Suppose four active nodes-nodes A. B. C and D-are competing for access to a chan
ID: 3571965 • Letter: S
Question
Explanation / Answer
a. (1 – p(A))4 p(A)
where, p(A) = probability that A succeeds in a slot
p(A) = p(A transmits and B does not and C does not and D does not)
= p(A transmits) p(B does not transmit) p(C does not transmit) p(D does
not transmit)
= p(1 – p) (1 – p)(1-p) = p(1 – p)3
Hence, p(A succeeds for first time in slot 5)
= (1 – p(A))4 p(A) = (1 – p(1 – p)3)4 p(1 – p)3
a. p(A succeeds in slot 4) = p(1-p)3
p(B succeeds in slot 4) = p(1-p)3
p(C succeeds in slot 4) = p(1-p)3
p(D succeeds in slot 4) = p(1-p)3
p(either A or B or C or D succeeds in slot 4) = 4 p(1-p)3
(because these events are mutually exclusive)
a. p(some node succeeds in a slot) = 4 p(1-p)3
p(no node succeeds in a slot) = 1 - 4 p(1-p)3
Hence, p(first success occurs in slot 3) = p(no node succeeds in first 2 slots) p(some node
succeeds in 3rd slot) = (1 - 4 p(1-p)3)2 4 p(1-p)3
a. efficiency = p(success in a slot) =4 p(1-p)3