M/G/1 queue

Weiqiang Sun

M/G/1 QUEUEShanghai Jiao Tong University

Weiqiang SunWeiqiang Sun

M/G/1 queue

2

The M/G/1 queue

General service time distribution- i.i.d (identical independent distribution)- Service time is independent from arrival- = 1/μ, average service time- : Second moment of service time, <∞

X2X

Poisson Arrival rate λ: Markovian

M/G/1Poisson arrivals

General independent service times


Pollaczek-Khinchin (P-K) formula

Applying Little’s Theorem

3

2

2(1 )XW

X

,

QN WT W X

Number of customers in queue

W: Average waiting time in queue

Average system time: queueing delay + service time

Again Little’s Theorem, we get the number of customers in system

QN T W N


M/G/1 examples• Example 1: M/M/1

4

• Example 2: M/D/1

22

1 2; X X

2 (1 ) (1 )W

Deterministic service time: 1/μ2

2

1 1; X X

22 (1 ) 2 (1 )W

• Waiting time of M/D/1 is half of M/M/1, in fact the results of M/G/1 is a lower bound for all M/G/1 with same λ and μ


Proof of P-K formula• Suppose customer i arrives to the system and finds– Ni customers waiting in queue– Up to one customer receiving service

• And also define:– Ri: the residual service time seen by i– Wi: the waiting time in queue of customer i

5

Customer i arrives

Xi-3Xi-2Xi-1

Xi

Xi-4

Ni =3


Proof of P-K formula (Cont.)

6

Xi-3Xi-2Xi-1Xi Xi-4

Customer i arrivesCustomer i starts to receive serviceWi

RiNi =3 1

i

ii i jj i N

W R X

[ ] [ ] [ ] [ ]i i i QE W E R E X E N R X N

By Little’s Theorem, QN W

QW R X N

/(1 )W R X

,


R – The average residual service time

• Average residual service time is the sum of area of the triangles divided by time t

7

R(t)residual service time

t

Let M(t) = the number of customers served by time t

X1

X1 X2 X3 X4

X2

X3

X4

2( )

10

1 1( ) ( )2

t M t ii

XR t R d

t t

( ) 21( )

2 ( )

M tii

XM tt M t

As ( ), M ttt

is the average departure rage, and is equal to arrival rate λ

Hence,

2X

2

2iX

R

The P-K formula is then proved.


R in M/M/1• For M/M/1, we already know that

8

22

1 2; X X

• So, R for M/M/1 is:2

2

12

iXR

• Why it is not equal to 1/μ, given the PASTA property? R(t)residual service time

t

X1

X1 X2 X3 X4

X2

X3

X4


M/G/1 with vacations• Once the system is empty, the server takes a vacation

– If system is still empty after the vacation, the server takes another vacation– Useful in analyzing some polling and reservation systems

• Vacation times are i.i.d and independent of service times and arrival times

• The only impact on analysis is that a customer may enter to find the server on vacation, and must wait until end of that vacation

9

Xi-3Xi-2Xi-1Xi

Customer i arrivesWi

ViNi =3 1

i

ii i jj i N

W R X


R(t) with vacations

• A customer will always experience some residual time, either because server is on vacation, or is serving a customer

10

R(t)residual service time

t

X1

X1 X2 X3 X4

X2

X3

X4

V2V1 V3

Let M(t) = the number of customers served by time tL(t) = the number of vacations taken by time t

2 2( ) ( )

1 1

1( )2 2

M t L ti ii i

X VR t

t


R(t) with vacations (cont.)

• From the server’s point of view,

11

( )limt

M tt

( )lim vt

L tt

And we have ,

1v V (1 ) /v V

• Hence, 22 1

2 2VXR

V

And

2 2

/(1 )2 1 2

X VW RV

?


Example: Slotted M/D/1 system

• Fixed slot duration 1/ μ• Service can begin only at start of a slot• If a customer misses the slot start, it must wait until next start

(slot in vacation)

12

1/μ

E[X] = E[V] = 1/μE[X2] = E[V2] = 1/μ2

2 2 / 12 1 2 22

X VWV

/ /11

2M DW

• This part of time is spent waiting for the slot (syncronizing)


Example. Text problem 3.29

13

M/G/1 queue

Documents

Transcript of M/G/1 queue