Why does the maximum packet lifetime have to be large enough to ensure that not only the packet but also its acknowledgments have disappeared? Give one potential disadvantage when Nagle's algorithm is used on a badly congested network. Give two examples of cases where TCP sends data-less packets on an established connection
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- Why does the maximum packet lifetime have to be large enough to ensure that not only the packet but also its acknowledgments have disappeared?
- Give one potential disadvantage when Nagle's
algorithm is used on a badly congested network. - Give two examples of cases where TCP sends data-less packets on an established connection (which is not being torn down).
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- 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?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.Asap
- 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 a long-lived TCP session with an end- to-end bandwidth of 1 Gbps (= 10⁹ bits-per- second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again isIt is possible for a series of packets to be sent from one host to another over the same connection. Please break down the time it takes for a single package to go from beginning to conclusion. Is it expected that one of the delays would last a certain period of time while the other delay's length will be more malleable?
- TCP is a connection-oriented protocol. This means that... а. there is a direct physical connection between the two endpoints of a TCP session. O b. both of the endpoints communicating over TCP keep information about the state of the connection. с. both endpoints of a TCP session have to connect to a third party before data can be sent or received. O d. a TCP sender can start sending data to the receiver before the three-way handshake takes place.The filed “window size” in TCP header is in length of 16 bit. It allows 64KB as the maximal size for a TCP segment by default. Under some circumstances, the sending rate is pretty high and it results in a very short sending time (<2ms). But the end to end transmission delay is 50ms. Therefore the idle rate of a channel will be more than 2/3 . How does TCP solve this problem?Consider four Internet hosts, each with a TCP session. These four TCP sessions share a common bottleneck link - all packet loss on the end-to-end paths for these four sessions occurs at just this one link. The bottleneck link has a transmission rate of R. The round trip times, RTT, for all fours hosts to their destinations are approximately the same. No other sessions are currently using this link. The four sessions have been running for a long time. i) What is the approximate throughput of each of these four TCP sessions? Explain your answer briefly. ii) What is the approximate size of the TCP window at each of these hosts? Explain briefly how you arrived at this answer.
- TCP sessions are full-duplex, which means that data can be sent in either direction during the lifetime of the session. Consider a session in which the connection is established, the client sends 100 data segments, all of them are ACKed, and all of the ACKS are received by the sender, then the session is ended by both sides closing the connect. How many segments in total have the SYN bit of the header set to 1? а. 209 O b. 2 6 с. d. 4 О е. 1TCP sessions are full-duplex, which means that data can be sent in either direction during the lifetime of the session. Consider a session in which the connection is established, the client sends 100 data segments, all of them are ACKed, and all of the ACKS are received by the sender, then the session is ended by both sides closing the connect. How many segments in total have the SYN bit of the header set to 1? а. 4 O b. 2 с. O d. 1 е. 209Consider a long-lived TCP session with an end- to-end bandwidth 500 Mb/s. The session starts with a sequence number of 3465. The minimum time (in seconds) before this sequence number can be used again is