GATE | GATE CS Mock 2018 | Set 2 | Question 62

Consider a link with packet loss probability of 0.2. What is the expected number of transmissions it would take to transfer 200 packets given that the stop and wait protocol is used?

(A) 125
(B) 250
(C) 225
(D) 150


Answer: (B)

Explanation: For N packets, Np packets will be lost(Since p is the loss probability). So they must be resent.
For Np packets, Np^2 packets will be lost and must be resent.
.
.
and so on.
So, Expected number of transmissions
= N + Np + Np^2 + ...
= N(1+p+p^2+...)
= N*\frac{1}{1-p}
For N = 200 and p = 0.2
Expected number of transmissions = 200/0.8 = 250

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