Let
-
λ
i
,
j
=
rate at which jobs leaving queue i go to queue j
λ
i
,
j
rate at which jobs leaving queue i go to queue j
-
q
i
,
j
=
probability that a job departing queue i goes
directly to queue j
q
i
,
j
probability that a job departing queue i goes
directly to queue j
During an interval of length
Δt
Δ
t
, only four possible events may occur:
- a job arrives from the outside world
- a job departs to the outside world
- a job leaves on queue and enters another
- none of the above
These four possibilities are incorporated into the following
equation:
P
k
1
k
2
…
k
M
t+Δt=∑j=1MP
k
1
k
2
…
k
j
-
1
k
j
−1
k
j
+
1
…
k
M
t
λ
0
,
j
Δt+∑i=1MP
k
1
k
2
…
k
i
-
1
k
i
+1
k
i
+
1
…
k
M
t
u
i
q
i
,
0
Δt+∑i=1M∑j=1MP
k
1
k
2
…
k
i
-
1
k
i
+1
k
i
+
1
…
k
j
-
1
k
j
−1
k
j
+
1
…
k
M
t
u
i
q
i
,
j
Δt+P
k
1
k
2
…
k
M
t(1−Δt∑j=1M
λ
0
,
j
+
μ
j
)
P
k
1
k
2
…
k
M
t
Δ
t
j
1
M
P
k
1
k
2
…
k
j
-
1
k
j
1
k
j
+
1
…
k
M
t
λ
0
,
j
Δ
t
i
1
M
P
k
1
k
2
…
k
i
-
1
k
i
1
k
i
+
1
…
k
M
t
u
i
q
i
,
0
Δ
t
i
1
M
j
1
M
P
k
1
k
2
…
k
i
-
1
k
i
1
k
i
+
1
…
k
j
-
1
k
j
1
k
j
+
1
…
k
M
t
u
i
q
i
,
j
Δ
t
P
k
1
k
2
…
k
M
t
1
Δ
t
j
1
M
λ
0
,
j
μ
j
(1)
Moving the
P
k
1
k
2
…
k
M
t
P
k
1
k
2
…
k
M
t
term to the left hand side of the equation,
dividing both sides by
Δt
Δ
t
, and taking the limit as
Δt→0
Δ
t
0
gives us
ddtP
k
1
k
2
…
k
M
t=∑j=1MP
k
1
k
2
…
k
j
-
1
k
j
−1
k
j
+
1
…
k
M
t
λ
0
,
j
+∑i=1MP
k
1
k
2
…
k
i
-
1
k
i
+1
k
i
+
1
…
k
M
t
u
i
q
i
,
0
+∑i=1M∑j=1MP
k
1
k
2
…
k
i
-
1
k
i
+1
k
i
+
1
…
k
j
-
1
k
j
−1
k
j
+
1
…
k
M
t
u
i
q
i
,
j
−∑j=1MP
k
1
k
2
…
k
M
t(
λ
0
,
j
+
μ
j
)=0
t
P
k
1
k
2
…
k
M
t
j
1
M
P
k
1
k
2
…
k
j
-
1
k
j
1
k
j
+
1
…
k
M
t
λ
0
,
j
i
1
M
P
k
1
k
2
…
k
i
-
1
k
i
1
k
i
+
1
…
k
M
t
u
i
q
i
,
0
i
1
M
j
1
M
P
k
1
k
2
…
k
i
-
1
k
i
1
k
i
+
1
…
k
j
-
1
k
j
1
k
j
+
1
…
k
M
t
u
i
q
i
,
j
j
1
M
P
k
1
k
2
…
k
M
t
λ
0
,
j
μ
j
0
(2)
