PART II

(3) Continuous Time Markov Chains : Theory and Examples-Pure Birth Process with Constant Rates-Pure Death Process-More on Birth-and-Death Process-Statistical Equilibrium

(4) Introduction to Queueing Systems-Basic Elements of Queueing Models-Queueing Systems of One Server-Queueing Systems with Multiple Servers-Little’s Queueing Formula-Applications of Queues-An Inventory Model with Returns and Lateral Transshipments -Queueing Sys-tems with Two Types of Customers-Queues in Tandem and in Parallel

“All models are wrong / inaccurate, but some are useful.”George Box (Wikipedia).

http://hkumath.hku.hk/∼wkc/course/part2.pdf1

3 Continuous Time Markov Chains : Theory and Examples

We discuss the theory of birth-and-death processes, the analysis of which isrelatively simple and has important applications in the context of queueing theory.

• Let us consider a system that can be represented by a family of random variables{N(t)} parameterized by the time variable t. This is called a stochastic process.

• In particular, let us assume that for each t, N(t) is a non-negative integral-valuedrandom variable. Examples are the followings.

(i) a telephone switchboard, where N(t) is the number of calls occurring in aninterval of length t.

(ii) a queue, where N(t) is the number of customers waiting or in service at timet.

We say that the system is in state Ej at time t if N(t) = j. Our aim is to computethe state probabilities P{N(t) = j}, j = 0, 1, 2, · · · .

2

Definition 1 A process obeying the following three postulates is called abirth-and-death process:

(1) At any time t, P{Ej → Ej+1 during (t, t + h)|Ej at t} = λjh + o(h) ash → 0 (j = 0, 1, 2, · · · ). Here λj is a constant depending on j (State Ej).

(2) At any time t, P{Ej → Ej−1 during (t, t + h)|Ej at t} = µjh + o(h) ash → 0 (j = 1, 2, · · · ). Here µj is a constant depending on j (State Ej).

(3) At any time t, P{Ej → Ej±k during (t, t + h)|Ej at t} = o(h) as h → 0 fork ≥ 2 (j = 0, 1, · · · ,).

�

-

µi−1

λi−2· · · � -

µi

λi−1����Ei−1

�

-

µi+1

λi����

Ei ����Ei+1

�

-

µi+2

λi+1· · ·

Figure 2.1: The Birth and Death Process.

Notation: Let Pj(t) = P{N(t) = j} and let λ−1 = µ0 = P−1(t) = 0.

3

• Then it follows from the above postulates that (where h → 0; j =0, 1, . . ..)

Pj(t + h) = (λj−1h + o(h))︸ ︷︷ ︸an arrival

Pj−1(t) + (µj+1h + o(h))︸ ︷︷ ︸a departure

Pj+1(t)

+ [1− ((λj + µj)h + o(h))]︸ ︷︷ ︸no arrival or departure

Pj(t)

• Therefore we have

Pj(t + h) = (λj−1h)Pj−1(t) + (µj+1)hPj+1(t) + [1− (λj + µj)h]Pj(t) + o(h) .

• Re-arranging terms, we havePj(t + h)− Pj(t)

h= λj−1Pj−1(t) + µj+1Pj+1(t)− (λj + µj)Pj(t) +

o(h)

h.

4

• Letting h → 0, we get the differential-difference equationsd

dtPj(t) = λj−1Pj−1(t) + µj+1Pj+1(t)− (λj + µj)Pj(t). (3.1)

At time t = 0 the system is in state Ei, the initial conditions are

Pj(0) = δij where δij =

{1 if i = j0 if i ̸= j.

Definition 2 The coefficients {λj} and {µj} are called the birthand death rates respectively.

•When µj = 0 for all j, the process is called a pure birth process;

• and when λj = 0 for all j, the process is called a pure deathprocess.

• In the case of either a pure birth process or a pure death process,the equations (3.1) can be solved by using recurrence relation.

5

3.1 Pure Birth Process with Constant Rates

We consider a Pure birth process (µi = 0) with constantλj = λ and initial state E0.

• Equations in (3.1) becomed

dtPj(t) = λPj−1(t)− λPj(t) (j = 0, 1, · · · )

where P−1(t) = 0 and Pj(0) = δ0j.

• Herej = 0, P ′0(t) = −λP0(t),

henceP0(t) = a0e

−λt.

From the initial conditions, we get a0 = 1.

6

• Inductively, we can prove that if

Pj−1(t) =(λt)j−1

(j − 1)!e−λt

then the equation

P ′j(t) = λ

((λt)j−1

(j − 1)!e−λt

)− λPj(t)

gives the solution

Pj(t) =(λt)j

j!e−λt. (3.2)

Remark 1 Probabilities (3.2) satisfy the normalization condition∞∑j=0

Pj(t) = 1 (t ≥ 0).

Remark 2 For each t, N(t) is the Poisson distribution, given by {Pj(t)}. We saythat N(t) describes a Poisson process.

Remark 3 Since the assumption λj = λ is often a realistic one, the simple formula(3.2) plays a central role in queueing theory.

7

3.1.1 Generating Function Approach

Here we demonstrate using the generating function approach for solving thepure birth problem.

• Let {Pi} be a discrete probability density distribution for a random variable X ,i.e.,

P (X = i) = Pi, i = 0, 1, . . . ,

Recall that the probability generating function is defined as

g(z) =

∞∑n=0

Pnzn.

Let the probability generating function for Pn(t) be

g(z, t) =

∞∑n=0

Pn(t)zn.

• The idea is that if we can find g(z, t) and obtain its coefficients when expressedin a power series of z then one can solve Pn(t).

8

From the differential-difference equations, we have∞∑n=0

dPn(t)

dtzn = z

∞∑n=0

λPn(t)zn −

∞∑n=0

λPn(t)zn.

Assuming one can inter-change the operation between the summation and the dif-ferentiation, then we have

dg(z, t)

dt= λ(z − 1)g(z, t)

when z is regarded as a constant.

• Then we haveg(z, t) = Keλ(z−1)t.

Since

g(z, 0) =

∞∑n=0

Pn(0)zn = 1

we have K = 1. Hence we have

g(z, t) = e−λt(1 + λz +

(λtz)2

2!+ . . .+

)=

(e−λt + e−λtλz +

e−λt(λt)2

2!z2 + . . .+

).

Then the result follows.

9

3.2 Pure Death Process

We then consider a pure death process with µj = jµ and initial stateEn.

• Equations in (3.1) becomed

dtPj(t) = (j + 1)µPj+1(t)− jµPj(t) j = n, n− 1, · · · , 0 (3.3)

wherePn+1(t) = 0 and Pj(0) = δnj.

• We solve these equations recursively, starting from the case j = n.d

dtPn(t) = −nµPn(t) , Pn(0) = 1

implies that

Pn(t) = e−nµt.

10

• The equation with j = n− 1 isd

dtPn−1(t) = nµPn(t)− (n− 1)µPn−1(t) = nµe−nµt − (n− 1)µPn−1(t).

• Solving this differential equation and we get

Pn−1(t) = n(e−µt)n−1(1− e−µt).

• Recursively, we get

Pj(t) =

(n

j

)(e−µt)j(1− e−µt)n−j for j = 0, 1, · · · , n. (3.4)

Remark 4 For each t, the probabilities (3.4) comprise a binomialdistribution.

Remark 5 The number of equations in (3.3) is finite in number. Fora pure birth process, the number of equations is infinite.

11

3.3 More on Birth-and-Death Process

A simple queueing example is given as follows (An illustration of birth-and-deathprocess in queueing theory context).

• We consider a queueing system with one server and no waiting position, with

P{one customer arriving during (t, t + h)} = λh + o(h)

and

P{service ends in (t, t + h)| server busy at t} = µh + o(h) as h → 0.

• This corresponds to a two state birth-and-death process with j = 0, 1. The arrivalrates are λ0 = λ and λj = 0 for j ̸= 0 (an arrival that occurs when the server isbusy has no effect on the system since the customer leaves immediately); and thedeparture rates are µj = 0 when j ̸= 1 and µ1 = µ (no customers can completeservice when no customer is in the system).

�

-

µ

���Idle ��

��Busy

Figure 3.1. The Two-state Birth-and-Death Process.

12

The equations for the birth-and-death process are given by

d

dtP0(t) = −λP0(t) + µP1(t) and

d

dtP1(t) = λP0(t)− µP1(t). (3.5)

One convenient way of solving this set of simultaneous linear differential equations(not a standard method!) is as follows:

• Adding the two equations in (3.5), we getd

dt(P0(t) + P1(t)) = 0,

hence P0(t) + P1(t) = constant.

• Initial conditions are P0(0) + P1(0) = 1; thus P0(t) + P1(t) = 1. Hence we get

d

dtP0(t) + (λ + µ)P0(t) = µ.

The solution (exercise) is given by

P0(t) =µ

λ + µ+

(P0(0)−

µ

λ + µ

)e−(λ+µ)t.

13

Since P1(t) = 1− P0(t),

P1(t) =λ

λ + µ+

(P1(0)−

λ

λ + µ

)e−(λ+µ)t. (3.6)

• For the three examples of birth-and-death processes that we have considered, thesystem of differential-difference equations are much simplified and can therefore besolved very easily.

• In general, the solution of differential-difference equations is no easy matter. Herewe merely state the properties of its solution without proof.

Proposition 1 For arbitrarily prescribed coefficients λn ≥ 0, µn ≥ 0 therealways exists a positive solution {Pn(t)} of differential-difference equations(3.1) such that ∑

Pn(t) ≤ 1.If the coefficients are bounded, this solution is unique and satisfies the regu-larity condition

∑Pn(t) = 1.

Remark 6 Fortunately in all cases of practical significance, the regularity condi-tion

∑Pn(t) = 1 and uniqueness of solution are satisfied.

14

3.4 Statistical Equilibrium (Steady-State Probability Distribution)

Consider the state probabilities of the above example when t → ∞, from (3.6) wehave

P0 = limt→∞

P0(t) =µ

λ + µ

P1 = limt→∞

P1(t) =λ

λ + µ.

(3.7)

We note that P0 + P1 = 1 and they are called the steady-state probabilitiesof the system.

Remark 7 Both P0 and P1 are independent of the initial values P0(0) and P1(0).If at time t = 0,

P0(0) =µ

λ + µ= P0

P1(0) =λ

λ + µ= P1,

(come from Eq. (3.6)) clearly show that these initial values will persist for ever.

15

• This leads us to the important notion of statistical equilibrium. Wesay that a system is in statistical equilibrium (or the state dis-tribution is stationary) if its state probabilities are constant in time.

• Note that the system still fluctuate from state to state, but there isno net trend in such fluctuations.

• In the above queueing example, we have shown that the systemattains statistical equilibrium as t → ∞.

• Practically speaking, this means the system is in statistical equi-librium after sufficiently long time (so that initial conditions have nomore effect on the system). For the general birth-and-death processes,the following holds.

16

Proposition 2 (a) Let Pj(t) be the state probabilities of a birth-and-deathprocess. Then

limt→∞

Pj(t) = Pj

exist and are independent of the initial conditions; they satisfy the system oflinear difference equations obtained from the difference-differential equationsin previous chapter by replacing the derivative on the left by zero.(b) If all µj > 0 and the series

S = 1 +λ0µ1

+λ0λ1µ1µ2

+ · · · + λ0λ1 · · ·λj−1µ1µ2 · · ·µj

+ . . .+ (3.8)

converges, then

P0 = S−1 and Pj =

λ0λ1 · · ·λj−1µ1µ2 · · ·µj

S−1 (j = 1, 2, · · · ).

If the series in Eq. (3.8) diverges, then

Pj = 0 (j = 0, 1, · · · ) .

17

Proof: We shall NOT attempt to prove Part (a) of the above proposition, butrather we assume the truth of (a) and use it to prove Part (b).

By using part (a) of the proposition, we obtain the following linear difference equa-tions.

(λj + µj)Pj = λj−1Pj−1 + µj+1Pj+1(λ−1 = µ0 = 0 ; j = 0, 1, · · · ) P−1 = 0 .

(3.9)

Re-arranging terms, we have

λjPj − µj+1Pj+1 = λj−1Pj−1 − µjPj . (3.10)

If we letf (j) = λjPj − µj+1Pj+1

then Equation (3.10) simply becomes

f (j) = f (j − 1) for j = 0, 1, · · ·

as f (−1) = 0. Hencef (j) = 0 (j = 0, 1, · · · ).

18

This impliesλjPj = µj+1Pj+1.

By recurrence, we get (if µ1, · · · , µj > 0)

Pj =λ0λ1 · · ·λj−1µ1 · · ·µj

P0 (j = 1, 2, · · · ). (3.11)

Finally, by using the normalization condition∑

Pj = 1 we have the result in part(b).

Remark 8 Part (a) of the above proposition suggests that to find the statisticalequilibrium distribution

limt→∞

Pj(t) = Pj.

We set the derivatives on the left side of difference-differential equations to be zeroand replace Pj(t) by Pj and then solve the linear difference equations for Pj. Inmost cases, the latter method is much easier and shorter.

Remark 9 If µj = 0 for some j = k (λj > 0 for all j), then, as equation (3.11)shows,

Pj = 0 for j = 0, 1, · · · , k − 1.In particular, for pure birth process, Pj = 0 for all j.

19

Example 1 Suppose that for all i we have

λi = λ and µj = jµ

then

S = 1 +λ

µ+

1

2!

(λ

µ

)2+

1

3!

(λ

µ

)3+ . . .+ = e

λµ .

Therefore we have the Poisson distribution

Pj =1

j!

(λ

µ

)je−λµ .

Example 2 Suppose that for all i we have

λi = λ and µj = µ

such that λ < µ then

S = 1 +λ

µ+

(λ

µ

)2+

(λ

µ

)3+ . . .+ =

1

1− λµ.

Therefore we have the Geometric distribution

Pj =

(λ

µ

)j(1− λ

µ).

20

3.5 A Summary of Learning Outcomes

• Able to give the definition of a birth-and-death process.

• Able to derive the general solutions for the pure birth process, the pure deathprocess and the two-state birth-and-death process.

• Able to compute and interpret the steady-state solution of a birth-and-death pro-cess.

• Able to give the condition for the existence of the steady-state probability of abirth-and-death process.

21

3.6 Exercises

1. For the pure death process with death rate µj = jµ, prove that

pj(t) =

(n

j

)(e−µt)j(1− e−µt)n−j (j = 0, 1, · · · , n)

where n is the state of the system at t = 0 and pj(t) is the probability that thesystem is in State j at time t.

2. In a birth and death process, if

λi = λ/(i + 1) and µi = µ

show that the equilibrium distribution is Poisson.

3. Consider a birth-death system with the following birth and death coefficients:{λk = (k + 2)λ k = 0, 1, 2, . . .µk = kµ k = 0, 1, 2, . . . .

All other coefficients are zero.(a) Solve for pk. Make sure you express your answer explicitly in terms of λ, kand µ only.(b) Find the average number of customers in the system.

22

3.7 Suggested Solutions

1. In the assumed pure death process, actually the lifetime of each of the individualfollows the exponential distribution µe−µx. The probability that one can surviveover time t will be given by ∫ ∞

t

µe−µxdx = e−µt.

Therefore the probability that one can find j still survive at time t is given by(n

j

)(e−µt)j(1− e−µt)n−j (j = 0, 1, · · · , n)

2. Let Pi be the steady-state probability and

Pj =λ0λ1 · · ·λj−1µ1 · · ·µj

P0 (j = 1, 2, · · · )

=λjP0j!µj

and P0 = (∑∞

i=0 Pi)−1 = (eλ/µ)−1. Hence we have a Poisson distribution

Pn =λje−λ/µ

j!µj.

23

3. (a) Because λk = (k + 2)λ and µk = kµ, we can get that

s = 1 + 2λ

µ+ · · · + (j + 1)(λ

µ)j + · · ·+ =

∞∑k=1

kρk−1 =1

(1− ρ)2

soP0 = 1/s = (1− ρ)2

andPi = (i + 1)ρ

i(1− ρ)2.(b)

E =∞∑k=0

iPi = 0 · P0 +∞∑k=1

k(k + 1)ρk(1− ρ)2 = 2ρ1− ρ

.

Note: ∞∑k=1

k(k + 1)ρk−1 =d

dρ

∞∑k=1

(k + 1)ρk =d

dρ[d

dρ

∞∑k=1

ρk+1]

24

4 Introduction to Queueing Systems

A queueing situation 1 is basically characterized by a flow of customers arriving ata service facility. On arrival at the facility the customer may be served immediatelyby a server or, if all the servers are busy, may have to wait in a queue until a serveris available. The customer will then leave the system upon completion of service.The following are some typical examples of such queueing situations:

(i) Shoppers waiting in a supermarket [Customer: shoppers; servers: cashiers].

(ii) Diners waiting for tables in a restaurant [Customers: diners; servers: tables].

(iii) Patients waiting at an outpatient clinic [Customers: patients; servers: doctors].

(iv) Broken machines waiting to be serviced by a repairman [Customers: machines;server: repairman].

(v) People waiting to take lifts. [Customers: people; servers: lifts].

(vi) Parts waiting at a machine for further processing. [Customers: parts; servers:machine].

1Queueing theory is the mathematical study of waiting lines, or queues. In queueing theory a model is constructed so that queue lengths and waitingtimes can be predicted. Queueing theory is generally considered a branch of operations research because the results are often used when making businessdecisions about the resources needed to provide a service. Queueing theory has its origins in research by Agner Krarup Erlang when he created models todescribe the Copenhagen telephone exchange. The ideas have since seen applications including telecommunications, traffic engineering, computing and thedesign of factories, shops, offices and hospitals. (Taken from From Wikipedia)

25

• In general, the arrival pattern of the customers and the service time allo-cated to each customer can only be specified probabilistically. Such service facilitiesare difficult to schedule “optimally” because of the presence of randomness elementin the arrival and service patterns.

• A mathematical theory has thus evolved that provides means for analyzing suchsituations. This is known as queueing theory (waiting line theory, congestiontheory, the theory of stochastic service system), which analyzes the operating char-acteristics of a queueing situation with the use of probability theory.

• Examples of the characteristics that serve as a measure of the performance of asystem are the “expected waiting time until the service of a customeris completed” or “the percentage of time that the service facility isnot used”.

• Availability of such measures enables analysts to decide on an optimal way ofoperating such a system.

26

4.1 Basic Elements of Queueing Models

A queueing system is specified by the following elements.

(i) Input Process: How do customers arrive? Often, the input pro-cess is specified in terms of the distribution of the lengths of timebetween consecutive customer arrival instants (called the inter-arrival times ). In some models, customers arrive and are servedindividually (e.g. supermarkets and clinic). In other models,customers may arrive and/or be served in groups (e.g. lifts) andis referred to as bulk queues.

• Customer arrival pattern also depends on the source from whichcalls for service (arrivals of customers) are generated. The callingsource may be capable of generating a finite number of cus-tomers or (theoretically) infinitely many customers.

27

• In a machine shop with four machines (the machines are the cus-tomers and the repairman is the server), the calling source beforeany machine breaks down consists of four potential customers (i.e.anyone of the four machines may break down and therefore callsfor the service of the repairman). Once a machine breaks down, itbecomes a customer receiving the service of the repairman (untilthe time it is repaired), and only three other machines are capablegenerating new calls for service.

This is a typical example of a finite source, where an arrivalaffects the rate of arrival of new customers.

• For shoppers in a supermarket, the arrival of a customer normallydoes not affect the source for generating new customer arrivals, andis therefore referred to as an input process with infinite source.

28

(ii) Service Process: The time allocated to serve a customer (ser-vice time) in a system (e.g. the time that a patient is served bya doctor in an outpatient clinic) varies and is assumed to followsome probability distribution.

• Some facility may include more than one server, thus allowing asmany customers as the number of servers to be serviced simulta-neously (e.g. supermarket cashiers). In this case, all servers offerthe same type of service and the facility is said to have parallelservers .

• In some other models, a customer must pass through a series ofservers one after the other before service is completed (e.g. pro-cessing a product on a sequence of machines). Such situations areknown as queues in series or tandem queues.

29

(iii) Queue Discipline: The manner that a customer is chosenfrom the waiting line to start service is called the queue disci-pline .

• The most common discipline is the first-come-first-served rule(FCFS). Service in random order (SIRO), last-come-first-serve (LCFS)and service with priority are also used.

