Consider a modification to TCP's congestion control algorithm. Instead of additive increase, we can use multiplicative increase. A TCP sender increases its window size by a small positive constant a(0

Consider a modification to TCP's congestion control algorithm. Instead of additive increase, we can use multiplicative increase. A TCP sender increases its window size by a small positive constant a(0

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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Suppose a TCP connection, with window size 1, loses every other packet. Those that do arrive have RTT: 1 second. What = happens? What happens to TimeOut? Do this for two cases: (a) After a packet is eventually received, we pick up where we left off, resuming with Estimated RTT initialized to its pre-timeout value, and TimeOut double that. (b) After a packet is eventually received, we resume with TimeOut initialized to the last exponentially backed-off value used for the timeout interval. In the following four exercises, the calculations involved are straightforward with a spreadsheet.
In TCP connection, congestion occurs if the load on the network is greater than the capacity of the network. TCP uses the following algorithm to handle congestion. The algorithm is based on the size of the congestion window (cwnd) which starts with one maximum segment size (MSS). The MSS is determined during connection establishment. At each time the whole window of segments is acknowledged (one transmission), the size of the window grows exponentially. To stop this exponential growth, the sender keeps the track of a threshold. The threshold is set to 16 MSS. After reaching the threshold, the size of the congestion window is increased by 1. After how many transmissions the congestion window size will be 22 MSS?
Consider a TCP connection betweek two hosts between A and B. The first data byte sent by A is numbered 1. The LastByteRcvd at B is 500. Then, the range of values SendBase at A may take is from to 500
COMPUTER NETWORKING
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.
Consider a TCP connection with congestion window size 40 KB, maximum segment size is 2 KB, if time taken by TCP connection to get 37 KB congestion window is 351 ms, round trip time of the connection is __(ms).
Consider the figure below in which a TCP sender and receiver communicate over a connection in which the segments can be lost. The TCP sender wants to send a total of 10 segments to the receiver and sends an initial window of 5 segments at t = 1, 2, 3, 4, and 5, respectively. Suppose the initial value of the sequence number is 151 and every segment sent to the receiver each contains 612 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. As shown in the figure, 1 of the 5 segments is lost between the sender and the receiver, but none of the ACKs are lost. Assume there are no timeouts and any out of order segments received are thrown out.
Consider the figure below in which a TCP sender and receiver communicate over a connection in which the segments can be lost. The TCP sender wants to send a total of 10 segments to the receiver and sends an initial window of 5 segments at t = 1, 2, 3, 4, and 5, respectively. Suppose the initial value of the sequence number is 151 and every segment sent to the receiver each contains 612 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. As shown in the figure, 1 of the 5 segments is lost between the sender and the receiver, but none of the ACKs are lost. Assume there are no timeouts and any out of order segments received are thrown out.
In TCP Fast Retransmit, if we receive three duplicate ACKS after sending out a packet that is not incrementally ACK-ed but before the time-out associated with that packet, we immediately retransmit the packet without waiting for the timeout to occur. Why do we wait for three duplicates instead of just one?
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Consider the figure below, which plots the evolution of TCP's congestion window at the beginning of each time unit (where the unit of time is equal to the RTT; i.e. a transmission round). TCP Reno is used here. In the abstract model for this problem, TCP sends a "flight" of packets of size cwnd (the congestion window) at the beginning of each time unit. The result of sending that flight of packets is that either (i) all packets are ACKed at the end of the time unit, (ii) there is a timeout for the first packet, or (iii) there is a triple duplicate ACK for the first packet. римә 2228ENNOSSANO 24 20 18 16 14 12 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Transmission round Which of the following events occurred at the 26th transmission round and triggered a new TCP phase? triple duplicate ACK timeout slow start threshold reached none of the mentioned corrupted ACK