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450_HW_4_solutions

450_HW_4_solutions - EE 450 Homework 4 Summer 09 Nazarian...

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EE 450 Homework 4 Summer ’09 ● Nazarian Name: _________________________________________ Lecture 8:00 Assigned: Friday July 10, 2009 Score: ________ Due: Thursday July 16, 2009, 8am 1)
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2. We have t 1 = 0 and t2 = 3 μ s a. t 3 t 1 = (2000 m) / (2 × 10 8 m/s) =10 μ s t3 = 10 μ s + t1 = 10 μ s b. t 4 t 2 = (2000 m) / (2 × 10 8 m/s) =10 μ s t4 = 10 μ s + t2 = 13 μ s c. d trans-frame (A) = t 4 t 1 = 13 0 = 13 μ s BitsA = 100 Mbps × 13 μ s = 1300 bits d. d trans-frame (C) = t3 t2 = 10 3 = 07 μ s BitsC = 100 Mbps × 07 μ s = 700 bits 3. Bit duration is 1/100Mbps =10nsec However the question wants us to measure the times in terms of number of bit durations not in seconds. a) A needs to have enough number of bits such that d trans-frame equal to or greater than 2d prop =10000bit durations, this means that A should have at least 10000 bits, because each bit takes one bit duration to be transmitted, so we need at least 10000 bits in A’s frame to make the transmission time 10000 times higher b) This part is tricky, only because it is too simple! It has nothing to do with Ethernet,
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