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ps5sol_ee550_fall08

# ps5sol_ee550_fall08 - UNIVERSITY OF SOUTHERN CALIFORNIA...

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UNIVERSITY OF SOUTHERN CALIFORNIA, FALL 2008 1 EE 550: Problem Set # 5 Solution I. B OOK P ROBLEM 3.14 (a) The average message transmission time is 1 = L/C so the service rate is μ = C/L . When the number of packets in the system is large than K , the arrival rate is λ 1 . We must have 0 λ < μ and 0 λ 2 in order for the arrival rate at node A to be less than the service rate for large state values. For these values, therefore, the average number of packets in the system will stay bounded. (b) The corresponding Markov chain is as shown in the figure below. 0 1 2 K-1 K K+1 λ 1 + λ 2 λ 1 + λ 2 λ 1 + λ 2 λ 1 λ 1 ...... μ μ μ μ μ μ μ The steady state probabilities satisfy p n = ρ n p 0 , for n k, p n = ρ n - k 1 ρ k p 0 , for n > k , where ρ = ( λ 1 + λ 2 ) , ρ 1 = λ 1 . Together with n =0 p n = 1 , we get p 0 = (1 - ρ )(1 - ρ 1 ) 1 - ρ 1 - ρ k ( ρ - ρ 1 ) , for ρ < 1 , p 0 = 1 - ρ 1 1 + k (1 - ρ 1 ) , for ρ = 1 . For packets of source 1 the average time in A is T 1 = 1 μ (1 + N ) , where N = n =0 np n is the average number of packets in the system seen by a type 1 arrival (by PASTA). The average number in A from source 1 is N 1 = λ 1 T 1 . For packets of source 2 the average time in A is T 2 = 1 μ (1 + ˜ N ) , where ˜ N = k - 1 n =0 np n k - 1 n =0 p n is the average number in the system found by an accepted packet of source 2. To find the average number in the system from source 2 we must find the arrival rate into node A of packets from source 2. This is

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