Analysis of open queueing network with dependent on time parameters of input flowand servicing

(1)

ANALYSIS OF OPEN QUEUEING NETWORK

WITH DEPENDENT ON TIME PARAMETERS OF INPUT FLOW AND SERVICING

Michal Matalycki¹, Andrzej Pankow²

1 Institute of Mathematics and Computer Science, Czestochowa University of Technology, Poland

2 Faculty of Mathematics and Computer Science, Grodno State University, Belarus

Abstract. The multivariate generating functions method for obtaining of the open queueing network state probabilities, functioning with heavy load conditions, with dependent on time parameters of input flow and servicing is considered in this paper.

1. The system of Kolmogorov difference-differential equations for queueing network state probabilities

Let’s consider open queueing network (QN), consisting of n queueing systems (QS), each of which has mi identical service lines. We will examine the case, when parameters of input flow and servicing parameters depends on time, i.e. in the time interval [t,t+∆t) message enters to the network with probability λ(t)∆t+o(∆t), and if at the time moment t there is an message being servicing at the i-th QS then it’s servicing will be finished in the time interval [t,t+∆t) with the probability

), ( )

(t t o t

i ∆ + ∆

µ i=1 n, . With the probability p₀_i message will come to peripheral system Si from the outside, i=1,n−1, 1,

1

1 0 =

∑

⁻

= n

i

p i after finished servicing in there this message will transfer to the central system Sn, and then will leave the network.

Service disciplines in network systems are FIFO. Each system has unlimited number of waiting places. Service time in queueing system’s lines have exponential distribution with various parameters for various systems. As a

networks state at

the moment of time t we will understand vector k⁽t⁾=⁽k^,t⁾=

(

k1^,k2^,Κ^,k_n^,t

)

^,

where ki - number of messages in QS Si, i=1 n, . The analysis of such network with central QS by multivariate generating functions method, when λ(t)=λ,

, )

( _i

i t =µ

µ i=1 n, , was carried out in paper [1]. Let us designate through Ii - vector of dimension n with zero-components, except for i-th, which is equal to 1.

(2)

Lemma. State probabilities of considered network satisfy to set of difference- -differential equations:

( )

⁺^λ ⁻ ⁺







λ + µ

−

=

∑ ∑

⁻

=

1

1 0 1

) , ( ) ( )

( ) , ( , min ) ( )

) ( ,

( ⁿ

i

i i

i n

i

i i

i t m k P k t t p u k P k I t

dt t t k dP

(

^, ¹

)

⁽ ^, ⁾ ⁽ ⁾^min

(

^, ¹

)

⁽ ^, ⁾

min ) ( ) (

1

t I k P k m t

t I I k P k m k

u

t _n _n _n _n

n

i

n i i

i n

i + + − +µ + +

µ +

∑

⁻

=

(1)

Proof. In view of servicing time exponentiality the random process k(t)=(k,t) is Markovian with countable states number. Following transitions to the state

) ,

( tk during period of time t∆ are possible:

a) from the state (k+I_i−I_n,t) with probability

(

^, ¹

)

⁽ ^),

min ) ( )

(t u k_n m_i k_i t o t

i + ∆ + ∆

µ i=1,n−1,

where u(x) - Heaviside function;

b) from the state (k−I_i,t) with probability λ(t)p₀_iu(k_i)∆t+o(∆t), i=1,n−1; c) from the state (k+I_n,t) with probability µn⁽^t⁾^min

(

^mn^,^kn+¹

)

∆^t+^o⁽∆^t^);

d) from the state ( tk, ) with probability

(

^,

)

⁽ ⁾

min ) ( )

( 1

1

t o t k m t

t

n

i

i i

i ∆ + ∆



 



λ + µ

−

∑

=

e) from other states with probability o( t∆ ).

