Quantitative Aids to Decision Making

1

QUANTITATIVE AIDSTO

DECISION MAKING

FOREWORD

People in every walk of life are required to make decisions,irrespective of their level or function and an officer in thedefence services, is no exception. In planning of an operationwhich includes selection of objectives, allocation of resourcesto tasks and scheduling of activities against a time frame, heis to select the most suitable option amongst the variousalternatives. Further, during the implementation freshdecisions in respect of readjustments, reallocations and useof reserves are to be made.

An important and essential ingredient to effective decisionmaking is accurate and timely information. Statistical tech-niques offer powerful tools for sifting raw data and extractingprecise and relevant information. When quantified, theinformation is precise and less vulnerable to distortions andmisinterpretations. Armed with options generated throughquantitative techniques, a decision-maker taking into accountsuch qualitative information that has a bearing on thedecision would make an effective and right decision.

The increasing application of quantitative methods todaymakes it imperative for anyone in the military or otherwiseto have a basic understanding of the available moderndecision-making tools. A comprehension of the fundamentals

enables him to identify, situations in which quantitativemethods might be usefully applied, as well as to appreciatethe type and quality of informational inputs required forvarious decision-making models. With increasing availabilityof analytic software and computer awareness, organisationshave ample opportunity to apply quantitative models in theirdecision making.

Operations Research, as a problem solving technique mustbe viewed as both, a science and an art. The science aspectlies in providing mathematical techniques and algorithms forsolving appropriate decision problems. It is an art becausesuccess of the solution of a mathematical model in all itsphases depends on creativity and personal ability of thedecision-making analyst.

The handbook, ‘Quantitative Aids to Decision Making,attempts to give the reader an exposure to various ORtechniques and their application through few Defence relatedexamples. Deliberate effort has been made not to delve intocomplex mathematical derivations, and to focus on theconcepts and formulation of the problems. The handbook, in1982, was initially prepared by Col MR Narayanan of thefaculty. Gp Capt SK Manocha of R&C Faculty has developedthe additional chapter on network flows. Credit is also due tohim for the illustrative presentation and graphics included inthis edition.

College of Defence ManagementBolaram PostSecunderabad - 500 010

6th October 1997

(NP Sambhi)Brigadier

Officiating Commandant

PREFACEPREFACEPREFACEPREFACEPREFACE

The handbook on ‘Quantitative Aids to Decision Making’ isdesigned to provide the reader with an introduction to someof the important quantitative methods that are available toaid the decision making. The illustrative problems used toexplain the techniques have been deliberately kept simplewith sole purpose of putting them in a straightforward andunderstandable manner. Many of the problems and examplesin the handbook are adopted for military application, but areequally relevant to management of resources anywhere. Someof the examples, to the doubting Thomas, regarding thepotential of the quantitative techniques may appear to be toosimplistic and not real-life solutions. But these should be takenas the springboard to tackle much larger and more complexproblems encountered in real life. However, for solving reallife problems an integral and systems approach is required.

First section of the book has six chapters on statistics.Starting with basic concepts of the discipline, this sectionthrough examples, explains different facets of datapresentation and analysis field of descriptive statistics. In thearea of inferential statistics fundamentals of probability theoryand some important distributions have been dealt. Thesection finally gives a brief exposure on sampling andhypothesis testing.

Diversity of the situations where Operations Research canbe applied is unfathomable. Specific approach adopted andtechnique selected to solve any complex problem dependson the creativity and ingenuity of the analyst. Therefore, it isimportant that the nuances and limitations of the mathemati-cal model are well understood by both the users and ORpractitioners. The viewpoint that an OR user need not learnabout the mathematics of OR because computer can ‘take care’of solving the problems, is dangerous. Computer solves themodel as presented by the user. Therefore, the user must be

aware of the process of formulation and conceptualisation ofthe problem. Second section of the handbook attempts toprovide this basic knowledge.

The section on Operations Research opens with a brief overviewof the subject. Some of the commonly applicable techniques havebeen covered without going Into anycomplex mathematical treatment. That is why Simplex method oflinear Programming has not been covered, but the problem formu-lation has been taken up. A new chapter on Network Flows has beenincluded in this edition to present thepotential of some simple algorithms which can be applied to reallifeproblems and affect appreciable saving in resources.

It is hoped that this handbook will be of value to thosegetting first exposure to OR as well as to others as a readyreference for the basics on the subject.

CONTENTSCONTENTSCONTENTSCONTENTSCONTENTS

SECTION - ISECTION - ISECTION - ISECTION - ISECTION - IChapterChapterChapterChapterChapter

1. Statistics for Management - Basic Concepts

2. Data Presentation and Analysis:

Frequency Distribution

3. Data Presentation and Analysis:

Measures of Central Tendency and Dispersion

4. Probability Concepts

5. Probability Distributions

6. Inferences Based on Sample Information

SECTION - 2SECTION - 2SECTION - 2SECTION - 2SECTION - 2

7. Operations Research: Introduction

8. Linear Programming

9. Transportation Model

10. Decision Theory

11. Queueing (Waiting Lines)

12. Monte Carlo Simulation

13. Network Flows

Bibliography

SECTION 1SECTION 1SECTION 1SECTION 1SECTION 1

CHAPTER 1

STATISTICS FOR MANAGEMENT -BASIC CONCEPTS

IntroductionIntroductionIntroductionIntroductionIntroduction

Decision making, at commanders level, is becoming moreand more complex, involving comprehension of a largenumber of interrelated/interdependent factors impinging onthe decision process. Simultaneously, the commanders arealso faced with the problem of ever increasing number ofalternatives to choose from. Under the circumstances,decision making, purely based on past experience, judgementand intuition, has become rather difficult. Moreover, theresearchers are of the view that the human mind, generally,is not capable of perceiving in all details more than sevenparameters, on an average, at a time.

Statistical techniques offer a useful and powerful tool inthe decision making process, by their ability to analyse andinterpret quantitative information in a scientific and objectivemanner as also in providing certain conceptual framework tothe decision maker, to enable him to comprehend qualitativeinformation in a more objective way.

Scope of StatisticsScope of StatisticsScope of StatisticsScope of StatisticsScope of Statistics

Statistics is concerned with scientific methods of collecting,organising, summarising, presenting and analysing data, and what

is even more important, is drawing valid conclusions andmaking reasonable decisions based on such analysis.

Importance of StatisticsImportance of StatisticsImportance of StatisticsImportance of StatisticsImportance of Statistics

Whatever be the field of human endeavour, complete in-formation can seldom be obtained due to high costs and timefactor.

In real life, partial information forms the basis of most ofour decisions. Statistical techniques enable us to:

q Identify what information or data is worth seeking.

q Decide when and how judgements may be made onthe basis of partial information.

q Measure the extent of doubt and uncertainty.

The key distinction between normative techniques andstatistical techniques lies in the fact that statistical methodsare explicit in nature and provide a clearly defined measure oferror. For example, when we say, ‘With 90% confidence it canbe said that the mission will succeed”, thereby implying thatnine out of ten times the mission will succeed, we arequantifying the error that one out of ten times the statementmay turn out to be false.

CLASSIFICATION OF STATISTICAL METHODSCLASSIFICATION OF STATISTICAL METHODSCLASSIFICATION OF STATISTICAL METHODSCLASSIFICATION OF STATISTICAL METHODSCLASSIFICATION OF STATISTICAL METHODS

Statistical methods may be classified as Descriptive orInductive, the latter pertaining to statistical inference.

Descriptive StatisticsDescriptive StatisticsDescriptive StatisticsDescriptive StatisticsDescriptive Statistics

When statistical methods are used, a problem is always formu-lated in terms of ‘population’ or ‘universe’- defined as all the ele-ments about which conclusions or decisions are made. For example,suppose we want to know the ‘satisfaction’ level of our naval officersin the present appraisal system, all the naval officers of ournavy would represent the population about whom conclusion

is to be drawn. Similarly, if general opinion of the officers onthe usefulness of a particular course is required, then all theofficers on that course would constitute the universe or thepopulation.

Census.Census.Census.Census.Census. If information or data is taken from. each and ev-ery element of the universe to reach a conclusion about thepopulation we are dealing with ‘Descriptive Statistics’. Insurvey language, this process of eliciting data or informationfrom every element of the population or universe is called‘census’.

Statistical InferenceStatistical InferenceStatistical InferenceStatistical InferenceStatistical Inference

Another way conclusions or decisions are made, is throughthe use of a portion or sample of elements from the universe.The sample data is analysed, and based on the sampleevidence, conclusions are generalised about the target popu-lation. For example, to assess the satisfaction level of the navalofficers, we may take a representative sample of officers atrandom and based on the limited observations (opinions) takenfrom this sample, we would draw conclusions about the popu-lation of all the naval officers of the Indian Navy. This methodis referred to as ‘Statistical Inference’ ie estimation of thecharacteristics of a large group (or population), based onlimited observations of a small portion of it.

Sampling.Sampling.Sampling.Sampling.Sampling. Since only a portion of the population isstudied, this method is called sampling as against censuswhere each and every element of the population is studied.

Merits and Demerits of Sampling. Merits and Demerits of Sampling. Merits and Demerits of Sampling. Merits and Demerits of Sampling. Merits and Demerits of Sampling. Some of the importantmerits and demerits of sampling are given below:

q Saving in Cost and Time. Cost of a census is oftenprohibitive, both in terms of rupee cost and time Forexample, if the Personnel Branch has to collectdata from each and every officer of the serviceconcerned, regarding their confidence level in theexisting appraisal system, it is obvious that the

process will be tedious, time-consuming and costly. Byusing sampling techniques, however, the sameconclusion may be obtained by interviewing a relativelysmall representative sample of officers at much lowercost and time.

q Limited Information. During war, time may not beavailable to obtain full information, and decisions are tobe taken based on partial or sample information.

q Destructive Testing. In a number of problemsinvolving quality control, destructive testing of arepresentative sample is resorted to for ascertainingthe quality of a batch. For example, while ascertainingthe quality of a batch of ammunition manufactured inan ordnance factory, destructive testing of only a ran-dom sample of cartridges is carried out. Obviously onecannot carry out destructive testing of each and everycartridge.

q Accuracy. When we draw conclusions about thepopulation based on sample evidence, there will be anelement of uncertainty or error in our conclusion. Thisis called sampling error. Statistical methods have beengenerated for deciding:

q What would be the sample Size consistent withthe population pattern /dispersion and predeter-mined acceptable error in the estimate of theparameter related to the problem.

q What is the degree of uncertainty due to sampling.

Many of the methods of Statistical Inference are based uponresults that have been taken from the field of probabilitytheory, for coping with the sampling error.

Statistical InvestigationStatistical InvestigationStatistical InvestigationStatistical InvestigationStatistical Investigation

Statistical Investigations may be broken down into following stages:

q Formulation or definition of the problem.

q Collection and validation of data.

q Classification, tabulation and description of results.

q Analysis and interpretation.

q Formulation or Definition of the Problem. This is the mostimportant stage in the statistical investigation of aproblem as perceived, with a view to arrive at the realproblem (based on cause-effect chain) and then to decideon the information and/or data required.

q Collection of Data. The next step in statistical investiga-tion is the collection and validation of data. It is a timeconsuming and costly component of the statistical process.Validation of data is a pre-requisite for problem definitionand solution.

q Classification, Tabulation and Description of Results.Collected data to be of any use must be classified in a sys-tematic manner and presented in the form of tables or asgraphs and charts to enable the decision maker tocomprehend the information quickly.

q Analysis and interpretation. If the universe or populationhas been studied’ and data collected, decisions can bearrived at using techniques of descriptive statistics. Ifgeneralisations or inferences are to be made for thepopulation based on the study of a sample or samples,theories and techniques of inductive statistics apply.

CHAPTER 2CHAPTER 2CHAPTER 2CHAPTER 2CHAPTER 2

DATA PRESENTATION AND ANALYSIS:FREQUENCY DISTRIBUTION

Data and Information

Data or quantitative information is normally obtained as under:

q Continuous DataContinuous DataContinuous DataContinuous DataContinuous Data. Obtained as a result of measurementeg, distance, timings, weights of men and material, dimen-sion of obstacles - both artificial and natural. The data ob-tained as thus is referred to as ‘continuous’ as it can as-sume all possible values, subject to the limitations of theaccuracy of the measurement ordered or the measuringdevice used.

q Discrete Data.Scoring- eg range classification results, appraisal data ofofficers based on an accepted scoring norm.

Counting - eg number of accidents, missile kills, failures ofweapons sub-systems, number of mines in a strip, number ofprisoners of war.

q Ranking - eg listing in order of preference the variouscourses of actions based on qualitative criteria.

Counted, or ranked data is called ‘discrete’ as it canassume certain whole number values only and not thefractional values.

Frequency Distribution

If we want to obtain any meaningful conclusions from data,discrete or continuous, we must first organise the data properly

and then be in a position to describe the data, its pattern ordistribution, so as to render the large number of observaionsreadily understandable. Since this is the first step with anyset of data, we will consider in detail this aspect of organisingdata with an example.

Monthly Flying Hours. Let us say that it is required to findthe monthly flying hours achieved by a squadron. For thispurpose, the monthly flying hours achieved by fivesquadrons during 20 months from November 1995 to June1997 was obtained. Table 2-1 gives the figures as recorded inthe sequence in which they were obtained.

Table 2-1: Number of Monthly Flying Hours Achieved by FiveTable 2-1: Number of Monthly Flying Hours Achieved by FiveTable 2-1: Number of Monthly Flying Hours Achieved by FiveTable 2-1: Number of Monthly Flying Hours Achieved by FiveTable 2-1: Number of Monthly Flying Hours Achieved by FiveSquadrons During 20 MonthsSquadrons During 20 MonthsSquadrons During 20 MonthsSquadrons During 20 MonthsSquadrons During 20 Months

A scrutiny of the table at first reveals very little. We mayperhaps be able to locate the maximum and minimum valueswith a little difficulty.

To get more information from these observations, it mustbe organised in some systematic fashion. The simplestdevice is to form an array which is an orderly arrangement ofvalues of items regarding the magnitude. These values caneither be arranged in an ascending or descending order. Theobservations shown in Table 2-1 has been re-arranged inascending order in Table 2-2

Table 2-2: Array of Flying Hours Achieved - Ascending OrderTable 2-2: Array of Flying Hours Achieved - Ascending OrderTable 2-2: Array of Flying Hours Achieved - Ascending OrderTable 2-2: Array of Flying Hours Achieved - Ascending OrderTable 2-2: Array of Flying Hours Achieved - Ascending Order

Such an array gives a slightly better picture of the datathan the original set of observations. For example, a quicklook at the array shows that the minimum flying hoursachieved is 124 and the maximum 268. Thus we draw theuseful conclusion that the range of the set of observations is144 hours. When crucial decisions have to be made on statis-tical studies, the number of observations is usually large andarranging them in an array is time consuming and unwieldy.To grasp the significance or understand the underlying pat-tern inherent in such massive data, the usual technique is tocondense such data by suitably dividing the range of theobservations. Such a tabulation of observed data into classes

is referred to as a ‘Frequency Distribution’ Each group ofobservations is, called a class, which is bound by an upperand a lower limit. By judicious selection of the number ofclasses, the inherent pattern of the data is revealed. Thepattern is generally of great interest to the decision maker.An example of a tabulated frequency distribution relating tothe number of monthly flying hours achieved in 100squadron months is given in Table, 2-3.

Table 2-3: Frequency Distribution of Monthly Flying HoursTable 2-3: Frequency Distribution of Monthly Flying HoursTable 2-3: Frequency Distribution of Monthly Flying HoursTable 2-3: Frequency Distribution of Monthly Flying HoursTable 2-3: Frequency Distribution of Monthly Flying HoursAchieved by SquadronsAchieved by SquadronsAchieved by SquadronsAchieved by SquadronsAchieved by Squadrons

Graphical Representation of the frequency TableGraphical Representation of the frequency TableGraphical Representation of the frequency TableGraphical Representation of the frequency TableGraphical Representation of the frequency Table

Frequency distributions may be presented graphically soas to give an effective presentation of the pattern of the data.Histogram and frequency polygon are the two common meth-ods used for this purpose. For the data above, histogram andfrequency polygon are shown in the Figures 2-1 and 2-2,respectively. Histogram, which is like a bar chart shows thefrequency of each class clearly, whereas the frequencypolygon presents a more effective picture of the variation inthe distribution of data. The latter can also be used for a moremeaningful comparison of related sets of data, eg: range

Figure 2-2: Frequency PolygonFigure 2-2: Frequency PolygonFigure 2-2: Frequency PolygonFigure 2-2: Frequency PolygonFigure 2-2: Frequency Polygon

classification performance of two units, rocket firing resultsof various air force squadrons etc. If instead of the absolutevalue of the frequency, the relative frequencies are used, theshape of the histogram will be identical except that the scaleon the Y axis will be reduced by n times, ie to 1/n, where n isthe total number of observations.

Figure 2-1: HistogramFigure 2-1: HistogramFigure 2-1: HistogramFigure 2-1: HistogramFigure 2-1: Histogram

Frequency Curve.Frequency Curve.Frequency Curve.Frequency Curve.Frequency Curve. We may also draw a smooth curve.instead of the frequency polygon. This is known as a frequencycurve (Figure 2-3) and is extensively used to analyse andinterpret data mathematically. When the curve is drawn forrelative frequency instead of absolute values, the area underthe curve is treated as unity and we may obtain the relativefrequency between any two values on the X axis by the areabounded by the two ordinates corresponding to these points.

Figure 2-3: Frequency CurveFigure 2-3: Frequency CurveFigure 2-3: Frequency CurveFigure 2-3: Frequency CurveFigure 2-3: Frequency Curve

Skewness and KurtosisSkewness and KurtosisSkewness and KurtosisSkewness and KurtosisSkewness and Kurtosis

The frequency curve is characterised by skewness andkurtosis. These are explained below:

Skewness.Skewness.Skewness.Skewness.Skewness. Frequency distribution that is not symmetricalie extreme values at one end and not balanced at the otherend. The distribution may be right skewed or left skewed.

q Right Skewed(+ve).

Extreme values in the direction of higher measurementtail to right, as shown in Figure 2-4.

q Left Skewed (-ve)

Extreme values in the direction of lower measurement Tailto left as also shown in Figure 2-4.

