QUEUING. Chapter CHAPTER OUTLINE

Transcription

1 15 Chapter QUEUING CHAPTER OUTLINE 15.1 Introduction 15.2 The Basic Model 15.3 A Taxonomy of Queuing Models 15.4 Little s Flow Equation and Related Results 15.5 The M/G/1 Queue 15.6 Model 1: An M/M/s Queue (Hematology Lab) 15.7 Economic Analysis of Queuing Systems 15.8 Model 2: A Finite Queue (WATS Lines) 15.9 Model 3: The Repairperson Model Transient Versus Steady-State Results: Order Promising The Role of the Exponential Distribution Queue Discipline Notes on Implementation Summary KEY TERMS SELF-REVIEW EXERCISES PROBLEMS CASE 1: HOW MANY OPERATORS? REFERENCES

2 CD15-2 C D C H A P T E R S APPLICATION OUTLINE Shortening the New York City Police Department s Arrest-to-Arraignment Time In 1988, New York City (NYC) s arrestees were in custody waiting to be arraigned for an average of 44 hours, occasionally for more than 72 hours. Moreover, they were held in crowded, noisy conditions that were emotionally stressful, unhealthy, and often physically dangerous. In March 1990, the New York Times ran a front-page story on a woman who spent 45 hours in pre-arraignment detention in the Bronx with the headline Trapped in the Terror of New York s Holding Pens. Arrestees were being denied a speedy court appearance and the lengthy delays greatly diminished the efficiency of the justice system. That same year, the NY Supreme Court ruled that the city was to attempt to arraign within 24 hours or release the prisoner. Under these circumstances, NYC undertook the single most ambitious management science project to date in its history with the goal of reducing the arrest-to-arraignment (ATA) time. There were basically four different boroughs (Manhattan, Bronx, Brooklyn, and Queens), each with its own idiosyncratic ways of doing things. The basic process included the following basic steps, which comprise a large queuing system: arrest by an officer of NYPD; taken to precinct where prisoner is searched, fingerprinted, and detained while an arrest report is completed; taken to central booking where fingerprints are faxed to state capital for identification and report of criminal history, the arresting officer fills out more paperwork including the sworn complaint with the assistant district attorney, and the arrestee is lodged to await arraignment. In 1988, in these four boroughs alone, over 325,000 arrests were made for which the defendants could be detained awaiting arraignment (i.e., more serious crimes). Unlike many other jurisdictions in the United States, in NYC felonies predominate, many of them involving violence or illicit drugs. Thus, an arrestee might find himself in the same holding cell with violent repeat offenders or defendants in drug episodes. The project team mounted an extensive two-year effort. While the lengthy delays in ATA time were the key factor to study, the high costs associated with the current ATA process were an additional item of study. One of the contributors to these high costs was that arresting officers were spending an average of more than 8 hours from the time they departed central booking until they swore out the complaint. Much of this time was overtime and the task itself only required 30 minutes! They were waiting in line for 7 1 \ 2 hours! The whole process was modeled as a series of stages. Some of the stages were modeled as single-server queues, others as multiple-server queues, and some were even more complex. The statistical distributions and their corresponding parameters for each stage had to be determined. The overall model could then look at various different what-if scenarios involving combinations of workloads and arrest processing policies. The model generated several types of output including average overall ATA time and average times for completing individual stages of the process. The cost of each of the scenarios could also be generated by a companion spreadsheet model. NYC could then choose between several different alternatives, each with its own cost and performance measure. In May 1990, Mayor David Dinkins released the findings of this project at a press conference with his strong endorsement of its recommended changes. The model saved the city over $10 million per year in police overtime costs alone. The city has reduced the average ATA time delay from 44 hours to about 24 hours citywide. Arrestees gain the right to a speedier trial and are no longer warehoused under horrible conditions for longer than absolutely necessary. The city has also greatly reduced its costs for prisoner supervision and transportation by about $11 million per year. One final recommendation was the elimination of a single arraignment courtroom. This resulted in additional savings of $9.5 million for the city and state. (See Larson et al.) 15.1 INTRODUCTION Queuing models are everywhere. This fact is obvious even to the most casual observer. Airplanes queue up in holding patterns, waiting for a runway so they can land, and then they line up again to take off. People line up for tickets, to buy groceries, and, if they happen to live in England, for almost everything else. Jobs line up for machines, orders line up to be filled, and so on. As you can probably tell, queue is the British term for any type of line for waiting. The Danish engineer A. K. Erlang is credited with founding queuing theory by studying telephone switchboards in Copenhagen for the Danish Telephone Company. He developed many of the queuing results used today. One of the greatest uses of queuing theory in the United States is for analyzing automobile traffic flow studying how many lanes to have, how to regulate the traffic lights, and so forth in order to maximize the flow of traffic.

3 C H A P T E R 1 5 Queuing CD15-3 FIGURE 15.1 General Queuing Models Arrivals Service facility Monte Jackson might not subscribe to the notion that all of life is a queue, but as administrative director of St. Luke s Hospital in Philadelphia, he must deal with a number of situations that can be described as queuing models. Briefly, a queuing model is one in which you have a sequence of items (such as people) arriving at a facility for service, as shown in Figure At this moment, Monte is concerned about three particular queuing models. Model 1: St. Luke s Hematology Lab St. Luke s treats a large number of patients on an outpatient basis; that is, there are many patients who come to the hospital to see the staff doctors for diagnosis and treatment but who are not admitted to the hospital. Outpatients plus those admitted to the 600-bed hospital produce a large flow of new patients each day. Most new patients must visit the hematology laboratory as part of the diagnostic process. Each such patient has to be seen by a technician. The system works like this: After seeing a doctor, the patient arrives at the laboratory and checks in with a clerk. Patients are assigned on a first-come, first-served basis to test rooms as they become available. The technician assigned to that room performs the tests ordered by the doctor. When the testing is complete, the patient goes on to the next step in the process (perhaps X-ray), and the technician sees a new patient. Monte must decide how many technicians to hire. Superficially, at least, the trade-off is obvious. More technicians means more expense for the hospital, but quicker service for the patients. Model 2: Buying WATS Lines As part of its remodeling process, St. Luke s is designing a new communications system. Monte must decide how many WATS lines the hospital should buy. WATS (Wide Area Telephone Service) is an acronym for a special flat-rate, long-distance service offered by some phone companies. When all the phone lines allocated to WATS are in use, the person dialing out will get a busy signal, indicating that the call can t be completed. Monte knows that when people pick up the phone, they want to get through without having to try several times. How many lines he needs to achieve that result at a reasonable cost is not so clear. Model 3: Hiring Repairpeople St. Luke s hires repairpeople to maintain 20 individual pieces of electronic equipment. The equipment includes measuring devices such as the electrocardiogram machine, small dedicated computers like the one used for lung analysis, and equipment such as the CAT scanner. If a piece of equipment fails and all the repairpeople are occupied, it must wait to be repaired. Monte must decide how many repairpeople to hire. He must balance their cost against the cost of having broken equipment. As Table 15.1 indicates, all three of these models fit the general description of a queuing model. Monte will resolve these models by using a combination of analytic and simulation models. However, before we reach the level of sophistication required to deal with Monte s specific models, it is necessary for us to spend some time with the basic queuing model. In the process we will learn some terminology, and we will see the type of analytic results that are available. Table 15.1 Some Queuing Models PROBLEM ARRIVALS SERVICE FACILITY 1 Patients Technicians 2 Telephone Calls Switchboard 3 Broken Equipment Repairpeople

4 CD15-4 C D C H A P T E R S 15.2 THE BASIC MODEL Consider the Xerox machine located in the fourth-floor secretarial service suite. Assume that users arrive at the machine and form a single line. Each arrival in turn uses the machine to perform a specific task. These tasks vary from obtaining a copy of a 1-page letter to producing 100 copies of a 25-page report. This system is called a single-server (or single-channel) queue. Questions about this or any other queuing system center on four quantities: 1. The number of people in the system: the number of people currently being served, as well as those waiting for service. 2. The number of people in the queue: the number of people waiting for service. 3. The waiting time in the system: the interval between when an individual enters the system and when he or she leaves the system. Note that this interval includes the service time. 4. The waiting time in the queue: the time between entering the system and the beginning of service. ASSUMPTIONS OF THE BASIC MODEL 1. Arrival Process. Each arrival will be called a job. Since the time between arrivals (the interarrival time) is not known with certainty, we will need to specify a probability distribution for it. In the basic model a particular distribution, called the exponential distribution (sometimes called the negative exponential distribution), is used. This distribution plays a central role in many queuing models. It provides a reasonable representation of the arrival process in a number of situations, and its so-called lack of memory property makes it possible to obtain analytic results. The exponential distribution is not symmetric, a fact that bothers people who think that an average must have as many values above the mean as below it. For example, if customers arrive, on the average, every 5 minutes according to an exponential distribution, then approximately 2 \ of them will have interarrival times less than 5 minutes, and only about 1 3 \ 3 of them longer than 5 minutes (but some of those may be very long and thus skew the average). The exponential distribution describes many services (bank tellers, postal clerks). About 2 \ 3 of the service times will be below the mean time (a lot of short, quick transactions) and 1 \ 3 of the service times will be above the mean (someone with the cash receipts from his or her business, or a person mailing a package overseas). The words Poisson input are also used to describe the arrival process when the time between arrivals has an exponential distribution. This is because of the relationship between the exponential distribution and the Poisson distribution. In particular, if the interarrival time has an exponential distribution, the number of arrivals in a specified length of time (say, three hours) has a Poisson distribution. The exponential distribution and its relationship to the Poisson is discussed in some detail in Section At this point, it is only necessary to understand that the exponential distribution is completely specified by one parameter. This parameter, called, is the mean arrival rate; that is, how many jobs arrive (on the average) during a specific period of time. In a moment we will consider an example in which = 0.05 jobs per minute. This implies that on the average 5 \ 100 of a job arrives every minute. It is probably more natural to think in terms of a longer time interval. An equivalent statement is that on the average one job arrives every 20 minutes. Using more technical terms, we say that the mean interarrival time is 20 minutes. Mean interarrival time is the average time between two arrivals. Thus, for the exponential distribution 1 average time between jobs = mean interarrival time = (15.1) Thus, if = 0.05, 1 mean interarrival time = = 20 =

5 C H A P T E R 1 5 Queuing CD Service Process. In the basic model, the time that it takes to complete a job (the service time) is also treated with the exponential distribution. The parameter for this exponential distribution is called. It represents the mean service rate in jobs per minute. In other words, T is the number of jobs that would be served (on the average) during a period of T minutes if the machine were busy during that time. In the upcoming example we will assume that = This implies that on the average 0.10 of a job is completed each minute. An equivalent statement is that on the average one job is completed every 10 minutes. The mean, or average, service time (the average time to complete a job) is 1/. When, the mean service rate, is 0.10, the average service time is 10 since 1/ = 1/0.10 = Queue Size. There is no limit on the number of jobs that can wait in the queue. The queue is said to be infinite. 4. Queue Discipline. Jobs are served on a first-come, first-served basis; that is, in the same order as they arrive at the queue. 5. Time Horizon. The system operates as described continuously over an infinite horizon. 6. Source Population. There is an infinite population available to arrive. Consider these assumptions in the context of the Xerox model. Suppose that the average arrival time between jobs is 20 minutes. As we have seen, the fact that the interarrival time has an exponential distribution means that 1/ = 20, and thus = 0.05, or that the jobs arrive at the rate of 0.05 job per minute. Similarly, if the average time to complete a job is 10 minutes, we know that 1/ = 10, and thus = 0.10, or that jobs are completed at the rate of 0.10 job per minute when the machine is operating. CHARACTERISTICS OF THE BASIC MODEL The values of these two parameters (together with the assumptions) are all that is needed to calculate several important operating characteristics of the basic model. The necessary formulas are presented in Table 15.2 WARNING! The formulas in Table 15.2 hold only if <. If this condition does not hold (i.e., if ), the number of people in the queue will grow without limit. Consider, for example, a specific case where = 0.25 and = Remember that 1/ is the average interarrival time. Thus, since 1/ = 1/0.25 = 4, on the average a job arrives every 4 minutes. Similarly, 1/ is the average time it takes to complete a job. Since 1/ = 1/0.10 = 10, on the average it takes 10 minutes to complete a job. It seems clear that in this case the service operation will get further behind (the queue will grow longer) as time goes by. Now return to the Xerox model, in which < and the formulas in Table 15.2 hold. Spreadsheets are ideal for crunching the numerical results from such formulas. We will use Table 15.2 Operating Characteristics for the Basic Model CHARACTERISTIC SYMBOL FORMULA Utilization / Expected Number in System Expected Number in Queue Expected Waiting Time (Includes Service Time) Expected Time in Queue L L q W W q 2 ( ) 1 ( ) Probability that the System is Empty P 0 1 /

6 CD15-6 C D C H A P T E R S FIGURE 15.2 Introductory Page of Queuing Workbook an Excel spreadsheet (Q.XLS) that was originally developed by Professor David Ashley and that already has these formulas entered. When you first open the spreadsheet you see the introductory page as shown in Figure 15.2 (note that there are four different worksheets [MMs, MG1, finiteq, finitepopulation] to be used as indicated by the tabs at the bottom of the spreadsheet). Plugging the numerical values from the Xerox model, = 0.05 and = 0.10, into the appropriate cells (E2 and E3) of the appropriate worksheet ( MMs in this example) yields the results presented in Figure We also have to tell it that we have only one server (i.e., one copy machine) and that our time unit is minutes. Steady-State Results These numbers require some interpretation. L, for example, is the expected number of people in the system (those being served plus those waiting) after the queue has reached steady state. In this sentence, steady state means that the probability that you will observe a certain number of people (say, 2) in the system does not depend on the time at which you count them. If a steady state has been achieved, the probability that there are two people using and/or waiting for the Xerox machine should be the same at 2:30 P.M. and at 4:00 P.M. The other characteristics presented in Figure 15.3 have a similar interpretation. Thus, in a steady state, (1) the system is empty with a probability of one-half (cell F7 shows that FIGURE 15.3 Evaluating the Operating Characteristics of the Basic Model

7 C H A P T E R 1 5 Queuing CD15-7 P 0 = 0.5); (2) on the average there is 0.5 person in the queue (cell F8 shows that L q = 0.5); (3) on the average an arrival must wait 10 minutes before starting to use the machine (cell F10 shows that W q = 10); and (4) on the average an arrival will spend 20 minutes in the system (cell F11 shows that W = 20). Remember that these values are averages and, as such, may have the same characteristics as the exponential distribution ( 2 \ 3 of observations below mean, 1 \ above). Thus, 2 of the customers will spend less than 10 minutes in line, while 1 3 \ 3 \ 3 will spend more than 10 minutes in line A TAXONOMY OF QUEUING MODELS Using the Results These results hold for the basic model and the particular values for the parameters ( = 0.05 and = 0.10). They provide information that is useful to management in analyzing this service facility. Suppose, for example, that management makes the following calculations: Since = 0.05, on the average 5 \ 100 of a job arrives each minute. During each 8-hour day there are 8 60 = 480 minutes. Thus, during each day there is on the average a total of (0.05)(480) = 24 arrivals. From the calculations in Figure 15.3 we know that on the average each person spends 20 minutes in the system (W = 20). Thus, a total of (24 arrivals per day) (20 minutes per arrival) = 480 minutes, or 8 hours, is spent at this facility. Management might well feel that this is too long. A variety of steps might be taken: 1. A new machine might be purchased with a smaller mean service time. 2. Another machine might be purchased and both machines used to satisfy the demand. This would change the system to a two-server queue. 3. Some personnel might be sent to a different and less busy copying facility. This would change the arrival process. Management might select one of these alternatives, or perhaps some other option. But in any case, management must balance the cost of providing service against the cost of waiting. The results in Figure 15.3 and similar results for other systems would be a central part of the analysis. These ideas will be developed in more detail in the context of Monte Jackson s models. There are many possible queuing models. For example, if the interarrival time in the basic model had been given a different distribution (not the exponential) we would have had a different model, in the sense that the previous formulas for L, L q, and so on, would no longer hold. To facilitate communication among those working on queuing models, D. G. Kendall proposed a taxonomy based on the following notation: A/B/s where A = arrival distribution B = service distribution s = number of servers Different letters are used to designate certain distributions. Placed in the A or the B position, they indicate the arrival or the service distribution, respectively. The following conventions are in general use: M = exponential distribution D = deterministic number G = any (a general) distribution of service times GI = any (a general) distribution of arrival times We can see, for example, that the Xerox model is an M/M/1 model; that is, a singleserver queue with exponential interarrival and service times.

