In modeling real-world phenomena, there are few situations where the actions ofthe entities within the system under study can be predicted completely. The world the model builder sees is probabilistic rather than deterministic. There are many causes of variation. The time it takes a repairperson to fix a broken machine is a function of the complexity of the breakdown, whether the repairperson brought the proper replacement parts and tools to the site, whether another repairperson asks for assistance during the course ofthe repair, whether the machine operator receives a lesson in preventive maintenance, and so on. To the model builder, these variations appear to occur by chance and canno be predicted. However, some statistical model might well describe the time to make a repair.

An appropriate model can be developed by sampling the phenomenon of interest. Then, through educated guesses (or using software for the purpose), the model builder would select a known distribution form, make an estimate of the parameter(s) of this distribution, and then test to see how good a fit has been obtained. Through continued efforts in the selection of an appropriate distribution form, a postulated model could be accepted.

Section 1 contains a review of probability terminology and concepts. Some typical applications of statistical models, or distribution forms, are given in Section 2. Then, a number of selected discrete and continuous distributions are discussed in Sections 3 and 4. The selected distributions are those tha describe a wide variety of probabilistic events. Additional discussion about the distribution forms appearing in this chapter, and about distribution forms mentioned but not described, is available from a number of sources [Devore, 1999; Hines and Montgomery, 1990; Law and Kelton, 2000; Papoulis, 1990; Ross, 2002; Walpole and Myers, 2002]. Section 5 describes the Poisson process and its relationship to the exponential distribution. Section 6 discusses empirical distributions.

1 Review of Terminology and Concepts

1.1 Discrete random variables

Let X be a random variable. If the number of possible values of X is finite, or countably infinite, X is called a discrete random variable. The possible values of X may be listed as $x_{1} , x_{2} , \ldots$. In the finite case, the list terminates; in the countably infinite case, the list continues indefinitely.

Example 1

The number of jobs arriving each week at a job shop is observed. The random variable of interest is X, where

The possible values of X are given by the range space of X, which is denoted by $R_{X}$. Here $R_{X} =$$\{0 , 1 , 2 , \dots \} .$ Let X be a discrete random variable. With each possible outcome $x_{i}$ in $R_{X}$, a number $p ( x_{i} ) =$$P ( X \ = \ x_{i} )$ gives the probability that the random variable equals the value of x . The numbers $p ( x_{i} ) , i = 1 , 2 , . . . ,$ must satisfy the following two conditions:

a.$p ( x_{i} ) \geq 0$, for all i

The collection of pairs $( x_{i} , p ( x_{i} ) ) , i = 1 , 2 , . . .$. is called the probability distribution of X, and $p ( x_{i} )$ is called the probability mass function (pmf) of X.

Example 2

Consider the experiment of tossing a single die. Define X as the number of spots on the up face of the die after a toss. Then $R_{X} = \{1 , 2 , 3 , 4 , 5 , 6 \}$. Assume the die is loaded so that the probability that a given face lands up is proportional to the number of spots showing. The discrete probability distribution for this random experiment is given by

The conditions stated earlier are satisfied—that is,$p ( x_{i} ) \geq 0$ for $i = 1 , 2 , \dots , 6$ and $\textstyle \sum_{i = 1} ^ {\infty} p ( x_{i} ) =$$1 / 21 + \cdot \cdot \cdot + 6 / 21 = 1$. The distribution is shown graphically in Figure 1.

Figure 1 Probability mass function for loaded-die example.

1.2 Continuous random variables

If the range space $R_{X}$ of the random variable X is an interval or a collection of intervals, X is called a continuous random variable. For a continuous random variable X, the probability that X lies in the interval [a, b] is given by

The function $f ( x )$ is called the probability density function (pdf) of the random variable X. The pdf satisfies the following conditions:

a.$f ( x ) \geq 0$ for all x in $R_{X}$ b.$\textstyle \int _ {R_{X}} f ( x ) d x = 1$ c.$f ( x ) = 0$ if x is not in $R_{X}$ As a result of Equation (1), for any specified value $x_{0} , P ( X = x_{0} ) = 0$, because

Figure 2 Graphical interpretation of $P ( a < X < b )$$P ( X = x_{0} ) = 0$ also means that the following equations hold:

The graphical interpretation of Equation (1) is shown in Figure 2. The shaded area represents the probability that X lies in the interval [a, b].

Example 3

The life of a device used to inspect cracks in aircraft wings is given by X, a continuous random variable assuming all values in the range $x \ge 0$. The pdf of the device’s lifetime, in years, is as follows:

This pdf is shown graphically in Figure 3. The random variable X is said to have an exponentia distribution with mean 2 years.

The probability that the life of the device is between 2 and 3 years is calculated as

Figure 3 pdf for inspection-device life.

1.3 Cumulative distribution function

The cumulative distribution function (cdf), denoted by $F ( x )$, measures the probability that the random variable X assumes a value less than or equal to x, that is,$F ( x ) = P ( X \leq x )$.