• If all servers are busy, in some models an arriving customer mayleave immediately (Blocked Customers Cleared: BCC),or in some other models may wait until served (Blocked Cus-tomers Delay: BCD).

• In some facility, there is a restriction on the size of the queue. Ifthe queue has reached a certain size, then all new arrivals will becleared from the system.

30

4.1.1 Some Simple Examples

(i) (Input process) If the inter-arrival time of any two customersis a constant, let say one hour then at the end of the second hourthere will be 2 arrived customers.

Suppose that customers only arrive at the end of each hour andthe probability that there is an arrival of customer is 0.5.

Let x be the number of customers arrived at the end of the secondhour. Then by the end of the second hour, we won’t know thenumber of customers arrived.

However, we know the probability that there are x arrived cus-tomers is given by (why?)

P (x = 0) = 0.25, P (x = 1) = 0.5 and P (x = 2) = 0.25.

31

(ii) (Service Process) Suppose that there is a job to be processedby a machine. The job requires a one-hour machine time. For areliable machine, it takes one hour to finish the job.

If the machine is unreliable and it may break down at the begin-ning of every hour with a probability of p. Once it breaks down ittakes one hour to fix it. But it may break down immediately afterthe repair with the same probability p(0 < p < 1). Clearly it takesat least one hour to finish the job but it may take much longer time.

Let x be the number of hours to finish the job. Then the probabilitythat the job can be finished at the end of the nth hour is given bythe Geometric distribution

P (x = k) = pk−1(1− p), k = 1, 2, . . . .

32

(iii) (Queueing Disciplines) Suppose there are three customers A,B and Cwaiting at a counter for service and their service times are in the following order10 minutes, 20 minutes and 30 minutes.Clearly it takes 10 + 20 + 30 = 60 minutes to finish all the service. However,the average waiting time before service for the three customers can be quitedifferent for different service disciplines.

Case 1: (FCFS): The waiting time for the first customer is zero, the waitingtime for the second customer is 10 minutes and the waiting time for the thirdcustomers is 10 + 20 = 30 minutes. Therefore the average waiting time beforeservice is

(0 + 10 + 30)/3 = 40/3.

Case 2: (LCFS): The waiting time for the first customer is zero, the waitingtime for the second customer is 30 minutes and the waiting time for the thirdcustomers is 30 + 20 = 50 minutes. Therefore the average waiting time beforeservice is

(0 + 30 + 50)/3 = 80/3

minutes which is twice of that in Case 1!

33

4.1.2 Definitions in Queueing Theory

To analyze a queueing system, normally we try to estimate quantities such as theaverage number of customers in the system, the fluctuation of the number of cus-tomers waiting, the proportion of time that the servers are idle, . . . etc.

• Let us now define formally some entities that are frequently used to measure theeffectiveness of a queueing system (with s parallel servers).

(i) pj = the probability that there are j customers in the system (waiting or inservice) at an arbitrary epoch (given that the system is in statistical equi-librium or steady-state). Equivalently pj is defined as the proportion oftime that there are j customers in the system (in steady state).

(ii) a = offered load = mean number of requests per service time. (In a systemwhere blocked customers are cleared, requests that are lost are also counted.)

(iii) ρ = traffic intensity = offered load per server = a/s (s < ∞).

34

(iv) a′ = carried load = mean number of busy servers.

(v) ρ′ = server occupancy (or utilization factor) = carried load per server =a′/s.

(vi) Ws = mean waiting time in the system,i.e the mean length of timefrom the moment a customer arrives until the customer leaves the system (alsocalled sojourn time).

(vii) Wq = mean waiting time in the queue, i.e. the mean length of timefrom the moment a customer arrives until the customer’ service starts .

(viii) Ls = mean number of customers in the system, i.e. including allthe customers waiting in the queue and all those being served.

(ix) Lq = mean number of customers waiting in the queue.

35

Remark 10 If the mean arrival rate is λ and the mean service time is τ then theoffered load a = λτ .

Remark 11 For an s server system, the carried load

a′ =

s−1∑j=0

jpj + s

∞∑j=s

pj.

Hence

a′ ≤ s∞∑j=0

pj = s and ρ′ =

a′

s≤ 1.

Remark 12 If s = 1 then a′ = ρ′ and a = ρ.

Remark 13 The carried load can also be considered as the mean number of cus-tomers completing service per mean service time τ . Hence in a system whereblocked customers are cleared, clearly the carried load is less than the offered load.On the other hand, if all requests are handled, then the carried load = the offeredload. In general

a′ = a(1−B)where B = proportion of customers lost (or requests that are cleared).

36

4.1.3 Kendall’s Notation

It is convenient to use a shorthand notation (introduced by D.G.Kendall)of the form a/b/c/d to describe queueing models, where a specifiesthe arrival process, b specifies the service time, c is the number ofservers and d is the number of waiting space. For example,

(i) GI/M/s/n : General Independent input, exponential (Markov)service time, s servers, n waiting space;

(ii) M/G/s/n : Poisson (Markov) input, arbitrary (General) servicetime, s servers, n waiting space;

(iii) M/D/s/n : Poisson (Markov) input, constant (Deterministic)service time, s servers, n waiting space;

(iv) Ek/M/s/n: k-phase Erlangian inter-arrival time, exponential(Markov) service time, s servers, n waiting space;

(v) M/M/s/n : Poisson input, exponential service time, s servers, nwaiting space.

37

Here are some examples.

(i) M/M/2/10 represents

A queueing system whose arrival and service process are randomand there are 2 servers and 10 waiting space in the system.

(ii) M/M/1/∞ represents

A queueing system whose arrival and service process are randomand there is one server and no limit in waiting space.

38

4.2 Queueing Systems of One Server

In this section we will consider queueing systems having one server only.

4.2.1 One-server Queueing Systems Without Waiting Space (Re-visit)

• Consider a one-server system of two states: 0 (idle) and 1 (busy).

• The inter-arrival time of customers follows the exponential distribution with pa-rameter λ.

• The service time also follows the exponential distribution with parameter µ.There is no waiting space in the system.

• An arrived customer will leave the system when he finds the server is busy (AnM/M/1/0 queue). This queueing system resembles an one-line telephone systemwithout call waiting.

39

4.2.2 Steady-state Probability Distribution

We are interested in the long-run behavior of the system, i.e., when t → ∞. Why?

Remark 14 Let P0(t) and P1(t) be the probability that there is 0 and 1 customerin the system. If at t = 0 there is a customer in the system, then

P0(t) =µ

λ + µ(1− e−(λ+µ)t)

and

P1(t) =1

λ + µ(µe−(λ+µ)t + λ).

Here P0(t) and P1(t) are called the transient probabilities. We have

p0 = limt→∞

P0(t) =µ

λ + µ

and

p1 = limt→∞

P1(t) =λ

λ + µ.

Here p0 and p1 are called the steady-state probabilities.

40

• Moreover, we have∣∣∣∣P0(t)− µλ + µ∣∣∣∣ = µe−(λ+µ)tλ + µ → 0 as t → ∞

and ∣∣∣∣P1(t)− λλ + µ∣∣∣∣ = µe−(λ+µ)tλ + µ → 0 as t → ∞

very fast.

• This means that the system will go into the steady state very fast.

• Therefore, it will be a good idea if we focus on the steady-stateprobability instead of the transient probability. The former is easierto be obtained.

41

4.2.3 The Meaning of the Steady-state Probability

The meaning of the steady-state probabilities p0 and p1 is as follows.

• In the long run, the probability that there is no customer in the system is p0 andthere is one customer in the system is p1.

For the server: In other words, in the long run, the proportion of time that theserver is idle is given by p0 and the proportion of time that the server is busy isgiven by p1.

For the customers: In the long run, the probability that an arrived customercan have his/her service is given by p0 and the probability that an arrived customerswill be rejected by the system is given by p1.

Remark 15 The system goes to its steady state very quickly. In general it ismuch easier to obtain the steady-state probabilities of a queueing system than thetransient probabilities. Moreover we are interested in the long-run performance ofthe system. Therefore we will focus on the steady-state probabilities of a queueingsystem.

42

4.2.4 The Markov Chain and the Generator Matrix

A queueing system can be represented by a continuous timeMarkov chain (statesand transition rates). We use the number of customers in the system to representthe state of the system. Therefore we have two states (0 and 1). The transitionrate from State 1 to State 0 is µ and The transition rate from State 0 to State 1 is λ.

• In State 0, change of state occurs when there is an arrival of customers and thewaiting time is exponentially distributed with parameter λ.

• In State 1, change of state occurs when the customer finishes his/her service andthe waiting time is exponentially distributed with parameter µ.

• Recall that from the no-memory (Markov) property, the waiting timedistribution for change of state is the same independent of the past history (e.g.how long the customer has been in the system).

�

-

µ

���0 ��

��1

Figure 4.1. The Markov Chain of the Two-state System.

43

• From theMarkov chain, one can construct the generator matrix as follows:

A1 =

(−λ µλ −µ

).

What is the meaning of the generator matrix? The steady-state probabilities willbe the solution of the following linear system:

A1p =

(−λ µλ −µ

)(p0p1

)=

(00

)(4.1)

subject to p0 + p1 = 1.

Remark 16 Given a Markov chain (generator matrix) one can construct the cor-responding generator (Markov chain). To interpret the system of linear equations.We note that in steady state, the expected incoming rate and the expected outgoing rate at any state must be equal. Therefore, we have the followings:

• At State 0: expected out going rate = λp0 = µp1 = expected incoming rate;

• At State 1: expected out going rate = µp1 = λp0 = expected incoming rate.

44

4.2.5 One-Server Queueing System with Waiting Space

�

-

µ

���0

�

-

µ

���1

�

-

µ

���2 ��

��3 ��

��4

�

-

µ

λ

Figure 4.2. The Markov chain for the M/M/1/3 queue.

• Consider an M/M/1/3 queue. The inter-arrival of customers and the service timefollow the exponential distribution with parameters λ and µ respectively. Thereforethere are 5 possible states. Why?

• The generator matrix is a 5× 5 matrix.

A2 =

−λ µ 0λ −λ− µ µ

λ −λ− µ µλ −λ− µ µ

0 λ −µ

. (4.2)

45

Let the steady-state probability distribution be

p = (p0, p1, p2, p3, p4)T .

