Effective Management Decision Making Chapter 4 PDF

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This document is chapter 4 of a textbook titled "Effective Management Decision Making". It covers topics like distribution functions, queuing theory, discrete and continuous variables, and the binomial function. The chapter uses examples to illustrate these concepts.

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Chapter 4 Chapter 4 After studying this chapter, you should be able to understand: 1. Understand and apply distribution function and queuing theory. 2. Understand and apply the mathematical function: discrete and continuous variables. 3. Understand and apply the discrete distribution as well as the...

Chapter 4 Chapter 4 After studying this chapter, you should be able to understand: 1. Understand and apply distribution function and queuing theory. 2. Understand and apply the mathematical function: discrete and continuous variables. 3. Understand and apply the discrete distribution as well as the binomial function. 4.1 Developing rational models with quantitative methods and analysis: Distribution Functions and Queuing Theory For a majority of Business / Management students, the preceding chapter on probability is usually manageable and accessible to them, as they are able to relate and visualize the range of outcomes that might follow from a decision making context. However, that discussion represents a narrow and atypical environment, within which probability can be studied. It is one extreme and simplified interpretation of a much wider set of relationships that can be called distribution functions. The easiest method of considering this is to reflect on how useful plotting a decision tree would be if the outcome of a decision was a hundred (or more) possible pathways. This rational approach to determining decision and outcome nodes, then rapidly becomes cumbersome. Yet, the majority of decision situations for individuals and organizations to manage are of this type. Consider for example how you would plot the likelihood of a given customer attending your retail store on a given day? If your business was highly unique with only a few known customers world-wide, you might be able to construct a probability table based upon your knowledge of the market and customer needs therein. However, for the vast majority of organizations, this is simply not feasible. We are therefore left in a quandary about how to manage this kind of decision making problem. To address this, we need to consider some more mathematics. 4.2 The mathematical function: discrete and continuous variables Mathematicians categorize data distinctively, depending upon its origin and scope. For example, in the preceding chapter on probability, all the data we were presented with described fully all possible potential outcomes. We could therefore know exactly and describe fully a given decision context. When this is the situation, the data values used are called discrete random variables. Usually, this data is presented (and comes from decision contexts) as positive integer (whole number) data. For example, in a sample of 6 students chosen at random from a class of 30, how many of the size might have black hair? Clearly, the answer must be whole number and as we are concerned with the sample of 6, we can know and describe all outcomes. The use of discrete random variables is (probably) familiar to you from you research methods courses and studies (or similar). Other examples would be determining the number of defects in products produced on a production line, from a sample of 100, or interviewing 50 people at random in the street about who they voted for in the last national election. One easy way to think about this type of data, is that it is usually used to count, not measure items. 4.2 The mathematical function: discrete and continuous variables Continuous random variables however are the opposite of this, in that they can take any value between two interval points – i.e. they measure rather than count. For example, in the same class of 30 students, now studying in an exam, the time taken to complete the exam by each student, can be any measured value from the start (time =0) to the scheduled end of the exam (t=end). Thus many measurements of a potentially infinite level of accuracy can be taken. Data gathered in this way and data gathered for the sampled black haired students, constitute different distributions of data. Hence: Data gathered from discrete random variables construct discrete distributions of data Data gathered from continuous random variables construct continuous distributions of data 4.2 The mathematical function: discrete and continuous variables The phrase ‘distribution’ often generates fear for students who have not met it before. It should not. It is simply a word used to describe how the data is arranged and presented. For example, let’s go back to our class of 30 and their exam and assume that the exam was such that each question answered could only be right or wrong. You ask one of the students to predict how likely it is that they will get X questions wrong and they generate this table for you: Number of questions answered incorrectly Predicted probability 0 0.02 1 0.23 2 0.39 3 0.26 4 0.10 4.2 The mathematical function: discrete and continuous variables Clearly, from the above table, the interviewed students