1. 1 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management CSE4884 Network Design and Management Lecturer: Ken Fletcher Lecture 3 (part 2 of 2) Network Dimensioning NOTE: This presentation consists of two files.

2. 2 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Transaction Arrivals “Input Characteristics” Storage Hopper Queue in a Buffer Pool Behaviour of this area is called “Queue Characteristics” Metering Device “Service Facility” Characteristics here determine the queue size (and variability), and hence Queue Delays of items in the Queue or Buffer Pool or Hopper Conveyer Belt Output Rate On average will not exceed the slower of: a.arrivals; or b.metering rate. Service Facility Characteristics

3. 3 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Service Facility (1) n This is the area where most design work is possible. input characteristics are determined by external issues - –traffic loads and distribution characteristics are determined by the work to be done queuing characteristics may be design issues if you are in control of the software, for example –traffic queuing discipline may be a design parameter, but is usually dictated by the customer’s work or application software ‘Service facility characteristics’ for communications systems are: –the design line speed for output; and –number of output lines. Some factors for consideration...

4. 4 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Service Facility (2) - “Utilisation” n Average Utilisation = (time facility in use) / (time available) or (actual load) / (maximum possible load) eg. Shop with one server. –average time to serve each customer is 3 minutes, and –average of 12 customers per hour server is busy for a total of 36 minutes in the hour, Then utilisation (of the server) = (36) / (60) = 0.6 = 60%

5. 5 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Service Facility (3) - Service Times n Service times may be : –Constanteg Fixed length packet processing Time slices in a time sliced processor or –Variableeg Variable length messages or packets Disk I/O times Constant gives most consistent queue behaviour ‘Variable’ generally considered as random Most are somewhere between these two extremes, that is: most communications servers are smoother than random Treat them as if Random - this is a conservative approach

6. 6 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Service Facility (4) - Number of servers n May be any integer (or whole) number n Communications systems usually have single servers, but multiple server situations are not uncommon. n Consider multi server systems: When higher speed lines are needed, but not available When higher speed interface boxes are needed, but not available (or too expensive), such as gateways, encryptors, format converters; or In the checkout line at supermarkets, banks, and McDonalds.

7. 7 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Which is Better System & Why? A B C Multiple Single Server Queues NOTE: Moving between queues is not allowed! (Difficult with a heavy shopping trolley in a congested area) X Y Z Single Queue with Multiple Servers

8. 8 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management “All Servers Busy” n The formula for “all servers busy” in a multi-server system, and the graphs plotted by the spreadsheet Delays.XLS (in ‘Tools and Toys’ on the web page). (Assumes a random system, FIFO discipline, all servers equally loaded etc) n This function is the heart for all queueing work in this subject. (It is not necessary to remember or evaluate this formula)

9. 9 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Queuing Formulas n The formula ‘all servers busy’ is the heart of many other formulae used to estimate delays, queue sizes, etc. These are quite complex. –They are derived and discussed in Martin - ‘Systems Analysis for Data Transmission’ Chapter 31 Queuing Calculations. n For our approach, pre-plotted graphs are used rather than complex calculations. n n The spreadsheet Delays.XLS can be used to plot the most commonly used graphs. Others are in the handouts from Martin - Chapter 31

10. 10 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Graphs n The main graphs are of the form: Average q size per server See the class work sheets for other graphs. The model “Delays_?.XLS” (where ? is a version number) is an Excel spreadsheet which will calculate and print the three most common multi-server graphs. Reference this file through the ‘Tools and Toys’ area of the web page Utilisation

11. 11 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Some other graphs n The handouts from James Martin contain many other graphs. The main graphs are also available from the tool “Delays.xls” available via the Tools and toys, or under ‘Queueing Graphs’ on the lectures and tutorials page Eg Mean number of items in a queue (waiting and being served) Mean queueing times Effect of different dispatching disciplines Standard deviations of queueing times etc n Some of these will be shown on the OHP projector – but you can look these up for yourself on the references given.

