Assignment 5 Solutions

Assignment 5 Solutions - ECSE 414 – Intro. to Telecom....

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Unformatted text preview: ECSE 414 – Intro. to Telecom. Networks Assignment #5 - Fall 2009 ECSE 414 - Homework Assignment #5 – Link Layer Solutions Problem 1 (6 marks, 2 for each part) a. Since initially all switching tables are empty, when A sends the message to H, all nodes will receive this message because it will be broadcast on all links. Similarly, when H sends a message to C, no switches will have entries in their tables for H, so all nodes in the LAN will also receive this message as it will be broadcast. However, when F sends a message to A, because the switches received a message from A previously, they will have entries in their table for A and will be able to route directly to A. Thus, only S2, S4, S1, and A receive the message from F to A. S1’s table: MAC Interface A 1 H 4 F 4 A H F A H 4 4 3 4 2 (NOTE: S3 doesn’t receive the message from F) b. S2’s table S3’s table S4’s table c. A 1 H 3 F 2 In order for S3 to add an entry to its table for D, D must transmit a message to any host except for A or F. If it transmits to A or F then the message will not reach S3. If D transmits to H, it will pass through S3 on the way, so this is ok. If D transmits to any other host, then the message will be broadcast to all hosts, so S3 will also add an entry for D. 25/11/09 M. Rabbat ECSE 414 – Intro. to Telecom. Networks Assignment #5 - Fall 2009 Problem 2 (6 marks, 4 for part a and 2 for part b) a. For G=11001, we have r=4 bits. 1101001001010111  ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ 11001| 10110110010110010000 11001  ­ ­ ­ ­ ­ 011111 11001  ­ ­ ­ ­ ­ 001101 00000  ­ ­ ­ ­ ­ 011010 11001  ­ ­ ­ ­ ­ 000110 00000  ­ ­ ­ ­ ­ 001101 00000  ­ ­ ­ ­ ­ 011010 11001  ­ ­ ­ ­ ­ 000111 00000  ­ ­ ­ ­ ­ 001111 00000  ­ ­ ­ ­ ­ 011110 11001  ­ ­ ­ ­ ­ 001110 00000  ­ ­ ­ ­ ­ 011101 11001  ­ ­ ­ ­ ­ 001000 00000  ­ ­ ­ ­ ­ 010000 11001  ­ ­ ­ ­ ­ 010010 11001  ­ ­ ­ ­ ­ 10110 11001  ­ ­ ­ ­ ­ 1111 = R Thus, the transmitted message is 1011 0110 0101 1001 1111. b. The four bit errors are guaranteed to be detected if they occur in a burst (i.e., all four in a row). This is because CRC codes are guaranteed to detect all bursts of r bit errors. If they do not occur in a burst we cannot be guaranteed whether or not the error will be detected, as it 25/11/09 M. Rabbat ECSE 414 – Intro. to Telecom. Networks Assignment #5 - Fall 2009 depends on the actual data transmitted, the generator used, and the locations of the errors. CRC codes can only be used to detect errors, not correct them, so the four bit errors cannot be corrected. Problem 3 (K&R Ch 5, P13) (3 marks) The length of a polling round is . The number of bits transmitted in a polling round is . The maximum throughput therefore is 25/11/09 M. Rabbat ...
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This note was uploaded on 05/09/2011 for the course ECSE 414 taught by Professor Rabbat during the Fall '10 term at McGill.

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