is 100% confident they will not score 5 or more questions incorrectly, but is less confident that they will score less than 5 questions incorrectly, as per the probabilities given. A plot of this discrete distribution function would then be: 4.2 The mathematical function: discrete and continuous variables Figure 4.1 illustrates the plot of the discrete distribution function of the data gathered. We would then be able to for example, determine the mean probability of the long run average of occurrences. This is where we move beyond asking one person their view and ask everyone in class – i.e. we have more than one trial (this is what is meant by the statement ‘ long run average’). So – the mean (μ or expected value) is given by the sum of the product of the number of possible incorrect answers and the likelihood of achieving those incorrect answers, or: μ= Σ [(number of incorrect answers given in exam) x (probability of achieving that many incorrect answers)] OR More generallyμ= Σ [(outcome of interest) x (probability of that outcome of interest)] 4.2 The mathematical function: discrete and continuous variables You can also easily work out the variance and standard deviation of the mean of a discrete distribution. If you are unfamiliar with these terms, then any introductory statistical text will guide you, although a short definition here would be: variance (σ2) – the measurement of how far numbers in a given dataset, are spread out from one another. Standard deviation (σ) is a closely associated term and describes how much spread there is in the data from the expected mean (determined above as μ). So: σ2 = Σ [( number of incorrect answers given in exam – μ)2 x (probability of achieving that many incorrect answers in the exam)] OR σ2 = Σ [( x – μ)2 x (P(x))] And the standard deviation is the square root of σ2 or: σ =√( Σ [( x – μ)2 x (P(x))] ) 4.2 The mathematical function: discrete and continuous variables You can therefore easily construct a table of this data (using that from table 4.1) as: Number of Predicted x questions probabilit multiplie answered y (P(x)) d by P(x) incorrectl y (x) (x-mean) squared (x-mean) squared multiplied by P(x) 0 0.02 0.00 4.80 0.10 1 0.23 0.23 1.42 0.33 2 0.39 0.78 0.04 0.01 3 0.26 0.78 0.66 0.17 4 0.10 0.40 3.28 0.33 Table 4.2: Calculations for discrete distributions Hence mean= 2.19 Hence variance = 0.93 Hence standard deviation = 0.97 4.3 The Discrete Distribution The Binomial Function Over the centuries, certain types of discrete distributions of data have been identified and used professionally as they have patterns that can be used for modelling. These types of discrete distribution are more useful for the manager in particular to become familiar with. Let’s begin to investigate these firstly with the roulette wheel. some important implications of discrete distribution functions. 4.3 The Discrete Distribution The Binomial Function A gambling roulette wheel has a fixed number of slots (numbered) into which the spun ball can fall. When the ball falls into one of those slots, it cannot occupy any other (i.e. these are mutually exclusive events). The number of numbers which can be selected will not vary (i.e. there are only 37 numbers so the probability of a number’s slot appearing or not, will not change in a single trial) and therefore in subsequent spins of the wheel, whatever number appeared previously, is equally likely to appear again (subsequent events are independent and they are states of nature as we have previously discussed). So far, so good. 4.3 The Discrete Distribution The Binomial Function What if we were looking to determine how often a given number may appear in N spins of the roulette wheel? We can rephrase this slightly (in a more mathematical sense) and instead ask what is the probability of R successes (our number appearing) in N attempts or trials (N spins)? I.e. How many times would the number 3 appear (for example) in 5 spins? So – we can write this as: P(exactly R successes)=P (R successes AND (N-R) failures) OR P(exactly R successes)=P( R successes) x P(N-R failures) 4.3 The Discrete Distribution The Binomial Function If we now start to generate the simplest case here and assume that P (our number appearing) =p and therefore P (our number not appearing)=q, we must have that for any single event (i.e. one spin of the wheel) p+q=1. We spin the wheel 5 times (N=5) and we have asked to find out the probability of seeing the desired number (R times) in N spins. Now one possible outcome of these five spins is that we ONLY see our number appear (of the 37 possible choices). In you recall from Chapter 4, we know that the relationship between subsequent independent events is calculated using the logical operator AND – so that the P (3 in the first spin, 3 in the second spin, 3 in the third spin, 3 in the fourth spin and 3 in the fifth spin) is going to be given by the value: P (3 in ALL spins (or R=5)) = p5 (i.e. our chosen number probability multiplied by itself 5 times) OR P (R successes in N trials) = pR 4.3 The Discrete Distribution The Binomial Function But we might also see failures (when our chosen number does not appear (which is much more likely!1)) hence the probability of failure in N trials is going to be given by: P((N-R) trials being failures) = qN-R. If we now combine these probabilities and ask the question of what is the probability of seeing