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King Fahd University of

Petroleum & MineralsComputer Engineering Dept

COE 540Computer Networks

Term 082

Courtesy of:

Dr. Ashraf S. Hasan Mahmoud

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Queuing Model

Consider the following system:

A(t)number of arrivals in (0, t]

D(t)number of departures in (0, t]

N(t)number of customers in system in (0,t]

Siservice time for ith customer

Tiduration of time spent in system for ith

customerWiduration of time spent waiting for service for ith customer

Wi = TiSi= DiAiSi

Queueing System

ith customer

arrives at time Ai

ith customer

departs at time Di

A(t) N(t) = A(t)D(t) D(t)

Ti = DiAi

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Example 1: Queueing System

Problem: A data communication line delivers a block ofinformation every 10 microseconds. A decoder checkseach block for errors and corrects the errors if necessary.It takes 1 microsecond to determine whether the blockhas any errors. If the block has one error it takes 5microseconds to correct it and if it has more than 1 error

it takes 20 microseconds to correct the error. Blocks waitin the queue when the decoder falls behind. Suppose thatthe decoder is initially empty and that the number oferrors in the first 10 blocks are: 0, 1, 3, 1, 0, 4, 0, 1, 0, 0.

a) Plot the number of blocks in the decoder as a function oftime.

b) Find the mean number of blocks in the decoder

c) What percent of the time is the decoder empty?

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contd

Solution:Interarrival time = 10 sec

Service time = 1 if no errors

1+5 if 1 error

1+20 if more than 1 errorThe queue parameters (A, S, D, and W) are shown

below:

Block #: 1 2 3 4 5 6 7 8 9 10

Arrivals: 10 20 30 40 50 60 70 80 90 100Errors: 0 1 3 1 0 4 0 1 0 0Service: 1 6 21 6 1 21 1 6 1 1Departs: 11 26 51 57 58 81 82 88 91 101

Waiting: 0 0 0 11 7 0 11 2 0 0

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contd

Solution:Using the previous results and knowing that

N(t) = A(t) D(t)One can produce the following results

The following Matlab code can be used tosolve this queue system (Note the codeis general it solves any systemprovided the Arrivals vector A, and theservice vector S)

Average no of customers in system = 0.950Average customer waiting time = 3.100 microsecMaximum simulation time = 101.000 microsecDuration server busy = 65.000 microsecServer utilization = 0.6436Server idle = 0.3564

0 20 40 60 80 100 1200

1

2

3

4

5

6

7

8

9

10Queue sysystem simulation

Noofcustomers

Time

A(t)D(t)N(t)

( )

1

1Average # of customers

A t

i

i

Tt

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contd0001 %

0002 % Problem 9.3 - Leon Garcia's book0003 clear all0004 A = [10:10:100];0005 Errors = [0 1 3 1 0 4 0 1 0 0];0006 S = zeros(size(A));0007 D = zeros(size(A));0008 %0009 % this loop to computes service times0010 for i=1:length(A);0011 if (Errors(i)==0) S(i) = 1;0012 else0013 if (Errors(i)==1) S(i) = 6;0014 else

0015 S(i) = 21;0016 end0017 end0018 %0019 % this section computes the departure time forthe ith user0020 if (i>1) % this is not the first user0021 if (D(i-1) < A(i)) D(i) = A(i) + S(i);0022 else0023 D(i) = D(i-1) + S(i);0024 end0025 else0026 D(i) = A(i)+S(i);

0027 end0028 %0029 % compute waiting time0030 W(i) = D(i) - A(i) - S(i);0031 end0032 %

0033 % Compute N(t)

0034 T = []; % time axis0035 T(1) = 0; % time origin0036 N = []; % number of cutomers0037 N(1) = 0; % initial condition0038 k = 2; % place for next insert0039 A_max = A(length(A)); % last arrival instant0040 i = 1; % index for arrivals0041 j = 1; % index for departures0042 t = 0; % system time00430044while (t < A_max)0045 t = min(A(i), D(j));0046 if (t == A(i))

0047 N(k) = N(k-1) + 1;0048 T(k) = t;0049 k = k + 1;0050 i = i + 1; % get next arrival0051 else% departure occurs0052 N(k) = N(k-1) - 1;0053 T(k) = t;0054 k = k + 1;0055 j = j + 1; % get next departure0056 end0057 end0058 %0059 % record remaining departure instants

