Queueing Models 1 Performance Engineering Prof. Jerry Breecher QUEUEING

Queueing Models 1

Performance Engineering

Prof. Jerry Breecher

QUEUEING

Queueing Models 2

WHAT WE ARE DOING HEREWe are about to embark on:

Queueing Lingo - all the definitions you’ll ever need!

Queueing - especially the Single Queue in DETAIL

Analytical Models - how to solve a number of complex problems using the equations we already know and love. Works especially well for open models.

Operational Laws - drawing conclusions about limits - what to do when we can’t solve the math exactly.

Mean Value Analysis - Using the equations in an iterative fashion to solve closed models.

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Queueing Models

This section is about being able to describe the behavior of queues. Queues are certainly a prevalent object in computer systems, and our goal here is to write the equations that describe them. The language we’ll use here is mathematical, but nothing really more complicated than algebra.

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Queueing Lingo

Goals:

– To understand the random/statistical nature of computer data. We will emphasize the non-deterministic.

– To understand distributions for simulation purposes.

– To impress your friends.

For an observation, two things matter:

– The value measured.

– When measured.

The occurrence of an event can give us the "when".

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Queueing Lingo

A STOCHASTIC PROCESS is a mechanism that produces a collection of measurements which all occur, randomly, in the same range of values. It applies to the VALUE measured for an observation. The dictionary says, "random, statistical".

Stochastic processes are well behaved phenomena which don't do things which are unpredictable or unplanned for.

Examples:

Throwing 2 dice always gives numbers in the range 2 - 12.

Actions of people are unpredictable ( unless the range of values is made very large.) Someone can always respond in a way you haven't predicted.

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Queueing Lingo

THE POISSON PROCESS applies to WHEN an observation is made. It looks random; the arrival points are uniformly distributed across a time interval. Poisson processes can be defined by:

Event counting The distribution of the number of events occurring in

a particular time is a Poisson distribution.

Time between events The distribution of times between event occurrences is

exponential.

Example: Show how a random "look" leads to an exponential distribution. See the next page for a picture of these distributions.

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Queueing LingoExponential Curve

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 1 1.5 2 2.5 3

t

F(t

)

F(t)

This is a simple exponential curve. What properties can you identify from it?

F(t) = exp(-t)

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Queueing Lingo

Example of the Poisson Probability Density Function.

Poisson Probability Density Function

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0 2 4 6 8 10 12

k

F(k

)

F(k)

F(k) = ( 5k / k! ) exp( -5 )

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Lab - You Get To See It Happen Before Your Very Eyes!

This is a group lab designed to demonstrate that the number of random events in a particular interval is a Poisson distribution, and that the distribution of spaces between events is exponential.

Using your laptop or Smartphone, go to random.org. Your goal is to get 20 random numbers in the range 0 – 49. Each time, write down the number presented beside the appropriate Segment in column I.

Count how many of the segments in Column I have 0 samples in them. Put this number in the first row of Column II. Repeat for 1, 2, 3, ... samples.

For each of the samples in Column I, determine the distance or separation between that sample and the next higher sample. If the distance is 0 ( the two samples are the same ) put a tick in the 0-interval row of column 3, if the separation is 3, tick the interval = 3 row, etc.

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Segment How Many

Samples

0 – 4

5 – 9

10 - 14

15 - 19

20 - 24

25 - 29

30 - 34

35 - 39

40 - 44

45 - 49

Column INumbers falling

in each segment.

Column IINumber of segments

containing N samples

Column III Number of instances of

intervals between samples.

N Number of Segments

0

1

2

3

4

5

6

7

8

9

Interval Number of Instances

0

1

2

3

4

5

6

7

8

9

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// Generate Poisson.// This program generates random numbers, puts them into buckets,// and calculates the distance between numbers.//// Inputs:// N - number of samples to generate

#include <time.h>#define MAX_DATA_VALUE 100#define MAX_BUCKETS 100#define MAX_DATA 1000

// Compare routine used in sortingint compare (const void * a, const void * b) { return ( *(int*)a - *(int*)b );}

int main( int argc, char *argv[]) { int Data[MAX_DATA]; int Bucket[MAX_BUCKETS]; int NumberOfSamples; int NumberOfBuckets = 10; int LargestBucketFill = 0; int Index, Temp, i;

if ( argc <2 ) { printf( "Usage: GeneratePoisson <number_of_samples>\n"); exit(0); } NumberOfSamples = atoi( argv[1] ); srand((unsigned)time(NULL)); for ( Index = 0; Index < NumberOfBuckets; Index++ ) Bucket[Index] = 0; for ( Index = 0; Index < NumberOfSamples; Index++ ) { Data[Index] = rand() % MAX_DATA_VALUE;

Temp = (NumberOfBuckets*Data[Index])/MAX_DATA_VALUE;Bucket[Temp]++;if ( Bucket[Temp] > LargestBucketFill ) LargestBucketFill = Bucket[Temp];

}

printf("\n Raw Random Numbers\n"); for ( Index = 0; Index < NumberOfSamples; Index++ ) printf( "%d ", Data[Index] ); printf("\n"); qsort (Data, NumberOfSamples, sizeof(int), compare); printf( "\n Sorted Random Numbers\n"); for ( Index = 0; Index < NumberOfSamples; Index++ ) printf( "%d ", Data[Index] ); printf("\n"); printf( "\n Range Number\n"); printf( " in Range\n"); for ( Index = 0; Index < NumberOfBuckets; Index++ ) printf( "%2d - %2d, %d\n", (MAX_DATA_VALUE/NumberOfBuckets * Index),

(MAX_DATA_VALUE/NumberOfBuckets * (Index+1)) - 1, Bucket[Index] );

// Calculate distribution of items in each bucket printf("\n"); for ( Index = 0; Index < LargestBucketFill; Index++ ) { Temp = 0;

for ( i = 0; i <NumberOfBuckets; i++ ) if ( Bucket[i] == Index ) Temp++;

printf( "Number of buckets with %d items, %d\n", Index, Temp ); } // Determine the distance between the random items. for ( Index = 0; Index < MAX_BUCKETS; Index++ ) Bucket[Index] = 0; LargestBucketFill = 0; for ( Index = 0; Index < NumberOfSamples - 1; Index++ ) { Temp = Data[Index+1] - Data[Index];

Bucket[Temp]++;if ( Temp > LargestBucketFill ) LargestBucketFill = Temp;

} printf( "\n Distance Number\n"); printf( " Between with this\n"); printf( " Samples distance\n"); for ( Index = 0; Index <= LargestBucketFill; Index++ ) printf( "%2d, %d\n", Index, Bucket[Index] );}

Code is here so it doesn’t get lost!!

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The results of running this code can be seen here This is the result of running 4 experiments of 20 samples each.The results aren’t pretty!! Perhaps by running the tests many times, everything would look nice and smooth. You can see the results you would expect, but there’s lots of jitter.

Experimental Results

http://www.cs.wpi.edu/~jb/perf/Lectures/Poisson.xls

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Queueing Lingo

Examples:Suppose that a piece of software has an expected lifetime of 5 years, and

that the average bug rate for this type of product is one bug/year.

• What is the bug expectation rate per year at the start of the five years, assuming this code is "average"?

• After two years, four bugs have been found. The code is still considered "average". How many bugs/year can be expected for the remaining three years?

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Queueing LingoMEMORYLESS means that the probability of an event doesn't depend on its past.

The above case highlights an example where the past does matter.

Examples: Which depend on the past, and which don't?

– Throwing dice?

– A disk seek distance?

– The address of an instruction execution?

– The measurement of the length of a table?

Prepare to Have Your Mind Bent In An

Unexpected Way!!

Example:Consider a bus stop where the time between bus arrivals is exponentially distributed with a rate L. Thus the bus arrivals form a Poisson process. If you walk up to the bus stop, how long do you have to wait until the next bus arrives?

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Queueing Lingo

Example:

Consider a bus stop where the time between bus arrivals is exponentially distributed with a rate L. Thus the bus arrivals form a Poisson process. If you walk up to the bus stop, how long do you have to wait until the next bus arrives?

1. Possible solution: Since buses arrive at a rate L, the average time between arrivals is 1/L. But since we walk up at random, we would wait for only half an interval on the average. So we would wait 1/(2L) for the next bus.

2. Possible solution: Since the time between buses is exponentially distributed, it is memoryless. So the residual lifetime for any time that I arrive should be distributed exponentially the same way as the original distribution. Since the average time between buses is 1/L, the average time ( residual ) to wait should also be 1/L.

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Queueing Lingo

// GenerateRandomBusArrivals

#include <time.h>#define MAX_DATA 1000// Hours in a week#define MAX_DATA_VALUE 168

// Compare routine used in sortingint compare (const void * a, const void * b) { return ( *(int*)a - *(int*)b );}int main( int argc, char *argv[]) { int Data[MAX_DATA]; int NumberOfSamples = 14; int Index;

printf( "This program generates 1 week's worth of bus arrivals.\n"); printf( "We assume that the average bus arrival rate is every 12 hours,\n"); printf( "The arrival rate is 1/12 per hour. We'll take 1 week's worth\n"); printf( "of arrivals - we will assume that there are 14 arrivals in a week \n"); printf( "- that's how we set the average arrival rate.\n"); printf( "Oh - and we assume the buses arrive on the hour (it's simpler that way).\n");

srand((unsigned)time(NULL)); for ( Index = 0; Index < NumberOfSamples; Index++ ) { Data[Index] = rand() % MAX_DATA_VALUE; } qsort (Data, NumberOfSamples, sizeof(int), compare); printf( "\n Sorted Random Numbers\n"); for ( Index = 0; Index < NumberOfSamples; Index++ ) printf( "%d ", Data[Index] ); printf("\n");}

This program generates 1 week's worth of bus arrivals.We assume that the average bus arrival rate is every 12 hours,The arrival rate is 1/12 per hour. We'll take 1 week's worthof arrivals - we will assume that there are 14 arrivals in a week- that's how we set the average arrival rate.Oh - and we assume the buses arrive on the hour (it's simpler that way).

Sorted Random Numbers2 5 73 81 100 100 102 109 134 136 145 148 152 154

So given these actual bus arrivals, how long on average must you wait for a bus?