R successes in a given number of N spins, we are then exposed to both successes and failures in those N spins – or: P( The first R trials being successful and the next N-R trials being failures)= (pR * qN-R). Although this calculation will generate a probability of seeing a single given sequence of successes and failures in N spins, we must also recognise that this sequence (say of seeing our number 3 times in 5 spins) can be achieved by a variety of combinational possibilities. Hence we must also ‘count’ this number of potential other valid combinations of our ‘success’ (i.e. seeing our number 3, 3 times in 5 spins). Fortunately, another branch of mathematics allows you to easily determine the number of possible combinations of number orders we could see (when in this case, we are not concerned with what particular order our number appears in those 5 spins). This is given by the calculation: NCR= N! R!(N-R)! 4.3 The Discrete Distribution The Binomial Function NCR= N! R!(N-R)! So- in the equation above, there are R things (successful appearances of our number 3) to chose from in N possible outcomes and if the order of that selection does not matter (as is the case here), then clearly, there are many potential ways of presenting that selection (these are called combinations). The calculation above divides the factorial of N (the number of spins) by the product of the factorial of R (the number of successes we need (i.e. our number to appear three times) and the factorial of the difference between N and R. Thus for our scenario, the calculation will be: 4.3 The Discrete Distribution The Binomial Function NCR= 3!(2)! 5! = (5x4x3x2x1) (3x2x1)(2x1) OR NCR = 10 Hence there are 10 possible combinations where we will see our number appear 3, appear 3 times in 5 spins – e.g. (3,3,3,X,X), (3,3,X,X,3), (3,X,X,3,3), (X,X,3,3,3) (X,3,3,3,X), (3,3,X,X,3), (3,3,X,3,X), (X,3,X,3,3) (3,X,3,X,3), (3,X,3,3,X) 4.3 The Discrete Distribution The Binomial Function Hence there are NCR possible sequences of R successes and (N-R) failures, each with a probability of pR * qN-R. The overall probability of R successes in N trails is therefore: P (R successes in N trials) = NCR * pR * qN-R This type of distribution, where the range of number outcomes is extracted from a discrete (finite) set of possibilities, is known as the Binomial Probability Distribution. It is a very important number distribution in mathematics that has applications for managers. Binomial means there are only two possible state of nature outcomes to consider and that each trial (attempt or spin of our roulette wheel, is independent of whatever outcome has been previously observed). Consider for example the following problem: 4.3 The Discrete Distribution The Binomial Function 1) 2) P (1 sale OR 2 sales OR 4 sales OR 5 sales OR 6 sales) P (1-(no sales OR 3 sales)) Let’s work out (1) and (2) to illustrate they are in fact the same. P(R=1) = 6C1*(0.5)1*(0.56-1) = (6)(0.5)(0.03125) = 0.09375 (or 9.4%) P(R=2) = 6C2*(0.5)2*(0.56-2) = (15)(0.25)(0.0625)=0.23475 (or 23.5%) P(R=4) = 6C4*(0.5)4*(0.56-4) = (15)(0.0625)(0.25) = 0.23475 (or 23.5%) P(R=5) =6C5*(0.5)5*(0.56-5) = (6)(0.5)(0.03125) = 0.09375 (or 9.4%) P(R=6) = 6C6*(0.5)6*(0.56-6) = (1)(0.015625)(1) = 0.015625 (or 1.6%) 4.3 The Discrete Distribution The Binomial Function Hence the probability of making any number of sales EXCEPT exactly 3 will be the summation of the above probabilities (0.6726 or 67.3%). We arrive at the same solution if we now solve (2) above: P(no sales) = P(no success hence R=0) = 6C0*(0.5)0*(0.56-0)= (1)(1)(0.015625)=0.015625. We know the probability of exactly 3 sales successes as 0.3125. Hence we add these two vales together (in other words (1- (P(no success)+P(3 sales exactly)) = 1-0.328 = 0.67 (or 67%)). Hence the final question above, is simply determined by evaluating P(no sales OR 1 sale OR 2 sales). 4.3 The Discrete Distribution The Binomial Function Question - One final example to consider is to determine the following solution: in 2010, students on the business degree programme were sampled about their views on the support they had received during their studies of ‘Management Decision Making’. At that time 75% said they thought they had received ‘good’ support. Assuming this figure remains valid for 2011, 2012 and 2013 – let’s consider that if the class of 2011 is sampled (say 34 out of a class of 54) that they will also express the view that they have received ‘good’ support during their studies? 4.3 The Discrete Distribution The Binomial Function Solution P(success (or that the student felt they received ‘good’ support)) = 0.75 = R P(q failures (i.e. students commenting another view of their module support) = 0.25 For the 2011 class, we have 54 trials and wish to know the probability that 34 comments will be successful – hence we can use the equation: 54C34*(0.75)34*(0.25)20)=3.213*1014 x 5.6504*10-5 x 9.0949*10-13 = 0.0165 (or 1.65%) Hence it is very unlikely that for 2011, the tutor would see exactly 34 of the student’s commenting on the ‘good’ support received during the module of study. On average (in the long run (i.e. with repeated sampling)), the tutor can expect to see (0.75)(54) or 41 students that comment on their module’s ‘good’ support. 4.5 The Poisson Sequence This sequence of probabilities (numbers) is named after the French statistician Simeon-Denis Poisson, who published the essence of the formulation in 1837 The difficulty of explaining and discussing this number sequence and its value to managers and students of business, is that most texts on this topic, do not explain clearly how the Poisson distribution function is derived from the Binomial as the mathematics tends to be relegated to more rigorous mathematical texts. There is instead, a focus upon the value of the application of the sequence in practical organizational contexts and in the management and allocation of resources. This latter point is appropriate and most students can make the transition between the two distribution functions without needing to understand their linkage. For some, it is simply a matter of faith and then matching contexts to which interpretation of the distribution function to adopt. However, a detailed explanation of the origin of the Poisson distribution helps to understand its value to managers and so the following videos will present this. An Introduction to Poisson Distribution (https://www.youtube.com/watch?v=jmqZG6roVqU) Proof that the Binomial Distribution tends to the Poisson Distribution (https://www.youtube.com/watch?v=ceOwlHnVCqo) 4.5 The Poisson Sequence P( Of observing R successes give a rate of success λ in a given period) = (λR * e –λ)/ R! (j) This is known as the Poission Distribution Function – and can be derived from the binomial distribution function – by relaxing some of the key assumptions of that distribution (namely that N is very large and we only know the RATE of success achieved in a given time period). Phew! Now you may be wondering that this is interesting, but what is the value of it? How does it help you as a student of Business or a professional manager? The answer is that unlike the binomial function, this distribution allows you to view the ‘success’ of an event happening not from a probability calculation but only from its frequency in the past. It therefore can describe rare events or unlikely events against a ‘normal’ context of them not occurring. Put another way, you can use the number sequence described by (j) to help you understand discrete events occurring against a continuous background of them being very unlikely to occur. So it can be used for a manager to help them forecast demand (and hence resourcing needs). It will help you answer for example the following situations: 4.5 The Poisson Sequence How many customers will call in for Pizza at the local Pizza delivery shop on any given evening – (i.e. customers calling = success against an almost constant background of customers not calling (if all potential customers did call in – you would need a very large shop!)) The number of paint flaws evident of new cars sold in a given period The rate of arrival of aeroplanes at an airport requiring landing The incidents of accidents recorded at a factory over a given period. The distribution characteristics described by the Poisson function – where you have a low observed success rate against a background of (much more likely) non success rates – can be used in many areas of Management. Black (2004) advises that in using the Poisson distribution function – you must however be confident that the rate of success λ (e.g. customer arrival, paint flaws observed etc) remains relatively constant over a given time period (or at least the period in which you are concerned). This may mean for example that λ would vary between Monday’s and Friday’s for the Pizza Manager and any staffing rota thus produced to manage demand – should reflect this variation in λ. 4.9 Queuing Theory – Modelling reality The key features of understanding and modelling queues – reduces to a number of model parameters (if you assume certain probability distributions for the arrival of an event of interest and the rate at which that event is serviced. Thus recalling key parameters as: 1. the rate at which ‘customers’ arrive to be served (A) 2. the time taken to serve that customer (B) We can note that (1) and (2) will be shaped by how many staff are available to service the arrival of events of interest. Hence we can also add, that any model we develop for managing a queue process should also include: 3. the number of outlets / servers available to service a customer.(k) 4.10 Defining the Queuing Characteristics The discussion in this chapter so far has attempted to outline the broad rationales for the development of probability analysis when we move away from discrete quantifiable events, to random events occurring against a background of ‘non events’ (or at least events not of interest to us or the manager). This has been undertaken with a clear focus upon supporting the derivation of the mathematical models that result from this. However, the next stage of this chapter does make a number of assumptions regarding what are termed as ‘queuing characteristics’ when we begin to consider the A/B/k model formats for queues in general. You will find this is a common focus on the majority of undergraduate texts on quantitative decision making – where the derivation of the formulae used in queuing A/B/k models is presented rather than derived. Firstly, whilst of interest, the derivation of these formulae does not aid deeper understanding of these methods of decision making (that has already been presented through this chapter with a focus upon the fundamental conceptual building blocks), secondly – the derivations