Figure 2-4: Skewness - Left and RightFigure 2-4: Skewness - Left and RightFigure 2-4: Skewness - Left and RightFigure 2-4: Skewness - Left and RightFigure 2-4: Skewness - Left and Right

Kurtosis. Kurtosis. Kurtosis. Kurtosis. Kurtosis. Refers to degree of peakedness. There are threetypes as under and are depicted in Figure 2-5:

q Platy-Kurticq Meso-Kurticq Lepto-Kurtic

Figure 2-5: KurtosisFigure 2-5: KurtosisFigure 2-5: KurtosisFigure 2-5: KurtosisFigure 2-5: Kurtosis

Ogives (Cumulative Frequency Diagrams)Ogives (Cumulative Frequency Diagrams)Ogives (Cumulative Frequency Diagrams)Ogives (Cumulative Frequency Diagrams)Ogives (Cumulative Frequency Diagrams)

Very frequently, it is desired to show in diagrammatic form, notthe frequency or relative frequency of occurrence in the variousintervals, but the cumulative frequency above or below a givenvalue. For example, we may wish to be able to read off from the chartthe number of squadron months where number of flying hours

achieved was not less than a given number of flying hours.Charts of this type are called ‘Cumulative FrequencyDiagrams’ or ‘Ogives’, pronounced as ‘Ohjives’. These are oftwo types, viz ‘Less Than’ curves and ‘Or More’ curves. Thetable and chart for the two curves are shown at Table 24 andFigure 2-6.

Table 2-4: Cumulative Relative FrequencyTable 2-4: Cumulative Relative FrequencyTable 2-4: Cumulative Relative FrequencyTable 2-4: Cumulative Relative FrequencyTable 2-4: Cumulative Relative Frequency

Figure 2-6: Cumulative Frequency Diagram(Ogives)Figure 2-6: Cumulative Frequency Diagram(Ogives)Figure 2-6: Cumulative Frequency Diagram(Ogives)Figure 2-6: Cumulative Frequency Diagram(Ogives)Figure 2-6: Cumulative Frequency Diagram(Ogives)

CHAPTER 3

DATA PRESENTATION AND ANALYSIS:MEASURES OF

CENTRAL TENDENCY AND DISPERSION

In Chapter 2, we have seen how a mass of unconnecteddata can be transformed into a more readily comprehensibleform by means of frequency distribution. But even this maynot be adequate. We should be able to present a gist of thedata in a nutshell. A set of observations can be described sat-isfactorily by

q Measures of central tendency.

q Measures of dispersion.

Measures of Central TendencyMeasures of Central TendencyMeasures of Central TendencyMeasures of Central TendencyMeasures of Central Tendency

Generally, all data tend to ‘cluster’ around a point in themiddle region of its range. The measures by which this cen-tral tendency is described are:

q Mean.

q Mode.

q Median.

Mean (m). Is also referred to as Arithmetic Mean orAverage. Given a number of values ‘Mean’ is easily obtainedby adding the individual values and dividing the total by thenumber of values.

Similarly, mean for the squadron monthly flying hours worksout to 205.

q Merits and Demerits of Mean. It is the most familiarconcept and is easily understood. One major advantageof mean is that it is the most reliable measure when wedeal with samples for estimating the average value ofthe population. However, one has to be careful in usingand interpreting average (mean) as representative valuefor drawing conclusions. The pitfalls of using mean valuealone for decision making purposes can be grasped fromthe following examples:

q For a beach landing operation, a site was selectedcarefully where average depth of water was 110cm. The troops selected for this operation had anaverage height of 150 cm. When the troopsactually landed at this site, more than 40% had tobe rescued from drowning.

q There are number of examples where the staff haveworked out the logistics provisioning based onaverage requirements, resulting in stockouts on50% of the occasions.

Reasons for the above can easily be understood. Average valuesvery often can completely distort reality if extreme low or highvalues occur in the data. While arriving at average depth of 110 cm

with 15 or 20 readings in the proposed landing site, even threeor four values less than 20 cm would have pulled the averageto 110 cm although a large number of readings could havebeen 150 cm or more.

Median.Median.Median.Median.Median. Median is the value of the middle item of theseries of observations when arranged in any array.

Examples:

q Number of values is odd 3, 4, 4, 5, 6, 8, 8, 8, 10

Median value = 6

q Number of values is even 5, 5, 7, 9. 11, 12, 15, 16

Median value = = 102

q Merits and Demerits of Median.

q It is not affected by extreme values.

q It is easy to determine because it is the mid value.

q It may be graphically arrived at.

q Even for open ended distributions, median can bedetermined.

q It is a positional measure.

ModeModeModeModeMode. Mode is the value that occurs most often and aroundwhich the values of other items tend to cluster. When data isgrouped and represented pictorially, mode will stand-out hav-ing the maximum height.

q Merits and Demerits of Mode.

q It is also not affected by extreme values and is easilydetermined.

q It is the least precise of the three measures of centraltendency

q It is difficult to calculate unless data is grouped.

q It is the least understood and has limited statistical use.

9+ 11

Measures of DispersionMeasures of DispersionMeasures of DispersionMeasures of DispersionMeasures of Dispersion

Measures of central tendency like mean, are onlyrepresentative single value of the data or observations. Theydo not indicate how the data is dispersed. When we say thataverage height of a jawan in a Gurkha unit is 140 cm, it is onlya typical figure. But it does not tell us how the heights of allthe jawans are distributed relative to this figure. It could bebetween 120 cm and 170 cm or perhaps 135 cm and 155 cm.

Let us consider another example. Consider two units, eachwith say 100 firers. Mean score of both may be 20, out of amaximum possible 30. While in one unit there may be a widedifference in scores ranging from 10 to 30, in the other, thedispersion may be less - say from 15 to 25. Describing theperformance of the unit on mean value alone will not berealistic, as it conceals the difference in the individualsstandards. If the desired minimum score is 18, there shouldbe more number of failures in the unit where dispersion inscores is from 10 to 30.

We, therefore, require a measure of dispersion. Smallerits value would imply greater uniformity in the data. Twomeasures of dispersion will be considered here:

q Range

q Standard Deviation.

Range.Range.Range.Range.Range. It is the difference between the highest andlowest values in a set of data.

q Merits and Demerits of Range.

q It is well understood and easily computable.

q Its main disadvantage is the emphasis on the extremevalues only, not being, based on each and every item.

q It cannot tell about the character of the distribution.

q It cannot be computed for open ended distributions.

q Uses. Being simple, it is extensively used in statistical qual-ity control in industry for Process control’. Another appli-cation area is for computing share values in the stock mar-ket.

Standard Deviation.Standard Deviation.Standard Deviation.Standard Deviation.Standard Deviation. This is the most important and math-ematically precise measure of variability or dispersion. It isalso called ‘Root Mean Square Deviation’ because it is thesquare root of the squared deviations from the mean. Let ustry to understand this definition with an example. Considerthe data shown at Table 3-1.

Table 3-1: Calculation of Standard DeviationTable 3-1: Calculation of Standard DeviationTable 3-1: Calculation of Standard DeviationTable 3-1: Calculation of Standard DeviationTable 3-1: Calculation of Standard Deviation

In the above example if the values were, say 30, 35, 40, 45and 50, mean (m) would be 40. But We would intuitively realisethat this data has less dispersion or variation with respect tomean. Standard deviation (s) in this case works out to 7.07 or7 approximately.

q Advantages of Standard Deviation. The standarddeviation is a comprehensive measure of the variability,since it takes into account every item in the series. Besides,it has mathematical properties and is used extensively inadvanced statistical theory.

q Disadvantages of Standard Deviation. The major disad-vantage is that it is an abstract measure and a little difficultto understand. However, one needs to realise that itrepresents a kind of average of the individual amounts ofvariability and greater the average, greater is the variabilityin the series. As in the case of the mean, the standarddeviation cannot be calculated for an openendeddistribution because of the lack of a mid-point in theopenended interval.

The exact significance of the standard deviation willbecome clear after a discussion of the ‘Normal Distribution’considered in Chapter 5.

Convention of SymbolsConvention of SymbolsConvention of SymbolsConvention of SymbolsConvention of Symbols

The statistical symbols used for different parameters inrespect of population and samples are at Table 3-2.

Table 3-2: Conventional Statistical SymbolsTable 3-2: Conventional Statistical SymbolsTable 3-2: Conventional Statistical SymbolsTable 3-2: Conventional Statistical SymbolsTable 3-2: Conventional Statistical Symbols

Data Analysis - Flying Hours ExampleData Analysis - Flying Hours ExampleData Analysis - Flying Hours ExampleData Analysis - Flying Hours ExampleData Analysis - Flying Hours Example

Let us take the flying hours data given earlier at Table 2-1as an example and based on various descriptive measures ofthat data set, examine various conclusions that could be made.

Treating all 100 observations as a data set, we can attemptto describe the overall data behaviour through the followingnumerical measures:

q Mean = 204.00 hrs.q Median = 205.13 hrs.q Mode = 206.56 hrs.q Range = 144.00 hrs.q Standard = 31.00 hrs

Deviation

The mean, median and mode being very close to eachother, the distribution could be treated as symmetrical (infact,however it is slightly left-skewed). As will be seen later inexplanation of the normal distribution, one can constructintervals about the mean in terms of standard deviations andconclude about the proportion of flying hours lying within suchintervals. For example, within an interval 204 ±31 hrs ie,between 173 hrs and 235 hrs, approximately 68% of theobservations would lie. And an interval of 204 ± 62 hrs wouldcontain about 95.4% of the observations.

If one has to evaluate the overall performance of all thesquadrons, one would like to question whether the meanmonthly flying hours achieved under the existing conditionsis acceptable? If so, the next question would be whether the‘dispersion’ as dictated by a standard deviation of 31 hrs isacceptable? The first question pertains to the mean perfor-mance standards while the second to the variability or con-sistency. If the performance is deemed to be unacceptable,measures could be thought of to enhance the ‘mean’ anddecrease the ‘standard deviation’. After implementation of such

measures, again similar data could be obtained and analysed.The new mean and standard deviation could then becompared with the original values to see if the differencesare significant or not ie, whether the differences are due torandom variability or due to the effect of the measures takento improve the performance. The above analysis, however, iscarried out using techniques of inferential statistics.

If one has to compare the performance of the fivesquadrons with one another, the approach will be different.The mean flying hours and the standard deviations for eachsquadron (ie for 20 data items each) have to be worked out.The results are shown in Table 3-3, here:

Table 3-3: Flying Effort Comparison of Five SquadronTable 3-3: Flying Effort Comparison of Five SquadronTable 3-3: Flying Effort Comparison of Five SquadronTable 3-3: Flying Effort Comparison of Five SquadronTable 3-3: Flying Effort Comparison of Five Squadron

One should not judge the squadrons’ performance bylooking at the mean alone. The standard deviation also has tobe examined simultaneously. In this example, Squadron 3appears to have the best performance with the highest meanand the lowest standard deviation. This combination may notalways happen. Therefore, both these measures are combinedinto one measure known as Coefficient of Variation.

Coefficient of Variation = ( ) X 100Standard Deviation

Mean

and wherever a higher mean value dictates better performance(this is not so if the variable under consideration is ‘time’ forperforming a specific task), the lower the value of thiscoefficient, the better is the performance.

The conclusion, based on the coefficient of variation, thatSquadron 3 has performed the best may not really be validbecause of random variability present in the measurementand/or observation. Secondly, a total of 20 observations incase of each squadron may not be adequate to provide us withreliable inference. Strictly speaking, the variations amongstthe individual means and standard deviations may not bestatistically significant. Or they may be. To test for this, inferential statistics provides suitable techniques and thestatistician carries out suitable tests before concluding aboutthe superiority of the squadron in question.

CHAPTER 4

PROBABILITY CONCEPTS


Uncertainty is an important element in most decisionmaking situations. There is uncertainty when we deal withhappenings in the future, for it is difficult to forecast the fu-ture with complete certainty. Again, there is uncertainty wheninformation is not complete, and decisions have to be takenwith limited information. The effect of uncertainty is all themore pronounced at higher levels because decision makingat higher levels involves looking deeper into the future,usually with less information.

Our need to cope with uncertainty leads us to the studyand use of probability theory - In fact, usages of the conceptsof the probability is part of our every day life. In personal aswell as managerial decisions, we face uncertainty and useprobability concepts, whether we admit the use of somethingso sophisticated or not. Perhaps we do so intuitively, largelysub-consciously. When we hear of a weather forecastindicating high likelihood of rain, when looking at thesky, our subjective assessment indicates such a likelihood,we drop the idea of a party out in the lawns and plan to organisethe same indoors. While playing bridge, we do make someprobability estimates say, before. attempting a finesse. Again,before putting in an attack, we do make a deliberate attemptto consider the chance of success of own forces. In fact, intactical appreciation, consideration of factors such as groundand relative strength and that of courses of action, is largelydevoted to estimating the likelihood of own success andevolving a plan that attempts tomaximise the same.

In so doing, what we are really delving in is the domain ofprobability.

Probability may be defined as likelihood of occurrence ofan event. It is represented on a zero-to-one interval scale.Probability of one indicates that an event is sure to take placeand zero, that its occurrence is impossible. Probabilities as-sociated with some events are shown in Figure 4.1. Amongthese, probabilty of 0.5 for a coin coming up heads on a tossis easily recognisable, for, there is a fifty percent chance ofexperiments leading to heads or tails.

Probability Event Probability Event Probability Event Probability Event Probability Event

Figure 4- 1: Probabilities of Some EventsFigure 4- 1: Probabilities of Some EventsFigure 4- 1: Probabilities of Some EventsFigure 4- 1: Probabilities of Some EventsFigure 4- 1: Probabilities of Some Events

Probability concepts help us in:

q Expressing ourselves more precisely, through a universallyaccepted language, thus enabling better communicaion.

q Finding answers to certain decision situations by use ofsimple probability laws.

q Inferring the characteristics of the whole populationfrom the data obtained from a sample.

Approaches to ProbabilityApproaches to ProbabilityApproaches to ProbabilityApproaches to ProbabilityApproaches to Probability

There are two approaches to probability estimation andinterpretation - objective and subjective. The objectiveapproach may be further sub-divided into the classical ap-proach and the empirical approach.

Classical Theory of Probability.Classical Theory of Probability.Classical Theory of Probability.Classical Theory of Probability.Classical Theory of Probability. The classical theorydefines probability (p) as:

Number of favourable outcomes

Total number of equally likely outcomes

Let us consider the probability of a coin coming up headson a toss. There are just two possible outcomes, either headsor tails. Assuming that the coin is unbiased, there is nosufficient reason to infer that either of these outcomes is morelikely to take place than the other. We thus have a situationwherein there is a total of two equally likely outcomes andthe probability of occurrence of one among these, ie heads,(or alternatively, tails) is thus given by p = 1/2 or 0.5. Simi-larly, the probability of a die showing a onespot on a roll is 1/6 as there is just one favourable outcome among a total of sixequally likely outcomes which are thrown up by thegeometry of symmetry or construction. Although theclassical theory helps in defining probability precisely, it can-not be used in most practical situations because it is notpossible to talk of total number of equally likely outcomes forwant of symmetry. The probability of an RCL gun shot hittinga tank at a specific range is one such case.

Empirical Methods.Empirical Methods.Empirical Methods.Empirical Methods.Empirical Methods. This approach is based on experimen-tation and helps in overcoming the limitation of the classicaltheory. Empirical probability may be defined as:

P =

Number of favourable outcomes

Total number of trials

To determine probability by this method, the experimenthas to be repeated until a stabilised value is reached. A largenumber of trials are thus required to be carried out. Thisaspect is illustrated by the results of an experiment shown atTable 4-1 wherein data obtained through repeated trials hasbeen tabulated and the cumulative probability of an ENTACmissile hitting a Patton tank at a range of 1500 mtrs has beenworked out as 0.7.

Table 4-1: Cumulative Probability of anTable 4-1: Cumulative Probability of anTable 4-1: Cumulative Probability of anTable 4-1: Cumulative Probability of anTable 4-1: Cumulative Probability of anENTAC Missile Hitting a Patton Tank at 1500 mtrsENTAC Missile Hitting a Patton Tank at 1500 mtrsENTAC Missile Hitting a Patton Tank at 1500 mtrsENTAC Missile Hitting a Patton Tank at 1500 mtrsENTAC Missile Hitting a Patton Tank at 1500 mtrs

Figure 4-2 depicts the cumulative probability obtained fromthe experiment in which trials have been repeated until astabilized value of probability is reached. In classical as wellas empirical approach, probability may be interpreted as thelong run relative frequency. Thus probability of 0.7 wouldindicate that if the experiment is repeated sufficiently largenumber of times, 70 percent of favourable outcomes orsuccesses may be expected on an average.

p =

Figure 4-2: Cumulative Probability of HitsFigure 4-2: Cumulative Probability of HitsFigure 4-2: Cumulative Probability of HitsFigure 4-2: Cumulative Probability of HitsFigure 4-2: Cumulative Probability of Hits

Subjective Probabil ity.Subjective Probabil ity.Subjective Probabil ity.Subjective Probabil ity.Subjective Probabil ity. There are situations whereexperimentation too is not feasible; take for example, theprobability of success in an attack that own forces areplanning to launch. In such cases, estimation of probability issubjective, based on informed judgement of the commander.In such an assessment, knowledge of whatever informationis available, experience of similar happenings in the past and,of course, good judgement on the part of the commander formimportant constituents. It thus represents the strength ofbelief of the decision maker that a given event would occur.Subjective probability estimates, although approximate, areuseful in employing analytical methods and can certainly berefined through practice and experience and through use ofthe Standard Gamble Technique (a formal method designedfor this purpose). An excellent example of subjectiveprobability is provided by the bookmakers odds on arace course.

Probability LawsProbability LawsProbability LawsProbability LawsProbability Laws

These help us in determining the overall probability of aset of events, once that of the constituent events are knownor established. There are two probability laws - multiplicativeand additive. The meaning of these terms and the usefulnessof these laws can be best illustrated by an example.

