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?

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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:

enter image description here

Let a frame be transmitted at time t0 and t be the time required for its transmission. If, any other station sends a frame between t0 and 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+t and 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:

Let 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.

Therefore, S = G x P(0), where P(0) is the probability of zero frames being transmitted during the vulnerable period, 2t.

According to Poisson's Distribution formula:

Probability of 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 k=0

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 e=2.718
S = 0.183958793

Hence through Poisson's Distribution we have derived that the maximum efficiency/throughput of Pure ALOHA protocol is approximately 18%

Graph below shows the relationship between S and G.

enter image description here


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