In steady state we have A2p = 0.

• We can interpret the system of equations as follows:

At State 0: the expected out going rate = λp0 = µp1 = expected incoming rate;

At State 1: the expected out going rate = (λ + µ)p1 = λp0 + µp2 = expectedincoming rate.

At State 2: the expected out going rate = (λ + µ)p2 = λp1 + µp3 = expectedincoming rate;

At State 3: the expected out going rate = (λ + µ)p3 = λp2 + µp4 = expectedincoming rate.

At State 4: the expected out going rate = µp4 = λp3 = expected incoming rate.

46

We are going to solve p1, p2, p3, p4 in terms of p0.

• From the first equation −λp0 + µp1 = 0, we have

p1 =λ

µp0.

• From the second equation λp0 − (λ + µ)p1 + µp2 = 0, we have

p2 =λ2

µ2p0.

• From the third equation λp1 − (λ + µ)p2 + µp3 = 0, we have

p3 =λ3

µ3p0.

• Finally from the fourth equation λp2 − (λ + µ)p3 + µp4 = 0, we have

p4 =λ4

µ4p0.

The last equation is not useful as A2 is singular (Check!).

47

• To determine p0 we make use of the fact thatp0 + p1 + p2 + p3 + p4 = 1.

• Therefore

p0 +λ

µp0 +

λ2

µ2p0 +

λ3

µ3p0 +

λ4

µ4p0 = 1.

• Let ρ = λ/µ, we have

p0 = (1 + ρ + ρ2 + ρ3 + ρ4)−1

andpi = p0ρ

i, i = 1, 2, 3, 4.

•What is the solution of a general one-server queueing system (M/M/1/n)? We shall discuss it in the next section.

48

4.2.6 General One-server Queueing System

Consider a one-server queueing system with waiting space. The inter-arrival of cus-tomers and the service time follows the Exponential distribution with parametersλ and µ respectively.

• There is a waiting space of size n−2 in the system. An arrived customer will leavethe system only when he finds no waiting space left. This is an M/M/1/n−2 queue.

• We say that the system is in state i if there are i customers in the system. Theminimum number of customers in the system is 0 and the maximum number ofcustomers is n − 1 (one at the server and n − 2 waiting in the queue). Thereforethere are n possible states in the system. The Markov chain of the system is shownin Figure 4.3.

�

-

µ

���0

�

-

µ

λ����1 · · · � -

µ

λ����s · · · ��

��n− 1

�

-

µ

λ

Figure 4.3. The Markov Chain for the M/M/1/n-2 Queue.

49

• If we order the state from 0 up to n−1, then the generator matrix for the Markovchain is given by the following tridiagonal matrix A2:

0 1 2 3 · · · n− 3 n− 2 n− 1012.........

n− 2n− 1

−λ µ 0λ −λ− µ µ

. . . . . . . . .

λ −λ− µ µλ −λ− µ µ. . . . . . . . .

λ −λ− µ µ0 λ −µ

.

(4.3)

• We are going to solve the probability distribution p. Let

p = (p0, p1, . . . , pn−2, pn−1)T

be the steady-state probability vector. Here pi is the steady-state probability thatthere are i customers in the system and we have also

A2p = 0 andn−1∑i=0

pi = 1.

50

• To solve pi we begin with the first equation:

−λp0 + µp1 = 0 ⇒ p1 =λ

µp0.

We then proceed to the second equation:

λp0 − (λ + µ)p1 + µp2 = 0 ⇒ p2 = −λ

µp0 + (

λ

µ+ 1)p1 ⇒ p2 =

λ2

µ2p0.

Inductively, we may get

p3 =λ3

µ3p0,

p4 =λ4

µ4p0,

, . . . ,

pn−1 =λn−1

µn−1p0.

• Let ρ = λ/µ (the traffic intensity), we have

pi = ρip0, i = 0, 1, . . . , n− 1.

51

• To solve for p0 we need to make use of the conditionn−1∑i=0

pi = 1.

Therefore we getn−1∑i=0

pi =

n−1∑i=0

ρip0 = 1.

One may obtain

p0 =1− ρ1− ρn

.

• Hence the steady-state probability distribution vector p is given by1− ρ1− ρn

(1, ρ, ρ2, . . . , ρn−1)T .

52

4.2.7 Performance of a Queueing System

Remark 17 Using the steady-state probability distribution, one can compute

(a) the probability that a customer finds no more waiting space left when he arrives

pn−1 =1− ρ1− ρn

ρn−1.

(b) the probability that a customer finds the server is not busy (he can have theservice immediately) when he arrives

p0 =1− ρ1− ρn

.

(c) the expected number of customer at the server:

Lc = 0 · p0 + 1 · (p1 + p2 + . . . + pn−1)=

1− ρ1− ρn

(ρ + ρ2 + . . . + ρn−1)

=ρ(1− ρn−1)

1− ρn.

(4.4)

53

(d) the expected number of customers in the system is given by

Ls =

n−1∑i=0

ipi =

n−1∑i=1

ip0ρi

=ρ− nρn + (n− 1)ρn+1

(1− ρ)(1− ρn).

(4.5)

(e) the expected number of customers in the queue

Lq =

n−1∑i=1

(i− 1)pi

=

n−1∑i=1

(i− 1)p0ρi

=

n−1∑i=1

ip0ρi −

n−1∑i=1

p0ρi

=ρ2 − (n− 1)ρn + (n− 2)ρn+1

(1− ρ)(1− ρn).

(4.6)

We note that Ls = Lq + Lc.

54

Remark 18 To obtain the results in (d) and(e) we need the following results.

n−1∑i=1

iρi =1

1− ρ

n−1∑i=1

(1− ρ)iρi

=1

1− ρ

(n−1∑i=1

iρi −n−1∑i=1

iρi+1

)=

1

1− ρ(ρ + ρ2 + . . . + ρn−1 − (n− 1)ρn

)=

ρ + (n− 1)ρn+1 − nρn

(1− ρ)2.

(4.7)

Moreover if |ρ| < 1 we have∞∑i=1

iρi =ρ

(1− ρ)2.

55

4.3 Queueing Systems with Multiple Servers

Now let us consider a more general queueing system with s parallel and iden-tical exponential servers. The customer arrival rate is λ and the service rateof each server is µ. There are n− s− 1 waiting space in the system.

• The queueing discipline is again FCFS. When a customer arrives and finds all theservers busy, the customer can still wait in the queue if there is waiting space avail-able. Otherwise, the customer has to leave the system, this is an M/M/s/n− s− 1queue.

• Before we study the steady-state probability of this system, let us discuss thefollowing example (re-visited).

• Suppose there are k identical independent busy exponential servers, let t be thewaiting time for one of the servers to be free (change of state), i.e. one of thecustomers finishes his service.

56

• We let t1, t2, . . . , tk be the service time of the k customers in the system. Thenti follows the Exponential distribution λe

−λt and

t = min{t1, t2, . . . , tk}.

We will derive the probability density function of t.

• We note thatProb(t ≥ x) = Prob(t1 ≥ x) · Prob(t2 ≥ x) . . .Prob(tk ≥ x)

=

(∫ ∞x

λe−λtdt

)k= (e−λx)k = e−kλx.

(4.8)

• Thus ∫ ∞x

f (t)dt = e−kλx and f (t) = kλe−kλt.

Therefore the waiting time t is still exponentially distributed with parameterkλ.

57

m��p pm��p pm��p p...

m��p pm��p pm��p p

1 2 3 · · ·k · · · n− s− 1p p p p p p · · · p p · · · � λ

�µ�µ�µ

�µ�µ�µ

: empty buffer in queuep p : customer waiting in queuem��p p : customer being served

Figure 4.4. The multiple server queue.

• To describe the queueing system, we use the number of customers in the queueingsystem to represent the state of the system.

• There are n possible states (number of customers), namely 0, 1, . . . , n− 1.

58

The Markov chain for the queueing system is given in the following figure.

�

-

µ

���0

�

-

2µ

λ����1 · · · � -

sµ

λ����s · · · ��

��n− 1

�

-

sµ

λ

Figure 4.5. The Markov chain for the M/M/s/n− s− 1 queue.

• If we order the states of the system in increasing number of customers the it isnot difficult to show that the generator matrix for this queueing system is given bythe following n× n tri-diagonal matrix:

A3 =

−λ µ 0λ −λ− µ 2µ

. . . . . . . . .

λ −λ− sµ sµ. . . . . . . . .

λ −λ− sµ sµ0 λ −sµ

. (4.9)

59

4.3.1 A Two-server Queueing System

• Let us consider a small size example, a M/M/2/2 queue.

�

-

µ

���0

�

-

2µ

���1

�

-

2µ

���2 ��

��3 ��

��4

�

-

2µ

λFigure 4.6. The Markov Chain for the M/M/2/2 Queue.

• The generator matrix is an 5× 5 matrix.

A4 =

−λ µ 0λ −λ− µ 2µ

λ −λ− 2µ 2µλ −λ− 2µ 2µ

0 λ −2µ

. (4.10)Let the steady-state probability distribution be

p = (p0, p1, p2, p3, p4)T .

In steady state we have A4p = 0.

60

• From the first equation −λp0 + µp1 = 0, we have

p1 =λ

µp0.

• From the second equation λp0 − (λ + µ)p1 + 2µp2 = 0, we have

p2 =λ2

2!µ2p0.

• From the third equation λp1 − (λ + 2µ)p2 + 2µp3 = 0, we have

p3 =λ3

2 · 2!µ3p0.

• Finally from the fourth equation

λp2 − (λ + 2µ)p3 + 2µp4 = 0,

we have

p4 =λ4

22 · 2!µ4p0.

The last equation is not useful as A4 is singular.

61

To determine p0 we make use of the fact that

p0 + p1 + p2 + p3 + p4 = 1.

Therefore

p0 +λ

µp0 +

λ2

2!µ2p0 +

λ3

2 · 2!µ3p0 +

λ4

222!µ4p0 = 1.

Letτ = λ/(2µ)

we have

p0 =

(1 +

λ

µ+ (

λ2

2!µ2)1− τ 3

1− τ

)−1and

p1 =λ

µp0

and

pi = p0

(λ2

2!µ2

)τ i−2, i = 2, 3, 4.

