In this question we consider a router that has to fragment an IP datagram to make it fit into a link with an MTU smaller than the datagram's original size. Assume the MTU for an outgoing link is 1,740 bytes and the datagram we want to transmit has a total length of 4,500 bytes (including 20 bytes of IP header!). What is the offset value for the second datagram sent by the router? Vour Answer:
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- Consider two hosts P and Q connected through a router R. The maximum transfer unit (MTU) value of the link between Pand Ris 1500 bytes, and between Rand Qis 820 bytes. A TCP segment of size 1400 bytes was transferred from P to Q through R, with IP identification value as 0x1234. Assume that the IP header size is 20 bytes. Further, the packet is allowed to be fragmented, i.e., Don't Fragment (DF) flag in the IP header is not set by P, Which of the following statements is/are correct? (a) Two fragments are created at R and the IP datagram size carrying the second fragment is 620 bytes. (b) If the second fragment is lost, P is required to resend the whole TCP segment. (c) TCP destination port can be determined by analysing only the second fragment. (d) If the second fragment is lost, R will resend the fragment with the IP identification value 0x1234Consider sending a 3500-byte datagram that has arrived at a router R₁ that needs to be sent over a link that has an MTU size of 1000 bytes to R2. Then it has to traverse a link with an MTU of 600 bytes. Let the identification number of the original datagram be 465. How many fragments are delivered at the destination? Show the parameters associated with each of these fragments.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? .
- The maximum packet size (Maximum Transmission Unit or MTU) of an IP packet including IPv4 header on 100 Mbps Ethernet is usually set at 1500 bytes. A typical IPv4 header consists of 20 bytes, and a UDP header consists of 8 bytes. If we split up a file of 20,000,000 bytes so we can send it as a series of UDP payloads, how many IP packets do we have to send in order to transfer the entire file? Enter an integer number without formatting (no commas). Answer:A 3200 bit long TCP message is transmitted to the IP layer and becomes a datagram after adding a 160 bit header. The following Internet is connected by two LANs through routers. But the data part of the longest data frame that the second LAN can transmit is only 1200 bits. Therefore, datagrams must be segmented in the router. How many bits of data does the second LAN transmit to its upper layer?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.
- The maximum packet size (Maximum Transmission Unit or MTU) of an IP packet including IPv4 header on 100 Mbps Ethernet is usually set at 1500 bytes. A typical IPv4 header consists of 20 bytes, and a UDP header consists of 8 bytes. If we split up a file of 20,000,000 bytes so we can send it as a series of UDP payloads, how many IP packets do we have to send in order to transfer the entire file? Enter an integer number without formatting (no commas). Answer: The MTU for IP packets on 100 Mbps Ethernet is typically set to 1500 bytes. A typical IPv4 header consists of 20 bytes, and a UDP header consists of 8 bytes. If we split up a file of 25,000,000 bytes so we can send it as a series of UDP payloads, how many bytes do we have to send at the network layer in order to transfer the entire file? Enter an integer number (no commas). Answer: You are sending a 27,000,000 byte file using UDP over IP over an Ethernet with MTU 1500 bytes. The Ethernet header is 14 bytes and the frame checksum is 4…Suppose Host A sends two consecutive TCP segments to Host B over a TCP connection. The sequence number of the first segment is 748, and the sequence number of the second one is 995. Determine the size of the payload carried by the first segment. Suppose that the first segment is lost but the second segment arrives at host B. What will be the acknowledgment number of the acknowledgment segment that Host B sends to Host A?Suppose Host A sends two consecutive TCP segments to Host B over a TCP connection. The sequence number of the first segment is 748, and the sequence number of the second one is 995. Determine the size of the payload carried by the first segment.Suppose that the first segment is lost but the second segment arrives at host B. What will be the acknowledgment number of the acknowledgment segment that Host B sends to Host A?
- Host A and B are communicating over a TCP connection, and Host B has already received from A all bytes up through byte 126. Suppose Host A then sends two segments to Host B backto-back. The first and second segments contain 80 and 40 bytes of data, respectively. In the first segment, the sequence number is 127, the source port number is 302, and the destination port number is 80. Host B sends an acknowledgment whenever it receives a segment from Host A. d. Suppose the two segments sent by A arrive in order at B. The first acknowledgment is lost and the second acknowledgment arrives after the first timeout interval. Draw a timing diagram, showing these segments and all other segments and acknowledgments sent. (Assume there is no additional packet loss.) For each segment in your figure, provide the sequence number and the number of bytes of data; for each acknowledgment that you add, provide the acknowledgment number.Host A and B are communicating over a TCP connection, and Host B has already received from A all bytes up through byte 126. Suppose Host A then sends two segments to Host B backto-back. The first and second segments contain 80 and 40 bytes of data, respectively. In the first segment, the sequence number is 127, the source port number is 302, and the destination port number is 80. Host B sends an acknowledgment whenever it receives a segment from Host A.a. In the second segment sent from Host A to B, what are the sequence number, source port number, and destination port number?b. If the first segment arrives before the second segment, in the acknowledgment of the first arriving segment, what is the acknowledgment number, the source port number, and the destination port number?c. If the second segment arrives before the first segment, in the acknowledgment of the first arriving segment, what is the acknowledgment number?d. Suppose the two segments sent by A arrive in order at B. The…For a given an IP v4 datagram, with header fields: HLEN - 5. Total length - 2220 and flag field (RDM) - 000. If this datagram is required to pass through a network of MTU - 1164 byte and need to be fragmented, choose the correct answer 1_For the given packet, the header size is 10 bytes 20 bytes 40 bytes 30bytes 2_The number of fragments to which the given packet is fragmented to pass through the network of MTU: 164: three Tour two Six Network Ineed answer after 30 min