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Which of the following statements is/are correct?I. The fragmentation is applicable for data in the datagram but not for header.
II. Reassembly of the fragments must be performed at the destination because, the intermediate
networks may have different maximum transmission unit (MTU) sizes.
III. In the IP header, the time-to-live (TTL) field is invariable.
(A) I and II
(B) I and III
(C) II and III
(D) All are correct
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- Which of the following statements is/are correct?I. The fragmentation is applicable for data in the datagram but not for header.II. Reassembly of the fragments must be performed at the destination because,theIntermediate networks may have different maximum transmission unit (MTU) sizes.III. In the IP header, the time-to-live (TTL) field is invariable. (A) I and II(B) I and III(c) II and III(D) All are correctSelect the FALSE statements as shown below: i) Two distinct Web pages (e.g., www.uthm.edu.my/research.html and www.uthm.edu.my/students.html) can be sent over the same persistent connection. ii) With non-persistent connections between browser and origin server, it is possible for a single TCP segment to carry two distinct HTTP request messages. iii) The Date: header in the HTTP response message indicates when the object in the response was last modified. iv) HTTP response messages never have an empty message body. a) i, ii, and iii b) i, ii, and iv c) i, iii, and iv d) ii, iii, and ivASN.1 is an ISO standard used in many internet‑related protocols, especially those in the area of network management. In ASN.1, BER (Basic Encoding Rule) uses a so‑called TLV (Type, Length, Value) approach to encoding data for transmission. According to this BER encoding rule, what would be the encoding of {weight, 178} {lastname, “Philips”}?
- Assume the maximum transmission unit (or MTU) of an IP packet on 100 Mbps Ethernet is set at 1500 bytes. Also, assume we are sending our file using IPv6 at the Network layer and UDP at the Transport layer. A typical IPv6 header consists of 40 bytes, a UDP header consists of 8 bytes. Answer the following three questions based on the information provided above. For all answers, enter a decimal integer value without formatting (no commas). 1. How many packets do we have to send in order to transfer a file of 24KB over 100 Mbps Ethernet? packets 2. How many bytes do we have to send at the network layer in order to transfer the above file entirely? bytes 3. How many bytes do we have to send at the data link layer in order to transfer the above file entirely? Assume that the Ethernet header is 14 bytes and the frame checksum is 4 bytes. bytesConsider sending a 2000-byte datagram into a link with a MTU of 980 bytes. Suppose the original datagram has the identification number 227. How many fragments are generated? For each fragment, what is its size, what is the value of its identification number, fragment offset, and fragment flag?5. Expand the information on the Transmission Control Protocol for this packet in the Wireshark “Details of selected packet” window (see Figure 3 in the lab writeup) so you can see the fields in the TCP segment carrying the HTTP message. What is the destination port number (the number following “Dest Port:” for the TCP segment containing the HTTP request) to which this HTTP request is being sent?
- The message to be sent from source to destination is 4000B. The link MTU's from source to destination are 1500, 1000, and 750. If we DO NOT use Path MTU, what are the respective fragment sizes to be reassembled? If we DO use Path MTU, what are the respective fragment sizes to be reassembled? Remember, assume that the L3 and L4 headers have length=0.assume that the link layer uses the flag bytes with byte stuffing approach for framing, with the following details: The beginning of a data frame is indicated by the special flag bytes: DLESTX The end of a data frame is indicated by the special flag bytes: DLEETX The special escape bytes DLE are used for escaping accidental occurrences of either the flag bytes or the escape bytes within the data. For simplicity, assume that no other header/trailer information is added to the data. 1. The following byte stream represents data that needs to be framed by the link layer on the sender’s side. Derive the resultant byte stream by adding necessary flag bytes and performing byte stuffing. ASTXDLELEFTDAOY I have this answer but it is wrong DLESTXASTXDLEDLELEFTDAODLEETXassume that the link layer uses the flag bytes with byte stuffing approach for framing, with the following details: The beginning of a data frame is indicated by the special flag bytes: DLESTX The end of a data frame is indicated by the special flag bytes: DLEETX The special escape bytes DLE are used for escaping accidental occurrences of either the flag bytes or the escape bytes within the data. For simplicity, assume that no other header/trailer information is added to the data. 1. The following byte stream represents data that needs to be framed by the link layer on the sender’s side. Derive the resultant byte stream by adding necessary flag bytes and performing byte stuffing. PARSDLEETXZKPUM I have this answer but it is wrong DLESTXPARDLEETXZKPUMDLEETX
- assume that the link layer uses the flag bytes with byte stuffing approach for framing, with the following details: The beginning of a data frame is indicated by the special flag bytes: DLESTX The end of a data frame is indicated by the special flag bytes: DLEETX The special escape bytes DLE are used for escaping accidental occurrences of either the flag bytes or the escape bytes within the data. For simplicity, assume that no other header/trailer information is added to the data. 1. The following byte stream represents data that needs to be framed by the link layer on the sender’s side. Derive the resultant byte stream by adding necessary flag bytes and performing byte stuffing. PZDLEAFRGSTXGYK I have this answer but it is wrong DLESTXPZDLEAFRGSTXGYKDLEETXConsider 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…2- If we have an IP packet of size 4000 bytes will be fragmented equally in two (equally )fragments a. the values of M bit of the flag field first fragment b. the values of M bit of the flag field Second fragment C. the values of Offset field on the first fragment. d. the values of Offset field on the Second fragment. e. the Total length field on the first fragment (if max options) f. the Total length field on the first fragment (if no options)