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Queueing Lingo

Types of Stochastic ProcessesType Event Counting Value Time Between

Events Example

Stochastic Arbitrary - any new state is possible

given the current state. Arbitrary Cars on a highway (highly

interactive.) Random Walk Next state may depend on current &

previous states. Arbitrary Monopoly Game.

Markov Future states depend only on

current state, not on previous state. Arbitrary Monopoly Game.

Birth-Death Next state depends only on current

state. Memoryless Event driven scheduler.

Poisson Memoryless Memoryless Bus problem above. Arrival rates are often Poisson.

Service times are often exponential.

( In other words, they're both random. )

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PROPERTIES OF QUEUES

The device doing the actual service of the customers.

Customer Departures

The queue – A place where

customers are stored before

being serviced.

Customer Arrivals

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PROPERTIES OF QUEUESHow do we describe a queue? These are the important aspects:

Arrival process:

Service Time Distribution:

Number of Servers:

System Capacity:

Population Size:

Service Discipline:

The shorthand for queue description is thus A / S / m / B / K / SD.

The inter-arrival and service times are typically of the following types:

M Exponential – Memoryless, the distribution we’ve just been talking about.

D Deterministic – the times are constant and there is no variance.

G General –distribution is not specified and the results are valid for all distributions

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THE SINGLE QUEUEIf we have a single queue obeying certain properties, we can get all kinds of

nice metrics. But, it must have those required properties!!

REQUIRED PROPERTIES:

• Arrivals are random with a rate of X per time. ( Poisson – when we say this, we mean the inter-arrival time is exponentially distributed. ) [ Note that

in steady state, throughput = arrival rate.] Many texts use for this.

• Service times are random with a value of D. (Exponential ) [ Note this is

the Demand we've seen before.] Many texts use for this. The rate of service is = 1/D.

• There's the possibility of an infinite number of customers.

• There's a single server. [ So the derivation we’re about to do doesn't work for a multiprocessor CPU.]

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THE SINGLE QUEUE

These are general requirements and hold for many practical applications. Other analysis can be done for cases outside these requirements, but we don't do it here.

We will apply a method called local balance to a system like that pictured on the next page:

The queue is of type M / M / 1.

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THE SINGLE QUEUEX = X =

= 1/D = 1/D

State with 0 in Queue



For simplification, in this particular case, the utilization U is related to throughput and demand by

U = X D (Remember N = XS)

Note: U

pi = U , p0 = ( 1 – U )N

1

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THE SINGLE QUEUEBy Definition: A queue is defined to contain customers that are both waiting and being

serviced.

In an equilibrium state, from the picture below, the following equations can be formed:

pi = p i-1

pp i

pp 0Up 0

The probability of having i customers in the queue is

pi = ( 1 – U ) U i

[ Note that p0 = ( 1 - U ) so p i > 0 = U. But this is just the utilization we defined before.]

X = X =

= 1/D = 1/D




Queueing Models 24

THE SINGLE QUEUEThe average number of customers in the queue (waiting and being serviced) is

From Little's Law ( N = X T ) in steady state, we can derive the average time spent at the queueing center ( both in the queue and being serviced ). Note what happens to this response time as the utilization increases!

)1( UUN

)1( UDRT k

Queueing Models 25

THE SINGLE QUEUEExample:

Pat is designing a communications server that receives requests from "higher level" routines. The requests are collected by a Request Handler that does nothing but put them into buffers. These requests are removed from the buffers by the Request Processor on a first come first serve basis.

requests -> Request Handler -> Buffers[n] -> Request Processor ->

The requests arrive randomly at a rate of 5/second. The Request Processor can service 10 items per second from the buffer.

Since allocating buffers is an expensive business, Pat wants to preallocate an adequate number of buffers so none need be allocated 99% of the time. Clearly there's a tradeoff here between memory usage and time-to-allocate.

How many buffers should be preallocated?

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THE SINGLE QUEUE

X =

= 1/D



This is the setup for the case of M / M / 2. The probability of transition from a lower population to a higher one is the same as before ( arrivals are the same.) But the probability of one of the TWO servers finishing is twice as great when both of them are filled.

X =


X =


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THE SINGLE QUEUEThe following equations hold for the single queue.

Utilization: U = X D

Probability of n customers in thesystem:

pn = ( 1 – U ) Un

Mean number of customers in thesystem:

N = U / ( 1 – U )

Mean time spent in the system: T = D / ( 1 – U )

Probability of finding n or more jobs inthe system:

Un

Mean number of jobs served in onebusy period:

1 / ( 1 – U )

Mean busy time duration: D / ( 1 – U )

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ANALYTICAL MODELGoals:

• You should be able to create, use, and understand a simple analytical model.

• You should have a general idea of when such models are applicable and should understand some of the buzzwords.

PERSPECTIVE:An Analytical Model uses mathematical operations to capture the relationships between observable quantities. The computations don't necessarily mimic real actions as they do for simulations.

Examples:• The equations we've been using such as Little's Law.

• Local Balance Equations - these enumerate the states the system can be in and then determine the transitions between states. (This is what we just did with single queues.)

• Mean Value Analysis - an iterative approach using the equations we've already learned.

