Discussion 9 EE450 Sample Problems CSMA CD 1 Problem 1 Description n n n n n n Two nodes A and B on the same 10Mbps Ethernet Segment The propagation delay between them is equivalent to 225 bit times 225 bit duration 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 2 Problem 1 Questions n n n n n 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 3 Problem 1 Solution BW 10Mbps A B Tprop 225 bit times 1 bit time 1 bit 10 Mbps 0 1 microsec Tprop 225 bit times 22 5 microsec T 0 sec A Data A Data B B Tprop 22 5 microsec 4 Collision Collision T 0 22 5 2 11 25 microsec A Data A T 11 25 microsec B T 11 25 microsec Collision is detected Collision is detected T 11 25 22 5 2 22 5 microsec Data B A B T 11 25 microsec T 11 25 microsec 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 5 Jamming Signal T 22 5 4 8 27 3 microsec Jam B A B TJam 48 bit times 4 8 microsec B s Jamming signal is transmitted Jam B T 27 3 22 5 49 8 microsec A B Tprop 22 5 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 6 A s Retransmission T 49 8 22 5 72 3 microsec A Data A B Tprop 22 5 microsec The first bit of A s retransmitted frame is received at B at T 72 3 microsec T 72 3 51 2 123 5 microsec Data A B 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 7 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 Mbps 8 Problem 2 Description n n n Consider a 100Mbps 100BaseT Ethernet Assume propagation speed is 1 8 108 m sec Assume a frame length of 72 bytes and no repeaters n n n In order to have an efficiency of 0 5 what should be the maximum distance between the nodes 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 How does your maximum distance compare with the actual 100 Mbps standard 9 Problem 2 Solution n n n n n n We want 1 1 5a 0 5 or Equivalently a 0 2 tprop ttrans I tprop d 1 8 108 m sec II ttrans Frame size BW 576 bits 108 bits sec 5 76 sec III Substitute II III in I Solve for d and we obtain d 207 meters 10 Problem 2 Solution n n n n For the 100 Mbps Ethernet standard the maximum distance between two hosts is 200 m For transmitting station A to detect whether any other station transmitted during A s interval ttrans must be greater than 2tprop Therefore 2 207 m 1 8 108 m sec 2 30 sec Because 5 76 2 30 A will detect B s signal before the end of its transmission 11