EE450-Discussion9-Fall2010

EE450-Discussion9-Fall2010 - Discussion #9 EE450,...

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1 Discussion #9 EE450, 10/20/2010 Sample Problems   -CSMA/CD   -Token ring
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2 Problem#1: Description Two nodes A and B on the same 10Mbps Ethernet Segment. The propagation delay between them is equivalent to 225 bit times. Both nodes start to transmit at the same time , t=0. Upon detecting the collision, each node transmit a jamming signal equivalent to 48 bit times. Node “A” will retransmit immediately after it senses the medium is idle (not after it detects a collision). Station B will schedule its retransmission 51.2 microsec after it senses the medium is idle (not after it detects a collision).
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3 Problem#1 : Questions? Construct a timeline diagram to indicate all the events involved in the question. At what time will node “A” start retransmission? At what time will the frame from “A” be completely delivered to “B”? Will there be a collision the second time? 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
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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
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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.
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This note was uploaded on 06/25/2011 for the course EE 450 taught by Professor Zahid during the Fall '06 term at USC.

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EE450-Discussion9-Fall2010 - Discussion #9 EE450,...

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