Say there are two firers, A and B, and their respective singleshot probability of hitting the target is p (A) = 0.8 and p (B). =0.7. Since a given firer can either hit or miss the target andthe sum of these probabilities must equal unity, the probabil-ity of the two firers missing the target would be p (A) = 0.2and p (B) = 0.3. If both A and B fire one shot each, what is theprobability of

q Both the shots hitting the target, ie p(A and B)?

q The target being hit, or in other words, at least oneshot hitting the target, ie p (A or B)?

p(A) = 0.8; p(A) = 0.2; p(B) = 0.7, p(B) = 0.30.7, p(B) = 0.30.7, p(B) = 0.30.7, p(B) = 0.30.7, p(B) = 0.3

Figure 4-3: Probability LawsFigure 4-3: Probability LawsFigure 4-3: Probability LawsFigure 4-3: Probability LawsFigure 4-3: Probability Laws

Both these situations are illustrated on the probabilityspace shown in Figure 4-3. The probability of both A and Bhitting the target is given by the Multiplicative Law, that is:

p(A and B) = p(A) x p(B) = 0.8 x 0.7 = 0.56

Similarly, the probability of at least one firer hitting thetarget is given by the Additive Law, that is:

p(A or B) = p(A) + p(B) - p(A and B)

= 0.8 + 0.7 - 0.56 = 0.94

An alternative approach to the Additive Law is provided by thefact that the requirement corresponding to p (A or B) is met unlessneither firer hits the targets. Thus

p (A or B) = 1 - P(A and B) = 1 - p(A) x p(B) = 1 - 0.2 x 1-0.06

= 0.94

A useful method of pictorially representing the variousevents in a sequential manner, so as to assist comprehen-sion, is provided by the probability tree. The situationrepresenting the above example in the form of a probabilitytree is depicted at Figure 4-4 here.

Figure 4-4: Probability TreeFigure 4-4: Probability TreeFigure 4-4: Probability TreeFigure 4-4: Probability TreeFigure 4-4: Probability Tree

The method indicated in alternative approach to theAdditive Law given earlier, is particularly useful when anexperiment is repeated a large number of times. We shallillustrate this with an example. Let us say an anti-aircraft gunfires 1000 rounds per minute and has a single shot hitprobability (SSHP) of 0.005. Two such guns, sited together toprotect a vulnerable point, engage an aircraft for 3 seconds.What is the probability of the aircraft being hit?

As long as even one shot hits the aircraft, the purpose isachieved.Thus, the required probability may be determinedas follows:

Number of rounds fired per gun in 3 seconds = 1000 60

= 50

Total number of rounds fired (by 2 guns) = 2 x 50 = 100,

SSHP = 0.005,

Probability of a shot not hitting the aircraft

= 1 - 0.005 = 0.995

Probability of all 100 shots missing the aircraft

= 0.995 x 0.995 x 0.995 ..... (100 times)

= (0.995)100 = 0.59

Hence, probability of the aircraft being hit = 1 - 0.59 = 0.41

In actual practice SSHP is much less than the figureindicated here. It is little wonder then that rarely is anaircraft downed by anti-aircraft guns. Of course, they doact as deterrents.

A General ApplicationA General ApplicationA General ApplicationA General ApplicationA General Application

It is seen from the previous examples that in order tocalculate ‘at least one hit one has to understand the event asa ‘complementary event’ to the event ‘no hit’or ‘all misses’and compute the probability by subtracting the probability of‘all misses from 1. This concept helps us to calculate theeffectiveness of number of weapons delivered and converselyto calculate the number of weapons to be delivered to achievea desired effectiveness. The example given belowillustrates this.

Let us consider a situation where ‘n’ missiles are fired, each having an SSKP (single shot kill probability) of 0.7. Thus the miss

x 3

probability is (1 - 0.7) ie 0.3. In general. the probability ofsuccess can be obtained by subtracting the total missprobability (ie probability of all misses) from 1. We cantabulate the results progressively as in Table 4-2.

Table 4-2: Calculation of Success ProbabilityTable 4-2: Calculation of Success ProbabilityTable 4-2: Calculation of Success ProbabilityTable 4-2: Calculation of Success ProbabilityTable 4-2: Calculation of Success Probability

The results can be graphed as shown in Figure 4-5.

The important conclusion from the above analysis is thatwith every additional launch, the increase in overall killprobability gradually diminishes. After a certain point anadditional missile does not increase the overall effectivenesssubstantially. In this example, while second missiles bringsabout an increase of 0.21 kill probability over a single missile,fourth missile results in an increase of only 0.019 over threemissiles. A military commander, therefore, has to view theproblem in this manner and decide whether the additionaleffectiveness is worth the expenditure of an

Figure 4-5: Number of Missiles vs Success ProbabilityFigure 4-5: Number of Missiles vs Success ProbabilityFigure 4-5: Number of Missiles vs Success ProbabilityFigure 4-5: Number of Missiles vs Success ProbabilityFigure 4-5: Number of Missiles vs Success Probability

additional weapon. In this case, expenditure on morethan 3 missiles may seem to be infructuous.

Similar arguments help us to calculate the number ofweapons necessary to achieve a desired overall killprobability. Suppose the SSKP is given as 0.5 and we desireto achieve an overall kill probability of 0.95 ie 95% satisfactionlevel, then we could proceed as follows:

Given, SSKP = 0.5; Thus, Miss probability = (1 - 0.5) = 0.5

If n is the number of missiles necessary,

then we may write,

1 - (0.5) = 0.95, and solve for n.

(0.5) = 1 - 0.95 = 0.05

Taking logarithms on both sides.

n log (0.5) = log (0.05)

log (0.05)n = = 4.32, log (0.5)

= 5.

n

n

Thus a total of 5 missiles would give us a total kill probabilityof more than 0.95, ie 96.875%. (1 -0.5n = 0.96875).

Dependent EventsDependent EventsDependent EventsDependent EventsDependent Events

So far, the discussion has been restricted to independentevents, that is, such events where occurrence (ornon-occurrence) of one does not affect the probability ofoccurrence of the other. Thus, the probability of a firerhitting the target is independent of the result of the other’sfire. In case of dependent events, however, occurrence of oneevent affects the probability of occurrence of the other. Forexample, the probability of an aircraft destroying a groundtarget would be dependent on whether the said target hasbeen provided air defence cover based on missiles. Let usassume the probability of destruction, when there are nomissiles, is 0.6 and that the same reduce to half thatvalue 0.3 when the target is protected by missiles. Also, say,our subjective assessment indicates a probability of 0.7 thatthe air defence is missile based. What then is the probabilityof destroying such a target?

Such a situation may be viewed as that composed of twoevents. Event A representing that the defence is missile based(A that it is not based on missiles) and Event B indicatingdestruction of the target (B that the target is not destroyed).The probability space corresponding to this situation is shownat Figure 4-6. The two areas shown shaded represent theprobability of destruction depending on whether the target isprotected by missiles or not. Each of these probabilities isobtained through the use of the Multiplicative Law. The overallprobability of destruction, whether missiles are there are not,is thus given by adding the two shaded areas and so equals0.39. This may also be called Marginal Probability ofdestruction.

An interesting off-shoot with regard to dependent events isprovided by Bayes’ Theorem which deals with updating the

Total p(Destruction) = 0.21 + 0.18 = 0.39Total p(Destruction) = 0.21 + 0.18 = 0.39Total p(Destruction) = 0.21 + 0.18 = 0.39Total p(Destruction) = 0.21 + 0.18 = 0.39Total p(Destruction) = 0.21 + 0.18 = 0.39

Fligure 4-6: Dependent Events - Probability ofFligure 4-6: Dependent Events - Probability ofFligure 4-6: Dependent Events - Probability ofFligure 4-6: Dependent Events - Probability ofFligure 4-6: Dependent Events - Probability ofDestruction of Ground Target by an AircraftDestruction of Ground Target by an AircraftDestruction of Ground Target by an AircraftDestruction of Ground Target by an AircraftDestruction of Ground Target by an Aircraft

probability of occurrence of one event based on informationobtained regarding another related event. Although thisconcept is of particular interest in defence applications, anadvanced treatment of this nature is beyond the scope of thishandbook.

ConclusionConclusionConclusionConclusionConclusion

In this chapter, we have dealt with some fundamentalprobability concepts including approaches to probability andprobability laws. The concept of dependent events has alsobeen briefly brought out. Although this knowledge would behelpful in tackling relatively simple problems, the reader isadvised to refer to any standard book on statistics for a morecomplete treatment. A broad understanding of these conceptsis also a pre-requisite for dealing with probability distributions that follow in the next chapter.

CHAPTER 5

PROBABILITY DISTRIBUTIONS


The probability of getting a head on toss of a coin is 0.5. Ifthe coin is tossed ten times, we should generally expect fiveheads. But we know that the outcome of such an experimentcould well be anything from zero at an extreme to ten at theother. Intuitively we could also say that the probability of anoutcome in the extreme ends of the range would be rathersmall, its value increasing as the number of heads approachesfive. We also know that the sum total of the probabilities overthe entire range of outcomes must equal unit. A plot of eachof these probabilities would tell us how the Probabilities aredistributed with respect to outcomes. Such a plot is called aprobability distribution.

Most of our planning is based on averages or means. Thetime to lay a bridge, useful life of an equipment and thepercentage of casualties to be expected in an operation of agiven intensity are just some examples. However, bydefinition, the mean implies a value that would be exceededabout half the time. In all such planning, therefore, ourassurance level is of the order of 50 per cent only. Surely, incritical areas we need a much higher assurance level thanthat. This, and many other useful aspects of planning can beeffectively dealt with through a knowledge of probabilitydistribution. Two distributions of particular interest to us arethe Normal and Poisson.

Normal DistributionNormal DistributionNormal DistributionNormal DistributionNormal Distribution

Perhaps the most important probability distribution instatistics is the normal distribution. This is because it isapplicable to a great many situations including humancharacteristics (weights, heights and lQs), outputs fromphysical processes (dimensions and yields) and othermeasures of interest to management. Its physical appearanceis that of a symmetrical bellshaped curve, extending infinitelyfar into positive and negative directions. Of course, mostreal-life populations do not extend for ever in both directions,but for such populations the normal distribution is aconvenient approximation.

Parameters: Mean and Standard Deviation.Parameters: Mean and Standard Deviation.Parameters: Mean and Standard Deviation.Parameters: Mean and Standard Deviation.Parameters: Mean and Standard Deviation. In fact, thereis a family of normal curves. To define a particular normalcurve we need two parameters; the mean (m) and thestandard deviation (s ). Because of the symmetry of thenormal distribution, its mean (as also the median and mode)lies at the centre of the normal curve. In other words, themean places the centre of the horizontal axis, and thestandard deviation governs, the dispersion of values. Figure5-1 shows three normal distributions, for the time taken forbridge laying by three; A, B and C engineer regiments. Eachof them has the same mean ie 30 mins but a differentstandard deviation of 1, 5 and 10 mins.

Figure 5-1: Time for Bridge Laying - A, B and C Engineer RegimentsFigure 5-1: Time for Bridge Laying - A, B and C Engineer RegimentsFigure 5-1: Time for Bridge Laying - A, B and C Engineer RegimentsFigure 5-1: Time for Bridge Laying - A, B and C Engineer RegimentsFigure 5-1: Time for Bridge Laying - A, B and C Engineer Regiments

If we were to take a large enough population of soldiersand draw the frequency curve of their heights, we shouldexpect a bellshaped curve. In a comparison of height betweenSikh and Gurkha troops, the following should be intuitivelyvisualised:

q Sikhs being generally tall, would have a higher mean.

q The curve for Gurkhas would show smaller dispersion asvariation in height of Gurkhas is relatively less.

Area Under the Curve.Area Under the Curve.Area Under the Curve.Area Under the Curve.Area Under the Curve. No matter what the values of mand s are, for a normal probability distribution the total areaunder the curve is taken as unity; thus we may think of areasunder the curve as probabilities. Mathematically this meansthat 68.3 per cent of all values in a normally distributedpopulation be within one standard deviation (plus and minus)from the mean, 95.4 percent within two standard deviationsand 99.7 per cent within three standard deviations. This isshown graphically in Figure 5-2. Practically, 100 per cent ofthe area may be considered to lie within three standarddeviations (plus and minus) from the mean. The graph clearlyshows the relationship between the area under the curve anddistance from the mean measured in Standard Deviations, fora normal proability distribution.

Fiiiiigure 5-2: Relationship Between the Area Under the Curvegure 5-2: Relationship Between the Area Under the Curvegure 5-2: Relationship Between the Area Under the Curvegure 5-2: Relationship Between the Area Under the Curvegure 5-2: Relationship Between the Area Under the Curve

and the Distance from the mean measured inand the Distance from the mean measured inand the Distance from the mean measured inand the Distance from the mean measured inand the Distance from the mean measured in

Standard Deviation, for a Normal Probibility DistributionStandard Deviation, for a Normal Probibility DistributionStandard Deviation, for a Normal Probibility DistributionStandard Deviation, for a Normal Probibility DistributionStandard Deviation, for a Normal Probibility Distribution

It is neither possible nor necessary to have different tablesof areas for every possible normal curve. Instead, we can usea standard table of areas, under the normal curve. With thistable, we can determine the area; or probability, that the val-ues of a normally distributed variable will he within certaindistances from the mean. These distances are defined in termsof Standard Deviations.

Z Table.Z Table.Z Table.Z Table.Z Table. The table at Appendix ‘B’ at the end of thishandbook shows the area under the normal curve betweenthe mean (m) and any other value (X) of the variable. Thevalue X of the variable with which we are concerned isrequired to be converted into another variable Z for using thetable. The value of Z represents’the distance from the mean(m) in terms of standard deviations, and so is derived fromthe formula:

X - mZ =

s

Thus, if m = 50 and s = 10, the probability of the variable lyingbetween 50 (ie the mean) and 60 (ie Z = 1) will be 0.3413, asindicated in the table for a value of Z equal to 1.00.

Assurance LevelAssurance LevelAssurance LevelAssurance LevelAssurance Level Let us say past data tells us that EngineerRegiment ‘A’ takes a mean time of 210 min to lay a bridge and that

Figure 5-3: Time Required for 95% Assurance LevelFigure 5-3: Time Required for 95% Assurance LevelFigure 5-3: Time Required for 95% Assurance LevelFigure 5-3: Time Required for 95% Assurance LevelFigure 5-3: Time Required for 95% Assurance Level

the standard deviation is 10 min. We wish to lay down asuitable time for bridge laying in the operation order so as to be 95per cent confident that the task will be completed in time.

The normal curve for Regiment A’s timings is shown asFigure 5-3. As the curve is symmetrical, only half the areaunder the curve would correspond to timings upto 210min. In other words, a time span of 210 min provides only 0.5probability, or 50 per cent assurance of the task beingcompleted in time. To increase the assurance level we mustcater for more time.

The required assurance level being 95 per cent, an increaseof 45 per cent, or 0.45 in terms of area, is desired. Referringto the table we find that this (0.4495) = 0.45 corresponds to avalue of Z = 1.64. Thus time is required to be increased by1.64 (s) or approximately 16 min. Therefore 226 min shouldbe laid down for completion of this task.

Comparison of an Assurance Level.Comparison of an Assurance Level.Comparison of an Assurance Level.Comparison of an Assurance Level.Comparison of an Assurance Level. With a similartreatment we can also compare the performance of tworegiments. Let us assume that similar data for ‘B’ EngineerRegiment indicates X = 200 min and s = 20 min. Althoughthe mean time taken by ‘ ‘B’ Regiment is less, the time thisunit requires to provide 95 per cent assurance is 200 + 1.64 x20 ie 233 min. The case for both the regiments is shown inFigure 5-4. on next page. As can be seen, the choice is infavour of ‘A’ Regiment - an alternative which is not obviouswithout this analysis.

Dispersion in Mean Expected Life Dispersion in Mean Expected Life Dispersion in Mean Expected Life Dispersion in Mean Expected Life Dispersion in Mean Expected Life : An Application. An Application. An Application. An Application. An Application. Let usconsider yet another application. Say, a new vehicle isbeing introduced, its mean expected life being 6 years andStandard Deviation one year. We wish to determine the year wisewastages, so that their replacement can be planned ahead. We knowthat the entire useful life may be assumed to spread over 3 StandardDeviations either side of the mean. Thus, wastages will take place

Figure 5-4: Comparison of Two Normal DistributionsFigure 5-4: Comparison of Two Normal DistributionsFigure 5-4: Comparison of Two Normal DistributionsFigure 5-4: Comparison of Two Normal DistributionsFigure 5-4: Comparison of Two Normal Distributionsat Given Assurance Levelat Given Assurance Levelat Given Assurance Levelat Given Assurance Levelat Given Assurance Level

between the fourth and the ninth year. Taking approxima-tions for the values indicated earlier to be 68, 95 and 100 percent respectively for 1,2 and 3 Standard Deviations (plus andminus) away from the mean, mean, mean, mean, mean, the wastages can be easilyworked out as percentages of the number introduced. This isillustrated in Figure 5-5.

Figure 5-5: Wastage of Equipment DispersionFigure 5-5: Wastage of Equipment DispersionFigure 5-5: Wastage of Equipment DispersionFigure 5-5: Wastage of Equipment DispersionFigure 5-5: Wastage of Equipment Dispersion

Application Areas.Application Areas.Application Areas.Application Areas.Application Areas. Typical application areas of normaldistribution are:

q Performance of man and machine, eg: Time taken tocomplete an activity or mission (bridge laying /crossing, movement).

- Distribution of rounds and/or bombs on target.

q Equipment life cycle, wastages and replacements.

q Equipment Mean Time Between Failure (MTBF) andreliability.

q Wastage pattern of general stores.

Poisson DistributionPoisson DistributionPoisson DistributionPoisson DistributionPoisson Distribution

Poisson Distribution deals with events that take place in acontinuum of time or space. An accident or a flat wheel ofvehicle is a typical example. It can take place at any momentor point in a continuum of time or space, given a certainenvironment - say, a given highway. The probability of itsoccurrence in a short interval (say, 1 min or 100m) is smalland remains constant, and although we can easily count thenumber of occurrences over an interval, it makes little senseto talk of non-occurrence of such an event. Basically suchevents are also rare events.

Parameter Parameter Parameter Parameter Parameter : The Mean. The Mean. The Mean. The Mean. The Mean. Unlike the normal distribution,Poisson distribution is fully defined by just one parameter,the mean (m). To take an example, let us assume thatvehicles plying between Chennai and Delhi have, on anaverage, two flat tyres. Then, although the mean equals two,there would surely be vehicles that have no flat tyres at all,those that have just one and others that have two, three ormore flat tyres. There would also be cases where vehicleshave as many as, say six flat tyres. Of course, we can sayintuitively that the proportion of vehicles with six or moreflat tyres would be extremely small. Poisson distribution tables(not attached here, but available in any standard book onStatistics) give us the probability associated with each of theseoccurrences. The case for m = 2 is graphically represented atFigure 5-6. As can be seen there is a sharp fall in probabilitiesbeyond two occurrences.

