337 Simulation and Modelling

8/13/2019 337 Simulation and Modelling

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There are many application areas, e.g. Computer architecture

Networks/distributed systems

Software performance engineering

Telecommunications

Manufacturing

Healthcare

Transport

Finance

Environmental science/engineering

Here, well focus on the important common principles, ratherthan become entrenched in specic application areas

5

Some Important Questions

What are we trying to measure? How do we measure it? What sorts of model can we build? Is the model well-dened? How can we solve the model? What formalisms can we use to help dene a model? How does the model compare to the real system?

6

The Focus...

Were going to focus on discrete event systems, with continuous time and discrete states

Our models will essentially be stochastic transition systemswhere delays are associated with each state

Different high-level formalisms can be used to dene thesemodels well study some popular approaches

These models can variously be solved using Simulation Numerical algorithms Analytically

Broadly, these are in increasing order of efficiency and decreasingorder of generality

7

This course...

Part I: Models

1. Modelling overview (ajf )

Examples and motivation Performance measures and their relationships

2. Timed transition systems (ajf)

Markov Processes (MPs)

Generalised Semi-Markov Processes (GSMPs)

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3. Solution methods (ajf)

Numerical solution of MPs Analytical solution of MPs Simulation of MPs and GSMPs

Part II: High-level formalisms, tools and techniques

1. Discrete-event Simulation (a jf)

Libraries for discrete-event simulation (Java)

Random number generation & distribution sampling Output analysis Applications

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2. Stochastic Petri Nets (jb)

Notation Generation of underlying MP Applications

3. Stochastic Process Algebras (jb)

Introduction to PEPA Generation of underlying MP

Applications3. Markovian Queueing Networks (jb)

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Modelling Overview

Example 1: A Simple TP server

A TP system accepts and processes a stream of transactions,mediated through a (large) buffer:

...

Transactions arrive randomly at some specied rate The TP server is capable of servicing transactions at a given

service rate

Q: If both the arrival rate and service rate are doubled, whathappens to the mean response time?

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Example 2: A simple TP server

Consider the same system as above:

...

The arrival rate is 15tps The mean service time per transaction is 60ms The current mean response time is around 350ms

Q: What happens to the mean response time if the arrival rateincreases by 10%?

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Example 3: Heathrow airport

There are ve security scanners (X-ray) between the check-in anddepartures lounges at Terminal 4. One of them is broken:

...

BROKEN

DEPATURES LOUNGE

Around 3.8 customers pass through the terminal each second It takes about 30 minutes on average to get through securityQ: How long would it take on average if all 5 scanners were working?

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Example 4: A central server

Jobs are submitted to a central server

2

3

1

Completions

Arrivals

Disk2

Disk 1

On average, each submitted job makes 121 visits to the CPU, has70 disk 1 accesses and 50 disk 2 accesses

The mean service times are 5 ms for the CPU, 30 ms for disk 1and 27ms for disk 2Q: How would the system response time change if we replace the

CPU with one twice as fast?

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Example 5: A manufacturing system

A factory has 5 assembly areas

jobs

Queue

QueueType 0

0

3

1

2

4

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There are three job types, numbered 0, 1 and 2, which occurwith probability 0.3, 0.5 and 0.2 respectively

Jobs arrive at the rate of 4/hr and visit the stations in the order:0 2, 0, 1, 4

1 3, 0, 22 1, 4, 0, 3, 2

The mean time for each job type at each station in visit order is0 0.50, 0.60, 0.85, 0.50

1 1.10, 0.80, 0.75

2 1.10, 0.25, 0.70, 0.90, 1.00

Q: If you were to invest in one extra machine, where would you putit?

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There are many techniques one can use to model such systems Simple models involve nothing more than back-of-the-envelope

calculations

The most complex model is a full-blown simulation A tiny taster...

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Two Simple Experiments with Random Numbers

Example 1: The Birthday Bet

You meet n people at a party youve never met before

You bet one of them that at least two of the n people have thesame birthday (e.g. 5th March) (Assume birthdays are uniformlydistributed and ignore leap years)

Question: How big should n be for your winning chances to be> 50%?

Alternatively, what is the probability , p

n say, that two or more

people in a random group of n people have the same birthday?

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Under the assumptions of the model, there is an analyticalsolution:

pn = 1 n 1

i =1

365 i365

Amazingly, p22 = 0 .476 and p23 = 0 .507, so the break-even pointis 23

If we couldnt do the math, we could instead simulate thescenario by generating m sets of n random birthdays (randomintegers between 0 and 364) and counting the number of sets cthat contain a duplicate

If the result of each of the m experiments is 0 or 1, then eachoutcome is an observation or estimate of pn , albeit a very badone!

Of course, a better estimate is c/m19

public class birthday {public static void main( String args[] ) {

int n = Integer.parseInt( args[0] ) ;int k = Integer.parseInt( args[1] ) ;i n t b = 0 , c = 0 ;

fo r ( i n t i = 0 ; i < m ; i++ ) {int tot[] = new int[365] ;for ( in t j = 0 ; j < n && to t [b] < 2 ; j++ ) {

b = (int)( Math.random() * 365 ) ;tot[b]++ ; if ( tot[b] == 2 ) c++ ;

}}

System.out.println( (float)c / m ) ;} }

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Here are the averages of 5 runs of the program for differentvalues of n and m

m n

5 10 15 20 25 30

100 0.000 0.000 0.200 0.200 0.400 0.200

101 0.000 0.080 0.300 0.460 0.520 0.840

102 0.028 0.100 0.236 0.482 0.580 0.660

103 0.024 0.117 0.250 0.402 0.567 0.716

104 0.028 0.117 0.255 0.414 0.565 0.710

105 0.027 0.117 0.254 0.412 0.569 0.707

For each n there is a noticeable convergence as m increases

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Lets home in on m = 22, 23 and 24 and look at the 95%condence intervals (see later) for the mean using m = 10 5

m Mean 95% Condence Interval

22 0.476 (0.475,0.478)

23 0.508 (0.506,0.510)24 0.538 (0.537,0.539)

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Example 2: Come, Friendly Bombs

It is required to drop bombs on a small island that is being usedas an illegal arms stash

Each of 10 bombers will aim one bomb at the centre of theisland (assume normalised to be the origin (0,0))

From a safe height there is a bombing error (due to uctuatingair temperature/density/velocity etc.) which is approximatelynormally distributed with standard deviation 180 m in thehorizontal; similarly 100 m in the vertical

Q: How many bombs will hit the island, on average?

Q: At greater risk, the bombers can y lower and reduce thestandard deviations by 30%. What is the hit rate now?