8 CD15-8 C D C H A P T E R S APPLICATION CAPSULE Merging Traffic: A Queuing Simulation Helps Eliminate a Costly Bottleneck The Westinghouse Hanford Company in Richland, Washington, is a secured work facility: All vehicles and passengers are checked at a guard station before being allowed onto the premises. This security checkpoint created enormous traffic backups during shift changes, when the volume of entering vehicles was greatest. The result was a severe hazard for the workforce and a major loss of productivity for the company as personnel were detained in long lines. An in-house engineering group was therefore asked to study the problem and make recommendations for changes. The study group found that each workday morning an average of 7 buses and 283 cars and vans arrived at the plant. Upon approaching the entrance gate, the vehicles formed one line to pass through the checkpoint, which was normally manned by two guards during rush periods. The line extended past the available queue space (which could accommodate only 40 cars) and spilled out onto the adjacent highway, causing a major safety problem. Because of the long line, drivers of other vehicles often elected to continue down the highway to a second gate. This option meant additional time and distance for the employees, as well as an unknown wait at the other gate. The standard analytical queuing model predicted correctly that because the service rate at the checkpoint was equal to the arrival rate, the queue would continue to grow without limit as long as cars kept arriving. This, however, merely confirmed what had already been observed. A simulation was therefore developed. The model was run to reproduce the current situation and then to try out alternatives. The first alternative scenario increased the number of guards to three while keeping the single lane of traffic. This approach reduced the maximum queue length from 45.5 to 28, but increased costs. The second scenario had vehicles forming two lines, with a security guard assigned to each line. When a bus arrived, it was routed around the two vehicle lines and serviced immediately by one guard while the other guard temporarily worked both lines. This solution produced a maximum queue length of 14 vehicles and a waiting time of only about 15 minutes, compared to over 30 minutes for the existing configuration. The second scenario appeared to be a good solution involving no additional cost. When it was implemented on a trial basis, the queue length was indeed drastically reduced. The biggest surprise was that the number of vehicles using the gate rose from 285 to 345. Obviously vehicles that had been regularly bypassing the main gate had started using it again. Thanks to the shorter queues, the new system easily handled the increased traffic load. (See Landauer and Becker.) 15.4 LITTLE S FLOW EQUATION AND RELATED RESULTS It can be proven that in a steady-state queuing process L = W (15.2) This result states that L, the expected number of people in the system, equals, the arrival rate, times W, the expected waiting time. To perform a quick numerical check, see if the numbers derived for the Xerox model (Figure 15.3) satisfy (15.2). The calculation is shown in (15.3). L = 1.0 = = W (15.3) To understand the intuitive foundation for this result, consider the diagram in Figure In Scene 1 our hero arrives and joins the queue. In Scene 2 he has just completed ser- FIGURE 15.4 Our hero arrives Little s Flow Equation Scene 1: Scene 2: He completes service Arrived during the time our hero waited and was served

9 vice. Assume the system is in steady state. Since in this case the average number of people in the system is independent of time, let us measure this quantity when our hero completes being served. At this time, the number of people in the system is precisely the total number who arrived after he did (i.e., the individuals who arrived during his waiting time). Therefore, if W is his waiting time and people arrive at a rate of, we would expect L,the average number in the system, to equal W. Equation (15.2) is often called Little s flow equation. Note that it applies to any steadystate queuing process and is thus applicable to a wide variety of models. The proof used to establish (15.2) also shows that L q = W q (15.4) A numerical check for the Xerox model shows that L q = 0.5 = = W q which again agrees with the result in Figure One must take some care in applying this result in more complicated cases. It is essential that represents the rate at which arrivals join the queue. This may be different from the rate at which people actually arrive. Consider, for example, a queue with an upper limit on the number of items that can wait in the queue (called a finite queue). A modern phone system that will hold a certain number of calls (say, 10) in a queue until a service representative becomes available provides a good example of such a queue. In such a system a person who calls and finds the system full simply receives a busy signal in other words, is sent away. He or she does not join the queue. This is called a balk. Thus, if = 0.25 (the arrival rate) and the mean time between calls is 4 minutes, this is not the rate at which people join. Thus, the relationship L = 0.25W will not hold for this system. Similarly, a customer may tire of waiting in line (or being on hold) and leave without being served. This is called reneging. Here again, L = 0.25W will not hold for this system. Another important general result depends on the observation that expected waiting time = expected waiting time in queue + expected service time For the basic model we have already made use of the fact that 1 expected service time = Putting the general result in symbols yields For the Xerox model we have C H A P T E R 1 5 Queuing CD15-9 W = W q W = 20 = = W q + 1 (15.5) Not only does this hold for the basic model, but the general result (equation [15.5]) holds for any queuing model in which a steady state occurs. Equations (15.2), (15.4), and (15.5) make it possible to compute the four operating characteristics L, L q, W, and W q once one of them is known. To illustrate this fact, let us start the Xerox model all over again. We begin as last time using the second formula in Table 15.2 to calculate L: L = = = 1 Now rather than using the other formulas in Table 15.2 that are specifically for the basic model, we will use the two general results that we have just presented. First, from Little s flow equation (15.2) we know that L = W

10 CD15-10 C D C H A P T E R S Thus, knowing L = 1 and = 0.05, we obtain W = L/ = 20. Then, turning to (15.5), we see that W = W q THE M/G/1 QUEUE Finally, (15.4) shows that W q = W 1 = = 10 L q = W q = = 0.5 This alternative method of obtaining numerical results will turn out to be most useful when analyzing more complicated systems than the basic model. While the exponential distribution accurately describes the arrival process in many situations, it may not fit the service process very well. Fortunately, there is a generalization of the basic model that permits the distribution of the service time to be arbitrary. It is not even necessary to know the service time distribution, only its mean, 1/, and its variance, 2. The operating characteristics for the generalized model are given in Table Note that we have made use of the results of Section 15.4 in obtaining all the operating characteristics except for L q. As a check on the validity of these formulas, suppose that the arbitrary service time distribution is exponential. The variance of an exponential distribution is (1/ ) 2 if the mean is 1/. Therefore, L q = 2 (1/ ) 2 + ( / ) 2 = 2 2(1 / ) ( ) which is the same result as in the basic model. As 2 increases, L, L q, W, and W q all increase. This means that the consistency of a server may be as important as the speed of the server. Suppose you must hire a secretary, and you have to select one of two candidates. Secretary 1 is very consistent, typing any document in exactly 15 minutes. Secretary 2 is somewhat faster, with an average of 14 minutes per document, but with times varying according to the exponential distribution. The average workload in the office is three documents per hour, with interarrival times varying according to the exponential distribution. Which secretary will give you shorter average turnaround times on documents? This can be easily solved with the MG1 and MMs worksheets of SECRETRY.XLS shown in Figures 15.5 and Since Secretary 1 types every document in exactly 15 minutes, 2 is equal to 0. The values of the other parameters are = 3 per hour (or 0.05 per minute) and = 1 \ 15 per minute. These values are entered in the input parameters section of the MMs spreadsheet (cells E3:E6), which automatically incorporates the appropriate values into the MG1 worksheet. The results are shown in Figure Table 15.3 Operating Characteristics for the Generalized Model CHARACTERISTIC SYMBOL FORMULA Utilization / Expected Number in System L L q + Expected Number in Queue L q ( / ) 2 2(1 / ) Expected Waiting Time W 1 W q + Expected Time in Queue W q L q Probability that the System is Empty P 0 1 /

11 C H A P T E R 1 5 Queuing CD15-11 FIGURE 15.5 Turnaround Time for Secretary 1 FIGURE 15.6 Turnaround Time for Secretary 2 We could also verify this manually by using the formulas shown in Table 15.3: L q = (0.05)2 (0) + [0.05/( 1 \ 15 )] 2 2[1 0.05/( 1 = 9 \ \ 15 )] 8 W q = ( 9 \ 8 )/0.05 = 45 \ 2 = 22.5 minutes W = 45 \ = 37.5 minutes average turnaround time Again, using the spreadsheet model for Secretary 2, we enter the parameters as = 0.05, = 1 \ 14 per minute, and = 14 minutes. The results are shown in Figure We could also use either the basic model (Table 15.2) or the generalized model (Table 15.3) to verify the spreadsheet answer for Secretary 2. Using the generalized model we get L q = (0.05)2 (14) 2 + [0.05/( 1 \ 14 )] 2 2[1 0.05/( 1 = 49 \ \ 14 )] 30 = minutes W q = ( 49 \ 30 )/0.05 = 98 \ 3 = minutes W = 98 \ = minutes average turnaround time Even though Secretary 2 is faster, her average turnaround times are longer because of the high variability of her service times.