62

• The result above can be further extended to the M/M/2/k queue as follows:

p0 =

(1 +

λ

µ+ (

λ2

2!µ2)1− τ k+1

1− τ

)−1, p1 =

λ

µp0, and pi = p0(

λ2

2!µ2)τ i−2, i = 2, . . . , k+2.

The queueing system has finite number of waiting space.

• The result above can also be further extended to M/M/2/∞ queue when τ =λ/(2µ) < 1 as follows:

p0 =

(1 +

λ

µ+ (

λ2

2!µ2)

1

1− τ

)−1, p1 =

λ

µp0, and pi = p0(

λ2

2!µ2)τ i−2, i = 2, 3, . . . .

or

p0 =1− τ1 + τ

, and pi = 2p0τi, i = 1, 2, . . . .

The queueing system has infinite number of waiting space.

• We then derive the expected number of customers in the system.

63

4.3.2 Expected Number of Customers in the M/M/2/∞ Queue

• The expected number of customers in the M/M/2/∞ queue is given by

Ls =

∞∑k=1

kpk =1− τ1 + τ

∞∑k=1

2kτ k.

Now we let

S =

∞∑k=1

kτ k = τ + 2τ 2 + . . .+

and we haveτS = τ 2 + 2τ 3 + . . . + .

Therefore by subtraction we get

(1− τ )S = τ + τ 2 + τ 3 + . . .+ = τ1− τ

andS =

τ

(1− τ )2.

We have

Ls =2τ

1− τ 2. (4.11)

64

4.3.3 Multiple-Server Queues

Now we consider general queueing models with Poisson input, in-dependent, identically distributed, exponential service times and sparallel servers.

Specifically, we shall consider two different queue disciplines, namely

- Blocked Customers Cleared (BCC), and

- Block Customers Delayed (BCD).

In the following, we assume that the Poisson input has rate λ and theexponential service times have mean µ−1 .

65

4.3.4 Blocked Customers Cleared (Erlang loss system)

The queueing system has s servers and there is no waiting spaceand we assume blocked customers are cleared.

Total possible number of states is s+ 1 and the generator matrix forthis system is given by

A5 =

−λ µ 0λ −λ− µ 2µ

λ −λ− 2µ. . . . . . . . .

λ −λ− (s− 1)µ sµ0 λ −sµ

.

66

Let pi be the steady-state probability that there are i customers inthe queueing system. Then by solving

A5p = 0 withs∑

i=0

pi = 1

one can get

pj =(λ/µ)j/j!s∑

k=0(λ/µ)k/k!

(j = 0, 1, · · · , s)

=aj/j!s∑

k=0ak/k!

(4.12)

and pj = 0 for j > s ; where a = λ/µ is the offered load.

• This distribution is called the truncated Poisson distribution(also called Erlang loss distribution).

67

• On the other hand, one can identify this system as a birth-and-death process, weproceed to find pj. Since customers arrive at random with rate λ, but affect statechanges only when j < s (BCC), the arrival rates (the birth rates) are

λj =

{λ when j = 0, · · · , s− 10 when j = s

Since service times are exponential, the service completion rates (the death rates)are

µj = jµ (j = 0, 1, 2, · · · , s).

Remark 19 The proportion of customers who have requested for ser-vice but are cleared from the system (when all servers are busy) isgiven by ps which is also called the Erlang loss formula and isdenoted by

B(s, a) =as/s!

s∑k=0

(ak/k!)

.

68

Remark 20 The mean number of busy servers, which is also equal tothe mean number of customers completing service per mean servicetime, is given by the carried load

a′ =s∑

j=1

jpj.

An interesting relation can be derived between the Erlang loss formulaand the carried load:

a′ =s∑

j=1

j(aj/j!)/s∑

k=0

(ak/k!)

= a

s−1∑j=0

(aj/j!)/s∑

k=0

(ak/k!)

= a (1−B(s, a)) .

This shows that the carried load is the portion of the offered load thatis not lost (captured) from the system.

69

4.3.5 Blocked Customers Delayed (Erlang delay system)

The queueing system has s servers and there is no limit in waiting spaceand we assume blocked customers are delayed.

• In this case we have the arrival rates λj = λ (j = 0, 1, · · · ), and the servicecompletion rates

µj =

{jµ (j = 0, 1, · · · , s)sµ (j = s, s + 1, · · · ).

• Hence we have

pj =

aj

j!p0 (j = 0, 1, · · · , s)aj

s!sj−sp0 (j = s + 1, · · · )

where a = λ/µ and

p0 =

(s−1∑k=0

ak

k!+

∞∑k=s

ak

s!sk−s

)−1.

70

• If a < s, the infinite geometric sum on the right converges, and

p0 =

s−1∑k=0

ak

k!+

as

(s− 1)!(s− a)

−1 .If a ≥ s, the infinite geometric sum diverges to infinity. Then p0 = 0and hence pj = 0 for all finite j.

• For a ≥ s, therefore the queue length tends to infinity with prob-ability 1 as time increases. In this case we say that no statisticalequilibrium distribution exists.

71

Remark 21 The probability that all servers are occupied (as observedby an outside observer) is given by the Erlang delay formula

C(s, a) =

∞∑j=s

pj =as

(s− 1)!1

s− ap0 =

as/[(s− 1)!(s− a)]

(s−1∑k=0

ak/k!) + as/[(s− 1)!(s− a)].

Since the arriving customer’s distribution is equal to the outside ob-server’s distribution, the probability that an arriving customer findsall servers busy (equivalently the probability that the waiting time inthe queue w > 0) is also given by C(s, a).

Remark 22 The carried load is equal to the offered load since norequest for service has been cleared from the system without beingserved. In fact, this equality holds for BCD queues with arbitraryarrival and service time distributions.

72

Remark 23 Suppose that an arriving customer finds that all the servers are busy.What is the probability that he finds j customers waiting in the ‘queue’ ?

• This is equivalent to find the conditional probability P{Q = j|w > 0} where Qdenotes the number of customers waiting in the queue.

• By the definition of conditional probability,

P{Q = j|w > 0} = P{Q = j, w > 0}P{w > 0}

.

Thus

P{Q = j and w > 0} = Ps+j =as

s!

(as

)jp0,

we get the Geometric distribution

P{Q = j|w > 0} =as

s! (as)

jp0as

s! (s

s−a)p0= (1− ρ)ρj (j = 0, 1 . . .).

where ρ = a/s is the traffic intensity.

73

4.4 Little’s Queueing Formula

If λ is the mean arrival rate, W is the mean time spent in the system(mean sojourn time) and L is the mean number of customers present,J.D.C. Little proved in 1961 that

L = λW.

This result is one of the most general and useful results in queueingtheory for a blocked customer delay queue.

The formal proof of this theorem is too long for this course. Let usjust formally state the theorem and then give a heuristic proof.

74

Proposition 3 (Little’s Theorem) Let L(x) be the number ofcustomers present at time x, and define the mean number L ofcustomers present throughout the time interval [0,∞) as

L = limt→∞

1

t

∫ t0L(x)dx;

let N(t) be the number of customers who arrive in [0, t], anddefine the arrival rate λ as

λ = limt→∞

N(t)

t;

and let Wi be the sojourn time of the ith customer, and definethe mean sojourn time W as

W = limn→∞

1

n

n∑i=1

Wi.

If λ and W exist and are finite, then so does L, and they arerelated by λW = L.

75

Proof: Let us follow the heuristic argument suggested by P. J. Burke.

Assume that the mean values L and W exist, and consider a long time interval(0, t) throughout which statistical equilibrium (steady state) prevails.

The mean number of customers who enter the system during this interval is λt.Imagine that a sojourn time is associated with each arriving customer; i.e., eacharrival brings a sojourn time with him. Thus the average sojourn time brought intothe system during (0, t) is λtW .

On the other hand, each customer present in the system uses up his sojourn timelinearly with time. If L is the average number of customers present throughout(0, t), then Lt is the average amount of time used up in (0, t).

Now as t → ∞ the accumulation of sojourn time must equal the amount of sojourntime used up; that is,

limt→∞

λtW

Lt= 1.

76

With the help of Little’s formula, we get the following useful results:

(a) λ, the average number of arrivals entering the system,

(b) Ls, the average number of customers in the queueing system,

(c) Lq, the average number of customers waiting in the queue,

(d) Lc, the average number of customers in the server,

(e) Ws, the average time a customer spends in the queueing system,

(f) Wq, the average time a customer spends in waiting in the queue,

(g) Wc, the average time a customer spends in the server.

then the Little’s formula states that if the steady-state probability distributionexists, we have

Ls = λWs, Lq = λWq, and Lc = λWc.

77

4.4.1 Little’s queueing Formula for the M/M/1/∞ Queue

In the following, we are going to prove Little’s queueing formula for the case ofM/M/1/∞ queue. We recall that

Ls =ρ

1− ρ, Lq =

ρ2

1− ρ, Lc = ρ, Ls = Lq + Lc, ρ =

λ

µ.

We first note that the expected waiting time Wc at the server is 1/µ.

Therefore we have

Wc =1

µ=

λ

λµ=

Lcλ.

Secondly we note that when a customer arrived, there can be i customers alreadyin the system. The expected waiting time before joining the server when there arealready i customers in the system is of course i/µ. Because there is only server andthe mean service time of each customer in front of him is 1/µ.

Therefore the expected waiting time Wq before one joins the server will be∞∑i=1

pi(i

µ) =

1

µ

∞∑i=1

ipi =Lsµ

=ρ

(1− ρ)µ.

78

Since i can be 0, 1, 2, . . ., we have

Wq =ρ

(1− ρ)µ=

ρ2

(1− ρ)µρ=

Lqλ

The expected waiting time at the server Wc will be of course 1/µ. Thus we have

Ws = Wq +Wc

=Lqµ

+1

µ

=1

µ(

ρ

1− ρ+ 1)

=1

µ(1− ρ)=

ρ

λ(1− ρ)=

Lsλ.

Hereρ = λ/µ

andLs = ρ/(1− ρ).