An Application.An Application.An Application.An Application.An Application. We shall now illustrate a typical application ofPoisson distribution. Let us say that at an aircraft repair depot, the

Figure 5-6: A Plot of Poisson” Distribution for m - 2Figure 5-6: A Plot of Poisson” Distribution for m - 2Figure 5-6: A Plot of Poisson” Distribution for m - 2Figure 5-6: A Plot of Poisson” Distribution for m - 2Figure 5-6: A Plot of Poisson” Distribution for m - 2

requirement of aircraft engines over a period of ten monthsis 32, ie 3.2 is monthly mean requirement. Stocks arereplenished monthly. We wish to determine the minimumstock holding to provide an assurance level of 95 per cent.

The values of probability as tabulated in Table 5-1 can beobtained from Poisson tables for m = 3.2:

Table 5-1: Poisson ProbabilitiesTable 5-1: Poisson ProbabilitiesTable 5-1: Poisson ProbabilitiesTable 5-1: Poisson ProbabilitiesTable 5-1: Poisson Probabilities

As can be seen, the minimum stock holding for 95% assur-ance level is six engines. If the stock held is five, the assur-ance provided is 89.46% ie approximately 90%.

Application Areas.Application Areas.Application Areas.Application Areas.Application Areas. Typical application areas for Poissondistribution are in the estimation of :

q Accidents, battle casualties.

q Failure of components and consequently assessment of demands for spares and the stocking policy.

q Arrivals at servicing facilities such as workshops andairfields. Applications pertaining to the field of arrivals havebeen dealt with separately under Queueing (Chapter 11:Waiting Lines).

CHAPTER 6

INFERENCES BASED ON SAMPLEINFERENCES BASED ON SAMPLEINFERENCES BASED ON SAMPLEINFERENCES BASED ON SAMPLEINFERENCES BASED ON SAMPLEINFORMATIONINFORMATIONINFORMATIONINFORMATIONINFORMATION


Decision making could be based on qualitative aspects suchas intuition, hunch, knowledge, experience etc orquantitative inputs in the form of data. As statistical techniquesbasically handle quantitative data, qualitative data is requiredto go through a certain process of quantification for anyworthwhile analysis and interpretation. When qualitativeinformation is in the form of claims, guesses, bunches,opinions or,unauthenticated authoritative statements,acceptance or rejection of such information is vital for adecision situation. How does one go about the process ofquantification and subsequent analysis under thesecircumstances in order to determine the validity of theseclaims? Let us consider a few examples of such claims indefence situations:

q A manufacturer supplying vehicles to the Army claimsthat his vehicles can meet the cross-country capability as perthe General Staff Qualitative Requirements (GSQR).

q A foreign manufacturer claims that ‘Single Shot KillProbability’ (SSKP) of Anti-Tank Guided Missile (ATGM)developed by them is better than 0.7.

q There is no difference in the firing performance ofdifferent types of RCL Guns.

q ‘A’ type of troops have better physical endurance inmountainous terrain than ‘B’ type.

q It is claimed that difference in firing performance of gunsis due to capabilities of firers and not due tocharacteristics of the guns.

For normal conversational purposes, such statements maybe acceptable but when decisions are to be made inoperational situations, the decision process would pivot onthe true values of their capabilities or characteristics. It wouldbe apparent that it is not possible to verify the accuracy orthe truthfulness of the claims which are under scrutiny.Conclusion or inference, perforce has to be made based ontesting a portion or sample selected randomly from theentire lot. In statistical terminology, making an inference aboutthe population based on the sample evidence is calledInductive Reasoning and a statistical technique called‘Hypothesis Testing’ is used to test the accuracy of such judgements which are based on sample information.

HYPOTHESIS TESTINGHYPOTHESIS TESTINGHYPOTHESIS TESTINGHYPOTHESIS TESTINGHYPOTHESIS TESTING

Hypothesis means a law which is not universally accepted.A claim may, therefore, be considered as a hypothesis. InHypothesis Testing, the mean value of randomly selectedsample elements is compared with hypothesised mean value.If the difference is statistically significant, the claim isrejected, otherwise the claim is accepted to be true. This isthe process of inductive reasoning which implies that, basedon sample’s information, the accuracy of the hypothesisedvalue of a lot or population is inferred. In this process, thereis bound to be certain degree of error which may bequantified to assist the decision maker. Let us consider a fewexamples to understand the technique.

Vehicle Manufacturer’s ClaimVehicle Manufacturer’s ClaimVehicle Manufacturer’s ClaimVehicle Manufacturer’s ClaimVehicle Manufacturer’s Claim

Say, the Service HQ have evolved a QR for 2.5 Ton Vehicle foruse in the services and have laid down 400 km cross-country performancewith dispersion 370-430 km. Vehicle manufacturer’s claim is thathis vehicle has a mean cross - country radius of action

of 400 km. Army is planning to procure, say, 500 vehicles.The manufacturer also has indicated that the radius of actionwould meet the QR and has offered a sample of 36 vehiclesfor trial. The performance of the vehicle may be expected tofollow normal distribution as shown in Figure 6-1 with themean value of 400 km and Standard Deviation (or dispersionfactor) of 10 km assuming that the manufacturer’s claim iscorrect.

Figure 6-1: Cross Country Radius of ActionFigure 6-1: Cross Country Radius of ActionFigure 6-1: Cross Country Radius of ActionFigure 6-1: Cross Country Radius of ActionFigure 6-1: Cross Country Radius of Action(Normal Distribution)(Normal Distribution)(Normal Distribution)(Normal Distribution)(Normal Distribution)

Individual vehicle’s cross country performance would bebetween 370 and 430 km km km km km The implication of performance ofvehicles following normal distribution is that if there are 100vehicles, 68 would have mean value of cross-country radiusof action between 390 and 410 km, 95 vehicles between 380and 420 km and 99 vehicles between 370 and 430 km. Themean value of radius of action is 400 km and its standarddeviation is 10 km which is within the specifications as perthe QR

But how do we validate the claim of the manufacturer?Obviously, we cannot test the performance of each and everyvehicle in the total population of 500 vehicles supplied bythe manufacturer. He has offered a sample of 36 vehicles fortrials. Let us assume that the mean obtained with thissample was 380 km. With this information should themanufacturer’s claim be accepted? On the face of it,

it appears that 380 km is close to 400 km claimed by themanufacturer and hence the claim could be accepted.

Based on theory of sampling, individual values of the sampleshould also vary between 370 and 430 km and their meanshould be close to 400 km but in the present case it is 380 kmwhich means individual values could vary between 350 and410 km assuming a dispersion factor of 10 km. Therefore themanufacturer’s claim of mean value of 400 km could turn outto be wrong as may be seen in Figure 6-2.

Figure 6-2: Manufacturer’s Claim and Sample ResultsFigure 6-2: Manufacturer’s Claim and Sample ResultsFigure 6-2: Manufacturer’s Claim and Sample ResultsFigure 6-2: Manufacturer’s Claim and Sample ResultsFigure 6-2: Manufacturer’s Claim and Sample Results

Let us statistically analyse further. Based on samplingtheory and discussion earlier, individual values of performanceof the 36 vehicles which we had taken as sample, could varybetween 370 and 430 km and the sample mean close to 400km. It is quite possible that mean value of 380 km obtainedcould be the actual mean of the population with individualvalues varying between 350 km and 410 km or perhaps, inreality the mean value was 390 km with values varyingbetween 360 and 420 km. See Figure 6-2.

Let us view the problem a little differently, with thefollowing information:

q As per manufacturer’s claim, the population has a meancrosscountry radius of action of 400 km.

q The sample of 36 vehicles after trials, yields a mean of 380 Km.

The question is whether this sample under considerationcould have come from the population whose mean radius ofaction is 400 km, or not.

To answer this question, we must study the statisticalbehaviour of all possible samples which could be drawn fromthe population in question, and check whether our sampleconforms or does not conform to such behaviour. If we exam-ine all possible samples of size ‘n’ from a population of size ‘N’and study the distribution of all the sample means, we findthat such distribution known as ‘Sampling Distribution’ ischaracterised by a mean which equals to the population meanand a Standard Deviation which is the population standarddeviation of the sampling distribution is known as StandardError, (SE).

Let us understand above by using our example:

q Population Parameters.

q Range of cross country radius of action = 370 to 430 km

q Population Mean = 400 km

q Population Standard Deviation = 10 km

q Sampling DistributionSample Size (n) = 36

Mean of the sampling distribution = Population Mean

= 400 km

Standard Error = = = = 1.66 Population SD 10 10

36 6n

The Sampling Distribution of Means of all samples of size 36drawn from a Population with its mean equal to 400 kms isshown in Figure 6-3.

Figure 6-3: Sampling Distribution of Sampling MeansFigure 6-3: Sampling Distribution of Sampling MeansFigure 6-3: Sampling Distribution of Sampling MeansFigure 6-3: Sampling Distribution of Sampling MeansFigure 6-3: Sampling Distribution of Sampling Means

If we construct an interval of ± 3 SE about the mean of 400km, we know that almost all sample means (ie 99.7%) wouldbe within this interval. Thus, we may consider 395.02 km asthe lowest acceptable limit below which we will not acceptany sample mean. In other words, if any sample mean isbelow 395.02 km, it is highly unlikely that this sample couldhave come from a population whose mean is 400 km. In thepresent case, the sample mean is 380 km and thus themanufacturer’s claim is to be rejected. Sometimes, depend-ing on the tolerance level of the decision maker, the criticallimit could even be placed at 396.68 ie at - 2 SE from themean. In this case in 95.5% instances, we will not go wrong ifwe reject the claim. However, we will commit an error of2.25% in we demarcate our rejection region as shaded in Fig-ure 6-3. In other words, there will be 2.25% instances whendue to randomness, typical sample from the claimedpopulation may yield a mean which could be very close to396.68 but fall in the rejection region. In such cases becauseof our decision rule, we shall reject the claim even if the

claim is justified and true. This error is known as the ‘level ofsignificance’ and is always decided in advance beforeconducting the hypothesis test.

Manufacturer’s Claim that SSKP of ATGM is 0.7Manufacturer’s Claim that SSKP of ATGM is 0.7Manufacturer’s Claim that SSKP of ATGM is 0.7Manufacturer’s Claim that SSKP of ATGM is 0.7Manufacturer’s Claim that SSKP of ATGM is 0.7

There are many situations where due to reasons of costand complexity, verification of claim has to be carried out withsmall samples. Consider the case where a manufacturer hasoffered only 10 ATGMs for testing and we shall assume thatthere were 5 hits. Do we accept the claim that the SSKP is0.7. If we do not accept, what is the degree of risk we arecommitting in rejecting the claim (or hypothesis) which inactual fact was true? Such cases involve small sample sizes.Depending on the nature of investigation, a binominal or at-distribution would need to be used for the hypothesistesting. These cases, however, are not discussed as they arebeyond the scope of this elementary handbook.

Checking the Firing Performance of GunsChecking the Firing Performance of GunsChecking the Firing Performance of GunsChecking the Firing Performance of GunsChecking the Firing Performance of Guns

The claims so far discussed could be verified withquantified data available. However, certain claims could be inthe form of qualitative statements and their accuracy wouldrequire to be validated. For example, verifying the claim thatthere is no difference in the performance of four types ofrecoilless (RCL) guns.

The guns are taken on the range and made to fire same ordifferent number of rounds on a standard target at a giveneffective range. The results obtained are shown in Table 6-1.

Table 6-1: Gun Performance DataTable 6-1: Gun Performance DataTable 6-1: Gun Performance DataTable 6-1: Gun Performance DataTable 6-1: Gun Performance Data

Looking at the performance of these guns, one finds thatthe results are quite close to each other but can one say withconfidence that there is no difference in their performance?

A statistical technique called the Chi Square (X ) test isused to verify such cases. This technique is applicable to testwhether all the four guns belong to a population of similarcharacteristics or not and whether the difference betweentheir firing performance is significant or not. This implies thatvariation in the firing could be due to characteristics of gunsor chance element Depending on the cumulative value of thedifference, it can be verified whether these samples belongto the same population or not.

The values given in the tables are the observed values ‘F0’of hits and misses obtained out of total round fired. Theseobserved values are compared with expected values. The con-cept of expected ‘Fe’ is that assuming that gun’s kill probabil-ity is 50% and if all guns are absolutely identical, thenproportion of hits to misses will be same in all cases. If killprobability is 60% then for all guns the proportion of hits tomisses will be in the proportion of 60:40 in all cases. Anydifference in the firing characteristics will create moredeviation from the expected values. In the given data, theexpected values are calculated based on this principle,except that the value of kill probability is not known atthis point in time because its value at the induction stage mayhave changed over a period of time. Kill probability and ex-pected values in such cases may be calculated based on thecollected data.

Say, there were 102 hits out of 192 fired.

Probability of hit p(Hit) =102 192

In case of Gun A, 40 rounds are fired

Expected value of hit = 40 x 102 = 21.3 192

2

X Tables are available for various levels of significance. Alevel of significance of 5% gives the extent of risk of commit-ting an error of rejecting a true hypothesis - called Type 1error. Degrees of freedom is obtained by the formula (m - 1)x (n - 1) where m is number of rows and n is number ofcolumns in the table. In the above case degrees of freedom is(2 - 1) x (4 - 1) = 3.

At 5% level of significance and 3 degrees of freedom, theval of X from the table is 7.81.

The computed value which in this case is 1.76 is comparedwith the table value. If it is less, then there is no difference inthe firing performance of these guns. If the computed valueis more, then there is a difference. Conclusion in this case isthat there is no significant difference in the firing performanceof these guns.


An attempt has been made in this chapter to bring outrelevance of some of the useful statistical techniques such asHypothesis Testing and Chi-Square Test in validating andinterpreting qualitative information when they are in the formof claims, opinions or authoritative statements. These arerather important in the defence context as there arecircumstances where accuracy of the statements need to beverified at a given measure of error. Awareness of relevantstatistical techniques is essential for commanders at alllevels when faced with such decision situations.

2

2

SECTION 2SECTION 2SECTION 2SECTION 2SECTION 2

CHAPTER 7

OPERATIONS RESEARCH INTRODUCTION


Operations Research (OR) may be defined as theapplication of the scientific method to provide the decisionmaker with a quantitative basis for decisions regardingcomplex operations under his control. OR quantitatively,evaluates alternative courses of action which the decisionmaker can take, and then provides this information to him sothat he can choose that alternative which will be in the bestinterest of the organisation as a whole.

There are basically four distinguishing characteristics ofOR. First, it is broad in scope. It is concerned with theorganisation as a whole rather than with sub-organisationunits. Furthermore, it is applicable to any organisationhaving definable objectives, where the attainment of theseobjectives is subject to limiting constraints and there arealternative ways of pursuing them.

Second, it emphasises the development of an analyticaltechnique for solving individual problems rather than theapplication of general models. No two problems are exactlyalike. Therefore, it attempts to develop a model to fit the prob-lem rather than modify the problem to fit a model. In thedevelopment of models, OR leans heavily on mathematics andother basic sciences.

Third, it uses a team effort in dealing with problems. Whileit may often be possible for one person to do OR work, manyproblems are of such dimension that the team approachfrequently results in faster and more efficient method ofobtaining a solution. The team should bring together personsfrom various disciplines and backgrounds and with differentexperiences.

Fourth, it is concerned with the practical managementof an organisation. When a scientist tackles a real worldproblem, he seeks a solution in the light of his ownbackground and experience. But when the solution is obtainedit must be presented in the language of the real world. Inaddition to the use of the scientific method,ORrequires that definite conclusions be drawn, and that theseconclusions be communicated to the decision maker.

Operations Research, neither is, nor does it claim to be amagic formula which can be manipulated to yield the ultimateunique solution, , , , , it is merely a logical systematic approach toprovide a rational basis for the decision maker’s decision.

Specifically, OR applies the scientific method which issimply a logical systematic approach for conducting researchor solving problems.

The scientific method for solving OR problems consists ofthe following steps:

q Formulate the problem.

q Construct a model.

q Obtain a solution by the model.

q Validate the solution given by the model.

q Establish controls over the model for an implementable solution.

q Implement the solution.

Some of the techniques of OR are discussed in thesucceeding chapters.

CHAPTER 8CHAPTER 8CHAPTER 8CHAPTER 8CHAPTER 8

LINEAR PROGRAMMING


Linear Programming (LP) is concerned with the study ofinterrelated components of a complex system so that limitedresources can be used as efficiently as possible in attaining adesired objective.

It was during the World War II that the theory of LP wasdeveloped. In order to maintain the huge war effort, it wascrucial that troops be recruited, trained, and deployed to battleon schedule and the vital equipments be produced as needed.All of this, of course, required extensive planning. An USAFresearch group, later known as Project SCOOP (ScientificComputation of Optimum Programs), developed elaborateplanning techniques, which shortly after the war resulted inthe theory of LP. George Dantzig, one of the leaders in thisresearch group, is credited with being the father of LP

In peacetime economy, an effort was made to apply LP tononmilitary uses. It soon became clear that the sophisticatedorganisational procedures developed by Project SCOOP hadwide application in industry also. Significant uses of LP weremade in operating petroleum refineries, exploring andproducing oil, planning efficient transportation routes ofmanufactured goods, producing steel, awarding contracts etc.

In fact, LP is so widely used today in military, government, com-merce and industry that it would be impossible to list all its manyapplications. If the value of a theory is measured by its ability

to solve problems, then surely LP must be considered one ofthe most important areas of OR.

Requirements of LP ProblemsRequirements of LP ProblemsRequirements of LP ProblemsRequirements of LP ProblemsRequirements of LP Problems

The following requirements must be met before the LPtechnique can be applied:

There must be an objective to achieve and which could be interms of optimum resources allocation, maximise profit,minimise cost, maximise operational effectiveness orminimise casualties.

q There must be alternative course of action - one of whichwill achieve the objective.

q Resources must be in limited supply.

The objective and constraints must be expressed inmathematical linear equations or inequalities.

An Example - LP GraphicalAn Example - LP GraphicalAn Example - LP GraphicalAn Example - LP GraphicalAn Example - LP Graphical

Two types of aircraft are available for a bombing mission,against dug-in and evenly dispersed enemy troops. Bombsare plentiful but aircraft, fuel and maintenance time arelimited. The data is available in Table 8-1.

Table 8-1: Mission ParametersTable 8-1: Mission ParametersTable 8-1: Mission ParametersTable 8-1: Mission ParametersTable 8-1: Mission Parameters

Only 400 KL of fuel is available for this mission. No more than30 planes can receive the required maintenance prior to departuretime. How many planes of each type should be sent on the mis-sion so as to maximise bomb tonnage delivered?

