The efficiency of an ALOHA system is that fraction of all transmitted frames which escape collisions that is which do not get caught in collisions.

Consider an infinite number of transmitters. Users of these transmitters are in one of two states, typing or idling. Initially all users are in the typing state. When a message is composed, the user sends it and stops typing – waiting for a response. The transmitter transmits the frame containing the message and checks the channel for success (no collision) or failure (collision). If successful, the user goes back to typing otherwise the user waits till a retransmission of the frame succeeds.

Frame Time: Frame Time is the amount of time to transmit a fixed length frame (Frame Time= frame length/bit rate of channel).

Assume that ∞ number of users generates new frames according to Poisson’s distribution with an average N frame per frame time. If N > 1 then the transmitters are generating frames at a rate that is greater than the channel capacity and therefore nearly every frame will suffer a collision. Assume that the probability of k transmission attempts per frame time (old transmissions and new transmissions combined) is also Poisson with mean G per frame time.

Then G ≥ N. At low load, (N ≅0), there will be a small # of collisions and therefore a correspondingly low number of retransmissions. Therefore G ≅ N. At high load we will have many collisions and therefore G > N.

After combining all these we can say that for all the loads the throughput is given by:

P0=S/G (Fraction of attempted frames that are transmitted successfully )

S = GP (throughput per frame time)

Let t be the time to send a frame – if another

user has sent a frame between t0 and t0 + t, the end of the frame will collide with the shaded one. Similarly on the right side of the shaded frame – any frame that began between t0 + t and t0 + 2t will have an overlap with the end of the shaded frame. Therefore the vulnerable period is 2t – or two frame times.

As per Poisson’s distribution, the Probability of Generating k frames during a given frame time is given by,

P[k] = [ Gk x e(-G) ]/k! (Poisson’s distribution)

Where G = Number of stations willing to transmit data.

The probability of generating zero frames i.e. k= 0 is 3.

Therefore,

P[k]=[ G0 x e(-G) ]/0!

P[k]= e-2G If an interval is two frame time long, the mean number of frames generated during that interval is 2G. The probability that another frame is transmitted during the Vulnerable period is,

P0= e-2G

Fig: Throughput of pure ALOHA:

The given above figure shows the relation between the offered traffic G and the throughput S. It shows that the maximum throughput occurs at G = 0.5

Therefore,

S= 0.5 x e-2x 0.5

= 0.5 x e-1

= 0.184

= 18.4%

Therefore Maximum efficiency of Pure ALOHA = 18.4%

Thus, S(max) = 0.184. And the best possible channel utilization is 18.4%. It means 82% of frames end up in collisions and are therefore lost.