EE450-Discussion9-Spring09

# EE450-Discussion9-Spring09 - Discussion#9 EE450 Sample...

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1 Discussion #9 EE450, 3/13/2009 Sample Problems -CSMA/CD -Token ring

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2 Problem#1: Description s Two nodes A and B on the same 10Mbps Ethernet Segment. s The propagation delay between them is equivalent to 225 bit times. s Both nodes start to transmit at the same time , t=0. s Upon detecting the collision, each node transmit a jamming signal equivalent to 48 bit times. s Node “A” will retransmit immediately after it senses the medium is idle (not after it detects a collision). s Station B will schedule its retransmission 51.2 microsec after it senses the medium is idle (not after it detects a collision).
3 Problem#1 : Questions? s Construct a timeline diagram to indicate all the events involved in the question. s At what time will node “A” start retransmission? s At what time will the frame from “A” be completely delivered to “B”? s Will there be a collision the second time? s What is the effective throughput for station “A” assuming that the frame length is the minimum allowed which is 512 bits?

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4 Problem#1: Solution A B Tprop=225 bit times BW=10Mbps A B Data[A] Data[B] T= 0 sec 1 bit time = 1 bit / 10 Mbps = 0.1 microsec Tprop = 225 bit times = 22.5 microsec Tprop=22.5 microsec
5 Collision A B T=11.25 microsec T= 11.25 + 22.5/2 =22.5 microsec T=11.25 microsec Collision is detected Collision is detected A transmitting data station that detects another signal while transmitting a frame, stops transmitting that frame, transmits a jam signal, and then waits for a random time interval (known as "backoff delay" and determined using the truncated binary exponential backoff algorithm) before trying to send that frame again. A B T=11.25 microsec T= 0 + 22.5/2 = 11.25 microsec T=11.25 microsec Data[A] Data[B] Collision

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6 Jamming Signal T= 27.3 + 22.5 = 49.8 microsec The last bit of B’s Jamming signal is received at A. A now senses the medium as idle so it starts retransmission. B schedules its retransmission for 51.2 microsec later, i.e. at T= 49.8 + 51.2 = 101 microsec A B T Jam =48 bit times= 4.8 microsec T= 22.5 +4.8 = 27.3 microsec Jam[B] B’s Jamming signal is transmitted. A B Jam[B] Tprop=22.5 microsec
7 A’s Retransmission T= 49.8 + 22.5 = 72.3 microsec The first bit of A’s retransmitted frame is received at B at T= 72.3 microsec. A B Data[A] Tprop=22.5 microsec T= 72.3 + 51.2 = 123.5 microsec A B Data[A] Tprop=22.5 microsec Frame size =512 bits , Frame transmission time= 512/10 Mbps=51.2 microsec The last bit of A’s retransmitted frame is received at B at T= 123.5 microsec.

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A second Collision? • There won’t be a second collision, because at T=101 microsec, B senses the medium to be busy so it doesn’t transmit and reschedules its retransmission. • Throughput for A :
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## This note was uploaded on 03/26/2009 for the course EE 450 taught by Professor Zahid during the Spring '06 term at USC.

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EE450-Discussion9-Spring09 - Discussion#9 EE450 Sample...

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