Modeling and simulation of systems Introduction to queuing theory Slovak University of Technology Faculty of Material Science and Technology in Trnava
Modeling and simulation of systems
Introduction to queuing theory
Slovak University of TechnologyFaculty of Material Science and Technology in Trnava
The system of queuing model
Service Queue Arriving customer
Serviced customer
The system of queuing model is an arbitrary service where the service of a particular kind is provided. The customers (demands) who require service arrive into this system. The element of the system which provides the service is called serviced channel or line.
The system of queuing model
Important qualities of elements of queuing model :
Arrivals of customersintervals between arrivals are important
run of serviceone or more serviced channel (links)period of service
behaviour of customers when they cannot be serviced immediately systems without queue with limited length of queue with unlimited length
Qualification of queuing models
Kandal´s qualification which divides queuing models according to three criteria is used the most frequently: Arrival of customers
is described by random process
The division of probabilities of period of service The number of servicing channels
Used shape of qualification X/Y/c
Qualification of queuing models
Examples of marking: M/M/n n = 1,2,3 ....
Arrival – Poisson´s process Service – exponential divisionn – number of serviced channels
D/D/cArrival – regular intervalsService – constant period
G/G/cgeneral case it means no assumptions about arrivals and general division of period of service
Arrivals of customers (stochastic process) The customers arrive individually For each time t we get number N(t) – number of
customers who came at time (0,t The result is non falling function N(t), t0, which gains
the whole non negative values and N(0) = 0
t0 t1 t2 t3t4 t5
N(t)
1
2345
Arrivals of customers(stochastic process)
Arrivals into the system are random then for given t is N(t) random quality what is division of probabilitiespk(t) = P{N(t) = k}, k = 0,1,2,...
Arrival of customer is defined as the system of random variables N(t), t0 and is edited{N(t)}tT T0,
Such system is called stochastic (random) process
Poisson´s homogenous process
Is given random process {N(t)}t0 with independent
growths it means random qualities N(b1) - N(a1), N(b2) -
N(a2), .., N(bn) - N(an) are independent, if intervals
(a1,b1, (a2,b2, .., (an,bn are disjunctive when all
random qualities N(t+) - N(t) have Poisson´s division of probability.
!k
etNtNPk
The middle value of number of customers who come in time is
Qualities of Poisson´s process
Homogeneity The probability of it that in the time comes to
customers does not depend on the beginning of interval (t, t+It means that arrival of customers is regular for the whole period.
Independence the number of customers who came during each period does not depend on the number of customers who came into the system in other disjunctive time period.
Qualities of Poisson´s process
Ordinary
0
1)(0
lim
NP
It means that more than one customer comes in a very short time interval with insignificant probability smaller that the length of this interval.It is not probable that more customers come at the same time.
The relation between Poisson´s division and exponential division
The division of random quantity which is the interval between arrival of two customers that isk= tk - tk-1 k = 1,2,...
Distributive function F() of random variable k
F() = P{k } = 1 - P{k } P{k } = P{N() = 0} = e-
Then:
F() =1 - e-, 0
0, 0
Exponential distribution
If is the mean value of the number of customers arriving into thesystem for time unit, then.... is the mean value of time between arrivals of customers
The number of customers in the systemMarkovov´s homogeneous process
The system is characterized by the number of customers X(t), who are in the time t in the system.
If arrivals of service period is random then the number of customers X(t) at the moment t is random.
The function X(t) can fall as well as rise in dependence on how customers arrive and how fast are they served
The run of the number of customers in the system
t0 t1 t2 t3 t4 t5 t6 t7
1
23
4
X(t)
Random value X(t) gains values k = 0,1,2,... with probabilitiespk(t) = PX(t) = k
0
1)(j
j tpwhere for each t0
Markovov´s random process
Definition:the random process X(t)t0 is called Markovov´s,
if is in force:P X(t) = kX(s1) = j1, X(s2) = j2, ..., X(sn) = jn = P
X(t) = k X(s1) = j1 for ts1 s2 ... sn 0
It means that the future X(t) does not depend on the past X(si), i1, but only on the presence X(s1)
Homogeneous process
If is in forceP X(t+) = j X(t) = i == P X(s+) = j X(s) = i = = P X() = j X(0) = i , then the process is homogeneous.
It means that there will be j of customers in the system after the time , if there were i of customers in the time t it depends only on length of time , and not on since when it is monitored.
Probability of transition from the state j into the state i
Probability of transition from j into i is a probability by which the system transits from the state i into the state j and is marked as pij().
Then pij() = P X(t+) = j X(t) = i = P N(t+) - N(t) = j - i
Poisson´s process is also Markovov´s and is in force:
)!()(
)(ij
epij
ij
Intensity of transition
The intensity of transition of Markovov´s homogeneous process X(t)t0 is called number
)(lim
0
ijij
p
ij
Intensity of transition
SHO is given with one service channel without waiting. The arrivals with Poisson´s division with the middle value , period of service – exponential division with the middle value 1/. Random process (then the number of customers in the system gains the values 0 a 1.
It is necessary to determine 01
0 1
01
10
Intensity of transition
)(lim 01
001
p
It is necessary to determine for the calculation p01.
The system transits from the state 0 into the state 1:A: For the time just one customer arrives and his
service after the time does not end.B: For the time more that one customer arrives and the
service of one of them after the time does not endIt means that p01( ) = PA + PBa
}{lim
}{lim
0001
BPAP
Intensity of transition
Event B:
PN() 1
0}1)({
lim}{
lim00
NPBP
Ordinary ofPoisson´s random process
Intensity of transition Event A:Probability A is given by product of probability that for
the time just one customer arrives and probabilities that his service after does not end.
PA= PN() =1. Ptobs PN() =1=e-.Distributive function of period of service tobs
F() = Ptobs = 1- e- , 0then
Ptobs = 1- Ptobs = e-
eeAP00
lim}{
lim
Intensity of transition
Intensity of transition 01= +0= It is possible to determine in similar way
10= Then it is necessary to determine for the
system with one service channel without queue probabilities of the fact that there is in the system in the time t k of customers:
pk(t)= PX(t) = k
Kolmogorov´s differential equations
Determination of previous probabilities leads to the following equations:
)()( 10'0 tptpp
)()( 01'1 tptpp
These equations are Kolmogorov´s differential equations. They are not ependent.It is force:p0(t) + p1(t) = 1
Kolmogorov´s differential equations
For transcript of Kolmogorov´s equations is in force:
n
kjjjkk
n
kjjkjkk tptptp
,0,0
' )()()(
Derivation p’k of probability that the system
in the time t in the state k equals to summation of probabilities, that in the state k multiplied by sum of negatively outstanding intensities of transition getting out of state k and probabilities of all other states multiplied by intensities of transition which getting out into the state k
Kolmogorov´s differential equations
3221100 1 2 3
01 12 23
Kolmogorov´s differential equations:p’
0 = - 01p0 + 10p1
p’1 = -(10 + 12) p1 + 01p0 + 21p2
p’2 = -(21 + 23) p2 + 12p1 + 32p3
p’3 = - 32p3 + 23p2
Example of transcript of equations: