Download - End sem solution
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EE 333, Communication Networks Solutions Final Exam (2014-15S)
Maximum Marks = 50 Time = 3 hours
1. (a)
* 2 * 2
1 2 1 2
* 1 2
1 2
1 1 1 11 (1 ) 0.5 (1 )
(2 ) 1 (1 0.5 )
N N
N
* 1 2
1 2
1 1
(2 ) 2N
(b)
* 2 * 2
1 1 2
*
1 2
1 1 11 (1 ) (1 )
1(2 ) 1 (1 )
N N
N
*
1 2
1 1 1 1
(2 ) (2 ) 2N
(c) When both A and B have the packet and both transmit, the average total number of transmissions required (from both A and B) would be –
1 2 1 2 1 2
2 2
1 (1 )(1 )
Using this,
* 2 * 2
1 2 1 2 1 2
1 1 21 (1 ) (1 )N N
* 2
1 2 1 2 1 2
1 1 2(2 ) 1 (1 )N
Therefore, * 1 2
1 2 1 2 1 2
1 1 2
(2 ) 2 2N
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2. (a) The State Transition Diagram is given below
(b) Balance Equations
0 1 1 0
1 2 2 1 0 0 1
2
2 1 3 3 1 2 0
( ) (1 ) (1 ) ( )
( ) ( ) (2 )
p p p p
p p p p p p p
p p p p p p p
3 2 4 4 2 3
1 2 1 2
( ) ( )
................................................................................
( ) ( )
..................................................................
n n n n n n
p p p p p p
p p p p p p
..............
(c) Summing both sides of all the equations, we get
1 0 0
0 01 2 1 2
k k k
k k k
p p p
p p
(d) Generating Function Multiplying LHS and RHS of the ith equation by zi and summing both sides, we get
2 1
1 0 0
i i i
i i i
i i i
p z p z p z
Using 0
( ) i
i
i
P z p z
and simplifying, we get
2 00 2
( ) ( ) ( ) ( )1
pP z p z P z zP z P z
z z
The Normalization Condition requires that P(1)=1
Therefore, 0 2
1 21 2 ( )
1p P z
z z
(e) Mean 1
( )
z
dP zN
dz
2 2
(1 2 )( 2 )( )
(1 )
zP z
z z
2
3 (1 2 ) 3
(1 2 ) 1 2N
(f) Using Little’s Formula with 2eff , we get 3
2 (1 2 )eff
NW
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3(a)
N DB DC DD DE DF DG DH DI DJ
A 2 A-B
3 A-C
7 A-D
∞ 7 A-F
∞ ∞ ∞ ∞
A,B 2 A-B
3 A-C
7 A-D
4 A-B-E
7 A-F
∞ ∞ ∞ ∞
A,B,C 2 A-B
3 A-C
7 A-D
4 A-B-E
5 A-C-F
∞ ∞ ∞ ∞
A,B,C,E 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
∞ ∞
∞
A,B,C,E,F 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
∞ ∞ ∞
A,B,C,E,F,G 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
7 A-B-E-G-H
9 A-B-E-G-I
∞
ABCEFGD 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
7 A-B-E-G-H
7 A-B-E-D-I
13 A-D-J
ABCEFGDH 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
7 A-B-E-G-H
7 A-B-E-D-I
8 ABEGHJ
ABCEFGDHI 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
7 A-B-E-G-H
7 A-B-E-D-I
8 ABEGHJ
ABCEFGDHIJ 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
7 A-B-E-G-H
7 A-B-E-D-I
8 ABEGHJ
Forwarding Table at Node A Destination B C D E F G H I J
Next Node B C B B C B B B B
(b) Multicast Cost = Sum of Link Costs in above tree = 16 (c) Delete all the links of the first path in the network graph and apply Dijkstra’s Algorithm once again
4. Refer to the textbook and the lecture notes for the answers
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3(a) ALTERNATE SOLUTION
N DB DC DD DE DF DG DH DI DJ
A 2 A-B
3 A-C
7 A-D
∞ 7 A-F
∞ ∞ ∞ ∞
A,B 2 A-B
3 A-C
7 A-D
4 A-B-E
7 A-F
∞ ∞ ∞ ∞
A,B,C 2 A-B
3 A-C
7 A-D
4 A-B-E
5 A-C-F
∞ ∞ ∞ ∞
A,B,C,E 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
∞ ∞
∞
A,B,C,E,G 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
7 A-B-E-G-H
9 A-B-E-G-I
∞
A,B,C,E,G,F 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
7 A-B-E-G-H
9 A-B-E-G-I
∞
ABCEGFD 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
7 A-B-E-G-H
7 A-B-E-D-I
13 A-D-J
ABCEGFDH 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
7 A-B-E-G-H
7 A-B-E-D-I
8 ABEGHJ
ABCEGFDHI 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
7 A-B-E-G-H
7 A-B-E-D-I
8 ABEGHJ
ABCEGFDHIJ 2 A-B
3 A-C
6 A-B-E-D
4 A-B-E
5 A-C-F
5 A-B-E-G
7 A-B-E-G-H
7 A-B-E-D-I
8 ABEGHJ
Slightly different sequence if the other “7” is chosen first Forwarding Table at Node A Destination B C D E F G H I J
Next Node B C B B C B B B B
(b) Multicast Cost = Sum of Link Costs in above tree = 16 (c) Delete all the links of the first path in the network graph and apply Dijkstra’s Algorithm once again 4. Refer to the textbook and the lecture notes for the answers