The above problem can now be formulated as follows:

These inequalities can now be plotted as shown at Figure 8-1.

Figure 8-1: Linear Programming - Graphical Solution

Using the Extreme Point Theorem (proof not beingprovided here), it is known that in a convex polygon OPQRTS,the optimum solution lies at one of the vertices. Thereforethe bomb loads delivered at points, P, Q,R, S, T bysubstituting the values of variables A and B (two types ofaircraft) at these points, will be:

Pt T(26, 0)Pt T(26, 4)Pt R(20, 10)Pt Q(10, 15)Pt Q(0, 15)

The optimum solution is at R ie we must send 20 of Type Aand 10 of Type B aircraft for this mission to deliver amaximum of 70 tons bomb load. This solution as you wouldhave noticed does not violate any of our constraints.

Summary of the Solution ProcedureSummary of the Solution ProcedureSummary of the Solution ProcedureSummary of the Solution ProcedureSummary of the Solution Procedure

The solution by graphical method involves three steps:

q Step 1 - Transform the given technical specifications of theproblem into inequalities and make a precise statement ofthe objective function.

q Step 2 - Graph each inequality in its limiting case andobtain a region of possible solution ie the solution space.

q Step 3 - Note the values of variables at all vertices andselect that vertex which gives the highest value of Z(objective function).

Comments on the Graphical MethodComments on the Graphical MethodComments on the Graphical MethodComments on the Graphical MethodComments on the Graphical Method

The graphical method of solving LP problem is limited tocases in which the variables are two or three. Even for threevariables, we move into a three dimensional space and thevisualisation of the problem becomes more difficult than incase of two variables.

Z = 2 x 26 + 3 x 0 = 52Z = 2 x 26 + 3 x 4 = 64Z = 2x 2O + 3x 10 = 70 OptimumZ = 2 x 10 + 3 x 15 = 65Z = 2 x 0 + 3 x 15 = 45

But what about problems involving variables greater thanthree? Most problems in actual practice do fall under thiscategory. These problems can be solved by the SimplexMethod which is the most general and more powerful. It isnot being presented in this handbook, as it is fairly involvedtechnique to be solved manually, but can be quite easily solvedwith the help of commercially available OR computersoftware packages.

Some examples of military problems which can be solvedusing the LP technique are given below along with theformulation of the LP equations, which constitute the crux ofthe problem solution. Once these have been correctlyformulated, the final solution can be easily arrived at usingany computer software for LP like QS, CMMS, TORA etc.

AIRCRAFT ALLOCATIONAIRCRAFT ALLOCATIONAIRCRAFT ALLOCATIONAIRCRAFT ALLOCATIONAIRCRAFT ALLOCATION

A decision has been taken, as a strategic move to strike afatal blow to the enemy’s capacity for aircraft production. Theenemy has four major plants located at different cities, andthe destruction of any one plant will effectively halt theproduction of aircraft by the enemy. The Bomber Commandis faced with shortage of fuel and the fuel supply for thisparticular mission is limited to 240 KL. Any bomber sent toany particular city, must atleast have enough fuel for the roundtrip plus a reserve of 500 litres. The number of bombersavailable to the command their description and fuelconsumption are given at Table 8-2 here.

Table 8-2: Data on AircraftTable 8-2: Data on AircraftTable 8-2: Data on AircraftTable 8-2: Data on AircraftTable 8-2: Data on Aircraft

Information about the location of enemy production plantsand their vulnerability to attack by bombers is as in Table 8-3.

Table 8-3: Enemy’s Plants and VulnerabilityTable 8-3: Enemy’s Plants and VulnerabilityTable 8-3: Enemy’s Plants and VulnerabilityTable 8-3: Enemy’s Plants and VulnerabilityTable 8-3: Enemy’s Plants and Vulnerability

The requirement is to plan the number of bombers requiredto maximise probability of destruction of enemy targets.

ARMY TRANSPORTATIONARMY TRANSPORTATIONARMY TRANSPORTATIONARMY TRANSPORTATIONARMY TRANSPORTATION

Army is planning to carry out the trials of the newly devel-oped main battle tank at five areas (operational bases) spreadover west, north and east India. These tanks can be suppliedfrom three depots. The requirements of tanks for trials at fivebases, the availability of tanks at the three depots and thedistances in km from the bases to the depots are given inTable 8-4 as follows.

Table 8-4: Depot-Base Distance, Availability and Requirement of TanksTable 8-4: Depot-Base Distance, Availability and Requirement of TanksTable 8-4: Depot-Base Distance, Availability and Requirement of TanksTable 8-4: Depot-Base Distance, Availability and Requirement of TanksTable 8-4: Depot-Base Distance, Availability and Requirement of Tanks

Find a schedule to transport the tanks from depots tobases that will minimise the total distance involved inmeeting the requirements of tanks.

NAVY WEAPON MIXNAVY WEAPON MIXNAVY WEAPON MIXNAVY WEAPON MIXNAVY WEAPON MIX

Indian Navy in considering three types of attack aircraft - asupersonic type, a transonic type, and V/STOL type for itsaircraft carrier. The effectiveness of the aircraft to the carrieris determined by the relative expected military value oftargets that it can destroy during a military engagement of acertain length. These have been estimated for the three typesand mentioned in first row of Table 8-5.

Table 8-5: Expected Values and Requirements of theTable 8-5: Expected Values and Requirements of theTable 8-5: Expected Values and Requirements of theTable 8-5: Expected Values and Requirements of theTable 8-5: Expected Values and Requirements of theThree AircraftsThree AircraftsThree AircraftsThree AircraftsThree Aircrafts

If the entire deck space available for aircraft on the carrier isallocated to one specific type, the numbers of aircraft that could beaccommodated are given in Table 8-5, second row. Personnelrequirements and costs are also included in the table.

Considering that a carrier has facilities for 1500 attack aircraft personneland the Navy’s planned expenditure for attack aircraft for the carrier is 325crores, the problem is to determine how many of each of the three types ofaircraft should be purchased for the carrier in order to maximise the value ofits attack capability.

SOLUTIONS : FORMULATION OF LP EQUATIONS

Aircraft Allocation

CHAPTER 9

TRANSPORTATION MODELTRANSPORTATION MODELTRANSPORTATION MODELTRANSPORTATION MODELTRANSPORTATION MODEL


A typical Transportation Model deals with a special classof linear programming problems in which the objective is totransport a single commodity from various origins todifferent destinations at a minimum total cost. The totalsupply available and the total requirement are given. Furthereach allocation has a known origin and a known destination.Also all relationships are assumed to be linear.

Transportation problems being a sub-set of the general LPproblems can be solved by the simplex method. However,the transportation algorithm provides a much more efficientmethod of handling such a problem.

To state explicitly, before transportation model can beapplied, following assumptions/conditions must be made:

q All relationships are assumed to be linear.

q It deals with only one (single) commodity which isrequired to be moved from different origins/depots, whereit is available, to various destinations/units where it isrequired.

A limiting factor is that all demands at destinations mustequal total availability at various origins. This, however, doesnot pose any problem as the two aspects can be made equalby a simple method which is described in paragraph titledBalancing Demand and Supply in this chapter.

q

q like the general LP model, we can solve the problem for asingle objective at a time eg, minimise cost, distance ortime etc.

Basic transportation model deals with a minimisation case.In order to handle a maximisation case (say the objective maybe to maximise profits) the approach has, therefore, to beslightly modified which is described in paragraphs titledMaximisation Case in this chapter.

Making Initial Transportation TableauMaking Initial Transportation TableauMaking Initial Transportation TableauMaking Initial Transportation TableauMaking Initial Transportation Tableau

A pictorial representation of the mathematical descriptionis available through a transportation tableau. The transporta-tion tableau is a matrix with rows (representing sources ofsupply) and columns (representing destinations to satisfydemands). Therefore, each cell in the matrix represents aunique routing for shipments from sources to destinations.The figures on the right and bottom sides of the matrix show,respectively, the amount of supplies available at each sourceand the amount required at each destination. The objective ofthe problem is to identify the amount to be shipped from eachsource to satisfy the demands at each destination with theminimum transportation costs.

Let us see this with the help of an example. Total of 21tanks are available at three origins and are required at fivedestinations and the cost of transportation from each originto various destinations is known and the data is arranged inTable 9-1.

Table 9-1: Transportation Cost-Per Tank (Rs thousands) withTable 9-1: Transportation Cost-Per Tank (Rs thousands) withTable 9-1: Transportation Cost-Per Tank (Rs thousands) withTable 9-1: Transportation Cost-Per Tank (Rs thousands) withTable 9-1: Transportation Cost-Per Tank (Rs thousands) withAvailability and DemandAvailability and DemandAvailability and DemandAvailability and DemandAvailability and Demand

Balancing Demands and AvailabilityBalancing Demands and AvailabilityBalancing Demands and AvailabilityBalancing Demands and AvailabilityBalancing Demands and Availability

In the above example, the demand and availability are bothequal so the tableau prepared above is ready for furtheroperation. However, in real life such situations would rarelyoccur. An example of this is at Table 9-2. The procedureadopted for balancing the demand and availability is as under:

Table 9--2: Unbalanced MatrixTable 9--2: Unbalanced MatrixTable 9--2: Unbalanced MatrixTable 9--2: Unbalanced MatrixTable 9--2: Unbalanced Matrix

q Demand Exceeds Supply. For purposes of solution of the problem, we create a fictitious source or resource where thesupply available is just sufficient to meet the excess of thedemand over the actual supply. This source of supply is calleda dummy supply. Transportation cost from a dummy supplyto any destination is naturally zero because no actualtransfers to any destination can take place. Also if in thesolution a given destination receives a supply of ‘d’ units fromthe dummy and ‘s’ units from an actual supply, it has been inreality short-supplied to the extent of ‘d’ units. As is evidentfrom balanced matrix at Table 9-3, there would be a short fallof 50.

q Supply Exceeds Demand. Similarly as in the above case afictitious consumption point is created for the sake ofbalancing and solving the problem. The requirement atthe ‘dummy’ destination is exactly equal to the differencebetween actual supply and demand. Once again the cost ofthis factitious transfer is zero. If a source is deemed tohave transferred ‘d’ units to the dummy, it is interpretedthat source has a left over stock of ‘d’ units.

Table 9-3: Balanced MatrixTable 9-3: Balanced MatrixTable 9-3: Balanced MatrixTable 9-3: Balanced MatrixTable 9-3: Balanced Matrix

VOGELS APPROXIMATION METHOD (VAM)

Vogel’s ApproximationVogel’s ApproximationVogel’s ApproximationVogel’s ApproximationVogel’s Approximation Method gives near optimal oroptimal solutions. VAM is based on the use of the differenceassociated with each row and column in a matrix giving theunit cost. A row or column difference is calculated as thearithmetic difference between the smallest and the nexthigher cost element in that row or column. This valueprovides a measure of the priorities for making allocations tothe respective rows and columns, since it indicates theminimum unit penalty incurred by failing to make anallocation to the smallest cost cell in that row or column. Thus,this procedure repeatedly makes the maximum feasibleallocation in the smallest cost cell of the row or column withthe largest difference. It never changes an allocationpreviously made, so the procedure stops as soon as the totaldemand and supply are exhausted. The detailed steps involvedin Vogel’s approximation method are summarised here:

q Step 1 - Construct the cost-data table like Table 9-2 forthe problem. Following is the detailed structure of this table:

q One column listing the sources (at the far left of thetable), one column for each of the (n) destinations, andone column for the supply entries (at the far rightof the table).

One row identifying the destination points (at the top of thetable), one row for each of the (m) sources, and one rowsummarising the demand values (at the bottom of the table).

q Small inset boxes for entering the costs associated witheach source-destination pairing or cell.

q The table is complete if the sum of the supply columnequals the sum of the demand row. If the sum of the supplycolumn does not equal the sum of the demand row, adjustthe table as follows. For total demand greater than the sup-ply; add a dummy row with cost entries equal to zero and therow requirement set equal to the excess demand. For totaldemand less than total supply, add a column with cost entriesequal to zero and the column requirement set equal to theexcess supply.

q Step 2 - For each row, calculate the difference betweenthe lowest and next higher cost entries. (If the two lowestcost elements are equal, the difference is zero).

q Step 3 - Similarly, for each column calculate the differencebetween the lowest and next higher entries (If the twolowest cost elements are equal the difference is zero).

q Step 4 - Select the row or column that has the greatestdifference. In the event of a tie, the selection is arbitrary.

q Step 5 - In the row or column identified in Step 4 selectthe cell that has the lowest cost entry.

q Step 6 - Assign the maximum possible number of units tothe cell selected in Step 5 (ie, the smaller of the two be-tween the availability row and demand column). This willcompletely exhaust a row or a column (perhaps both).

q Step 7 - Reapply Steps 2 to 6 using the remaining rowsand columns. Repeat the process until total supply has been

exhausted. If the calculations are correct, this will alsoexhaust total demand.

Illustrative Problem-SolutionIllustrative Problem-SolutionIllustrative Problem-SolutionIllustrative Problem-SolutionIllustrative Problem-Solution

The above method is applied to the tank transportationproblem whose data has been given in Table 9-1 earlier. Theiterations are explained below in reference to Table 9-4.

Note: Note: Note: Note: Note: For ease of understanding, the iteration number isindicated on left hand top comer of the cell in whichallotment is made.

Table 9-4: Transportation Table - VAM SolutionTable 9-4: Transportation Table - VAM SolutionTable 9-4: Transportation Table - VAM SolutionTable 9-4: Transportation Table - VAM SolutionTable 9-4: Transportation Table - VAM Solution

Iteration 1. The highest difference is 10 in column two.Hence allot the maximum feasible amount to the lowest costcell (02, D2) in this column. The availability in row two being4, only 4 can be allotted to cell (02, D2). The balance require-ment in column two is also amended to 1, quantity 4 havingbeen allotted. Row two is now eliminated from further con-sideration and the remaining matrix is operated upon as perabove procedure in the next step.

Iteration 2. Against maximum column difference 9, maxi-mum feasible quantity 3 is allotted to cell (03 , D1,) and zeroput in requirement of column one. Column one is noweliminated from further consideration. The above procedureis continued till all allocations have been made.

The total cost of transportation works out to 150 units ofmoney. It might appear that the procedure is tedious butafter a little practice, one could probably execute this entireprocedure on the original cost and requirements table. Onewould need merely to cross out rows and columns as theyare completed and revise the supplies and demands.

Salient FeaturesSalient FeaturesSalient FeaturesSalient FeaturesSalient Features

q Follows principle of ‘opportunity costs’ ie, it examinesthe cheapest route from each origin to differentdestinations and to reach the destinations from differentorigins.

q Compares opportunity cost or penalty for not followingthe cheapest route.

Makes maximum allocation to that route which has highpenalty.

q Gives a solution which is optimal or very close to it.

q Procedure is tedious but not difficult.

q Being repetitive in nature lends itself to easy handling bycomputers.

Maximisation CaseMaximisation CaseMaximisation CaseMaximisation CaseMaximisation Case

Though solution methods for transportation basically ca-ter to minimisation of costs, occasionally, a problem is mostnaturally stated in terms of expected profit instead of cost oftransportation. This situation is accommodated in the trans-portation method by treating profits as negative losses andthen applying minimisation techniques.

A convenient method for converting a profit statement tocost relationships is to subtract all stated profits from the highest profit; the resulting numbers are relative costs andare treated as regular costs in solving the problem. For ex-ample, if the largest profit in any resource transfer is Rs 100(or 100 units), all transfer profits would be subtracted from100. This policy gives 100 - 100 = 0 as the relative cost for themore profitable route, which of course, makes it the mostattractive route, for cost minimisation. Similarly the lowestprofit route would have the highest relative cost. After allsource-to-destination ‘transfer charges’ are calculated, themethods described for a minimisation case can be used todevelop a solution. From the preferred routes indicated inthe solution, the total profit expectation can be calculated byusing original profit figures.

Modified Distribution Method (MODI)Modified Distribution Method (MODI)Modified Distribution Method (MODI)Modified Distribution Method (MODI)Modified Distribution Method (MODI)

Transportation problems are a special class of linearprogramming problems. VAM gives a solution which is veryclose (or even in some cases equal) to the optimal solution.For an optimal answer one may make use of MODI. Ititeratively improves on an initial basic feasible solution -obtained by any one of the approximate methods, includingVAM. It is based on a cell evaluation method of an empty cell,keeping the rim requirements ie supply and demand intact.Readjustments to previous allocations are made iterativelytill finally optimal solution is obtained. The commerciallyavailable OR computer software packages give optimalsolutions.

CHAPTER 10

DECISION THEORY


In every walk of life, be it at individual level or extremelycomplex modern organisational level, we are concerned withtaking effective decisions. Administrators in the government,commanders in the services, managers in the public orprivate sectors, irrespective of their functions or level, are allinvolved in the decision making process.

Managerial Decision MakingManagerial Decision MakingManagerial Decision MakingManagerial Decision MakingManagerial Decision Making

It is a process whereby a commander or a manager whenconfronted with a problem, chooses a course of action from aset of possible alternative courses available. Decision makingis a response to a problem situation which generally arises asa result of a discrepancy between prevailing conditions andorganisation goals. Uncertainty will always be there about thefuture since one cannot be absolutely sure of the ultimateconsequences of the decision made.

Elements of a DecisionElements of a DecisionElements of a DecisionElements of a DecisionElements of a Decision

In a problem situation where decisions are to be taken,the following elements can be identified:

q Decision maker - who may be an individual or a group.

q A set of possible and viable alternative courses of action

q A set of states of nature or occurances which may or maynot be predictable.

q A set of consequences or effects related to each possiblecourse of action depending upon the state of nature (cause-effect relationship).

The complexity of a decision problem would depend uponthe nature of decision element It is extremely difficult for adecision maker to intuitively take into consideration, theimplications of consequences of each possible course ofaction vis-a-vis states of nature. Decision analysis providesthe necessary base for defining, analysing and solving a prob-lem in a rational, logical and scientific manner based on data,facts, information and logic when dealing with complexdecision problems. The focus of such an approach is to studyas to how decisions should be made rather than how peopleactually make decisions.

Pay Off MatrixPay Off MatrixPay Off MatrixPay Off MatrixPay Off Matrix

Pay Off.Pay Off.Pay Off.Pay Off.Pay Off. Decision making is choosing a course of actionfrom among the alternatives. Effect or consequence of a courseof action would depend upon the state of nature.Consequences are; that could be measured like profitnumber of targets destroyed, number of casualties or losses.Such outcomes could be viewed as pay off of a decision. Butbefore taking a decision, pay off values could be assigned forthe different courses of action to facilitate choice among thealternatives.

