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In slotted ALOHA, the maximum efficiency is 36.8% when only one station transmits in one time slot . My question is, if only one station is transmitting in one time slot, then there will be no collisions and since we are talking about maximum efficiency, all time slots will be utilized. If propagation time is negligible, then shouldn't the efficiency be nearly 100%?

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The reason it cannot be 100% utilised is that there is access contention. The stations are not all coordinating with each other (through a side channel) to use the slots one after another.

You say "if only one station is transmitting in one time slot then there will be no collisions .." No, there will be collisions. Stations will try, and there will be many collisions since they are not coordinating with other through some side channel. Only in the case that one station tries to transmit in a slot and succeeds, then that is when it is successful.

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Generally, only one station can - successfully - transmit in one time slot. Concurrent transmission attempts cause collisions which in turn cause delays.

Statistically, when there is access contention, the probability is 1/e ≈ 36.8% that a frame is successfully transmitted in any given time slot, so that's also the efficiency of the channel. With no contention and just a single transmission attempt per slot, every slot can be used and the efficiency would be 100%.

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  • You're saying that throughput is 36.8% when there is access contention . And if only one frame is generated per time slot(no access contention) then throughput will be 100%(Max). But the text I'm following says something different : "The throughput for slotted ALOHA is S = G*e^-2G. The maximum throughput Smax=0.368 when G =1." Here G is average frames generated by system during one time slot . – Bisma May 2 '20 at 11:51

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