Notes Chapter No.4 PDF

UNIVERSITY OF BELIZE FACULTY OF MANAGEMENT & SOCIAL SCIENCE COURSE: Decision Making in Management (MGMT 4023) LECTURER: Dr. Romaldo Isaac Lewis (DBA) Chapter No.4; Developing rational models with quantitative methods and analysis: OBJECTIVES After studying this chapter, you should be able to underst...

UNIVERSITY OF BELIZE FACULTY OF MANAGEMENT & SOCIAL SCIENCE COURSE: Decision Making in Management (MGMT 4023) LECTURER: Dr. Romaldo Isaac Lewis (DBA) Chapter No.4; Developing rational models with quantitative methods and analysis: OBJECTIVES 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) 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 a typical 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 organisations 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 organisations, 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 categories 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. 1|Page 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. 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: 1. Data gathered from discrete random variables construct discrete distributions of data 2. Data gathered from continuous random variables construct continuous distributions of data 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 Table 4.1 2|Page 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: Figure 4.1 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)] 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: 3|Page 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 [( x – μ)2 x (P(x))] ) (x-mean) squared multiplied by P(x) (x-mean) squared x multiplied by P(x) Predicted probability (P(x)) Number of questions answered incorrectly (x) You can therefore easily construct a table of this data (using that from table 4.1) as: 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 This all seems good and interesting (!) – but 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. 4|Page 4.3) The discrete distribution – the binomial function Before we delve into the depths of mathematics further, let’s use an example you may be familiar with, to explore some important implications of discrete distribution functions. 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. 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) 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: 5|Page 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 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: \endash You can work out that the probability of seeing the number three appear, each time, in 5 spins is 0.0000000144 (or (1/37)5). Hence it is much more likely that you will not see our number 3 at all in any of the spins. 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: N! NCR = R!(N-R)! 6|Page So- in the equation above, there are R things (successful appearances of our number 3) to choose 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: NCR = 3!(2)! 5! = (5x4x3x2x1) 3!(2)! (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) 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: 7|Page A salesman knows historically he has a 50% chance of making a sale when calling on a customer. If he arranges 6 visits determine the following: What is the probability of making exactly three sales? What are the probabilities of making any other numbers of sales? What is the probability of making fewer than three sales? We have a discrete distribution profile (6 events) with 2 states of nature outcomes possible per event (a sale or no sale) and we know that, based upon historical records, the salesman has a 50% chance of securing a sale (one state of nature) per event (the visit). So we know we can apply the binomial distribution to understand this salesman’s endeavours. Let’s assume that the success of each visit (P(success)=0.5 or P(p)=0.5). We know N=6 (i.e. 6 trials (or visits for this example). The first question asks to determine the probability of securing 3 sales from those 6 visits. This answer is not, as you perhaps might first think just 50%, but is determined by P(R=3)= 6C3*(0.5)3*(0.56-3)=0.3125 or in longhand ((720/36)*0.125*0.125). Similarly, the probability of making any other number of sales means we would need to determine either the following: 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%) 8|Page 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). 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? 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. 9|Page 4.4) More on probabilities: Expected Monetary Values (EMVs) As a mathematical probability distribution of discrete numbers, the Binomial number sequence can be expanded. Let’s take now the generic example from above and expand this: P (R successes in N trials) = NCR * pR * qN-R P(R=0)+P(R=1)+P(R=2)+P(R=3)+P(R=4)+P(R=5)+P(R=6) This describes the full discrete distribution and can therefore be written as: P(discrete distribution) = 1P0Q6+6P1Q5+15P2Q4+20P3Q3+15P4Q2+6P5Q1+1P6Q0 (Which, you may have noted, is the expansion of (p+q)6) Clearly, this is useful providing we have a manageable limit on the value of N (and subsequently R). Let’s look at some further examples, before we consider what happens when N becomes very large. Suppose you are a quality inspector working on a production line and you are sampling the finished items at random for defects. You know from your historical data (which could be from yesterday for example) that the probability of selecting a faulty item at random is 5%. If you select an item and then don’t put it back into the sample, can you be said to still be following a binomial distribution