midterm_sol - Prob. 1 (20 points) Let be the expected...

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Prob. 1 (20 points) Let γ be the expected number of transmitted frames from node A to node B per successfully accepted packet at node . Let β be the expected number of transmitted frames from node to node between the transmission of a given frame and the reception of feedback about that frame (including the frame in transmission when the feedback arrives). Let p be the probability that a frame arriving at node contains errors (with successive frames assumed independent). Assume that node i s a lway s busy transmitting frames, that n is large enough that node never goes back in the absence of feedback, and that node always goes back on the next frame after hearing that the awaited frame contained errors. Find as a function of and . Sol. 1 1.) The success probability in 1st packet transmission = 1( 1 ) p × = The expected number of transmitted frames. 2.) The success probability in 2 nd packet transmission: (1 ) p p The expected number of transmitted frames: (1 1 ) (1 ) p p + +− 3.) The success probability in ith packet transmission: 1 ) i p p The expected number of transmitted frames: 1 ((1 ) ) ( i ii p p + −− (10 points) 1 1 1 {( ( 1) ) )} 1 i i p p p p γβ = + =+ = (10 points) Sol. 2 [ ] [#of transmitted frames when an error occurs] + (1- [ ]) [#of transmitted frames when an error does not occurs] P error E P error E × (10 points) ) ) p p βγ + + 1 1 p p + = (10 points)
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Prob. 2. (15 points) Consider the M/G/1 system with the difference that each busy period is followed by a single vacation interval. Once this vacation is over, an arriving customer to an empty starts service immediately. Assume that vacation intervals are independent, identically distributed, and independent of the customer interarrival and service times. Prove that the average waiting time in queue is 22 2(1 ) 2 X V W I λ ρ =+ , where I is the average length of an idle period, and show how to calculate . We have 1 R W = (2 points) where () 11 1 1 lim Mt Lt ii t R XV tt →∞ == ⎧⎫ ⎨⎬ ⎩⎭ ∑∑ (4 points) where is the number of vacations (or busy periods) up to time t . The average length of an idle period is 00 V V I pv v e d v e d dv λτ λττ ∞∞ −− ⎡⎤ ⎢⎥ ⎣⎦ ∫∫
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This note was uploaded on 04/29/2010 for the course CSE 4541.525A taught by Professor Choisunghyun during the Spring '09 term at Seoul National.

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midterm_sol - Prob. 1 (20 points) Let be the expected...

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