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ANALYTICAL MODELAnalytical models can be applied to:

• Single Queues• Queueing Networks

Queueing Networks are networks of queues; two or more queues tied together. They can be:

• Open: - Typical of transaction processing. Jobs enter and leave the system being studied.

• Closed: - typical of batch or terminal systems. Jobs always remain somewhere within the system.

• Single Class - the customers are indistinguishable from each other; they have similar service demands and routing characteristics.

• Multiple Class - several categories of customers can be identified. A batch class, for example, might be heavily CPU bound while a terminal class is I/O bound. To use a multiple class model requires determining the characteristics for EACH of the classes involved.

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ANALYTICAL MODELSINGLE CLASS OPEN QUEUEING NETWORK MODEL SOLUTIONS:

This is EASY!! It's simply an extension of the equations we used for the single queue.

Utilization: Uk = X Dk {Dk is the service time)

Throughput: Xk = X Vk (Throughput = arrivals)

Max. Throughput: Xmax =

Residence Time: Rk = (delay centers)

(queueing centers)

max

1D

kD

)1( k

kU

D

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ANALYTICAL MODELSINGLE CLASS OPEN QUEUEING NETWORK MODEL SOLUTIONS:

Queue Length: Qk = (delay centers)

(queueing centers)

System Response T = R =Time:

Average Number N = Q =In System:

Remember,N = Number of requests in the "system"X = ThroughputR = Residence time per request.S = Service Time

Little's law is 90% of all you'll ever need.

kU

)1( k

kU

U

kR

kQWe'll be using the FIGURE on

the next page and applying these Laws to the system

shown there, at a number of different levels.

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ANALYTICAL MODELTerminals4

3

2

1

CPU

DISK ADISK BDISK C

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ANALYTICAL MODELBox 1 in the Figure: A single resource, not including the queue.

Here the population, N, is either 1 or 0 ( in use or not ).

Utilization is equal to the average number of requests present, or the average N.

Throughput X is the number of requests serviced per time.

Residence time R is, for this case, the service time.

[NOTE: Little's Law reduces to Utilization Law.]

1

DISK C

Example: Suppose a disk, serves 100 requests/second, with the average request needing 0.008 seconds of disk service.

What is the utilization of the disk?N = X * RU = X * SR = S/(1 – U )

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ANALYTICAL MODELBox 2 in the Figure: A single resource, including the queue.

Now the population includes both those requests in the queue and in service.

Throughput remains the rate that the resource satisfies requests.

Residence time is the sum of queueing and service times.

Example: You will need to use the result from the previous slide.What is the time a request spends at the disk subsystem ( spindle + queue)?

What is the average number of requests in “Box 2”?

For one of these requests, what is the average queueing time, and what is the average service time?

2

DISK C

N = X * RU = X * SR = S/(1 – U )

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ANALYTICAL MODEL

Box 3 in the Figure: Central Subsystem, not including terminals.

The population now is the number of users with activity in the system - those not "thinking".

Throughput is the rate that requests flow between system and users.Residence time is now what we call response time.

Example: The average system throughput (entering & leaving Box3) is 0.5/sec. There are an average of 7.5 "ready" (waiting) users in the Box.

What is the average response time?

3

2

CPU

DISK ADISK BDISK C

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ANALYTICAL MODELBox 4 in the Figure: Entire system, including terminals.

The population is the total number of users, both waiting and thinking.

Throughput is the rate that requests flow between system and users. (same as box 3).

Residence time is the sum of response time and think time.

Example: There are 10 users with average think time of 5 seconds, and the system has average response time of 15 seconds.

What is the throughput?

Terminals4

3

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ANALYTICAL MODELFor system wide applications of Little's Law, since time represents both thinking and waiting for a response,

NSS = X R for Subsystem or lower.NES = X ( R + Z ) for the Entire system.

Example:A System has 64 interactive users, with an average think time of 30 seconds.

Two interactions complete on average each second.

What is the average response time for this system?

How many interactions are in the system at any time?

What sized System ( how many CPU's ) would be needed for this application?

RESPONSE TIME LAW: R = N/X - Z

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ANALYTICAL MODELExample Problems You Should Now Be Able To Do:

The average delay experienced by a packet when traversing a computer network is 100 msec. The average number of packets that cross the network is 128 packets/sec.

What is the average number of packets in transit in the network?

Example Problems You Should Now Be Able To Do:Measurements taken during one hour from a Web server indicate that the utilization of the CPU

and the two disks are: UCPU = 0.25, Udisk1 = 0.35, and Udisk2 = 0.30. The Web server log shows that 21,600 requests were processed during the measurement interval.

What are the service demands (the time used by each request) at the CPU and both disks?What is the maximum throughput, and what was the response time of the Web server during the measurement interval?

Example Problems You Should Now Be Able To Do:A computer system is measured for 30 minutes. During this time, 5,400 transactions are completed

and 18,900 I/O operations are executed on a certain disk that is 40% utilized.

What is the average number of I/O operations per transaction on this disk? What is the average service time per transaction on this disk?

Solutions on a later page

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ANALYTICAL MODELExample Problems You Should Now Be Able To Do:

A file server is monitored for 60 minutes, during which time 7,200 requests are completed. The disk utilization is measured to be 30%. The average service time at this disk is 30 msec per IO.