0060 for i=j:1:length(D)0061 t = D(i);0062 N(k) = N(k-1) - 1;0063 T(k) = t;0064 k = k + 1;0065 end00660067 k = k - 1; % decrement k to get real size of N and T0068 %0069 % compute means0070 MeanW = mean(W);0071 T_Intervales = T(2:k)-T(1:k-1);0072 MeanN = sum(N(1:k-1).*T_Intervales) / T(k);0073 IdleDurationsIndex = find(N(1:k-1) ~= 0);0074 Utilization = sum(T_Intervales(IdleDurationsIndex))/T(k);0075 %

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contd

0076 % Display results0077 fprintf('Block #: '); fprintf('%3d ', [1:1:length(A)]); fprintf('\n');0078 fprintf('Arrivals: '); fprintf('%3d ', A); fprintf('\n');0079 fprintf('Errors: '); fprintf('%3d ', Errors); fprintf('\n');0080 fprintf('Service: '); fprintf('%3d ', S); fprintf('\n');0081 fprintf('Departs: '); fprintf('%3d ', D); fprintf('\n');0082 fprintf('Waiting: '); fprintf('%3d ', W); fprintf('\n');0083 fprintf('\n\n');0084 fprintf('Average no of customers in system = %7.3f\n', MeanN);0085 fprintf('Average customer waiting time = %7.3f microsec\n', MeanW);0086 fprintf('Maximum simulation time = %7.3f microsec\n', T(k));0087 fprintf('Duration server busy = %7.3f microsec\n', ...0088 sum(T_Intervales(IdleDurationsIndex)));0089 fprintf('Server utilization = %7.4f\n', Utilization);0090 fprintf('Server idle = %7.4f\n',1.0-Utilization);0091 %0092 % Plot results0093 figure(1)0094 h = stairs(T, N); grid0095 set(h, 'LineWidth', 3);0096 xlabel('Time');0097 ylabel('No of customers in system, N(t)');00980099 figure(2);0100 [AT, AA] = stairs(A, cumsum(ones(size(A))));0101 [DT, DD] = stairs(D, cumsum(ones(size(D))));0102 [NT, NN] = stairs(T, N);

0103 h = plot(AT, AA, '-', DT, DD,'--r', NT, NN,'-.'); grid0104 set(h, 'LineWidth', 3);0105 title('Queue sysystem simulation');0106 ylabel('No of customers');0107 xlabel('Time');0108 legend('A(t)', 'D(t)', 'N(t)', 0);01090110 figure(3);0111 h = stem(W); grid0112 set(h, 'LineWidth', 3);0113 ylabel('Waiting time');0114 xlabel('Customer index');0115 LegendStr = ['MeanW = ' num2str(MeanW)];

0116 legend(LegendStr, 0);

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0 5 10 15 20 250

1

2

3

4

5

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8A(t) and D(t) for a queueing system

time

Number of Customers in System

Blue curve:A(t)

Red curve:D(t)

Total timespent in thesystem for all

customers =area inbetween twocurves

T1

T2

T3

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Littles Formula

Littles formula:

E[N] = E[T]

i.e. For systems reaching steady state, the averagenumber of customers in the system = averagearrival rate average time spent in the system

Holds for many service disciplines and for systemswith arbitrary number of servers. It holds formany interpretations of the system as well.

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Example 2:

Problem: Let Ns(t) be the number ofcustomers being served at time t, and lett denote the service time. If wedesignate the set of servers to be thesystem then Littles formula becomes:

E[Ns] = E[t]Where E[Ns] is the average number of busy

servers for a system in the steady state.

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Example 2: contd

Note: for a single server Ns(t) can be either 0 or 1 E[Ns]represents the portion of time the server is busy. If p0 =Prob[Ns(t) = 0], then we have

1 - p0 = E[Ns] = E[t], Orp0 = 1 - E[t]

The quantity E[t] is defined as the utilization for a singleserver. Usually, it is given the symbol r

r = E[t]For a c-server system, we define the utilization (the fraction of

busy servers) to be

r = E[t] / c

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Example 3: Applications on Littles

Formula

Refer to the slides for Dr. Waheed.

The slides have 8 good examples!

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Queueing Jargons

Queueing system

Customers

Queue(s) (waiting room)

Server(s)

Kendalls notation Standard notation to describe queueing containing single queue:

X/Y/m/n/y/SD

InterArrival process

Service processServers #

Queue capacity

Population size

X/Y/m/n/y/SDService discipline

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Courtesy of Dr. Abdul Waheed

Common Distributions

G = general distribution of inter-arrival times or service times

GI = general distribution of inter-arrival time with the restrictionthat they are independent

M = negative exponential distribution (Poisson arrivals)

D = deterministic arrivals or fixed length service

M/M/1? M/D/1?

