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Consider the use of 10 K-bit size frames on a 10 Mbps satellite channel with 270 ms delay. What is the link utilization for stop-and-wait ARQ technique assuming P = 10-3?

 

My doubt - In frame size 10 K-bit, what should you take value of K? i.e K as 1000 bits or K = 1024 bits? I'm more inclined about taking K = 1024 bits, but the question's solution has K = 1000 bits!

 

 

Given solution - Link utilization = (1-P) / (1+2a)
Where a = (Propagation Time) / (Transmission Time)
Propagation time = 270 msec
Transmission time = (frame length) / (data rate)
= (10 K-bit) / (10 Mbps)
= 1 msec
Hence, a = 270/1 = 270
Link utilization = 0.999/(1+2*270) ≈0.0018 =0.18%

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In storage K= 1024 like in Operating System and CO

in transmission K = 1000 like in Computer Networks

 

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GateAspirant999 asked Jan 18, 2017
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