Homework 2 Solutions

Homework 2 Solutions - CS421/S03 Chapter 6 (7/e) 1. Problem...

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CS421/S03 Solutions to Homework Two Chapter 6 (7/e) 1. Problem 6.9 a. We have: Pr [single bit in error] = 10 -3 Pr [single bit not in error] = 1 – 10 -3 = 0.999 Pr [8 bits not in error] = (1-10 -3 ) 8 = 0.992 Pr [at least one error in frame] = 1-(1-10 -3 ) 8 = 0.008 b. Pr [at least one error in frame] = 1- (1-10 -3 ) 10 = 1-(0.999) 10 = 0.01 2 . Problem 6.12 _____ 10110110__________ 110011/ 1110001100000 110011 101111 110011 111000 110011 101100 110011 111110 110011 CRC = 11010 Chapter 7 . 3. Problem 7.2 Let L be the number of bits in a frame. Then, using Equation 7.5 of Appendix 7A: a = (Propagation Delay)/(Transmission Time) = ) 10 4 /( 10 20 3 3 × × - L = 80/ L Using equation 7.4 of Appendix 7A: 5 . 0 ) / 160 ( 1 1 2 1 1 + = + = L a U L 160 Therefore an efficiency of at least 50% requires a frame size of at least 160 bits.
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4. Problem 7.9 5. Problem 7.12 Let t 1 = time to transmit a single frame t 1 = 1024 bits/10 6 bps = 1.24 msec The transmitting station can send 7 frames without an acknowledgment. From the
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Homework 2 Solutions - CS421/S03 Chapter 6 (7/e) 1. Problem...

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