M/G/1 and M/G/1/K queuing systemsM/G/1 and M/G/1/K queuing ... · 4.2. How to choose the state of...

M/G/1 and M/G/1/K systems

Dmitri A. Moltchanov

[email protected]

http://www.cs.tut.fi/kurssit/ELT-53606/

Network analysis and dimensioning I D.Molthchanov, TUT, 2013

OUTLINE:

• Description of M/G/1 system;

• Methods of analysis;

• Residual life approach;

• Imbedded Markov chain approach.

Lecture: M/G/1 and M/G/1/K systems: part I 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: PDF or pdf of waiting time!



1.1. Arrival and service processes

Arrival process:

• Poisson with parameter (mean value) λ

• interarrival times are exponential with mean 1/λ;

• CDF, pdf and LT are:

A(t) = 1− e−λt, a(t) = λe−λt, A(s) =

∫ ∞0

a(t)e−stdt =

∫ ∞0

λe−λte−stdt =λ

s+ λ. (1)

Service process:

• service times are arbitrary 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:

• 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 direct integration.

• method of supplementary variables: the most complicated one:


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



3. Residual life approach: mean waiting timeNote the following:

• we want to find mean queuing time (mean time spent waiting for service);

• no need to assume FCFS service discipline:

– service order does not influence mean waiting time;

– we have to require that the server does not stay idle when there are waiting customers;

– mean values are the same as for LCFS, RANDOM, PS.

...

Arrivals:1

m

1

lService:

Figure 1: M/G/1 queuing system.

Note: we are looking for E[WQ]: mean waiting time in the buffer.



Consider arriving customer, we have two choice:

• customer arrives to non-empty system or;

• it arrives to empty system.

When the customer arrives to non-empty system it finds:

• nq customers waiting in the buffer;

• customer which is currently in service:

– it has R residual service time left to complete its ongoing service;

– residual service time is the time required to complete the ongoing service.

When the customer arrives to empty system:

• no customers waiting nq = 0;

• residual service time R is zero;

• customer can enter service immediately.



Let us denote:

• RV R be the RV of residual service time:

– in case of exponential service times R has exponential distribution.

• E[R] be it mean.

Note the following:

• PASTA property holds for M/G/1 system;

• distribution that arrival sees is the same as time-averaged (at arbitrary time);

• using Little’s result we have for the mean waiting time in the buffer, E[WQ]:

E[WQ] = E[NQ]E[X] + E[R]. (3)

– E[NQ] - mean number of packets waiting in the buffer, E[X] mean service time;

– E[R] - mean residual service time.

Using Little’s result E[NQ] = λE[WQ] we get:

E[WQ] = λE[WQ]E[X] + E[R]. (4)



Denoting ρ = λE[X] and solving for E[WQ]:

E[WQ] = ρE[WQ] + E[R], E[WQ] =E[R]

(1− ρ). (5)

• to find E[W ] we, therefore, have to find mean residual service time E[R].

• in what follows we take graphical approach.

ô

)(ôr

Figure 2: Residual service times.

• R jumps by the amount of service time X when a new customer enters the service;

• decreases until the customer completes service, duration is X;

• R = 0 when customer completes its service.



Time averaged of r(τ) in (0, t) is:

E[r[0, t)] =1

t

∫ t

0

r(τ)dτ. (6)

Approximate previous equation by the following sum:

E[r[0, t)] =1

t

∫ t

0

r(τ)dτ ≈ 1

t

M(t)∑i=1

1

2X2i . (7)

• X2/2 is the square of a service triangle, M(t) – arrivals in (0, t).

Note the following:

• we have an error in the last residual service time in [0, t);

• approximation becomes better as t→∞ (number of components increases!).

