Question

5. (10 points) Multiple Access Protocols Suppose nodes A and B are on the same 10 Mbps Ethernet segment and the propagation delay between the two nodes is 225 bit times. Suppose A and B send frames at the same time, the frames collide, and then A and B choose their own values of K in the CSMA/CD algorithm. Assuming active, no other nodes are can retransmissions 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- 225 bit times. They finish transmitting a jam signal at t-225+48-273 bit times. Suppose KA-1 and KB-0. Hints: Note that the random backoff component of scheduling delay is K* 512 bit times. Also note that the minimum Ethernet frame size is 512 bits. a) (5 points) At what time doesB schedule its retransmission (in bit times)? b) (5 points) At what time does A begin transmission (in bit times)?

Answer 1

given that

-> the ethernet segment bandwidth is 10Mbps

           i.e   per one second both nodes will send 10Mb of data to other end

-> propagation delay between 2 nodes is 225 bit times

        i.e the time taken to send data frame completely to other end is 225 bit times

given two stations A and B start sending frames at same time that is at t=0.

after sometime while they sending frames they colloid each other and we don't know at what time they colloid

whenever they colloid the collision signal will be sent to those nodes then they find out that their data is changed

the time taken to know that their data got corrupted is the propagation time because the time to send the information to other end is propagation time .

whenever collision is detected the nodes send the 48 bit jamming signal and wait for random amount of time to re transmits the frame

after jamming signal sent nodes A and B chooses different K value

   in the given question

       A chooses K value as 1

      B chooses K value as 0

which means B can re transmits data frames immediately

But A wait for 512 bit times to re transmit their data frames because scheduling delay is 512 bit times

for B

at t=0    B starts transmitting the data frame

at t=225                            B find out that collision is detected

at t=225+48=273            B sends the jamming signal

at t=273+225=498       B re transmits the data frame

for A

at t=0                                          A starts transmitting the data frame

at t=225                                     A find out that collision is detected

at t=225+48=273                     A sends the jamming signal

at t=273+512=785                  A starts re transmitting the data frames

at t=785+225=1010              A re transmits it's data frames

A)    B schedule it's re transmission at t=273 and completes it's re transmission at t=478

B)   A schedule it's re transmission at t=785 and completes it's re transmission at t=1010

so while A starts re transmitting the data B can completes it's re transmission

so no collision will occur wile re trasmission