12 CD15-12 C D C H A P T E R S 15.6 MODEL 1: AN M/M/s QUEUE (HEMATOLOGY LAB) Recall that as we started this chapter, our stated goal was to attack three particular models at St. Luke s Hospital with queuing models. In the preceding sections we have laid the groundwork for this process. We have introduced, defined, and illustrated the characteristics of the systems that we will consider (e.g., expected number in queue, expected waiting time, etc.) as well as developed a spreadsheet that can automatically calculate such characteristics. We have also made some general results such as Little s flow equation available for use in future analysis. We are now in a position to turn our attention to Monte Jackson s models. The system described in Model 1 of Section 15.1, the blood-testing model, is illustrated in Figure Note that each patient joins a common queue and, on arriving at the head of the line, enters the first examining room that becomes available. This type of system must not be confused with a system in which a queue forms in front of each server, as in the typical grocery store. Assume that the interarrival time is given by an exponential distribution with parameter = 0.20 per minute. This implies that a new patient arrives every 5 minutes on the average, since 1 mean interarrival = = = 5 Also, assume that each server is identical and that each service time is given by an exponential distribution with parameter = per minute. This implies that the mean service time is 8 minutes, since 1 mean service time for an individual server = = = 8 Note that if there were only one server, the queue would grow without limit, since > (0.20 > 0.125). For a multiserver queue, however, a steady state will exist as long as < s, where s is the number of servers. For example, if we have two servers, we will achieve a steady state because 0.20 < 0.25 (= 2 * 0.125). The Key Equations As before, we want to find values L, L q, W, and W q. However, since this is a multiserver queue (not a single-sever queue as in the Xerox model), we must use different formulas. To evaluate these formulas it is convenient to start with the expression for P 0, the probability that the system is empty. For this model P 0 = s 1 n=0 ( / ) n and L q, the expected number of people in the queue, is expressed as ( / ) L q = P 0 s+1 2 (s 1)!(s / ) n! 1 + ( / )s s! 1 1 ( /s ) (15.6) (15.7) FIGURE 15.7 Multiserver Queue Arrivals Server Server 2 Server n

13 C H A P T E R 1 5 Queuing CD15-13 FIGURE 15.8 Results for Hematology Lab with 2 Servers Equations (15.6) and (15.7) and the general results in (15.2), (15.4), and (15.5) make it possible to calculate values for W q, W, and L for any specified parameter values ( and ) and any number of servers (value of s). Again, these new formulas are already entered in the MMs worksheet of our queuing template in our workbook (HEMATLGY.XLS). Example Calculations Assume, for example, that Monte decided to hire two technicians. Then, since s = 2, = 0.20, and = 0.125, we can put these values into the MMs worksheet of HEMATLGY.XLS and get the following results shown in Figure We see that the utilization ( /s ) = 0.8 (cell F6) and the probability that the system is empty is 0.11 (cell F7). These two values can be used in equation (15.7) to find L q = 2.84 (cell F8). That is, the expected number of people in the queue is somewhat less than 3. Using equation (15.4), L q = W q, we see that, on the average, a patient waits for minutes (cell F10) before entering an examining room. Lastly, the template uses the general observation that expected waiting time = expected waiting time in queue + expected service time to calculate that the expected waiting time (W) = minutes (cell F11). On the average, then, a patient spends minutes in the hematology area, waiting for a technician and having tests. Monte now wants to check what happens if he adds a third or fourth technician (server). These results are shown in Figures 15.9 and respectively. FIGURE 15.9 Results for Hematology Lab with 3 Servers

14 CD15-14 C D C H A P T E R S FIGURE Results for Hematology Lab with 4 Servers 15.7 ECONOMIC ANALYSIS OF QUEUING SYSTEMS There is a dramatic shortening of waiting time (W q ) with a third server (down to 1.57 minutes), but at the cost of having an extra server. Adding a fourth server doesn t make such a dramatic difference, as it reduces waiting time in the queue to 0.30 minutes. Another factor to take into account is how busy the servers would be in each scenario. We see in Figures 15.8 to that the utilization drops from 80% to 53.3% to 40%. The more servers added, the higher the percentage of idle time for the technicians, which could lead to boredom and sloppy work. These calculations provide lots of information to help Monte make his decision. With one technician, since >, the system is unstable and the queue will steadily grow. This could be considered irresponsible. With two technicians, the average waiting time in the queue is less than 15 minutes. By current hospital standards, this is a small and acceptable value. Obviously by adding more servers, Monte can reduce the average waiting time but at significant expense to St. Luke s. If, in some cases the queue gets uncomfortably long with two servers (remember that W q is an expected value, and the actual time in the queue will vary), the supervisors of the hematology laboratory can temporarily move one of the blood analysts to a technician s position. Monte thus feels comfortable with the idea of hiring two full-time technicians without performing a detailed cost analysis. We note that this example model is identical to the models that are faced by fast-food franchise managers: how many people to put on a shift to keep the average customer wait below a certain value. McDonalds reportedly figures it will lose a customer if the total wait is more than five minutes. Monte selected the number of lab technicians to hire by looking at the operating characteristics and using his judgment. This is not an unusual approach in queuing models and is especially common in the not-for-profit sector. Monte realizes that he is balancing the cost of hiring more technicians against the costs he incurs by forcing the patients to wait. The cost of hiring additional technicians is fairly clear. The waiting cost is not. Monte first notes that the cost to the patient is irrelevant to his decision, except as it affects the patient s willingness to use the hospital. It really does not matter who is waiting a consultant who charges $250 per hour for his services or an unemployed person with no opportunity cost unless the waiting time persuades the patient to use another health facility. This observation explains why certain monopolies like government agencies and utilities can be so casual about your waiting time. There is no place else to go! Besides the possible effect on demand, the hematology lab could cost the hospital money if it reduced the output of the hospital. Suppose, for example, that the outpatient

15 C H A P T E R 1 5 Queuing CD15-15 clinics could process 50 new patients each day, but that the hematology lab could handle only 10 patients. (This is clearly an extreme example to establish a point.) In this case, the hospital would be wasting a valuable resource, the doctors and other staff in the clinics, because of a bottleneck in the hematology lab. However, having stated this, it still is not easy to assess an explicit cost of a patient waiting. Cost Parameters If you are willing and able to estimate certain costs, you can build expected cost models of queuing systems. Consider, for example, the hematology lab model (in general terms any multiserver queue with exponential interarrival and service times), and suppose the manager is willing to specify two costs: C s = cost per hour of having a server available C w = cost per hour of having a person wait in the system (a very fuzzy or qualitative cost) With these it is possible to calculate the total costs associated with the decision to use any particular number of servers. Let us start by calculating the total cost of hiring 2 servers for an 8-hour day. There are two components: server cost = (C s )(2)(8) where C s is the cost per hour for one server, 2 is the number of servers, and 8 is the number of hours each server works, and waiting cost = (C w )(L 2 )(8) where L 2 is the number of people in the queue when there are 2 servers. This second calculation may not be as obvious, but the rationale is the same as for the server cost. If there are, on the average, L 2 people waiting when the system has 2 servers, then L 2 times 8 is the average number of waiting hours per day. Hence, (C w )(L 2 )(8) is the average waiting cost for the 8-hour day. If we wanted to calculate the total cost of using 4 servers for a 6-hour day, we would take (C s )(4)(6) + (C w )(L 4 )(6) or [(C s )(4) + (C w )(L 4 )]6 The term in square brackets, [(C s )(4) + (C w )(L 4 )], then, is the total cost per hour of using 4 servers. The Total Cost per Hour We now define TC(s) = total cost per hour of using s servers and we see that TC(s) = (C s )(s) + (C w )(L s ) Our goal is to choose s, the number of servers, to minimize this function. We can see that as s increases, the waiting cost will decrease and the server cost will increase. The idea is to find that value of s that minimizes the sum of these two costs. Figure shows the worksheet Monte created called Econ. Analysis in his HEMATLGY.XLS workbook to determine the optimal value of s. Unfortunately it is not possible to derive a formula that gives the optimal value of s. (This is in contrast to the EOQ model, where we can find the optimal order quantity, Q*, with the equation Q* = 2DC 0 /C h, as in Chapter 7.) In this example, let s put a relatively large cost on waiting and see if the decision changes from Monte s original decision of choosing two servers. We establish C s = $50/server/hour and C w = $100/customer/hour (see cells B1 and B2), and then we can calculate the server cost and waiting cost for 2, 3, and 4 servers. We ll assume we want to com-

16 CD15-16 C D C H A P T E R S FIGURE Economic Analysis for Hematology Lab with 2, 3, or 4 Servers Cell Formula Copy To C6 $B$1*A6*$E$1 C7:C8 D6 $B$2*B6*$E$1 D7:D8 E6 SUM(C6:D6) E7:E8 pare the cost over an 8-hour shift (cell E1), and we must enter in cells B6:B8 the values for L (expected number in system) for each value of s we want to explore (obtained from Figures 15.8 to 15.10). We can see that 3 servers minimizes the Total Cost at $2,730 (cell E7). Next Monte creates a data table to determine the sensitivity of this decision to the fuzzy cost, C w. He decides he wants to explore values for C w from 0 to $180. The steps for Monte to do this in his spreadsheet are: 1. Enter the initial value of 0 in cell A Click back on cell A11, then choose Edit, Fill, and then Series. 3. Click on Series in Columns, enter a step value of 20 and a terminal value of 180. Click on OK. 4. Enter the formulas for the quantities we want to track (total cost with 2 servers, total cost with 3 servers, total cost with 4 servers) in cells B10:D10. These formulas are =E6, =E7, and =E8, respectively. 5. Highlight the range A10:D20, and click on Data, then Table. 6. Enter the Column Input Cell as B2. Click on OK. 7. Excel automatically fills in the table as shown in cells A10:D20 of Figure Finally, Monte wants to graph the results of this sensitivity analysis to look for patterns and trends. To do this, he highlighted the range A11:D20, clicked on the Chart Wizard, and then followed the steps to generate the graph shown in Figure We can see that 2 servers is optimal for C w = 0, while 3 servers is optimal from C w = $20 up to $180. It does look as though 4 servers will become the optimal decision for values of C w $200. Let us complete our examination of the hematology lab model with these observations: We have seen how to find values for L, L q, W, and W q. These values were then used to select the appropriate number of technicians (servers). This decision might be made on intuitive grounds or on the basis of an explicit economic analysis. Sensitivity analysis is also important to perform, especially on harder-to-quantify parameters like C w. We now move on to Monte s second model.

17 C H A P T E R 1 5 Queuing CD15-17 FIGURE Graph of Sensitivity Analysis on Cost of Waiting for Hematology Lab 15.8 MODEL 2: A FINITE QUEUE (WATS LINES) Do not be misled by the title of this section. It is devoted to Model 2, Monte s attempt to select the appropriate number of WATS lines for St. Luke s. Fortunately, in this case he can expect help from the telephone company. It has a great deal of expertise in such matters, since queuing models have found extensive use in the field of telephone traffic engineering. The problem of how many lines are needed by a switchboard was typically attacked by using the M/G/s model, with blocked customers cleared. You already know that this model is a multichannel queue with s servers (s lines), exponential interarrival times for the calls, and a general distribution for the service time, which in this case is the length of each call. The phrase blocked customers cleared is queuing jargon. It means that when an arrival finds all of the servers occupied (all of the lines busy), he or she does not get in a queue but simply leaves. This phrase clearly describes the behavior of the traditional telephone switchboard. More sophisticated systems now provide for queuing of a finite number of customers, in some cases even providing the lucky customer the opportunity to enjoy a Muzak version of Elton John s Can You Feel the Love Tonight? or the Macarena. Probability of j Busy Servers The problem of selecting the appropriate number of lines (servers) is attacked by computing the steady-state probability that exactly j lines will be busy. This, in turn, will be used to calculate the steady-state probability that all s lines are busy. Clearly, if you have s lines and they are all busy, the next caller will not be able to place a call. The steady-state probability that there are exactly j busy servers given that s lines (servers) are available is given by the expression P j = s k=0 where = arrival rate (the rate at which calls arrive) 1 = mean service time (the average length of a conversation) s = number of servers (lines) ( / ) j /j! ( / ) k /k! (15.8) The expression is called the truncated Poisson distribution or the Erlang loss distribution. It is noteworthy that although we are considering a general service-time distribution, the value P j defined by (15.8) depends only on the mean of this distribution. Consider a system in which = 1 (calls arrive at the rate of 1 per minute) and 1/ = 10 (the average length of a conversation is 10 minutes). Here / = 10. Suppose that we have

18 CD15-18 C D C H A P T E R S five lines in the system (s = 5) and want to find the steady-state probability that exactly two are busy (j = 2). From (15.8) we see that P 2 = 5 k=0 ( / ) 2 /2! ( / ) k /k! (10) 2 /(2 1) = / /(2 1) /(3 2 1) /( ) /( ) 50 = = = In other words, on the average, two lines would be busy 3.4% of the time. An alternative way of obtaining P j that is easy to implement in a spreadsheet (because of its sequential formulation) is as follows: P i = P i 1 ( / )/i So, for example once we know P 2, we can calculate P 3 as: Likewise, P 4 is found as P 3 = P 2 (10)/3 = (0.034)(10)/3 = P 4 = P 3 (10)/4 = (0.1133)(10)/4 = Each successive P i 1 is multiplied by ( / ) and divided by i to achieve the new P i. The more interesting question is: What is the probability that all of the lines are busy? since in this case a potential caller would not be able to place a call on the WATS lines. To find the answer to this question, we simply set j = s (in our example s = 5) and we obtain P 5 = P 4 (10)/5 = (0.2833)(10)/5 = or on the average the system is totally occupied 56.4% of the time. Again, it is easy enough to implement all these formulas in a spreadsheet. The probability that the system is totally occupied (all servers are busy) is calculated in a new worksheet called finiteq in the workbook WATS.XLS. We can see that the probability that a customer balks with 5 servers is much more easily calculated in a spreadsheet, as shown in Figure Of course, we get the same value (0.564 in cell F13). We can then build a data table to determine this value for several different values of s. Figure shows such a data table that Monte built for examining the possibility of between 0 and 10 phone lines. The steps for Monte to do this in the spreadsheet are: 1. Enter the initial value of 0 in cell A Click back on cell A23, then choose Edit, Fill, and then Series. 3. Click on Series in Columns, enter a step value of 1 and a terminal value of 10. Click on OK.

19 C H A P T E R 1 5 Queuing CD15-19 FIGURE Finite Queue Spreadsheet Calculation of Probability that Customer Balks FIGURE Data Table for Probability that Customer Balks for Different Values of s 4. Enter the formulas for the quantity we want to track (probability that a customer balks) in cell C22. The formula is =F Highlight the range B22: C33, and click on Data, then Table. 6. Enter the Column Input Cell as E4. Click on OK. Excel automatically fills in the table as shown in Figure Next we can create column D that calculates the marginal improvement in this probability as we add servers. This is also shown in Figure Here it is clear that the marginal effect of adding more servers decreases. For example, adding a second line when there was one in service decreases the probability of the system being busy by 0.089, whereas adding the tenth line when there were already nine in service decreases this probability by Average Number of Busy Servers Another interesting and useful quantity in the design of phone installations is the average number of busy lines. This quantity is called the carried load in queuing jargon. If we define N as the average number of busy servers, then N = [1 Prob. That a Customer Will Balk] (15.9) Assume now that in Monte s model with WATS lines for St. Luke s, = 1 and 1/ = 10. Thus, if he purchases 10 lines, we see in Figure that the probability that all 10 are busy =0.215 (cell C33). It follows from (15.9) that N = 10( ) = 7.85

20 CD15-20 C D C H A P T E R S 15.9 MODEL 3: THE REPAIRPERSON MODEL In other words, the entire system will be busy with probability or about one fifth of the time and, on the average, almost 8 lines will be busy. After N has been calculated, the server utilization can be calculated by dividing N by s (the number of servers). Thus, for the situation at St. Luke s, the server utilization is 7.85/10 = 78.5%, which means that each server (on average) is busy 78.5% of the time and idle 21.5% of the time. Monte feels that ten lines is a reasonable compromise. There does not seem to be a great deal of excess capacity, but, on the other hand, the probability of finding the system busy is in a region (70% to 80%) that he feels is appropriate for the hospital. If he is uncomfortable with this solution, based on a subjective balancing of the number of lines and the probability of finding the system busy, and is willing to specify a cost for each time a caller finds the system busy, he can select the number of lines to minimize the expected cost per hour. He would proceed in the same manner as in the M/M/s system of Sections 15.6 and In this model Monte must decide how many repairpersons to hire to maintain 20 pieces of electronic equipment. Repairpersons deal with machines on a first-come (perhaps firstfailed is more accurate), first-served basis. A single repairperson treats each broken machine. You can thus think of the failed machines as forming a queue in front of multiple servers (the repairpersons). This is another M/M/s model, but it differs in a fundamental way from the M/M/s system (the blood-testing model) considered in Section In this model there is a limited number of items (20) that can join the queue, whereas in the hematology lab model an unlimited number could potentially join the queue. A queuing model, like the repairperson model, in which only a finite number of people are eligible to join the queue is said to have a finite calling population.models with an unlimited number of possible participants are said to have an infinite calling population. Consider the model with 20 machines and 2 repairpersons. Assume that when a machine is running, the time between breakdowns has an exponential distribution with parameter = 0.25 per hour; that is, the average time between breakdowns is 1/ = 4 hours. Similarly, assume that the time it takes to repair a machine has an exponential distribution and that the mean repair time is 0.50 hour (i.e., 1/ = 0.50). This model is an M/M/2 model with a maximum of 18 items in the queue (20 including the 2 in service) and a finite calling population. In this case the general equations for the steady-state probability that there are n jobs in the system is a function of,, s (the number of repairpersons), and N (the number of machines). In particular, We also know that N! P n = n!(n n)! ( / )n P 0 for 0 n s N! P n = (N n)!s!s n s ( / )n P 0 for s < n N (15.10) N P n = 1 (15.11) n=0 We thus have N + 1 linear equations (N from equation [15.10] and 1 from equation [15.11]) in the N + 1 variables of interest (P 0, P 1,...,P n ). This makes it possible (if painful) to calculate values of P n for any particular model. One can see that each model becomes a bit more complex than the previous one, and that the formulas of P n become more complex. There are, however, no simple expressions (even by these standards) for the expected number of jobs (broken machines) in the system or for waiting. If the values for P n are

