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241-460 Introduction to QueueingNetworks : Engineering Approach
Chapter 10 M/M/S and M/G/1 Queue
Assoc. Prof. Thossaporn KamolphiwongCentre for Network Research (CNR)
Department of Computer Engineering, Faculty of EngineeringPrince of Songkla University, Thailand
Chapter 10 M/M/S and M/G/1 Queue
Email : kthossaporn@coe.psu.ac.th
Outline
M/M/sAnalysis M/M/sAnalysis M/M/s
Average Number of customer (N) in System
Average Number of Customer in Queue (Nq)
Waiting Time (T)
Waiting Time (Tq) in queueq
Example
Chapter 10 : M/M/s and M/G/1 Queue
2
M/M/s Queue
M/M/s Customers arrive according to a Poisson processCustomers arrive according to a Poisson process
with rate Service time for each entity is exponential with
rate .
There are s servers.
Chapter 10 : M/M/s and M/G/1 Queue
M/M/s Queue
Poisson Arrival =
There are s servers. Customers arrive according to a Poisson process with
rate
Queue
Server
Service rate =
Service time for each entity is exponential with rate .
Chapter 10 : M/M/s and M/G/1 Queue
sn
sn
s
nn
3
Example
Queue
(Infinite buffer) Server = 3
Chapter 10 : M/M/s and M/G/1 Queue
State transition Diagram : M/M/s
s-12 s10 s+1… n… …
µ 2µ 3µ (s-1)µ sµ sµ sµ
Chapter 10 : M/M/s and M/G/1 Queue
4
State Diagram for M/M/s
s 12 s10 s+1
n
P0 = P1
P1 = 2P2
Ps-1 = sPs
Ps = sPs+1
s-12 s10 s+1
sµ2µ sµµ
…
(s-1)µ3µ
n…
sµ
…
P2 = 3P3
…
Ps-1 = sPs
Chapter 10 : M/M/s and M/G/1 Queue
…
Pn-1 = sPn
(Continue)
s 12 s10
n
For n < s
P0 = P1
P1 = 2P2
s-12 s10sµ2µµ
…
(s-1)µ3µ
nP0 = 12 3… (n-1)n nPn
n …
(n-1)µnµ
P1 2P2
P2 = 3P3
…
Pn-1 = nPn
Chapter 10 : M/M/s and M/G/1 Queue
s ,321 0
nP
nP
n
n
n
5
(Continue)
s-12 s10 s+1
…
n…
…
For n > s
P0 = P1
P1 = 2P2
…nP0 = 12 3… ssn-s nPn
n
s 12 s10 s+1
sµ2µ sµµ (s-1)µ3µ
nsµ
Ps-1 = sPs
Ps+1 = sPs+1
…
Pn-1 = sPn
Chapter 10 : M/M/s and M/G/1 Queue
snPss
Pnsnn
,
321 0
M/M/s
For n < s For n > s
s ,321 0
nP
nP
n
n
n s ,
321 0
nPss
Pnsn
n
n
s ,!
10
nP
nP
n
n
s ,!
10
nP
ssP
sn
n
n
Chapter 10 : M/M/s and M/G/1 Queue
6
M/M/s
snP
n
n
0!
1
snP
ss
nP
sn
nn
0!
1
!
snP
s n
0
Chapter 10 : M/M/s and M/G/1 Queue
s
snPs
s
PnP
nsn
0
0
!
!
M/M/s
P0 + P1 + P2 + P3 + … = 1 10
nnP
1
!! 0
1
00
sn
nss
n
n
Ps
sP
n
s
1!!
11
0
0
ss
ns
P ss
n
n
1
!! 0
1
00
sn
nss
n
n
s
sP
n
sP
1
Chapter 10 : M/M/s and M/G/1 Queue
1
1!! 0
1
00
sss
n
n
s
sP
n
sP
7
Probability of All servers are busy
If the number of customers in the system is less than or equal to s then all customers will be at athan or equal to s then all customers will be at a server.
The probability of arriving and finding all servers busy (and therefore having to queue for a server) :
ns
Pq
Chapter 10 : M/M/s and M/G/1 Queue
sn sn
ns
nq Ps
sPPP 0!
)FullServer All(
(Continue)
sn
ns
q Ps
sP 0!
Because
sn
1
1
0i
isn
sn
sns
q Ps
sP
0!
Prob. of all
Chapter 10 : M/M/s and M/G/1 Queue
10isn
01!
Ps
sP
s
q
servers are busy
8
Average # of packet in M/M/s queue
0i
sisn
nq iPPsnN
0
0!i
sis
q iPs
sN
,i = n-s
Chapter 10 : M/M/s and M/G/1 Queue
0
0! i
is
iPs
s
M/M/s
is
iPs
N 1
di ii
0
0! iq iP
sN
0
00
1
1
d
d
d
d
di
i
i
ii
0
10! i
is
iPs
s
002 1!11!