of faults found in your selection? (i.e. of locating faulty ‘successes’ in your N selections (trials)?). Well no and yes. The ‘no’ response comes from the fact that by not replacing the item back into the population, the subsequent trial is not now independent of previous trials (as that trial event has affected the probability distribution of the remaining items in the sample and whether they may or may not be faulty). However, if your population sample is large (say you are selecting buttons at random from a large box of finished buttons) the impact of not replacing that one button when there are thousands still to be trialled in the box, is not significant. Black (2004) recommends that as a general guide, when the sample size n is less than 5% of the population size, we should not be concerned about the dependency of events (if the sample if not returned to the population) and can then view sampling as a series of independent events. If, however, the sample was say 20% of the population – then we can no longer assume that the sampling events are independent. The impact of a sample being taken will therefore affect the likelihood of a faulty part (or the event of interest) being observed in the next subsequent sampling undertaken. In addition, of course, we are also implicity concerned with very large values of N (and hence continuous distributions). In these situations, we need to reconsider the binomial sequence of numbers – and generate the Poisson sequence. 10 | P a g e 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 (citing Napoleonic Cavalry survival (as returns to their stables) as empirical data!). 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 organisational 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, in the author’s experience, a detailed explanation of the origin of the Poisson distribution helps to understand its value to managers and so the following section will present this. You are of course welcome to skip ahead to the section afterwards if you feel familiar with this derivation. Deriving Poisson from the Binomial distribution function You may recall that we can present the binomial function as the following generalized relationship: NCR * pR * qN-R Some texts present this as: Where the first bracket of n over k represents the combinational value previously discussed, k represents R (the success events of interest to us) and (n-k) is the same as N-R (i.e. number of trials less the number of successes of interest to us). Let’s break one of the conditions in which the binomial function works by saying we DON’T know what N will be (i.e. we don’t know how many trials or sampling we will undertake). We do know however; the average success rates we see over a time period (i.e. we know from our records that a shop may sell 20 pairs of shoes in a day). We therefore know the rate of success per day (in this example) but we don’t know the original probability of success occurring in a given number of trials (i.e. we don’t know R (or K above) or N). One method of progressing this uncertainty is to define a new variable to reflect the RATE of successes we see in the uncertain sampling we undertake. So if we have N trials and the probability of success in each trial is p, then the rate of success(λ) we see will be: λ = N*p 11 | P a g e If you have difficulty seeing where this equation came from, consider the simple example of tossing a coin. If you toss a coin N times and a success is when you see a ‘head’ then your rate of success is N*0.5. Tossing a coin 10 times in succession and seeing a ‘head’ ten times is then given by the value of (10*0.5=0.05). In other words you only have a 5% chance of this sequence occurring, so that λ is then the success rate of this sequence (but don’t forget you may also be concerned with other ‘success rates’!- hence the need for the combinational calculation to be retained in the Poisson distribution). If we go back to our binomial equation and now insert this new relationship, we see that: NC R * (λ /N)R * (1- (λ /N))N-R In the above equation, we have just eliminated p and q through substitution. Now let’s also go back to the implicit assumption in the above equation and our earlier statement that we don’t know what N is – but we are assuming it is a very large number – so large that we are not concerned with discrete data, but with continuous data. In this situation we assume that N tends towards being infinitely large. If you are familiar with your mathematics, you may have come across this concept before that we take N to the limit. This is written in the following way (and by expanding the combination calculation): Now as with any equation where only one variable is changing (here N), the other values are in effect constants (i.e. as we increase N, the values of R and λ in effect, remain unchanged. We can therefore reorder the equation and extract those values which are not affected by the limit (as they are regarded as constants). In doing this our equation changes to: The expansion of the RHS of equation (a) – is shown in (b) – if you doubt this – substitute some numbers into (a) for λ, N, and R and then for (b). You will find they give the same solution. To progress this further and work towards the next probability distribution by changing the assumptions in the binomial distribution, we need to finish the expansion of (b). If you expand the following portion of the RHS of (b): OR (as a majority of the numerator values cancel out the denominator values): OR by expanding the equation in (d) – through recognising that both numerator and denominator are simply powers of R to: If you now imaging that N becomes infinitely large – then each value in (e) will tend towards 1. 