23

(0,0)

(180,60)

(170,110)

MISS

MISS

HIT

HITBombers

y

x

(280,80)

(140,30)

(180,100)

(40,200)

(10,140)

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In general, this cannot be solved analytically we must resort tosimulation

The island will be modelled as a polygon

We will simulate a number, m, of missions and will estimatethe average number of hits per mission The impact point of each bomb is simulated by sampling the two

Normal distributions (details later)

The hit/miss status reduces to a point in polygon test...

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public class Bombers {static boolean isOnTarget( double x, double y ) {

double[] xs = {-280,10,180,170,40,-180,-140} ;double[] ys = {-80,-140,-60,110,200,100,-30} ;

int ncrosses = 0, n = xs.length - 1 ;for ( int i = 0 ; i < xs.length ; i++ ) {

if ( ( ( ys[i] y && y >= ys[n] ) ) &&

x < xs[i] + ( xs[n] - xs[i] ) * ( y - ys[i] ) /( ys[n] - ys[i] ) )

ncrosses++ ;

n = i ;}return ( ncrosses % 2 == 1 ) ; }

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public static void main( String args[] ) {int m = Integer.parseInt( args[0] ) ;int hits = 0 ;fo r ( i n t i = 0 ; i < m ; i++ ) {

fo r ( i n t j = 0 ; j < 10 ; j++ ) {

double x = Normal.normal( 0, 180 ) ;double y = Normal.normal( 0, 100 ) ;if ( isOnTarget( x, y ) )

hits++ ;}

}System.out.println( (float)hits / m ) ;

}}

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Here are the results of 5 runs of the program for differentnumbers of simulated missions ( m):

m Run 1 Run 2 Run 3 Run 4 Run 5

101 5.800 4.900 5.100 5.800 5.500

102 5.110 4.930 5.610 5.330 5.010

103 5.357 5.310 5.326 5.249 5.284

104 5.359 5.319 5.336 5.345 5.374

105 5.352 5.346 5.343 5.340 5.354

The ve results are noticeably more consistent as m increases

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This is what happens if the standard deviations are reduced by30% (to 126m and 70m respectively), for 10 5 missions:

m Run 1 Run 2 Run 3 Run 4 Run 5

105 7.679 7.675 7.672 7.676 7.675

By averaging the results for m = 10 5 an estimate of theimprovement in hit rate is (approximately) 44%

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Some observations so far...

Mathematical models can be really simple Under the given assumptions, mathematical models deliver exact

answers

Mathematical models can only be used in restricted cases Simulation models are extremely exible (just change the code!) Simulation models produce estimates , not exact answers Each run of a simulation model produces a different estimate

The longer we run a simulation the better the estimatesbecome

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Fundamental Laws

Consider the following:

SystemArrivals Completions

Lets assume we observe the system for time T and let The number of arrivals be A

The number of completions be C

From these we can dene a number of useful performancemeasures and establish relationships between them

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Some Direct Measures

The Arrival Rate is = A/T The Throughput is = C/T

If the system comprises a single resource whose total busy time is Bthen,

The Utilisation is U = B/T The average Service Demand of each job (or average service

time ) is S = B/C

Note: The net service rate is = 1 /S = C/B

Now some Fundamental Laws (or Operational Laws )...

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The Utilisation Law

Since U = B/T = C/T B/C we have:

U =

S Sometimes we work with service rates rather than times , in

which case we have U = /

Importantly, note that U < 1 so we require = 1 /S > Q: What if = ?

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Littles Law

Suppose we plot ( A C ) over the observation time ( T ):

t1 t2 t4 t11t3 t5 t6 t7 t8 t9 t10 time

1

4

3

2

T

Total area

0

= I

0

AC

Let the accumulated area be I (in request-seconds)

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The average number of jobs in the system is N = I /T The average time each job spends in the system (the average

waiting time ) is W = I /C

Since I /T = C/T I/C we have:

N = W Note: Assumes ( A C ) is zero at either end of the observation

period, or that is much larger than the residual populations ateither end

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Littles Law can be applied to any system (and subsystem) inequilibrium, e.g.

...

Clients

CPU

Subservers

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The Response Time Law

The special case of a population of users in think/computemode is ubiquitous and has its own law

Let Z be the think time, i.e. the time between completion of one request and the submission of the next N is now the total population of users and the request

submission rate

The average total time for each cycle is W + Z , so from LittlesLaw, N = (W + Z ), or

W = N/ Z 37

The Forced Flow Law

Suppose the system in question comprises a number of resourcesthat a job can visit during its time in the system

Let C k be the number of job completions at, resource k The average number of visits each job makes to resource k is

then V k = C k /C

Rearranging: C k = V k C so, dividing both sides by T ,C k /T = V k C /T , i.e.

k =

V k where k is the throughput of resource k38

The Flow Balance Asuumption

It is often useful to assume that A = C in which case,A/T = C/T , i.e. =

In particular, systems that are in equilibrium (or steady state )must satisfy this property in the long term

Note: If the ow balance assumption does not hold in the longterm (as time ), then the system is fundamentally unstable

Motto: What goes in must come out!

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Timed State Transition Systems

We now focus attention on modelling systems with state Time is continuous, but the states are assumed to be discrete and

nite, or countably innite, in number

E1 E2 E4t1 t2 t3

E3t4 t5

E5etc.

Times1 s2 s3 s4 s5Initial state s0

State transitions

State holding times

These are called discrete-event systems Note that the state doesnt change between events (transitions)

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We need to know the distribution of the time the system spendsin a given state

timeE E

t t

(t) (t)t = t t

Density function

The state holding times might bebased on measurement, or assumed

distributionto have a known mathematical

State s

Measured/assumed distribution

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In order to understand the key issues, consider a very simpleexample: a single server with a FIFO queue; arrival rate andservice rate :

Population n Population n1 2

... ...

Infinite capacity Infinite capacity

1 2

The simplest way to represent the state of the system is toidentify a state with each queue population (0 , 1, 2..., N ):

0 1

...

...arrival arrival arrival arrival

com plet ion completion completion completion

NN1

Transitions are triggered by events, here job arrival and jobcompletion

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With no time delays, this is just a Labelled Transition System orLTS, cf FSP. Its not useful for performance analysis!

To add state holding times in a completely general way weassociate clocks with events

aC

C a(set )arrival

aCC c

C a

C ccompletion

(set )C ccompletion

(set )

C a C c(set , )

C a C c(set , )

C aC c

C aC c

C c

...

...