Then using composite probability function it is possible to write

(

+

)

∆ + − +λ ∆ − +

µ

=

∆

+

∑ ∑

⁻

=

−

=

1

1 0 1

1

) , ( ) ( ) ( ) , (

1 , min ) ( ) ( )

, (

n

i

i i

i n

i

n i i

i n

i tuk m k tPk I I t t p uk tPk I t

t t k P

(

^, ¹

)

⁽ ^, ⁾ ¹ ⁽⁾ ⁽⁾^min

(

^,

)

⁽ ^, ⁾ ⁽ ⁾

min ) (

1

t o t k P t k m t t

t I k P t k m t

n

i

i i i

n n

n

n + ∆











 ∆



 



λ + µ

− + +

∆ + µ

+

∑

=

Dividing both left and right parts of this relation by t∆ and passing to limit where ,

→0

∆t we will obtain a set of equations for network state probabilities (1).

2. Network state probabilities for the case n = 3

(3)

Let n = 3 and all queueing systems are single-line, i.e. m_i =1, i=1 n, . Let’s suppose also, that all systems functioning in heavy load conditions, i.e.

0 ) (t >

k_i ∀t>0, i=1,n (there are any servicing messages at any moment of time). Then set (1) in the lemma will look like

+

− λ

 +





λ + µ

−

=

∑ ∑

=

2

1 0 3

1

) , ( ) ( )

( ) , ( ) ( )

) ( , (

i

i i

i t P k t t p u k P k I t

dt t t k dP

) , ( ) ( ) , (

) ( )

( ₃ ₃

2

1

3 P k I I t t P k I t

k u t

i

n i

i + − +µ +

µ +

∑

=

Let’s define three-dimensional generating function. We will assume for )

, , (z₁ z₂ z₃ z=

∑ ∑ ∑

^∞

=

∞

=

∞

=

∞

=

∞

=

∞

=

= Ψ

0 0 0

3 2 1

0 0 0

3 2 1 3 2 1 3

1 2 3

3 2 1

1 2 3

3 2

1 ( , )

) , , , ( )

, (

k k k

k k

k z z P k t z z z

z t k k k P t

z (2)

Multiplying each equation of set (1) by z₁^k¹z₂^k²z₃^k³ and summing up by all possible values k1, k2, k3 we can obtain:

 +





λ + µ

−

∑ ∑ ∑

^∞ =

∑ ∑ ∑ ∑

=

∞

=

∞

=

∞

=

∞

=

∞

=

0 0 0 0 0 0

3 2 1 3

1 3

2 1

1 2 3 1 2 3

2 3 3 1

2

1 ( ) ( ) ( , )

) , (

k k k k k k

k k k i

i k

k

k z z t t P k t z z z

dt z t k dP

+

− + µ

+

− λ

+

∑ ∑∑∑ ∑ ∑∑∑

=

∞

=

∞

=

∞

=

∞

=

∞

=

∞

=

2

1 0 0 0

3 2 1 3 3

3 2 1 2

1 0 0 0

0

1 2 3

3 2 1 3

2 1

1 2 3

) , (

) ( ) ( )

, ( )

( ) (

i k k k

k k k i

i k k

k

i k k k

i i

iuk P k I t z z z tuk P k I I t z z z

p t

∑ ∑ ∑

^∞

=

∞

=

∞

=

+ µ

+

0 0 0

3 2 1 3 3

1 2 3

3 2

) 1

, ( )

(

k k k

k k

k z z

z t I k P t

We will examine some sums from the right side. Let

2 3 1

1 2 3

3 2 1 2

1 0 0 0

2( , ) 0 ( ) ( , ) ^k ^k ^k

i k k k

i i

iu k P k I t z z z

p t

z

∑ ∑ ∑ ∑

∑

=

∞

=

∞

=

∞

=

−

=

Then

+

−

=

∑ ∑∑

∑

^∞

=

∞

=

∞

=

−

1 0 0

3 2 1 1 3 2 1 1

2 01

1 2 3

3 2

) 1

, , , 1 ( )

, (

k k k

k k

k z z

z t k k k P z

p t z

(4)