If X is discrete, then

If X is continuous, then

Some properties of the cdf are listed here:

a. F is a nondecreasing function. If $a < b ,$, then $F ( a ) \leq F ( b )$).

b. lim $_ {x \longrightarrow \infty} F ( x ) = 1$ c. lim $x {\longrightarrow} - \infty \overset {} {F} ( x ) = 0$ All probability questions about X can be answered in terms of the cdf. For example,

For continuous distributions, not only does Equation (5) hold, but also the probabilities in Equation (2) are equal to $F ( b ) - F ( a )$

Example 4

The die-tossing experiment described in Example 2 has a cdf given as follows:

where [a,$b ) = \{a \leq x < b \}$. The cdf for this example is shown graphically in Figure 4.

If X is a discrete random variable with possible values $x_{1} , x_{2} , \dotsc ,$, where $x_{1} < x_{2} < . . . .$, the cdf is a step function. The value of the cdf is constant in the interval $[ x_{i - 1} , x_{i} )$ and then takes a step, or jump, of size $p ( x_{i} )$ at $x_{i} .$. Thus, in Example 4,$p ( 3 ) = 3 / 21$, which is the size of the step when $x = 3$ Figure 4 cdf for loaded-die example.

Example 5

The cdf for the device described in Example 3 is given by

The probability that the device will last for less than 2 years is given by

The probability that the life of the device is between 2 and 3 years is calculated as

as found in Example 3.

1.4 Expectation

An important concept in probability theory is that of the expectation of a random variable. If X is a random variable, the expected value of X, denoted by $E ( X )$, for discrete and continuous variables, is defined as follows:

and

The expected value E(X) of a random variable X is also referred to as the mean,$\mu .$, or the first moment of X. The quantity $E ( X^{n} ) , n \geq 1$, is called the nth moment of X, and is computed as follows:

and

The variance of a random variable X denoted by V(X) or var(X) or $\sigma^{2}$, is defined by

A useful identity in computing V(X) is given by

The mean $E ( X )$ is a measure of the central tendency of a random variable. The variance of X measures the expected value of the squared difference between the random variable and its mean. Thus, the variance,$V ( X )$, is a measure of the spread or variation of the possible values of X around the mean $E ( X )$. The standard deviation, σ, is defined to be the square root of the variance,$\sigma^{2}$. The mean,$E ( X )$, and the standard deviation,$\sigma = \sqrt {V ( X )}$, are expressed in the same units.

Example 6

The mean and variance of the die-tossing experiment described in Example 2 are computed as follows:

To compute $V ( X )$ from Equation (10), first compute $E ( X^{2} )$ from Equation (8) as follows:

Thus,

and

Example 7

The mean and variance of the life of the device described in Example 3 are computed as follows:

To compute $V ( X )$ from Equation (10), first compute $E ( X^{2} )$) from Equation (9) as follows:

Thus,

giving

Statistical Models in Simulation

and

With a mean life of 2 years and a standard deviation of 2 years, most analysts would conclude tha actual lifetimes, X, have a fairly large variability.

1.5 The mode

The mode is used in describing several statistical models that appear in this chapter. In the discrete case, the mode is the value of the random variable that occurs most frequently. In the continuous case, the mode is the value at which the pdf is maximized. The mode might not be unique; if the modal value occurs at two values of the random variable, the distribution is said to be bimodal.

2 Useful Statistical Models

Numerous situations arise in the conduct of a simulation where an investigator may choose to introduce probabilistic events. In a queueing system, interarrival and service times are often probabilistic. In an inventory model, the time between demands and the lead times (time between placing and receiving an order) can be probabilistic. In a reliability model, the time to failure could be probabilistic. In each of these instances, the simulation analyst desires to generate random events and to use a known statistical model if the underlying distribution can be found. In the following paragraphs, statistical models appropriate to these application areas will be discussed. Additionally, statistica models useful in the case of limited data are mentioned.

2.1 Queueing systems

In queueing examples, interarrival- and service-time patterns, the times between arrivals and the service times are often probabilistic. However, it is possible to have a constant interarrival time (as in the case of a line moving at a constant speed in the assembly of an automobile), or a constant service time (as in the case of robotized spot welding on the same assembly line). The following example illustrates how probabilistic interarrival times might occur.

Example 8

Mechanics arrive at a centralized tool crib as shown in Table 1. Attendants check in and check out the requested tools to the mechanics. The collection of data begins at 10:00 A.M. and continues unti 20 different interarrival times are recorded. Rather than record the actual time of day, the absolute time from a given origin could have been computed Thus the first mechanic could have arrived at time $0 ,$ the second mechanic at time 7:13 (7 minutes, 13 seconds), and so on.