12 | P a g e We can now consider the remaining part of the equation in (b) i.e. the calculations now after the ‘limit’ operation and firstly, the equation part of: (f) - (1- (λ /N))N. To resolve this – we need to use a piece of mathematical sleight of hand – by recognising that the part equation given above in (f) looks similar to the following derivation of ‘e’ (to which the solution is known (NB – we will meet ‘e’ again shortly, but for a fuller discussion please review Chapter 2 and see later in this chapter): Now recall that we initially set up a value to describe our RATE of success (R) in N trials as: λ = N*p If we rearrange this slightly – so that we obtain and equivalent expression: -N/λ = -1/p and let this equal x If we then substitute this equation for x (and hence p) into (g): Now if you compare (h) with (g) it is easy to see that: Hence equation (b) has now become: We now need to expand the final part of the limit in the equation above i.e. Clearly, as N approaches infinity then this equation reduces to (1-R) = (1). If we now substitute this final part of the expansion, back into equation (b) – we derive the following: OR 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 as you have seen, 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: 1. 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!)) 2. The number of paint flaws evident of new cars sold in a given period 13 | P a g e 3. The rate of arrival of airplanes at an airport requiring landing 4. The incidents of accidents recorded at a factory over a given period. You can see that 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. 4.6) Queuing Theory Now much of the use of Poisson distribution function in Business – is categorized as Queuing Theory – in other words as managers, we are concerned with the effective use of organisational resources to manage demand, quality or shipping for example – where allowing arriving customers or departing goods to wait – both costs money and potential results in lost future business. We therefore need to manage arrival queues or departing goods. The Poisson distribution function therefore allows us to visualize customers arriving or goods departing as ‘successes’ against an almost continuous background of their non-arrival or non-departure. It has found particular value in minimizing the occurrence of ‘bottlenecks’ in organisational functions (such as manufacturing, customer service etc). Let’s take a simple example: 1) A construction company has had 40 accidents in the past 50 weeks. In what proportion of weeks (i.e. how many weeks) would you expect 0,1,2,3 and more than 4 accidents to occur? Y A small number of accidents occur, at random (this is our assumption and these are our ‘successes’). We are not interested in the accidents that did *not* occur, so we can use the Poisson distribution function. (B) We can determine that the Mean number of accidents/week (λ = rate of accidents (i.e. our ‘successes’)) = 40/50=0.8 per week. 14 | P a g e So if P(0) is the probability of seeing a week with no accidents (i.e. no success), we substitute R=0 and λ =0.8 in equation (j), which will give 0.449 (satisfy yourself of this) You can then repeat this to determine P(1 accident ‘success’), P(2 accident ‘successes’) ,P(3 accident ‘successes’) and P(4 accident ‘successes’) – (NB your answers should be 0.359,0.143, 0.038,0.0076). This use of the Poisson Distribution function was first proposed by Danish engineer A.K.Erlang at the turn of the 19th century when the problem of telephone line congestion was analysed (and the implications of requiring a customer to wait for service, what a reasonable waiting time would be for a given customer in an organisation and how fast a customer can be served etc). Each of these problems has a natural trade off in terms of resourcing and service – and this is the heart of the issue for management decision making. Should an organisation invest in large plant and machinery to ensure it can produce – with minimal waiting times – a given product for a consumer? It would for example be attractive to a consuming customer to be able to always walk into a pizza retailer and – regardless of the time of day, day of the week or time in the year – receive instant service. This is possible and could be done – but the resourcing required to deliver this is not attractive enough for the retailer (unless they then charge a price large enough to support this level of service). Would you pay >£100 for a pizza with this guarantee of service? If we now start to bring together some of these threads discussed so far – we find we have a small number of fundamental concerns to work with. Fundamentally, the length of any queue for service (when the context of that service satisfies the use of the Poisson distribution function (i.e. there are a very large number of ‘trials’ and the probability of a success in any one of those trials is very small)) is dependent upon: B The rate at which ‘customers’ arrive to be served – called the ARRIVAL RATE (λ) C The time taken to serve that ‘customer’ – called the SERVICE RATE (μ) D The number of outlets / servers available to service a customer- called the number of SERVICE CHANNELS (this can be labelled as k or s (depending upon the materials you are reading)). 