6. Datagram fragmentation and reassembly are handled by IP and are invisible to TCP. Does this mean that TCP does not have to worry about data arriving in the wrong order?
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cs network
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- - Suppose a TCP message that contains 1024 bytes of data and 20 bytes of TCP header is passed to IP for delivery across two networks interconnected by a router (i.e., it travels from the source host to a router to the destination host). The first network has an MTU of 1024 bytes; the second has an MTU of 576 bytes. Each network's MTU gives the size of the largest IP datagram that can be carried in a link-layer frame. Give the sizes and offsets of the sequence of fragments delivered to the network layer at the destination host. Assume all IP headers are 20 bytes.1. Suppose a TCP connection is transferring a file of 5550 bytes. The first byte is numbered 10001. What are the sequence numbers for each segment if each carries 500 bytes? How many Segments will be there ?TCP is a lightweight Transport-layer protocol that allows programmes to interact with one another. Many of the same properties of UDP, such as reliability and flow control, are absent. Is it correct or incorrect?
- Below are the two questions, so make sure to answer each part carefully and label the response for each question: 1) Why does TCP implement congestion control if it already has flow control to manage the sender's window? 2) Consider our recent reading [Chiu+89] Analysis of the Increase and Decrease algorithms for congestion avoidance in computer networks. How does this article showcases that TCP is fair?1. What is the difference between packet fragmentation (i.e., at network layer) and frame frag- mentation (i.e., at link layer) in terms of purpose? 2. Suppose that host A is connected to a router R1, R1 is connected to another router, R2, and R2 is connected to host B. Suppose that a TCP message that contains 800 bytes of data and 20 bytes of TCP header is passed to the IP function at host A for delivery to B. Show the Total length, DF, MF, and Fragment offset fields of the IP header in each packet transmitted over the three links. (Assume that link A-R1 can support a maximum frame size of 1024 bytes including a 14-byte frame header, link R1-R2 can support a maximum frame size of 512 bytes, including an 8-byte frame header, and link R2-B can support a maximum frame size of 432 bytes including a 12-byte frame header.) (*hint: the Fragment offset field is denominated by 8-bytes, not bytes) 3. What is the purpose of the path MTU discovery process (see textbook Figure 5-42) and why does…Explain the flow control of TCP protocol when the size of the sliding window will be 500 and the first client sends 201, 301, 402, 455, 330, and once acknowledgments will receive client will send 602. Make a flow control of these packets
- Below are the two questions, so make sure to answer each part carefully: Why does TCP implement congestion control if it already has flow control to manage the sender's window? Consider our recent reading [Chiu+89] Analysis of the Increase and Decrease algorithms for congestion avoidance in computer networks. How does this article showcases that TCP is fair?Suppose an extension of TCP allows window size much larger than 64 KB. Assume the extended TCP runs over a 100-Mbps link with RTT 100 ms, segment size is 1 KB, and receiving window size is 1 MB. How long does it take to send a 200 KB file?Is there a distinction between TCP's Full-duplex service and its Connection-oriented service?
- 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.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 session sends 50 packets per second over an Ethernet Local Area Network (LAN). Each packet consists 2280B (excluding the preamble and cyclic redundancy check (CRC)). Calculate the size of the headers, and hence the TCP payload data. What therefore is the TCP throughput of the session?