21 C H A P T E R 1 5 Queuing CD15-21 FIGURE Finite Population Spreadsheet for the Repairperson Model TRANSIENT VERSUS STEADY-STATE RESULTS: ORDER PROMISING computed, then it is (truly) a simple task to find a numerical value for the expected number in the system. You must just calculate expected number in system = L = N np n If God had intended humans to perform this kind of calculation by hand, He would not have let spreadsheets be invented. Figure shows the setup of the finite Population worksheet in the REPAIR.XLS workbook that can be used to compute values of P n, the expected number in the system, and the expected number waiting, for a variety of different systems. As shown, the user enters the typical system parameters (,, and s) in the MMs worksheet, then clicks on the new worksheet finitepopulation and adds the size of the finite population (cell E5), and the spreadsheet does the rest. In this case, the rest consists of numerically evaluating the equations for P n and using the results to find the expected number in the system. There is one trick with this kind of model, when you enter the value for the arrival rate ( ) in the MMs worksheet, you actually need to enter N*, or the entire population s arrival rate. As you see in Figure 15.15, for this system, L q, the average number of machines waiting for service, is (cell F8) and W, the expected time in the system, is hours (cell F11). The utilization is high at 92.5% and we see from the histogram that there s a 3.3% chance that the system is empty, an increasing probability of there being one machine (8.3%), and the probabilities keep growing for increasing numbers of machines until it reaches a maximum at four and five machines (they re tied at 11.8%). Then the probabilities decline for machines beyond five. It is not always the case that we are interested in steady-state results, or that an analytical model is available to predict the behavior of the queuing system of interest. In this section we will consider a situation in which we are interested in the transient behavior of the system and must use simulation to obtain the desired answers. Manufacturing processes can be viewed as complex queuing systems. Probably the most frequently used management science tool in manufacturing is queuing systems simulation. Larry Lujack, a production planner at SONOROLA and a recent graduate of the University of Chicago s Graduate Business School, is wondering if what he learned about queuing models in his last semester of school could help him decide when to promise a new customer order. The order is for 20 units of an item that requires sequential processing at 2 n=0

22 CD15-22 C D C H A P T E R S work stations. The average time to process a unit at each work station is 4 hours. Each work station is available for 8 hours every working day. By considering when the last of the 20 units will be completed, Larry initially estimates that it will take 10.5 days to process the order. The last unit must wait at Work Station 1 for the first 19 units to be completed, then it must be processed at Work Station 1 and then at Work Station 2. Assuming that it does not have to wait when it gets to Work Station 2, Larry calculates as follows: (19 units 4 hours/unit + 4 hours + 4 hours) 8 hours/day = 10.5 days However, this analysis is somewhat simplistic. It ignores the variability of the processing times and the possibility of queuing at Work Station 2. Larry feels that the exponential distribution is an appropriate distribution for processing times because the 4-hour figure was arrived at by averaging many processing times that were less than 4 hours with a few processing times that were significantly longer than 4 hours (see Section 15.11). These few long processing times were due to equipment failures at a work station while processing a unit. Next Larry checks to see whether the assumptions of the basic queuing model are met. The output from Work Station 1 are the arrivals to Work Station 2, and the time between arrivals is exponential because the processing time at Work Station 1 is exponential. The service time at Work Station 2 is exponential because it is the same as the processing time. The units are processed on a first-come, first-served basis at Work Station 2, and there is sufficient buffer capacity between the work stations so that the queue size is for all practical purposes infinite. However, the assumption of an infinite time horizon is not met. Larry is interested only in the behavior of the system until customer 20 ends its processing. Larry decides to apply the basic model anyway and use it as an approximation. He approximates the time it takes to process 20 units as follows. First, he estimates that the last unit in the batch of 20 will leave Work Station 1 after 20 4 = 80 hours. Then this unit will wait in the queue in front of Work Station 2. Finally, it will complete processing at Work Station 2, at which time all 20 units will have been completed. The total time that the last unit spends at Work Station 2 is W. Thus, Larry s estimate is W. In the basic model, W is given by the formula 1/( ). Larry now realizes his dilemma: and are equal (1 unit per 4 hours) and the formula is valid only when is greater than. Larry decides to set up a spreadsheet (ORDER.XLS) to stimulate the flow of the 20 units through the 2 work stations. His spreadsheet is shown in Figure Larry assumes that raw material is always available at Work Station 1 (WS1) so that the next unit at Work Station 1 can start as soon as the current unit is finished. This means that for Work Station 1 the start time of a unit is the stop time of the previous unit. For example, the formula in cell B8 of the spreadsheet is =C7. The start time of a unit at Work Station 2 is either the stop time of that unit on Work Station 1 or the stop time of the previous unit on Work Station 2, whichever is larger. So, for example, the formula in cell D8 is =MAX(C8,E7). The stop time of a unit is just the start time plus the processing time. The finish time in days is shown in cell F2 and is calculated by dividing the stop time at Work Station 2 of the last unit by the number of hours/day (8 in cell F1). The spreadsheet calculates this time to be 10.5 days if every unit takes exactly 4 hours at every work station. To analyze the impact of processing time variability, Larry replaces the constant processing time of 4 at WS1 in his spreadsheet with the appropriate random distribution (in Crystal Ball, he chooses the exponential distribution from the gallery; he enters =RiskExpon($B$1) ), which samples from an exponential distribution with a mean of 4. This makes the finish time a random variable. Larry would like to know the 99th percentile of this random variable (the number that 99% of the time the random variable will be less than or equal to). He could then promise the order in that number of days and be 99% sure that it would actually be completed on time.

23 C H A P T E R 1 5 Queuing CD15-23 FIGURE Order Promising Spreadsheet Cell Formula Copy To C7 $B$1 + B7 C8:C26 D7 C7 E7 B8 D8 F2 D7+$B$2 C7 MAX(C8, E7) E26/F1 E8:E26 B9:B26 D9:D26 Using Crystal Ball it is a simple matter to find any percentile of any random variable in the spreadsheet. Figure shows the 99th percentile for cell F2 (the finish time in days) based on 1000 iterations (1000 sets of 40 random processing times 20 for Work Station 1, 20 for Work Station 2). First, note that the expected finish time ( Mean for cell F2 in Figure 15.17) is days, 2 days longer than Larry s initial calculation. The extra FIGURE Statistical Output for Order Completion Model

24 CD15-24 C D C H A P T E R S FIGURE Histogram of Finish Times for Order Completion Model THE ROLE OF THE EXPONENTIAL DISTRIBUTION 2 days is the average queuing delay caused by the variability of the processing times. If Larry wants to be 99% sure of having the order completed by the time he promises, he should set the due date to be days ( Target #1(Value)= in Figure 15.17) after the material becomes available at Work Station 1. The queuing that takes place at Work Station 2 has increased the lead time (time from the start of the order to its completion) by nearly 8 days (18.28 less 10.5) over what it would be if there were no variability in the processing times. Figure shows the histogram of the finish time, and we can see that the time can vary from 6.5 days up to 21 days. Even though the basic model was not applicable in this case, it helped Larry to think about the model and to understand this answer from his spreadsheet simulation. Problem explores a slightly different situation in which the basic model can be used to estimate the average finish time and the 99th percentile. There is an enormous body of literature concerning queuing systems, and it is virtually impossible for a manager to be aware of all the results. There are, however, some general considerations that are useful in helping a manager think about the use of queuing models. One such consideration is the role of the exponential distribution in analytic queuing models. There are essentially no analytic results for queuing situations that do not involve the exponential distribution either as the distribution of interarrival times or service times or both. This fact makes it important for a manager to recognize the set of circumstances in which it is reasonable to assume that an exponential distribution will occur. The following three properties of the exponential distribution help to identify it: 1. Lack of memory: In an arrival process this property implies that the probability that an arrival will occur in the next few minutes is not influenced by when the last arrival occurred; that is, the system has no memory of what has just happened. This situation arises when (1) there are many individuals who could potentially arrive at the system, (2) each person decides to arrive independently of the other individuals, and (3) each individual selects his or her time of arrival completely at random. It is easy to see why the assumption of exponential arrivals fits the telephone system so well.