79

4.4.2 Applications of the Little’s Queueing Formula

Arrival rate λService rate µTraffic intensity ρ = λ/µProbability that no customer in the queue p0 = 1− ρProbability that i customers in the queue pi = p0ρ

i

Probability that an arrival has to wait for service 1− p0 = ρExpected number of customers in the system Ls = ρ/(1− ρ)Expected number of customers in the queue Lq = ρ

2/(1− ρ)Expected number of customers in the server Lc = ρExpected waiting time in the system Ls/λ = 1/(1− ρ)µExpected waiting time in the queue Lq/λ = ρ/(1− ρ)µExpected waiting time in the server Lc/λ = 1/µ

Table 4.1. A summary of the M/M/1/∞ queue.

80

Example 3 Consider the M/M/2/∞ queue with arrival rate λ andservice rate µ. What is the expected waiting time for a customer inthe system?

We recall that the expected number of customers Ls in the system isgiven by

Ls =2ρ

1− ρ2.

Here ρ = λ/(2µ). By applying the Little’s queueing formula we have

Ws =Lsλ

=1

µ(1− ρ2).

Example 4 On average 30 patients arrive each hour to the healthcentre. They are first seen by the receptionist, who takes an averageof 1 min to see each patient. If we assume that the M/M/1 queueingmodel can be applied to this problem, then we can calculate theaverage measure of the system performance, see Table 4.2.

81

Arrival rate λ = 30 (per hour)Service rate µ = 60 (per hour)Traffic intensity ρ = 0.5Probability that no customer in the queue p0 = 0.5

Probability that i customers in the queue pi = 0.5i+1

Probability that an arrival has to wait for service 0.5Expected number of customers in the system Ls = 1Expected number of customers in the queue Lq = 0.5Expected number of customers in the server Lc = 0.5Expected waiting time in the system Ls/λ = 1/30Expected waiting time in the queue Lq/λ = 1/60Expected waiting time in the server Lc/λ = 1/60

Table 4.2. A summary of the system performance

82

4.5 Applications of Queues

4.5.1 Allocation of the Arrivals in a System of M/M/1 Queues

�µ1��

��1 Queue 1

�λ1

•••�µn��

��s Queue n �

λn

�Allocation M

Figure 4.7. The Queueing System with Allocation of Arrivals.

•We consider a queueing system consisting of n independent M/M/1queues. The service rate of the serve at the ith queue is µi.

• The arrival process is a Poisson process with rate M .

83

• An allocation process is implemented such that it diverts an arrivedcustomers to queue i with probability

λiλ1 + . . . + λn

=λiM

.

• Then the input process of queue i is a Poisson process with rate λi.

• The objective here is to find the parameters λi such that some sys-tem performance is optimized.

• We remark that we must have λi < µi.

84

4.5.2 Minimizing Number of Customers in the System

The expected number of customers in queueing system i:

λi/µi1− λi/µi

.

The total expected number of customers in the system isn∑i=1

λi/µi1− λi/µi

.

• The optimization problem is then given as follows:

minλi

n∑i=1

λi/µi1− λi/µi

.subject to

m∑i=1

λi = M and 0 ≤ λi < µi for i = 1, 2, . . . , n.

85

• By consider the Lagrangian function

L(λ1, . . . , λn,m) =

n∑i=1

λi/µi1− λi/µi

−m

(n∑

i=1

λi −M

)and solving

∂L

∂λi= 0 and

∂L

∂m= 0

we have the optimal solution

λi = µi

(1− 1√

mµi

)< µi

where

m =

n∑i=1

õi

n∑i=1

µi −M

2

.

86

4.5.3 Minimizing Number of Customers Waiting in the System

• The expected number of customers waiting in queue i is(λi/µi)

2

1− λi/µi.

• The total expected number of customers waiting in the queues isn∑

i=1

(λi/µi)2

1− λi/µi.

• The optimization problem is then given as follows:

minλi

{n∑

i=1

(λi/µi)2

1− λi/µi

}.

subject tom∑i=1

λi = M

and0 ≤ λi < µi for i = 1, 2, . . . , n.

87

By consider the Lagrangian function

L(λ1, . . . , λn,m) =

n∑i=1

(λi/µi)2

1− λi/µi−m

(n∑

i=1

λi −M

)and solving

∂L

∂λi= 0 and

∂L

∂m= 0

we have the optimal solution

λi = µi

(1− 1√

1 +mµi

)< µi

where m is the solution ofn∑i=1

µi

(1− 1√

1 +mµi

)= M.

88

4.5.4 Which Operator to Employ?

• We are going to look at one application of queueing systems. In alarge machine repairing company, workers must get their tools fromthe tool centre which is managed by an operator.

• Suppose the mean number of workers seeking for tools per hour is5 and each worker is paid 8 dollars per hour.

• There are two possible operators (A and B) to employ. In averageOperator A takes 10 minutes to handle one request for tools is paid5 dollars per hour. While Operator B takes 11 minutes to handleone request for tools is paid 3 dollars per hour.

• Assume the inter-arrival time of workers and the processing timeof the operators are exponentially distributed. We may regard therequest for tools as a queueing process (M/M/1/∞) with λ = 5.

89

• For Operator A, the service rate is µ = 60/10 = 6 per hour.Thus we have

ρ = λ/µ = 5/6.

The expected number of workers waiting for tools at the tool centrewill be

ρ

1− ρ=

5/6

1− 5/6= 5.

The expected delay cost of the workers is

5× 8 = 40dollars per hour and the operator cost is 5 dollars per hour. Thereforethe total expected cost is

40 + 5 = 45.

90

• For Operator B, the service rate is µ = 60/11 per hour. Thuswe have

ρ = λ/µ = 11/12.

The expected number of workers waiting for tools at the tool centrewill be

ρ

1− ρ=

11/12

1− 11/12= 11.

The expected delay cost of the workers is

11× 8 = 88dollars per hour and the operator cost is 3 dollars per hour. Thereforethe total expected cost is

88 + 3 = 91.

Conclusion: Operator A should be employed.

91

4.5.5 Two M/M/1 Queues Or One M/M/2 Queue ?

• If one more identical operator can be employed, then which offollowings is better? (In our analysis, we assume that λ < µ).

(i) Put two operators separately. We have two M/M/1/∞queues. In this case, we assume that an arrived customer caneither join the first queue or the second with equal chance.

(ii)Put two operators together. We have an M/M/2/∞ queue.

����1 · · ·�

µ

����2 · · ·�

µ@

@@I

��

�

λ2

λ2

Figure 4.8. Case (i) Two M/M/1/∞ Queues.

����1�

µ

����2

· · ·�

µ �λ

Figure 4.9. Case (ii) One M/M/2/∞ Queue.

92

• To determine which case is better, we calculate the expected number of customers(workers) in both cases. Clearly in our consideration, the smaller the better (Why?).

In case (i), the expected number of customers in any one of the queues will be givenby

( λ2µ)

1− ( λ2µ).

Hence the total expected number of customers (workers) in the system is

S1 = 2×( λ2µ)

1− ( λ2µ)=

(λµ)

1− ( λ2µ).

In case (ii), the expected number of customers in any one of the queues will begiven by (see previous section)

S2 =(λµ)

1− ( λ2µ)2.

Clearly S2 < S1.

Conclusion: Case (ii) is better. We should put all the servers (operators) together.

93

4.5.6 One More Operator?

• Operator A later complains that he is overloaded and the workers have wastedtheir time in waiting for a tool. To improve this situation, the senior managerwonders if it is cost effective to employ one more identical operator at the toolcentre. Assume that the inter-arrival time of workers and the processing time ofthe operators are exponentially distributed.

• For the present situation, one may regard the request for tools as a queueingprocess (An M/M/1/∞) where the arrival rate λ = 5 per hour and the service rateµ = 60/10 = 6 per hour. Thus we have ρ = λ/µ = 5/6.

• The expected number of workers waiting for tools at the tool centre will beρ

1− ρ=

5/6

1− 5/6= 5.

The expected delay cost of the workers is 5×8 = 40 dollars per hour and the opera-tor cost is 5 dollars per hour. Therefore the total expected cost is 40+5 = 45 dollars.

• When one extra operator is added then there are 2 identical operators at the toolcenter and this will be an M/M/2/∞ queue.

94

• The expected number of workers in the system is given by (c.f. (4.11))

1− ρ1 + ρ

∞∑i=1

2iρi =2ρ

1− ρ2

where

ρ =λ

2µ=

5

12.

• In this case the expected delay cost and the operator cost will be given respectivelyby

8× 2ρ1− ρ2

=8× 120119

= 8.07 and 2× 5 = 10 dollars.

• Thus the expected cost when there are 2 operators is given by 18.07 dollars.

• Conclusion: Hence the senior manager should employ one more operator.

• How about employing three operators? (You may consider M/M/3/∞ queue).

• But it is clear that there is no need to employ four operators. Why?

95

4.6 An Unreliable Machine System

• Consider an unreliable machine system. The normal time of themachine is exponentially distributed with mean λ−1. Once the ma-chine is broken, it is subject to a n-phase repairing process.

• The repairing time at phase i is also exponentially distributed withmean µ−1i (i = 1, 2, . . . , n). After the repairing process, the machineis back to normal. Let 0 be the state that the machine is normal andi be the state that the machine is in repairing phase i. The Markovchain of the model is given by

-

���0 -

µ1����1 -

µ2����2 -

µn−1· · · ��

��n

?�6

µn

Figure 4.10. The Markov Chain for the Unreliable Machine System.

96

• Let the steady-state probability vector be

p = (p0, p1, . . . , pn)

satisfiesA6p = 0

where

A6 =

−λ 0 µnλ −µ1

µ1 −µ2. . . . . . 0

0 µn−1 −µn

.

97

• From the first equation −λp0 + µnpn we have

pn =λ

µnp0.

From the second equation λp0 − µ1p1 we have

p1 =λ

µ1p0.

From the third equation µ1p1 − µ2p2 we have

p2 =λ

µ2p0.

We continue this process and therefore

pi =λ

µip0.

Since p0 + p1 + p2 + . . . + pn = 1, we have

p0

(1 +

n∑i=1

λ

µi

)= 1.

Therefore

p0 =

(1 +

n∑i=1

λ

µi

)−1.

98

4.7 A Reliable One-machine Manufacturing System

• Here we consider an Markovian model of reliable one-machine manufacturingsystem. The production time for one unit of product is exponentially distributedwith a mean time of µ−1.

• The inter-arrival time of a demand is also exponentially distributed with a meantime of λ−1.

