1. 371-1-0291: An Introduction to Computer Networks
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Handout #13: Error Detection and Correction
Additional Reading
Text book: Chap. 2.4
Lect-11-1: Error Control
Computer Networks

2. Outline Basic ideas: BER, PER and Hamming Distance Parity
Error Detection: Cyclic Redundancy Codes
Error Correction
Error Detection versus Correction
Lect-11-1: Error Control
Computer Networks

3. Errors
A bit error occurs when a source sends a bit, b, and the destination receives NOT b.
The error can take place on the link (e.g. interference, or signal loss), in a (malfunctioning) switch or router along the path, or in the source or destination (e.g. failed hardware, or bit errors in memories).
The bit error rate (BER) tells us the probability of any given bit being in error. Typical values are BER = 10-9 for an electrical link, and for an optical link.
Lect-11-1: Error Control
Computer Networks

4. Error Detection D H H ACK An example:
Assume BER and independent errors,
Packet Error Rate = PER = 1 – (1 – BER)N
PER ~= N (BER) if N (BER) << 1
e.g. N = 104, BER = 10-7 = PER = 10-3
Lect-11-1: Error Control
Computer Networks

5. Error Detection In practice, bit errors occur in BURSTS:
Perhaps caused by mechanical switches.
Time constants.
Therefore, PER is lower for a given BER
Lect-11-1: Error Control
Computer Networks

6. Encoding to detect errors
We use codes to help us detect errors.
The set of possible messages is mapped by a function onto the set of codes.
We pick the mapping function so that it is easy to detect errors among the resulting codes.
Example: Consider the function that duplicates each bit in the message. E.g. the message would be mapped to the code , and then transmitted by the sender. The receiver knows that bits always come in pairs. If the two bits in a pair are different, it declares that there was a bit error.
Of course, this code is quite inefficient (doubles #bits, un-detects if both bits are corrupted)
Lect-11-1: Error Control
Computer Networks

7. Hamming Distance Number of bits that differ between two codes
e.g
In our example code (replicated bits), all codes have at least two bits different from every other code. Therefore, it has a Hamming distance of at least2.
HD=3
Lect-11-1: Error Control
Computer Networks

8. Hamming Distance Set of codes
- To reliably detect errors in d bits: HD > d
- To reliably correct errors of d bits : HD > 2d 4 3
Example
Suppose we encode “0” by “0” and “1” by “1”
A corrupted single bit leads to a decoding
error
If we encode “0” –> “00” ; “1” –> “11”, then a
single corrupted bit is detected, but from
receiving “01” we don’t know if “1” or “0”
has been transmitted
If we encode “0” –> “000” ; “1” –> “111” and
one bit is corrupted, we reliably detect the
xmitted by the majority rule d 23 1 2
HD = min (dij ) ij
Lect-11-1: Error Control
Computer Networks

9. Parity: A simple error detection code
We look for to add redundancy efficiently
Parity: A simple error detection code
Parity added to
make # 1s even/odd
If parity is wrong ERROR
If parity is right NO ERROR
(or an even number of errors
has occurred)
Q: What is the minimum Hamming distance for this code?
Lect-11-1: Error Control
Computer Networks

10. Error Detection Cyclic Redundancy Code (CRC)
Goal: Maximize the error detection prob using only a small # of redundant bits
Technique comes from Finite Fields Algebra
A (n+1) bit message is represented by a polynomial with degree n
MSB
e.g.,
For CRC, the XMT and RCV agree on a divisor polynomial , of degree k, e.g., for k=3
Lect-11-1: Error Control
Computer Networks

11. T(X) is divisible by C(X) if deg[T(X)] >= deg[C(X)]
The selection of C(X) has a crucial impact on what errors can be detected. E.g., for Ethernet the C(X) is a 32 degree polynomial
Given M(X) and C(X), the transmitted message is (n+1)+k bits long and corresponds to a polynomial T(X) that is exactly divisible by C(X) (using binary polynomial arithmetic):
T(X) is divisible by C(X) if deg[T(X)] >= deg[C(X)]
If deg[T(X)] = deg[C(X)], T(X) is divisible once by C(X), and if deg[T(X)] < deg[C(X)] it is not
The remainder of T(X)/C(X) is [T(X) XOR C(X)]
Lect-11-1: Error Control
Computer Networks

12. A division example will be given later
A remainder computational example
Example
C(X) ~
T(X) ~
XOR
Lect-11-1: Error Control
Computer Networks