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Courtesy of Dr. Abdul Waheed

General Characteristics of NetworkQueuing Models

Customer population Generally assumed to be infinite arrival rate is persistent

Queue size

Infinite, therefore no loss Finite, more practical, but often immaterial

Dispatching discipline FIFO typical

LIFO Relative/Preferential, based on QoS

Processor sharing (PS) discipline Useful for modeling multiprogramming

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Queue System and Parameters

Queueing system with m servers When m = 1 single server system

Input: arrival statistics (rate ), service statistics (rate ),number of customers (m), buffer size

Output: E[N], E[T], E[Nq], E[W], Prob[buffer size = x],Prob[W

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The M/M/1 Queue

Consider m-server system wherecustomers arrive according to a Poissonprocess of rate inter-arrival times are iid exponential r.v.

with mean 1/ Assume the service times are iid

exponential r.v. with mean 1/m Assume the inter-arrival times and

service times are independent Assume the system can accommodate

unlimited number of customers

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The M/M/1 Queuecontd

What is the steady state pmf of N(t), thenumber of customers in the system?

What is the PDF of T, the total customerdelay in the system?

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The M/M/1 Queuecontd

Consider the transition rate diagram for M/M/1system

Note: System state number of customers in systems is rate of customer arrivals

m is rate of customer departure

0 1

l

m

l

m

2 3

l

m

j j+1

l

mm

l l

m

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The M/M/1 QueueDistribution of

Number of Customers

Writing the global balance equations forthis Markov chain and solving forProb[N(t) = j], yields (refer to previousexample)

pj = Prob[N(t) = j]

= (1-r)rjfor r = /m < 1Note that for r = 1 arrival rate = service

rate m

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The M/M/1 QueueExpected

Number of Customers

The mean number of customers is givenby

E[N] = j Prob[N(t) = j]j=0

= r/(1-r)Note: Nhas geometric distribution withp= (1-r) (i.e.pj= Prob[N(t)

= j] =p(1-p)j = (1-r)rjfor j = 0, 1, ) E[N] = (1-p)/p= r/(1-r)

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The M/M/1 QueueMean Customer

Delay

The mean total customer delay in thesystem is found using Littles formula

E[T] = E[N]/ = /[ (1- )]= 1/ (1-)

= 1/()

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The M/M/1 QueueMean Queueing

Time

The mean waiting time in queue is givenby

E[W] = E[T] E[t]= [1/()] E[t]= [(1/ )/(1r)] E[t]= [E[t]/(1r)] E[t]= [r/(1r)] E[t]

Note: E[t] = mean service time for a customer = 1/

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The M/M/1 QueueMean Number in

Queue

Again we employ Littles formula:

E[Nq] = E[W]

= r2 / (1-r)Remember:

server utilization r = /m = 1-p0All previous quantities E[N], E[T], E[W], and

E[Nq] as r 1

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Scaling Effect for M/M/1 Queues

Consider a queue of arrival rate whoseservice rate is m r = /m, The expected delay E[T] is given by

E[T] = (1/m) / (1-r) If the arrival rate increases by a factor of

K, then we either1. Have K queueing systems, each with a

server of rate m2. Have one queueing system with a server of

rate Km Which of the two options will perform

better?

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Example 4: Scaling Effect for M/M/1

Queues

Example: K = 2: M/M/1 and M/M/2systems with the same arrival rate andthe same maximum processing rate

Case1:

K independent

queueing systems

2

2Case 2:ONE queueing

system

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Example 4: Scaling Effect for M/M/1

Queuescontd

Case 1: K queueing systems Identical systems

E[T] is the same for all E[T] = (1/m) / (1-r)

Case 2: 1 queueing system with server of rateKm r for this system = (K) /(Km) = /m same as the

original system

E[T] = (1/(Km)) / (1-r) = (1/K) E[T]

Therefore, the second option will provide a lesstotal delay figure significant delayperformance improvement!

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M/M/1/KFinite Capacity Queue

Consider an M/M/1 with finite capacity K<

For this queue there can be at most Kcustomers in the system

1 being served

K-1 waiting

A customer arriving while the system hasK customers is BLOCKED (does not wait)!

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contd

Transition rate diagram for this queueingsystem is given by:

N(t) - A continuous-time Markov chainwhich takes on the values from the set {0,

1, , K}

0 1

l

m

l

m

2 3

l

m

K-1 K

l

mm

l

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contd

The global balance equations: p0 = mp1

( + m)pj = pj-1 + mpj+1 for j=1, 2, , K-1m pK= pK-1

Prob[N(t) = j] = pj j=0,1, , K; r

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M/M/1/KMean Number of

Customers

Mean number of customers, E[N] is givenby:

K

j

jtNjNE0

])(Pr[][

)

12/

11

1

1 1

1

r

rr

r

r

r

K

KK

K

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M/M/1/KBlocking Rate

A customer arriving while the system isin state K is BLOCKED (does not wait)!

Therefore, rate of blocking, b, is given by

b = pK

The actual (i.e. serviced) arrival rate into the

system is a given

a = - b

= (1 - pK)

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M/M/1/KBlocking Ratecontd

Queueing System

a: actual

(serviced) arrivals

b: blocked (not

serviced) arrivals

: total

arrivals

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M/M/1/KMean Delay

The mean total delay E[T] is given by

E[T] = E[N] / a

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Multi-Server Systems: M/M/c

The transition rate diagram for a multi-server M/M/c queue is as follows:

Departure rate = km when k servers are busy We can show that the service time for a

customer finding k servers busy isexponentially distributed with mean 1/(k)

0 1

l

m

c-1 c

l

cm

j j+1

l

cmcm

l l

cm

l

2m (c-1)m

l

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contd

Writing the global balance equations: p0 = mp1jm pj = pj-1 for j=1, 2, , ccm pj = pj-1 for j= c, c+1,

pj= (a

j/j!) p0 (forj=1, 2, , c) andpj= (rj-c/c!) ac p0 (forj=c, c+1, )

where a = /m and r = a/c From this we note that the probability of system being in

state c, pc, is given by

pc = ac/c! p0

Note this distribution is the

same as that for M/M/1 when

you set c to 1.

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contd

To find p0, we resort to the fact that pj = 1

The probability that an arriving customer has to wait

Prob[W > 0] = Prob[N c]

= pc + r pc+1 + r2pc+2+ = pc/(1-r)

Question: What is Prob[W>0] for M/M/1 system?

11

0

01

1

!!

r

c

j

cj

c

a

j

ap

Erlang-C

formula

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contd

The mean number of customers in queue(waiting):

)

cj

q jtNcjNE ])(Pr[][

)

cj

c

cjpcj r

) cp2

1 r

r

]0Pr[1

Wr

r

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contd

The mean waiting time in queue:

The mean total delay in system:

The mean number of customers insystem:

l/][][ qNEWE

][][][ tEWETE

m/1][ WE

][][ TENE l

aNE q ][ Why?

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Example 5:

A company has a system with four privatetelephone lines connecting two of its sites.Suppose that requests for these lines arriveaccording to a Poisson process at rate of one

call every 2 minutes, and suppose that calldurations are exponentially distributed withmean 4 minutes. When all lines are busy, thesystem delays (i.e. queues) call requests until aline becomes available.

Find the probability of having to wait for a line.

What is the average waiting time for anincoming call?

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Example 5: contd

Solution: = , 1/m = 4, c = 4 a = /m = 2

r = a/c = p0 = {1+2+2

2/2!+23/3!+24/4! (1/(1-r))}-1= 3/23

pc = ac

/c! p0= 24/4! X 3/23

(1) Prob[W > 0] = pc/(1-)= 24/4! 3/23 1/(1-1/2)= 4/23

0.17

(2) To find E[W], find E[Nq] E[Nq] = /(1- ) * Prob[W>0] = 0.1739E[W] = E[Nq]/ = 0.35 min

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Multi-Server Systems: M/M/c/c

The transition rate diagram for a multi-server with no waiting room (M/M/c/c)queue is as follows:

Departure rate = k when k servers are

busy

0 1

l

m

c-1 c

l

cm

l

2m (c-1)m

l

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PMF for Number of Customers for

M/M/c/c

Writing the global balanceequations, one can show:

pj= aj/j! p0 (forj=0, 1, , c)

where a = / (the offered load)

To find p0, we resort to the fact that pj

= 1 1

0

0!

c

j

j

j

ap

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Erlang-B Formula

Erlang-B formula is defined as theprobability that all servers are busy:

cpcN

]Pr[

2

/ !

1 / 2! ... / !

c

c

a c

a a a c

E t d N b f t i

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Expected Number of customers in

M/M/c/c

The actual arrival rate intothesystem:

Average total delay figure:

Average number of customers:

)1( ca p ll

][][ tETE Why?