What is the average number of accesses to this disk per file request?

Example Problems You Should Now Be Able To Do:A computer system has one CPU and two disks: disk 1 and disk 2. The system is monitored for one hour and the utilization of the CPU and of disk 1 are measured to be 32% and 60%, respectively. Each transaction makes 5 I/O requests to disk 1 and 8 to disk 2. The average service time at disk 1 is 30 msec and at disk 2 is 25 msec.

Find the system throughput.Find the utilization of disk 2.Find the average service demands at the CPU, disk 1, and disk 2.

Example Problems You Should Now Be Able To Do:An interactive system has 50 terminals and the user's think time is equal to 5 seconds. The utilization of one of the system's disk was measured to be 60%. The average service time at the disk is equal to 30 msec. Each user interaction requires, on average, 4 I/Os on this disk.

What is the average response time of the interactive system?

Solutions on a later page

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ANALYTICAL MODELProblem Solution:

The average delay experienced by a packet when traversing a computer network is 100 msec. The average number of packets that cross the network is 128 packets/sec.

What is the average number of packets in transit in the network? Straight usage of Little’s Law: N = XS = 128 packets/sec * 0.1 sec = 12.8 packets

Problem Solution:Measurements taken during one hour from a Web server indicate that the utilization of the CPU

and the two disks are: UCPU = 0.25, Udisk1 = 0.35, and Udisk2 = 0.30. The Web server log shows that 21,600 requests were processed during the measurement interval.

What are the service demands (the time used by each request) at the CPU and both disks?What is the maximum throughput, and what was the response time of the Web server during the measurement interval? There are 21,600 requests/hour = 60 requests/sec. During each second, the CPU is used 250

milliseconds – so each of the 60 requests is using 4.16 milliseconds. Similarly the disks use 5.83 milliseconds and 5 milliseconds of service per transaction.

Maximum throughput is determined by the device having the highest utilization, Udisk1 = 0.35. When that disk is max’d out, there will be 1 / 0.35 more traffic – or 2.86 more. Thus the maximum throughput will be 2.86 * 60 transactions/second = 171 transactions/second.

You can determine the total response time by calculating the response time at each queueing center. This is 4.16 msec/(1 – 0.35) + 5.83 msec / ( 1 – 0.35) + 5.0 msec / ( 1- 0.3) = 6.4 + 8.97 + 7.1 = 22.5 msec

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A computer system is measured for 30 minutes. During this time, 5,400 transactions are completed and 18,900 I/O operations are executed on a certain disk that is 40% utilized.

What is the average number of I/O operations per transaction on this disk? What is the average service time per transaction on this disk? Put everything into the same time units – (seconds usually work best) 5,400 transactions / 30 minutes = 3 transactions/second18,900 IO / 30 minutes = 10.5 IOs / secondSo the number of IOs/transaction = 10.5 / 3 = 3.5 IOs / transactionIn each second, this disk is busy 400 milliseoncds of time. During a second, 3 transactions

complete. So the disk service time per transaction is 133 milliseconds.

Problem Solution:A file server is monitored for 60 minutes, during which time 7,200 requests are completed. The disk

utilization is measured to be 30%. The average service time at this disk is 30 msec per IO.

What is the average number of accesses to this disk per file request? The idea is that a file server request may result in multiple IOs to the disk – it’s not a 1 to 1 match

necessarily. 7,200 requests / 3,600 seconds = 2 requests/second. The disk is 30% busy so it runs for 300 milliseconds each second. The service time is 30

milliseconds, so 10 IOs are completed each second. 10 IOs for 2 requests means there are on average 5 IOs / request.

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A computer system has one CPU and two disks: disk 1 and disk 2. The system is monitored for one hour and the utilization of the CPU and of disk 1 are measured to be 32% and 60%, respectively. Each transaction makes 5 I/O requests to disk 1 and 8 to disk 2. The average service time at disk 1 is 30 msec and at disk 2 is 25 msec.

Find the system throughput.Find the utilization of disk 2.Find the average service demands at the CPU, disk 1, and disk 2.It turns out here that disk 1 is where we have the most information: For this disk, it’s busy 600

milliseconds out of each second; it takes 30 milliseconds for each IO; so there are 20 IOs/second. Since each transaction makes 5 IO requests to disk 1, that means there are 4 transactions/second.

Once we have this answer, we know there are 32 IOs to disk 2 per second. Each of those IOs takes 25 milliseconds, giving a total time usage of 800 milliseconds. That means this disk is 80% utilized.

Problem Solution:An interactive system has 50 terminals and the user's think time is equal to 5 seconds. The utilization of one of the system's disk was measured to be 60%. The average service time at the disk is equal to 30 msec. Each user interaction requires, on average, 4 I/Os on this disk.

What is the average response time of the interactive system? The disk is doing 20 IOs / second. Each transaction is 4 IOs, so the throughput is 5 trans/sec.To get this throughput requires each of the 50 users is executing a transaction every 10 seconds. Thus the time in the system must be 5 seconds (because the user is already thinking for 5 secs.)

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ANALYTICAL MODELFORCED FLOW LAW:

The Forced Flow Law states that the flow in all parts of a system must be consistent.