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arrival arrival

completion

Clock for arrival eventClock for completion event

N1 N(set )arrival

completion

Events that are active in a given state compete to trigger thenext transition, via expiry of their clock

Events may schedule other events by setting their clocks duringthe transition

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aC

C a(set )arrival

C a

C ccompletion

(set )C ccompletion

(set )

C a C c(set , )

C a C c(set , )

C aC c

C aC c

C c

...

...

10

arrival arrival

completion

N1 N(set )arrival

completion

Clocks run down (to 0) at a state-dependent rate The event whose clock reaches 0 rst in any state triggers the

next transition

A clock setting function sets the clocks for new events that arescheduled during the transition

The clocks of old events, i.e. that are active in both the old andnew states, continue to run down

Events that were active in the old state, but not the new, arecancelled

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The Generalised Semi-Markov Process (GSMP)

A formal system that encapsulates this behaviour is theGeneralised Semi-Markov Process ; the components are: A set S of states (nite or countably innite) A set of events E = {e1 ,...,e N } E (s) the set of events schdeuled to occur in state sS p(s ; s, e ) the probability that the next state is s when event

e occurs in state s F (x; s , e , s , e ) the distribution function used to set the clock

for new event e when event e triggers a transition from s to s

r(s, e ) the rate at which the clock for event eE runs downin state s S 0 , F 0 the distribution functions used to assign the initial

state and to set the clocks for all active events in that state.

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It is important to understand the concept of old , new andcancelled events

0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111

000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111

0 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 0

1 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 1

0 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 0

1 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 11 1 1

000000000111111111 000000000111111111000000000000000000000000000000000000111111111111111111111111111111111111

000000000000000000000000000000000000111111111111111111111111111111111111

E(s) E(s)Old eventsO(s; s, e ) = E(s) (E(s){e})

New events N(s; s, e ) = E(s)(E(s){e

Cancelled events(E(s){e})E(s)

We talk of scheduling new events and descheduling cancelledevents

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Example: consider the transition from s to s via event e2 below

C 1C 2C 3

C 5

C 4C 3

NewOldCancelled

4(e , e )53(e )

1(e )

s

se

e

e

e

e

e

3

1

2

4

5

3

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Note: This only allows one event to trigger a transition GSMPsin general allows multiple events to occur simultaneously

Back to the example. We have... S = {0, 1, 2..., N } E =

{a, c

} where a=arrival, c=completion

E (0) = {a}, E (0 < a < N ) = {a, c}, E (N ) = {c} p(s + 1; s, a ) = 1 , p(s 1; s, c) = 1; 0 elsewhere F (x; s ,a ,s ,e ) = F a (x)

F (x; s ,c ,s ,e ) = F c(x)

For all sS , r (s, e ) = 1 for all e in E (s)

S 0 (0) = 1 , S 0 (s = 0) = 0, F 0 (x; a, 0) = F a (x)

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The state at time t denes a continuous-time process, X (t) say;formally, this is the GSMP

Note that the clocks in state sS determine how long theprocess remains in state s, prior to the next transition, e.g.

= time of ith transitioni

4

3

2

1

0 Time 1 2 3 4 5 6 7 8 9

X(t) = 3t

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A sample path of a GSMP is a particular (random) traversal of the transition system

To start it off, we use S 0 to select an initial state sS ; for eachevent e

E (s) use F 0 (x; s, e ) to initialise C e

A sample path is then generated by repeating the following steps:1. Find tnext = min {C e /r (s, e ) | eE (s)} and let e be the

winning event

2. Determine the next state s using p(.; s, e )

3. For each oO(s ; s, e ), reduce C n by tnext r (s, o)

4. For each n

N (s ; s, e ), set C n using F (x; s ,n ,s ,e )

5. Set s = s

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Example: The innite capacity single-server queue in Java (i.e.as above but with N = )

Assume the inter-arrival times are uniform-random between 0and 1 (i.e. U (0, 1)) and that service times are xed at 0.25...

class ssqClocks2 {public static void main( String args[] ) {

int s, sPrime ;double now = 0.0 ; // Keeps track of timedouble stopTime ;double arrClock = 0.0 ;double compClock = 0.0 ;double stopTime = Double.parseDouble( args[0] ) ;

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s = 0 ;arrClock = Math.random() ;while ( now < stopTime ) {

i f ( s = = 0 ) {// E(0) = {arrival}

now += arrClock ; // Note: all rates are 1// p(1;0,arrival) = 1sPrime = 1 ;// O(1;0,arrival) = {}// N(1;0,arrival) = {arrival,completion}arrClock = Math.random() ;compClock = 0.25 ;

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} else {if ( arrClock < compClock ) { // Arrival event

now += arrClock ;// p(s+1;s,arrival) = 1

sPrime = s + 1 ;// O(s;s,arrival) = {completion}compClock -= arrClock ;// N(s+1;s,arrival) = {arrival}, s>0arrClock = Math.random() ;

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} else { // Completion eventnow += compClock ;// p(s-1;s,completion) = 1sPrime = s - 1 ;

// O(s;s,completion) = {arrival}, s>0arrClock -= compClock ;// N(s-1;s,completion) = {completion}, s>0// = {}, s=0if ( sPrime > 0 ) {

compClock = 0.25 ; ;}

}}s = sPrime ; }}}

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The above code is an example of a discrete event simulation With completely general clock setting functions, simulation is the

only way to analyse a GSMP

The code above doesnt accumulate any measures, but we caneasily keep track of, e.g.

The time the system spends in each state

The area under some function of the state (see the exampleunder Littles Law earlier)

The average time between leaving one set of states andentering another

and so on

From these, useful performance measures can be computed

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Another example: The Machine repair Problem There are N machines in a factory that are subject to periodic

failure

Machines that fail are serviced in fail-time order by a repairer

The individual time-to-breakage ( B ) and repair times ( R) areeach assumed to be independent and identically distributed This can be modelled as a tandem queueing system:

... ...

Total population N

Working machines Broken machines

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Because the population ( N ) is xed the state can be describeduniquely by the population of one queue (broken machines, say)

We need only two events: break and repair , say

Let F b(x) = P (B

x) be the distribution function for the time

to breakage for a given machine, and F r (x) = P (R x) be thatfor the repair time

A GSMP for this system is as follows...

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S = {0, 1,...,N }E = {break , repair }

E (0) = {break }E (N ) = {repair }

E (0 < s < N ) = {break,repair } p(s 1; s, repair) = 1 , 0 < s N p(s + 1; s, break) = 1 , 0 s < N

p(s ; s, e ) = 0 for all other s , s , e

F (x; s , break , s , e ) = F b(x) = P (B x)F (x; s , repair , s , e ) = F r (x) = P (R x)

r (s, e ) = 1 , 0 s N, eE (s)S 0 (0) = 1 , S 0 (s > 0) = 0 , F 0 (x; break , 0) = F b(x)

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Q: What are the new events, N (s ; s, e ), in each state s whenevent e triggers a transition from state s?

