M/G/1 and M/G/1/K systems: another lookTeletra c theory I: queuing theoryD.Molthchanov, TUT, 2011 3....

M/G/1 and M/G/1/K systems: another look

Dmitri A. Moltchanov

[email protected]

http://www.cs.tut.fi/kurssit/TLT-2716/

Teletraffic theory I: queuing theory D.Molthchanov, TUT, 2011

OUTLINE:

• Description of M/G/1 system;

• Methods of analysis;

• Choosing state of the system;

• Method of supplementary variables;

• Imbedded Markov chain approach;

• Description of M/G/1/K queuing system;

• Imbedded Markov chain for M/G/1/K system.

Lecture: M/G/1 and M/G/1/K systems: part II 2


1. Description of M/G/1 queuing systemM/G/1 queuing system stands for:

• single server;

• infinite number of waiting positions;

• Poisson arrival process;

– interarrival times are exponentially distributed.

• generally distributed service times:

– practically, any distribution.

What we also assume:

• ’first come, first served’ (FCFS);

• what it may affect: waiting time!



1.1. Arrival and service processes

Arrival process:

• Poisson with parameter (mean value) λ

• interarrival times are exponential with mean 1/λ;

• PDF, pdf and LT are:

A(t) = 1− e−λt, a(t) = λe−λt, A(s) =λ

s+ λ. (1)

Service process:

• service times are exponential with mean 1/µ;

• PDF, pdf and LT are:

B(t), b(t), B(s). (2)



2. Methods of analysisThere are a number of methods. We consider four most popular methods:

• residual life approach: simple to understand:

– only mean values can be obtained.

• transform approach based on imbedded Markov chain: harder to deal with;

– distributions of desired performance parameters can be obtained;

– the idea: find points at which Markov property holds and use transforms.

• direct approach based on imbedded Markov chain: harder to deal with;


– the idea: find points at which Markov property holds and use convolution.

• method of supplementary variables: the most complicated one:


– the idea: look at arbitrary points and make them Markovian.



3. Choosing state of the systemLet N(t) be the number of customers at time t:

• do we know how the system evolves in time after t in ∆t?

• arrival may occur with rate λ∆t:

– it is independent from previous arrival: memoryless property (A1 = A2)!

• departure may occur with probability µ∆t:

– D2 is not the same as D1!

• this process is no longer Markovian!

1 1 2( ) : ( ) ( )D t D t D t¹

t

A2(t)

D2(t)

t

A1(t)

Figure 1: State of M/G/1 queuing system.



3.1. Transform method based on imbedded Markov chain

State of the system: M/G/1 queuing system

• number of customers in the system at special time instants d+1 , d+2 , . . . :

– choose these instants such that we should not track time since previous service started;

– which instants: just after service completion of customers (D(t+ x|x) = 0)!

t

d1

+d2

+

( ) ( ) ( 0) (0)( | ) 0

1 ( ) 1 (0)

D t x D x D t DD t x x

D x D

+ - + -+ = = =

- -

0x ®

( ) 1 tA t e

l-= -

( )D t

( ) , 0,1,...Q

S t k k= =

Figure 2: State of M/G/1 queuing system.



3.2. Alternative state description

State of the system: M/G/1 queuing system

• number of customers in the system and time since previous service started (N(t), D2(t)):

– in this case we know how the system evolves in time;

– we know the distribution of time till the next arrival: it is the same as initial one;

– we know the distribution of time till the next departure: we track it:

D(t+ x|x) = Pr{(T ≤ t+ x)|T > x} = (D(t+ x)−D(x))/(1−D(x)). (3)

t

t

( ) 1 tA t e

l-= -

( ) ( )( | )

1 ( )

D t x D xD t x x

D x

+ -+ =

-( ) , 0,1,...

QS t k k= =

Figure 3: State of M/G/1 queuing system given by (N(t), D2(t)).



4. Method of supplementary variablesNotes about the method:

• proposed by Kleinrock and Takagi in the middle of 70th;

• very powerful:

– can be used for a number of variations of M/G/1 queuing systems;

– can be used for G/M/1 and its variations too.

• allows to better understand operation of the M/G/1 system;

• one of the most complicated among available for M/G/1 system.

What is the approach:

• track the number of customers and service time that already achieved.

• this process is Markovian: treat it using some approach.



