Forouzan question:- Expected no of empty slots in slotted aloha

Question

A slotted aloha network is working with maximum throughput.How many slots, n, on average, should pass before getting an empty slot?

Comments

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=$\sum_{k=1}^{\infty}k(1-e^{-G})^{k-1}e^{-G}$

$S=e^{-G}+2(1-e^{-G})e^{-G}+3(1-e^{-G})^{2}e^{-G}+.........$

$(1-e^{-G})S=(1-e^{-G})e^{-G}+2(1-e^{-G})^{2}e^{-G}+3(1-e^{-G})^{3}e^{-G}+.........$

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$e^{-G}S=e^{-G}+(1-e^{-G})e^{-G}+(1-e^{-G})^{2}e^{-G}+.....$

$e^{-G}S=\frac{e^{-G}}{1-(1-e^{-G})}$

$S=\frac{1}{e^{-G}}=e^{G}$

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.