Exercise: sketch the Java code for generating sample paths fromthe GSMP in this case

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Distribution Sampling

Sample path generation (discrete event simulation) depends onthe ability to sample distributions, e.g. of

Inter-arrival times

Service/repair times Message lengths

Batch size

etc.

E.g. in computing a sample path of a GSMP we need to samplesuch distributions in order to set the clocks

Where possible, we sample from a mathematical distribution, butif this is not possible we must use empirical data

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Example Uses of Mathematical Distributions

Exponential Approximates inter-arrival times well in somesystems, e.g. independently arriving jobs/customers

Weibull Fits well time-to-failure of certain components/servicesNormal Often accurate when a time delay is the sum of number

of independent sub-delays, e.g. product assembly (BEWARE of negative observations!)

Gamma Very versatile: can be parameterised to model manynon-negative r.v.s

Cauchy A heavy-tailed distribution; ts well Linux le sizes,for example

Geometric (Discrete) Sometimes used to model (integer) batchsizes

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Sampling Continuous Distributions

There are a number of methods; most importantly:1. The inverse tranform method

2. The rejection method3. The convolution method

and various special methods for particular distributions (e.g. seetutorial exercise)

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Short Diversion: Sampling U(0,1)

All distribution samplers assume the ability to generateuniform-random numbers in the interval (0 , 1)

The robust RNGs exploit number theory and are based on theLinear Congruential Generator which generates a sequencex0 , x 1 , x 2 ... via the rule

xn +1 = ( ax n + c) mod m

where a is a multiplier , c is an increment and m is the modulus

Samples from U (0, 1), viz. u0 , u 1 , u 2 ,... are obtained by settingu i = xi /m,i = 0 , 1, 2,... .

Note that 0 x i < m so the maximum period is m An important objective is to maximise the period

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Multiplicative Congruential Generators

A Multiplicative Congruential Generator (MCG) is obtained bysetting c = 0 in an LCG (but note now that x i must be non-zero)

It can be shown that a multiplicative generator has (maximum)period m

1 if m is prime and if the smallest integer k for which

(ak 1) mod m = 0 is k = m 1 Example: a = 3 , m = 7 whereupon, (only) when k = m 1 = 6,

ak 1 = 728 = 104 7; starting with x 0 = 1, we obtain:x0 = 1 , x 1 = 3 , x 2 = 2 , x 3 = 6 , x 4 = 4 , x 5 = 5

Period =6, x 6 = 1 , x 7 = 3 ...

Javas Math.random() uses an MCG with a 48-bit seed

Modern generators combine several MCGs or generalise theMCG approach e.g. the Mersenne Twister (period 2 19937 1)!

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Well assume the ability to generate U (0, 1) samples by anystandard method (well use Javas Math.random() ) and take itfrom there

Recall that a continuous random variable, X say, can becharacterised by its density and/or distribution functions:

P (x < X x+ dx) = f (x)dx F (x) = x f (y)dy = P (X x)

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1. The (Inverse) Transform Method

Suppose X is a continuous r.v. with cdf F (x) = P (X x) andthat we are trying to sample X

A curious but useful result is that if we dene R = F (X ) thenRU (0, 1); OR... given RU (0, 1) sample X from F

1 (R):

x

F(x)

X

r

R

x

1

Note that F (x) increases monotonically, so:P (X x) = P (F 1 (R) x) = P (R F (x)) = F (x),

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Algorithm: Sample U (0, 1) giving some value 0 R 1, thencompute F 1 (R)

Of course, this only works if F (x) has an inverse!

Example: U(a,b)

If X U (a, b) thenF (x) =

x aba

, a x b

Setting R = F (x) and inverting F givesX = F 1 (R) = R(ba) + a

This conrms what we (should!) already know: if RU (0, 1),then ( R(ba) + a)U (a, b)

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Example: exp( )

If X exp() thenF (x) = 1 exp x , x 0

Setting R = F (X ) and inverting, we get:

R = 1 exp X1 R = exp X

loge (1 R) = X loge (1 R)

= X

So, if R

U (0, 1), then

log

e(1

R)/

U (0, 1) (this is alsothe case for loge (R)/ since (1 R)U (0, 1))

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2. The Rejection Method

If F (x) cannot be inverted we can sometimes work with thedensity function f (x)

The rejection method is a Monte Carlo algorithm (i.e. basedon random sampling): we enclose f (x) in a box and then throwdarts(!)

f(x)

xa b

m

Y

Y

1

2

X X 1 2(Accept) (Reject)

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The procedure is as follows:1. Find an a, b and m that bounds all (or most of) f (x)

2. Pick two random samples X U (a, b) and Y U (0, m)3. If the point ( X, Y ) lies beneath the curve for f (x) accept X ;

otherwise reject X and repeat from step 2

Intuitively, the method works because the smaller f (X ) is theless likely you are to accept X

More rigorously, we need to show that the density function forthose values of X that we accept is precisely f (x), i.e. we need toshow that:

P (x < X x + dx | Y < f (X )) = f (x) dx

70

xa + x b x

m

f(x)

X

Y

Using the formula for conditional probability, this becomes:P (x < X x + dx,Y < f (X )) /P (Y < f (X ))

= dx/ (b

a)

f (x)/m

1/ (m(ba))= f (x) dx

71

The efficiency depends on the number of rejections R beforeaccepting a value of X

The probability of accepting X in any one experiment, p say, issimply the ratio of the area of the box to that under the curve,i.e.

p = 1

m(ba) Since each experiment is independent, R is geometrically

distributed:P (R = r ) = p(1 p)r

If f has very long tails, p will be small, although it is possible todivide the x axis into regions and treat each separately

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3. The Convolution Method

A number of distributions can be expressed in terms of the(possibly weighted) sum of two or more random variables fromother distributions (The distribution of the sum is the

convolution of the distributions of the individual r.v.s)

Example: Erlang( k, )

A r.v. X Erlang( k, ) is dened as the sum of k r.v.s each withdistribution exp( k)

We can think of X being the time taken to pass through a chainof k servers, each with an exp( k) service time distribution:

k k k

1 2 k

73

Notice thatE [X ] =

1k

+ 1k

+ . . . 1k

= 1

We can generate Erlang( k, ) samples using the sampler for theexponential distribution: if X i

exp( k) then

X =k

i =1X i Erlang( k, )

If R i U (0, 1) then X i is sampled using log R i / (k) We can save the expensive log calculations in the summation by

turning the sum into a product:

X =k

i =1

log R ik

= 1k

logk

i =1R i

74

Sampling Discrete Distributions

All discrete distribution sampling is based on the inversetransform method, either through table look-up or algebraicallyin some cases

We work with the cumulative distribution function F (x), whichfor a discrete r.v. is a step function, e.g.

