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Or, if we consider that
λ
i
=
λ
i
1
+
λ
i
2
, the equation above can be re-written as
follows.
1
μ
i
=
(5)
(
λ
i
1
1
μ
i
1
)+(
λ
i
2
1
μ
i
2
λ
i
×
λ
i
×
)
These relationships can be easily established by considering the machine with
the two failure modes as two serial machines with single failure mode for each
machine.
If we take into account the assumption that the two failure modes cannot
occur simultaneously the equivalent failure rate will be the following.
λ
i
=
λ
i
1
+
λ
i
2
−
λ
i
1
×
λ
i
2
(6)
The equivalent repair rate will be changed consequently.
1
μ
i
=
(7)
λ
i
1
1
μ
i
1
λ
i
2
1
μ
i
2
(
λ
i
1
+
λ
i
2
−λ
i
1
×λ
i
2
×
)+(
λ
i
1
+
λ
i
2
−λ
i
1
×λ
i
2
×
)
4 Equivalent Machine Method Formulation
The main idea of the proposed approach is to replace each machine by an equiv-
alent one that has only up and down states. The blockage and starvation are
integrated in the up state of the machine. This formulation is a generalization
of two-machine-one-buffer production line proposed by Ouazene et al. [13,11]. In
this paper, we adapt the equivalent machine method proposed by the authors
[12] to deal with discrete flow of material to the continuous case.
Based on the analysis of the buffers steady states using birth-death Markov
processes, we determine the probabilities of starvation and blockage of each
buffer. Then, these probabilities are used to model and analyze the system
behavior in its steady state.
4.1 Two-machine-one-buffer Model
Before detailing the general formulation, we introduce the simple system which
consists of two machines separated by one buffer. This model is used as a building
block to construct the general model. To analyze the steady states of the buffer,
we consider a birth-death Markov process with (
N
+ 1) states
{
0
,
1
...N}
such as
N
is the capacity of the intermediate buffer and
ω
1
and
ω
2
are, respectively, the
birth and death transition rates.
The differential equations for the probability that the system is in state
j
at
time
t
are:
⎧
⎨
∂P
0
(
t
)
∂t
=
−ω
1
× P
0
(
t
)+
ω
2
× P
1
(
t
)
∂P
j
(
t
)
∂t
(8)
=
ω
1
×
P
j−
1
(
t
)
−
(
ω
1
+
ω
2
)
×
P
j
(
t
)+
ω
2
×
P
j
+1
(
t
)
⎩
∂P
N
(
t
)
∂t
=
ω
1
×
P
N−
1
(
t
)+
ω
2
×
P
N
(
t
)
