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IERG4020 Tutorial 3
FENG Shen
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View Full Document QUIZ 1 DISSCUSION
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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View Full Document 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
+⋅
⋅
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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View Full Document Problem 2
Assume in the worst case, there are (n1)
existing calls for A and (n1) existing calls for B.
1 call to block the n x n switch. 2 calls to block
at
most
1of the 2n x 2n secondstage switch.
A
B
n
2n
n
n
2n
n
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 (n1) existing calls on A can block
at
most
secondstage modules
(1 pt)
1
(
2) / 2
n
+−
(
1) / 2
n
−
1
(
2) / 2
n
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View Full Document Problem 2
B can block at most
secondstage
modules.
(1 pt)
Totally, A & B can block
secondstage 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.
 Spring '10
 FENGShen

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