Consider a simplified time slotted MAC protocol, where each host always has data to send and transmits with probability p = 0.2 in every slot. There is no backoff and one frame can be transmitted in one slot. If more than one host transmits in the same slot, then the transmissions are unsuccessful due to collision. What is the maximum number of hosts which this protocol can support, if each host has to be provided a minimum through put of 0.16 frames per time slot?
(A) 1
(B) 2
(C) 3
(D) 4
Answer: (B)
Explanation: Here we are talking about slotted MAC protocol where when one station transmits, no other station can transmit.Now, assume that the probability of transmitting data by a single station be p and n be the number of stations that can transmit.
Since, when one station is transmitting no other station can transmit. So there will be n-1 stations that are silent with the probability of 1-p.
Now, for one station a minimum throughput of 0.16 frames per slot of time is to be given. So, for n stations the throughput is given by 0.16*n, and every station always has some data to transmit with the probability of p = 0.2 in every time slot.
Hence,
0.16 * n = n * 0.2 * (0.8)^(n -1)
0.8 = 0.8 ^ (n-1)
Comparing both the sides we get
1 = n-1
This means n = 2
This explanation has been contributed by Namita Singh.
Quiz of this Question