Do the following:

• consider M/G/1 queuing system at arbitrary time t;

• N(t) be the number of customers:

– evolution of N(t) is no longer Markovian;

– that is why we had to imbed Markov chain at departures.

Let us assume the following:

• X0(t), t ≥ 0 is the time already received by the customer which is in service;

• process {N(t), X0(t), t ≥ 0} is Markovian:

– X0(t) is the supplementary variable that helps in analysis.

Note that by definition for {N(t), X0(t), t ≥ 0} process we assume:

X0(t) = 0, N(t) = 0. (4)



t

...X0(t) = 0

t

...

observation point: tiX0(t)

observation point: ti

Figure 4: Service time X0 already received by the customer: X0 6= 0 and X0 = 0.



Define the steady-state probability that there are k customers in the system:

pk = limt→∞

Pr{N(t) = k}, k = 0, 1, . . . . (5)

Note the following:

• state is now given a couple at time t: {N(t), X0(t), t ≥ 0};

• we have to determine jpdf:

fk(t, x)dx = Pr{N(t) = k, x < X0(t) < x+ dx}. (6)

Considering this fk(t, x) under equilibrium conditions (t→∞):

fk(x) = limt→∞

Pr{N(t) = k, x < X0(t) ≤ x+ dx} = limt→∞

fk(t, x)dx,

f0(x) = 0. (7)

• the latter one is by definition:

– time already received by customer is zero when there are no customers in the system;

– recall we required that the server does not stay idle when there are customers in the system.



Consider a customer that requires a service time of duration X:

• it has pdf b(x) and PDF B(x);

• bc(x) is the pdf of service time X given that the certain age has already provided (X > x):

bc(x)dx = Pr{x < X < x+ dx |X > x} =b(x)

1−B(x). (8)

– this is a death rate of service time at the age X.

departuredepartureservice time

t

...

X0(t)

Pr{x<X<=x+dx|X>x}

Figure 5: Illustration of the death rate bc(x)dx = Pr{x < X ≤ X + dx|X > x}.



Consider steady-state operation:

• balance equations for state 0: flow out = flow in:

λp0 =

∫ ∞0

f1(x)bc(x)dx. (9)

– f1(x): jpdf that the state is 1 and service X, X ≥ 0 already provided;

– bc(x): death rate given that age X, X ≥ 0 already provided.

0 1 2 ...

ò¥

01 )()( dxxbxf c

0pl

Figure 6: Applying global balance principle for state 0.



Consider time instants t and (t+ ∆x):

• at t: {N(t) > 0, X0(t) = x};

• at (t+ ∆x): {N(t+ ∆x) = k,X0(t+ ∆x) = x+ ∆x}.

These event may occur in two ways:

• at t the state was {N(t) = k,X0(t) = x}:– there were no arrival in dx: (1− λ∆x);

– there were no service completion in dx: (1− bc(x)∆x).

• at t the state was {N(t) = k − 1, X0 = x}:– there was an arrival in dx: λ∆x;

– there were no service completion in dx: (1− bc(x)∆x).

• we assume that multiple events do not occur;

• combining these two we have:

fk(x+ ∆x)dx = (1− λ∆x)(1− bc(x))fk(x)dx+ λ∆x(1− bc(x)∆x)fk−1(x)dx. (10)



t

...

X0(t)

xDlone arrival:

xxbc D- )(1no completion:

})(,)({: 0 xxxtXkxtNxt D+=D+=D+D+})(,1)({: 0 xtXktNt =-=

dxxfxxbx kc )())(1( 1-D-Dl

......

dxxfxxbx kc )())(1( 1-D-Dl

,k xD1,k x-

Figure 7: Transition from {N(t) = k − 1, X0 = x} to {N(t+ ∆x) = k,X0(t+ ∆x) = x+ ∆x}.



t

...

