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In collision free protocol channel efficiency is given by-

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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))

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This interesting question is from Layers topic in division Cryptography Overview, TCP/IP and Communication Networks of Cryptograph & Network Security

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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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