Pure ALOHA has a vulnerable time of
2T where T is the transmission time for a single frame. Pure aloha has a maximum throughput of
18% but how is it actually derived from Poisson's distribution?
Pure ALOHA is a random access protocol. It controls the multiple access over a shared medium. In this, the stations are allowed to transmit data when available.
In this there's a central controller station that governs the communication. Whenever a station wants to communicate data with another station, it first sends it to the central controller station, which then broadcasts it to the other stations.
In Pure ALOHA a station sends a frame and waits for an
ACK from the receiver to proceed with the sending of the next frames. If another station transmits a frame during the same period then collision occurs. If the sender doesn't receive
ACK signal then it re-transmits the frame/data using the Back-off Algorithm.
To avoid collision, no other station must transmit data during the same period.
Consider the following example:
Let a frame be transmitted at time
t be the time required for its transmission. If, any other station sends a frame between
t0+t then the end of the frame will collide with that earlier sent frame. Similarly, if any other station transmits a frame between the time interval
t0+2t again, it will result in a garbage frame due to collision with the reference frame. Hence, 2t is the vulnerable interval for the frame.
Maximum Throughput / Efficiency Derivation:
S be the maximum throughput of the channel and
G be the rate of arrival of new frames or the number of stations that wish to transmit the data.
The throughput of the system
S is equal to total arrival rate
G times the probability of successful transmission with no collision.
S = G x P(0), where
P(0) is the probability of zero frames being transmitted during the vulnerable period,
According to Poisson's Distribution formula:
k frames transmitted in time
t is given by
P(k) = G^k x e^(-G) / k!, k=0,1,2,3
For pure ALOHA,
P[k=0, during vulnerable time 2t] = (2G)^0 x e ^(-2G) / 0! = e^(-2G), where
Substituting the above in the equation
S = G x P(0)
S = G x e^(-2G)
From the figure attached above, it can be seen that
S will be at peak only when
G = 1/2
S = 1/2 x e^(-2*1/2)
S = 1/2 x e^(-1)
S = 1/(2 x e)
S = 1/2 x 2.718, where
S = 0.183958793
Hence through Poisson's Distribution we have derived that the maximum efficiency/throughput of Pure ALOHA protocol is approximately