MARKOVIAN QUEUES IN TANDEM

Open Queueing Networks

6.1 MARKOVIAN QUEUES IN TANDEM

This is the situation where two queues are joined in series, as shown in Figure 6.3. Customers arrive at Queue 1 according to a Poisson process of mean l and are served by an exponential server with mean m1−1

. After completion of service, customers join the second queue and are again served by an exponen- tial server, who is independent of the server of Queue 1, with mean m2−1

. As usual, we assume that the arrival process is independent of any internal

Figure 6.2 An example of closed queueing networks

Queue 1 Queue 2

λ µ₁ µ₂

Figure 6.3 Markovian queues in tandem

processes in both queues. The waiting time at both queues are assumed to be infi nite so that no blocking occurs. This is an example of a simple feedback open network where no feedback path exists.

The analysis of this system follows the same approach as that of a single Markovian queue except now we are dealing with a two-dimensional state space.

Let us focus our attention on the system state (k1, k2), where k1 and k2 are the numbers of customers in Queue 1 and Queue 2, respectively. Since the arrival process is Poisson and the service time distributions are exponential, we can have only the following events occurring in an incremental time interval ∆t:

•

an arrival at Queue 1 with probability l∆t;

•

a departure from Queue 1, and hence an arrival at Queue 2, with probability m1∆t;

•

a departure from Queue 2 with probability m2∆t;

•

no change in the system state with probability [1 − (l + m1+ m2)∆t].

Using the same technique that we employed in studying birth-death pro- cesses, by considering the change in the joint probability P(k1, k2) in an infi ni- tesimal period of time ∆t, we have

P k k t t P k k t t P k k t t

P k k

( , ; ) ( , ; ) ( , ; )

( ,

1 2 1 2 1 2 1

1 2

1 1 1

1

+ = − + + −

+ +

∆ λ∆ µ∆

;; )t µ²∆t+P k k t( ,^{1 2}; )[1− + +(λ µ µ^{1 2})∆t] (6.1) Rearranging terms and dividing the equation by ∆t and letting ∆t go to zero, we arrive at a differential equation for the joint probability:

′ = − + + −

+ + −

P k k t P k k t P k k t

P k k t

( , ; ) ( , ; ) ( , ; )

( , ; )

1 2 1 2 1 2 1

1 2 2

1 1 1

1

λ µ

µ PP k k t( ,1 2; )[λ µ µ+ +1 2] (6.2) where P′(k1, k2; t) denotes the derivative. If l < m1 and l < m2, the tandem queues will reach equilibrium. We can then obtain the steady-state solution by setting the differentials to zero:

µ µ λ µ µ

1 2 1 2 1λ 1 2 2 1 2

1 2

1 1 1

1 + +

( ) = + − + +

+ −

P k k P k k P k k

P k k

( , ) ( , ) ( , )

( , ) (6.3)

Using the same arguments and repeating the process, we obtain three other equations for the boundary conditions:

λP( , )0 0 =µ2P( , )0 1 (6.4) (µ λ²+ ) ( ,P 0 k²)=µ¹P( ,1k²− +1) µ²P( ,0 k²+1) k²>0 (6.5) (µ λ¹+ ) ( , )P k¹ 0 =µ²P k( , )¹1 +λP k( ¹−1 0, ) k¹>0 (6.6)

MARKOVIAN QUEUES IN TANDEM 173

Students will be quick to notice that this set of equations resembles that of the one-dimensional birth-death processes and can be interpreted as fl ow bal- ancing equations for probability fl ow going in and out of a particular state. The state transition diagram which refl ects this set of fl ow equations is shown in Figure 6.4.

Similar to the one-dimensional case, this set of equations is not independent of each other and the additional equation required for a unique solution is pro- vided by the normalization condition:

P k k

k k

( ,1 2)

2 1

∑

1

∑

^{= (6.7)}

To solve Equations (6.3) to (6.6), let us assume that the solution has the so- called Product form, that is

P k k( ,1 2)=P k P k1( ) ( )1 2 2 (6.8) where P1(k1) is the marginal probability function of Queue 1 and is a function of parameters of Queue 1 alone; similarly P2(k2) is the marginal probability function of Queue 2 and is a function of parameters of Queue 2 alone.

Substituting Equation (6.8) into Equation (6.4), we have

λP2( )0 =µ2 2P( )1 (6.9) Using it together with Equation (6.8) in Equation (6.6), we have

µ1 1P k( )1 =λP k1( 1−1) (6.10)

P k P k

kP

1 1 1

1 1

1 1 1

1

1 0

1

( ) ( )

( )

= −

= =

λ µ

ρ ρ λ

where µ (6.11)

0,0 1,0

0,1

2,0

1,1

0,2

k₁,0

k₁–1,1

λ λ λ

λ µ2

µ2 µ2

µ1 µ1

µ1

λ λ

µ2

µ2

Figure 6.4 State transition diagram of the tandem queues

Since the marginal probability should sum to one, we have

P k P k

k k

1 1 2 2

1 2

1 1

∑

^{( )= and}

∑

^{( )= (6.12)}

Using the expression (6.12), we obtain P1(0) = (1 − r1), therefore

P k1( )1 = −(1 ρ ρ1) 1^k1 (6.13) Now, substituting Equations (6.10), (6.8) and (6.9) into Equation (6.3) and after simplifying terms, we arrive at

