Pacation 2) [pt]Suppose that TCP's current estimated values for the round trip time (estimated RTT) and deviation inthe RTT (DeVRTT) are 500 msec and 50 msec, respectively. Suppose that the next one measuredvalues of the RTT are 100.Compute TCP's new value of estimated RTT, DeVRTT, and the TCP timeout value after the measuredRTT values is obtained. Use the values of a = 0.125 and ẞ = 0.25Estimated RTT = (1-a)*Estimated RTT + a*SampleRTTDevRTT(t+1)(t+1)(t)= (1-3)*DevRTT + B*|SampleRTT(t)TimeoutInterval = Estimated RTT + 4*DevRTT(t+1)-Estimated RTT(t+1)(t+1)

To compute TCP's new value of estimatedRTT, DevRTT, and the TCP timeout value after the measured RTT…

Answered: Pacation 2) [pt] Suppose that TCP's…

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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3.5-1a. TCP RTT estimation and timeout value. Suppose that TCP's current estimated values for the round trip time (estimated RTT) and deviation in the RTT (DevRTT) are 300 msec and 13 msec, respectively. Suppose that the next two measured RTTS are 330 msec and 240 msec respectively. We want to calculate TCP's RTT estimate, and the value of TCP's timeout interval. Note that given a new measured RTT, you should first compute devRTT, then estimated RTT (the textbook incorrectly reverses those computations), and then (lastly) the timeout interval. Use the values of a = 0.125, B = 0.25. Following the newly measured RTT of 330 msec, what is the new value for devRTT in msec? [Note: round your answer to the nearest msec - enter an integer value, do not include any decimal places/point or any leading zeros, but keep your full resolution value of devRTT handy and use it in any later calculations you perform using devRTT.
a) Complete the missing sequence numbers (SEQ), acknowledgment numbers (ACK), and segment length (LEN) in the following TCP connection. We assume: • No timeouts occur at the receiver. • The sender starts the timer at tl. • The connection is full duplex (bi-directional data flow in same connection). • The sender and the receiver have always data to transmit. • There are no delayed acknowledgements at the sender or the receiver. Sender Receiver t1 SEO:190 ACKA LEN40 12 SEO B ACK20LEN C A B SEO F ACK=G LEN SEO:160 ACK25LEN= SEQD ACKE LEN E: F: G: J SEQ HACKLLEN= t4 SEO=140 ACKSLEN=20 b) Calculate the vahue of timeout for the segment with sequence number 140.
Suppose that TCP's current estimated values for the round trip time (estimatedRTT) and deviation in the RTT (DevRTT) are 300 msec and 13 msec, respectively. Suppose that the next two measured RTTs are 330 msec and 240 msec respectively. We want to calculate TCP's RTT estimate, and the value of TCP's timeout interval. Note that given a new measured RTT, you should first compute devRTT, then estimatedRTT (the textbook incorrectly reverses those computations), and then (lastly) the timeout interval. Use the values of α = 0.125, β = 0.25.Following the newly measured RTT of 330 msec, what is the new value for devRTT in msec? [Note: round your answer to the nearest msec - enter an integer value, do not include any decimal places/point or any leading zeros, but keep your full resolution value of devRTT handy and use it in any later calculations you perform using devRTT.
5. What can you say about the TCP segment in which the value of the control field isone of the following? a) 000000 b) 000001 c) 010001 Solving with explanation
Q3) A. Suppose TCP operates over a 1-Gbps link. Assuming TCP could utilize the full bandwidth continuously, how long would it take the sequence numbers to wrap around completely? B. fill in the blanks 1. The packets exchanged between TCP peers are called 2. In TCP Header The Acknowledgment, SequenceNum, and AdvertisedWindow fields are all involved in TCP's 3. In TCP Header The SYN flag is used when 4. Required window size for 100-ms RTT and bandwidth 1.5Mbps is ---- 5. In Karn/ Partridge algorithm If the variance among Sample RTTS is small Then the Estimated There is no need to 6. АТM (-- ---> 7. Instead of using coax cable, an Ethernet can be constructed from a thinner cable known as 10Base2, thus 10 means Base means 2 means that a
Consider a TCP connection with the following parameters: a) Initial ssthreshold = 8b) Round Trip Time (RTT) = 10 millisecondsc) Timeout Interval = 40 milliseconds d) Segment size = 2800 bytes The TCP connection was used to transfer 16 segments. While sending the segments, Segment 5 and Segment 15 were lost the first time they were sent (but arrived at the receiver properly when retransmitted). There were no other segment losses apart from the aforementioned. Assume that segments received out-of-order are buffered at the receiver. What is the average throughput obtained in sending all the segments (in bytes/sec)? Draw a diagram to show your work in detail.
4) Host A and B are communicating over a TCP connection, and Host B has already received from A all bytes up through byte 130. Suppose Host A then sends two segments to Host B back-to-back. The first and second segments contain 80 and 40 bytes of data, respectively. In the first segment, the sequence number is 131, the source port number is 301, 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 are the sequence number, source port number, and destination port number? b. If the first segment arrives before the second segment, in the acknowledgment of the first arriving segment, what is the 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?
(a) If the second segment arrives before the first segment, in the ACK of the first arriving segment, what is the ACK number? (b) Draw and explain the 3-Way Handshake in TCP protocol.
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.
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 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.
SInternet Check sum Consider the figure below (simplex communication scenario). The TCP sender sends an initial window of 4 segments. Suppose the initial value of the sender ན༽)། sequence number, SEQ1, is 100 and the first 4 segments each contain 400 bytes of payload. As you can see ACK 3 is lost. What is the SEQ number (SEQ2, SEQ3, SEQ4, SEQ5) from the Sender and ACK Number (ACK1, ACK2, ACK3, ACK4, ACK5) from the receiver for each segment? Sender Seq1 Seq2 Seq3 Seg4 ACK1 ACK2 ACK3 ACK4 Seq5 ACK5 Receiver
Consider a TCP connection with the following parameters: a) Initial ssthreshold = 4 segmentsb) Round Trip Time (RTT) = 10 millisecondsc) Timeout period = 40 millisecondsd) Segment size = 3900 bytesThe TCP connection was used to transfer 17 data segments. While sendingthe segments, data segment 7 and data segment 15 were lost the first timethey were sent (but arrived at the receiver properly when retransmitted).There were no other segment losses apart from the aforementioned. Assumethat data segments received out-of-order are buffered at the receiver. Whatis the average throughput obtained in sending all the data segments (inbytes/sec)? Draw a diagram to show the communication process in detail.

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