An IPv4 datagram is fragmented into three fragments. The second fragment carries bytes 2000 to 5000. What is the fragmentation offset value for the second fragment?
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An IPv4 datagram is fragmented into three fragments. The second fragment carries bytes 2000 to 5000. What is the fragmentation offset value for the second fragment?
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- Most IP datagram reassembly algorithms have a timer to avoid having a lost fragment tie up reassembly buffers forever. Suppose that a datagram is fragmented into four fragments. The first three fragments arrive, but the last one is delayed. Eventually, the timer goes off and the three fragments in the receiver's memory are discarded. A little later, the last fragment stumbles in. What should be done with it?Suppose Alice and Bob are going to communicate using AES in CBC mode. Unfortunately Alice's message length (in bytes) is not a multiple of 16. Suppose the last block of her message is just a single zero byte. How can she pad out the last block so that she can use CBC mode? Since this needs to be a reversible operation, how does Bob recognize the padding and remove it?Maximum Transmission Unit (MTU): Longest prefix matching: IP Fragmentation: Consider sending a 2,400-byte datagram into a link with a MTU of 900 bytes. Suppose the original datagram has the identification number 422. (a) How many fragments are generated? (b) For each fragment, what is its size, what is the value of its identification, fragment offset, and fragment flag? Length ID Fragment Flag Offset
- 2) In an IPV4 packet, the value of the HLEN field is 1100 in binary. How many bytes of options field is being carried by this packet? a) 28 byte b) 20 byte c) 8 byte d) 40 byteIn a TCP data packet, if the sequence number has the value 001111, ACK number has the value 010110 and the urgent pointer has the value 001100 assuming that URG flag is set. What is the last urgent byte in the data section of the packet? All numbers are in binary.If there is wide variation in the TCP segment size , it is natural to organize the buffers as a pool of fixed-sized buffers, with one segment per buffer TRUE OR FALSE
- In IPV4, consider sending a 4,000 byte datagram (20 bytes of IP header) into a link that has an MTU of 1,500 bytes. The datagram will be allocated to fragments, and the offset value of the third segment is 3, 370 3, 185 4, 185 4, 370A 4480-octet datagram is to be transmitted and needs to be fragmented because it will pass through an Ethernet with a maximum payload of 1500 octets. Show the Total Length, More Flag, and Fragment Offset values in each of the resulting fragments.UDP and TCP use one’s complement for the checksum. Suppose you have the flowing four 8-bit data in a datagram: 10011011, 01011100, 11001101. 10101001 a) What is the one’s complement of the sum of the four? Show your work. b) Is it possible that a 1-bit error will go undetected? Is it possible that a 2-bit error will go undetected?
- Consider an IP packet with a data length of 4400 bytes. TCP header is of 40 bytes while that of IPv4 header is 20 bytes. The packet is forwarded to an IPv4 router that supports MTU of 900 bytes. Length of the IP header for outgoing fragments is 20 bytes. Assume that the fragment offset value stored in the first fragment is 100. The fragmentation offset value of the penultimate fragment isR6Fragmentation of an IP datagram takes place if its size is larger than the MTU of the subnet over which the datagram will be routed. Most IP datagram reassembly algorithms have a timer to avoid having a lost fragment tie up reassembly buffers forever. Suppose a datagram is fragmented into four fragments. The first three fragments arrive, but the last one is delayed. Eventually the timer goes off and the three fragments in the receiver's memory are discarded. A little later, the last fragment stumbles in. What should be done with it?Which of the following instructions falls in the category of data transfer instructions? JUMP LOAD ROTATE AND