GI/M/1 queuing systemGI/M/1 queuing system

GI/M/1 queuing system

Dmitri A. Moltchanov

[email protected]

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

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

OUTLINE:

• GI/M/1 queuing system;

• Methods of analysis;

• Imbedded Markov chain approach;

• Waiting time in FCFS GI/M/1;

• GI/M/m queuing system;

• Note on results for GI/M/-/-/-;

• Rate conservation law for GI/M/-/-.

Lecture: GI/M/1 queuing system 2


1. GI/M/1 queuing systemGI/M/1 queuing system is characterized by:

• generally distributed interarrival times.

• exponentially distributed service times;

• single server;

• infinite capacity (therefore, infinite number of waiting positions);

We also assume that:

• FCFS service discipline when needed:

– mean waiting time is the same for all disciplines;

– PDF and pdf may be different!

Note: GI/M/1 is exactly opposite to what we had for M/G/1 queuing system.



1.1. Arrival and service processes

Arrival process:

• renewal arrivals with mean value λ;

• interarrival times are 1/λ;

• PDF, pdf and LT are:

A(t), a(t), A(s). (1)

Service process:

• service times are exponential with mean 1/µ;


B(t) = 1− e−µt, b(t) = µe−µt, B(s) =µ

s+ µ. (2)



2. Methods of analysisThere are a number of methods:

• transform approach based on imbedded Markov chain:

– 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:


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

• method of supplementary variables:


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

Note: we consider only the second.



3. Direct approach based on 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 (B1 = B2)!

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

t

A1(t)

A2(t)

B1(t)=B2(t)

B2(t)t

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



3.1. Problem with GI/M/1 queuing system

Let N(t) be the number of customers at time t:

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

• service completion may occur with rate λ∆t:

– it is independent from previously received service time: memoryless property (B1 = B2)!

• arrival may occur with probability µ∆t:

– A2 is not the same as A1!

• this process is no longer Markovian!

1 1 2( ) : ( ) ( )B t B t B t=

t

A2(t)

B2(t)

t

1 1 2( ) : ( ) ( )A t A t A t¹

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



3.2. How to choose the state of GI/M/1 system

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

• number of customers in the system and time since previous arrival: (N(t), A(t+ x|x)):

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

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

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

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

tt

( ) 1 tB t e

l-= -

( ) ( )( | )

1 ( )

A t x A xA t x x

A x

+ -+ =

-( ) , 0,1,...N t k k= =

Figure 3: State of GI/M/1 queuing system given by (N(t), A(t+ x|x)).



3.3. Steady-state distribution as seen by arrival

Let ρ be offered traffic load to G/M/1 queuing system:

ρ =λ

µ. (4)

Note: for GI/M/1 queuing system to be stable we require ρ < 1.

State of the system:

• we consider number of customers just before arrivals;

• note that arriving customer is not included!

...

arrivals

time instant tistate

Figure 4: Illustration of the system state in GI/M/1 queuing system.



Define the following:

• ni, i = 0, 1, . . . , be the number of customers in the system just before arrival i;

• consider two successive arrivals to the system ith arrival and (i+ 1)th arrival:

– si+1 be the number of served customers between ith and (i+ 1)th arrivals.

si+1 customers served in (i+1)th

interarrival time

(i+1)th

arrivalith

arrival

(i+1)th

interarrival time

t

...

ni customers just before ni+1 customers just before

Figure 5: Time diagram of GI/M/1 queuing system.



...Arrivals Server

ti ti+1

just before just before

nini+1

ni+1

(i+1)th

interarrival time

si+1

ni+1+1

Figure 6: Better view of time diagram of GI/M/1 queuing system.



We can relate ni and ni+1 using si+1 as:

ni+1 = ni + 1− si+1, ni = 0, 1, . . . . (5)

• note that by definition:

si+1 ≤ ni + 1. (6)

– since there are not more than (ni + 1) customers at time point t+i .

Note the following:

• stochastic process {ni, i = 0, 1, . . . } given by (5) and (7) is a continuous-time Markov chain;

• it defines number of customers seen by arrival.

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

• we will be looking for steady-state probabilities as seen by arrival;

• to find them we have to determine transition probabilities of the imbedded Markov chain:

pjk = limt→∞

Pr{ni+1 = k∣∣ni = j}, j, k ∈ {0, 1, . . . }, (7)

– pjk is the probability that (j+1−k) customers served between consecutive arrival instants.



