T305: DIGITAL COMMUNICATIONS
T305: Digital Communications
Block II – Modelling Activities
Probability & Reliability
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Introduction:
Probability and Reliability used in communication
Technology (e.g. Network Traffic)

The various topics covered
Topic 1: Probability
Topic 2: Reliability
Topic 3: Preparation for Next Tutorial

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Topic 1: Probability
If a process or measurement can have more than one
outcome, then the frequency of the ith possible outcome,
referred to as the occurrence of event Ei, is the number
of times, Ni, that Ei occurs out of a total number, N, of
events. The value of Ni may vary each time the
experiment is repeated.
Example of Ei are bits in error arrived at
communicatioon channel
The frequency of occurrence of an event may be
expressed as a fraction of the total number of events.
This is called a relative frequency, and the relative
frequency of event Ei is given by
Ni /N.

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Example Relative freq.
(Probability)
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A set of events which includes every possible
outcome of a process or measurement is said to
be exhaustive. The events are mutually
exclusive if no two of them can occur
simultaneously. An exhaustive list of all the events
that may occur is known as the event space.

The relative frequency can be used to estimate
the probability, P(Ei), of event Ei , provided that
N is sufficiently large such that any further
increases in the value of N do not significantly
affect the value of the relative frequency of event
Ei.

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If the result of an experiment or measurement can be one of a set of r
exhaustive and mutually exclusive events, E1, E2, E3, …, Ei,…, Er, and
if, as the result of a large number, N, of measurements, the frequencies
are found to be N1, N2, N3, …, Ni , …, Nr , respectively, then the
probability of event Ei is:
P(Ei ) = Ni / N
where N = N1 + N2 + N3 + … + Ni + … + Nr
It follows directly from the above that:
P( E1 ) + P( E2 ) + … + P( Ei ) + … + P( Er ) = 1
and that the value of the probability of event Ei occurring is in
the range from zero to one:
0 ≤ P( Ei ) ≤ 1

Probabilities can be represented by the
use of a Venn diagram as shown below.
Fig. Venn diagram

for event E.

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Example : All pattern of 4 bits would
be 16 patterns
Probability of any one of those
patterns would be 1/16=0.0625 or
6.25%
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Question
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Question
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Question
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The total area covered by a Venn diagram (the
rectangular area labelled S) represents all the events
which can occur. The frequency of a specific event
corresponds to a given area of the diagram (such as area
E). The ratio (area of E)/(area of S) is the probability of
the event E occurring. Instead of representing frequencies,
areas on a Venn diagram can represent probabilities.

The sum of the probabilities of a set of exhaustive and
mutually exclusive events is 1. If P(E) is the probability
that event E occurs, then P( E ) is used to represent the
probability that the event does not occur, and: P( E ) +
P( E ) = 1

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The notation P(AB) is used for the joint probability that
events A and B will both occur. The notation P(A OR B) is
used for the probability that event A, or event B, or both
event A and event B will occur. The probabilities are
related by the equation:
P( A OR B) = P( A) + P( B) - P(AB)
P(AB) = 0 for mutually exclusive events, in which case:
P(A OR B)= P(A) + P(B)

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Question
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
The notation P(A | B) is used for the conditional
probability (a method to check if two events are dependent)
that A should occur, given that B has definitely occurred. For
any two events, A and B:
P(AB) = P( A | B) P( B) = P( B | A) P( A)
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
Independent events are ones that have no influence
on each other –that is, the occurrence of one event does
not affect the probability of occurrence of the other.
P( A | B) = P( A) for independent events A and B
If A and B are independent events, the equation for the
joint probability of the two events reduces to the
multiplication rule:
P(AB) = P( A) P( B)
If A and B are independent, the expression for P(A OR B)
reduces to the addition rule:
P( A OR B) = P( A) + P( B) − P( A) P( B)
Generally, if the occurrence of an event A affects the
probability of occurrence of an event B, then they are said
to be dependent events.
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Question
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binomial distribution:

If a joint event can be represented as the outcome of n
identical elementary events, each capable of having only one
of two outcomes (outcome X, say, with probability p and
outcome Y with probability q = 1 - p) then the probability,
P(r), that the outcome of the joint event will contain r
occurrences of outcome X, and therefore n - r occurrences of
outcome Y, is given by the binomial distribution:
where is called a binomial coefficient and is given by:
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The binomial coefficients can be conveniently tabulated using
Pascal’s triangle as is shown below. The top row of Pascal’s
triangle corresponds to the binomial coefficient 0C0. Row 2
corresponds to 1C0 and 1C1. Row 3 corresponds to 2C0, 2C1
and 2C2, and so on.
Fig. Pascal’s triangle.
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The binomial distribution is an example of a
probability distribution, which is in general a
function that expresses the probability of an event
occurring. A probability distribution may be for either
discrete events (as in the binomial distribution) or for
continuous events (such as the probability of a
continuous variable taking a value within a given
range).

