Given that you are to transfer a total datagram of size 28939 bytes over a network that has an MTU of 3038 bytes. The given header size is 38 bytes. (a) Calculate the total number of packets required to send this data. (b) What is the data size of the last packet? (c) What would be the header size of the 8th packet and the offset value of the 4th packet?
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Given that you are to transfer a total datagram of size 28939 bytes over a network that has an MTU of 3038 bytes. The given header size is 38 bytes.
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- "An application message is segmented into 20 packets, each of which is 2500-byte in size. Calculate the end-to-end delay in milliseconds (ms) for the message when the bit rate for any node is 10 Mbps, the optical path total distance is 1200-km, and the queuing and processing delays are 1-ms and 3-ms, respectively. There are 2 intermediate routers in the path, each of which uses the store and-forward approach. Round your answer to the nearest integer. Question 16 of 37Q1.Suppose a client, say C, has established a TCP connection with a server, say S. After establishing the connection, the sender sends two TCP messages with data back-to-back ( the first TCP message with sequence number 32 and the second TCP message with sequence number 100). a. What is the sequence number chosen by the client? What could be the reason for choosing this value of sequence number?b. What is the data size of first and second TCP data messages?c. Suppose the server receives the first and second TCP messages with data with the difference of 100 milliseconds delay. What could be the ACK number for both the TCP data messages from the server side as per the TCP RFC.Assume now that each packet includes not only payload data, but also a header portion, which contains information like sender and receiver addresses. Furthermore, assume that the maximum payload size of a packet is 1480 bytes, and that the packet header is always 40 bytes long. What is now the maximum packet size (given in bytes)?
- A 4000 octet user data is to be transmitted over a network which supports a maximum user data size of 536 octets. Assuming the header in each IP data gram requires 20 octets, derive the number of datagrams (fragments) required and the contents of the following fields: Identification Total length Fragment offset More Fragments flagSelect the correct option: Given, you are to transfer a total datagram of size 28939 Bytes over a network that has an MTU of 3038 Bytes. The given header size is 38 Bytes. (a) Calculate the total number of packets required to send this data. 1) 9 2) 10 3) 11 4) 8 (b) What is the data size of the last packet? 1) 1900 2) 1938 3) 1939 4) 1901 5) 1940 (c) What would be the header size of the 8th packet? 1) 0 2) 8 3) 38 4) 39 5) depends on data sizeA file of size 20 KiloBytes is transmitted to a destination over a 10 Megabit/s network link (1 Mega = 10^6, 1 Kilo = 10^3, 1 byte = 8 bits). The propagation delay to the destination is 40 milliseconds (1 milli = 10^−3). Assume the queueing delay encountered by packets of the file is negligible. What is the total delay (in milliseconds) for the file to get to its destination?
- 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 ReceiverAssume that there is a shared connection (for example, an Ethernet bus) with a speed of 1 Mbps. How much time (in microseconds) is required to send a frame with a length of 1000 bits over this link?Given a M/M/1 queue, which has packets arriving at a rate 2 packets/sec, an output link rate R bps, and an average packet size of 1500 bytes, write down: the expression for T, the average delay for a packet (i.e., from arrival till completing transmission at the output). b. the expression for the output utilization c. the expression for average number of packets in the queue d. the expression for average number of packets in the output NIC
- Host A and B are communicating over a TCP connection, and Host B has already received all bytes up through byte 100 (including Byte 100) from A. Suppose Host A then sends two packets to Host B back-to-back. The first and second packets contain 40 and 80 bytes of data, respectively. Host B sends an acknowledgment whenever it receives a packet from Host A. a. What is the sequence number of the first packet sent from Host A to Host B? What is the sequence number of the second packet sent from Host A to Host B? b. If the second packet arrives before the first packet, in the acknowledgment of the first arriving packet, what is the acknowledgment number? .Consider a half-duplex point-to-point link using a stop-and-wait scheme, in which a series of messages is sent, with each message segmented into a number of frames. Ignore errors and frame overhead. A. What is the effect on line utilization of increasing the message size so that fewer messages will be required? Other factors remain constant. B. What is the effect on line utilization of increasing the number of frames for a constant message size? C. What is the effect on line utilization of increasing frame size?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.