in steady state, which will exist as long as
λ
j
<
μ
j
λ
j
μ
j
,
1≤j≤M
1
j
M
. Hence,
∑j=1MP
k
1
k
2
…
k
M
t(
λ
0
,
j
+
μ
j
)=∑j=1MP
k
1
k
2
…
k
j
-
1
k
j
−1
k
j
+
1
…
k
M
t
λ
0
,
j
+∑i=1MP
k
1
k
2
…
k
i
-
1
k
i
+1
k
i
+
1
…
k
M
t
u
i
q
i
,
0
+∑i=1M∑j=1MP
k
1
k
2
…
k
i
-
1
k
i
+1
k
i
+
1
…
k
j
-
1
k
j
−1
k
j
+
1
…
k
M
t
u
i
q
i
,
j
j
1
M
P
k
1
k
2
…
k
M
t
λ
0
,
j
μ
j
j
1
M
P
k
1
k
2
…
k
j
-
1
k
j
1
k
j
+
1
…
k
M
t
λ
0
,
j
i
1
M
P
k
1
k
2
…
k
i
-
1
k
i
1
k
i
+
1
…
k
M
t
u
i
q
i
,
0
i
1
M
j
1
M
P
k
1
k
2
…
k
i
-
1
k
i
1
k
i
+
1
…
k
j
-
1
k
j
1
k
j
+
1
…
k
M
t
u
i
q
i
,
j
(3)
Assume that the network of queues has a product-form
solution; i.e., assume
P
k
1
k
2
…
k
M
=∏i=1M(1−
ρ
i
)
ρ
i
k
i
P
k
1
k
2
…
k
M
i
1
M
1
ρ
i
ρ
i
k
i
Substitute this solution into the steady state equation and
cancel common terms:
∑j=1M
λ
0
,
j
+
μ
j
=∑j=1M
λ
0
,
j
ρ
j
+∑i=1M
ρ
i
μ
i
q
i
,
0
+∑i=1M∑j=1M
ρ
i
ρ
j
μ
i
q
i
,
j
j
1
M
λ
0
,
j
μ
j
j
1
M
λ
0
,
j
ρ
j
i
1
M
ρ
i
μ
i
q
i
,
0
i
1
M
j
1
M
ρ
i
ρ
j
μ
i
q
i
,
j
(4)
Looking at the individual terms,
∑j=1M
λ
0
,
j
ρ
j
=∑j=1M
λ
0
,
j
μ
j
λ
j
j
1
M
λ
0
,
j
ρ
j
j
1
M
λ
0
,
j
μ
j
λ
j
∑i=1M
ρ
i
μ
i
q
i
,
0
=∑i=1M
λ
i
(1−∑i=1M
q
i
,
j
)=∑i=1M
λ
i
−∑i=1M∑j=1M
λ
i
q
i
,
j
=∑j=1M
λ
0
,
j
i
1
M
ρ
i
μ
i
q
i
,
0
i
1
M
λ
i
1
i
1
M
q
i
,
j
i
1
M
λ
i
i
1
M
j
1
M
λ
i
q
i
,
j
j
1
M
λ
0
,
j
∑i=1M∑j=1M
ρ
i
ρ
j
μ
i
q
i
,
j
=∑j=1M
μ
j
λ
j
∑i=1M
λ
i
q
i
,
j
=∑j=1M
μ
j
λ
j
(
λ
j
−
λ
0
,
j
)=∑j=1M
μ
j
−∑j=1M
μ
j
λ
0
,
j
λ
j
i
1
M
j
1
M
ρ
i
ρ
j
μ
i
q
i
,
j
j
1
M
μ
j
λ
j
i
1
M
λ
i
q
i
,
j
j
1
M
μ
j
λ
j
λ
j
λ
0
,
j
j
1
M
μ
j
j
1
M
μ
j
λ
0
,
j
λ
j
Substituting these terms back into the previous equation
gives us
∑j=1M
λ
0
,
j
+∑j=1M
μ
j
=∑j=1M
λ
0
,
j
μ
j
λ
j
+∑j=1M
λ
0
,
j
+∑j=1M
μ
j
−∑j=1M
λ
0
,
j
μ
j
λ
j
j
1
M
λ
0
,
j
j
1
M
μ
j
j
1
M
λ
0
,
j
μ
j
λ
j
j
1
M
λ
0
,
j
j
1
M
μ
j
j
1
M
λ
0
,
j
μ
j
λ
j
(5)
proving the theorem.