Table 1 Mechanics’ Arrival Data

Example 9

Another way of presenting interarrival data is to find the number of arrivals per time period. Here, such arrivals occur over approximately 1.5 hours; it is convenient to look at 10-minute time intervals for the first 20 mechanics. That is, in the first 10-minute time period, one arrival occurred at 10:05::03. In the second time period, two mechanics arrived, and so on. The results are summarized in Table 2. This data could then be plotted in a histogram, as shown in Figure 5.

The distribution of time between arrivals and the distribution of the number of arrivals per time period are important in the simulation of waiting-line systems. “Arrivals” occur in numerous ways; as machine breakdowns, as jobs coming into a job shop, as units being assembled on a line, as orders to a warehouse, as data packets to a computer system, as calls to a call center, and so on.

Service times could be constant or probabilistic. If service times are completely random, the exponential distribution is often used for simulation purposes; however, there are several other possibilities. It could happen that the service times are constant, but some random variability causes fluctuations in either a positive or a negative way. For example, the time it takes for a lathe to traverse a 10-centimeter shaft should always be the same. However, the material could have slight differences in hardness or the tool might wear; either event could cause different processing times. In these cases, the normal distribution might describe the service time.

Table 2 Arrivals in Successive Time Period

A special case occurs when the phenomenon of interest seems to follow the normal probability distribution, but the random variable is restricted to be greater than or less than a certain value. In this case, the truncated normal distribution can be utilized.

The gamma and Weibull distributions are also used to model interarrival and service times. (Actually, the exponential distribution is a special case of both the gamma and the Weibull distributions.) The differences between the exponential, gamma, and Weibull distributions involve the location of the modes of the pdfs and the shapes of their tails for large and small times. The exponential distribution has its mode at the origin, but the gamma and Weibull distributions have their modes at some point $( \geq 0 )$ that is a function of the parameter values selected. The tail of the gamma distribution is long, like an exponential distribution; the tail of the Weibull distribution can decline more rapidly or less rapidly than that of an exponential distribution. In practice, this means that, if there are more large service times than an exponential distribution can account for, a Weibul distribution might provide a better model of these service times.

Number of arrivals in 10-minute period Figure 5 Histogram of arrivals per time period.

2.2 Inventory and supply-chain systems

In realistic inventory and supply-chain systems, there are at least three random variables: (1) the number ofunits demanded per order or per time period, (2) the time between demands, and (3) the lead time. (The lead time is defined as the time between the placing of an order for stocking the inventory system and the receipt of that order.) In very simple mathematical models of inventory systems, demand is a constant over time, and lead time is zero, or a constant. However, in most real-world cases, and, hence, in simulation models, demand occurs randomly in time, and the number of units demanded each time a demand occurs is also random, as illustrated by Figure 6.

Distributional assumptions for demand and lead time in inventory theory texts are usually based on mathematical tractability, but those assumptions could be invalid in a realistic context. In practice, the lead-time distribution can often be fitted fairly well by a gamma distribution [Hadley and Whitin, 1963]. Unlike analytic models, simulation models can accommodate whatever assumptions appear most reasonable.

The geometric, Poisson, and negative binomial distributions provide a range of distribution shapes that satisfy a variety of demand patterns. The geometric distribution, which is a special case of the negative binomial, has its mode at unity, given that at least one demand has occurred. If demand data are characterized by a long tail, the negative binomial distribution might be appropriate. The Poisson distribution is often used to model demand because it is simple, it is extensively tabulated, and it is well known. The tail of the Poisson distribution is generally shorter than that of the negative binomial, which means that fewer large demands will occur if a Poisson model is used than if a negative binomial distribution is used (assuming that both models have the same mean demand).

Figure 6 Random demands in time.

2.3 Reliability and maintainability

Time to failure has been modeled with numerous distributions, including the exponential, gamma, and Weilbull. If only random failures occur, the time-to-failure distribution may be modeled as exponential. The gamma distribution arises from modeling standby redundancy, where each component has an exponential time to failure. The Weibull distribution has been extensively used to represent time to failure, and its nature is such that it can be made to approximate many observed phenomena [Hines and Montgomery, 1990]. When there are a number of components in a system and failure is due to the most serious of a large number of defects, or possible defects, the Weibull distribution seems to do particularly well as a model. In situations where most failures are due to wear, the normal distribution might very well be appropriate [Hines and Montgomery, 1990]. The lognormal distribution has been found to be applicable in describing time to failure for some types of components.

2.4 Limited data

In many instances, simulations begin before data collection has been completed. There are three distributions that have application to incomplete or limited data. These are the uniform, triangular, and beta distributions. The uniform distribution can be used when an interarrival or service time is known to be random, but no information is immediately available about the distribution [Gordon, 1975]. However, there are those who do not favor using the uniform distribution, calling it the “distribution of maximum ignorance” because it is not necessary to specify more than the continuous interval in which the random variable may occur. The triangular distribution can be used when assumptions are made about the minimum, maximum, and modal values of the random variable. Finally, the beta distribution provides a variety of distributional forms on the unit interval, ones that, with appropriate modification, can be shifted to any desired interval. The uniform distribution is a special case of the beta distribution. Pegden, Shannon, and Sadowski [1995] discuss the subject of limited data in some detail.