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Topic 2: Reliability
Basic concepts
The reliability function, R(t), is the probability that
a component will survive to time t. This can be
estimated by a test of N0 components:

where Ns(t) is the number of components surviving at time t.
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accumulate
d failures reliability
at start of
R(t)
day
day
failures
in day
survivors
at start
of day
1
2214
10000
0
1
2
1422
7786
2214
0.7786
3
976
6364
3636
0.6364
4
812
5388
4612
0.5388
5
686
4576
5424
0.4576
6
576
3890
6110
0.389
7
497
3314
6686
0.3314
8
410
2817
7183
0.2817
9
358
2407
7593
0.2407
10
300
2049
7951
0.2049
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failure function (Unreliability)

The failure function, Q(t), is the probability that a
component will have failed by time t. The failure function,
in terms of the results of a test on N0 components, is
defined as:
where Nf (t) is the number of components which have failed
by time t.
Because survival and failure are mutually exclusive and
exhaustive events:
R(t ) + Q(t ) = 1

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accumulat
ed failures failure function
at start of
Q(t)
day
day
failures
in day
survivors
at start of
day
1
2214
10000
0
0
2
1422
7786
2214
0.2214
3
976
6364
3636
0.3636
4
812
5388
4612
0.4612
5
686
4576
5424
0.5424
6
576
3890
6110
0.611
7
497
3314
6686
0.6686
8
410
2817
7183
0.7183
9
358
2407
7593
0.7593
10
300
2049
7951
0.7951
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failure density function
It is often useful to know the probability of failure during a small
interval around time t. This can be found from the failure density
function, ƒ(t). The probability of failing, Q(t), during a very
small interval t at t can be expressed as:
Q(t) = ƒ(t)  t
Fig. Failure densityFunction.

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What is the probability of failure at day 10?
day
failures
in day
surviv accumulate probabi
ors at
d failures
lity of
start
at start of
failure
of day
day
in day
1
2214
10000
0
0.2214
2
1422
7786
2214
0.1422
3
976
6364
3636
0.0976
4
812
5388
4612
0.0812
5
686
4576
5424
0.0686
6
576
3890
6110
0.0576
7
497
3314
6686
0.0497
8
410
2817
7183
0.041
9
358
2407
7593
0.0358
10
300
2049
7951
0.03
11
262
1749
8251Arab Open University-Lebanon
0.0262
Tutorial 7
Prob of failure by day 10
Prob of failure by day 11
Q(t)=0.8251-0.7951
=0.03=Prob of failure
ON day 10
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Note: prob of failure on each day decreases with
time reaching zero on day 32
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
For a continuous reliability function R(t) and failure
function Q(t), the relationship to the failure density
function ƒ(t) can be found by differentiation:

Equivalently, Q(t) and R(t) can be found by
integrating ƒ(t):

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The integrals can be thought of as areas under
the curve of ƒ(t). The relationship between the
failure density function, ƒ(t), and the reliability
function, R(t), and failure function, Q(t), is
shown below.

Fig. Relationship between failure
density function,ƒ(t), reliability
function, R(t), and
failure function, Q(t).
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The assumption that all components will eventually fail
implies that:


This is known as the normalization condition.
The reliability function, R(t), and failure function,
Q(t), are known as cumulative probability
distributions because they give the probability of
something happening up to a specified time.

The failure density function, ƒ(t), is a probability
density function because it is used to give the
probability of failure during an interval of time.

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Hazard rate
The hazard rate, h(t), or failure rate, is
the ratio of the number of failures per
unit time, at time t, to the number of
components exposed to failure at that
time:

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Question
day
failures in day
survivors at
start of day
1
2214
10000
2
1422
7786
3
976
6364
4
812
5388
5
686
4576
6
576
3890
7
497
3314
410
2817
358
2407
300
2049
262
1749
218
1487
184
1269
157
1085
137
928
How many components
survived to start of day
10?___________
How many failed during
day 10?________
8
Given a component
9
survived until day 10.
10
(i.e. this is conditional
11
probability) What is the
probability it might fail
12
DURING day 10?_____
13
Calculate the same for
14
day 16?__________ Arab Open University-Lebanon
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Plots of hazard rate against time often take the form
of a ‘bath-tub’ curve shown below, with a steep fall in
the burn-in period (early life and it is when faulty
components are weeded out), a level useful life period
(or normal operating period) when h(t) is effectively
constant, and wear-out (or end-of-life period), when the
hazard rate rises steeply.
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The hazard rate is also related to the failure
density function and the reliability function by
the equation:
During the useful life period, the hazard rate,
h, is independent of time. The failure process
during the useful life period is known as
memoryless, because the probability of
failure is independent of previous history.
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Question
ON Day 14
What is hazard rate
h(t)=_________
What is the failure
density Function
f(t)=___________
day
failures in day
survivors at
start of day
1
2214
10000
2
1422
7786
3
976
6364
4
812
5388
5
686
4576
6
576
3890
7
497
3314
8
410
2817
9
358
2407
10
300
2049
11
262
1749
12
218
1487
13
184
1269
14
157
1085
15
137
928
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Exponential probability distribution
During useful life, when the hazard rate, h, is
constant, the reliability function and failure
density function are simple exponential functions
of time:
R(t ) = exp(-ht )


and
f (t ) = h exp(−ht )


These functions represent examples of the
exponential probability distribution. Figure
below shows three examples of reliability
functions with constant hazard rates.
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Fig. Plots of reliability function for constant hazard
rate h, where the units of h are months-1.
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Reliability
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reliability R(t)
1.000
0.900
0.800
0.700
0.600
0.500
0.400
0.300
0.200
0.100
0.000
day
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
Day
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failures
in day
survivor
s at
start of
day
reliabil
ity R(t)
1
200
1000
1.000
2
160
800
0.800
3
128
640
0.640
4
102
512
0.512
5
82
410
0.410
6
66
328
0.328
7
52
262
0.262
8
42
210
0.210
9
34
168
0.168
10
27
134
0.134
11
21
107
0.107
12
17
86
0.086
13
14
69
0.069
14
11
55
0.055
15
9
44
0.044
16
7
35
0.035
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Figure below shows the corresponding
failure density functions.
Fig. Plots of failure density function for constant
hazard rate h, where the units of h are months-1.
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failu survivor
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COMMUNICATIONS
day
res
in
day
s at
start of
day
reliability
R(t)
failure
density
exp(-ht)
function
f(t)
1
200
1000
1.000
1.000
0.200
2
160
800
0.800
0.819
0.160
3
128
640
0.640
0.670
0.128
4
102
512
0.512
0.549
0.102
5
82
410
0.410
0.449
0.082
6
66
328
0.328
0.368
0.066
7
52
262
0.262
0.301
0.052
8
42
210
0.210
0.247
0.042
9
34
168
0.168
0.202
0.034
10
27
134
0.134
0.165
0.027
11
21
107
0.107
0.135
0.021
12
17
86
0.086
0.111
0.017
13
14
69
0.069
0.091
0.014
14
11
55
0.055
0.074
0.011
15
9
44
0.044
0.061
0.009
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0.035
0.050
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0.007
Note that
beginning of the
test, the value of
failure density
function is equal
to the hazard rate
h=0.2.
This is because at
the start of 1st
interval, the
number of
survivals is the
number of
components at
the start of the
45
test.
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Average lifetimes
The average lifetime of a batch of components is called
the mean time to failure (MTTF). The MTTF of a batch
of components can be found by recording the time at
which each component failed. For instance, if N1
components failed at time t1, N2 components at time t2,
and so on until all the components have failed, the MTTF
is given by the equation:

where  is the summation sign and the summations
extend over all the relevant values of i.
In the case of an exponential distribution with constant
hazard rate h, the MTTF is simply:

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15 components tested in 5 days What is
the MTTF?
Failure
time(days
1
Number of
components
0
2
3
4
5
1
3
5
6
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Confidence levels
The degree to which an MTTF
estimate for a type of component can
be trusted depends on the number of
components used in obtaining it. This
degree of trust is often expressed in
terms of confidence levels.

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MTBF, MTTR and availability
For repairable components, the mean time
between failures (MTBF) is used instead of the
mean time to failure (MTTF). It is defined as the
average time from when a component starts to be
operational until it fails– that is, it does not include
the time when the component is waiting to be
repaired.


The mean time to repair (MTTR) of a system is
the mean time taken to repair a fault. The system
availability, A, is the proportion of time for which
a system is operating correctly:
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Topic 3: Preparation for Next Tutorial
Please inform the students to do the following
activities before coming to the next tutorial:
Overview the Contents of Block

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