Decision Situations.Decision Situations.Decision Situations.Decision Situations.Decision Situations. All decision situations fall under one ormore of the following categories:

q Certainty. For a set of given initial conditions, future eventscan be predicted with certainty.

q Risk. Although future cannot be predicted, it is possible toassess or predict the likelihood (or probability) of futureevents.

q Uncertainty. Where probability of future events cannotbe assessed.

Assignment of Pay Offs.Assignment of Pay Offs.Assignment of Pay Offs.Assignment of Pay Offs.Assignment of Pay Offs. Assignment of pay off values forcertainity situation is comparatively simple. For example, inan investment decision of whether to invest in a bank,reputed company or in shares, the return on investment couldbe viewed as pay offs. The choice as to where to invest would,however, depend upon the risk profile of the decision maker.Similarly, measures of effectiveness such as target destruc-tion potential (number of targets that can be destroyed) of aweapon or fire unit can be assigned with reasonable level ofassurance which could be viewed as direct measures of thepay off values. But in many of the defence situations, wherefuture events cannot be predicted, we will be dealing withsituations of risk and uncertainty. In such situation pay offsmay have to be assigned subjectively. In order to reduceindividual subjectivity, the concept of participative decisionmaking may be resorted to, wherein personnel involved inthe decision making process and in its subsequent implementation are associated in the assignment of pay off values.

Decision MatrixDecision MatrixDecision MatrixDecision MatrixDecision Matrix

Decision matrix is an extension of pay off matrix and thereare two main aspects involved:

q Assigning of probabilities to the occurrence of differentstates of nature.

q Assigning of pay offs to various cells in the matrix.

For assigning probabilities, methods available are:

q Determining how frequently similar events have occurredin the past, in case such data is available and thus calculatethe probability.

q Available intelligence about the states of nature.

q Informed judgement of the decision maker based on hisknowledge, professional skill and experience.

Basic Steps for Decision MatrixBasic Steps for Decision MatrixBasic Steps for Decision MatrixBasic Steps for Decision MatrixBasic Steps for Decision Matrix

q Listing of all viable alternatives. These are also calledstrategies as these are under the direct control of thedecision maker.

q Listing of events affecting these alternatives. There arecalled states of nature and are beyond the control ofdecision maker.

q Construction of pay off matrix by assigning pay off values.

q Assigning probabilities to the states of nature.

q Calculation of expected value of each alternative.

q Presenting the expected value of each alternative to thedecision maker for taking the decision.

In normal military appreciation, although there may be alarge number of own and enemy courses of action, byapplying value judgements, the choices are narrowed down.However, decision matrix poses no such restrictions and cansimultaneously analyse a number of own and enemy coursesof action. In fact, more complex the situation greater could bethe advantage of using a decision matrix.

Methodology of analysing a risk decision situation usingDecision Matrix will now be considered with the help of anexample. Sketch at Figure 10-1 shows three differentapproaches for attack and capture of an enemy post.

Figure 10-1: Alternate Approaches for Attack

q Approaches for Attack (Alternatives)

q A3 - Longer than Al. Enemy position tackled from flank.No minefield encountered. Cover available. Takes less timethan A2. Enemy may withdraw.

q Al - Shortest approach. It is across a minefield. Maxi-mum casualties expected. Enemy can withdraw.

q A2 - Longest approach. Enemy tackled from the flank.No minefield encountered. Cuts off enemy route of with-drawal. Limited cover available.

q Courses Open to the Enemy (States of Nature)Courses Open to the Enemy (States of Nature)Courses Open to the Enemy (States of Nature)Courses Open to the Enemy (States of Nature)Courses Open to the Enemy (States of Nature)

q S1 S1 S1 S1 S1 - Enemy upsticks and withdraws before ownassault is mounted.

q S2 - Enemy gives a limited fight and withdrawsunder pressure.

q S3 - Enemy fights to last man, last round.

The next step is to construct pay off matrix by assigningpay off values that may accrue by adopting a particular strat-egy and matching it against a probable state of nature. Thisnecessitates finding a measure for the pay offs, or a measureof effectiveness for the goal to be achieved. In any militaryoperation results may be measured in terms of attrition (ieown and enemy casualties), time taken and spatial gain. Inthis particular case, the factors considered are:

q Enemy casualties.

q Own casualties.

q Speed of operations.

Consideration of these factors would lead to a descriptiveanalysis as shown in Table 10-1 and these can be quantifiedas given in Table 10-2.

Table 10-1: Evaluation of StrategiesTable 10-1: Evaluation of StrategiesTable 10-1: Evaluation of StrategiesTable 10-1: Evaluation of StrategiesTable 10-1: Evaluation of Strategies

The factors can be weighted, if required. Qualification ofthe descriptive standards has been done on a scale of 0 to 10to give the values as given in Table 10-2 on next page.

Probabilities of occurrence are next assigned to the statesof nature and a pay off matrix constructed as shown at Table10-3. ‘Expected value’ of pay off is worked out for eachalternative, by adding product of probability associated to stateof nature and pay off for each strategy for respectivealternative. For example, expected value for alternative Al isworked out as; (28 x 0.5) + (22 x 0.3) + (18 x 0.2) = 24.2.

Decision Matrix in Tactical AppreciationDecision Matrix in Tactical AppreciationDecision Matrix in Tactical AppreciationDecision Matrix in Tactical AppreciationDecision Matrix in Tactical Appreciation

In the normal tactical appreciation, courses of action available tothe enemy are considered and the course most likely to be adopted

Table 10-2: Pay Offs

(Alternative Al, is selected since it has the highest expected value)

Table 10-3: Calculation of Expected Value

by him arrived at. Thereafter own plans are evaluated only in thelight of the enemy’s most likely course. Thus a situation of risk is

considered as one of certainity by reducing the problem to asingle state of nature. The decision matrix permits evaluationof own courses against all states of nature that may possiblyoccur. In case the states of nature are likely to occur withdifferent probabilities, according to the strategy adopted, thesituation can be easily handled as shown in Table 10-4 below.

Note: Figures in the inset boxes indicate probabilities ofoccurance of states of nature

Table 10-4: Decision Matrix for Tactical AppreciationTable 10-4: Decision Matrix for Tactical AppreciationTable 10-4: Decision Matrix for Tactical AppreciationTable 10-4: Decision Matrix for Tactical AppreciationTable 10-4: Decision Matrix for Tactical Appreciation

The strategy yielding the highest ‘expected value’ should beadopted which is Al in this example.

Decision TreeDecision TreeDecision TreeDecision TreeDecision Tree

In decision matrix, one adopts the best choice based oninformation and criteria, information being probability associ-ated with states of nature while criteria being expected value.

In analysing decision situations, the decision maker, apart fromconsidering all the choices open to him as well as the uncertaintiesassociated with each, has to identify the time sequence in whichvarious actions and consequent events would occur. Decision mak-ing may involve several stages and choices at each stage; each of

the choices open will result in a different outcome and pay-off.It Is possible to depict the situation graphically by branchesof events. Each branch represents a possible event resultingfrom a potential choice. The combinations of all the possibleactions and potential events are depicted in the form of adecision tree consisting of nodes and branches where eachbranch represents an alternative course of action and/or anevent.

Decision trees are basically a combination of decision ma-trix and probability tree as it represent both the strategies aswell as states of nature. The technique of decision treeenables commanders to take more direct account of:

q Impact of possible future decisions.

q Impact of uncertainity.

q Relative importance of present and future pay-offs.

This technique has the following deficiencies:

q Unable to display the logical inter-relationship be-tween the different steps of a complicated situation.

q Difficult to introduce time relationship especially inmultidecisional problem.

q Difficult to depict sequential decisions as Decision Treerepresents a single decision problem at a Given point in time.

Characteristics.Characteristics.Characteristics.Characteristics.Characteristics. Decision tree is a pictograph of the wholesituation as it chronologically depicts the sequence of actionsand the outcomes as they unfold. It has a tremendouspotential as a decision making tool for long range planningand multi-state decision as it can clarify for the commanderthe choice, the objectives, the monetary gains and theinformation needs. it mainly consists of nodes and branches.

A Square box represents decision/ alternative node or allalternative courses of actions. The branches end either in chance

node or to the final pay-off-value. Chance/Event node rep-resenting states of nature or alternative outcomes of thechances event, are represented by a circle as shown here.

Decision Node Chance Node

Selection of Best Alternative.Selection of Best Alternative.Selection of Best Alternative.Selection of Best Alternative.Selection of Best Alternative. All possible decision pathsare analysed, and depending on the criteria optimal strategyis selected. Having identified the possible courses of action,the potential states of nature and the outcome for each courseof action in a sequence, the decision maker evaluates the ex-pected value of the pay-off’s of each branch and its associatedprobability. The path with highest expected value is identi-fied as the best alternative.

In a single stage problem, there will be one decision nodeand number of branches indicating alternatives and then thatmany chance nodes from which all possible events will ema-nate. In a multistage problem, there may be two, three orfour sets of decision nodes, the branches of which ending inthat many number of chances.

Laying the Tree.Laying the Tree.Laying the Tree.Laying the Tree.Laying the Tree. It implies spreading the problem intodecision nodes, chance nodes, alternatives, events and thepay-off values and the probability indicated.

Folding Back the Tree.Folding Back the Tree.Folding Back the Tree.Folding Back the Tree.Folding Back the Tree. It means calculating expectedvalue at the chance node and annotating the highest value atthe decision node. In case there is another chance node, newexpected value with the highest expected value of each deci-sion node is calculated. The branch to which the highest ex-pected value refers is left unmarked, while other branchesare slashed.

The methodology for laying and folding back a tree is ex-plained with help of an example of air attack.

Example Example Example Example Example : Air AttackAir AttackAir AttackAir AttackAir Attack

It is desired to carry out an air attack on an enemy posi-tion. For this purpose three alternative are available to thestation commander.

q Use bombers by day.

q Use bombers by night

q Use fighter-bombers by day.

There is a 0.6 probability that the target will be welldefended and 0.4 probability that main target will not be welldefended. The aircraft on arrival at the target area maydecide to attack the main target or take alternative target.

The decision tree of the problem is drawn at Figure 10-2on the foldout sheet. Probability of total/partial destructionand expected pay-offs in each contingency are as shown onthe decision tree. These probabilities and pay-offs have beenassigned by the station commander.

Methodology of Constructing a Decision TreeMethodology of Constructing a Decision TreeMethodology of Constructing a Decision TreeMethodology of Constructing a Decision TreeMethodology of Constructing a Decision Tree

It will be seen that there are two sequential decisionsinvolved. The decisions required and their associatedalternatives and state of nature are as under:

Decision 1

q Alternatives

q Use bombers by day.

q Use bombers by night.

q Use fighter-bomber by day.

q States of Nature Probability

q Target well defended ... 0.6

q Target not well defended ... 0.4

Decision 2q Alternatives

q Attack main target.q Attack alternative target.

qqqqq States of Nature States of Nature States of Nature States of Nature States of Natureq Target may be fully destroyed.q Target may be partially destroyed.

Folding back of the tree will be done from right to left Start-ing from last chance node, calculate expected value of eachpath and add. Taking the topmost branch, for example; theexpected value for last chance node 2 on this branch worksout to be (0.4 x 10 + 0.6 x 2) = 5.2. The added value is put ontop of each chance node. Then move left to the decision nodeand place the value highest of the branches on it. Slash thebranches of the node other than with the highest value. Simi-larly, the highest value for each of the decision nodes is an-notated. Move left again to the next chance node. Computethe expected value for each branch and annotate the highestone for the node. Slash the other branches. Repeat the pro-cess till the first decision node is reached. Mark the highestexpected value. The path left inslashed would give the optionfor maximum expected value. Continuing with the examplefor the topmost branch; moving left to last decision node 2 the value annotated is 5.2, higher of the two (5.2 and 3.1).Following the unfolding procedure, finally value 7.06 for firstdecision node l is annotated. The analysis of decision treetells us to use fighterbomber by day and attack the main tar-get irrespective of the fact whether it is well defended or not.Generally optimal decisions are shown by dark branches/lines emanating from decision nodes.


Decision tree is a pictorial presentation of a decision situa-tion where each possible action choice, related risks and pos-sible outcomes are identified easily. It helps to sharpen thethinking process of the decision maker and assist him in fo-cusing on key issues. This structured decision process canhelp in making better, orderly and effective decisions.

CHAPTER 11

QUEUEING (WAITING LINES)

GeneralGeneralGeneralGeneralGeneral

A queue forms when existing demand exceeds existingcapacity of the service facility and customers do not receiveservice immediately on arrival but have to wait, as servicefacility is temporarily busy providing service to those arrivedearlier. In other words, when the number of units arriving ismore than the capacity of the service facility for a particularperiod, a queue will start forming whose length will increaseor decrease depending on the rate of subsequent arrivals. Theobvious answer to the reduction of the queue length is toincrease the service facilities but then the facilities may alsohe idle for longer durations. The concept of queueing theoryhelps us to determine optimum trade-off between theminimum waiting time and maximum utilisation of servicefacility. The service facility could be a bridge, workshop,machine, service station, dockyard, airport etc. while therespective customers could be vehicles, defective radio sets,vehicles, ships, aircraft and so on.

ConceptConceptConceptConceptConcept

The basic cause for long queue lengths or long waiting times of thecustomers is because neither the arrivals nor time taken for servicingfollow a fixed pattern. A fixed pattern is practical only in fullyautomated systems and not in most real lifesystems. In fact, long waiting occurs because of randomness or variabilityin the arrival pattern of customers as well as in the time required in the

service to customers. In real life, it is extremely difficult toexactly predict the arrival pattern and the pattern of time takenfor service. Because of the randomness in the arrival andservice patterns, probabilistic models have been developedbased on which mathematical theory of queues has beenformulated. These models enable us to calculate for a givensituation, what kind of a queue would result and how long thecustomers may have to wait for the service.

In order to make use of the queueing concepts and theassociated formulae, values of two basic parameters must beavailable. These are:

q Average rate of arrival : l (lamda)

q Average rate of servicing: m (Mu)

If the average rate of arrival (l ) is less than the averagerate of service and both are stabilized values, the system willeventually settle down to a steady state. In such a situation,probability of a particular length of queue forming will be thesame at any time. If the rates are not constant, the systemwill not reach a steady state. If the rate of arrival is equal orless than rate of servicing, the system is unstable and hasvery high probability of a long queue whose length is steadilyincreasing. Let us explain this concept with some situations:

q Situation OneSituation OneSituation OneSituation OneSituation One

q Arrivals are at a constant rate of 10 per hour whichmeans that in every six minutes one unit arrives.

q Servicing is at a constant rate of 12 per hour which meansthat every arrival takes exactly five minutes to service.

A queue will not form but after one hour it will be realisedthat service facility was lying idle for 10 minutes.

q Situation TwoSituation TwoSituation TwoSituation TwoSituation Two

q Arrivals are at a constant rate of 10 per hour.

q Servicing at a constant rate is also 10 per hour.

In this case there is a perfect match between the require-ments of customers and pattern of service. In other wordsdemands equal service facility, therefore, neither a queue willform nor service facility will be idle.

q Situation ThreeSituation ThreeSituation ThreeSituation ThreeSituation Three

q Arrivals has a constant rate of 10 per hour.

q Servicing is at a constant rate of 8 per hour whichimplies that it takes 7.5 minutes to service thecustomer.

In one hour there will be 10 arrivals and only 8 can beserviced, therefore, there will be two units waiting for servicingafter the end of the hour. If the situation continues, afterevery one hour the length of the queue will go on increasing by 2.

In all the above situations, a very basic assumption has beenmade of constant arrivals and constant servicing andcalculation of whether a queue will form or not seems veryeasy. However, we all know that it is difficult to maintainarrivals at a constant rate and it is also not possible to ensurethat time taken for servicing each arrival is exactly the same.Because of this variability in these two parameters, queuingtheory tells us that a queue of infinite length will occur inSituations Two and Three as any arrival later than theconstant rate will make the service facility idle and this lossof servicing time will never be recouped. Further, if servicingtime is more than the constant rate, it will further aggravatethe queue situation. Even in Situation One, the queue occursdue to this variability and queueing theory helps us todetermine the queue lengths for the situation. Therefore, toapply queuing concepts, it must be ensured that averagerate of arrivals is less than average rate of servicing.

Mathematical FormulaMathematical FormulaMathematical FormulaMathematical FormulaMathematical Formula

The formulae associated with a single service station aregiven:

Utilisation FactorUtilisation FactorUtilisation FactorUtilisation FactorUtilisation Factor

For a single service station having queue discipline of‘first come first service’, one can completely predict thebehaviour of the queue as it is totally dependent on theratio of average arrival rate and average service rate. Thisratio is called utilisation factor and is denoted by p (Rho),and mathematically;

A graph indicating the relationship-utilisation factor andmean queue length is shown at Figure 11-1. It will be seenthat queues start forming for a utilisation factor of 0.5 andboth the queue length and waiting time tend to becomeinfinite for utilisation factor of 1.

Figure 11-1: Exponential Relationship between Utilisation FactorFigure 11-1: Exponential Relationship between Utilisation FactorFigure 11-1: Exponential Relationship between Utilisation FactorFigure 11-1: Exponential Relationship between Utilisation FactorFigure 11-1: Exponential Relationship between Utilisation Factorand Mean Queue Lengthand Mean Queue Lengthand Mean Queue Lengthand Mean Queue Lengthand Mean Queue Length

It will be further seen that for even a slight increasebeyond the utilisation factor of 0.75, there is disproportion-ate increase in the queue length and waiting time. Therefore,if any system is planned to function with p values of morethan 0.75, the system will become inefficient therebyadversely affecting the operational efficiency.

Let us consider an example to illustrate this. In an opposedriver/canal crossing operation, one vehicle takes one minuteto cross the bridge, and if it is planned to release one vehicleevery minute for arrival at the bridge entry point, it may causea lot of congestion of vehicles at the bridge during executionstage. As the average bridge crossing time is one minute, itis essential to have a planning figure of approximately 1.2minutes for arrival of vehicles at the bridge entry point sothat value of p is 0.75. This will, of course, reduce number ofvehicles which could be crossed in one hour thus affectingthe total number of vehicles which could be planned for agiven time of availability of bridge before first light.