2.5 Other distributions

Several other distributions may be useful in discrete-system simulation. The Bernoulli and binomia distributions are two discrete distributions which might describe phenomena of interest. The hyperexponential distribution is similar to the exponential distribution, but its greater variability might make it useful in certain instances.

3 Discrete Distributions

Discrete random variables are used to describe random phenomena in which only integer values can occur. Numerous examples were given in Section 2—for example, demands for inventory items. Four distributions are described in the following subsections.

3.1 Bernoulli trials and the Bernoulli distribution

Consider an experiment consisting of n trials, each of which can be a success or a failure. Let $X_{j} = 1$ if the jth experiment resulted in a success, and let $X_{j} = 0$ if the jth experiment resulted in a failure.

The n Bernoulli trials are called a Bernoulli process if the trials are independent, each trial has only two possible outcomes (success or failure), and the probability of a success remains constant from trial to trial. Thus,

and

For one trial, the distribution given in Equation (11) is called the Bernoulli distribution. The mean and variance of $X_{j}$ are calculated as follows:

and

3.2 Binomial distribution

The random variable X that denotes the number of successes in n Bernoulli trials has a binomia distribution given by $p ( x )$, where

Equation (12) is motivated by computing the probability ofa particular outcome with all the successes, each denoted by S, occurring in the first x trials, followed by the $n - x$ failures, each denoted by an F—that is,

where $q = 1 - p$. There are

outcomes having the required number of $S \mathrm{s}$ and $F \mathbf{s} .$. Therefore, Equation (12) results. An easy approach to calculating the mean and variance of the binomial distribution is to consider X as a sum of n independent Bernoulli random variables, each with mean $p$ and variance $p ( 1 - p ) = p q$. Then,

and the mean, E(X), is given by

and the variance V(X) is given by

Example 10

A production process manufactures computer chips on the average at $2 \%$ nonconforming. Every day, a random sample of size 50 is taken from the process. If the sample contains more than two nonconforming chips, the process will be stopped. Compute the probability that the process is stopped by the sampling scheme.

Consider the sampling process as $n = 50$ Bernoulli trials, each with $p = 0 . 02$; then the total number of nonconforming chips in the sample, X, would have a binomial distribution given by

It is much easier to compute the right-hand side of the following identity to compute the probability that more than two nonconforming chips are found in a sample:

The probability $P ( X \leq 2 )$ is calculated from

Thus, the probability that the production process is stopped on any day, based on the sampling process, is approximately 0.08. The mean number of nonconforming chips in a random sample of size 50 is given by

and the variance is given by

The cdf for the binomial distribution has been tabulated by Banks and Heikes [1984] and others. The tables decrease the effort considerably for computing probabilities such as $P ( a < X \leq b )$). Under certain conditions on n and $p ,$ both the Poisson distribution and the normal distribution may be used to approximate the binomial distribution [Hines and Montgomery, 1990].

3.3 Geometric and Negative Binomial distributions

The geometric distribution is related to a sequence of Bernoulli trials; the random variable of interest, X, is defined to be the number of trials to achieve the first success. The distribution of X is given by

The event $\{X = x \}$ occurs when there are $x - 1$ failures followed by a success. Each of the failures has an associated probability of $q = 1 - p_{;}$, and each success has probability $p .$ Thus,

The mean and variance are given by

and

More generally, the negative binomial distribution is the distribution of the number of trials unti the kth success, for $k = 1 , 2 , \ldots . \mathrm{f f} Y$ has a negative binomial distribution with parameter $p$ and k, then the distribution of Y is given by

Because we can think of the negative binomial random variable Y as the sum of $k$ independen geometric random variables, it is easy to see that $E ( Y ) = k / p$ and $V ( X ) = k q / p^{2}$

Example 11

Forty percent of the assembled ink-jet printers are rejected at the inspection station. Find the probability that the first acceptable ink-jet printer is the third one inspected. Considering each inspection as a Bernoulli trial with $q = 0 . 4$ and $p = 0 . 6$ yields

Thus, in only about 10% of the cases is the first acceptable printer the third one from any arbitrary starting point. To determine the probability that the third printer inspected is the second acceptable printer, we use the negative binomial distribution Equation (18),

3.4 Poisson distribution

The Poisson distribution describes many random processes quite well and is mathematically quite simple. The Poisson distribution was introduced in 1837 by S. D. Poisson in a book concerning criminal and civil justice matters. (The title of this rather old text is Recherches sur la probabilit´e desjugements en mati`ere criminelle et en mati`ere civile. Evidently, the rumor handed down through generations of probability theory professors concerning the origin of the Poisson distribution is jus not true. Rumor has it that the Poisson distribution was first used to model deaths from the kicks o horses in the Prussian Army.)