∑∑∑

^∞

=

∞

=

∞

=

− =

− +

0 1 0

3 1 2 1 3 2 1 2

02

1 2 3

3 2

) 1

, , 1 , (

k k k

k k

k z z

z t k k k P z

p

∑ ∑ ∑

^∞

=

∞

=

∞

=

∞

=

∞

=

∞

=

= +

=

0 0 0

3 2 1 3 2 1 2

02

0 0 0

3 2 1 3 2 1 1

01

1 2 3

3 2 1

1 2 3

3 2

1 ( , , , )

) , , , (

k k k

k k

kz z p z P k k k t z z z

z t k k k P z

p

∑

=

Ψ

= Ψ

+

= ²

1

3 0 3

2 02 1

01 ) ( , ) ( , )

(

i i

iz z t

p t

z z p z p

For the sum ¹ ² ³

1 2 3

3 2 1 2

1 0 0 0

3

3( , ) ( ) ( 3) ( , ) ^k ^k ^k

i k k k

i

i t u k P k I I t z z z

t

z

∑ ∑ ∑ ∑

∑

=

∞

=

∞

=

∞

=

− + µ

= we get:

+

− +

µ

=

∑ ∑ ∑

∑

^∞

=

∞

=

∞

=

0 0 0

3 2 1 3

2 1 3

3 1

1 2 3

2 3

) 1

, 1 , , 1 ( )

( ) ( ) , (

k k k

k k

kz z

z t k k k P k

u t t z

=

− + µ

+

∑ ∑ ∑

^∞

=

∞

=

∞

=

0 0 0

3 2 1 3

2 1 3

2

1 2 3

) , 1 , 1 , ( )

( ) (

k k k

r m lz z z t k k k P k

u t

∑ ∑∑

^∞

=

∞

=

∞

=

+ − +

− +

µ

=

0 0 1

1 3 2 1 1 3

2 1 1

3 1

1 2 3

3 2

) 1

, 1 , , 1 ( )

(

k k k

k k

k z z

z t k k k z P

t z

=

− + µ

+

∑ ∑∑

^∞

=

∞

=

∞

=

− +

0 0 1

1 3 1 2 1 3

2 1 2

3 2

1 2 3

3 2

) 1

, 1 , 1 , ( )

(

k k k

k k

kz z

z t k k k z P

t z

= µ

+ µ

=

∑ ∑ ∑ ∑∑ ∑

^∞

=

∞

=

∞

=

∞

=

∞

=

∞

= 0 1 0

3 2 1 3 2 1 2

3 2

1 0 0

3 2 1 3 2 1 1

3 1

1 2 3

2 3 1

1 2 3

2 3

1 () ( , , , )

) , , , ( )

(

k k k

k k

k P k k k t z z z

z t z z

z z t k k k z P

t z

− µ

−

Ψ







µ +µ

=

∑ ∑

^∞

=

∞

=

0 0

3 2 3 2 1

3 1 3

2 2 1 1 3

2 3

3

) 2

, , , 0 ( )

( ) , ) ( ( ) (

k k

k k z z t k k z P

t z t z z

t z

z t

3

1 3

0 0

1 3 1 3

3

2( ) ( ,0, , ) ^k

k k

k z z t k k z P

t z

∑ ∑

^∞

=

∞

=

µ

−

And finally for the last sum

∑ ∑ ∑ ∑

^∞

=

∞

=

∞

=

+ µ

=

0 0 0

3 2 1 3

2 1 4 3

1 2 3

3 2

) 1

, 1 , , ( )

( ) , (

k k k

k k

kz z

z t k k k P t

t z

we can have:

∑ ∑ ∑

∑

^∞

=

∞

=

∞

=

+ = µ +

=

0 0 0

1 3 2 1 3

2 1 3

3 4

1 2 3

3 2

) 1

, 1 , , ) (

) ( , (

k k k

k k

k z z

z t k k k z P

t t z

(5)

∑ ∑

∑ ∑∑

^∞

=

∞

=

∞

=

∞

=

∞

=

−µ µ Ψ

µ =

=

0 0

2 1 2 1 3

3 3

3 3 3 2 1

0 0 1

3 2 1 3

3

1 2

2 1 3

2 1

1 2 3

) , 0 , , ) (

) ( , ) ( ) (

, , , ) (

(

k k

k k k

k k

k k k

z z t k k z P

t t z z

z t z z t k k k z P

t

Thus, taking into account designation (2), we obtain heterogeneous linear differential equation for generating function