25 C H A P T E R 1 5 Queuing CD15-25 FIGURE A High Probability of Short Service Times Prob S t t QUEUE DISCIPLINE 2. Small service times: With an exponential distribution, small values of the service time are common. This can be seen in Figure This figure shows the graph of the probability that the service time S is less than or equal to t (Prob {S t}) if the mean service time is 10; that is = 0.1 and 1/ = 10. Note that the graph rises rapidly and then slowly approaches the value 1.0. This indicates a high probability of having a short service time. For example, when t = 10, the probability that S t is In other words, more than 63% of the service times are smaller than the average service time. This compares to a normal distribution where only 50% of the service times are smaller than the average. The practical implication of this fact is that an exponential distribution can best be used to model the distribution of service times in a system in which a large proportion of jobs take a very short time and only a few jobs run for a long time. It is said that engineers think that the whole world is exponentially distributed, while social scientists think that the whole world is normally distributed. A quick and dirty way to check what kind of distribution you have is to see if the average of the data is close to its standard deviation. If it is, then most likely the numbers are exponentially distributed. If it is 1 \ 3 or less of the mean, it is probably normally distributed. 3. Relation to the Poisson distribution: While introducing the basic model (Section 15.2) we noted the relationship between the exponential and Poisson distributions. In particular, if the time between arrivals has an exponential distribution with parameter, then in a specified period of time (say, T) the number of arrivals will have a Poisson distribution with parameter T. Then, if X is the number of arrivals during the time T, the probability that X equals a specific number (say, n) is given by the equation Prob{X = n} = e T ( T) n n! This equation holds for any nonnegative integer value of n (i.e., n = 0, 1, 2, and so on). The relationship between the exponential and the Poisson distributions plays an important role in the theoretical development of queuing theory. It also has an important practical implication. By comparing the number of jobs that arrive for service during a specific period of time with the number that the Poisson distribution suggests, the manager is able to see if his or her choices of a model and parameter values for the arrival process are reasonable. In the previous sections we specified the arrival distribution, the service distribution, and the number of servers to define a queuing system. Queue discipline is still another characteristic that must be specified in order to define a queuing system. In all of the models that we have considered so far, we have assumed that arrivals were served on a first-come, firstserved basis (often called FIFO, for first-in, first-out ). This is certainly an appropriate

26 CD15-26 C D C H A P T E R S NOTES ON IMPLEMENTATION SUMMARY assumption for telephone systems and for many systems where people are the arrivals. This is not necessarily the case for other systems, however. In an elevator the last person in is often the first out (LIFO). And in the repairperson model, there is really no reason to fix the machines in the same order as they break down. If a certain machine can be returned to production in 5 minutes, it seems like a good idea to do it first, rather than making it wait until a 1-hour job on a machine that broke down earlier is completed. Adding the possibility of selecting a good queue discipline makes the queuing models more complicated. Models of this sort are often referred to as scheduling models, and there is an extensive literature that deals with them, which is left to more advanced courses. The models discussed in this chapter are useful representatives of only a small portion of the broad expanse of queuing models. The results presented here in general require that either the time between arrivals, the service time, or both have an exponential distribution. They are important because they yield tight analytic results and because, in many circumstances, it is reasonable to assume that the arrival process is a Poisson process. In particular, we have noted that a large (essentially infinite) calling population in which individual members of the population decide at random to arrive at service facilities generates an exponential distribution for the time between arrivals. It is not surprising, then, that the analytic models are often used on systems with this type of arrival mechanism. Communication networks (especially the telephone system) and traffic control systems are two important examples of such systems. Discrete event dynamic simulation (DEDS) is a popular approach for studying queuing models that do not fit the analytic mold. Indeed, current programs such as Arena (by Systems Modeling), Extend, and Alpha/SIM have been created to facilitate the simulation process. This topic is now covered in detail in Chapter 10. This chapter provided an introduction to the subject of queuing. It pointed out that many interesting models can be cast in the arrival/service mode of a queuing model. Section 15.2 was devoted to the basic model, a single-channel queue with exponential interarrival times and service times. Four system characteristics expected number in system, L; expected number in queue, L q ; expected waiting time, W; and expected time in queue, W q were defined. Formulas were presented for these characteristics as a function of the parameters of the arrival and service processes. A numerical example was presented. Section 15.3 briefly introduced a system of notation for describing queuing systems. Little s flow equation, L,= W, was presented in Section This equation plus the general fact that W = W q + expected service time were offered as alternative means for computing queue characteristics. Section 15.5 generalized the basic model to allow for an arbitrary service time distribution. Section 15.6 considered a multiserver queue. Some new formulas were presented. These formulas were combined with results from Section 15.4 to compute numerical results in an Excel spreadsheet for a staffing model in a hematology lab. Section 15.7 was devoted to an economic analysis of the staffing model in the hematology lab. Sections 15.8 and 15.9 continued the consideration of multiserver queues. Section 15.8 was devoted to an M/G/s system, in which customers who arrive and find all the servers occupied do not wait, but simply leave. This model is particularly useful in the design of telephone systems. A specific example of this type was presented. Section 15.9 considered the repairperson model, an M/M/s system with a finite calling population. It also illustrates the use of spreadsheets in obtaining numerical results for a particular model.

27 C H A P T E R 1 5 Queuing CD15-27 Section showed how a spreadsheet simulation can be used to explore the transient rather than the steady-state behavior of a system. Section described the importance of the exponential distribution in the analytic analysis of queuing systems. It also presented two characteristics of the exponential distribution: the lack of memory property and the high probability of small values. Section briefly considered the topic of queue discipline. Key Terms Queuing Model. A model that involves waiting in line. Channel. A synonym for server in queuing jargon (e.g., a single-channel queue is a queue with a single server). Arrival Process. That part of a queuing model that determines the arrival pattern. Interarrival Time. The amount of time between two consecutive arrivals at a service facility. Typically a random quantity. Lack of Memory. A characteristic of the exponential distribution that makes it possible to derive analytic results for many queuing models. Service Process. That part of a queuing model that determines the service time for each item. Service Time. The amount of time that it takes an item to pass through the service facility. Typically a random quantity. Queue Size. The limit on the number of items that are permitted to wait in line for service. Queue Discipline. The rule used by the service facility to determine which items to serve. First-come, first-served is a typical example. Operating Characteristics. Quantities such as the expected number in queue that describe the operation of the queuing system. Steady State. A condition in which the probability of viewing a certain situation (e.g., an empty queue) does not depend on the time at which you look. Finite Queue. A queue with an upper limit on the number of items that can wait in the queue. Balk. A balk occurs when a customer arrives at a finite queue that is fully occupied. Reneging. Reneging occurs when a customer leaves a system without being served. Calling Population. The number of items that might call on the system for service: thus, a factor in determining the arrival process. Self-Review Exercises True-False 1. TF The number of people in the system means the number waiting in line. 2. TF The waiting time includes the service time. 3. TF The exponential distribution is a two-parameter distribution that is defined by a mean and a standard deviation. 4. TF The mean interarrival time is the reciprocal of the mean arrival rate, and the mean service time is the reciprocal of the mean service rate. 5. TF The basic model is M/M/1. 6. TF As the number of servers increases the cost of waiting generally increases. 7. TF The assumption that the mean service rate is less than the mean arrival rate is enough to eliminate the formation of infinitely long queues. 8. TF Little s flow equation states a directly proportional relationship between expected waiting time and expected number of people in the system. 9. TF The notation G/M/2 means the service distribution is general, the arrival distribution is exponential, and there are two parallel servers. Multiple Choice 10. Which of the following does not apply to the basic model? a. exponentially distributed arrivals b. exponentially distributed service times c. finite time horizon d. unlimited queue size e. the discipline is first-come, first-served 11. A major goal of queuing is to a. minimize the cost of providing service b. provide models that help the manager to trade off the cost of service c. maximize expected return d. optimize system characteristics

