c 4 pt In CSMACD after the fifth collision what is the probability that a node

# C 4 pt in csmacd after the fifth collision what is

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(c)(4 pt.) In CSMA/CD, after the fifth collision, what is the probability that a node chooses K=4? The result K=4 corresponds to a delay of how many seconds on a 10Mbps Ethernet? (d)(6 pt.) Suppose nodes A and B are on the same 10 Mbps Ethernet bus, and the propagation delay between the two nodes is 245 bit times (i.e. time to transmit 245 bits). Suppose A and B send frames at the same time, the frames collide, and then A and B choose different values of Kin the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmission from A and B collide? For our purposes, it suffices to work out the following example. Suppose A and B begin transmission at t=0 bit times. They both detect collisions at t=245 bit times. They finish transmitting a jam signal at 245+48=293 bit times. Suppose KA=0 and KB=1. At what time does B schedule its retransmission? At what time does A begin transmission? (Note: The nodes must wait for an idle channel after returning to Step 2 – see protocol. Also, the node must sense idle channel for 96 bit times before it transmits.) At what time does A’s signal reach B? Does B refrain from transmitting at its scheduled time? ririrpqppqpqpipT-+=--+=å¥=-11)1)(1(22
Chuah and Sabidur ECS152A/EEC173A 4 Time, Event 0 and begin transmission 245 and detect collision 293 and finish transmitting jam signal 293+245 = 538 538+96=634 's last bit arrives at ; detects an idle channel A starts transmitting 293+512 = 805 (805+96=901) 634+245=879 B returns to Step2 B must sense idle channel for 96 bit times before it transmits A’s transmission reaches B Because 's retransmission reaches before 's scheduled retransmission time (805+96), refrains from transmitting while retransmits. Thus and do not collide. Thus the factor 512 appearing in the exponential backoff algorithm is sufficiently large. t A B A B A B B A A A B B B A A B