Suppose we count both system completions and also completions at each resource. The visit count ( visit ratio ) is defined as:

Resource Completion: Ck

System Completion: C

Visit Count: Vk = Ck / C

THE FORCED FLOW LAW: Xk = Vk X

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ANALYTICAL MODELExample:

Suppose a system has:30 terminals ( N = 30 ).18 seconds average think time ( Z = 18 ).20 visits to a specific disk/interaction (Vdisk = 20 ).30% utilization of that disk (Udisk = 0.30 ).25 millisecs is the average service required per visit to the disk (Sdisk =

0.025 sec.).

We want to know:Disk throughput =System throughput =Response time =

THE FORCED FLOW LAW: Xk = Vk XRESPONSE TIME LAW: R = N/X - Z

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ANALYTICAL MODELExample:

Consider the problem of a spy from Burger King trying to figure out how many people are at a McDonald's. The spy can't sit inside and watch all day, so must somehow calculate the number from information obtained from outside observations. Thirty customers/hour arrive on the average ( over a long period of time ) and the average customer exits after 12 minutes.

Assuming that all this time is spent standing in line, what is the mean queue length in the restaurant?

If the Standard Deviation of the 30 customers is 3, what is the uncertainty of the queue length?

What happens if both the arrival rate and the service time have uncertainties?

THE FORCED FLOW LAW: Xk = Vk XRESPONSE TIME LAW: R = N/X - Z

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ANALYTICAL MODELEXAMPLE OF THE SOLUTION OF AN OPEN MODEL:

Model Inputs: Vcpu = 121 Vdisk1 = 70Vdisk2 = 50Scpu = 0.005 Sdisk1 = 0.030 Sdisk2 = 0.027Dcpu = 0.605 Ddisk1 = 2.1 Ddisk2 = 1.35

= 0.3 jobs/sec

Model Structure:

Arrivals

Departures

CPU

Disk1

Disk2

Queueing Models 47

Solution Of An Open Model

Model Inputs: Vcpu = 121 Vdisk1 = 70 Vdisk2 = 50Scpu = 0.005 Sdisk1 = 0.030 Sdisk2 = 0.027Dcpu = 0.605 Ddisk1 = 2.1 Ddisk2 = 1.35

= 0.3 jobs/sec

Model Structure:

Arrivals

Departures

CPU

Disk1

Disk2

Model Outputs:

= 1 / Dmax = 1 / 2.1 = 0.476 jobs/sec

Xcpu (0.3) = Vcpu = (0.3)(121) = 36.3 visits/sec

Ucpu (0.3) = Dcpu = (0.3)(0.605) = 0.182

Rcpu (0.3) = Dcpu / ( 1 – Ucpu(0.3) ) = 0.605 / 0.818 = 0.740 secs

Qcpu (0.3) = Ucpu(0.3) / ( 1 – Ucpu(0.3) ) = 0.182 / 0.818 = 0.222 jobs

R(0.3) = Rcpu(0.3) + Rdisk1(0.3) + Rdisk2(0.3) = 0.740 + 5.676 + 2.269 = 8.685 secs

Q(0.3) = R() = (0.3) (8.685) = 2.606 jobs

EXAMPLE:

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SUMMARY OF PERFORMANCE METRICS

T is the length of TIME we observed the system.A is the number of request ARRIVALS observed. C is the number of request DEPARTURES observed. W is the ACCUMULATED TIME for all requests within the system – time

spent both waiting for and using resources. B is the length of time that the resource was observed to be BUSY. Z is the think time of a terminal user. Vk the visit ratio, is the number of times device k is visited per

transaction.

Arrival Rate Y = A / T

Throughput (Departure Rate) X = C / T

Utilization U = B / T

Service Requirement S = B / C

Requests in system N = W / T

Residence time R = W

/ C

UTILIZATION LAW U = X S

LITTLE'S LAW N = X R

RESPONSE TIME LAW R = N/X – Z


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OPERATIONAL LAWSGoals:

To increase facility using performance metrics.

To be able to calculate limits or boundaries on performance metrics.

To be able to do "back of the envelope" calculations.

UTILIZATION LAW U = X S

LITTLE'S LAW N = X R

RESPONSE TIME LAW R = N/X – Z


Queueing Models 50

OPERATIONAL LAWSBOUNDARY VALUES:

The goal here is to make estimations of the outside limits of a performance parameter. We do this by selective "blind" application of our simple laws to complex systems; our results give upper and lower bounds, not an exact answer.

Example:An editor program wants to read 100 disk pages into memory. The resources required for each disk read are:

5 milliseconds of CPU Processor time.4 milliseconds of Controller Processor time.2 milliseconds of SCSI Handshaking time.10 milliseconds of DISK seek/rotation/transfer time.

• What is the bottlenecking device? Warning!! This is a trick question!• What is the worst throughput for this system, assuming the editor single threads the

reads - waiting until each read completes before starting the next?• What is the best throughput for this system? How long will it take to read in the 100

disk blocks?

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Now that we know the limits for one user, let's expand this for N users.

Example:Multiple users, each with their own copy of the editor, are reading files from the

same disk (parameters are the same as the last example.)

• What is the bottleneck device in this case?

• Considering only the bottleneck device, what is the best throughput achievable?