Note the following:

• transition probabilities from state j to state k, k > j + 1, are not possible:

pjk = 0, k > j + 1. (8)

– state of the system cannot increase on more that one customer between imbedded points.

0 1 n-2 n-1 n n+1...

1nn, +p

n,1p

...1-nn,

p

2-nn,p

n,0p

Figure 7: State transition diagram of imbedded Markov chain.



Probabilities pik can be summarized in the compact matrix form:

P =

p00 p01 0 0 0 · · ·

p10 p11 p12 0 0 · · ·

p20 p21 p22 p23 0 · · ·

p30 p31 p32 p33 p34 · · ·...

......

......

. . .

. (9)

Let the steady-state probabilities be defined as follows:

pj, j = 0, 1, . . . . (10)

To determine pj, j = 0, 1, . . . we have to solve the following:

pk =∞∑j=0

pjpjk, k = 0, 1, . . . ,

∞∑k=0

pk = 1. (11)



Do the following:

• let αn, n = 0, 1, . . . : probabilities that there are n departures in interarrival time.

• service time is exponentially distributed with rate µ;

• number of service completions in a fixed interval is Poisson;

• get αn, n = 0, 1, . . . by integration over all possible durations of interarrival time:

αj =

∫ ∞0

(µx)j

j!e−µxa(x)dx, j = 0, 1, . . . . (12)

– this holds when the server is always busy during an interarrival time.

Use (12) to write balance equations for states of the system:

pj =∞∑k=0

αk+1pk, j = 0

pj = α0pj−1 +∞∑k=0

αk+1pj+k, j = 1, 2, . . . . (13)

• these have to be solved to get steady-state probabilities pj, j = 0, 1, . . . .



3.4. Solution

Note the following:

• to determine steady-state probabilities we use direct approach;

• we will try to find the solution in the form:

pn = σn, n = 0, 1, . . . . (14)

Do the following:

• substitute pn = σn into last equation of (13) and divide by σn−1:

σ =∞∑i=0

σiαi. (15)

• we already have expression for αi, i = 0, 1, . . . :

σ =∞∑i=0

σi∫ ∞0

(µt)i

i!e−µta(t)dt =

∫ ∞0

e−(µ−µσ)ta(t)dt. (16)

– the latter is Laplace transform of interarrival times in GI/M/1 queuing system.



Therefore, we get the following equation:

σ = A(µ− µσ), (17)

• which is LT of interarrival times with parameter (µ− µσ).

Finding the roots:

• σ = 1 is the simple root:

– it does not satisfy normalizing condition!

• we have to find a root in 0 < σ < 1 if ρ < 1:

– testing pn = σn in (13) shows that this is a root.

• normalizing pn = σn leads to the final result:

pn = (1− σ)σn, n = 0, 1, . . . . (18)

Conclusion: number of customers in the system as seen by arrival is the solution of:

σ = A(µ− µσ). (19)



3.5. Example: M/M/1 queuing system

LT of exponentially distributed interarrival times are given as follows:

A(s) =λ

λ+ s. (20)

Hence the equation from which we have to determine σ reduces to:

σ =λ

λ+ µ− µσ. (21)

Therefore:

σ(λ+ µ− µσ)− λ = (σ − 1)(λ− µσ) = 0. (22)

Finally,we get σ = ρ and the arrival distribution is:

pn = (1− ρ)ρn, n = 0, 1, . . . . (23)

• due to PASTA and Kleinrock this is the same as time-averaged and seen by departure!



3.6. Example: E2/M/1 queuing system

Assume, we are given:

• interarrival times are Erlang distributed with two phases and mean 2/3;

• Laplace transform of interarrival times is then:

A(s) = (3/(3 + s))2. (24)

• µ = 4 corresponding to stable system: ρ = 3/8 < 1.

Equation from which we have to determine σ reduces to:

σ = (3/(7− 4σ))2. (25)

Thus:

σ(7− 4σ)2 − 9 = (σ − 1)(4σ − 9)(4σ − 1) = 0. (26)

Therefore, the desired root is σ = 1/4:

pn = (3/4)(1/4)n, n = 0, 1, . . . . (27)



3.7. Example: E2/M/1 queuing system

Assume, we are given:

• interarrival times are generalized Erlang with two phases: µ and 2µ;

• Laplace transform of interarrival times are given by:

A(s) = (2µ2)/((µ+ s)(2µ+ s)). (28)

• service rate is µ leading to stable system.