12. 12 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Notes re Graphs n These graphs are based on assumptions – n General Assumptions (but check carefully on each graph): Random arrivals Random Service times All servers equally loaded in a multi server situation FIFO Dispatching. n Most of the graphs are of the same form: X axis is utilisation of the servers, Y axis is the number of items, usually on a ‘per server’ basis

13. 13 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Formulas for single server queues n These formulas apply to single server queues where the: input is random, ie random traffic arrivals service times are random, eg variable length transactions, and queuing discipline is First in-First out. n These formulas should be remembered as this is the most common queuing situation. n Average Queue Size (including the one being processed) = Utilisation / (1 - Utilisation) Low Utilisation indicates small average queue size High utilisation leads to large queue size Regardless of utilisation, –sometimes queue is empty, sometimes quite large –in theory could go to infinite length n Average queuing time (including processing)= Service time * (1 + number of queued items)

14. 14 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Effects of Utilisation & Multi Servers n Single Server Low utilisation (less than 0.2) –queue rarely builds, delays are minimal Medium utilisation (say 0.5) –av queue size, including one being served, = 1 element High utilisation (say 0.8) –average queue size = 4 elements n Multiple servers increases capacity and reduces variability of delays for any particular utilisation multiple servers reduce q size, and variability of delays formulae are frightening, (but surprisingly easy once you get into the style)

15. 15 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Simple Examples n For some mechanism Each action takes 0.05 seconds ie 20 actions per second Load is 15 actions per second Single Server n Find Utilisation, Average Queue Size, and Average Queue delay n Answers Utilisation (15/20 = 0.75) Av queue size (including element being serviced) –from graph (3 elements): by calculation 0.75/0.25=3 elements Av queue delay (including element being serviced) from graph –(4 service times = 4 * 0.05 = 0.2 seconds)

16. 16 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Simple Examples (2) n Now have 2 servers, each capable of 10 actions per second n Answers Find utilisation (15/20) = 0.75 Average queue size (including elements being serviced) =1.6 * 2 3.2 elements Average delay 2.3 service times = 2.3 * 0.1 =0.23 seconds n Summary delay is slightly longer, but variability is less. WHY?

17. 17 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Simple Examples (3) n Now 2 servers each able to handle 20 actions per second n Answers Find utilisation (15/2/20)=0.375 Average queue size 0.5 * 2 = 1 element Average delay 1.2 * (1/20) = 0.06 seconds n Summary Doubling number of high speed servers reduces average delay to one third.

18. 18 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Comparison of Mechanisms n TopicCase 1Case 2Case 3 n Load, actions per second151515 n Number of Servers1 22 fastslowfast n Service time on a server, seconds0.050.10.05 n Total Capacity actions per second202040 n Utilisation0.7500.7500.375 n Average Queue Size33.21 n Average Delay seconds0.20.230.06

19. 19 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management A Useful Formula for link Design. n The normal formulae and graphs for single server queues are adequate for most circumstances where analysis of an existing system is required. n During design work, we often do not know the utilisation until the end of the design, and hence have to design re-iteratively until a suitable solution is found. The following formula is useful in this circumstance. n Note: Assumptions of this derivation: Single server random arrivals FIFO queuing random service times

20. 20 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Derivation of Formula (1) n Let: D = Allowable average delay seconds L = Average Size of Data item (packet, block, etc) bits k = Line speed required to transmit L bits in D secondsbps  k = L/D bits per second n To allow for network congestion and the associated queuing delays: Let: Z = Average Line Loading bps K = Actual Line Speed that will be neededbps p = Utilisation of the line (when running at speed K) = actual/possible = Z/K Average Service Time = L/K seconds (Continued)

21. 21 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Derivation of Formula (2) Average queuing Time = Average Service Time * ( 1 + ( Utilisation / (1- Utilisation ) ) = ( ( L / K ) * ( 1 + ( p / ( 1 - p ) ) ) ) = ( ( L / K ) * ( 1 / (1 - p ) ) ) = ( L / (K ( 1 - p ) ) seconds Actual Line Speed Required K = k * ( 1 / ( 1 - p ) )  K = ( L / D ) * ( 1 / ( 1 - p ) )  K = ( L / D ( 1 - p ) )  K ( 1 - p ) = L / D  K - K p = L / D But p = Z / K  K - K * ( Z / K ) = L / D  K - Z = L / D  K = ( L / D ) +Z bits per second