X0(t)

})(,)({: 0 xxxtXkxtNxt D+=D+=D+D+

xD- l1

})(,)({: 0 xtXktNt ==

dxxfxxbx kc )())(1)(1( D-D- l

xxbc D- )(1no completion:

no arrival:

......,k xD,k x

dxxfxxbx kc )())(1)(1( D-D- l

Figure 8: Transition from {N(t) = k,X0 = x} to {N(t+ ∆x) = k,X0(t+ ∆x) = x+ ∆x}.



t→∞: retain ∆x terms and drop those with higher power of ∆x to get:

fk(x+ ∆x) = λ∆xfk−1(x) + (1−∆x(λ+ bc(x)))fk(x), k = 1, 2, . . . (11)

Take limits when ∆x→ 0 we get:

dfk(x)

dx+ (λ+ bc(x))fk(x) = λfk−1(x), k = 1, 2, . . . (12)

To solve for the jpdf fk(x), k = 0, 1, . . . , x > 0, use boundary conditions:

f1(0) = λp0 +

∫ ∞0

f2(x)bc(x)dx k = 1

fk(0) =

∫ ∞0

fk+1(x)bc(x)dx, k = 2, 3, . . . . (13)

Normalization condition is given by:

∞∑k=0

pk = p0 +∞∑k=1

∫ ∞0

fk(x)dx = 1. (14)



5. Direct approach based on imbedded Markov chainConsider again M/G/1 queuing system:

• consider imbedded Markov chain approach;

• Markov points are just after service completions.

ô

)(ôr

Figure 9: Imbedded Markov points at M/G/1 queuing system.

State of the system N(t): number of customers in the system just after departure.



Using the transform approach we proceeded as follows:

• obtain PGF of the number of customers just after departures;

• use Kleinrock’s result to get number of customers at arrival time instants;

• use PASTA property to get number of customers at arbitrary time instants;

In what follows, we will proceed as follows:

• obtain PF (not PGF) of the number of customers just after departures;

• use Kleinrock’s result to get number of customers at arrival time instants;

• use PASTA property to get number of customers at arbitrary time instants;

Let us define the following:

• pi, i = 0, 1, . . . : number of customers in the system at arbitrary time instants;

• qi, i = 0, 1, . . . : number of customers in the system just after departure;

• pi, i = 0, 1, . . . : number of customers in the system that arrival sees.



Let αk, k = 0, 1, . . . , be the probability of k arrivals in the service time.

arrivals

departuredepartureservice time

t

...ka

Figure 10: k arrivals in a service time.

Arrivals come from Poisson process αk, k = 0, 1, . . . , we can get αk:

αk =

∫ ∞0

(λx)k

k!e−λxb(x)dx, k = 0, 1, . . . . (15)



Consider imbedded Markov chain at steady-state t→∞:

• let qij, i, j = 0, 1, . . . , be transition probabilities of going from i to j;

• we can obtain qij, i, j = 0, 1, . . . as

qjk = αk, j = 0,

qjk = αk−j+1, j = 1, 2, . . . . (16)

0 1 2 ...

1a

2a

3

3a

0a 0a 0a 0a

2a

3a

2a2a

Figure 11: Transition probabilities of imbedded Markov chain at steady-state.



Steady-state probabilities are given by the solution of:

qk =∞∑j=0

qjqjk, k = 0, 1, . . . ,

∞∑k=0

qk = 1. (17)

In vector form it looks much better:

~qQ = ~q : (q0, q1, . . . )

q00 q01 q02 q03 . . .

q10 q11 q12 q13 . . .

0 q21 q22 q23 . . .

0 0 q32 q33 . . ....

......

......

= (q0, q1, . . . )

~q~e = 1 : (q0, q1, . . . )

1

1...

= 1 (18)



5.1. Solution: transform approach

Write the previous set of linear equations as follows:

q0 = q0α0 + q1α0, k = 0,

q1 = q0α1 + q2α0 + q1α1, k = 1,

. . .

qk = q0αk + qk+1α0 +k∑j=1

qjαk−j+1, k = 2, 3, . . . . (19)

Multiply kth equation by zk and summing all LHSs and RHSs from k = 0 up to ∞:

Q(z) = q0A(z) + q1A(z) + q2zA(z) + q3z2A(z) + · · · = q0A(z) +

A(z)

z(Q(z)− q0), (20)

• Q(z) is the PGF of the number of customers in the system just after departure.

Rearranging terms gives us:

Q(z) =q0(1− z)A(z)

A(z)− z. (21)



Note the following:

• if the steady-state exists (ρ = λE[X]) we have:

q0 = 1− ρ, when ρ = λE[X]. (22)

– we obtained this result previously;

– we can obtain it also noting the property Q(1) = 1.