( )

^Ψ ⁻











 −µ







µ +µ

− +

λ

− µ + λ

−

Ψ =

∑

=

) , ) ( ( ) ( ) ) (

( ) ( )

) ( , (

3 3

1 3

3 2

2 1 1 3 2 02 1 01

3 z t

z t z

t z

z t z p z p t t dt t

t z d

i i

− µ

−

∑ ∑ ∑ ∑

^∞

=

∞

=

∞

=

∞

= 0 0

3 1 3 1 2

3 2 3

0 0

2 3 2 1

3 1

1 3

3 1 3

2 3

2 ( ) ( ,0, , )

) , , , 0 ( )

(

k k

k k k

k k

k P k k t z z

z t z z

z t k k z P

t z

∑ ∑

^∞

=

∞

=

−µ

0 0

2 1 2 1 3

3

1 2

2

) 1

, 0 , , ) (

(

k k

k kz z t k k z P

t (3)

Since all QS functioning in heavy load conditions, last three addends in equation (3) are equal to zero and equation becomes homogeneous:

( )

⁽ ⁾ ⁽ ⁾ ⁽ ⁾ ⁽ ^, ⁾

) ( ) ( )

) ( , (

3 3

1 3

3 2

2 1 1 3 2 02 1 01

3 z t

z t z

t z

z t z p z p t t dt t

t z d

i

i Ψ











 −µ







µ +µ

− +

λ

− µ + λ

−

Ψ =

∑

=

(4) We will consider that QN was free at initial state, i.e. P(0,0,0,0)=1. Then starting condition for the last equation will be Ψ₃(z,0)=1. So, we get homogeneous equation (4) for the generating function. Solving it with starting condition we obtain:

(

^Λ ⁻^Λ

)

_×

=

Ψ

∑∑∑∑∑

^∞

=

∞

=

∞

=

∞

=

∞

=

+

0 0 0 0 0

02 01 0

3 ! !! ! !

) 0 ( ) ) (

( ) , (

i j l s q

j j i i

q s l j i

p p t t

a t z

(

^Μ ^t ⁻^Μ

) (

^l ^Μ ^t ⁻^Μ

) (

^s ^Μ ^t ⁻^Μ

)

^q^zⁱ⁻^l^z^j⁻^s^z^l⁺^s⁻^q

× ₁( ) ₁(0) ₂( ) ₂(0) ₃( ) ₃(0) ₁ ₂ ₃ (5)

where ^Λ(^t)⁼

∫

^λ(^t)^dt, ^Μⁱ⁽^t⁾⁼

∫

^µⁱ⁽^t⁾^dt^, ⁱ⁼¹^,³^,

( ) ( )

^



 



− Λ −Λ − Μ −Μ

=

∑

= 3

1

0() exp () (0) () (0)

i

i i t t

t a

Example 1. Let’s examine case when service rates λ(t), µ_i(t), i=1 n, , do not depend on time, i.e. λ(t)=λ, µ_i(t)=µ_i, i=1 n, . Then Λ(t)=λt, Μ_i(t)=µ_it,

(6)

. 3 ,

=1

i We will find, for example, probability state P(1,1,1,t), which equals to coefficient of z₁z₂z₃ in decomposition of Ψ₃(z,t). Thus degrees at z₁z₂z₃ should satisfy to relationships:







=

− +

=

−

=

−

, 1 , 1 , 1

q s l

s j

l i

i.e.











>

− +

=

−

=

−

=

. 0 ,

, 3 , 1 , 1

j i

j i q

j s

i l

So in this case, following from (5),

∑

^∞

>

+=

−

− + +

−

 +











λ+ µ

−

− +

−

µ µ µ

∑ λ

= ⁼

2 , 0 ,

5 3 3 3 3 1 2 1 1 02 01

)!

3 (

)!