• The demand is served in a first come first serve manner. In order to retain thecustomers, there is no backlog limit in the system. However, there is an upper limitn(n ≥ 0) for the inventory level.

• The machine keeps on producing until this inventory level is reached and theproduction is stopped once this level is attained. We seek for the optimal value ofn (the hedging point or the safety stock) which minimizes the expected running cost.

• The running cost consists of a deterministic inventory cost and a backlog cost.In fact, the optimal value of n is the best amount of inventory to be kept in thesystem so as to hedge against the fluctuation of the demand.

99

• Let us summarized the notations as follows.

I , the unit inventory cost;B, the unit backlog cost;n ≥ 0, the hedging point;µ−1, the mean production time for one unit of product;λ−1, the mean inter-arrival time of a demand.

• If the inventory level (negative inventory level means backlog) isused to represent the state of the system, one may write down theMarkov chain for the system.

�

-

µ

���n

�

-

µ

λ����n− 1 · · · � -

µ

���1 ��

��0

�

-

µ

λ· · ·

Figure 4.11. The Markov Chain for the Manufacturing System.

100

• Here we assume that µ > λ, so that the steady-state probabil-ity distribution of the above M/M/1 queue exists and has analyticsolution

q(i) = (1− p)pn−i, i = n, n− 1, n− 2, · · ·where

p = λ/µ

and q(i) is the steady-state probability that the inventory level is i.

• Hence the expected running cost of the system (sum of the inventorycost and the backlog cost) can be written down as follows:

E(n) = In∑i=0

(n− i)(1− p)pi︸ ︷︷ ︸inventory cost

+B∞∑

i=n+1

(i− n)(1− p)pi︸ ︷︷ ︸backlog cost

. (4.13)

101

Proposition 4 The expected running cost E(n) is minimized ifthe hedging point n is chosen such that

pn+1 ≤ II +B

≤ pn.

Proof: We note that

E(n− 1)− E(n) = B − (I +B)(1− p)n−1∑i=0

pi = −I + (I +B)pn

and

E(n+ 1)−E(n) = −B + (I +B)(1− p)n∑i=0

pi = I − (I +B)pn+1.

Therefore we have

E(n−1) ≥ E(n) ⇔ pn ≥ II +B

and E(n+1) ≥ E(n) ⇔ pn+1 ≤ II +B

.

Thus the optimal value of n is the one such that pn+1 ≤ II+B ≤ pn.

102

4.8 An Inventory Model with Returns and Lateral Transshipments

• In the world of limited resources and disposal capacities, there is aenvironmental pressure in using re-manufacturing system, a recyclingprocess to reduce the amount of waste generated.

• A major manufacturer of copy machines Xeron reported on annualsavings of several hundred million dollars due to the re-manufacturingand re-use of equipment.

• A return is first repaired/tested and then re-sell to the market. Theresult of re-manufacturing is that the manufacturers have to take intoaccount of returns in their production plans.

• M. Fleischmann (2001) Quantitative Models for Reverse Logis-tics, Lecture Notes in Economics and Mathematical Systems, 501,Springer, Berlin.

103

(i) λ−1, the mean inter-arrival time of demands,(ii) µ−1, the mean inter-arrival time of returns,(iii) a, the probability that a returned product is repairable,(iv) Q, maximum inventory capacity,(v) I , unit inventory cost,(vi) R, cost per replenishment order.

Returns-

µCheck/Repair -aµ

6

Disposal (1− a)µ?

Replenishment

-1 0 1 Q· · · · · ·Demands

- λ

figure 4.16. The Single-item Inventory Model.

• W. Ching, W. Yuen and A. Loh, An Inventory Model with Returns and LateralTransshipments, J. Operat. Res. Soc., 54 (2003) 636-641.

104

�

-

λ

aµ����0

�

-

λ

aµ����1 · · ·

�

-

λ

aµ����Q− 1 ��

��Q

6

λfigure 4.17. The Markov Chain.

• The (Q + 1)× (Q + 1) system generator matrix is given as follows:

A =

01......Q

λ + aµ −λ 0−aµ λ + aµ −λ

. . . . . . . . .

−aµ λ + aµ −λ−λ −aµ λ

. (4.14)• The steady state probability distribution p is given by

pi = K(1− ρi+1), i = 0, 1, . . . , Q (4.15)

where

ρ =aµ

λand K =

1− ρ(1 +Q)(1− ρ)− ρ(1− ρQ+1)

.

105

Proposition 5 The expected inventory level is

Q∑i=1

ipi =

Q∑i=1

K(i−iρi+1) = K

(Q(Q + 1)

2+QρQ+2

1− ρ− ρ

2(1− ρQ)(1− ρ)2

),

the average rejection rate of returns is

µpQ = µK(1− ρQ+1)and the mean replenishment rate is

λ× p0 ×λ−1

λ−1 + (aµ)−1=

λK(1− ρ)ρ(1 + ρ)

.

106

Proposition 6 If ρ < 1 and Q is large then K ≈ (1+Q)−1 and theapproximated average running cost (inventory and replenishmentcost)

C(Q) ≈ QI2

+λ(1− ρ)ρR

(1 + ρ)(1 +Q).

The optimal replenishment size

Q∗ + 1 ≈

√2λ(1− ρ)ρR(1 + ρ)I

=

√2aµR

I

(2λ

λ + aµ− 1). (4.16)

107

4.9 Queueing Systems with Two Types of Customers

In this section, we discuss queueing systems with two types of customers. Thequeueing system has no waiting space. There are two possible cases: infinite-servercase and finite-server case.

4.9.1 Infinite-Server Queue

Consider the infinite-server queue with two types of customers. The arrival processof customers of type i (i = 1, 2) is Poisson with rate λi and their service times areindependent, identically distributed, exponential random variables with mean µ−1i(i = 1, 2).

• We define the 2-dimensional states {Ej1,j2}, where ji is the number of customersof type i in the system, with corresponding equilibrium distribution {p(j1, j2)},then clearly the Markov property still holds.

• Here p(j1, j2) is the steady-state probability that there are j1 type 1 customersand j2 type 2 customers in the system.

108

• By equating expected rate out to expected rate in for each state, the equilib-rium state equations are

(λ1 + λ2 + j1µ1 + j2µ2)p(j1, j2) = λ1p(j1 − 1, j2) + (j1 + 1)µ1p(j1 + 1, j2)+λ2p(j1, j2 − 1) + (j2 + 1)µ2p(j1, j2 + 1)

and [p(−1, j2) = p(j1,−1) = 0 ; j1 = 0, 1, . . . ; j2 = 0, 1, . . . .]

Ej1,j2 Ej1+1,j2Ej1−1,j2

Ej1,j2+1

Ej1,j2−1

6

?

?

6

λ2

λ2 j2µ2

(j2 + 1)µ2

�- -

�

λ1λ1

(j1 + 1)µ1j1µ1

Figure 4.12. The Markov Chain of the System at State Ej1,j2.

109

• In addition, the probabilities must satisfy the normalization equation∞∑

j1=0

∞∑j2=0

p(j1, j2) = 1.

• In this case, we already know the answer. Since the number of servers is infinite,the two types of customers do not affect one another.

• Thus the marginal distribution of the number of customers of each type is thatwhich would be obtained by solving the corresponding one-dimensional problem,namely the Poisson distribution:

p1(j) =

∞∑k=0

p(j, k) =(λ1/µ1)

j

j!e−(λ1/µ1),

p2(j) =∞∑k=0

p(k, j) =(λ2/µ2)

j

j!e−(λ2/µ2) .

(4.17)

110

• Since the number of customers present of each type is independent of thenumber present of the other type, therefore

p(j1, j2) = p1(j1)p2(j2)

=(λ1/µ1)

j1

j1!

(λ2/µ2)j2

j2!e−[(λ1/µ1)+(λ2/µ2)].

(4.18)

• Solution of Product Form:

The fact that the solution p(j1, j2) can be decomposed into a product of twofactors has enabled us to solve the problem with ease.

111

• In fact one may try

p(j1, j2) =(λ1/µ1)

j1

j1!

(λ2/µ2)j2

j2!C (4.19)

where the constant C is determined from the normalization equation. In this case∞∑

j1=0

∞∑j2=0

p(j1, j2) =

∞∑j1=0

∞∑j2=0

(λ1/µ1)j1

j1!

(λ2/µ2)j2

j2!C

= Ceλ2/µ2∞∑

j1=0

(λ1/µ1)j1

j1!

= Ce[(λ1/µ1)+(λ2/µ2)].

HenceC = e−[(λ1/µ1)+(λ2/µ2)].

• In practice, a good strategy for finding solutions to equations of the form (4.17) isto assume a product solution of the form (4.19); and see if such a solution satisfies(4.17).

• If it goes, then the solution has been obtained. If it doesn’t, then try a differentapproach. In this case it works!

112

4.9.2 Multiple-server Queue with Blocked Customers Cleared

• The situation is similar to that of the previous example except nowthe system has finitely many servers.

• The system is again described by equation (4.17), which is validnow only for j1 + j2 < s.

• When j1 + j2 = s, then the states Ej1+1,j2 and Ej1,j2+1 cannotoccur and the equation becomes

(j1µ1 + j2µ2)p(j1, j2) = λ1p(j1 − 1, j2) + λ2p(j2, j2 − 1). (4.20)• Observe that (4.20) can be obtained from (4.17) by deleting thefirst two terms on the left and the last two terms on the right.

113

• Since the product-form solution (4.19) satisfies the equation (4.17) and also theequation with only the deleted terms

(λ1 + λ2)p(j1, j2) = (j1 + 1)µ1p(j1 + 1, j2) + (j2 + 1)µ2p(j1, j2 + 1),

therefore it also satisfies (4.20). Thus the product solution (4.19) is a solution ofthis problem.

• In particular, if we don’t distinguish the two types of customers, then the prob-ability p(j) that there are j customers (including type 1 and type 2) in service isgiven by

p(j) =∑

j1+j2=j

p(j1, j2) =

j∑j1=0

p(j1, j − j1).

• With the help of binomial theorem, we have

p(j) = C

j∑j1=0

(λ1µ1)j1

j1!

(λ2µ2)j−j1

(j − j1)!=

C

j!

j∑j1=0

j!

j1!(j − j1)!(λ1µ1

)j1(λ2µ2

)j−j1 = C1

j!