Multiplying and dividing it by M(t) we get:

E[r[0, t)] =M(t)

t

1

M(t)

1

2

M(t)∑i=1

X2i . (8)



Consider the last equation when t→∞ we get:

limt→∞

M(t)

t→ λ

limt→∞

1

2

1

M(t)

M(t)∑i=1

X2i →

1

2E[X2]. (9)

Therefore, we have the following expression for mean residual service time:

E[R] = E[r] =1

2λE[X2], (10)

Substituting this result in E[W ] = E[R]/(1− ρ) we get:

E[WQ] =λE[X2]

2(1− ρ). (11)

• this result is known as Pollaczek-Khinchine formula:

– they simultaneously got this result.



3.1. Performance parameters

Mean time spent in the buffer: E[WQ]

E[WQ] =λE[X2]

2(1− ρ). (12)

Mean number of customers in the buffer: E[NQ]

E[NQ] = λE[WQ] =λ2E[X2]

2(1− ρ). (13)

Mean time spent in the system: E[W ]

E[W ] = E[WQ] + E[X] =λE[X2]

2(1− ρ)+ E[X]. (14)

Mean number of customers in the system: E[NQ]

E[N ] = E[W ]λ =λ2E[X2]

2(1− ρ)+ ρ. (15)



4. Imbedded Markov chainRecall M/M/-/-/-: let N(t) be the number of customers at time t:

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

• arrival may occur with probability λ∆t:

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

• departure may occur with rate µ∆t:

– it is independent from previous departure: memoryless property (D1 = D2)!

• the resulting process is actually birth-death Markov one!

t

A1(t)=A2(t)

A2(t)

D1(t)=D2(t)

D2(t)t

Figure 3: State of M/M/1 queuing system.



4.1. Problem with M/G/1 queuing system

Let SQ(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:

– µ? 6= µ as 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 4: State of M/G/1 queuing system.



4.2. How to choose the state of M/G/1 system

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

• number of customers in the system AND time since the service started:

– 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)). (16)

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 5: State of M/G/1 queuing system given by (SQ(t), D(t+ x|x)).



Are there any problems with such description?

• two complicated to deal with;

• we better prefer to avoid it if more simple description is available!

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 completions as D(t+ x|x) = D(t) when 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 6: State of M/G/1 queuing system.



4.3. Steady-state distribution as seen by departure

Doe the following:

• consider Markov chain imbedded at ti, i = 1, 2, . . . when ith customer departs from the system;

• system state: number of customers at ti, i = 1, 2, . . . denoted by ni, i = 1, 2, . . . ;

• departing customer is not included in the system state!

...

Time instant ti-

Time instant ti+

...

ni

-= k

ni

+= k-1 Just departed at ti

Figure 7: Time instants t−i and t+i in M/G/1 queuing system.



Denote:

• ai+1 be the number of arrivals in (i+ 1)th service time.

t

0x ®( )D t

ti ti+1

ai+1

ith departure

Figure 8: Arrivals in (i+ 1)th service time.

Consider time diagram of the system and distinguish between:

• ith departure leaves non-empty system: ni > 0;

• ith departure leaves empty system: ni = 0.



Consider the case when ni > 0.

ai+1 arrivals in (i+1)th

service time

(i+1)th

departureith

departure

(i+1)th

service timet

...

(system is not empty;

ni>0 ni+1 = ni - 1 + ai+1

(system may be empty may be not;

Figure 9: Time diagram of the M/G/1 queuing system when ni > 0.

In this case we have the following relation between ni and ni+1:

ni+1 = ni − 1 + ai+1, ni = 1, 2, . . . . (17)



Consider the case when ni = 0.

(i+1)th

departureith

departure

(i+1)th

service timet

...

system is empty

ni = 0 ni+1 = ai+1

first arrival after empty period

system may be empty may be not

ai+1 arrivals in (i+1)th

service time

Figure 10: Time diagram of the M/G/1 queuing system when ni = 0.

