Tutorial 3

Tutorial 3 - IERG4020 Tutorial 3 FENG Shen QUIZ 1 DISSCUSION Quiz 1 You can get back your quiz paper during tutorial or at SHB 702 Appeal is only

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IERG4020 Tutorial 3 FENG Shen
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QUIZ 1 DISSCUSION
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Quiz 1 You can get back your quiz paper during tutorial or at SHB 702. Appeal is only available during the moment you get the paper. Once you leave with the quiz paper, NO APPEAL IS ALLOWED. AVERAGE 3.02 SD 2.07 MAX 8.5 MIN 1
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Problem 1 5 active conversations Each conversation has 0.4 probability to be in (0.6 probability to be not in) a talkspurt. When there is 0 or 1 existing talkspurt, the new talkspurt will not be clipped. (1 pt) Prob[not clipped] = = 0.4752 Prob[clipped] = 1 - 0.4752 = 0.5248 (1 pt) Note that when calculating, we exclude the conversation which is attempting for a new talkspurt. 43 0.6 4 0.6 0.4 +⋅
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Problem 2 Similar to Problem 2.8 in the second problem set, except for the appearance of n x n switch. The proof is similar to Problem 2.8.
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Problem 2 Assume in the worst case, there are (n-1) existing calls for A and (n-1) existing calls for B. 1 call to block the n x n switch. 2 calls to block at most 1of the 2n x 2n second-stage switch. A B n 2n n n 2n n
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Problem 2 WLOG, assume the n x n switch can only be blocked by one of the existing calls on A. Case 1: not blocked by A or B - Case 2: blocked by A - The (n-1) existing calls on A can block at most second-stage modules (1 pt) 1 ( 2) / 2 n +−   ( 1) / 2 n 1 ( 2) / 2 n
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Problem 2 B can block at most second-stage modules. (1 pt) Totally, A & B can block second-stage modules.
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This note was uploaded on 12/08/2010 for the course IEG IEG4020 taught by Professor Fengshen during the Spring '10 term at CUHK.

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Tutorial 3 - IERG4020 Tutorial 3 FENG Shen QUIZ 1 DISSCUSION Quiz 1 You can get back your quiz paper during tutorial or at SHB 702 Appeal is only

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