The Poisson probability mass function is given by

where α$> 0$. One of the important properties of the Poisson distribution is that the mean and variance are both equal to α, that is,

The cumulative distribution function is given by

The pmf and cdf for a Poisson distribution with $\alpha = 2$ are shown in Figure 7. A tabulation of the cd is given in Table A.4 in the Appendix.

(a)

(b) Figure 7 Poisson pmf and cdf. Example 12

A computer repair person is “beeped” each time there is a call for service. The number of beeps per hour is known to occur in accordance with a Poisson distribution with a mean of $\alpha = 2$ per hour. The probability of three beeps in the next hour is given by Equation (19) with $x = 3 ,$, as follows:

This same result can be read from the left side of Figure 7 or from Table A.4 by computing

Example 13

In Example 12, find the probability of two or more beeps in a 1-hour period.

The cumulative probability,$F ( 1 )$, can be read from the right side of Figure 7 or from Table A.4.

Example 14

The lead-time demand in an inventory system is the accumulation of demand for an item from the point at which an order is placed until the order is received—that $\mathrm{i} \mathbf{s} .$

where $L$ is the lead-time demand,$D_{i}$ is the demand during the ith time period, and T is the number of time periods during the lead time. Both $D_{i}$ and $T$ may be random variables.

An inventory manager desires that the probability of a stockout not exceed a certain fraction during the lead time. For example, it may be stated that the probability of a shortage during the lead time not exceed 5%.

If the lead-time demand is Poisson distributed, the determination of the reorder point is greatly facilitated. The reorder point is the level of inventory at which a new order is placed.

Assume that the lead-time demand is Poisson distributed with a mean of $\alpha = 10$ units and that 95% protection from a stockout is desired. Thus, it is desired to find the smallest value of x such that the probability that the lead-time demand not exceed x is greater than or equal to 0.95. Using Equation (20) requires finding the smallest x such that

The desired result occurs at $x = 15$, which can be found by using Table A.4 or by computation of $p ( 0 ) , p ( 1 )$

4 Continuous Distributions

Continuous random variables can be used to describe random phenomena in which the variable o interest can take on any value in some interval—for example, the time to failure or the length of a rod. Nine distributions are described in the following subsections.

4.1 Uniform distribution

A random variable X is uniformly distributed on the interval [a, b] if its pdf is given by

The cdf is given by

Figure 8 pdf and cdf for uniform distribution.

Note that

is proportional to the length of the interval, for all $x_{1}$ and $x_{2}$ satisfying $a \leq x_{1} < x_{2} \leq b .$ The mean and variance of the distribution are given by

and

The pdf and cdf when $a = 1$ and $b = 6$ are shown in Figure 8.

The uniform distribution plays a vital role in simulation. Random numbers, uniformly distributed between 0 and 1, provide the means to generate random events. Numerous methods for generating uniformly distributed random numbers have been devised. Uniformly distributed random numbers are then used to generate samples of random variates from all other distributions.

Example 15

A simulation of a warehouse operation is being developed. About every 3 minutes, a call comes for a forklift truck operator to proceed to a certain location. An initial assumption is made that the time between calls (arrivals) is uniformly distributed with a mean of 3 minutes. By Equation (25), the uniform distribution with a mean of 3 and the greatest possible variability would have parameter values of $a = 0$ and $b = 6$ minutes. With very limited data (such as a mean of approximately 3 minutes) plus the knowledge that the quantity of interest is variable in a random fashion, the uniform distribution with greatest variance can be assumed, at least until more data are available.

Example 16

A bus arrives every 20 minutes at a specified stop beginning at 6:40 A.M. and continuing unti 8:40 A.M. A certain passenger does not know the schedule, but arrives randomly (uniformly distributed) between 7:00 A.M. and 7:30 A.M. every morning. What is the probability that the passenger waits more than 5 minutes for a bus?

The passenger has to wait more than 5 minutes only if the arrival time is between 7:00 A.M. and 7:15 A.M. or between 7:20 A.M. and 7:30 A.M. If X is a random variable that denotes the number of minutes past 7:00 A.M. that the passenger arrives, the desired probability is

Now, X is a uniform random variable on (0, 30). Therefore, the desired probability is given by

4.2 Exponential distribution

A random variable X is said to be exponentially distributed with parameter $\lambda > 0$ if its pdf is given by

The density function is shown in Figures 9 and 3. Figure 9 also shows the cdf.

The exponential distribution has been used to model interarrival times when arrivals are completely random and to model service times that are highly variable. In these instances, λ is a rate: arrivals per hour or services per minute. The exponential distribution has also been used to model the lifetime of a component that fails catastrophically (instantaneously), such as a light bulb; then λ is the failure rate.