22. 22 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Derivation of Formula (3) n In narrative: –Without allowing for line errors, the notional line speed required is equal to the sum of the average line loading (bits per second) plus the line speed required to transmit an average block or packet within the required or allowable delay time. n In other words –The notional line speed required is: the line load (in bps) (AKA background load) plus the line speed (in bps) needed to process an average transaction within the desired time, assuming that there is no other traffic

23. 23 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Example usage of this Formula: n Example task Average line loading from all sources Z = 100 bytes per second = 800 bits per second D = Allowable delay is three seconds L = Average block size is 375 bytes = 3000 bits n Required line speed Min. Line Speed Required = ( L / D ) + Z bps = (3000/3) + 800 bps = 1800 bps Do you believe it could be this easy? see next page

24. 24 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Check Formula by Normal Approach. n Previous sheet estimated that 1800 bps was adequate Now that we know the ‘answer’ of 1800 bps, we can check it using normal formula or graphs n Utilisation p = 800/1800 = 0.444 n Average Service time = 3000/1800 seconds = 1.667 seconds n Queuing Delay= Service time * (1 + (p/(1-p))) = 1.667 * (1 + (0.444 / (1-0.444))) = 1.667 * (1 + 0.799) = 3 seconds (as is required) (See, delay systems queuing theory is easier than people think it is)

25. 25 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management In Practice n This formula can be applied to a communications design problem, in order to estimate the required minimum line speed. However, as lines and modems are normally usable only at certain discrete speeds (eg 2400, 4800, 9600, 14400, 19200, 28800, 33400, 48000, 64000 bps etc) the designer must use the next higher available speed. 1.First estimate the minimum line speed required using the formula or graphs, or other method 2.Then choose the next higher available speed, and 3.Finally, estimate the average delays inherent in the system by recalculating using the chosen line speed and estimated loadings.

26. 26 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Simple Queuing Exercises (1) n Problem: Total loading is 500 bytes per second Line speed available is 9600 bits per second Average transaction size is 1000 bytes Exercise: Write down your assumptions Estimate average delay A. Using the traditional methods B Using the derived formula discussed previously

27. 27 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Simple Queuing Exercises (2) n Problem: total load on a link is 4000 bits per second Average transaction size is 8000 bits Average transaction to be processed within 4 seconds. n Exercise: Estimate minimum line speed required Write down your assumptions

28. 28 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Simple Queuing Exercises (3) n Problem: Communications link supports 30 terminals Average transaction rate is 1 per minute per terminal Average transaction size is1000 bytes (includes all overheads) Each entire transaction to be handled within average of 4 seconds n Estimate and write down: minimum line speed required line speed selected for use average delay when using selected line speed

29. 29 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Simple Queuing Exercises (4) n Problem: A link has an average response of 5 seconds. Average transaction size is 1000 bytes Known nominal line speed is 9600 bps n Exercise: Estimate the utilisation

30. 30 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Simple Queuing Exercises (5) n Exercise 5a A link is operating a some speed, say 64Kbps, with a utilisation of 60%. What is the effect on response time if the network is taken to the next speed, say 128Kbps. (ie speed is doubled)? n Exercise 5b A link carrying a certain load, operates at 30% utilisation. What happens to response time when the load is doubled, but link speed remains the same? n Exercise 5c What conclusions can you draw from these two exercises

31. 31 Copyright Ken Fletcher 2004 Australian Computer Security Pty Ltd Printed 26-May-16 07:39 Prepared for: Monash University Subj: CSE4884 Network Design & Management Answers to the exercises n The answers to these exercises are on the web pages – see n ‘Examples’ under lecture 3 entry on ‘Lectures and Tutorials’ page