Ps
sP
s
s ss
Chapter 10 : M/M/s and M/G/1 Queue
21
1
qq PN
1
9
M/M/s
1qq PN
Wh ll h k h
1
1q
q
P
NN(t) (M/M/1)
• When all the servers are working, the system is equivalent to an M/M/1 queue with a service rate of s
Chapter 10 : M/M/s and M/G/1 Queue
Waiting Time
• Waiting Time in Queue
• Total Waiting Time in System
s
PPNW qqq
q 1
Chapter 10 : M/M/s and M/G/1 Queue
11
s
PWT q
q
10
Average # of Customer in system
TN Nq
s
Pq
sP
Chapter 10 : M/M/s and M/G/1 Queue
sPq
1
M/M/2
n 12 n
10
P0 = P1
P1 = 2P2
n-12 n
2µ
10
µ 2µ2µ
…
2µ
P2 = 2P3
...................
Pn-1 = 2Pn
Chapter 10 : M/M/s and M/G/1 Queue
11
M/M/2
nnn PP 2221 P0 = P1
P1 = 2P2nPP 0 221
0, 2
2 0
nPP
n
n
02 PP n
P1 2P2
P2 = 2P3
...................Pn-1 = 2Pn
Chapter 10 : M/M/s and M/G/1 Queue
2, 0, 2 0 nPP n
n
M/M/2
12 00
nPP 1
00 n
121
00
n
nPP
121
100
n
nPP 1, nm
Chapter 10 : M/M/s and M/G/1 Queue
1n
120
00
m
mPP
12
M/M/2
12 00
mPP
1 , 1
10
P11
210
P
0m
11
12 00
PP
Chapter 10 : M/M/s and M/G/1 Queue
1
11
210
P
Average # of customers in System
0
nnPN0n
0
02n
nPnN
0
02 nnPN 21
1
Chapter 10 : M/M/s and M/G/1 Queue
0
10
0
2n
n
n
nP
1
13
Average # of customers in System
12 PN
201
2
PN
22
1
1
1
12
N
Chapter 10 : M/M/s and M/G/1 Queue
21
2
Average # of customers in System
10in
M/M/1
23456789
num
ber
of c
ust
omer
s i
syst
em
M/M/1
1
N
M/M/2
Chapter 10 : M/M/s and M/G/1 Queue
012
0 0.2 0.4 0.6 0.8 1
/
Ave
rage
n
21
2
N
M/M/2
14
Example
A printer is attached to the LAN of the department. The printing jobs are assumed to arrive with aThe printing jobs are assumed to arrive with a Poissonian intensity and the actual printing times are assumed to obey the distribution Exp(µ). The capacity of the printer has become insufficient with regard to the increased load. In order to improve the printing service, there are three alternatives:
Chapter 10 : M/M/s and M/G/1 Queue
Old system
Chapter 10 : M/M/s and M/G/1 Queue
15
Case I : Solution
Replace the old printer by a new one twice as fast, i.e. with service rate 2µ.fast, i.e. with service rate 2µ.
2
Chapter 10 : M/M/s and M/G/1 Queue
Case II : Solution
Add another similar printer (service rate µ) and divide the users in two groups of equal sizedivide the users in two groups of equal size directing the works in each group to their own printer. The arrival rate of jobs to each printer is /2
Chapter 10 : M/M/s and M/G/1 Queue
16
(Continue)
/2
/2
Chapter 10 : M/M/s and M/G/1 Queue
Case III : Solution
• The same as alternative 2, but now there is a common printer queue where all jobs are takencommon printer queue where all jobs are taken and the job at the head of the queue is sent to which ever printer becomes free first.
Chapter 10 : M/M/s and M/G/1 Queue
= /2
17
(Continue)
Chapter 10 : M/M/s and M/G/1 Queue
Solution
Compare the performance of the alternatives at different loads. As measure of performance wedifferent loads. As measure of performance we use the average total waiting time of job (time in system, from the arrival of the printing job to the full completion of the job)
Case I : In this case we have an M/M/1 queue with parameter and 2p
Chapter 10 : M/M/s and M/G/1 Queue
2
2
1
1
1
2
11
T
18
Solution
Case II : Now we have two separate M/M/1 queues with parameters /2 and queues with parameters /2 and
The load per server is the same as before. Now j t thi h t ti l (b th
2
2
1
1
1
2
12
T
just everything happens two times slower (both arrivals and the service)
Chapter 10 : M/M/s and M/G/1 Queue
2
1
1
11
T
Solution
Case III : In the case of a common printing queue, an appropriate model is the M/M/2 queue withan appropriate model is the M/M/2 queue with parameters and
2
2
113
qPT
Chapter 10 : M/M/s and M/G/1 Queue
2
1
1
11
qP
19
Solution
M/M/22
1
1
113 qPT
2 213 q 0
2
1!2
2PPq
1
10P
11
1,1
3
T
Chapter 10 : M/M/s and M/G/1 Queue
1
2 2
qP
1,
2
1
1
13
Summary of Comparison
T1 T2 T3 T2/T1 T3/T1
<< 12 2
1
1
1
2
11
2
1
1
11
qP
2
1
1
1
1
12 1
Chapter 10 : M/M/s and M/G/1 Queue
2
1
1
1
2
1
1
1
2
1
1
1
20
Summary of Comparison
• Case I , one fast printer is the best one
• Case II, the total waiting time is twice as long as in case I
• Case III, the second printer does not help at all t l l d h j b i t k di tl i t that low loads: each job is taken directly into the
service (without waiting) but the actual printing takes twice the time as with the fast printer
Chapter 10 : M/M/s and M/G/1 Queue
M/G/1
Poisson
M/G/1 Server General independent
• Poisson arrivals at rate • Service time are independent, identically
distributed with general PDF
o sso
independent Service times
µFCFS service
g• Single Server queue
Chapter 10 : M/M/s and M/G/1 Queue
21
M/G/1
• Customers arrive according to a Poisson process with rate , but the customer service times havea general distribution. Customers are served in the order they arrive.
