender and receiver is 7 time units, and so the first segment arrives at the receiver at t = 8,and an ACK for this segment arrives at t = 15.→ Show Transcribed TextTCP sendert = 1t = 2t = 3-t = 4t=5+t = 15-t= 16-t = 17t = 18t = 19data segmentdata segmentdata segmentdata segment →→*datasegmentACK segmentACK segmentACK segmentACK segmentงTCP receivert = 8t = 9t = 10t = 11t = 12As shown in the figure above, 4th segment is lost between sender and receiver, but none of the ACKSare lost. Assume there are no timeouts and any out of order segments received are dropped. Answerthe following questions:a) What are the sequence numbers for each of the 20 data segments

For the given scenario, we need to find the sequence numbers for each of the 20 data segments.

4. Consider following figure. TCP sender wants to send a total of 20 segments to the receiver and sends an initial window of 5 segments at t = 1, 2, 3, 4, & 5, respectively. Suppose the initial value of the sequence number is 50 and every segment sent to the receiver each contains 512 bytes. The delay between the sender and receiver is 7 time units, and so the first segment arrives at the receiver at t = 8, and an ACK for this segment arrives at t = 15. Show Transcribed Text TCP sender t=2 t=3 t=4 t=5 t=15 t = 16 t = 17 t = 18 t = 19 data segment data segment data segment data segment* data segment ACK segment ACK segment ACK segment ACK segment 3 C TCP receiver t = 8 t=9 t = 10 t = 11 t = 12 As shown in the figure above, 4th segment is lost between sender and receiver, but none of the ACKS are lost. Assume there are no timeouts and any out of order segments received are dropped. Answer the following questions: a) What are the sequence numbers for each of the 20 data segments

4. Consider following figure. TCP sender wants to send a total of 20 segments to the receiver and sends an initial window of 5 segments at t = 1, 2, 3, 4, & 5, respectively. Suppose the initial value of the sequence number is 50 and every segment sent to the receiver each contains 512 bytes. The delay between the sender and receiver is 7 time units, and so the first segment arrives at the receiver at t = 8, and an ACK for this segment arrives at t = 15. Show Transcribed Text TCP sender t=2 t=3 t=4 t=5 t=15 t = 16 t = 17 t = 18 t = 19 data segment data segment data segment data segment* data segment ACK segment ACK segment ACK segment ACK segment 3 C TCP receiver t = 8 t=9 t = 10 t = 11 t = 12 As shown in the figure above, 4th segment is lost between sender and receiver, but none of the ACKS are lost. Assume there are no timeouts and any out of order segments received are dropped. Answer the following questions: a) What are the sequence numbers for each of the 20 data segments

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Host A sends the information of size 3000 bytes to Host B using TCP protocol. As long as Host B receives each packet in size 1000 bytes, Host B must acknowledge Host A. We assume the header size in each packet is 40 bytes, the starting sequence number used by Host A is 1000, and the window size of the sender side is 3000 bytes. What is the value of SND.NXT when the first packet is sent out, but not acknowledged? 1000 2000 3000 1999 If a host is monitored, and there are seven TCP packets captured: {S1, S2, S3, S4, S5, S6, E1}, where Si represents ith Send packet, and Ej represent jth Echo packet. E1 here acknowledges and echoes all the six Send packets cumulatively. Which of the follow matching can make more sense to represent the RTT of the connection chain? E1 matches with S1 E1 matches with S4 E1 matches with S5 E1 matches with S6
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.
If there is wide variation in the TCP segment size , it is natural to organize the buffers as a pool of fixed-sized buffers, with one segment per buffer TRUE OR FALSE
Consider one TCP connection on a path with a bottleneck link with capacity L. We have a file of size F(assume F is a very large file size). If TCP segment size is B bytes, the two-way propagation delay is tp and connection is always in congestion avoidance phase, find the following: A. Maximum window size that TCP can achieve B. Average window size and average throughput in this TCP connection C. Time it will take from connection establishment to reaching the maximum window size.
Is this the right answer or the wrong answer? Assume that the most recent SampleRTT for a TCP connection is one second. The TimeoutInterval setting for the connection must now be set to a value that is greater than one second.
TCP 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.
Consider a simple protocol for transferring files over a link. After some initial negotiation, A sends data packets of size 1 KB to B; B then replies with an acknowledgment. A always waits for each ACK before sending the next data packet; this is known as stop-and-wait. Packets that are overdue are presumed lost and are retransmitted. (a) In the absence of any packet losses or duplications, explain why it is not necessary to include any "sequence number" data in the packet headers. (b) Suppose that the link can lose occasional packets, but that packets that do arrive always arrive in the order sent. Is a 2-bit sequence number (that is, N mod 4) enough for A and B to detect and resend any lost packets? Is a 1-bit sequence number enough? (c) Now suppose that the link can deliver out of order and that sometimes a packet can be delivered as much as 1 minute after subsequent packets. How does this change the sequence number requirements?
TCP is a sliding window protocol. This means... а. the TCP retransmission timer slides up as ACKS are received. O b. there can be more than one segment "in flight" (sent but no ACK has been received) at a time. с. there can only be one segment outstanding (sent but no ACK has been received) at at time.
Computer A uses the Go-back-N ARQ protocol to send a 110 Mbytes file to computer B with a window size of 15. Given each frame carries 100K bytes data. How long does it take to send the whole file (the total time taken from A sending the first bit of the file until A receiving the last acknowledgment)? Given that the transmission rate of the link is 500 Mbps and the propagation time between A and B is 15ms. Assume no data or control frame is lost or damaged and ignore the overhead due to header and trailer.
In TCP Vegas, the calculation of Actual Rate is done by dividing the amount of data transmitted in one RTT interval by the length of the RTT. (a) Show that for any TCP, if the window size remains constant, then the amount of data transmitted in one RTT interval is constant once a full window-full is sent. Assume that the sender transmits each segment instantly upon receiving an ACK, packets are not lost and are delivered in order, segments are all the same size, and the first link along the path is not the slowest. (b) Give a timeline sketch showing that the amount of data per RIT above can be less than Congestion Window.
Given a M/M/1 queue, which has packets arriving at a rate 2 packets/sec, an output link rate R bps, and an average packet size of 1500 bytes, write down: the expression for T, the average delay for a packet (i.e., from arrival till completing transmission at the output). b. the expression for the output utilization c. the expression for average number of packets in the queue d. the expression for average number of packets in the output NIC
Host A sends the information of size 3000 bytes to Host B using TCP protocol. As long as Host B receives each packet in size 1000 bytes, Host B must acknowledge Host A. We assume the header size in each packet is 40 bytes, the starting sequence number used by Host A is 1000, and the window size of the sender side is 3000 bytes. What is the value of RCV.NXT in Host B when the second packet in Host A is sent out and acknowledged? Question options: 1000 2000 3000 4000