end sem solution

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

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

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

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