![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 n
P 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 2
1 ( 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 1n
x 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.](https://image.slidesharecdn.com/quiz1solution-150425103651-conversion-gate01/85/Quiz-1-solution-1-320.jpg)
Quiz 1 solution
- 1. 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 n P 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 2 1 ( 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 1n x 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.