In this case we have the following relation between ni and ni+1:

ni+1 = ai+1, ni = 0. (18)



Summarizing both cases (ni > 0 and ni = 0) we have:

ni+1 = ni − 1 + ai+1, ni = 1, 2, . . . ,

ni+1 = ai+1, ni = 0. (19)

Introducing unit step function U(ni) rewrite:

ni+1 = ni − U(ni) + ai+1, ni = 0, 1, . . . , (20)

• where unit step function is given by

U(ni) = 1, ni = 1, 2, . . . ,

U(ni) = 0, ni = 0. (21)

• this expressin gives us the idea how the system evolves between ti and ti+1;

• recall that ti, i = 0, 1, . . . are imbedded points of the Markov chain.

GENERAL RELATION:

ni+1 = ni − U(ni) + ai+1, ni = 0, 1, . . . , (22)



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

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

• let ak the number of arrivals in the service time;

qik = αk, i = 0, k = 0, 1, . . . ,

qik = αk−i+1, i = 1, 2, . . . , k = i− 1, i, i+ 1, . . . (23)

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.



Observe that αk, k = 0, 1, . . . fully determine the Markov chain.

arrivals

departuredepartureservice time

t

...ka

Figure 12: 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, . . . . (24)



4.4. Transform approach vs. direct integration

The difference is twofold:

• in the way we compute αk, k = 0, 1, . . .

αk =

∫ ∞0

(λx)k

k!e−λxb(x)dx, k = 0, 1, . . . . (25)

– first term – number of arrivals from Poisson process in time x (Erlang distribution);

– b(x) service time distribution equals to x;

– we integrate over all length of t: from 0 to ∞.

• in the way we solve linear equations of the Markov chain

– asking Mathlab to do it for us;

– trying to get closed-form analytical solution.

Let us define the following:

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



4.5. Brute-force direct approach

What we do:

• estimate αk directly via integration;

• use classic way to get steady-state solution for qi, i = 0, 1, . . . ;

• works when capacity is limited M/M/1/K.

In vector/matrix form:

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

q00 q01 q02 q03 . . .

q10 q11 q12 q13 . . .

0 q21 q22 q23 . . ....

......

......

= (q0, q1, . . . )

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

1

1...

= 1 (26)



4.6. Other steady-state distributions

What we get:

• PF and moments of the number as seen by departure...

Note: we are usually asked to get state distribution at arbitrary time!

Let us define the following:

• pk, k = 0, 1, . . . : time-averaged (arbitrary time) steady-state distribution of states;

• ak, k = 0, 1, . . . : steady-state distribution of states as seen by arrival;

• dk, k = 0, 1, . . . : steady-state distribution of states as seen by departure.

Question: how to find pk, k = 0, 1, . . . ?



Two next results gives us a way to use this PF:

• Kleinrock’s result (!!! remember it as well as PASTA):

– the following holds for those system when state can change at most ±1 in ∆t→ 0:

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

• PASTA property:

– the following holds when the arrival process is homogenous Poisson:

ak = pk, . . . , k = 0, 1, . . . . (28)

• combining both we have for M/G/1 queuing system:

ak = pk = dk, . . . , k = 0, 1, . . . . (29)

What it gives:

• we obtained steady-state distribution at any instant of time;

• qk = ak facilitates derivation of waiting time CDF.



4.7. Performance parameters

Mean time spent in the buffer: E[WQ]

E[WQ] =λE[X2]

2(1− ρ). (30)

Mean number of customers in the buffer: E[NQ]

E[NQ] = λE[WQ] =λ2E[X2]

2(1− ρ). (31)

Mean time spent in the system: E[W ]

E[W ] = E[WQ] + E[X] =λE[X2]

2(1− ρ)+ E[X]. (32)

Mean number of customers in the system: E[NQ]

E[N ] = E[W ]λ =λ2E[X2]

2(1− ρ)+ ρ. (33)



5. Transform approachConsider ni+1 = ni − U(ni) + ai+1 and take means of both sides to get:

E[ni+1] = E[ni]− E[U(ni)] + E[ai+1]. (34)

