Therefore, this is expressed mathematically:
PT
¼
P
ð
8
:
2
Þ
114
8
Markov Processes
where
P
is the limiting state probability vector and
T
is the transition matrix. This
may be rewritten as:
P
ð
T
±
I
Þ ¼
0
ð
8
:
3
Þ
where
I
is the identity matrix.
Substituting the transition matrix
T
into the above equation gives the following
expression.
P
1
P
2
P
3
P
4
½
²
±ð
k
1
þ
k
2
Þ
k
1
k
2
0
l
1
±ð
l
1
þ
k
2
Þ
0
k
2
l
2
0
±ð
l
2
þ
k
1
Þ
k
1
0
l
2
l
1
±ð
l
1
þ
l
2
Þ
2
6
6
6
4
3
7
7
7
5
¼
0 0 0 0
½
ð
8
:
4
Þ
By taking the transpose of the equation, the general form of the equation is
obtained.
±ð
k
1
þ
k
2
Þ
l
1
l
2
0
k
1
±ð
l
1
þ
k
2
Þ
0
l
2
k
2
0
±ð
l
2
þ
k
1
Þ
l
1
0
k
2
k
1
±ð
l
1
þ
l
2
Þ
2
6
6
4
3
7
7
5
P
1
P
2
P
3
P
4
2
6
6
4
3
7
7
5
¼
0
0
0
0
2
6
6
4
3
7
7
5
ð
8
:
5
Þ
8.2.4 Step 4: Full Probability Condition
The sum of all the individual probabilities is equal to 1.
P
1
þ
P
2
þ
P
3
þ
P
4
½
¼
1
This condition is required to be able to solve the above equation as it contains
only
n

1 independent equations and there are four state variables involved.
Therefore, any row within the above equation can be replaced with this condition,
such as the first row.
00
11
01
10
μ
1
μ
2
μ
1
μ
2
λ
1
λ
1
λ
2
λ
2
State 1
State 3
State 2
State 4
Fig. 8.1
Diagram for a two
component repairable system
for availability analyses
8.2
Systems Availability Analyses
115
1
1
1
1
k
1
±ð
l
1
þ
k
2
Þ
0
l
2
k
2
0
±ð
l
2
þ
k
1
Þ
l
1
0
k
2
k
1
±ð
l
1
þ
l
2
Þ
2
6
6
4
3
7
7
5
P
1
P
2
P
3
P
4
2
6
6
4
3
7
7
5
¼
1
0
0
0
2
6
6
4
3
7
7
5
ð
8
:
6
Þ
8.2.5 Step 5: Solving the Markov Matrix Equation
Using Linear Algebra
As equation now contains four independent equations, it may be solved using
linear algebra. In general, this yields the following expressions.
P
1
¼
l
1
l
2
ð
l
1
þ
k
1
Þð
l
2
þ
k
2
Þ
ð
8
:
7
Þ
P
2
¼
k
1
l
2
ð
l
1
þ
k
1
Þð
l
2
þ
k
2
Þ
ð
8
:
8
Þ
P
3
¼
l
1
k
2
ð
l
1
þ
k
1
Þð
l
2
þ
k
2
Þ
ð
8
:
9
Þ
P
4
¼
k
1
k
2
ð
l
1
þ
k
1
Þð
l
2
þ
k
2
Þ
ð
8
:
10
Þ
8.3 Example of Markov Chains for Reliability Analyses
A simple system with one active component named primary system and one
standby redundant component named secondary system is connected with another
component, i.e., switch [
7
]. Figure
8.2
shows this system.
When switch fails, it is unable to switch. The failure of the switch matters, if it
is required to switch from the primary to the secondary system. If the switch fails
after the standby spare is already in use, then the system can continue operation.
However, if the switch fails before the primary unit fails, then the secondary
system cannot be turned on and the system fails when the primary unit fails. The
order in which the primary system and switch fail determines whether the system
continues operation.