1 ( )!

1 (

! ) !

, 1 , 1 , 1 (

3 1

j i

j i j i j i j j j t i

j t i j i j i

p e p

t

P ⁱ

i

3. Network state probabilities with arbitrary n

Now we will generalize our results for the case of arbitrary number of queueing systems, but, like before, we will suppose that systems are single-line. Then set of equations for state probabilities (1) will look like:

+

− λ

 +





λ + µ

−

=

∑ ∑

⁻

=

1

1 0 1

) , ( ) ( )

( ) , ( ) ( )

) ( ,

( ⁿ

i

i i

i n

i

i t P k t t p u k P k I t

dt t t k dP

) , ( ) ( ) , (

) ( ) (

1

t I k P t t I I k P k u

t _n _n

n

i

n i n

i + − +µ +

µ +

∑

⁻

=

As before, let us assume that network systems functioning in heavy load conditions, i.e. k_i(t)>0 ∀t>0, i=1 n, . We will define n-dimensional generating function:

∑ ∑ ∑

^∞

∏

=

∞

=

∞

= =

= Ψ

0 0 0 1

2 1

1 2

) , ,.., , ( ...

) , (

k k k

n

i k i n

n

z i

t k k k P t

z (6)

Theorem. Expression for generating function (6) looks like

( )

_×

⋅

⋅ Λ

−

= Λ

Ψ

∑∑ ∑ ∑∑ ∑ ∑

^∞

=

∞

=

∞

=

∞

=

∞

=

∞

=

∞

= − −

+ − + +

− −

−

0 0 0 0 0 0 0 1 2 1 1 2 1

1 0 02

01 0

1 2 1 1 2 1

1 2

1 1 2 1

!

! ) 0 ( ) ) (

( ) , (

l l l r r r q n n

l n l

l l l l n

n n

q r r r l l l

p p

p t t

a t

z Κ Κ

Κ Κ

Κ ^Κ

(7)

(

^Μ ⁻^Μ

) (

^Μ ⁻^Μ

)

^⋅ ^⋅

(

^Μ ⁻^Μ

) (

^Μ ⁻^Μ

)

^×

× ₋ ₋ ⁻ n n ^q

r n n

r

r t t t

t) (0) ( ) (0) ( ) (0) ⁿ ( ) (0)

( ₁ ¹ ₂ ₂ ² ₁ ₁ ¹

1 Κ

q r r r n r l n r

l r

l _n _n _n

z z z

z ⁻ ⁻ ⋅ ⋅ ₋⁻⁻ ⁻ ⁺ ⁺ ⁺ ⁻⁻

× ₁¹ ¹ ₂² ² Κ ₁¹ ¹ ¹ ² ^Κ ¹ (7)

where ⁽ ⁾ ^exp

(

⁽ ⁾ ⁽⁰⁾

) (

⁽ ⁾ ⁽⁰⁾

)

^.

1

0 



− Λ −Λ − Μ −Μ

=

∑

= n

i

i i t t

t a

Proof. Doing investigations similar to the case n = 3, we obtain homogeneous linear differential equation for the function (6):

) , ) ( ( ) ) (

( ) ( )

) ( ,

( ¹

1 1

1 0 1

t z z

t z

z t z p t t dt t

t z d

n n n n

i i

i n n

i i i n

i i

n Ψ



 



λ + µ −λ − µ −µ

−

Ψ =

∑ ∑ ∑

⁻

=

−

=

The solution of this equation with starting condition Ψ_n(z,0)=1 is function

( ) ( ) ( )









Λ −Λ + Μ −Μ − Λ −Λ −

−

=

Ψ

∑ ∑

⁻

=

1

1 0 1

) 0 ( ) ( ) 0 ( ) ( )

0 ( ) ( exp ) , (

n

i i i n

i

i i

n z t t t t p z

( )







 Μ 

−

−Μ Μ

− Μ

−

∑

⁻

= n

n n

n

i i

n i

i z

t z

t z ( ) (0)

) 0 ( ) (

1

Let’s re-arrange (7) to form, convenient for state probabilities obtaining:

( )

=







Μ −Μ







 Μ −Μ







Λ −Λ

=

Ψ

∑ ∑

⁻

=

−

= n

n n

n

i i

i i n n

i i i

n z

t z

z t z

p t

t a t

z () (0)

) exp 0 ( ) exp (

) 0 ( ) ( exp ) ( ) , (

1

1 1

1 0 0

) , ( ) , ( ) , ( )

( ₁ ₂ ₃

0 t a z t a z t a z t

=a

where

( )

=







Λ −Λ

=

∑

⁻

= 1

1 0 1( , ) exp ( ) (0)

n

i i iz p t

t z a

( )

∑∑ ∑

^∞

=

∞

=

∞

= − −

− + + +

−

− −

⋅

⋅ ⋅

⋅ Λ

−

= Λ

0 0 0

1 2

1 1 0 02 01 1

2

1 2 1 1

1 2

1 1 2

1 1 2 1

!

! ) 0 ( ) (

l l l

l n l

l l

n l

l n

l l l

n

n n

n

z z z p p

l p l l

t Κ Κ

Κ Κ ^Κ

=





 Μ −Μ

=

∑

⁻

= 1

1 2

) 0 ( ) exp (

) , (

n

i i

n z

z t t

z a

(

Μ −Μ

) (

Μ −Μ

)

⋅ ⋅

(

Μ −Μ

)

×

⋅

=

∑∑ ∑

^∞ ⋅

=

∞

=

∞

= − −

− + + +

−

− −

0 0 0

1 1

2 2

1 1 1 2

1 2 1 1

1 2

1 1

2 1

) 0 ( ) ( )

0 ( ) ( ) 0 ( )

! (

!

r r r !

r n n

r r

n r r r n

n

n n

t t

r t r r

z Κ

Κ Κ^Κ

1 2

1

1 2

1 ⁻

−−

−

− ⋅ ⋅

× n^rⁿ

r

rz z

z Κ

(8)

( )

[ ]

∑

^∞

=

Μ −

−

= Μ







Μ −Μ

=

0

1

3 !

) 0 ( ) ( )

0 ( ) exp ( ) , (

q

q n n n

n n n

q z t

z t t

z a

Multiplying of these functions gives us required result.

Example 2. Let’s obtain, for example, in the case when λ(t)=λ, µ_i(t)=µ_i, ,

, 1 n

i= state probability P(1,1,1,...,1,t), which equal to coefficient of

∏

= n

i

zi 1

in decomposition of Ψ_n( tz, ). Let’s compose relationships for degrees:















=

− + + +

=

−

=

−

=

−

, 1 ...

, 1 . . . . . . . . . .

, 1

1 2

1 1 1

2 2

1 1

q r r

r r l

r l

n n n

i.e.







−

=

−

=

∑

⁻

=

. 1 , 1

1

1 n

i i i i

l q

l r

Then, according (7),

×

= ( ) ) , 1 , , 1 , 1

( t a₀ t

P Κ

− × + + +

−

⋅

−

⋅

µ µ

⋅

⋅ µ µ

⋅

×

∑∑ ∑

^∞ λ

=

∞

=

∞

= − − −

− + + +

−−

−

− − +

+ +

−

1 1 1 1 2 1 1 2 1 1 2 1

1 1 1

2 1 1 1 0 02 01

1 2 1

1 2 1 1 2

1 1 2

1 1 2 1

)!

( )!

1 ( )!

1 (

!

l l l ! n n n

n l l l n l n l

l l

n l

l l l l

n

n n

n l l

l l

l l l l l

p p

p

Κ Κ

Κ

Κ ^Κ Κ ^Κ

1 2 3 3

3₁+ ₂+ + ₋₁− +

×t ^l ^l ^Κ ^lⁿ ⁿ

References

[1] Matalytski M., Gomanchuk S., Pankow A., Analysis of open queueing networks using method of generating functions, Scientific Research of the Institute of Mathematics and Computer Science, Czestochowa University of Technology 2002, 1(1), 143-152.