Mid term solution

EE 333, Communication Networks
Mid-Term Exam (2014-15S)
Maximum Marks = 30 Time = 2 hours
1. For N-packet blocks, let XN be the time taken from the instant the transmission of the block begins for the first
time to the time when the transmission of the next block can be started.
(a) N=1
   1 1
1
1
1
1
2 2 (1 )
1 2
2
where
1 2
X T d T d X
d
XT d dTX b
T b T
 


 


     
 
  
            
 
[2]
(b) N=2
   
 
2 2
2 2
2
1
2
2
(2 2 ) 2 2 2 (1 ) 2 2 (1 )
(2 ) (2 2 ) 2( 2 )(1 ) 2 (2 ) 2 (3 2 )
2 2 (3 2 ) 1 (3 2 )
(2 ) (2 )
T d
X T d T d T d X
X T d T d T d
T d b
X
   

    
 

     

 
          
 
         
  
     
  
[4]
(c) N=3 with d=0 10 20
2T T
X X
 
 
     
   
 
 
3 2 2 3
30 10 20 30
3 2
30
2 3 2
30
2 2
30
30
30
3 3 3 (1 ) 3 3 (1 ) 3 (1 )
1 (1 ) 3 3 (1 ) 6(1 )
3 3 3 1 (1 ) 2(1 )
3 3 3 3 3
3 3
X T T X T X T X
X T T T
X T
X T
T T
X
X
     
   
     
    
 

         
        
        
      
   
[4]
We can logically argue that in this ARQ scheme, each packet is sent α-1
times as in a simple Stop-&-Wait ARQ
system. (This is because every packet is repeated until it is correctly received and only the packets in error are
repeated.) The differences between schemes with different block-lengths arise only because of the effects of the
propagation delay d. When d=0, all the schemes will perform similarly and will have the same efficiency as the
equivalent Stop-&-Wait approach.

2. (a) Consider the code word to be 1 2
1 2 1 0( ) ...........n n
n nT x T x T x T x T 
     
A circular right shift of T(x) by one position yields the polynomial T*(x) as follows –
1 2 3
0 1 2 2 1*( ) ...........n n n
n nT x T x T x T x T x T  
      
Note that T*(x) can also be obtained by circularly left shifting T(x) by n-1 positions.
We can show (given below) that circularly left shifting T(x) by one position also creates a code word.
Therefore, circularly left shifting by n-1 positions will also create a code word.
To show that circularly left shifting T(x) by one position creates another code word -
Circular Left shift by one position, 1 1 1
ˆ( ) ( ) ( ) (1 )n n
n n nT x xT x T x T xT x T x       
In ˆ( )T x , (a) T(x) is obviously divisible by g(x)
(b) (1+xn
) has (1+x) as a factor and (1+xn
) is also divisible by P(x). Therefore (1+xn
) is also
divisible by g(x)
Therefore ˆ( )T x is also a code word.
(b) Comparing ( )C x and ( )C x , it is easy to see that the error polynomial ( )E x is –
10 9 8 4 2
( ) 1E x x x x x x     
This error pattern will be detectable if ( )E x is divisible by ( )g x . We can check this by long division
and verify that Rem
10 9 8 4 2
8 2
1
0
1
x x x x x
x x x
     
 
   
Note: Actually 10 9 8 4 2 2 8 2
1 ( 1)( 1)x x x x x x x x x x           from the long division
Therefore, the error pattern ( )E x will not be detectable by CRC-8 and ( )C x will be accepted by the
receiver as a valid code word.
We can also verify that
17 14 12 11 10 9 7 5 4 2
9 6 4 8 2
( ) 1
( 1)( 1)
C x x x x x x x x x x x x
x x x x x x
           
      
Therefore ( )C x is indeed a valid code word and will be accepted by the receiver.
[By long division, we get
17 14 12 11 8 7 5 9 6 4 2 8 2
( ) ( )( 1)C x x x x x x x x x x x x x x x x x                 ]

3.
(a) State Transition Diagram
(b) 3 2 2 2 2 2 2
0 1 1 23 (1 ) (1 ) (1 ) 2 (1 ) 2 (1 )P q q q P q q P q q q q q P q q                       
Therefore,
2 2
1 0 2 1 02 2 4
(3 2 ) 1 (3 2 )(1 )
(1 ) 2(1 ) 2(1 )
q q q q q q q q
P P P P P
q q q
     
  
  
(c) It is easy to show that  2 2
0 1 2 33 (1 ) 3 (1 )S q q P P P P q q      
Differentiating this with respect to q, we find that the maximum occurs at
1
3
q  and the
maximum value of S is max
4
0.444
9
S  

Mid term solution

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Mid term solution