Several different exponential pdfs are shown in Figure 10. The value of the intercept on the vertical axis is always equal to the value of λ. Note also that all pdfs eventually intersect. (Why?)

Figure 9 Exponential density function and cumulative distribution function.

Figure 10 pdfs for several exponential distributions.

The exponential distribution has mean and variance given by

Thus, the mean and standard deviation are equal. The cdf can be exhibited by integrating Equation (26) to obtain

Example 17

Suppose that the life of an industrial lamp, in thousands of hours, is exponentially distributed with failure rate $\lambda = 1 / 3$(one failure every 3000 hours, on the average). The probability that the lamp will last longer than its mean life, 3000 hours, is given by $P ( X > 3 ) = 1 - P ( X \le 3 ) = 1 - F ( 3 )$ Equation (28) is used to compute F(3), obtaining

Regardless of the value of $\lambda ,$, this result will always be the same! That is, the probability that an exponential random variable is greater than its mean is 0.368, for any value o $\lambda .$ The probability that the industrial lamp will last between 2000 and 3000 hours is computed as

Again, from the cdf given by Equation (28),

One of the most important properties of the exponential distribution is that it is “memoryless,” which means that, for all $s \geq 0$ and $t \geq 0$

Let X represent the life of a component (a battery, light bulb, computer chip, laser, etc.) and assume that X is exponentially distributed. Equation (29) states that the probability that the component lives for at least $s + t$ hours, given that it has survived s hours, is the same as the initial probability that it lives for at least t hours. If the component is alive at time s (if $X > s )$, then the distribution of the remaining amount of time that it survives, namely $X - s ,$, is the same as the original distribution of a new component. That is, the component does not “remember” that it has already been in use for a time s. A used component is as good as new.

That Equation (29) holds is shown by examining the conditional probability

Equation (28) can be used to determine the numerator and denominator of Equation (30), yielding

Example 18

Find the probability that the industrial lamp in Example 17 will last for another 1000 hours, given that it is operating after 2500 hours. This determination can be found using Equations (29) and (28), as follows:

Example 18 illustrates the memoryless property—namely, that a used component that follows an exponential distribution is as good as a new component. The probability that a new component will have a life greater than 1000 hours is also equal to 0.717. Stated in general, suppose that a component which has a lifetime that follows the exponential distribution with parameter λ is observed and found to be operating at an arbitrary time. Then, the distribution of the remaining lifetime is also exponential with parameter λ. The exponential distribution is the only continuous distribution that has the memoryless property. (The geometric distribution is the only discrete distribution that possesses the memoryless property.)

4.3 Gamma distribution

A function used in defining the gamma distribution is the gamma function, which is defined for al $\beta > 0$ as

By integrating Equation (31) by parts, it can be shown that

If $\beta$ is an integer, then, by using $\Gamma ( 1 ) = 1$ and applying Equation (32), it can be seen that

The gamma function can be thought of as a generalization of the factorial notion to all positive numbers, not just integers.

A random variable X is gamma distributed with parameters $\beta$ and θ if its pdf is given by

$\beta$ is called the shape parameter, and $\theta$ is called the scale parameter. Several gamma distributions for $\theta = 1$ and various values of $\beta$ are shown in Figure 11.

The mean and variance of the gamma distribution are given by

and

The cdf of X is given by

Figure 11 pdfs for several gamma distributions with $\theta = 1$ When $\boldsymbol {\cdot} \beta$ is an integer, the gamma distribution is related to the exponential distribution in the following manner: If the random variable X is the sum of $\beta$ independent, exponentially distributed random variables, each with parameter $\beta \theta$, then X has a gamma distribution with parameters $\beta$ and θ. Thus, i

where the pdf of $X_{j}$ is given by

and the $X_{j}$ are mutually independent, then X has the pdf given in Equation (34). Note that, when $\beta = 1$, an exponential distribution results. This result follows from Equation (38) or from letting $\beta = 1$ in Equation (34).

4.4 Erlang distribution

The pdf given by Equation (34) is often referred to as the Erlang distribution of order (or number of phases) k when $\beta = k ,$ an integer. Erlang was a Danish telephone engineer who was an early developer of queueing theory. The Erlang distribution could arise in the following context: Consider a series of k stations that must be passed through in order to complete the servicing of a customer. An additional customer cannot enter the first station until the customer in process has negotiated all the stations. Each station has an exponential distribution of service time with parameter kθ. Equations (35) and (36), which state the mean and variance of a gamma distribution, are valid regardless of the value of $\beta .$. However, when $\beta = k ,$ an integer, Equation (38) may be used to derive the mean of the distribution in a fairly straightforward manner. The expected value of the sum of random variables is the sum of the expected value of each random variable. Thus,

The expected value of each of the exponentially distributed $X_{j}$ is given by $1 / k \theta$. Thus,

If the random variables $X_{j}$ are independent, the variance of their sum is the sum of the variances, or

When $\beta = k ,$, a positive integer, the cdf given by Equation (37) may be integrated by parts, giving

which is the sum of Poisson terms with mean $\alpha = k \theta x$. Tables of the cumulative Poisson distribution may be used to evaluate the cdf when the shape parameter is an integer.

