My logic :-
there are n slots .
We have to find expected no of pass before getting an empty slots where no station is transmitting.
Here in slotted aloha ,Probability of successful transmission =$P_{succ}=e^{-G}$.
Probability of collision=$1-e^{-G}$.[ each station can transmit independently in beginning of the slots.
So ,
Probability of successful transmission in first pass=$e^{-G}$
Probability of successful transmission in second pass= $(1-e^{-G})e^{-G}$
Probability of successful transmission in third pass=$(1-e^{-G})(1-e^{-G})e^{-G}=(1-e^{-G})^{2}e^{-G}$
Probability of successful transmission in nth pass=$(1-e^{-G})^{n-1}e^{-G}$
Expected no of passes before a successful transmission=$\sum_{k=1}^{\infty}k(1-e^{-G})^{k-1}e^{-G}$
$S=e^{-G}+2(1-e^{-G})e^{-G}+3(1-e^{-G})^{2}e^{-G}+.........$
$(1-e^{-G})S=(1-e^{-G})e^{-G}+2(1-e^{-G})^{2}e^{-G}+3(1-e^{-G})^{3}e^{-G}+.........$
$e^{-G}S=e^{-G}+(1-e^{-G})e^{-G}+(1-e^{-G})^{2}e^{-G}+.....$
$e^{-G}S=\frac{e^{-G}}{1-(1-e^{-G})}$
$S=\frac{1}{e^{-G}}=e^{G}$
here it is given slotted aloha is working is maximum capacity so the value of G=1.
So expected no of pass before getting empty slots=e.
Is it correct logic ? any senior can confirm it.