15 | P a g e Remember – we have assumed (in order to use this probability distribution function) that the arrival of customers/parts etc in a system is (largely) a random event. In stating this, we have therefore assumed therefore the probability of the arrival of a given or specific customer or part is very small (tending towards zero) whilst the number of potential customers / parts that could arrive is potentially infinite (or very large). We must also take into consideration that the service rate (how quickly a customer is served or a part dispatched/auctioned for example) is described by a continuous random distribution. We usually assume this is determined from the exponential distribution. This type of distribution is continuous and describes the times between random occurrences (rather than with the Poisson function which is discrete and describes random occurrences over a given time interval). Let me explain this further (and review the constant ‘e’ introduced earlier in this chapter). Anyone employed to complete a task that has some regularity to it – will take a similar but not always exact time, to do that task. The more tasks that are added to the work to be done, then the more likely a longer time will be needed. Clearly, if you were to undertake an experiment to time the duration needed to complete tasks – there would be some variation about a mean and some relationship between the passage of time and the number of tasks that could be undertaken in that time. Take for example, loading a car with luggage. Different people – because of natural variation – would take different – but similar lengths of time to do this. Adding more or less luggage would also vary the duration of time needed to complete the task. You could plot these variations in time on an x,y graph and you would be able to visualize a distribution function of the SERVICE rate for that task. Let’s say you are really bored one day and find a friend ‘willing’ to load (and then unload) your car. You give him a set of suitcases and time his loading. The mean time you measure over say 20 trials is 15 seconds (this would be the service rate (U) for this completion of the task). You know from your data that although the service rate had this mean, there would have been some variation in those times. Each different value obtained for μ therefore generates a different distribution function. So there would be (potentially) an infinite number of distribution functions for each task to be serviced (although some would be very very unlikely). The exponential function arises naturally when you therefore try to model the time between independent events that happen at a constant average rate (μ). We know that modelling, as discussed earlier, is a mathematical representation of an apparent set of relationships between data. 16 | P a g e There will always be some variation in the probability of a task needing to be serviced and it has been found that using the exponential function ‘e’ is a good way to model these observed relationships. So what is ‘e’? The meaning of ‘e’ ‘e’ is a rather unique mathematical constant. If you recall some basic calculus relationships - we could start to describe this uniqueness by saying that the derivative of the function of ex is itself – i.e. e. Think of it this way, velocity is nothing more than a measure of how fast the distance travelled changes with the passage of time. The higher the velocity, that more distance that can be travelled in a given time (and of course the reverse is also true). Velocity is then said to be the derivative of distance with respect to time (you may recall seeing this written as ds/dt at school?). All derivatives (of this first order) are therefore just measures of change in one variable with respect to another. However, amongst this range of derivatives, there is one unique value where the rate of change in the one variable (say distance) with respect to (say) time is the same as that variable (say distance). Look at this data of distance covered (metres) vs time taken (seconds): S 10 20 30 40 50 60 70 80 90 100 m 10 20 30 40 50 60 70 80 90 100 Clearly an x,y plot would give this graph as a straightline – i.e. the rate of change of distance with respect to time is constant. The gradient of this plot is fixed and is described by therefore a single constant velocity – which is the derivative of this data (i.e. in this case 1 metre per second (or change in distance/change in time)). Let’s now look at another set of data which shows a different relationship between distance (height in cm) and time (days) – this could be for example plant growth from a seed. S(cm) T (days) 10 0.1 20 0.2 30 0.4 40 0.8 50 1.1 60 2.3 70 4 This plot would look like: 17 | P a g e Clearly, we do not have a constant growth rate – but that the growth rate seems to be increasing with the passage of time. The derivative here (i.e. as a measure of the rate of change of height with respect to time) is not going to be a fixed value… or is it? Compare this data with some other data that seems to show a changing height vs time relationship: Both curves look similar – they both increase with time (y values become larger) – but they seem to be increasing at different rates – which is certainly clear from say the 50th day onwards. Both curves therefore are described by a different relationship between height and time. The bottom curve – if you consider the data – is described by the following: S(cm) 0.01 0.04 0.16 0.64 1.21 5.29 16 T (days) 0.1 0.2 0.4 0.8 1.1 2.3 4 By inspecting the data – you can see that s is simply the square of t – i.e. s=t2. Hence if this data did occur in reality, our mathematical model would be s=t2. We can work out the velocity by considering the rate of change in distance over any given period of time – this is (as noted above) often shortened to ds/dt (or more accurately