hw2key

hw2key - EEE459 Homework 2 answer key 5 points/review...

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EEE459 Homework 2 answer key 5 points/review question: Ch. 2, 6, 9. 10 points/problem: CH. 1 #14–17 Problems 14. R = 10 6 ( bits/sec. ), m = 10 7 ( meters ), s = 2 . 5 × 10 8 ( meters/sec. ), L = 4 × 10 5 ( bits ) (a) Bandwidth delay product = R · t prop = R m s = 40 000 = 40 kb (b) The maximum number of bits on the link is equal to the number of bits that can be transmitted from host A in the time it takes a bit to cross the link. In other words, it’s the transmit rate times the propogation time, = R · t prop = 40 kb (c) Since the answers to (a) and (b) are equal, the bandwidth delay product is the maximum number of bits in the link at any given time. (d) The width of a bit in the link is equal to the length of the link divided by the maximum number of bits in the link, 10 000 km/ 40 kb = 250 meters/bit . (This is nearly three times the length of an American football field.) (e) A general expression for the “bitwidth” is given by m R · t prop = m R · ( m/s ) = s R (this question is a backhanded way of asking you to show that bit length is
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This note was uploaded on 11/01/2009 for the course EEE 459 taught by Professor Martinreisslein during the Spring '09 term at ASU.

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hw2key - EEE459 Homework 2 answer key 5 points/review...

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