Example 19

A college professor of electrical engineering is leaving home for the summer, but would like to have a light burning at all times to discourage burglars. The professor rigs up a device that will hold two light bulbs. The device will switch the current to the second bulb if the first bulb fails. The box in which the light bulbs are packaged says, “Average life 1000 hours, exponentially distributed.” The professor will be gone 90 days (2160 hours). What is the probability that a light will be burning when the summer is over and the professor returns?

The probability that the system will operate at least x hours is called the reliability function $R ( x )$:

In this case, the total system lifetime is given by Equation (38) with $\beta = k = 2$ bulbs and $k \theta = 1 / 1000$ per hour, so $\theta = 1 / 2000$ per hour. Thus,$F ( 2160 )$ can be determined from Equation (39) as follows:

Therefore, the chances are about 36% that a light will be burning when the professor returns.

Example 20

A medical examination is given in three stages by a physician. Each stage is exponentially distributed with a mean service time of 20 minutes. Find the probability that the exam will take 50 minutes or less. Also, compute the expected length of the exam. In this case,$k = 3$ stages and $k \theta = 1 / 20$, so that $\theta = 1 / 60$ per minute. Thus,$F ( 50 )$ can be calculated from Equation (39) as follows:

The cumulative Poisson distribution, shown in Table A.4, can be used to calculate that

The probability is 0.457 that the exam will take 50 minutes or less. The expected length of the exam is found from Equation (35):

In addition, the variance of X is $V ( X ) = 1 / \beta \theta^{2} = 1200$ minutes^2—incidentally, the mode of the Erlang distribution is given by

Thus, the modal value in this example is

4.5 Normal distribution

A random variable X with mean $- \infty < \mu < \infty$ and variance $\sigma^{2} > 0$ has a normal distribution if it has the pdf

Figure 12 pdf of the normal distribution.

The normal distribution is used so often that the notation $X \sim N ( \mu , \sigma^{2} )$) has been adopted by many authors to indicate that the random variable X is normally distributed with mean $\mu$ and variance $\sigma^{2}$ The normal pdf is shown in Figure 12.

Some of the special properties of the normal distribution are listed here:

a. lim $1_{x} {\longrightarrow} - \infty f ( x ) = 0$ and lim $\begin{array} {r} {{1} _ {x \longrightarrow \infty} f ( x ) = 0 ;} \end{array}$ the value off(x) approaches zero as x approaches negative infinity and, similarly, as x approaches positive infinity.

b.$f ( \mu - x ) = f ( \mu + x ) ;$; the pdf is symmetric about $\mu$.

c. The maximum value of the pdf occurs at $x = \mu ;$ the mean and mode are equal.

The cdf for the normal distribution is given by

It is not possible to evaluate Equation (42) in closed form. Numerical methods could be used, but it appears that it would be necessary to evaluate the integral for each pair $( \mu , \sigma^{2} )$. However, a transformation of variables,$z = ( t - \mu ) / \sigma$, allows the evaluation to be independent of $\mu$ and $\sigma$. If $X \sim N ( \mu , \sigma^{2} )$, let $Z = ( X - \mu ) / \sigma$ to obtain

Figure 13 pdf of the standard normal distribution.

The pdf

is the pdf of a normal distribution with mean 0 and variance 1. Thus,$Z \sim N ( 0 , 1 )$ and it is said tha $Z$ has a standard normal distribution. The standard normal distribution is shown in Figure 13. The cdf for the standard normal distribution is given by

Equation (45) has been widely tabulated. The probabilities $\Phi ( z )$ for $z \geq 0$ are given in Table A.3. Several examples are now given that indicate how Equation (43) and Table A.3 are used.

Example 21

Suppose that it is known that $X \sim N ( 50 , 9 )$. Compute $F ( 56 ) = P ( X \leq 56 )$). Using Equation (43), we get

from Table A.3. The intuitive interpretation is shown in Figure 14. Figure $14 ( \mathrm{a} )$ shows the pdf of $X \sim N ( 50 , 9 )$ with the specific value,$x_{0} = 56$, marked. The shaded portion is the desired probability. Figure 14(b) shows the standard normal distribution o $Z \sim N ( 0 , 1 )$) with the value 2 marked;$x_{0} = 56$ is 2σ (where $\sigma = 3 )$ greater than the mean. It is helpful to make both sketches such as those in Figure 14 to avoid confusion in figuring out required probabilities.

(a)

Figure 14 Transforming to the standard normal distribution.