• Xi is the service time of the ith customer. (X1,X2, . . .) are identically distributed, mutually independent and independent of the interarrivalindependent, and independent of the interarrival times.
Chapter 10 : M/M/s and M/G/1 Queue
(Continue)
• Wi : waiting time in queue of the ith customer.
• Ri : residual service time seen by the ith customer. That is, if customer j is already being served when i arrives, Ri is the remaining time until customer j’s service time is complete.
• Ni : the number of customers found waiting in queue by the ith customer upon arrival.
Chapter 10 : M/M/s and M/G/1 Queue
22
M/G/1
2XW
22 XWN
12Wq
12
2XXT
12
WN qq
12
22 XTN
Chapter 10 : M/M/s and M/G/1 Queue
22
1
XR
M/G/1 Examples
• M/M/1 queue: 22 2
X
and
• M/D/1 queue:
1qW
22 1
X
and
Chapter 10 : M/M/s and M/G/1 Queue
12qW
23
M/D/1 Queues
M : Poisson arrival process with rate D : Service time is constantD : Service time is constant1 : single server, load = /
• Convenient way to model an Asynchronous Transfer Mode (ATM) node with fixed size cells
• Good model for packet switching node or router p gin a computer network where the packets are of fixed sized
Chapter 10 : M/M/s and M/G/1 Queue
M/D/1
• Average number of customer in system
1 2
• Average waiting time in system
12
1 2
N
11
Chapter 10 : M/M/s and M/G/1 Queue
1
12
11
T
24
M/D/1 Example
The output buffer of an ATM multiplexer can be modeled as an M/D/1 queue. Constant servicemodeled as an M/D/1 queue. Constant service time means now that an ATM cell has a fixed size (53 octets) and its transmission time to the link is constant. If the link speed is 155 Mbit/s.
What is the mean number of cells in the buffer (including the cell being transmitted) and the ( g g )mean waiting time of the cell in the buffer when the average information rate on the link is 124 Mbit/s?
Chapter 10 : M/M/s and M/G/1 Queue
(Continue)
155 Mbits/sATM M
124 Mbits/s.
Chapter 10 : M/M/s and M/G/1 Queue
ATM Mux.
25
Solution
Link speed = 155 Mbit/s = 155 Mbits/s = 155 Mbits/s1 octet = 8 bits transmission time = 53·8/155M = 2.7 s = 124 Mbits/s
The load (utilization) of the link is = 124/155
= 0.8
Chapter 10 : M/M/s and M/G/1 Queue
Average number of ATM cells
• M/D/1
4.28.01
8.0
2
18.0
2
N
12
1 2
N
Chapter 10 : M/M/s and M/G/1 Queue
8.012
26
• mean waiting time of the cell in the buffer
(Continue)
801
1
12
11
T
Chapter 10 : M/M/s and M/G/1 Queue
s1.8s 7.28.01
8.0
2
11
T
Summary of Comparison
• At heavy loads, the total waiting time of case 3 is the same as in case 1 (in both cases it consists
i l f h i i ) T l i f d bmainly of the waiting). Two slow printers fed by a common queue discharge the work in the queue as efficiently as one fast printer
• Case 2, when the queues are separate, it is possible that one printer stays idle while there are jobs waiting in the queue for the other printer. This deteriorates the overall performance in such a waydeteriorates the overall performance in such a way that also at high loads alternative 2 is on the average times slower than alternative 1
Chapter 10 : M/M/s and M/G/1 Queue
27
References
1. Robert B. Cooper, Introduction to QueueingTheory, 2nd edition, North Holland,1981.Theory, 2nd edition, North Holland,1981.
2. Donald Gross, Carl M. Harris, Fundamentals of Queueing Theory, 3rd edition, Wiley-Interscience Publication, USA, 1998.
3. Leonard Kleinrock, Queueing Systems VolumnI: Theory, A Wiley-Interscience Publication,I: Theory, A Wiley Interscience Publication, Canada, 1975.
Chapter 10 : M/M/s and M/G/1 Queue
(Continue)
4. Georges Fiche and Gerard Hebuterne, Communicating Systems & Networks: Traffic &Communicating Systems & Networks: Traffic & Performance, Kogan Page Limited, 2004.
5. Jerimeah F. Hayes, Thimma V. J. Ganesh Babu, Modeling and Analysis of Telecommunications Networks, John Wiley & Sons, 2004.
Chapter 10 : M/M/s and M/G/1 Queue
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