][][ tl ENE a

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Example 6:

A company has a system with fourprivate telephone lines connecting two ofits sites. Suppose that requests for theselines arrive according to a Poisson

process at rate of one call every 2minutes, and suppose that call durationsare exponentially distributed with mean4 minutes. When all lines are busy, the

system BLOCKS the incoming call andgenerates a busy signal.

Find the probability of being blocked.

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Example 6:

Solution: = 1/2, 1/ = 4, c = 4 a = / = 2

= a/c = 1/2

ac/c!pc = ------------------------------------

1 + a + a2/2! + a3/3! + a4/4!

24/4!

= ------------------------------------ = 9.5%1 + 2 + 22/2! + 23/3! + 24/4!

Therefore, the probability of being blocked is 0.095.

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M/G/1 Queues

Poisson arrival process (i.e. exponentialr.v. interarrival times)

Service time: general distribution ft(x) For M/M/1, ft(x) = me-mx for x > 0

The state of the M/G/1 system at time tis specified by

1. N(t)2. The remaining (residual) service time of the

customer being served

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The Residual Service Time

Mean residual time (see example andderivation in Chapter 9 of Garciastextbook) is given by

E[t2]E[R] = ---------

2E[t]

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Mean Waiting Time in M/G/1

The waiting time of a customer is thesum of the residual service time R of thecustomer (if any) found in service andthe Nq(t)= k-1 service time of the

customers (if any) found in queue

E[W] = E[R] + E[Nq] E[t]= E[R] + E[W]E[t]= E[R] + r E[W]

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Mean Waiting Time in M/G/1contd

But residual service time R (as observedby an arriving customers) is either 0 is the server is free

R if the server is busy

Therefore, mean of R is given by

E[R] = 0 P[N(t)=0] + E[R](1-P[N(t)=0])= E[t2]/(2E[t]) r= E[t2]/2

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Mean Waiting Time in M/G/1contd

Substituting back, yields

E[t2]E[W] = ----------

2(1-r)(2tE[t]2)

= ----------------

2(1-r)r (1 + Ct2)

= ---------------- E[t]2(1-r)

Note:

- E[t2

] = 2

t+E[t]2

- C2t = 2

t/E[t]2

= coefficient of variation

of the service time

Pollaczek-Khinchin (P-K)

Mean Value Formula

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Mean Delay in M/G/1contd

The mean waiting time, E[T] is found byadding mean service time to E[W]:

E[T] = E[t] + E[W]r (1 + Ct2)

= E[t] + ---------------- E[t]2(1-r)

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Example 7:

Problem: Compare E[W] for M/M/1 and M/D/1systems.

Answer:

M/M/1: service time, t, is exponential r.v. withparameter m

E[t] = 1/m , E[t2] = 2/m2 , 2t= 1/m2 , C2t= 1M/D/1: service time, t, is constant with value t =

1/m E[t] = 1/m , E[t2] = 1/m2 , 2t= 0 , C2t= 0

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Example 7: contd

Answer: contdSubstitute in P-K mean value formula

M/M/1:E[t2] r

E[WM/M/1] = ---------- = ---------- E[t]2(1-r) (1-r)

M/D/1: E[t2] rE[WM/D/1] = ---------- = ---------- E[t]

2(1-r) 2 (1-r)1

= -- E[WM/M/1]2

The waiting time in an

M/D/1 queue is half of

that of an M/M/1 system

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Example 8:

Problem: Assume traffic is arriving at the inputport of a router according to a Poisson arrivalprocess of rate = 100 packets/sec. If thetraffic distribution is as follows:30% of packets are 512 Bytes long,

50% of packets are 1024 Bytes long,20% of packets are 4096 Bytes longIf the transmit speed of the router output portis 1.5 Mb/s

a) What is the average packet transmit time?

b) What is the average packet waiting time beforetransmit?

c) What is the average buffer size in the router?

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Example 8: contd

Solution:a) Average packet size,

E[L] = 0.3512 + 0.51024 + 0.24096= 1484.8 Bytes

average transmit time = E[L]/R = 1484.88/1.5106 =0.0079 sec

b) E[L2] = 0.3(5128)2 + 0.5(10248)2 + 0.2x(40968)2 =2.5334e+008 Bits2

E[t2] = E[L2]/R2 = 1.1259e-004 sec2 = E[t] = 0.7919E[W] = () E[t2] /(1-)

= 0.0271 sec

c) E[Nq] = E[W]

= 2.705 packet

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