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Tutorial 7

# Tutorial 7 - IERG4020 Tutorial 7 FENG Shen 1 PROBLEM SET 4...

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IERG4020 Tutorial 7 FENG Shen 1

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PROBLEM SET 4 DISCUSSION 2
Problem 4.1 Four states: (1,1), (1,2), (2,1), (2,2) Assuming each input packet has the same probability to be destined for each output, we can draw the Markov chain: 3

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Problem 4.1 4 ( i,j ): HOL of queue 1 -> output 1, HOL of queue 2 -> output 2.
Problem 4.1 By symmetry, P (1,1) = P (2,2); P (2,1) = P (1,2) In stationary state, considering (1,1) As a consequence, P (1,1) = P (2,2) = P (2,1) = P (1,2). P (1,1) + P (2,2) + 2 × P (2,1) + 2 × P (1,2) = 1.5 packets are routed each time slot. Maximum throughput is 5 (1,1) 0.25 (2,1) 0.25 (1,2) 0.5 (1,1) 0.5 ( 0 1, 5 2) . PP P P   * 1.5 / 2 0.75 

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Problem 4.3 For an internally nonblocking input-buffered switch, the saturation throughput is * = 0.586. When 0 = 0.5 0 = 0.5 is smaller than 0.586. Hence = 0 = 0.5. 6
Problem 4.3 The loss probability of the waiting system is The loss probability of the loss system is 7 0 loss;wait 0 0.5 0.5 0 0.5 P 0 0 0 0 0 loss;los 0 s (since is large) 0. 1 21 11 3 N P N e N        

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Problem 4.3 When 0 = 1 Because 0 = 1 is larger than the saturation throughput * = 0.586, the waiting system is saturated and achieves a throughput = * = 0.586. Thus For the loss system, 8 0 0 loss;wait 1 0.586 0.414 1 P    0 loss;loss 0 0 1 0.368 e P 
Problem 4.3 9 loss;w 0 ait 0 0 0 0 0, .586, 0.586 0.58 for , 6, for P   0 loss;loss 0 0 1 P e  0 loss;wait

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Tutorial 7 - IERG4020 Tutorial 7 FENG Shen 1 PROBLEM SET 4...

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