A. During which of the following intervals of time is TCP performing slow start"? B. During which of the following intervals of time is TCP performing congestion avoidance*? C. At the end of which units of time does TCP detect a triple-duplicate-ACK*? D. At the end of which unit(s) of time does TCP detect a loss via timeout"?
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- Question 6 Phases of TCP congestion control. 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); see Figure 3.53 in the text. In the abstract model for this problem, TCP sends a "flight" of packets of size cwnd 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. Congestion window size (in segments) 40 35 30 25 20 15 10 5 0 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 Time unit (in RTT) A. During which of the following intervals of time is TCP performing slow start*? B. During which of the following intervals of time is TCP performing congestion avoidance*? C. At the end of which units of time does TCP detect a…AsapTCP 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 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 ACKTCP 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.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
- 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.Suppose two TCP connections share a path through a router R. The router's queue size is six segments; each connection has a stable congestion window of three segments. No congestion control is used by these connections. A third TCP connection now is attempted, also through R. The third connection does not use congestion control either. Describe a scenario in which, for at least a while, the third connection gets none of the available bandwidth. and the first two connections proceed with 50% each. Does it matter if the third connection uses slow start? How does full congestion avoidance on the part of the first two connections help solve this? 10:33 am Type a message Dorcon 1 tahir reIn 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?
- 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.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.In the generic SR protocol, the sender transmits a message as soon as it is available (if it is in the window) without waiting for an acknowledgment. Suppose now that we want an SR protocol that sends messages two at a time. That is, the sender will send a pair of messages and will send the next pair of messages only when it knows that both messages in the first pair have been received correctly.Supposethat the channel may lose messages but will not corrupt or reorder messages. i) Outline an error-control protocol for the unidirectional rdt of messages. ii) Give an FSM description of the sender and receiver. iii) Describe the format of the packets sent between sender and receiver and vice versa. iv) If you use any procedure calls other than those in SR protocol (for example, udt_send(), start_timer(), rdt_rcv(), and so on),clearly state their actions. v) Give an example (a timeline trace of sender and receiver) showing how your protocol recovers from a lost packet