13. T(X)= M(X) (adds k zeros at the message end)
To create T(X) do:
T(X)= M(X) (adds k zeros at the message end)
Divide T(X) by C(X) and find the remainder R(X)
T(X)=T(X) – R(X)
The resulting T(X) is exactly divisible by C(X) and contains M(X) followed by R(X) (recall that by binary subtraction 0 -b = b)
Lect-11-1: Error Control
Computer Networks

14. Example: M= 110101 C = 1001 M(X) T = 1 1 0 1 0 1 0 1 1 Xmitted message
M(X)
0 1 1
T =
Xmitted message R
Lect-11-1: Error Control
Computer Networks

15. To detect an error upon receiving T(X) do: Divide T(X) by C(X)
If the remainder R(X) is zero, no-error is concluded
If the remainder R(X) is not zero, an error has ocured
Lect-11-1: Error Control
Computer Networks

16. Cyclic Redundancy Code (CRC) single bit error detection
A single-bit error can be represented as:
T(X)+E(X), where E(X) = Xi (bit i corruption)
If C(X) = Xk + Xj , i.e. C(X) has any 2 terms we can detect a single bit error. Indeed,
Since C(X) divides T(X) (so generated) but can’t divide E(X), all single bit errors can be detected.
[ Check what happens when you divide (10000) by (101) ]
Lect-11-1: Error Control
Computer Networks

17. Cyclic Redundancy Code (CRC) Double bit error detection
(ii) Two isolated errors can be represented as:
E(x) = xi + xj = xj (xi-j + 1)
If (xk + 1) is not a factor of C(x) for any k then:
all two-bit errors are detected, or if C(x) has a
3-term factor. (In this cases, C(x) cannot divide
E(x) )
(iii) Detecting an odd number of errors
Theorem: If (x + 1) is a factor of C(x) then:
all odd errors are detected.
Proof by contradiction:
Assume E has odd # of 1’s and (x + 1) is a factor:
E(x) = (x +1)E’ (x) for some E’ (x)
Lect-11-1: Error Control
Computer Networks

18. Cyclic Redundancy Code (CRC)
(iii) Proof (cont.)
E(1) = (1 + 1) E’(1) = 0 E’(1) = 0
Evaluated at x=1
But E(x) has odd # of 1’s so : E (1) = 1
Therefore, if E has an odd # of 1’s,
it is not divisible by (x + 1)
Lect-11-1: Error Control
Computer Networks

19. Cyclic Redundancy Code (CRC)
(iv) Burst error of length < k (degree of C(x)) can
be detected
Burst of length k-1 is represented :
E(x)=xi (x k-1-i + …+1), 0 <= i < k
Thus, C(x) doesn't divide any factor of E(x)
(v) Longer bursts can not necessarily be guaranteed to be detected.
Lect-11-1: Error Control
Computer Networks

20. Cyclic Redundancy Code (CRC)
Examples of common CRC generators:
CRC-8: x8 + x2 + x1 + 1
CRC-10: x10 + x9 + x5 + x4 + x1 + 1
CRC-12: x12 + x11 + x3 +x2 +1
CRC-16: x16 + x15 + x2 +1
CRC-CCITT: x16 + x12 + x5 + 1
CRC-32: x32 + x26 + x23 + x22 + x16 + x12
+ x11 + x10 + x8 + x7 + x5 + x4 + x2 +x
+ 1
Lect-11-1: Error Control
Computer Networks

21. Error Correction Common code used: Bose-Chaudhuri-Hocquenghem (BCH) R
BCH (R + M, M, t)
e.g. BCH (1023, 923, 10)
Can detect all “t”
bit errors
If t=1 then the code is called Reed-Solomon
and is used in CD players
Lect-11-1: Error Control
Computer Networks

22. Most data networks today use error detection,
Detect or Correct?
Advantages of Error Detection
Requires smaller number of bits/overhead.
Requires less/simpler processing.
Advantages of Error Correction
Reduces number of retransmissions.
Most data networks today use error detection,
not error correction.
Lect-11-1: Error Control
Computer Networks

23. Detect or Correct? An example
Assume: 1. Packets are of lengths 923 bits
2. PER = 10-5
Overhead of Error Correction:
Assume we use: BCH (1023, 923, 10)
Therefore, we send 923 data bits as 1023 bits.
Overhead of Error Detection:
Assume we use: 32-bit CRC; one retransmission per error.
Therefore, we send 923 data bits as 955 bits.
100
923
Transmission Overhead = ~= 10%
Transmission Overhead = ~= 3%
( )
923
Lect-11-1: Error Control
Computer Networks