δs/δt – where the symbol δ means ‘small change in’). To save many different calculations being made by considering these small mutual changes, it has been proven that the derivative of this equation would be simply 2t (if you need refreshing on your basic differential calculus – please pick up a basic maths text book at this time!). What this means is that ds/dt (here velocity) can be modelled by the equation 2t – so if t was say 0.5 (half a day) then the growth rate here would be 1cm at that point in time (i.e. 2x 0.5). Now going back to the initial argument identified earlier about the value of ‘e’ – ask yourself what if rather than s=t2, the growth of the plant was shaped by its already evident growth – i.e. if we assume its growth at day 1 is set to 1 (arbitrarily), and its growth each subsequent day was determined by how much it grew on the previous day: 18 | P a g e s=Growth at time t = (1+ 1/n)n So – just like before, growth is some power relationship (here simply given the label of n) – but that the current observed growth is proportional to its existing growth. If you chose a fairly large value of n (say n=50) and plotted s (using (l)) – you end up with this graph: Increasing the value of n does not change the general shape of the curve (plot it yourself to see). The rate of change of s in (l) is s! So if we decided that ds/dt could be found by viewing s as equal to et, then ds/dt = et aswell. In other words, a mathematical modelled relationship where the rate of change is proportional to the relationship itself is called an exponential function and then only called ‘e’ where its derivate is the same as the function itself. It happens that ‘e’ has an indeterminate value of 2.71828.... (to 5 decimal points) and is itself derived by solving – at the limit equation (l). Now – after that mathematical interlude – if we return back to our car loading (generous!) friend, we stated that the service rate (μ) for this job was timed (mean) at 15 seconds – but that each different value obtained for μ – would generate a different probability distribution function (i.e. the probability that our friend could complete the loading task in a faster or slower time). Our friend works at a service rate of 15 seconds to load the car and suitcases can be placed in front of him (independent events) at a rate of λ. The exponential function ex (which is written in excel as EXP(X)), can be used to model this service relationship (where the rate of work (tasks served) is proportion to the tasks to be completed). If we let β = 1/ λ (otherwise described as μ), then we can write the probability distribution function (f(x)) as: (m) f(x) = β.e- βx 19 | P a g e Remember – in using this mathematical relationship to model service rates for a given event(s) occurrence, we have assumed a continuous distribution (i.e. a potentially unknown number of events could occur), there is an (infinite) family of service distributions that could occur (where we usually focus upon a mean μ to describe one service rate distribution), this equation (m) will generate x values that peak at the apex (when x=0) and which gradually rises (but proportionately so) as x increases). Review the following graphs (figure 4.2 and figure 4.3) to convince yourself of the validity of choosing this mathematical relationship to explain the probability of a waiting customer being served / actioned and the probability of the service being completed by some time period t. Figure 4.2 – Exponential probability plot (for an arbitrary service time) Figure 4.3 – Exponential probability plot (for an arbitrary service time) 20 | P a g e 4.7 Examples of poisson problems Consider this question - What is the probability that our generous friend can load the suitcases into the car in less than 10 seconds? The area under the curve described by (m) represent all the probabilities included between two x points (i.e. it includes all probabilities that generate that area of the curve in equation (m)). We also know that as this is probability relationship – the whole area under the curve must equal 1 - hence to answer the question, all we have to work out is the probability of 1 minus the sum of all other probabilities that result in the car being loaded in less than 10 seconds – i.e. ; Just as with the derivative of the exponential function, the integration of it (to determine the area under the probability curve that lies between time x=0 and time x=10) is also the same function (multiplied by any constant value in the power of the exponential) – hence the integration calculation reduces to: P(loading time < 10 seconds) = (1/ μ )( μ )(1-e –10/ μ) In other words, the first integration when x=0 has reduced to 1 and the second integration for x=10, has become e –10/ μ. You will find that most textbooks simply give you this formula as: P(event of interest< event maximum time (x)) = 1-e –x/ μ As we know μ=15, we can now substitute values so that we derive: P(loading time < 10 seconds) = 1-e –10/ 15 or 49% (if you evaluate the equation above). In other words there is a 49% chance that our generous friend will be able to load the car (complete the task) in less than 10 seconds (given a mean service rate of 15 seconds). Similarly, if you wanted to know the probability of our friend completing the task between 20 and 25 seconds (maybe he’s not feeling well...) you need to first determine the probability of completing the task in less then 20 seconds and then in less than 25 seconds. Subtracting one probability from the other will then be the area under the curve and be equal to the probability of the task being completed in 20