How to proceed further:

• use Klienrock’s principle to claim:

ak = dk, k = 0, 1, . . . . (23)

– where ak, k = 0, 1, . . . is steady-state distribution as seen by arrival.

• use PASTA property to claim:

ak = pk, k = 0, 1, . . . . (24)



5.2. Solution: direct approach

Using q0 = 1− ρ we can recursively evaluate qk, k = 1, 2, . . . :

q1 =1

α0

q0(1− α0), k = 1,

qk =1

α0

(qk−1 −

k−1∑j=1

qjαk−j − q0αk−1

), k = 2, 3, . . . . (25)

• it gives the steady-state distribution just after departure.

How to proceed further:

• use Klienrock’s principle to claim:

ak = qk, k = 0, 1, . . . . (26)

• use PASTA property to claim:

ak = pk, k = 0, 1, . . . . (27)

Note: we will obtain pk, k = 0, 1, . . . directly.



Consider the mean time interval between successive imbedded time points:

D = q0

(1

λ+ E[X]

)+ (1− q0)E[X] = E[X] + q0

1

λ. (28)

Probability p0: system is empty at arbitrary time t:

• fraction of time the system is idle between successive imbedded points;

• therefore, we can write:

p0 =q0(1/λ)

E[X] + q0(1/λ)=

q0q0 + ρ

. (29)

Substitute q0 = (1− ρ) to get:

p0 =1− ρ

(1− ρ) + ρ= 1− ρ = q0. (30)

We can continue to get pk = qk, k = 1, 2, . . . .



6. M/G/1/K queuing systemM/G/1/K queuing system is characterized by:

• arrivals come from the Poisson process with mean λ;

• generally distributed service times with PDF B(t), pdf b(t) and LT B(s);

• single server;

• limited capacity: limited number of waiting positions.

Note: M/G/1/K queuing system is a very powerful model for packet switching systems.

Departures

...

Arrivals

K-1 waiting positions

Server

Losses

Figure 12: Graphical representation of M/G/1/K queuing system.



Additional notes:

• K represents the maximum number of customers in the system where:

– one can be served;

– (K − 1) at maximum can wait in the buffer.

• M/G/1/K queue is the loss system:

– arrivals that arrive when K are in, just leave without service;

– these customers are often referred to as lost or blocked.

• let PB be the probability that the customer is lost:

– this is the same as probability that arrival sees K in the system;

– the following fraction of customers actually enters the system:

λA = λ(1− PB). (31)

Note the following:

• λA is the actual rate at which customers enter the system;

• λA must be used in Little’s result.



6.1. Steady-state distribution

Consider Markov chain imbedded at moments of customer departures:

• at times instants ti, i = 0, 1, . . . there are ni, i = 0, 1, . . . , K − 1 customers;

• the state space of the system is then:

ni ∈ {0, 1, . . . , K − 1}, i ∈ {0, 1, . . . }, (32)

– departure cannot leave the full system!

To get probabilities of transitions we have to distinguish:

• ith customers leaves an empty system ni = 0;

• ith customers leaves a non-empty system ni 6= 0.

In each case we have to distinguish:

• overflow;

• no overflow.



state of the system: ni+1 = K-1state of the system: ni = K-r > 0

service timet

...

K customers K-1 customers

1ira

+³

ith

departure (i+1)th

departure

state of the system: ni+1 = K - r - 1 + ai+1state of the system: ni = K-r > 0

service timet

...


1ira

+<

ith

departure (i+1)th

departure

ove

rflo

wn

oo

ve

rflo

w

Figure 13: The case ni > 0.



ith

departure(i+1)

thservice time

t

...


11

iKa

+< -

state of the system: ni = 0 state of the system: ni+1 = ai+1

(i+1)th

departure

ith

departure(i+1)

thservice time

t

...


11

iKa

+³ -

state of the system: ni = 0 state of the system: ni+1 = K-1

(i+1)th

departure

no

ove

rflo

wove

rflo

w

Figure 14: The case ni = 0.



We can write that:

• when ith customers leaves an empty system ni = 0:

ni+1 = min(αi+1, K − 1), ni = 0. (33)

• when ith customers leaves a non-empty system ni 6= 0:

ni+1 = min(ni − 1 + αi+1, K − 1), ni = 1, 2, . . . , K − 1. (34)

What we are doing next:

• let pd,k, k = 0, 1, . . . , K − 1 are steady-state probabilities as seen by departure;

• we can take transform approach;

• we are going to find pd,k, k = 0, 1, . . . , K − 1 directly.