Equation from which we have to determine σ reduces to:

σ =2µ2

(2µ− µσ)(3µ− µσ)=

2

(2− σ)(3− σ). (29)

It leads to following equation:

σ3 − 5σ2 + 6σ − 2 = (σ − 1)(σ − 2−√

2)(σ − 2 +√

2) = 0. (30)

The only root σ = 2−√

2 is acceptable leading to:

pn = (√

2− 1)(2−√

2)n, n = 0, 1, . . . . (31)



4. Waiting time in FCFS GI/M/1 systemWhat is nice about GI/M/1:

• we had to imbed Markov chain just before customer arrivals;

• steady-state distribution we found is what arrival sees;

• arrival has to wait until service completion of all customers it sees.

Let us define the following:

• T : waiting time in the system with PDF FT (t), pdf fT (t) and LT FT (s);

• Q: waiting time prior to service with PDF FQ(t), pdf fQ(t) and LT FQ(s);

• X: service time with PDF B(t), pdf b(t) and LT B(s).

We are looking for: distributions of T and Q.



ti

n

t

Q: waiting time prior to service

W: waiting time in the system

FCFS service discipline

Figure 8: Waiting time and time prior to service in GI/M/1 queuing system.

To determine total delay use the following:

• we know distribution as seen by arrival;

• we know that service times are exponential;

• use convolution to obtain time needed to serve customers which arrival sees.



Do the following:

• note that it is better to work in transform domain:

– sum of RVs is given by convolution;

– property: LT of the convolution is the product of individual LTs.

• we get for the total delay:

FT (s) =∞∑n=0

pn

(µ

µ+ s

)n+1

. (32)

Substituting pn = (1− σ)σ2 we get LT FT (s):

FT (s) =∞∑n=0

(1− σ)σn(

µ

µ+ s

)n+1

=

=µ(1− σ)

µ+ s

∞∑n=0

(µσ

µ+ s

)n=

=µ(1− σ)

µ(1− σ) + s. (33)



Observe the latter result:

FT (s) =µ(1− σ)

µ(1− σ) + s. (34)

• one may recognize LT of exponential with with parameter µ(1− σ):

FT (t) = Pr{T ≤ t} = 1− e−µ(1−σ)t, t ≥ 0. (35)

Note the following:

• total waiting time is almost the same as for M/M/1 (replace ρ by σ).

We may use the same approach to get FQ(t):

FQ(t) = Pr{Q ≤ t} = 1− σe−µ(1−σ)t, t ≥ 0. (36)

Note: probability that customer has to wait is (1− σ)!



4.1. Mean sojourn time

How to get:

• estimate directly from FT (t);

• compute using mean values approach.

Note that according to Little’s result we have:

E[N ] = λE[T ]. (37)

Note the following:

• arrivals are not Poisson and PASTA property does not hold:

E[NA] 6= E[N ]. (38)

For arbitrary arriving customer we have:

E[T ] = E[NA]1

µ+

1

µ, (39)

• E[NA] is the mean number of customers in the system as seen by arrival.



How to get NA?

Express E[NA] considering steady-state distribution at arrival time:

E[NA] =∞∑n=0

npn =∞∑n=0

n(1− σ)σn =σ

1− σ. (40)

Substitute this result into E[T ] = E[NA](1/µ) + 1/µ to get:

E[T ] =σ

(1− σ)µ+

1

µ=

1

(1− σ)µ. (41)

Now we can get the mean number of customers in the buffer as follows:

E[N ] =λ

(1− σ)µ=

ρ

(1− σ). (42)



5. GI/M/m queuing systemGI/M/m queuing system is characterized by:

• generally distributed interarrival times.

• exponentially distributed service times;

• multiple identical servers;

• infinite capacity: infinite number of waiting positions.

We also assume that:

• FCFS service discipline when needed:

– mean waiting time is the same for all disciplines;

– PDF and pdf may be different!

Note: GI/M/m is exactly opposite to what we had for M/G/m queuing system.



5.1. Arrival and service processes

Arrival process:

• renewal arrivals with mean value λ;

• interarrival times are 1/λ;


A(t), a(t), A(s). (43)

Service process:

• service times are exponential with mean 1/µ;


B(t) = 1− e−µt, b(t) = µe−µt, B(s) =µ

s+ µ. (44)



5.2. Direct approach based on imbedded Markov chains

Let ρ be offered traffic load to G/M/m queuing system:

ρ =λ

mµ. (45)

Note: for GI/M/m queuing system to be stable we require ρ < 1.