Example 22

The time in hours required to load an ocean-going vessel, X, is distributed as $N ( 12 , 4 )$. The probability that the vessel will be loaded in less than 10 hours is given by $F ( 10 )$, where

The value of $\Phi ( - 1 ) = 0 . 1587$ is looked up in Table A.3 by using the symmetry property of the normal distribution. Note that $\Phi ( 1 ) = 0 . 8413$. The complement of 0.8413, or 0.1587, is contained in the tail, the shaded portion of the standard normal distribution shown in Figure 15(a). In Figure 15(b), the symmetry property is used to work out the shaded region to be $\Phi ( - 1 ) = 1 - \Phi ( 1 ) = 0 . 1587$[From this logic, it can be seen that $\Phi ( 2 ) = 0 . 9772$ and $\Phi ( - 2 ) = 1 - \Phi ( 2 ) = 0 . 0228$. In general,$\Phi ( - x ) = 1 - \Phi ( x ) . ]$ The probability that 12 or more hours will be required to load the ship can also be discovered by inspection, by using the symmetry property ofthe normal pdfand the mean as shown by Figure 16. The shaded portion of Figure 16(a) shows the problem as originally stated [i.e., evaluate $P ( X < 12 ) .$]. Now,$P ( X \ > \ 12 ) \ = \ 1 - F ( 12 )$. The standardized normal in Figure 16(b) is used to evaluate $F ( 12 ) = \Phi ( 0 ) = 0 . 50$. Thus,$P ( X \ > \ 12 ) \ = \ 1 - 0 . 50 \ = \ 0 . 50$. (The shaded portions in both Figure 16(a) and (b) contain 0.50 of the area under the normal pdf.)

(a)

Figure 15 Using the symmetry property of the normal distribution.

Figure 16 Evaluation of probability by inspection.

Figure 17 Transformation to standard normal for vessel-loading problem.

The probability that between 10 and 12 hours will be required to load a ship is given by

using earlier results presented in this example. The desired area is shown in the shaded portion of Figure 17(a). The equivalent problem shown in terms of the standardized normal distribution is shown in Figure 17(b). The probability statement is $F ( 12 ) - F ( 10 ) = \Phi ( 0 ) - \Phi ( - 1 ) = 0 . 5000 - 0 . 1587 =$0.3413, from Table A.3.

Example 23

The time to pass through a queue to begin self-service at a cafeteria has been found to be $N ( 15 , 9 )$ The probability that an arriving customer waits between 14 and 17 minutes is computed as follows:

The shaded area shown in Figure 18(a) represents the probability $F ( 17 ) - F ( 14 )$. The shaded area shown in Figure 18(b) represents the equivalent probability,$\Phi ( 0 . 667 ) \mathrm{~ - ~} \Phi ( - 0 . 333 )$), for the standardized normal distribution. From Table A.3,$\Phi ( 0 . 667 ) \mathrm{~ = ~} 0 . 7476$. Now,$\Phi ( - 0 . 333 ) =$

(b) Figure 18 Transformation to standard normal for cafeteria problem.$1 - \Phi ( 0 . 333 ) = 1 - 0 . 6304 = 0 . 3696 .$. Thus,$( 0 . 667 ) - \Phi ( - 0 . 333 ) = 0 . 3780$. The probability is 0.3780 that the customer will pass through the queue in a time between 14 and 17 minutes.

Example 24

Lead-time demand, X, for an item is approximated by a normal distribution having mean 25 and variance 9. It is desired to compute the value for lead time that will be exceeded only 5% of the time. Thus, the problem is to find $x_{0}$ such that $P ( X > x_{0} ) = 0 . 05$, as shown by the shaded area in Figure 19(a). The equivalent problem is shown as the shaded area in Figure 19(b). Now,

or, equivalently,

From Table A.3, it can be seen that (1.645) 0.95. Thus,$x_{0}$ can be found by solving

(a)

Figure 19 Finding $x_{0}$ for lead-time-demand problem.

Therefore, in only 5% of the cases will demand during lead time exceed available inventory if an order to purchase is made when the stock level reaches 30.

4.6 Weibull distribution

The random variable X has a Weibull distribution if its pdf has the form

The three parameters of the Weibull distribution are ν$( - \infty < \nu < \infty )$, which is the location parameter; α$( \alpha > 0 )$, which is the scale parameter; and $\beta \left( \beta > 0 \right)$, which is the shape parameter. When $\nu = 0 ,$, the Weibull pdf becomes

^Figure ^20 ^{Weibull pdfs for} ν = 0; α = 1; β = ^1 , 1, 2, 4^.

Figure 20 shows several Weibull densities when $\nu = 0$ and $\alpha = 1$. When $\beta = 1$, the Weibul distribution is reduced to

which is an exponential distribution with parameter $\lambda = 1 / \alpha$ The mean and variance of the Weibull distribution are given by the following expressions:

where $\Gamma ( \cdot )$ is defined by Equation (31). Thus, the location parameter ν has no effect on the variance; however, the mean is increased or decreased by ν. The cdf of the Weibull distribution is given by