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ps4sol_ee550_fall08

# ps4sol_ee550_fall08 - UNIVERSITY OF SOUTHERN CALIFORNIA...

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UNIVERSITY OF SOUTHERN CALIFORNIA, FALL 2008 1 EE 550: Problem Set # 4 Solution I. S TOP AND W AIT E FFICIENCY a) Remaining time = n · RTT if, starting at the time the sender gets the “ACK” with an error, the sender gets the first correct “ACK” after n retransmissions, which happens with probability q n - 1 (1 - q ) . Obviously the remaining time is RTT times a geometric random variable with parameter 1 - q . Thus the expected remaining time equals to RTT/ (1 - q ) . b) Define X to be the time between first deliveries of two consecutive packets. The expectation of X equals to E { X } = RTT + p E { X } + (1 - p ) q RTT 1 - q + (1 - p )(1 - q ) · 0 On the right hand side, the first term is the time needed for one transmission. The second term represents, with prob. p , there are errors in the forward transmission; and after the first round trip time, the remaining time until the first delivery of the next packet is still X . The third term represents, with prob. (1 - p ) q , after the first round trip time during which a packets has been successfully delivered but there are errors in the ACK, by part a) the expected remaining time until the first delivery of the next packet is RTT/ (1 - q ) . The last term represents, with prob. (1-p)(1-q), a packet is successfully received and acked during the first round trip time. The above equation yields E { X } = 1 - pq (1 - p )(1 - q ) RTT, and the throughput equals to 1 E { X } = (1 - p )(1 - q ) (1 - pq ) RTT II. S TOP AND W AIT WITH E NERGY C ONSIDERATION a) The throughput equals to 1 - p RTT = 1 - 1 / 2 16 = 1 32 b) Let X be the time taken by a particular packet. Then E { X } = E { X | first transmission is loss-free, error-free } ( 3 4 )( 3 4 ) + E { X | error occurs, no loss } ( 1 4 )( 3 4 ) + E { X | first transmission is lost } ( 1 4 ) = 9 16 RTT + 3 16 ( RTT + E { X } ) + 1 4 ( TO + E { X } ) , where RTT = 16 and TO = 20 yield E { X } = 272 / 9 , and throughput equals to 9 / 272 . c) The throughput is low because the time between retransmissions is at least RTT . The throughput is maximized by setting TO = 1 , i.e., retransmit a packet every timeslot without waiting for ACK/NACKs until receiving the first ACK of the packet. Let X be the time until receiving the first ACK of a packet. Then X = RTT w.p. ( 3 4 )( 3 4 ) = 9 16 , q RTT + 1 w.p. (1 - q ) q RTT + 2 w.p. (1 - q ) 2 q . . . . . . , which yields E { X } = RTT + (1 - q ) /q = 151 / 9 . Thus the throughput equals to 9 / 151 .

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