F(x)

1

0 x

0.39

0.15

0 1 2 3 4 5

0.66

0.81

75

Here, we have:x p(x) F(x)

0 0.15 0.15

1 0.24 0.39

2 0.22 0.61

3 0.15 0.81

4 0.09 1.00

To generate a sample, we drive the table backwards We simply pick a value on the y-axis (a U (0, 1) sample) and then

map this back to an interval on the x-axis

Q This involves a linear-time lookup. Can we do better...?!A Yes! But how...? See the quiz during the tutorial!

76


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Steady State Output Analysis

Usually (but not always) were interested in measuring a systemat equilibrium , or in steady state

A system is in equilibrium if the state probabilites areindependent of time, i.e. for state variable S (t)

P (S (t) = s) = ps (t) ps as t

P(s(t) = s2 )

P(s(t) = s1 )

P(s(t) = s3 )

1

23

p

pp

Transient

Somewhere here

Probability

Equilibriumtime

77

To estimate equilibrium behaviour we must ignore thepost-initialisation (warm-up) transient as it may introduce bias

We must start any measurement (or reset ongoing measurement)at the end of the transient period

Note: this simply means that the state probabilities areapproximately those at equilibrium the state on an individualsample path at that time may assume any value

In general, we can only work out the approximate length of thewarm-up transient by doing pilot runs, e.g.

Plot running mean or moving average over one run

Plot measures having discarded p% of the initial observations Plot means/distributions of n th observation of many runs

78

Condence Intervals

Suppose we have n independent, identically-distributed (iid)observations X i , 1 i n each with mean and variance 2

Wed like to compute a point estimate for based on the X i thats easy:

X = 1n

n

i =1X i

Wed also like a measure of condence in this estimate We seek two numbers, a, b say, for which we can state:

P (a b) = pfor some probability p, e.g. 0.9, 0.95...

This is called a condence interval and is usually symmetricabout X , i.e. (a, b) = X h where h is the half-width

79

The Central Limit Theorem tells us that the sample mean of theX i approaches the N (, 2 /n ) distribution as n Normalising,

X N (,

2 /n )

X

2 /n N (0, 1) as n

Because we know the cdf of N (0, 1) we can easily locate twopoints z and z such that

P (z < X

2 /n z) = 1

for any given

z is messy to compute, so particular values have been tabulated(see handouts)

80


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Rearranging...P X

z n < X +

z n = 1

If we know we can nd X

z/ n by looking up z in tables of

N (0, 1)

/ 2 / 2

0z z

This is ne provided n is sufficiently large We can replace by S (sample standard deviation), again

provided n is sufficiently large

81

Q What does sufficiently large mean?

A It depends on the distribution of the X i

However, key fact: it can be shown that if the X i , 1 i n, arethemselves normally distributed then

X

S 2 /nhas the Students t distribution with parameter n 1 (degreesof freedom)

Provided the X i are normally distributed and independent, anexact 100(1

)% condence interval for the mean ( ) is

X tn 1 ,1 / 2 S

n n 2, tn 1 ,1 / 2 from tables

82

Example: Suppose the observations 1.20, 1.50, 1.68, 1.89, 0.95,1.49, 1.58, 1.55, 0.50, 1.09 are samples from a normal distributionwith unknown mean . Construct a 90% condence interval for

The sample mean is

X = 110

10

i =1X i = 1 .34

where X i is the i th observation

The sample variance is

S 2 = 1n 1

n

i =1(X i X )

2 = 0 .17

so S = 0 .41

83

From tables of the Students t distribution , t 9 ,0 .95 = 1 .83 so the90% condence interval is

X 1.83 0.41 10 = 1 .34 0.24

In the context of a simulation, how do we ensure in general thatour observations are a. independent and b. approximatelynormally distributed?

If the observations come from independent runs of thesimulation, then they will be independent

If the observations are themselves internally-computedpopulation means then they will be approximately normally

distributed (Central Limit Theorem) problem solved!

But what if theyre not?!

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Batched Means

A popular general approach seeks to achieve approximatenormality and independence simultaneously

Idea: Run the model once and divide the post-warm-up periodinto n batches (typically 10 n 30) where batch i computesthe average of m estimates Y i, 1 ,...,Y i,m

Compute X i = 1 j m Y i,j , i = 1 ,...,n Provided m is reasonably large, the X i will (hopefully) be

approximately normal and independent

...

Batc o servations

m individual observationsor state changes

1X X 2 3X nXTransient0

End SimulationReset measures

85

Independence

Independent replications guarantees independent observations atthe expense of a warm-up transient at the start of each run

The batched means method involves a single run and a singlewarm-up transient, but the observations might now be dependent(why?)

If the observations are dependent then the formula for the samplevariance S 2 will be wrong because the covariances are ignored

Consequence: The condence interval will be too narrow

86

Recall, covariance measures the dependence between two r.v.s For r.v.s X and Y with means X and Y , and variances 2X and

2Y respectively

COV (X, Y ) = = E ([X X ][Y Y ])= E (XY ) X Y

Note that covariance may be positive (positive correlation),negative (negative correlation) or zero (uncorrelated, whichimplies the r.v.s are independent)

Note thatCOV (X, X ) = E ([X X ]

2 ) = V AR(X )

87

Recall also,V AR(X + Y ) = E [((X + Y ) (X + Y ))

2 ]

= E [((X X ) + ( Y Y ))2 ]

= E [(X X )2 ] + E [(Y Y )

2 ]

+ 2 E [((X X )(Y Y ))]= 2X + 2Y + 2 COV (X, Y )

Of course, if X and Y are independent then C OV (X, Y ) = 0 andV AR(X + Y ) = 2X +

2Y

88


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To obtain our condence interval we assumed thatV AR(X ) = 2 /n , but if the X i , 1 i n are dependent then

V AR(X ) = V AR(1n

n

i =1X i )

= 1

n 2V AR(

n

i =1X i )

= 1

n 2E ([(

n

i =1X i ) n]

2 )

= 1

n 2E ([

n

i =1(X i )]

2 )

= 2

n +

1n 2

2n 1

i =1

n

j = i +1Cov( X i , X j )

89

The bottom line: Covariances are typically positive so if weignore them altogether S 2 /n becomes an under-estimate of V AR(X ) and the condence intervals will be too narrow

90

State Space Coverage

For any discrete-event model consider the underlying statetransition system

...

