Chapter18 - = 1 : 70 7. See Figure 18.1. SECTION 2 9. See...

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1 CHAPTER 18 Remote Login: TELNET Exercises 1. The pattern is: 11110011 00111100 11111111 11111111 Note : The last byte is duplicated because it is the same as IAC; it must be repeated to be interpreted as data. 3. To do the task in Exercise 1, we need to send: Client to Server : IAC DO BINARY (3 bytes) Server to Client : IAC WILL BINARY (3 bytes) Client to Server : 11110011 00111100 11111111 11111111 (4 bytes) If each transmission is encapsulated in a single TCP segment with 20 bytes of header, there will be 3 segments of 23, 23, and 24 bytes for the total of 70 bytes or 560 bits. 5. If we assume the useful bits are the 3 bytes of data from Exercise 1: 3 bytes of data / 216 bytes transmitted
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Unformatted text preview: = 1 : 70 7. See Figure 18.1. SECTION 2 9. See Figure 18.2. 11. See Figure 18.3. Figure 18.1 Exercise 7 Figure 18.2 Exercise 9 Client Server IAC WONT SGA IAC GA ECHO DONT IAC IAC WONT ECHO SGA DONT IAC GA: GO AHEAD SGA: SUPPRESS GO AHEAD Default mode Character mode Client Server IAC WILL LM ECHO DONT IAC IAC WONT ECHO LM DO IAC GA: GO AHEAD SGA: SUPPRESS GO AHEAD LM: LINE MODE Line mode Character mode SECTION 3 Figure 18.3 Exercise 11 Client Server IAC WONT LM IAC WONT SGA IAC GA ECHO DO IAC IAC WILL ECHO LM DONT IAC SGA DONT IAC GA: GO AHEAD SGA: SUPPRESS GO AHEAD LM: LINE MODE Line mode Default mode SECTION 4...
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Chapter18 - = 1 : 70 7. See Figure 18.1. SECTION 2 9. See...

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