EE 333, Communication Networks Solutions to Quiz-I (2014-15S)

Maximum Marks = 10 Time=45 minutes

1. Let M be the number of times A has to transmit to deliver the packet to D and let N be the number of times B has to transmit to deliver the packet to D.

Then 1 1{ } (1 ) { } (1 )m nP M m P P P N n P P

Let K be the number of times (both) A and B transmit for the packet to reach D where K=min (M, N). Therefore,

1 1 2( 1) 2

1

1 2( 1) 2

2( 1) 2 2

2 1 2

{ } 2 (1 ) (1 ) (1 )

2 (1 ) (1 )

[2 2 1 2 ]

( ) (1 )

k j k

j k

k k k

k

k

P K k P P P P P P

P P P P P

P P P P P

P P

This is the required distribution. Note that this could have been derived by simple logical arguments without doing the explicit calculation given above. I will accept that as a solution if you have clearly given your arguments – even though that is an answer that can be given in just a few lines!

Using this, E{K}=2

1

1 P

2. (a) Factoring, one can show that 8 2 7 6 5 4 3 21 ( 1)( 1)x x x x x x x x x x

We assume that 7 6 5 4 3 2( 1)x x x x x x is a Primitive Polynomial and will therefore not

divide any 1nx for n < 27-1=127 but will exactly divide 127 1x All single errors will be detected (more than one term) All error patterns with an odd number of errors will be detected (1+x is a factor) All double errors will be detected if the codeword has less than 127 bits (i.e. 118 bits of data and 8 check bits) (using the assumption of the other factor being a primitive polynomial) All burst errors with burst length of 8 or less will be detected. (b)

Feed in the polynomial to be divided, MSB first. The bits in the shift register after all the bits have been fed will be remainder.