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Consider sending a 2400-byte datagram into a link that has an MTU of 700 bytes. Suppose the original datagram is stamped with the identification number 422. How many fragments are generated? What are the values in the various fields in the IP datagram(s) generated related to fragmentation?
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- R6Fragmentation 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?arrow_forwardMost 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?arrow_forwardConsider an IP packet with a length of 4,500bytes that includes a 20-byteIPv4 header ans 40-byte TCP header. The packet is forwarded to an IPv4 router that supports a Maximum Transmission Unit (MTU) of 600bytes. Assume that the length of the IP header in all the outgoing fragments of this packet is 20bytes. Assume that the fragmentation offset value stored in the first fragment is 0. The fragmentation offset value stored in the third fragment isarrow_forward
- 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, 370arrow_forward1. 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…arrow_forwardConsider the following scenario in which host A is sending a file to host B over a TCP connection. Assuming that the sequence number of the first data byte sent by A is 0 and every segment always includes 1000 bytes of data, excluding the TCP header. At some point of time, bytes up to 6400 have been written into the sender’s buffer. Bytes up to 4999 have been sent out but the segment which contains bytes 2000~2999 has not arrived at host B yet. At the receiver’s side, all bytes up to 3999 have been received except for bytes 2000~2999. Bytes up to 499 have been read from the buffer by the application. Assume that the maximum size of the sender’s buffer is large enough. Consider the sliding window algorithm in TCP and answer the following questions. 1) What are the values for LastByteAcked, LastByteSent, and LastByteWritten? 2) What are the values for LastByteRead, NextByteExpected, and LastByteRcvd? 3) Assuming that the maximum size of the receiver’s buffer is 4000 byte, what would…arrow_forward
- The maximum transmission unit on an Ethernet link is 4500 bytes. This means that the IP packets sent over Ethernet cannot be larger than 4500 bytes including the IP header. Suppose the application layer sends a 6500-byte message. The transport layer uses TCP with no options. The network layer is using IP version 4. Obviously, the IP layer will have to fragment the data. Provide the length of new datagrams (after fragmentation). Provide the Flag and offset of each of the new datagrams.arrow_forwardKeep in mind that packet reordering and reassembly occur throughout the transport phase of the TCP/IP protocol suite's operation. It is possible for a firewall to operate at a lower abstraction level, such as the Internet or data layer. Any information about a traffic stream that is out of order or has been deleted will be lost by a stateful inspection firewall that is using stateful inspection.arrow_forwardWe have seen that Internet TCP sockets treat the data being sent as a byte stream but UDP sockets recognize message boundaries. What are one advantage and one disadvantage of byte-oriented API versus having the API explicitly recognize and preserve application-defined message boundaries?arrow_forward
- Consider 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.arrow_forwardData collision is caused when many senders access the media at the same time; in order to avoid this, what protocol(s) are available, and at what layer(s) do they operate? Are these protocols able to provide a transport that is free of collisions? In the event that this is not the case, what other potential solutions are available for fixing the issue?arrow_forwardIn a bit-oriented link-layer protocol, the start and end of a frame are each marked by a flag, which is a sequence of exactly 6 consecutive 1 bits. Bit stuffing is used during the rest of the frame: after sending 5 consecutive 1 bits, a 0 is added. The bits below contain an example frame, with the leftmost bit received first. The content of the frame is a sequence of bytes, sent most significant bit first. Put the values of the first four bytes of the frame content, in order, in the four answer boxes below. You may enter the values in either decimal (e.g. 76, 123) or hexadecimal (e.g. 2a, f7). 00101011111010001111110011100111110000111101111101010110100010010011111101101100100 57−240−123−229arrow_forward
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