Statetransition(Event)

State space

The accuracy (consistency) of a measure derived directly, orindirectly, from the (equilibrium) state probabilities is determinedby the average number of times the simulation visits each state

91

Many visits high accuracy; few visits low accuracy

...

Parameterisation 1: Simulation time Parameterisation 2: Simulation timespent in a small number of states spent covering a large number of states

...

State with equilibrium probability >

For the same simulator execution time we would expect narrowercondence intervals from the model on the left than the one onthe right

92


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Markov Processes

Note that the future history of a GSMP is determined entirely bythe current state

However, note that the current state includes clock settings thatare history dependent, e.g.:

C k

C k (set )

C k (set )

C k (set )s

e1

e2

e3

In general, the value of clock C k depends on how we entered thestate

93

Now... suppose that when entering state s ,1. Each transition into s initialises the clocks of the new events

in e E (s ) using the same distribution F (x; e ), i.e.

independently of how the state was entered

2. All clocks are set according to an exponential dsitribution, i.e.

F (x; e ) = 1 e rxfor some parameter r that depends only on e

The exponential distribution is memoryless , in that the future isindependent of the past, i.e. if X exp(r )

P (t < X t + s | X > t ) = P (X s) s, t 0

94

With the above assumptions the GSMP now assumes a verymuch simpler form:

1. All occurrences of event e are associated with the samedistribution, hence the same distribution parameter

2. There is no need to remember the time spent in any previous

state(s), e.g. throught the use of clocks (memorylessness)3. In state s we simply race the distribution samplers

associated with eE (s) and choose the event with thesmallest time

4. Transitions need only be labelled with the associated (rate)parameter, e.g. for our queue:

c0 1...

...

c1

95

This much simpler process is called a Markov Process Markov Processes are easily analysed by discrete-event

simulation (Exercise: sketch the algorithm)

Crucially, however, they can be analysed Numerically, provided the state space is nite

Analytically, in some special cases

provided the process is irreducible : every state can be reachedfrom any other

96


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Numerical Solution Methods (Equilbrium)

Let the state space be S and ps (t) be P (S (t) = s) where S (t)S is the state at time t

Let q

s,s be the transition rate from state s

S to states S, s

= s

At equilibrium, the process is time homogeneous : ps (t + ) = ps (t) = ps , say

It can be shown that every nite, irreducible, time-homogeneousMarkov process has a unique equilibrium probability distribution

ps , sS , which is also limt ps (t)

97

At equilibrium, the probability uxes (average number of transitions per unit time) into, and out of, each state must be inbalance, i.e. for each state sS

pss S,s = s

q s,s =s S,s = s

ps q s ,s

These (linear) ux equations are called the global balance equations and are subject to the normalising equation

sS ps = 1

98

Example:

s

q

s

ss

s,sq s,ss,s

q

q

s,s

The ux out is ps (q s,s + q s,s ); the ux in is ps q s ,s + ps q s ,s Thus, at equilibrium:

ps (q s,s + q s,s ) = ps q s ,s + ps q s ,s

We can use standard linear solvers to nd ps for any state s ,given the transition rates q s,s , s , s S

99

The problem is usually dened in matrix form, viz. the solutionto p Q = 0 where p = ( ps 1 , ps 2 ,...,p s N ), S = {s1 , s 2 ,...,s N } is thestate space, and Q is the Generator Matrix

Q =

q 1 q s 1 ,s 2 q s 1 ,s 3 . .. q s 1 ,s N q s 2 ,s 1 q 2 q s 2 ,s 3 . .. q s 2 ,s N

...

q s N ,s 1 q s N ,s 2 q s N ,s 3 ... q N

The diagonal terms q i , s i S encode the net output rate fromstate i in the balance equations:

q i = s j S,j = i

q s i ,s j

100


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For example, consider p applied to the rst column: ps 1 q 1 + ps 2 q s 2 ,s 1 + ... + ps N q s N ,s 1 = 0

which is the balance equation for state s 1

As an example, consider our single-server queue (the M/M/1/cqueue), state space {0, 1,...,c}

Q =

0 0 ... 0 ( + ) 0 ... 00 ( + ) ... 0...

...

Recall: is the service rate ; 1 is the mean service time , S say

101

Note that Q is singular and there are innitely many solutions topQ = 0 as written

In order to solve for the equilibrium distribution p it is necessaryencode the normalising condition:

sS ps = 1

Simple: pick a column, j , set all its elements to 1, and likewisethe j th element of the 0 vector, e.g. in our example we mightsolve pQ = 1 c , or

p

0 0 ... 1 ( + ) 0 ... 10 ( + ) ... 1...

... 1

= (0 , 0, ..., 1)

Now choose your favourite solver and click go!102

Analytical Solution Methods

For Markov processes with the right structure direct analyticalsolutions to the system of equations p Q = 0 can be obtained

Our M/M/1/c queue is an example; well consider the casec =

(the standard M/M/1 queue), although there is a

solution for nite c

The balance equations are:p 0 = p1

( + ) pi = pi 1 + pi +1 , i > 0

By inspection, the (unique) solution is pn = n p0 , where =

103

p0 is found from the normalising equation:n =0 pn =

n =0

n p0 = 1

Rearranging... p0 =

n =0n

1

= 1

Notice, importantly, that the sum only converges if < 1 i.e. < ; hence the case = has no solution (the system isunstable )

Putting this together, we obtain:

pn = (1

) n , n

0

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Performance Measures for the M/M/1 Queue

Throughput,

We can use the above result to determine the throughput : = p0 0 + (1 p0 ) = =

Server utilisation, U

The server is idle when n = 0; busy otherwiseU = 1

Pr

{ The server is idle

}= 1 p0 = 1 (1 )= =

= S

105

Mean queue length, N

Recall the denition of the mean queue length:N =

n =0npn

The n in the summation is ddly, but notice that:nn =

dd

n

Hence:N =

n =0

npn = (1 )

n =0

nn

= (1 ) dd

n =0n =

1

106

Mean Response ( Waiting) Time, W

Recall: the mean response time is the mean time that a jobspends in the system (queueing time + service time)

The easiest way to nd W is to use Littles Law

N = W = W Thus

W = N/

= (1 )

= 1

107

Mean Queueing Time

The time spent waiting to go into service ( W Q say) is easilycomputed once we know W :

W Q = W 1

=

Note: If a customer is removed from the queue before it goes intoservice then Littles Law applied to W Q delivers the meannumber of customers waiting to be served:

N Q = W Q

= 2

1

108


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Some Examples

U N W

2 4 2 1/ 2 1 1/ 2

2 5 2 2/ 5 2/ 3 1/ 32 10 2 1/ 5 1/ 4 1/ 8

3 4 3 3/ 4 3 1

5 10 5 1/ 2 1 1/ 5

8 12 8 2/ 3 2 1/ 4

Compare these with results from the GSMP simulation

109

More than one Queue

Consider two queues in tandem:Population n Population n1 2

... ...