Since we consider the system at steady-state E[ni+1] = E[ni] = E[n]:

E[U(ni)] = E[ai+1]. (35)

Note the following:

• assuming that the system is empty with probability d0 we have:

E[U(ni)] = 1− d0. (36)

• mean number of arrivals in (i+ 1)th service time is given by:

E[ai+1] =

∫ ∞0

λtb(t)dt = λE[X] = ρ. (37)

Therefore, in the steady-state we have:

d0 = 1− ρ. (38)



Introduce PGFs of ni and ni+1:

Pi(z) = E[zni ] =∞∑k=0

zkPr{ni = k},

Pi+1(z) = E[zni+1 ] =∞∑k=0

zkPr{ni+1 = k}. (39)

Considering ni+1 = ni − U(ni) + ai+1 as the sum of RVs ni − U(ni) and ai+1:

Pi+1(z) = E[zni−U(ni)]E[zai+1 ]. (40)

To simplify it, consider the following:

• since we consider system at steady-state we can drop subscripts i and write:

P (z) = E[zn−U(n)]E[za]. (41)

• defining PGF of number of arriving customers in service time by A(z) = E[za], then:

A(z) =

∫ ∞0

E[za|X = t]b(t)dt =

∫ ∞0

∑zn

(λt)n

n!e−λtb(t)dt =

∫ ∞0

e−λt(1−z)b(t)dt. (42)



Rewrite the last result as:

A(z) =

∫ ∞0

e−λt(1−z)b(t)dt =

∫ ∞0

b(t)e−(λ−λz)tdt. (43)

Recall the definition of the Laplace transform:

F (s) =

∫ ∞0

f(t)e−stdt, (44)

Thus, we have for PGF of arriving customers in service time:

A(z) = B(λ− λz). (45)

• where B(·) is the LT.

Now we can get moments of number of arrivals in service time:

A′(1) = λB′(0) = λE[X] = ρ;

A′′(1) = λ2B′′(0) = λ2E[X2]. (46)

• we used moments generating properties of PGF and Laplace transform.



Let us get back to P (z) = E[zn−U(n)]E[za]:

P (z) = A(z)E[zn−U(n)] = A(z)∞∑k=0

zk−U(k)Pr{n = k} =

= A(z)

(z0d0 +

∞∑k=1

zk−1pk

)= A(z)

(d0 +

1

z

∞∑k=0

zkpk −1

zd0

)=

= A(z)

(1

zP (z)− 1

zd0(1− z)

). (47)

Use d0 = (1− ρ) to get PGF of the number of customers as seen by departure:

P (z) =(1− ρ)(1− z)A(z)

A(z)− z, (48)

• where A(z) = B(λ− λz).

Substituting A(z) = B(λ− λz), we get Pollaczek-Khinchine transform equation:

P (z) =(1− ρ)(1− z)B(λ− λz)

B(λ− λz)− z. (49)



5.1. Mean number of customers as seen by departure

Using using moment generating properties of PGF and LT:

• we can get moments of the number of customers as seen by departure.

To get mean, rewrite (48) as follows:

P (z)(A(z)− z) = (1− ρ)(1− z)A(z). (50)

Taking two derivatives we get:

P ′(z)(A(z)− z) + P (z)(A′(z)− 1) = (1− ρ)(1− z)A′(z)− (1− ρ)A(z),

P ′′(z)(A(z)− z) + 2P ′(z)(A′(z)− 1) + P (z)A′′(z) = (1− ρ)(1− z)A′′(z)− (1− ρ)A′(z). (51)

Let z = 1 and use A(1) = 1, P (1) = 1, A′(1) = ρ and A′′(1) = λ2E[X2] to get:

−2P ′(1)(1− ρ) + λ2E[X2] = (1− ρ)(−2ρ) (52)

Mean number of customers as seen by departing customer:

E[N ] =λ2E[X2]

2(1− ρ). (53)


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