(λ µ+ ²) ( )P k^{2 2} =λP k²( ²− +1) P k²( ²+1)µ² (6.14) This is a recursive equation in P2(k2), which we solve by z-transform:

P k z P k z P k z

k

k k

k k 2 2

1

2 2

1

2 2

1

2 2

2

2 2

2 2

1 1

( )( ) ( ) ( )

=

∞

=

∞

=

∑

^{λ µ+ =}

∑

^{λ − +}

∑

∞ ^{+ µ kk2 (6.15)}

Defi ne the z-transform of P2(k2) as

P z P k z

k

k

2 2 2

2 0

( )= ( ) 2

=

∑

∞ ^(6.16)

We have from Equation (6.15):

(λ µ)[ ( ) ( )] λ ( ) µ ( ) ( ) λ ( )

+ − = +  − −µ



 P z P zP z 

z P z P P

P

2 2 2

2

2 2

2 2

2

0 0 0

(( )(z ρ2z−1)(z− =1) P2( )(0 1+ρ2)(1−z) where ρ2=λ µ/ 2

and

P z P

2 2 z

2 2

0 1 ( )= ( )1+

− ρ

ρ (6.17)

Similarly, since the marginal probability P2 should sum to 1, that is equivalent to P2(z)|z=1= 1, we arrive at

P2

2 2

0 1

( )= −1 + ρ

ρ (6.18)

and

P z^{2 2} 2z

1 1

( )= −( )1 ρ −

ρ (6.19)

MARKOVIAN QUEUES IN TANDEM 175 Inverting the z-transform, we have

P k2( )2 = −(1 ρ ρ2) 2^k2 (6.20) Therefore:

P k k( ,1 2)= −(1 ρ ρ1) 1^k1(1−ρ ρ2) 2^k2 (6.21) where ρ λ

µ ρ λ

1 µ

1 2

2

= and =

The expression of (6.21) holds for r0< 1 and r1< 1.

For an isolated M/M/1 queueing system, the probability that there are k customers in the system is P(k) = (1 − r)r^k, therefore

P(k1, k2) = (1 − r1)r1k₁

(1 − r2)r2k₂

= P1(k1)P2(k2)

We see that the joint probability distribution is the product of the marginal probability distributions, and hence the term Product-Form solution.

6.1.1 Analysis of Tandem Queues

The foregoing analysis provides us with a steady-state solution but fails to give us an insight into the interaction between the two queues. The fi nal expression seems to suggest that the two queues are independent of each other. Are they really so?

To answer that, let us examine the two tandem queues in more detail. First let us look at Queue 1. The customer arriving pattern to Queue 1 is a Poisson process and the service times are distributed exponentially, therefore it is a classical M/M/1 queue.

How about Queue 2, what is the customer arriving pattern, or in other words the inter-arrival time distribution? It is clear from the connection diagram that the inter-departure time distribution from Queue 1 forms the inter-arrival time distribution of Queue 2. It can be shown that the customer arriving pattern to Queue 2 is a Poisson process as follows.

When a customer, say A, departs from Queue 1, he/she may leave behind an empty system with probability (1 − r1) or a busy system with probability r1. In the case of an empty system, the inter-departure time between A and the next customer (say B) is the sum of B’s service time and the inter-arrival time between A and B at Queue 1. Whereas in the busy system case, the inter- departure time is simply the service time of B.

Therefore, the Laplace transform of the density function for the uncondi- tional inter-departure time (I) between A and B is given by

L f t

s s s

s [ ( )]I = −( )

+ ⋅

 +

 

+ +

= +

1 1

1 1

1 1

1

ρ λ

λ µ

µ ρ µ

µ λ

λ (6.22)

This is simply the Laplace transform of an exponential density function. Hence a Poisson process driving an exponential server generates a Poisson departure process. Queue 2 can be modelled as an M/M/1 queue.

6.1.2 Burke’s Theorem

The above discussion is the essence of Burke’s theorem. In fact, Burke’s theorem provides a more general result for the departure process of an M/M/m queue instead of just the M/M/1 queue discussed earlier. This theorem states that the steady-state output of a stable M/M/m queue with input parameter l and service-time parameter m for each of the m servers is in fact a Poisson process at the same rate l. The output is independent of the other processes in the system.

Burke’s theorem is very useful as it enables us to do a queue-by-queue decomposition and analyse each queue separately when multiple-server queues (each with exponential pdf service-times) are connected together in a feed forward fashion without any feedback path.

Example 6.1

Sentosa Island is a famous tourist attraction in Singapore. During peak hours, tourists arrive at the island at a mean rate of 35 per hour and can be approxi- mated by a Poisson process. As tourists complete their sightseeing, they queue up at the exit point to purchase tickets for one of the following modes of trans- portation to return to the mainland; namely cable car, ferry and mini-bus. The average service time at the ticket counters is 5 minutes per tourist.

Past records show that a tourist usually spends an average of 8 hours on sightseeing. If we assume that the sightseeing times and ticket-purchasing times are exponentially distributed; fi nd:

i) the minimum number of ticket counters required in operation during peak periods.