ILLUSTRATIVE EXAMPLEILLUSTRATIVE EXAMPLEILLUSTRATIVE EXAMPLEILLUSTRATIVE EXAMPLEILLUSTRATIVE EXAMPLE

Field Repair WorkshopField Repair WorkshopField Repair WorkshopField Repair WorkshopField Repair Workshop

Normal Load. Normal Load. Normal Load. Normal Load. Normal Load. A field repair workshops receives varyingnumber of vehicles each month. The monthly average worksout to 80 vehicles. Similarly, the average service capability is100 vehicles per month.

We observe that :

Input Increased. Input Increased. Input Increased. Input Increased. Input Increased. A couple of months before a formation’scollective training, the number of vehicles reporting toworkshop increases, and this trend continues till after a monthof the exercise. The average input thus rises to 95 vehiclesper month. Let us see its effect on waiting length of vehiclesand waiting time.

Observation. It may be noted that an average increase of 15vehicles a month has shot up the queue length from 3vehicles to 18 vehicles and the waiting time from 1 day to 6 days.

Increased Output.Increased Output.Increased Output.Increased Output.Increased Output. It is only natural that the user unitsand the OC field repair workshop would like to see allvehicles ‘on road’ or under active repairs with no waiting timeparticularly during the collective training period. OCworkshop declares a minor emergency, stops all sports andsocial activities and orders overtime in the evening. Thedouble shift increases his average output from 100 to 140vehicles a. month.

Let us see the effect on waiting length and waiting time for vehicles:

It means that on an average, there is one to two vehicle in thequeue waiting for less than a day.

Note: The increase of average output to 140 vehicles permonth achieved by overtime could also be achieved byproviding additional resources as an alternative toovertime and/or double shift.


In this chapter the basic concepts of queueing have beenexplained in very simple terms but today queue models andformulae for almost all contingencies are available. Thesemodels tell us, for example, if there are two bridges, what isan efficient system having two parallel lanes or one feedinglane which further distributes into two lanes, before theentry to bridges. Basic formulae are available for calculatingmean queue length for say a twin canal system, number offacilities required for a particular operating system, numberof runways required for an airport, number of servers requiredat ammunition and/or supply points etc. The option selectedusing queuing concepts will provide cost effective andefficient service.

CHAPTER 12

MONTE CARLO SIMULATION


During World War 11, the work of John von Neumann andUlam at the Los Almos Project encountered a major problememanating from the behaviour of neutrons as they passedthrough various materials. It was dangerous and hazardousto experiment on a trial and error basis, yet it was absolutelynecessary to determine the thickness of the appropriateshielding materials. Finding that pure analytical proceduresalone would not solve their problem, Neumann and Ulam‘simulated’ the environment by using fictitious neutrons andsending them through representations of various materialsabout which basic data relating to speed, distance travelled,probable absorption rates, etc, were known. The key to thewhole process was the use of a device to generate therandom and independent selection of the particles, their paths,speeds, absorption rates, and so forth. The device used was ahigh speed digital computer and millions of iterations of theexperiment were conducted. This experiment was given thecode name ‘Monte-Carlo’, the gambling casino connotationcoming naturally, and applications of the process have usedthe name ever since.

Today, when Monte-Carlo is associated with an experiment,some device is used that provides a subset of values from apopulation in such a way that each population element hasthe same chance of being selected.

Typical devices used to generate Monte-Carlo samples arerandom number tables and tapes, and digital computerprograms written to generate random numbers.

random number tables and tapes, and digital computerprograms written to generate random numbers.

Directly related to Monte-Carlo is the concept of ‘simula-tion’, and more specifically ‘Monte-Carlo simulation’. Anddirectly related to simulation is the process of modelling.Placing these processes and techniques in properperspective, requires a look at the basic abstraction processknown as model building.

ModelsModelsModelsModelsModels

A model is a representation of an object a system, or aprocess. A descriptive synonym for the word ‘representation’in the above definition is ‘abstraction’. Whether the model isphysical (a lookalike representation) or analogue (where oneset of properties is used to represent another set) orsymbolic (where the elements of what is being representedand their interrelationships are represented by mathemati-cal equations), abstraction is the key activity involved. InMonte-Carlo Simulation we primarily use mathematicalmodels for exploratory, predictive, or control purposes.

The mere formulation of mathematical models does notnecessarily ensure that valid predictions about theunderlying system can be made using either analyticalmethods or experimentation. Figure 12-1 illustrates thetypical modelling process.

Figure 12-1: A Typical Modelling ProcessFigure 12-1: A Typical Modelling ProcessFigure 12-1: A Typical Modelling ProcessFigure 12-1: A Typical Modelling ProcessFigure 12-1: A Typical Modelling Process

Simulation enters the scene when any of the implied stepsof modelling depicted in the diagram are either impossible ortoo costly to perform. For example, Block A may have insuf-ficient accumulated data or observations of the system toadequately specify the hypothesis which finally becomes thesymbolic model. Block B may be inadequately specifiedbecause of the complexity of the observed system. The stateof analytic technology, for symbolic manipulation (Block C)may be inadequate to provide solutions to the model, andtherefore to provide predictions (Block D) about the futurebehaviour of the system, even if sufficient information wereavailable from Block E. And finally, Block F, the validationand evaluation of the model, might be too costly to perform,or have too little test data available coming from Block G.

The paramount difficulty experienced about in Blocks A,E, F and G is the lack of sufficient data or information. It isbecause of these problems that Monte-Carlo Simulationoffers assistance to the modeler. Simulation can be definedsimply as a numerical technique for experimenting with theelements and relationships of mathematical models forpredicting most likely outcome of a situation.

Steps or Phases in a Simulation StudySteps or Phases in a Simulation StudySteps or Phases in a Simulation StudySteps or Phases in a Simulation StudySteps or Phases in a Simulation Study

Proper problem formulation is the first step in simulation(Figure 12-2). This involves translation of the vague idea thatthe organisation has, into an explicit problem statement bythe analyst through continual interactions, questionnaires, andestimations of results. Second step - the most timeconsuming - is of data collection. Quantitative data are requiredto describe the real system and for generation of variousstochastic variables in the system. Also, it is required fortesting and validation of the model.

Next step of data analysis leads to deriving the propergenerating functions, after determining the distributionthat describes the raw data. Model formulation is a difficultstep in the process as simulation model is an abstraction ofsome real phenomenon or system. It should adequatelydescribe the real system at a minimum cost of human and

Figure 12-2: Phases in Simulation StudyFigure 12-2: Phases in Simulation StudyFigure 12-2: Phases in Simulation StudyFigure 12-2: Phases in Simulation StudyFigure 12-2: Phases in Simulation Study

computer resources. Computer is anabsolute necessity of program generation in simulations,therefore, computer language used is a matter of someconsequence. Special purpose simulation languages, such asSIMSCRIPT, GPSS, DYNAMO and GASP simplifyprogramming with features like the simulation clock andnext-event logic being pre-programmed.

Model validation provides confidence in the model bymaking certain verifications. It involves testing for congruencebetween simulator design and program; outputs and real-worldvariables; subjective human validation and finally comparisonof output with historical data. Once a simulation model hasbeen implemented and validated, it can be used for itsoriginal purpose, experimentation, to gather theinformation necessary for decision making. Analysis ofsimulation results, is fairly straightforward drawing ofinferences for rational decision-making often using certainstatistical techniques.

When to SimulateWhen to SimulateWhen to SimulateWhen to SimulateWhen to Simulate

On formulation of the problem, the analyst must decidewhether or not to attempt to solve the problem using dis-crete digital simulation. Figure 12-3 depicts this decisionprocess. A solution technique other than simulation is oftenjudged to be more appropriate. The important point is,selection of solution methodology must follow problemformulation and not visa-versa. If a classical analyticaltechnique is not available then decision between simulationand making an intuitive, seat-of-the-pant decision be made.

Figure 12-3: When to SimulateFigure 12-3: When to SimulateFigure 12-3: When to SimulateFigure 12-3: When to SimulateFigure 12-3: When to Simulate

Purpose of SimulationPurpose of SimulationPurpose of SimulationPurpose of SimulationPurpose of Simulation

The advantages and purpose of simulation could beunderstood from the following:

q Conducting of exercise is a very costly method of testingin terms money, time and resources. Even physicalobservation of a system becomes too expensive, at times.

q A number of aspects may be missed out in a large scaleexercise and if it so to be investigated, repetition may notbe possible.

q Mathematical models give us greater insight into the sys-tem, are easier to handle and lend themselves tocomputerisation. However, it is not possible to developmathematical solutions for all kinds of problems.

q When dealing with future weapon systems andfuture combat environment, it is not possible toconduct exercises.

q At times, systems may be so complex with large numberof probabilistic elements which do not lend themselvesto easy handling by even mathematical models.

q Sufficient time is not available to allow the systemto operate extensively.

q Actual operation and observation of the systemmay be too disruptive.

Limitations of SimulationLimitations of SimulationLimitations of SimulationLimitations of SimulationLimitations of Simulation

Simulation is not without limitations and some of them are:

q Is not precise, or an optimisation process, or can ityield an exact answer; but merely provides a set ofsystem responses to different operating conditions.

q It may consume tremendous amount of time andmoney to develop and run the simulation.

q Not all situations can be evaluated by simulation.

q Simulation can generate a way to evaluatingsolutions but does not generate solutions themselves.

SIMULATION PROCESSSIMULATION PROCESSSIMULATION PROCESSSIMULATION PROCESSSIMULATION PROCESS

Bomber Mission.Bomber Mission.Bomber Mission.Bomber Mission.Bomber Mission. The process of simulation will now beexplained with an example of a bomber mission.

Scenario. A bombing mission is sent to bomb a military target,which is rectangular in shape and has the dimensions 75 mtr by

150 mtr. The bombers will drop 10 bombs altogether, allaimed at the geometric centre of the target. It is assumedthat the bombing run is made parallel to the longestdimension of the target, that deviation of the impact pointfrom the aiming point is normal with mean zero, andstandard deviation 50 mtr in each dimension and that thesetwo deviations are independent random variables. Use MonteCarlo simulation to estimate the expected number of bombhits, and compare the result with the exact value.

Solution. Solution. Solution. Solution. Solution. Let, X be the range error (undershoot orovershoot with respect to aiming point), and Y be the lineerror (left or right of flight line) as shown in Figure 124 here.

A hit results considering the target area when:

- 75 < X < 75 and - 37.5 < Y < 37.5.

If either of these conditions is violated the result would bea miss.

X Range Error ( Overshoot +ve Undershoot -ve) Y Line Error ( Right +ve Left -ve )

Figure 12-4: Bombing Mission - Target ParametersFigure 12-4: Bombing Mission - Target ParametersFigure 12-4: Bombing Mission - Target ParametersFigure 12-4: Bombing Mission - Target ParametersFigure 12-4: Bombing Mission - Target Parameters

Now if X and Y are measured in terms of their standarddeviates say, u and v respectively, and the standard deviationbeing 50 mtr in both direction, we can say,

(X - 0) (Y - 0)U = and v =

50 50

By substituting values for X and Y; the corresponding rangefor u and v for a hit can be found as

- 1.5 < u < 1.5 < and - 0.75 < v < 0.75

With help of a random Normal Number table (like oneextracted at Table 12-1), the results of bombing trials can be simu-lated. A Random Normal Number table, if not readily available,can be constructed with the help of random number table and the

Table 12.1: Random Normal Table in Units ofTable 12.1: Random Normal Table in Units ofTable 12.1: Random Normal Table in Units ofTable 12.1: Random Normal Table in Units ofTable 12.1: Random Normal Table in Units ofStandard Deviates (Z)Standard Deviates (Z)Standard Deviates (Z)Standard Deviates (Z)Standard Deviates (Z)

Standard Normal Probability Distribution table (similar toAppendix ‘A). The simple procedure is to pick one randomnumber and look for it by putting a decimal before it in thebody of normal probability table. Note down the correspond-ing Z value with its sign. If the table has maximum value upto0.4990 and the number is 0.5 or greater (random number5000 or more) then number is reduced by subtracting 0.5and then looking for the remainder, and a negative sign isassigned to Z value. For example, if random number happensto be 7612, look for 0.2612 (ie 0.7612 - 0.5) in the table. TheZ value corresponds to -0.71 and is noted in sequence, toconstruct the Random Normal Number table. Likewisereferring to several random numbers in sequence thetable can be constructed.

The simulation trials are carried out by reading standarddeviates from Table 12.1 sequentially either in rows orcolumns in the usual method of referring to random tables.The values, for example, we pick from this table are 0.80,-0.54, 0.42, -0.48, ... and so on. These are copied sequentiallyunder columns u and v. If both u and v lie within theirrespective ranges of ± 1.5 and ± 0.75, then only result is a hit,otherwise a miss. Result column of the table is accordinglycompleted, and total hits for the trial counted.

Results of the first three trials are given at Table 12-2. Thetrials are carried out by reading the standard deviates fromthe Table 12-1 sequentially either in rows or columns. In thistable, the standard deviates occur randomly within thenormal frequency distribution ie values around zero standarddeviation occur more frequently and values around ± 3standard deviations occur very rarely and normally reducingin between. These values which we get from the table whenwe read in columns are 0.80, - 0.54, 0.42, - 0.48, ... and so on,in this example.

These three trials yield 4.67 as the average number of hitsper mission. Many more trials should be conducted beforeone can have any real confidence in the results. One way ofestimating how many trials are necessary is to list thecumulated mean at the end of each trial and to stop the trialswhen the mean seems to settle down to a stable value. In thisexample we have:

After trial number : 1 2 3 4 5 6

Cumulated means as : - 5 4.67 - - -

So, more trials are necessary.

In this problem, unlike most Monte Carlo problems, an exactcalculation of the answer is much easier than the Monte Carlocalculation. The probability of a hit with a single bomb is equal to

Table 12-2: Results of Simulation TrialsTable 12-2: Results of Simulation TrialsTable 12-2: Results of Simulation TrialsTable 12-2: Results of Simulation TrialsTable 12-2: Results of Simulation Trials

area under the normal curve between -1.50 to 1.50 multipliedby area between -0.75 to 0.75

= 2 (0.4332) x 2 (0.2734) (from ‘Z’ Table at Appendix ‘A’)

= 0.866 x 0.547

= 0.474

The mean number of hits in a mission dropping 10 bombsis just 10 times this, or 4.74. This figure nearly agrees withthe result of simulation which is 4.67.


Monte Carlo Simulation is one of the most powerful andfrequently used technique in the discipline of OperationsResearch. It has been effectively used in many areas such asinventory, allocation, queueing, sequencing, replacement, wargaming, weapon systems analysis and strategic planning.

Simulation is most widespread as far as military problemsare concerned, and continues to be essential in the field ofwar gaming in most advanced countries, an area in which weare also now making a headway.

CHAPTER 13

NETWORK FLOWS

The network structure of some typical real world problemsthat deal with physical distribution systems, communicationroutes, pipelines, airline flight legs and so on, makes themamenable to solution by specific algorithms for Network Flows.Another reason for using network models is their solutionefficiency and capability to handle thousands of variables withhelp of computers in a reasonable time, where at times otherapproaches are impractical.

Before going into a step-by-step procedure or algorithmfor solution of network related problems, it would beappropriate to first familiarise with few terms associated withnetwork flows. Figure 13-1 shows a simple network toexplain these.

Fig 13-1: An Illustrative Simple NetworkFig 13-1: An Illustrative Simple NetworkFig 13-1: An Illustrative Simple NetworkFig 13-1: An Illustrative Simple NetworkFig 13-1: An Illustrative Simple Network

q Network. A set of nodes linked by branches or arcsassociated with flow of some type, is a network.

q Nodes. In a network, nodes represent beginning or end of an activity (or end points of a line representingdistance, time,

capacity etc). These are depicted by circles. A sourcenode is a point of entry in specific types of networks,and sink (or output) node is the final exit point.

q Branches. Lines joining nodes and referring toactivities, distances, costs, times etc as the case maybe are called branches or arcs.

q Path. A sequence of connected branches such that inthe alternation of nodes and branches no node isrepeated, is a path in a network.

q Acyclic and Cyclic Network. A network is said to beacyclic if it contains no loops; otherwise it is cyclic.

Most of the problems related to network flows also called routescheduling or net flows, generally fall under one of the generalcategories viz shortest path, minimal spanning tree, maximal flows,capacitated network (CNet) and ‘travelling salesman solution’. Fig-ure 13-2 depicts these diagrammatically.

q Shortest Route on a network depicts the connectedbranches (representing distance, cost, or time) in anetwork that

Fig 13-2: Some Typical Network Flow Problems

collectively comprise shortest ‘distance’ between asource and destination. The path may not necessarilypass through all nodes.

q Minimal Spanning Tree is a connected set of all nodes ofthe network containing no loops and connecting all nodesat minimum total value (distance, cost, time).

q Maximal Flow Network is a directed networkcharacterised by flow capacity of the branches anddetermines the maximum flow that can pass from oneinput node (source) through the network to one outputnode (sink) during a specified period.

q Travelling Salesman’s Problem is concerned with theminimisation of total travel cost, distance, time, orcombination of these. Where all nodes are required tobe connected only once in a continuous path and returnto the origin. There is no general algorithm availablefor this, apparently trivial, problem. There are a largenumber of solving methods but all are enumerative tosome degree or another and hence become impracticalfor large sized problems. This specific type of problemis not being covered here.

SHORTEST ROUTE PROBLEMSHORTEST ROUTE PROBLEMSHORTEST ROUTE PROBLEMSHORTEST ROUTE PROBLEMSHORTEST ROUTE PROBLEM

Ferry CrossingFerry CrossingFerry CrossingFerry CrossingFerry Crossing

No 543 Engr Regt operates four ferry services daily acrossBrahamputra river from Dibrugarh to Sikraghat. Due todirection of river flow, strength of currents, presence ofintervening islands and minimum water depth required forcruising; the route followed is circuitous and takes well overan hour. The CO of the regiment is concerned with highconsumption of diesel and also wants to reduce the ferrycrossing time for passengers. Over the period, a number ofpoints on various navigable routes have been identified andtravel times, in minutes, for each segment recorded. Theseare depicted besides the sketch at Figure 13-3.

Figure 13-3: Shortest Route Problem - Ferry CrossingFigure 13-3: Shortest Route Problem - Ferry CrossingFigure 13-3: Shortest Route Problem - Ferry CrossingFigure 13-3: Shortest Route Problem - Ferry CrossingFigure 13-3: Shortest Route Problem - Ferry Crossing

The problem is to locate that route which would takeminimum time for the crossing the river each way. Thesolution for one way crossing from Dibrugarh to Sikragbat isgiven here. Corollary of the problem with related data couldbe solved for the return journey in a similar way.

