This document discusses error detection in communication networks. It contains:
1) An equation to calculate the expected number of transmissions (K) needed for a packet to reach its destination from two possible transmitters (A and B). The expected value of K is shown to be 2/(1-P) where P is the probability of failure for a single transmission.
2) A discussion of a (127,118) BCH code for error detection. It is shown that this code can detect all single errors, errors with an odd number of errors, all double errors for codewords under 127 bits, and all bursts of 8 errors or fewer. The encoding works by dividing the data polynomial by a primitive polynomial.
A Primality test is an algorithm for determining whether an input number is Prime. Among other fields of mathematics, it is used for Cryptography. Factorization is thought to be a computationally difficult problem, whereas primality testing is comparatively easy (its running time is polynomial in the size of the input).
A Primality test is an algorithm for determining whether an input number is Prime. Among other fields of mathematics, it is used for Cryptography. Factorization is thought to be a computationally difficult problem, whereas primality testing is comparatively easy (its running time is polynomial in the size of the input).
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Permutations and Combinations IIT JEE+Olympiad Lecture 4Parth Nandedkar
Continues from PnC lecture 3. The series Follows the JEE Advanced syllabus, but this lecture goes beyond into Mathematical Olympiad territory. Covers the following more advanced topics:
Simple idea of Inclusion Exclusion principle,
Explanation through Venn Diagrams,
Application of I-E Principle,
Counting Derangements using I-E Principle,
Partitioning Indistinguishable Objects(+Comparison with Distinguishable Objects) ,
Problem Session
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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.