Infinite capacity Infinite capacity

1 2

Note: the arrival rate (throughput) is the same for both queues

Lets proceed as before - the state this time is the population of both queue 1 ( n 1 say) and queue 2 (n 2 say), i.e. a pair ( n1 , n 2 )

The queue length vector random variable is ( N 1 , N 2 ) and wellwrite pn 1 ,n 2 = P (N 1 = n1 , N 2 = n2 ) at equilibrium

110

Because the state is a pair (of queue populations) the statetransition diagram is now two dimensional:

1

1 1

2 2 2

2 2

2

0,0

0,11,0

1,1 0,2

2,0

111

Now a small problem: no closed contour gives us an immediatelysolvable balance equation, e.g. the dotted box:

p 0 ,0 = 2 p0 ,1

or the solid box:

( + 1 ) p2 ,0 = p1 ,0 + 2 p2 ,1

We can, however, come up with four equations which cover allstates and transitions. For n 1 , n 2 > 0:

p 0 ,0 = 2 p0 ,1

( + 1 ) pn 1 ,0 = pn 1 1 ,0 + 2 pn 1 ,1

( + 2 ) p0 ,n 2 = 1 p1 ,n 2 1 + 2 p0 ,n 2 +1

( + 1 + 2 ) pn 1 ,n 2 = pn 1 1 ,n 2 + 1 pn 1 +1 ,n 2 1 + 2 pn 1 ,n 2 +1

112


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One way to proceed is to guess a solution! Lets try: pn 1 ,n 2 = n 11 n 22 p0 ,0

where 1 = / 1 , 2 = / 2

We can test it by substitution into the balance equations: 01

02 p0 ,0 = 2

01

12 p0 ,0

by p0 ,0 =

( + 1 ) n 11 02 p0 ,0 = n 1

11

02 p0 ,0 + 2 n 11

12 p0 ,0

by n 11 p0 ,0 + 1 = / 1 + 2 2 = 1 +

and so on for the other cases (check them out!)

113

So it is a solution, and its unique! We nd p0 ,0 again using the fact the probabilities sum to 1:

n 1 , n 2

n 1 , n 2 0

pn 1 ,n 2 =n 1 , n 2

n 1 , n 2 0

n 11 n 22 p0 ,0 = 1

Hence

p0 ,0 =

n 1 =0

n 2 =0n 11 n

2

2

1

=

n 1 =0n 11

1

n 2 =0n 22

1

= (1 1 )(1 2 )

114

Incredibly, we arrive at:

pn 1 ,n 2 = (1 1 ) n 11 (1 2 ) n 22 , n1 , n 2 0

So, the joint probability (of being in state ( n 1 , n 2 ) is the productof the marginals (the probability of the rst queue being in staten 1 and similarly the second)

This is an example of the Product Form result which applies to anarbitrary queueing network, provided the initial assumptions hold

An extraordinary consequence of this is that each queue nowoperates as if it were an M/M/1 queue with an independentPoisson arrival stream

115

In particular, the M/M/1 queue results apply to each node inisolation, hence, for example:

The utilisations are U 1 = 1 and U 2 = 2 The mean population of queue 2 is

2

1 2 The mean time spent in queue 1 is1

2 The probability that queue 2 contains exactly 4 customers is

p(N 2 = 4) = (1 2 )42

and so on...Q: What about arbitrary networks? More later...

116

d h


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Part II: High-level Modelling Tools

GSMPs are a neat way to describe general discrete-event systems BUT - they are low level and hard to get right

Markov processes provide an elegant way to formulate and solvemathematically tractable discrete-event systems BUT - they are even lower level than GSMPs and hard to get

right

The same applies to other transition systems we havent seen yet(e.g. SMPs, STAs...)

In practice wed like high-level formalisms/tools that correspondto, or are directly translated into, these formal state transitionsystems

117

Some Targeted Approches

1. Discrete-event simulation languages/APIs

GPSS, Simula, Simscript, SimC++, SimPy, JINQS...

2. Markovian Stochastic Petri Nets Mobius, PRISM, DNAmaca...

3. Markovian Process Algebras

PEPA, EMPA, TIPP, SFSP...4. Generalised/extended Stochastic Process Algebras

SPADES, EMPA blah , SPADES(!), MoDeST...5. Queueing Networks

Athene, QNET, QNAT,...

118

1. Tools for Discrete-event Simulation

These all provide abstractions for encoding GSMP-likeoperations at a high level, e.g. for

Managing simulated (virtual) time

Dening and scheduling events

Measurement and analysis

Meaurement is typically built in to supporting libraries/classes,e.g. for modelling resources, queues

Explicit measurement classes are also made available, e.g. tocompute explicitly means, variances, bar charts...

Output analysis may support e.g. independent replications,batch-meanand calculation of condence intervals

119

In a GSMP the current state is dened by the value(s) of program variable(s), e.g. class instance variables

The set of states is implicitly dened in terms of these statevariables

New events in state s , N (s ; s, e ), are created by explicitly

scheduling them during the transition from s to s

Event scheduling involves sampling the appropriate clock settingfunction (distribution sampling)

Cancelled events in state s , K (s ; s, e ) say, are removed bydescheduling them during the transition from s to s

Probabilistic choice of next state is achieved by sampling aU (0, 1), i.e. via a RNG

120

A i l l ( h JINQS API) E l Th Si l Q ( )


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A simple example (the JINQS API)... now() delivers the current virtual time ( double )

schedule( e ) adds event e to an event set (aka calendar queue, event diary, future event list...)

deschedule( e ) removes event e from the event set simulate() invokes events in time order and advances time

Events are class instances, with associated code ( public voidinvoke() {... }) for effecting a state transition andscheduling/descheduling events

All events have an invocation time a superclass instancevariable

The event code models a transition in some underlying GSMP

121

Example: The Single-server Queue (no measures)

The state is the queue population ( int ) stop is an abstract (virtual) method in superclass Sim that

controls termination

import tools.* ;

class SSQSim extends Sim {in t n = 0 ;

public boolean stop() {return now() > 100000.0 ;

}

122

One arrival schedules the next. In state 1, we set the clock forthe completion event, as in the GSMP...

