Answers without showing work used to arrive at the answer will receive zero points. Answer the following questions: 1. Suppose that a packet's payload consists of 5 eight-bit values (e.g., representing HELLO ASCII-encoded characters) shown below. 01001000 01000101 01001100 01001100 01001111 Compute the two-dimensional parity bits for each of the five rows and eight columns assuming even parity. Assume that the parity bit in the lower right corner is computed so that the parity of the row parity bits in the last row has even parity. 2. Consider the Cyclic Redundancy Check (CRC) algorithm discussed in Section 6.2.3 of the text. Suppose that the 4-bit generator (G) is 1101, that the data payload (D) is 10101110 and that r=3. What are the CRC bits (R) associated with the data payload of D = 10101110, given that r=3?
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Give short answers to the following questions. Answer Q1-Q4 by using the following information. A data frame contains a text of 8 characters "ethernet" encoded using ASCII characters. NOTE: the required ASCII values are: e = 1100101 t = 1110100 h = 1101000 r = 1110010 n = 1101110 Assuming this, determine: Q1. Codewords for the text "ethernet" using even parity. Q2. Two-dimensional parity check bits for the text "ethernet" using even parity. Q3. Codeword at the sender site for the dataword "h" using the divisor x^4 + x^2 + x + 1. Q4. Checksum at the sender site for the text "ethernet". Hint: Use hexadecimal equivalents of the characters. e = 0x65 t = 0x74 h = 0x68 r = 0x72 n = 0x6E Q5. Show the checking of codeword 10010001101 at the receiver site by using the divisor x^4 + x^2 + x + 1. Is there an error? Q6. Find the checksum at the receiver site if the four data items are received as 0x6674, 0x7865, 0x726F, 0x5584, 0x5A43. Is there an error? Q7. Find the codeword at the sender site for dataword 1110 using Hamming code C(7,4). Q8. The codeword 1100010 is received. Using the Hamming code C(7,4), find if there is an error in the codeword. If there is, what is the correct codeword and dataword sent?
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Using the prescribed syntax of an Internet Protocol packet, construct an IP Version 4 TCP/IP transmission packet using the following particulars: The packet is sent from IP address 192.168.4.111 (MAC address 4c:79:6e:86:b0:2d) to IP address 3.208.72.68 (MAC address 8c:3b:ad:f0:dc:e0) with the following fields (You don't have to convert these values to binary, but realize that the packet will be a binary encoding of these. A bit is a digital 0 or 1. 0x00 is hexadecimal notation, where each digit 0 - F represents 4 bits. 0 = 0000, 1 = 0001 up to 9 = 1001 where we run out of decimal digits so continue with A=1010 through F=1111. A byte = 8 bits = 1 ASCII character): IP Header Length is counted as the number of 32-bit words from Diff Serv through the end of the checksum. Differentiated Services: 0x00 (equals 00000000 in binary bits). Total Length is the number of bytes in the payload (data field). Identification: 0xe0f9. Flags: 010 (First bit is always 0, Second bit is set to not fragment packet, Third bit is fragments). Fragment Offset: 0x00. Time to Live: 0x80 (=128 hops). Protocol: 0x06 (TCP). Header Checksum: 0x0000. Source Port: 0xd19d (=53661 in decimal). Destination Port: 0x01bb (=443 in decimal). Sequence Number: 0x4066faf8. Acknowledgment Number: 0x6993afc2. Data Offset: 0x5. TCP Flags: 0x010 (5th flag bit = Acknowledgment). Window Size: 0x01fd (=509 in decimal). Checksum: 0x1146. Urgent Pointer: 0x0000.
A simple parity-check bit, which is normally added at the end of the word, cannot detect even numbers of errors. A better solution is to organize the characters in a table and create row and column parities. The bit in the row parity is sent with the byte, the column parity is sent as an extra byte. Assuming that errors occur as in the following figures, can we detect those errors? P10-11. A simple parity-check bit, which is normally added at the end of the word (changing a 7-bit ASCII character to a byte), cannot detect even numbers of errors. For example, two, four, six, or eight errors cannot be detected in this way. A better solution is to organize the characters in a table and create row and column parities. The bit in the row parity is sent with the byte, the column parity is sent as an extra byte (Figure 10.23). Figure 10.23 P10-11 Show how the following errors can be detected: a. An error at (R3, C3). b. Two errors at (R3, C4) and (R3, C6). c. Three errors at (R2, C4), (R2, C5), and (R3, C4). d. Four errors at (R1, C2), (R1, C6), (R3, C2), and (R3, C6).
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