(λ1µ1

+λ2µ2

)j.

114

• The normalization equations∑

k=0

p(k) = 1

implies that

C =

{s∑

k=0

1

k!(λ1µ1

+λ2µ2

)k

}−1.

We conclude that

p(j) =aj/j!s∑

k=0

ak/k!(j = 0, 1, . . . , s)

where

a = (λ1µ1

) + (λ2µ2

).

• We note that for the case of infinite-server queue (we let s → ∞) we have

p(j) =aje−a

j!(j = 0, 1, . . . , ).

115

4.10 Queues in Tandem

• Consider two sets of servers arranged in tandem, so that the output from the firstset of servers is the input of the second set of servers.

• Assume that the arrival process at the first stage of this tandem queueing systemis Poisson with rate λ, the service times in the first stage are exponentially dis-tributed with mean µ−11 , and the queueing discipline is blocked customers delayed.

• The customers completing service in the first stage will enter the second stage(and wait if all servers in second stage are busy), where the service times are as-sumed to be exponentially distributed with mean µ−12 . The number of servers instage i is si(i = 1, 2) .

116

· · · · · ·... ...

� � �

Second Set of Servers First Set of Servers

Figure 4.13. Two Queues in Tandem.

• Let p(j1, j2) be the probability that there are j1 customers in stage 1 and j2customers in stage 2. Let

µi(j) =

{jµi (j = 0, 1, . . . , si)siµi (j = si + 1, si + 2, . . .)

be the departure rates. Then equating rate in to rate out for each state, weobtain

(λ + µ1(j1) + µ2(j2)) p(j1, j2) = λp(j1 − 1, j2) + µ1(j1 + 1)p(j1 + 1, j2 − 1)+µ2(j2 + 1)p(j1, j2 + 1)

(4.21)

[p(−1, j2) = p(j1,−1) = 0 ; j1 = 0, 1, . . . ; j2 = 0, 1, . . .]

117

• Since the first stage of this system is precisely an Erlang delay system, sothat the marginal distribution of the number of customers in stage one is given bythe Erlang delay probabilities.

• Let us try (hopefully it would work) a product solution of the form

p(j1, j2) = p1(j1)p2(j2) (4.22)

with the factor p1(j1) given by the Erlang delay probabilities:

p1(j1) =

C1

(λ/µ1)j1

j1!(j1 = 0, 1, . . . , s1 − 1) ,

C1(λ/µ1)

j1

s1!sj1−s11

(j1 = s1, s1 + 1, . . .) .(4.23)

118

• We shall substitute the assumed product solution (4.22) and (4.23) into theequilibrium state equations (4.21), with the hope that the system of equationswill be reduced to a one-dimensional set of equations that can be solved for theremaining factor p2(j2). Indeed, (4.21) is reduced to

[λ + µ2(j2)]p2(j2) = λp2(j2 − 1) + µ2(j2 + 1)p2(j2 + 1) (4.24)

[p2(−1) = 0 ; j2 = 0, 1, . . .](4.24) is the equilibrium state equations that define the Erlang delay probabilities.

• We conclude that p2(j2) is given by

p2(j2) =

C2

(λ/µ2)j2

j2!(j2 = 0, 1, 2, . . . , s2 − 1) ,

C2(λ/µ2)

j2

s2!sj2−s22

(j2 = s2, s2 + 1, . . . , ).(4.25)

119

• Using∞∑j=0

pi(j) = 1 (i = 1, 2)

we get

Ci =( si−1∑

k=0

ρkik!

+ρsii

si!(1− ρi/si)

)−1where ρi =

λ

µi(i = 1, 2).

This equation implies that a proper joint distribution exists only whenλ/µ1 < s1 and λ/µ2 < s2.

• It is a remarkable result that the number of customers in the secondstage also follows the Erlang delay distribution, that is, the distribu-tion of customers in the second stage is the same as if the first stagewere not present and the customers arrived according to a Poissonprocess directly at the second stage.

120

• It also suggests that the output from the first stage is a Poissonprocess. This is in fact true in general and we state without proof theBurke’s theorem as follows.

Proposition 7 (Burke’s Theorem) The statistical equilibrium out-put of an M/M/s queue with arrival rate λ and mean service timesµ−1 is a Poisson process with the same rate λ.

Remark 24 In 1956 Burke showed that the departure process of anM/M/s queue was Poisson.

P. Burke, (1956) The Output of a Queueing System, Oper. Res. (4)699-704.

121

4.11 Queues in Parallel

· · ·

· · ·

...

...

�

�

� λ1

� λ2

� λ1

� CCCCCCCCCCCO �

����������� λ2

Case 2: With OverflowCase 1: No Inter-action

Figure 4.14. Two Queues in Parallel.

Case (i) We assume there is no inter-action between the two queues.There are si(i = 1, 2) servers in queue i(i = 1, 2) and there is infinitemany waiting spaces in each queue and we assume blocked customersare delayed. The arrival rate of queue i is λi and the service comple-tion rate of a serve in queue i is µi.

122

• Hence we have the steady-state probability that queue i has j cus-tomers given by

pi(j) =

ρjij!Ci (j = 0, 1, · · · , si)

ρji

si!sj−sii

Ci (j = si + 1, . . .)

where ρi = λi/µi. Moreover if ρi < si then

Ci =

si−1∑k=0

ρkik!

+

∞∑k=si

ρki

si!sk−sii

−1 =si−1∑

k=0

ρkik!

+ρsii

(si − 1)!(si − ρi)

−1 .• If ρi ≥ si, the infinite geometric sum diverges then Ci = 0 andhence pi(j) = 0 for all finite j.

• If ρi < si then the steady-state probability that there are i cus-tomers in Queue 1 and j customers in Queue 2 is given by

p(i, j) = p1(i)p2(j) i, j = 0, 1, 2, . . . .

123

• Case (ii) There are si(i = 1, 2) servers in queue i(i = 1, 2) and there are finitemany waiting spaces in each queue and we assume blocked customers are cleared.The arrival rate of queue i is λi and the service completion rate of a serve in queuei is µi.

• We assume there is inter-action between the two queues as follows. WheneverQueue 2(1) is full, an arrived customer of type 2(1) will overflow to Queue 1(2) pro-vided that Queue 1(2) is not yet full. Let us consider a simple example as follows.We assume that Queue 1 and 2 are M/M/1/1 queue. The following figure gives theMarkov chain of the queueing system.

E1,1 E2,1E0,1

E1,2E0,2 E2,2

E0,0 E1,0 E2,0

?

6

6

?

µ2

µ2 λ2

λ2?

6

6

?

µ2

µ2 λ2

λ2?

6

6

?

µ2

µ2 λ̃ = λ1 + λ2

λ̃ = λ1 + λ2�

- -�

λ1λ1

µ1µ1

�- -

�

λ1λ1

µ1µ1

�- -

�

λ̃ = λ1 + λ2λ̃ = λ1 + λ2

µ1µ1

figure 4.15. The Markov Chain of the Two Queue Overflow System.

124

• The generator matrix for this queueing problem is given by

A8 =

(0, 0)(1, 0)(2, 0)(0, 1)(1, 1)(2, 1)(0, 2)(1, 2)(2, 2)

* µ1 0 µ2 0 0 0 0 0λ1 * µ1 0 µ2 0 0 0 00 λ1 * 0 0 µ2 0 0 0λ2 0 0 * µ1 0 µ2 0 00 λ2 0 λ1 * µ1 0 µ2 0

0 0 λ̃ 0 λ1 ∗ 0 0 µ20 0 0 λ2 0 0 * µ1 0

0 0 0 0 λ2 0 λ̃ * µ10 0 0 0 0 λ̃ 0 λ̃ *

. (4.26)

• Here “∗” is such that the column sum is zero.

• Unfortunately there is no analytic solution for the steady-state probabilities ofthis system. Direct method or numerical method are common methods for solvingthe steady-state probabilities.

125

4.12 A Summary of Learning Outcomes

• Able to use the Kendall’s notation to describe a queueing system.

• Able to compute and interpret an M/M/s/n queue including:- the Markov chain diagram, the generator matrix, the steady-state probability.- Erlang loss formula, Erlang delay formula.- waiting time distribution in an M/M/s/∞ queue.

• Able to state and show the Little’s queueing formula and Burke’s theorem.

• Able to give a system performance analysis of a Markovian queueing system:- Expected number of customers.- Expected number of busy servers.- Mean waiting time.

• Able to apply queues in tandem and in parallel to real problems such as:- Employment of operators.- Unreliable machine system.- Manufacturing system and inventory system.

126

4.13 Exercises

1. Customers arrive according to a Poisson process with rate λ at a single serverwith n waiting positions. Customers who find all waiting positions occupied arecleared. All other customers wait as long as necessary for service. The meanservice time is µ−1, and ρ = λµ. Making no assumption about the form of theservice time distribution function, show that

ρ =1− p01− pn+1

.

Here pi is the steady-state probability that there are i customers in the system.

2. A company has 3 telephone lines. Incoming calls are generated by the customersaccording to a Poisson process with a mean rate of 20 calls per hour. Calls thatfind all telephone lines busy are rejected and those calls that are able to getthrough will be served for a length of time that is exponentially distributedwith mean 4 minutes.

(a) Let the number of busy lines be the state of the system. Write down theMarkov chain and the generator matrix for this system. Hence solve the steady-state probability distribution for the system.

(b) Find the proportion of calls that are rejected.

127

3. An company offers services that can be modeled as an s-server Erlang loss sys-tem (M/M/s/0 queue).

Suppose the arrival rate is 2 customers per hour and the average service timeis 1 hour. The entrepreneur earns $2.50 for each customer served and the com-pany’s operating cost is $1.00 per server per hour (whether the server is busyor idle).

(a) Write down the expected hourly net profit C(s) in terms of s.

(b) Show that

lims→∞

C(s)

s= −1

and interprets this result.

(c) If the maximum number of servers available is 5, what is the optimal numberof servers which maximizes the expected hourly net profit ? What is the ex-pected hourly net profit earned when the optimal number of servers is provided?

128

4. Consider an Erlang loss system with two servers. The arrival rate is 1 and meanservice time is µ−1i for server i(i = 1, 2). When the system is idle, an arrivedcustomer