• Now taking into account all devices, what is the BEST throughput N users can achieve ( assuming the requests NEVER get in each other's way?)

• Now taking into account all devices, what is the WORST throughput N users can achieve ( assuming the requests ALWAYS get in each other's way?)

Queueing Models 52


We define DEMAND as the amount of a resource used for a transaction; this is just the number of visits to that resource during one transaction, and the service time used at each of those visits:

Dk = Vk * Sk

Dtotal = D = Dk

Dmax = Dbottleneck

SUMMARY: Best throughput overall: X = 1 / DBottleneck

Best throughput for N users: X = N / D

Worst throughput for N users: X = N / ( N * D ) = 1 / D

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OPERATIONAL LAWSExercise:Consider an interactive system with a CPU and two disks. The following measurement data was obtained by measuring the system:

----------- Observation Data ----------------Observation interval 30 minutesActive terminals 30Think time 12 secondsCompleted transactions 1,600Fast disk accesses 32,000Slow disk accesses 12,000CPU busy 1,080 secondsFast disk busy 400 secondsSlow disk busy 600 seconds

Determine the visit counts Vk, service times per visit Sk, and service demands Dk at each center.

Vk Sk Dk

CPUFast DiskSlow DiskD = Dk -- --

Data At Centers

Queueing Models 54

OPERATIONAL LAWSExercise: ----------- Observation Data ----------------

Observation interval 30 minutesActive terminals 30Think time 12 secondsCompleted transactions 1,600Fast disk accesses 32,000Slow disk accesses 12,000CPU busy 1,080 secondsFast disk busy 400 secondsSlow disk busy 600 seconds

Give optimistic and pessimistic asymptotic bounds on throughput and response time for 10 and 40 active terminals.

Metric Terminal Count -- 10 -- -- 40 --

Optimistic X 1 / D max

N / {D+Z}Pessimistic X N / { N * D + Z }Optimistic R D

N * D max - ZPessimistic R N * D

Bounds on Throughput and Response Time

Queueing Models 55

OPERATIONAL LAWSExercise: ----------- Observation Data ----------------

Observation interval 30 minutesActive terminals 30Think time 12 secondsCompleted transactions 1,600Fast disk accesses 32,000Slow disk accesses 12,000CPU busy 1,080 secondsFast disk busy 400 secondsSlow disk busy 600 seconds

Consider the following changes to the system:

• Move all files to the fast disk.

• Replace the slow disk by a second fast disk.

• Increase the CPU speed by 50% (with the original disks.)

• Increase the CPU speed by 50% and balance the disk load across 2 fast disks.

Rank the efficacy of these changes.

What changes should be made so response time will not exceed 10 seconds?

Queueing Models 56

MEAN VALUE ANALYSISThis is an extension of the Operational Analysis we did before. This is the method that should be used on closed systems.

Assumptions:• Applies to various service disciplines and service time distributions. (We will only do

fixed capacity centers here.)

• The arrival rate is not dependent on the load.

• Arrivals are a Poisson process.

We will use a technique called Mean Value Analysis that is based on the repetitive application of three equations:

1. Little's Law applied to the whole network:

2. Little's Law applied to each service center:

3. The service center residence time equations:

A great book on this, now out of print is athttp://www.cs.washington.edu/homes/lazowska/qsp/

Menasce, et. al. do this in Section12.3 of the text.

Queueing Models 57

MEAN VALUE ANALYSIS1. Little's Law applied to the whole network:

where X(N) is the system throughput, Z is the think time, and Rk (N) the residence time at center k when there are N customers in the network.

2. Little's Law applied to each service center:

3. The service center residence time equations:Delay CentersQueueing Centers

where Ak (N) is the average number of customers seen at center k when a new customer arrives.

The number seen on arrival (in equilibrium) is equal to the average queue length when there is one less customer in the system! So the equations bootstrap!!!

)1()( NQNA kk

K

kk NRZ

NNX

1

)()(

)()()( NRNXNQ kk

kk DNR )(

)(1)( NADNR kkk

Queueing Models 58

MEAN VALUE ANALYSIS////////////////////////////////////////////////////////////////////////////////// This program does very simple mean value analysis.////////////////////////////////////////////////////////////////////////////////#define MAX_NUMBER_OF_CENTERS 30

int number_of_customers;int number_of_centers; /* # of centers requested by the user. */int number_of_customers; /* The population in each class */float input;double think_time; /* The time spent at the terminal */double demand[MAX_NUMBER_OF_CENTERS];/* Service demand of a center. */double q_length[MAX_NUMBER_OF_CENTERS];double residence_time[MAX_NUMBER_OF_CENTERS+1]; // Residence time at a center.double throughput;double system_response_time;

// Prototypesvoid print_the_numbers(int n);

