Consider two active nodes, A and B, on an ethernet with lots of frames to send.
ID: 3589078 • Letter: C
Question
Consider two active nodes, A and B, on an ethernet with lots of frames to send. Each node tries to send its first frame and they collide, send their jamming signal and calculate a back-off time in the interval [0,1] slots. Let A be the node that chooses a back-off time 0 and B chooses a back-off time of 1 slot (if both pick the same slot, they collide again). So A sends its first frame successfully and B decides to wait 1 slot. However during the next time slot when B tries to send its first frame, A also tries to send its second frame, so they collide again. Answer the following questions
a) If A wins that round, do you think that B has a fair chance to win the next round? Explain why
Explanation / Answer
In the starting 1st time slot(t=0), collision number of A and B both is 1.
Now A has chosen a backoff time 0, this means A will not need to wait at all. It will keep transmitting. On the other hand, B has chosen a back off time as 0. This means during second time slot(t=1) A will send and B will wait. Thus A will be able to transmit successfully.
During 3rd time slot(t=2), B will have waited for 1 time and will send its frame. A also will send its frame leading to collision. Now A has collision number 1 as its last transmission was successful and B has collision number 2. They both need to choose a back off time between 0 and 2n-1 where n gives their collision number.
So A will chose between 0 and 1, while B will choose between 0 and 3 for their respective backoff times. There are 8 possibilities of backoff time TA and TB
Case 1: TA=0, TB=0. This will make sure of collision which can go either way
Case 2: TA=0, TB=1. This will again let A win
Case 3: TA=0, TB=2. A wins here as well
Case 4: TA=0, TB=3. A wins here as well
Case 5: TA=1, TB=0. In this case B will win
Case 6: TA=1, TB=1. This will make sure of collision which can go either way
Case 7: TA=1, TB=2. A wins here
Case 8: TA=1, TB=3. A wins here
So in 8 cases, A wins 5 times while B wins only once and rest 2 are collisions. So clearly A is more likely to win.
This does not give B a fair chance of winning.