State of the system:

• we consider number of customers just before arrivals at times ti, i = 0, 1, . . . ;

• note that arriving customer does not included!

...Arrivals

...

1

m

Figure 9: Graphical representation of GI/M/m queuing system.



Define the following:

• denote the number of customers in the system at time ti by ni, i = 0, 1, . . . ;

• process of ni changes constitutes a imbedded continuous-time Markov chain.

ti ti+1

just before just before

nini+1

ni+1

(i+1)th

interarrival time

si+1

ni+1+1

Figure 10: Interarrival time between two successive imbedded time points.

Note: the only difference compared to GI/M/1 is that customers are served by m servers.



Define steady-state probabilities as seen by arrival:

qj, j = 0, 1, . . . . (46)

These probabilities should satisfy the following balance equations at t→∞:

qj =∞∑

i=j−1

qiqij, j = 0, 1, . . . . (47)

• qij are transition probabilities from i to j between successive imbedded Markov points.

0 1 n-2 n-1 n n+1...

n,n 1q

+

n,1q

...n,n-1

q

n,n-2q

n,0q

Figure 11: Transition diagram of imbedded Markov chain.



Steady-state probabilities must also satisfy normalizing condition:

∞∑j=0

qj = 1. (48)

What we should do to get: qj, j = 0, 1, . . . :

• we have to get transition probabilities qij;

• to get qij we have to determine the number of service completions in interarrival time.

Consider two arrival time instants ti and ti+1:

• assume that i customers exist in the system just before time ti and (i+ 1) just after;

• probability that (i− j + 1) customers are served during interarrival time:

– integrate over all possible number of arrivals (Poisson: mµ) during interarrival time:∫ ∞0

(mµx)(i−j+1)

(i− j + 1)!e−mµxa(x)dx, m ≤ j ≤ i+ 1. (49)

– this results in exactly j customers prior to time instant ti+1.



5.3. Some performance parameters of GI/M/m queuing system

Probability that arriving customer has to wait:

PW =∞∑i=m

qi. (50)

• since this event may only occur when all servers are busy;

• note that we use steady-state distribution as seen by arrival.

PF of the number of customers ahead in the queue given that the customers waits:

• tag an arbitrary customer arriving to the queue;

• letting n be the number of customers prior to this arrival, we have:

Pr{n = j∣∣n ≥ m} =

Pr{n = j⋂n ≥ m}

Pr{n ≥ m}=qm+j

PW, j = 1, 2, . . . . (51)

• using previous result for probability that an arriving customer has to wait we get:

Pr{n = j∣∣n > m} =

qm+j∑∞i=m qi

, j = 1, 2, . . . . (52)



6. Note on results for GI/M/-/-/-Notes on results we got for GI/M/-/-/- queuing system:

• we got steady-state probabilities of the number of customers as seen by arrival;

• we got waiting time directly:

– thanks that imbedded points are just before arrivals.

• we got time-averaged parameters indirectly!

Observing GI/M/s queuing system one can claim that:

• Klienrock principle holds:

– obtained distribution is the same as the distribution seen by departure.

• PASTA property does not hold:

– at this time we cannot say what is the distribution at arbitrary time t.

Question: how to get distribution at arbitrary time t?



7. Rate conservation law for GI/M/-/-/- queuesConsider a system with following parameters:

• customers arrive in a renewal arrival process with rate λ;

• each customer require exponentially distributed service time with mean µ−1;

• system has m, m ≥ 1 servers.

Let us denote by:

• pj: steady-state probability that there are j customers in the system at arbitrary time;

• qj: steady-state probability that there are j customers in the system just before arrival.

For considered system the following rate conservation law holds:

λqj−1 = jµpj, j ≤ m, loss system,

λqj−1 = jµpj, j ≤ m, delay system,

λqj−1 = mµpj, j > m, delay system. (53)



7.1. Rate conservation law for loss system

Consider loss system:

• example: GI/M/1/K;

• example: GI/M/m/m, etc.

Think as follows:

• let Sj denote j customers in the system;

• qj−1 is the probability that j − 1 customers just prior to arrival and λ is the arrival rate:

Sj−1 → Sj. (54)

• at the same time RHS represents rate:

Sj → Sj−1 (55)

• both sides must be balanced at steady-state:

λqj−1 = jµpj. (56)


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