To do it, we have to find transition probabilities of imbedded Markov chain:

pd,jk = Pr{ni+1 = k|ni = j}, j, k ∈ {0, 1, . . . , K − 1}. (35)



Do the following:

• αk, k = 0, 1, . . . : probability of exactly k arrivals during a service time;

• the number of arrivals during service time is given by Poisson process;

• we can find these probabilities as follows by integration:

αk =

∫ ∞0

(λt)k

k!e−λtb(t)dt. (36)

• using αk you can find transition probabilities:

– case nj = 0:

pd,0k =

{αk, k = 0, 1, . . . , K − 2,∑∞

m=K−1 αm, k = K − 1,, (37)

– case nj > 1, 2, . . . , K − 1:

pd,jk =

{αk−j+1, k = j − 1, j, . . . , K − 2,∑∞

m=K−j αm, k = K − 1,. (38)



Using pd,jk you can solve the following:

pd,k =K−1∑j=0

pd,jpd,jk, k = 0, 1, . . . , K − 1,

K−1∑k=0

pd,k = 1. (39)

• and you’ll find pd,k, k = 0, 1, . . . , K − 1!

Note the following:

• we have K unknowns and therefore K independent equations are needed;

• take K − 1 equations out of (39) and normalizing condition:

pd,k = pd,0αk +k+1∑j=1

pd,jαk−j+1, k = 0, 1, . . . , K − 2,

K−1∑k=0

pd,k = 1. (40)



Considering system at steady-state, let:

• pa,k, k = 0, 1, . . . , K: probability that arrival finds k = 0, 1, . . . , K customers in the system:

– it does not matter whether this arrival actually enters the system.

• pk, k = 0, 1, . . . , K: probability that system has k = 0, 1, . . . , K customers at arbitrary time;

• pac,k, k = 0, 1, . . . , K − 1: probability that arrival sees k = 0, 1, . . . , K − 1 customers:

– we assume that this arrival joins the system.

Note the following:

• Klienrock’s result states:

pd,k = pac,k, k = 0, 1, . . . , K − 1. (41)

• PASTA property states:

pk = pa,k, k = 0, 1, . . . , K. (42)

Question: how to find pk?



Let PB be probability that the arrival is blocked:

• this is the probability that arrival observes K in the system;

• therefore, PB = pK and we can write:

pk = pa,k = (1− PB)pac,k = (1− PB)pd,k, k = 0, 1, . . . , K − 1. (43)

Note also the following:

K∑k=0

pa,k = 1,K−1∑k=0

pac,k = 1. (44)

Since the previous holds we can get:

K−1∑k=0

pa,k = 1− PB =K−1∑k=0

(1− PB)pac,k, (45)

Note: to find pk, k = 0, 1, . . . , K have to find PB!



Let us recall the following:

• E[X] be the mean service time of a customer;

• ρ = λE[X] be the offered traffic to the system.

The mean arrival rate of customers actually entering the system is:

λc = λ(1− PB), (46)

The throughput of the system is:

ρc = ρ(1− PB). (47)

These imply that the probability of finding empty system at arbitrary time is:

p0 = 1− ρc. (48)

Using the latter one and previous results we can write:

1− ρ(1− PB) = (1− PB)pd,0, (49)

• where pd,0 has been found earlier.



From the last equation we can find probability of blocking:

PB = 1− 1

pd,0 + ρ. (50)

Using pd,k, k = 0, 1, . . . , K−1, and previous results (43, 50) we get pk, k = 0, 1, . . . , K−1:

pk =1

pd,0 + ρpd,k, k = 0, 1, . . . , K − 1, (51)

Now, for example, mean number in the system:

E[N ] =K∑k=0

kpk =1

pd,0 + ρ

K−1∑k=0

kpd,k +K

(1− 1

pd,0 + ρ

). (52)

The effective arrival rate to the queue is given by:

λc = λ(1− PB) =λ

pd,0 + ρ. (53)

• that should be used to obtain other performance parameters.


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M/G/1 and M/G/1/K systems: another lookTeletra c theory I: queuing theoryD.Molthchanov, TUT, 2011 3....

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