class Arrival extends Event {public Arrival( double t ) {

super( t ) ;}public void invoke() {

schedule( new Arrival( now() + Math.random() ) ) ;n++ ;if ( n == 1 )

schedule( new Completion( now() + 0.25 ) ) ;

}}

123

The code for the completion event mirrors the equivalentGSMP...

class Completion extends Event {public Completion( double t ) {


n-- ;if ( n > 0 )

schedule( new Completion( now() + 0.25 ) ) ;}

}

124

The initial state is 0 (w p 1 in the GSMP); the only active Application: The Simple Computer Model


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The initial state is 0 (w.p. 1 in the GSMP); the only activeevent in state 0 is Arrival ...

public SSQSim() {schedule( new Arrival( now() + Math.random() ) ) ;execute() ;

}}

public class ssq1Ev {public static void main( String args[] ) {

new SSQSim() ;}

}

125

Application: The Simple Computer Model

Jobs are submitted to a job queue for execution by a CPU There are periodic page faults which are serviced by one of two

disks

Whilst a fault is being serviced the CPU switches to the next jobin the job queue; once serviced, the job re-enters the back of the job queue

2

3

1

Completions

Arrivals

Disk2

Disk 1

126

From careful observation it has been established that: Each submitted job makes 121 visits to the CPU, has 70 page

faults serviced by disk 1 and 50 by disk 2 on average

The mean service times are 0.005s for the CPU, 0.03s for disk1 and 0.027s for disk 2

The current arrival rate of jobs = 0.1/ s

Note: the CPU mean service time is the mean time a job spendsat the CPU (before leaving or page-faulting)

Q: How does the system response time ( W ) vary as the loadincreases?

For illustration purposes, well assume all time delays areexponentially distributed (this is easily changed)

127

The state of the underlying GSMP is the population of each of the three queues (( c,m,n ) say)

Here well use explicit Queues; each entry is a job, representedby its arrival time

Events: Arrival new job arrival

StopJob exit CPU

Resume( n, ... ) exit disk n

We need four distribution samplers for setting clocks:inter-arrival time, cpu time slice and two disk (page fault) servicetimes

Well measure response time W using a CustomerMeasure

128

i l * The arrival event ( Arrive ) generates the next arrival places the


33/35

import tools.* ;

class CPUSim extends Sim {Queue cpuQ = new Queue(),

diskQ[] = { new Queue(), new Queue() } ;Exp iaTime,sTime[] = { new Exp( 1000.0/30 ),

new Exp( 1000.0/27 ) },xTime = new Exp( 1000.0/5 ) ;

CustomerMeasure respTime = new CustomerMeasure() ;

int completions = 0 ;int stopCount = 100000000 ;

129

The arrival event ( Arrive ) generates the next arrival, places thenew job in the job queue ( cpuQ) and starts the CPU if it was idle(state (0 ,m,n ) in the GSMP)

Event StopJob may exit the job or pass it to one of thediskstransition from ( c,m,n ) to either ( c 1,m,n ) w.p. 1/121,(c 1, m + 1 , n ) w.p. 70/121, or ( c,m,n + 1) w.p. 50/121

In states ( c, 0, n ), (c,m, 0), c,m, n 0 we need to schedule aResume event at the corresponding disk

On exit from a server the next job in the job queue is started, if there is one

Well build in an EndOfWarmUp event to allow us to deletetransient measures

130

class Arrival extends Event {public Arrival( double t ) {


schedule( new Arrival( now() + iaTime.next() ) ) ;cpuQ.enqueue( new Double( now() ) ) ;if ( cpuQ.queueLength() == 1 )

schedule( new StopJob( now() + xTime.next() ) ) ;}

}

131

class StopJob extends Event {public StopJob( double t ) {


Double t = (Double) cpuQ.dequeue() ;if ( Math.random()


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// StopJob continued...else {

int i = (int) ( Math.random() + 50.0/120 ) ;diskQ[i].enqueue( t ) ;

if ( diskQ[i].queueLength() == 1 )schedule( new Resume( i, now() + sTime[i].next() ) ) ;}if ( cpuQ.queueLength() > 0 )

schedule( new StopJob( now() + xTime.next() ) ) ;}

}

133

class Resume extends Event {int d ;public Resume( int disk, double t ) {

super( t ) ;

d = disk ;}public void invoke() {

Double t = (Double) diskQ[d].dequeue() ;cpuQ.enqueue( t ) ;if ( cpuQ.queueLength() == 1 )

schedule( new StopJob( now() + xTime.next() ) ) ;if ( diskQ[d].queueLength() > 0 )

schedule( new Resume( d, now() + sTime[d].next() ) ) ;} }

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class EndOfWarmUp extends Event {public EndOfWarmUp( double t ) {

super( t ) ;}

public void invoke() {respTime.resetMeasures() ;completions = 0 ;stopCount = 5000 ; // Whatever...!

}}

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public boolean stop() {return completions == 3000 ;

}public CPUSim( double lambda ) {

iaTime = new Exp( lambda ) ;schedule( new Arrival( now() + iaTime.next() ) ) ;schedule( new EndOfWarmUp( now() + 15000 ) ) ; // Whatever!simulate() ;System.out.println( respTime.mean() ) ;

}}

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Some Results Now lets up the load to = 0 .4...


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Here are three example time series plots for the mean waitingtime (moving average) when = 0 .15:

CPU model - disk 1 mean queue length for =0.15(three trial runs)

0

2

4

6

8

10

12

14

16

0 4000 8000 12000 16000 20000 24000 28000

Time

M e a n

W a i

t i n g

T i m e ,

W

It looks as if the system is approaching equilibrium after around15000 time units

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pCPU model - disk 1 mean queue length for =0.4

(three trial runs)

0

5

10

15

20

25

30

0 4 000 80 00 12000 16000 20000 24000 28000

Time

M e a n W a i

t i n g

T i m e ,

W

As increases the system takes longer to reach equilibrium(why?)

An an example, wed expect measurements taken between 15000and 30000 time units, say, to have wider condence intervals for = 0 .4 than for = 0 .15. Lets see...

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W (exact) Point Estimate 90% Condence Interval

(10 independent replications)

0.1 4.86 4.839 (4.776, 4.902)

0.15 5.42 5.267 (5.044, 5.490)

0.20 6.16 6.198 (5.790, 6.607)

0.25 7.17 6.679 (6.158, 7.199)

0.30 8.68 8.662 (7.784, 9.541)

0.35 11.25 11.726 (10.767, 12.685)

0.40 16.86 16.295 (12.297, 20.293)

0.45 42.45 30.765 (18.621, 42.909)

Q: Whats going on? Q: Where did the exact results come from?

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337 Simulation and Modelling

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