are time consuming and lengthy and thirdly, that a managers skill derives from the correct assessment of different impacts of different queuing A/B/k models and subsequent selection of appropriate derived formulae to use to analyze a queue per se. Queuing Model essence, arrival & service rate, 8 formulae (https://www.youtube.com/watch?v=lEfXlksz4uw) 4.11 Examples of M/M/1 Models You’ve bought 4 tickets to attend a circus which comes to town. On the night of the show you arrive early and find yourself at the back of a long queue waiting to get into the big top tent. As a student of business, you quickly analyse the queuing characteristics. You note the following general information: There is only one server at the head of the queue checking and selling the tickets. You are waiting and counting how many new customers arrive and how many are served over 3 ten minute periods. You average those observations to reach a conclusion that the arrival rate seems to be 15 every 10 minutes and the service rate seems to be 18 every 10 minutes. As far as you can observe, there are no factors about the service and operation which lead you to think that the arrival rate is not Poisson distributed and the service rate not exponentially distributed. You feel confident therefore that you are dealing with an M/M/1 system where λ= 15 every 10 minutes (or 90 per hour) and μ = 18 every 10 minutes (or 108 every hour). Now as you arrived in good time to see the show, you joined an existing queue. Your friends seem bored and ask you how long you think they will have to wait in the queue. What do you say? Answer: As: W = Lq = 5 hours. q λ/μ The M/M/1 Queue (https://www.youtube.com/watch?v=TlHUxvCo3Sg&t=29s) Queuing Theory Tutorial - Queues/Lines, Characteristics, Kendall Notation, M/M/1 Queues (https://www.youtube.com/watch?v=SqSUJ0UYWMQ) 4.14 The Economic Analysis of Queues We can formalize some of the relationships revealed in the above discussion. This is sometimes labeled as the economic analysis of queues. This is an important focus in decision making for queues and waiting line structures because this allows the manager to make comparative decisions regarding the cost-benefit of different waiting line structures. However – you should not just consider the cost implications of different waiting line structures – there may be other in kind / goodwill benefits that may warrant support for less efficient (financially) types of structure. You can for example make the queuing time more engaging with entertainment whilst queuing, or presenting a virtual presence in a queue until you near the front of it (such as happens in large theme parks around the world), or offer other services such as drinks and food (complimentary?), improve the environment of the queue to help pass time or use the queuing time to complete any relevant documentation (again typical of customs declarations or landing card information at airports (Waters, 1998)). In other words, the management science of queuing should not just be thought of, or practiced as a numerical analysis – but one that also considers other aspects of effective management of resources. 4.14 The Economic Analysis of Queues Returning to the cost considerations at present, we need to be able to present an analysis that articulates the total cost of a given waiting line structure. This is called a total cost model (described by the notation TC). Hence: TC = CwL + CsK Where Cw is the waiting cost per unit/person/entity per time period, L is as before the average length of the queue (or the number of units/persons/entities in the queue), Cs is the service cost per time period for each channel of service that ‘serves’ the queue and K is also as before, the number of channels in the waiting line system. It is not easy obtaining or estimating some of these costs – as for example, the cost Cw is not a direct cost to an organisation, but represents a opportunity cost loss as customers may decide to take their business elsewhere if the waiting time is deemed unacceptable. Service costs are usually easier to identify and determine (as with our circus example) as they refer to a business function with costed inputs. The basic relationship between service cost and waiting cost is familiar to students of microeconomics as ‘cost curves’. 4.14 The Economic Analysis of Queues Figure 4.5 Comparative cost curves for Queue waiting structures 4.14 The Economic Analysis of Queues A simple example might be the following – consider first of all, a specialist retail store selling model aero planes. This store operates one cashier till with one server on that till. That server has a variable cost associated with their salary, benefits and other direct costs associated with overheads and cash till rental. Let’s assume this works out to be $10 / hour. As noted about, the cost associated with customer waiting is harder to estimate but it could be expected that in this context, for customers specifically locating and entering the shop (with little competition in the immediate vicinity as the shop is specialized) it could be argued that the waiting cost would be lower than say a café on the seafront. We could therefore assign a waiting cost of $5 hour. As: TC = 5L + 10(1) Where L could be determined as before, from the parameters of the service (waiting and service times).

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