This program is at: http://web.cs.wpi.edu/~jb/perf/Lectures/mva.c

http://web.cs.wpi.edu/~jb/perf/Lectures/mva.c

Queueing Models 59

MEAN VALUE ANALYSISmain() { char input_string[80]; int center, n; //////////////////////////////////////////////////////////////////////////// // Get the input information from the user. //////////////////////////////////////////////////////////////////////////// printf( "Number of users: "); gets( input_string ); number_of_customers = atoi( input_string );

printf( "Number of centers: "); gets( input_string ); number_of_centers = atoi( input_string );

printf( "Think Time: "); gets( input_string ); sscanf( input_string, "%f", &input ); think_time = (double)input;

for ( center = 0; center < number_of_centers; center++ ) { printf( " Demand for center %d: ", center); gets( input_string ); sscanf( input_string, "%f", &input ); demand[center] = (double)input; } //////////////////////////////////////////////////////////////////////////// // Now that the network is described, we perform the evaluation. // Begin by initializing to the trivial solution for 0 customers. //////////////////////////////////////////////////////////////////////////// for ( center = 0; center < number_of_centers; center++ ) { q_length[center] = 0; }

Queueing Models 60

MEAN VALUE ANALYSIS //////////////////////////////////////////////////////////////////////////// // The algorithm now iterates through each population value n. //////////////////////////////////////////////////////////////////////////// for ( n = 0; n <= number_of_customers; n++ ) { system_response_time = 0; for ( center = 0; center < number_of_centers; center++ ) { residence_time[center] = demand[center] * ( 1.0 + q_length[center] ); system_response_time += residence_time[center]; } //////////////////////////////////////////////////////////////////////////// // Next use Little's Law to compute the system throughput. //////////////////////////////////////////////////////////////////////////// throughput = n / ( think_time + system_response_time );

//////////////////////////////////////////////////////////////////////////// // Finally use Little's Law to compute center queue lengths. //////////////////////////////////////////////////////////////////////////// for ( center = 0; center < number_of_centers; center++ ) { q_length[center] = residence_time[center] * throughput; } print_the_numbers(n); } /* END OF LOOP THROUGH POPULATION */ exit(0);} /* END OF MAIN */

Queueing Models 61

MEAN VALUE ANALYSIS

////////////////////////////////////////////////////////////////////////////////// Print the results of each loop////////////////////////////////////////////////////////////////////////////////

void print_the_numbers(int n) { int i;

printf( "\n" ); printf( "Iteration: %d System Response Time: %f Throughput: %f\n", n, system_response_time, throughput ); for ( i = 0; i < number_of_centers; i++ ) printf( "Center: %d Residence Time: %f Queue Length: %f\n", i, residence_time[i], q_length[i] );} // END OF PRINT_THE_NUMBERS

Queueing Models 62

MEAN VALUE ANALYSISEXAMPLE OF THE SOLUTION OF A CLOSED MODEL:

Model Inputs: Vcpu = 121 Vdisk1 = 70Vdisk2 = 50Scpu = 0.005 Sdisk1 = 0.030 Sdisk2 = 0.027Dcpu = 0.605 Ddisk1 = 2.1 Ddisk2 = 1.35

= 0.3 jobs/sec

Model Structure:

Terminals

CPU

Disk1

Disk2

The INPUT numbers are the same as were used for the open model. We modify the picture so there are no “arrivals” or “departures”, but instead there are terminals – the jobs stay in the picture (which is what makes it closed.)

There are 3 terminal users with average think time of 15 seconds; N = 3, Z = 15.

Queueing Models 63



= 0.3 jobs/sec

To start with, we initialize the queue lengths to 0:

Qcpu = Qdisk1 = Qdisk2 = 0

Iteration Number 1:Rcpu = 0.605 sec Rdisk1 = 2.1 sec Rdisk2 = 1.35 sec.X = 1 / ( 15 + 4.055 ) = 0.0525Qcpu = 0.0318 Qdisk1 = 0.1102 Qdisk2 = 0.0708

Iteration Number 2:Rcpu = 0.624 sec Rdisk1 = 2.331 sec Rdisk2 = 1.446 sec.X = 0.1030Qcpu = 0.0643 Qdisk1 = 0.2403 Qdisk2 = 0.1490

EXAMPLE:

K

kk NRZ

NNX

1

)()(

)()()( NRNXNQ kk

kk DNR )(

)(1)( NADNR kkk

)1()( NQNA kk

Queueing Models 64



= 0.3 jobs/sec

Iteration Number 3:

Rcpu = 0.644 sec Rdisk1 = 2.605 sec Rdisk2 = 1.551 sec.X = 0.1515Qcpu = 0.0976 Qdisk1 = 0.3947 Qdisk2 = 0.2350

From these numbers we can calculate:

X(3) = 0.1515R(3) = ( N / X ) - Z = 4.74Q(3) = N - X(3) Z = 0.72

There’s a very nice example on Jain, pp 578-9.

EXAMPLE:

K

kk NRZ

NNX

1

)()(

)()()( NRNXNQ kk

kk DNR )(

)(1)( NADNR kkk

)1()( NQNA kk

Queueing Models 65

CONCLUSIONThis section has discussed:

Queueing Lingo - all the definitions you’ll ever need!

Queueing - especially the Single Queue in DETAIL

Analytical Models - how to solve a number of complex problems using the equations we already know and love. Works especially well for open models.

Operational Laws - drawing conclusions about limits - what to do when we can’t solve the math exactly.

Mean Value Analysis - Using the equations in an iterative fashion to solve closed models.

Documents

Queueing Models 1 Performance Engineering Prof. Jerry Breecher QUEUEING