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# sol6 - t trans&amp;gt 2 t prop Now 2 × t prop = 2.304...

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EE3015 Tutorial 6 (Solution) Question 1 y At t = 0, A transmits. y At t = 576, A would finish transmitting. (Note: minimum frame size is 72 bytes.) y In the worst case, B begins transmitting at t = 225. y At t = 225 + 225 = 450, B’s first bit arrives at A. y Because 450 < 576, A cannot finish transmitting before it detects that B transmitted. Question 2 a) It is important to note that in this question, a hub is used, instead of a switch. (why?) According to the formula, t prop / t trans = 0.2. t prop = 2 d / 1.8 × 108 t trans = 72 × 8 / (100 × 106) = 5.76 μ sec Solving for d , we obtain 103.7 m. b) To ensure collision can be detected, we must have
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Unformatted text preview: t trans &amp;gt; 2 t prop . Now 2 × t prop = 2.304 μ sec &amp;lt; t trans . Hence, collision can be detected. Question 3 Wait for 51,200 bit times. For 10 Mbps, the waiting time is 51200 bits / 10 Mbps = 5.12 msec. For 100 Mbps, the waiting time is 512 μ sec. Question 4 After the first event, there will be one entry in the switch table: (A, 1). The switch will forward the frame through its interfaces 2, 3, and 4. After the second event, there will be two entries in the switch table: (A, 1) and (D, 4). The switch will forward the incoming frame through its interface 1....
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