Solution.Solution.Solution.Solution.Solution. The algorithm to solve shortest route problemis iterative in nature; at each step determining the shortestdistance to one node, thus after (n-1) steps the procedureends for a network with n nodes. The network at Figure 13-4has been drawn from the information at the sketch. Segmenttimings which are not directly related to respectivedistances have been marked near each node

Figure 13-4: Segment Times of Ferry CrossingFigure 13-4: Segment Times of Ferry CrossingFigure 13-4: Segment Times of Ferry CrossingFigure 13-4: Segment Times of Ferry CrossingFigure 13-4: Segment Times of Ferry Crossing

corresponding to direction of travel. The iterative algorithmto find a route from Dibrugarh to Sikraghat with shortest timeis as under:

q Step 1. Construct a master-list, tabulating each node inthe first row (refers Table 13-1 Next page). Under eachnode, in ascending order of their values, list all thebranches incident on that node. Thus start node of eachbranch in a column is same as the one at the top of thecolumn. Omit in listing all those branches which havethe source (origin) as their endnodes or the sink(destination) as their start-nodes.

q Step 2. Mark a star on the source node and assign it avalue 0. Locate the branch with the smallest valueunder the source column and circle it Star the node thatmatches the end-node of this branch, and assign thisnode a value equal to that of the branch. Strike off fromthe master list all other branches that have the newlystarred node as their end-node.

q Step 3. If the newly starred node is the sink, go to step6, otherwise continue.

q Step 4. Consider all those starred node columns that have anyuncrossed or uncircled branches left in the master list. Foreach column, add to the assigned value of the node value ofthe uncircled branch with smallest value in that column.Denote the smallest of these sums as Z, and circle that branchwhose value contributed to Z. Star the end-node of this branchand assign it the value Z.

q Step 5. Delete from the master-list all other branches havingthis newly starred node as end-node. Go to Step 3.

q Step 6. Z* is the final value assigned to the sink.Identify the shortest route recursively.

Corresponding steps and iterations in above example are:

q Step 1. In the master-list BA, CA, GB, GD, GF and GEare omitted.

Fig 13-1: Master-Sheet at Successive IterationsFig 13-1: Master-Sheet at Successive IterationsFig 13-1: Master-Sheet at Successive IterationsFig 13-1: Master-Sheet at Successive IterationsFig 13-1: Master-Sheet at Successive Iterations

q Step 2. Node A (the source) is starred and assignedvalue 0. AC10 is circled. Node C is starred and assignedvalue 10. Branches DC and EC are crossed out.

q Step 3. Newly starred node is C, an intermediate node,hence go to next step.

q Step 4. Columns A and C are considered. Z is lesser of sumsfrom columns A and C, ie between 40 (0 +40) and 20 (10 + 10)respectively. D is starred and assigned value 20.

q Step 5. Branch BD is struck off.

q Continuing Step 3 through 4. Z being least of the sums;40, 30 and 40 for columns A, C and D respectively, equals30 at this stage. Therefore CE is circled, E is starredand assigned value 30. FE is struck off. Next Z = 35, viaEF is assigned to F. Least Z at this stage is obtained viaAB and DB, both being equal to 40. AB is arbitrarilyselected and circled. Column B is starred and assignedvalue 40. Next Z = 45 via FG takes us to the sink, G, andthe process completes.

q Step 6. Z* = 45. Beginning with sink, by including inthe path each circled branch whose end node belongsto the path. Starting at G, column F provides connectingbranch FG, column E gives next branch EF, C ProvidesCE and finally column A gives AC. The shortest path forthe ferry from Dibrugarh to Sikraghat thus identified isACEFG = 45 mins, sum of the branches on the path.

Figure 13-5 shows the route marked on the sketch. Readermay work out the return route.

Figure 13-5: Ferry Crossing - Shortest Route SolutionFigure 13-5: Ferry Crossing - Shortest Route SolutionFigure 13-5: Ferry Crossing - Shortest Route SolutionFigure 13-5: Ferry Crossing - Shortest Route SolutionFigure 13-5: Ferry Crossing - Shortest Route Solution

MINIMUM SPANNING TREEMINIMUM SPANNING TREEMINIMUM SPANNING TREEMINIMUM SPANNING TREEMINIMUM SPANNING TREE

A problem under this group involves selection of a set ofbranches that span (connect) all nodes of a network so thattotal value of connected branches is to be minimum. Thestructure resembles a ‘tree’ graph connecting all nodes andforming no closed loops. A tree with n nodes will containn-1 branches.

Laying of Underground (UG)CableLaying of Underground (UG)CableLaying of Underground (UG)CableLaying of Underground (UG)CableLaying of Underground (UG)Cable

With ensuing termination of long-term contract betweenDoT and the Army, 456 Signals Company is to plan laying ofunderground cable for telephone communication withinRampur Cantonment. On the sketch at Figure 13-6, locationof the Signal Centre (Y), civil exchange (Z) and all therequired distribution points (DPs) are marked. DPs have beenidentified taking into consideration requirements of load,location of subscribers, expansion plans, and other factors offeasibility of routings. The routings from DPs are to be inclusters and overground - however not a part of this project.

Lines cannot be laid along direct straight lines connecting anytwo DPs due to existing structures, natural obstructions, and at placesfor reasons of security, ownership of land and maintenance

Fig 13-6: laying UG Cable - Minimal Spanning Tree Problem

considerations. The costs to interconnect DPs on all possibleroutes have been worked out taking the facts on ground andall associated costs into account. These costs (in lacs ofrupees) have been indicated on the sketch. The problem isto identify all the branches which would interconnect all nodes(DPs) and the exchanges - military and civil - of the network,with the aim of minimising the overall cost.

Solution.Solution.Solution.Solution.Solution. The graphical algorithm for finding minimal span-ning tree is quite straight forward as given here.

q Step 1. Start at any node at random and join it to itsnearest neighbouring node.

q Step 2. Choose the unconnected node which is nearestto any of the connected nodes, and join it to that node.

q Step 3. If there are still any unconnected nodes, repeatStep 2, otherwise stop.

For a small network, solution can be obtained graphicallyas above. But for a large network the sequence should befollowed in a tabular format to avoid any errors of omission orincorrect selection.

Tabular Method.Tabular Method.Tabular Method.Tabular Method.Tabular Method. A table listing all nodes for columns androws in the same sequence is made. Body of the tableindicates branch values between the nodes. The branchesbetween nodes that cannot be connected, are left blankindicating non-availability. Such a table for UG Cable problemis at Table 13-2 (next page).

q Arbitrarily we select any node, say node Y and mark itby # sign. Cross out column Y. Circle the least value inthe row ie 8 which is in column F.

q Next we mark F row with #, cross out column F and findthe least value (among 17, 13, 22, 20, 14 and 22) inmarked rows (Y and F) and circle it. Thus value 13 incolumn I is circled.

Table 13-2: Minimal Spanning Tree TabularTable 13-2: Minimal Spanning Tree TabularTable 13-2: Minimal Spanning Tree TabularTable 13-2: Minimal Spanning Tree TabularTable 13-2: Minimal Spanning Tree Tabular

It takes us to I row for the third iteration. Crossing outcolumn I, and considering F, Y and I rows the least value11 from column H is circled. The process is continued.

q After tenth iteration we find that no cell is available hence, theprocess is complete. Each # marked or circled node as well ascross-out lines for columns have been appended with iterationserial numbers within small triangles, tofacilitate following the step-by-step procedure by the reader.

q The resulting tree is traced by highlighting the circledbranches on the sketch. Figure 13-7 (next page) showsthis for the UG Cable problem. Minimum cost, sum of thehighlighted branches, equals Rs 102 lacs. One could startfrom any other node and confirm the results.

Figure 13-7: Laying UG Cable - Minimal Spanning Tree SolutionFigure 13-7: Laying UG Cable - Minimal Spanning Tree SolutionFigure 13-7: Laying UG Cable - Minimal Spanning Tree SolutionFigure 13-7: Laying UG Cable - Minimal Spanning Tree SolutionFigure 13-7: Laying UG Cable - Minimal Spanning Tree Solution

MAXIMAL FLOW PROBLEM

When the objective is to determine the maximum amountof flow (ie fluid, traffic, information etc) that can betransmitted through a network, Maximal Flow algorithm canbe used. It concerns with a single input node (the source)and a single output node (the sink), and is helpful in capacityplanning of varied distribution systems.

The maximal flow problem is characterised by the branchesof the network having finite capacities which limit the amountof flow that can pass through the branch in a given amount oftime. The problem is to determine a ‘break-through’ path,subject to capacity limits, the amount of flow across eachbranch that will permit the maximal total flow from source tosink. The problem may pertain to networks consisting ofdirected or undirected branches. Physical interpretation of adirected branch is one-way street in city traffic system withits associated flow capacity. A network of gaspipelines,connected at valves, represents a network of undirectedbranches - gas may flow in either direction through a pipelineregardless of flow direction.

This special algorithm’s first guideline is that except forsource -and sink, whatever flows into a node must flow out ofthe node. Second, in a flow across a path from source to sink.the maximum amount that can flow equals the minimumcapacity of any’branch on that path. This is similar to theweakest link in the chain analogy. Finally a mechanism shouldbe provided in the search for optimal allocation along variouspaths for adjusting (or changing) them at a later stage in thesearch if they are not optimal. A two step mechanismenables this to be achieved. While assigning a flow to a branch,first the capacity in the direction of flow by the amount of theflow is reduced, and second the capacity in the oppositedirection of flow increased by the amount of flow. Second stepmight seem strange, but allows to ‘undo’ a flow that is notoptimal.

Maximal Flow Algorithm.Maximal Flow Algorithm.Maximal Flow Algorithm.Maximal Flow Algorithm.Maximal Flow Algorithm. Following the guidelines givenabove the steps in the procedure are:

q Step 1. Find a path from source node to sink node withpositive flow capacity. If no path with positive flowcapacity exists stop. The current flows are optimal. Thestart can be made from any path from the source node.

q Step 2. Find the branch in the path that has theminimum flow capacity, Cmin. Increase the flow along thepath by Cmin

q Step 3. Decrease the capacities in the direction of flowby Cmin for all branches along the path. Increase thecapacities in the opposite direction by Cmin for all thebranches in the path. Go to Step 1.

Different decision makers might make different choices in Step 1.However, properly executed algorithm will provide optimal solution.

Transportation of Helicopter AssembliesTransportation of Helicopter AssembliesTransportation of Helicopter AssembliesTransportation of Helicopter AssembliesTransportation of Helicopter Assemblies

HQ Eastern Air Command, IAF is faced with a situation where30 helicopters in completely knocked down (CKD) state, are to be

collected from Mumbai. The consignment is arriving by seathis week-end and should reach Gauhati within a fortnight forassembly there by the foreign supplier’s team. The cratescan be transported via Calcutta by three different means, vizby-road in special trailer vehicles, by-rail and by-air inservice aircraft only.

The consignment may be moved via the same mode oftransportation for the complete trip, or it may switch modesat Calcutta. There are constraints on all the three modes. Onlylimited numbers of specialised trailers are available at Mumbaiand Calcutta for transportation by road. The transportation byrail is restricted due to limited number of wagons readilyavailable at both places. The air-staff do not wish to launchspecial sorties for air-lift and would use the routine courierflights from both cities to Gauhati. Complete information onthe capacities of three modes available for moving theconsignment during next fortnight, is now available with thestaff. The information is at Table 13-3. Movements fromharbour to airport and railway-yard at Mumbai, and from rail-way yard to airport at Gauhati can be handled locally.

Table 13-3: Capacities of Various BranchesTable 13-3: Capacities of Various BranchesTable 13-3: Capacities of Various BranchesTable 13-3: Capacities of Various BranchesTable 13-3: Capacities of Various Branches

The air-staff wish to determine if the task can be achievedusing existing capabilities.

Solution. Solution. Solution. Solution. Solution. The data in Table 13-3 have been converted intoa network at Figure 13-8. The flow capacities along eachbranch are indicated next to each node. These may varyaccording to direction of flow which is also indicated. Thebranches labelled A, T and R represent air, trailer (road) andrail links respectively. Nodes CA, CT and CR represent thethree transhipment points; ie airport, road terminal andrailway yard respectively, at Calcutta.

Figure 13-8: Network for Transportation of Helicopter AssembilesFigure 13-8: Network for Transportation of Helicopter AssembilesFigure 13-8: Network for Transportation of Helicopter AssembilesFigure 13-8: Network for Transportation of Helicopter AssembilesFigure 13-8: Network for Transportation of Helicopter Assembiles

Referring to Figure 13-9 and following the algorithmpresented earlier under para titled Maximal Flow Algorithm,we identify path M-CA-G (the air route via Calcutta) as allow-ing positive flows through each branch on the path. The flowcapacity for branch M-CA is 8. The flow capacity for M-CA iscompared with C-AG which gives optimum flow = 8. Thus 8CKD helicopters are allocated along this path. Flowcapacities of all nodes are adjusted along the path asindicated in Figure 13-9(i).

Arbitrarily, we identify flow potential along path M-CT-G.Comparing M-CT = 20 and CT-G = 10, we allocate only 10CKD helicopters to this path and re-adjust the flowcapacities. Refers Figure 13-9 Q.

Further, flow potential is identified along the path M-CT-CA-G(ie by road-trailer to Calcutta and then by air to Gauhati). Optimal

Figure 13-9: Step-by-Step Allocations forTransportation of Helicopter Assemblies

which is 6, is allocated and all flow capacities along the pathadjusted as in Figure 13-9 (iii).

At this stage it can be seen that the air capacity has beenfully used. Continuing further, when all the three branches toGauhati get fully exhausted, as is the case depicted in Figure13-9(iv),the solution is obtained. It suggests a total capacityof 30, therefore, the task can be just met with the availablecapacity of the resources. The schedule according to thissolution is; send 8 CKD helicopters by air, 6 by road trailersand 16 by rail to Calcutta. From thereon the distribution couldbe 14 by air, 10 by road trailers and 6 by rail to Gauhati.

Arbitrary selection of paths gives different combinationsbut finally maximum capacity remains the same. In this case,for example, if air effort from Mumbai is not used, 30 CKDhelicopters could still be transported to Calcutta byroad-trailers and rail in various possible combinations, and thento Gauhati as suggested in the solution.

Problems with large number of nodes solved with help ofa computer will bring out various alternatives from whichdecision maker can select the most suitable one.


In this chapter three kinds of applications of network mod-elling were presented. Although special algorithms for eachnetwork model have been discussed, most of the specialalgorithms are subsume of Simplex method. Nevertheless,the specialised algorithms do offer computational advantage.Variations of tabular and graphical network algorithms arealso found with their merits. For example, different algorithmscan be used in finding shortest-route in acyclic and cyclic networks. The one for the cyclic network being moregeneral includes the acyclic case but is not as efficientbecause it entails more computation.

Applications of these techniques extend beyond theproblems dealing with distances, costs and time oftransportation, flow or capacity. For example, shortest-routealgorithm may be applied to decide on equipmentreplacement policy or viewed as a transhipment model - aspecial case of transportation model. Considering thebranches as probabilities, ‘the most reliable route’ can bedetermined. A variation of minimum spanning tree emergeswhen it is needed that all nodes are connected to one centralnode. If the branches have capacity limitations imposed, theproblem becomes a ‘capacitated minimum spanning tree’ -much more difficult to solve. However heuristic andoptimisation procedures do exist to solve it. Large andcomplex problems of networks as mentioned can be handledby various computer software packages.

APPENDIX ‘A’Areas under the Standard Normal Probability Distribution between the Mean andSuccessive Value of z.

Example: To find the area under the curve between the mean and a point 2.24 standarddeviations to the right of the mean look up the value opposite 2.2 and below 0.04 in thetable. 0.4875 of the area under the curve lies between the mean and a Z value of 2.24.

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .090.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .03590.1 .0398 .0438 .0478 .0517 .0.577 .0596 .0636 .0675 .0714 .07530.2 0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .11410.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .15170.4 .1554 .1591 .1628 .1664 1700 .1736 .1772 .1808 .1844 .18790.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .22240.6 .2257 .2291 .2324 .2357 .2389 .2454 .2486 .2486 .2517 .25490.7 .2580 .2611 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .28520.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .31330.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .33891.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .36211.1 .3643 .3665 .3686 .3708 .3729 .3749 .3770 .3790 .3810 .38301.2 .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .40151.3 .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .41771.4 .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .43191.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .44411.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .45451.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .46331.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .47061.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .47672.O .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .48172.1 .4821 .4826 .4830 .4834 .4838 .4842 .4846 .4850 .4854 .4857

2.2 .4861 .4864 .4868 .4875 .4871 .4878 .4881 .4884 .4887 .48902.3 .4893 .3896 .3898 .4901 .4904 .4906 .4909 .4911 .4913 .49162.4 .4918 .4920 .4922 .4925 .4927 .4929 .4931 .4932 .4934 .49362.5 .4938 .4940 .4941 .4943 .4945 .4946 .4948 .4949 .4951 .49522.6 .4953 .4955 .4956 .4957 .4959 .4960 .4961 .4962 .4963 .49462.7 .4965 .4966 .4967 .4968 .4969 .4970 .4971 .4972 .4973 .49742.8 .4974 .4975 .4976 .4977 .497 7 .4978 .4979 .4979 .4980 .49812.9 .4981 .4982 .4983 .4983 .4984 .4984 .4985 .4985 .4986 .49863.0 .4987 .4987 .4987 .4988 .4988 .4989 .4989 .4989 .4990 .4990

From Robert D Mason, Essentials of Statistics, Prentice Hall,Inc.1976.

BIBLIOGRAPHYBIBLIOGRAPHYBIBLIOGRAPHYBIBLIOGRAPHYBIBLIOGRAPHY

STATISTICSSTATISTICSSTATISTICSSTATISTICSSTATISTICS

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2. Simpson and Kafka. Basic Statistics, Oxford & IBH.

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4. Moroney. Facts from Figure, Pelican Books.

OPERATIONS RESEARCHOPERATIONS RESEARCHOPERATIONS RESEARCHOPERATIONS RESEARCHOPERATIONS RESEARCH

5. Levin and Kirkpatrick. Quantative Approaches toManagement, Mc Graw Hill.

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9. Vohra ND. Quantitative Techniques in Management,Tata Mc Graw-Hill, New Delhi.

10. Budnick Fronk S, Mcleavey & Mojena R. Principles ofOperations Research for Management; Richard D IrwinInc, Illinois.

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Quantitative Aids to Decision Making

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