Question

Two nodes (A and B) are on the same network, and distanced by 500-meter. The two nodes are connected via 3 switches, each with a delay of 5 µs. Suppose that both of A and B transmit 1500 bytes at the same exact time and their frames collide. A chooses to retransmit immediately after jam signal is received, and B chooses to retransmit after 50 µs of receiving the jam signal. Assume the transmission speed is 10 Mbps, the channel is CSMA/CD with backoff intervals that are multiples of, the jam signal takes 10 µs to transmit, and the inter-frame gap is 9.6 µs. Find the time required to deliver A’s packet to B

Answer 1

Transmission delay = 1500 * 8/ 10 *10^6 = 1.2 millisec = 1200 micros.

And if the jamming signal is 10 micros, the propagation delay will be = 2* 10 micros= 20 micros.

Assume at time T0 A and B start transmission.

at time T1= 20/2 = 10 micros(Exactly at the middle of the link) there will be collision, and the jamming signal will be initiated

at time T2 = 10 + 10 = 20 micros A will be aware that there is a collision.

At time T3, A will wait for one propagation delay just to make sure that there is no last bit present in the link and the link is clear = 20+20 =40 micros.

At time T4, A will re-transmit the packet in the link = 40 + 1200 = 1240 micros the entire packet will be in the link

at T5 = 1240 + 20(propagation delay)+15(switching delay of 3)= 1275 micros A’s packet will reach B.

So total time required to deliver A’s packet is= 1275 micros