1 Discussion #9 EE450 Sample Problems - CSMA/CD2 Problem#1: Description n Two nodes A and B on the same 10Mbps Ethernet Segment. n The propagation delay between them is equivalent to 225 bit times (225*bit duration). n Both nodes start to transmit at the same time , t=0. n Upon detecting the collision, each node transmit a jamming signal equivalent to 48 bit times. n Node “A” will retransmit immediately after it senses the medium is idle (not after it detects a collision). n 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? n Construct a timeline diagram to indicate all the events involved in the question. n At what time will node “A” start retransmission? n At what time will the frame from “A” be completely delivered to “B”? n Will there be a collision the second time? n What is the effective throughput for station “A” assuming that the frame length is the minimum allowed which is 512 bits?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 microsec5 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] Collision6 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 TJam =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 microsec7 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.8 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 : Frame size / Transfer time = 512 bits/123.5 microsec = 4.15 Mbps9 Problem#2: Description n Consider a 100Mbps 100BaseT Ethernet. n Assume propagation speed is 1.8 ×108 m/sec n Assume a frame length of 72 bytes and no repeaters. n In order to have an efficiency of 0.5, what should be the maximum distance between the nodes? n Does this maximum distance ensure that a transmitting node A will be able to detect whether any other node transmitted while A was transmitting? Why or why not? n How does your maximum distance compare with the actual 100 Mbps standard?10 Problem#2: Solution n We want 1/(1 + 5a) = 0.5 or n Equivalently a = 0.2 = tprop / ttrans (I) n tprop = d /(1.8 ×108 ) m/sec (II) n ttrans = Frame size/BW= (576 bits ) /(108 bits/sec ) = 5.76µ sec (III) n Substitute (II) & (III) in (I) n Solve for d and we obtain d = 207 meters.Problem#2: Solution n For the 100 Mbps Ethernet standard, the maximum distance between two hosts is 200 m. n For transmitting station A to detect whether any other station transmitted during A 's interval, ttrans must be greater than 2tprop. n Therefore 2×207 m/1.8 ×108 m/sec = 2.30µ sec. n Because 5.76 > 2.30, A will detect B 's signal before the end of its transmission.