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In collision free protocol channel efficiency is given by-(a) d/(d + log2(N))(b) d*(d + log2(N))(c) log2(N)(d) (d + log2(N))This question was posed to me in an online interview.This interesting question is from Layers topic in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

Answer»

The correct option is (a) d/(d + LOG2(N))

Best EXPLANATION: In collision FREE PROTOCOL CHANNEL efficiency is given by d/(d + log2(N)).



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