Example Question:
Consider the HEC mechanism. Let p be the probability that a bit is received in error.
a) With what probability a cell is rejected when the HEC state machine is in the
"Correction Mode"?
b) With what probability a cell is rejected when the HEC state machine is in the
"Detection Mode"?
c) Assume that the HEC state machine is in the "Correction Mode." What is the
probability that n successive cells will be rejected, where n >= 1 ?
d) Assume that the HEC state machine is in the "Correction Mode." What is the
probability p(n) that n successive cells will be accepted, where n >= 1 ?
Hint: Write down the expression for p(1) and p(2), and express p(3) as a function of
p(1) and p(2). Then write down the general expression for p(n) for any n as a function
of p(n-1) and p(n-2).
Answer:
a) p = probability that a bit is erroneous.
A cell is rejected if it has multiple-bit error in its header, when the HEC state
machine is the “Correction Mode”. In other words, for a cell not to be discarded,
it should have at most 1-bit error in its header. The header is 40 bits-long. Hence,
P = probability of a cell is rejected when HEC state machine is in the “Correction
Mode” can be computed as follows;
P  1-p(no bit error in header)-p(1 bit error in header)
 40 
p(no bit error)=   p 0 .(1  p) 40  (1  p) 40
0 
 40 
p(1-bit error)=   p1.(1  p)39  40. p.(1  p)39
1 
Thus,
P  1  (1  p)40  40. p.(1  p)39 .
b) When the HEC state machine is in “Detection Mode”, the cell is discarded if
any error is detected in its header. Hence, the probability of a cell being discarded
when the machine is in “Detection Mode” is equal to the probability of having
one or more bit errors in the header of a cell. Hence,
P  1- p(no bit error)
P  1  (1  p)40 .
c) The HEC state machine is in the “Correction Mode”. To reject first one of the
successive n cells, the cell must have multiple-bit errors with a probability derived
in part (a) as
1  (1  p)40  40. p.(1  p)39 .
After this point, the machine goes into “Detection Mode”. To reject remaining n-1
successive cells, it is sufficient that each of them has at least one bit error in their
headers. This probability was derived in part (b) as
1  (1  p) 40 .
Hence, the probability of rejecting n successive calls starting from the “Correction
Mode” is given as,


P   1  (1  p)40  40. p.(1  p)39 . 1  (1  p)40 .
d) The HEC state machine is in the “Correction Mode”. To accept a cell in this
mode, it must have either no error or at most 1 bit error in its header. If it has 1 bit
error in its header, the state machine goes to “Detection mode”, which means next
cell's header should be error-free to keep the n successive acceptance sequence.
n 1
Let m and s be the probabilities of the two cases that a cell can be accepted when
the machine is in “Correction Mode”, where
m=p(no error in cell header)= (1  p) 40
s=p(1-bit error in cell header)= 40. p.(1  p)39 ,
and
p(n)=p(n successive cells are accepted).
Thus, the first of these probabilities is
p(1)  m  s .
Accepting 2 consecutive cells can occur only under two circumstances:
 The second cell is error-free (with probability m) AND the first cell
had either 1-bit error or no errors (with probability (m + s)).
OR
 The second cell has 1-bit error (with probability s) AND the first cell
had no error (with probability m).
Hence,
p(2)  (m  s) p(1)  ms
Accepting 3 consecutive cells can occur only under two circumstances:
 The third cell is error-free (with probability m) AND the first two
consecutive cells were accepted (with probability p(2)).
OR
 The third cell has 1-bit error (with probability s) AND the previous
cell has no errors (with probability m) AND first cell was accepted
(with probability p(1)).
Hence,
p(3)  mp(2)  msp(1)
Therefore, with the same reasoning, accepting n (n>2) consecutive cells can occur
only under two circumstances:
 The last cell is error-free (with probability m) AND the first (n-1)
consecutive cells were accepted (with probability p(n-1)).
OR
 The last cell has 1-bit error (with probability s) AND the previous cell
has no errors (with probability m) AND the first (n-2) cells were
accepted (with probability p(n-2)).
Hence,
p(n)  mp(n  1)  msp(n  2)
p(n)  (1  p) 40 p(n  1)  (1  p) 40 40. p.(1  p)39 p(n  2)
p(n)  (1  p) 40  p(n  1)  40. p.(1  p)39 p(n  2)
