hw4_solution

hw4_solution - Transmission time per frame = x = 1000/R R =...

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6.12 7.3 a = Propagation Delay L R = 270 × 10 - 3 10 3 10 6 = 270 a. U = 1/(1 + 2a) = 1/541 = 0.002 b. Using Equation 7.6: U = W/(1 + 2a) = 7/541 = 0.013 c. U = 127/541 = 0.23 d. U = 255/541 = 0.47 7.4 A B: Propagation time = 4000 × 5 μ sec = 20 msec Transmission time per frame = 1000 100 × 10 3 = 10 msec The effective data rate: r1 = (3 frames)/(10 msec + 2x20 msec) = 60 kbps B C: Propagation time = 1000 × 5 μ sec = 5 msec
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Unformatted text preview: Transmission time per frame = x = 1000/R R = data rate between B and C (unknown) The effective data rate: r2= (1 frame)/(1000/R + 2x5 msec) To avoid buffer overflow at B, there must be r2 >= r1, thus, R>=150 kbps. So, the minimum data rate between B and C is 150 kbps. 7.9 a. 0 1 2 3 4 5 6 7 0 b. 0 1 2 3 4 5 6 7 0 c. 0 1 2 3 4 5 6 7 0...
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hw4_solution - Transmission time per frame = x = 1000/R R =...

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