28 CD15-28 C D C H A P T E R S 12. Characteristics of queues such as expected number in the system a. are relevant after the queue has reached a steady state b. are probabilistic statements c. depend on the specific model d. all of the above 13. In Little s flow equation, which of the following is not true? a. is the constant of proportionality between expected number in the queue and expected time in the queue. b. is the constant of proportionality between expected number in the system and expected time in the system. c. is the arrival rate, including those arrivals who choose not to join the system. 14. The most difficult aspect of performing a formal economic analysis of queuing systems is a. estimating the service cost b. estimating the waiting cost c. estimating use 15. In a multiserver system with blocked customers cleared, which one of the following does not apply? a. When all servers are busy, new arrivals leave. b. Interesting characteristics are the probability that all servers are busy and the average number of busy servers. c. One would never do an expected-cost-per-hour analysis. 16. For the exponential distribution, which of the following is not a characteristic? a. lack of memory b. typically yields service times greater than the mean c. a single parameter Answers 1. F, 2. T, 3. F, 4. T, 5. T, 6. F, 7. F, 8. T, 9. F, 10. c, 11. b, 12. d, 13. c, 14. b, 15. c, 16. b Skill Problems Barges arrive at the La Crosse lock on the Mississippi River at an average rate of one every 2 hours. If the interarrival time has an exponential distribution, (a) What is the value of? (b) What is the mean interarrival time? (c) What is the mean arrival rate? Cars arrive at Joe s Service Station for an oil change every 15 minutes, and the interarrival time has an exponential distribution. The service station is capable of serving up to 48 cars during an 8-hour period with no idle time. Assume that the service time is also a random variable with an exponential distribution. Estimate: (a) The value of. (b) The mean arrival rate. (c) The value of. (d) The mean service time. (e) The mean service rate An immigration agent at Heathrow Airport in London could on the average process 120 entrants during her 8 hours on duty if she were busy all of the time. If the time to process each entrant is a random variable with an exponential distribution. (a) What is the value of? (b) What is the mean service time? (c) What is the mean service rate? For the data in Problem 15-2, determine: (a) The expected number of cars in the system. (b) The expected number of cars in the queue. (c) The expected waiting time. (d) The expected time in the queue. (e) The probability that the system is empty.

29 C H A P T E R 1 5 Queuing CD Consider the immigration officer mentioned in Problem Assume that the basic model is a reasonable approximation of her operation. Recall that if she were busy all the time she could process 120 entrants during her 8-hour shift. If on the average an entrant arrives at her station once every 6 minutes, find: (a) The expected number in the system. (b) The expected number in the queue. (c) The expected waiting time. (d) The expected time in the queue. (e) The probability that the system is empty Consider a single-channel queue. Assume that the basic model is a reasonable approximation of its operation. Comment on the following scheme to estimate : 1. Let N equal the number of arrivals between 8:00 A.M. and 4:00 P.M. 2. Set = 8/N Consider the basic model. Let = 5, and plot the expected number in the system for = 6,7,..., Consider the basic model. Let = 10, and plot the probability that the system is empty for = 0, 1,..., Use Little s flow equation and the fact that L = /( ) in the basic model to derive the expression for W Use Little s flow equation, the expression for the mean service time, and the fact that L = /( ) in the basic model to derive the expression for W q A doctor spends, on average, 20 minutes with her patients. If the expected waiting time is half an hour, what is the expected time in the queue? Assume that it is stated in Problem that patients arrive at the rate of seven per hour. Comment on this problem Describe a M /D/3 queuing system in words. Application Problems Consider the La Crosse lock mentioned in Problem Assume that the basic model is a reasonable approximation of its operation. The new estimate of the mean interarrival time for the coming season is 60 minutes for barges, and on the average it takes 30 minutes to move a barge through the lock. Find: (a) The expected number in the system. (b) The expected number in the queue. (c) The expected waiting time. (d) The expected time in the queue. (e) The probability that the system is empty. (f) The longest average service time for which the expected waiting time is less than 45 minutes Use the answers to Problem to show that Little s law holds Solve (a) through (e) of Problem using the generalized model for the case in which the variance of the service time distribution is equal to its mean At the Homeburg Savings and Loan, customers who wish to buy certificates of deposit form a single line and are served on a first-come, first-served basis by a specific bank officer. Service time is normally distributed with a mean of 5 minutes and a standard deviation of 1 minute. Customers arrive at the rate of one every 8 minutes. A time study shows that customers spend an average of minutes in the system (i.e., waiting and being served). What is the average number of people in the system? The Homeburg Saving and Loan uses three tellers on Saturdays. The interarrival time and the service time for customers each has an exponential distribution. Customers arrive at the rate of 20 per hour, and the mean service time is 6 minutes. Customers form a single queue and are served by the first available teller. Under steady-state conditions, find: (a) The probability that no customers are waiting or being served. (b) The expected number of people in the queue. (c) The expected waiting time in the queue. (d) The expected waiting time. (e) The expected number of people in the system.

30 CD15-30 C D C H A P T E R S The Darden Business School reserves ten high-speed modems for faculty to use to connect to the school s network from home. If a faculty member attempts to log on and all the modems are busy, she is told that all modems are busy and must try again at a later time. To estimate the system characteristics, the director of information technology wants to know the steady-state values of the characteristics assuming a finite calling population of 100 and an infinite queue. (This is an approximation because faculty who are turned away must try again.) Each faculty member wants to dial up from home once every 8 hours on average, and the interarrival time is exponentially distributed. Faculty spend an average of 30 minutes connected from home once they get through, exponentially distributed. Find: (a) The probability that all ports are open. (b) The expected number of people in the queue. (c) The expected waiting time in the queue. (d) The expected waiting time. (e) The expected number in the system For Problem 15-19, estimate the probability that all connections are busy using the spreadsheet template ( finiteq ) shown in the chapter. (Assume an M/G/s model with blocked customers cleared, an infinite calling population, and an arrival rate 100 times that for a single faculty member.) STECO has 100 sales representatives in the United States. They call orders into a central office where an office worker using the central inventory control system confirms product availability, price, and delivery date. The representative calls directly from the customer s office before signing a contract. Calls are held in a queue and served by the first available office worker on a first-come, first-served basis. Calls arrive at the rate of 40 per hour, and the mean service time is 6 minutes. Management estimates that it costs $20 per hour to have a sales representative call in and order, and $12 per hour to employ an office worker. Model this situation as an M/M/s queue with an infinite calling population, and calculate the expected total cost per hour if STECO hires five office workers Find the expected total cost for the system in Problem if STECO hires six office workers Use the solutions to Problems and to determine the value for the ratio C s /C w for which STECO is indifferent between having five or six office workers In a particular manufacturing cell, one repairman has to maintain four machines. For the machines, the time between breakdowns is exponentially distributed with an average of four hours. On the average, it takes half an hour to fix a machine. (a) Find the probabilities that there are 0, 1, 2, 3, or 4 machines under repair. (b) Find the average number of machines under repair A telephone exchange has seven lines. Calls arrive at the rate of two per minute, and the interarrival time has an exponential distribution. Conversations have a normal distribution with a mean of 5 minutes and a standard deviation of 1 minute. When all seven lines are occupied, the caller simply receives a busy signal. (a) What is the probability that exactly three lines are busy? (b) What is the probability that the system is totally occupied? (c) What is the average number of busy servers? A market research group has three interviewers located in adjacent booths in a suburban shopping mall. A contact person meets people walking in the mall and asks them if they are willing to be interviewed. They estimate that customers willing to agree to the interview arrive at the rate of 15 per hour, and the interarrival time has an exponential distribution. On the average the interview takes 15 minutes. If all booths are occupied, a person who has agreed to be interviewed will not wait and simply goes about his or her business. (a) Comment on the following statement: Since > s, this system will grow without bound. (b) Calculate the probability that exactly one interviewer is occupied. (c) Find the probability that all three interviewers are occupied. (d) Find the average number of busy interviewers Consider again the La Crosse lock mentioned in Problems 15-1 and Suppose that the mean interarrival time is 60 minutes and that on the average it takes 30 minutes to move a barge through the lock, but that the standard deviation of this service time is 3 minutes. Re-answer (a) through (f) of Problem How did your answers change and why? Repair requests are handled by a handyman at an apartment complex on a first-come, first-served basis. Requests arrive at the rate of 1 per hour on average. The time it takes the handyman to make a

31 repair is normally distributed with a mean of 30 minutes and a standard deviation of 15 minutes. How long on the average is the time between when a repair request is made and the repair is completed? Larry Lujack is unhappy with the long lead times he is having to quote to customers (see Section 15.10). He feels that SONOROLA is going to start losing business to competitors who can quote shorter lead times. Larry initially assumed that processing times were exponentially distributed. After taking a closer look at the data, he discovers that 90% of the time it takes 3 hours to process a unit on either Work Station (WS) 1 or 2 and 10% of the time it takes 13 hours on either WS1 or WS2. Thus the average time is.9 * * 13 = 4.0 hours. After talking with the production supervisor of these stations he learns that the high time is due to equipment failure when processing a unit. Invariably, it takes 10 hours to repair the equipment once a failure occurs. Larry has heard that preventive maintenance can reduce the chance of equipment failure and wonders what the value would be of decreasing the chance of failure from 10% to 1%. (a) Use Crystal Ball to find the average time to complete 20 units and the 99th percentile of the time if at each work station there is a 10% chance of equipment failure while processing a unit. How do your answers compare to those of Section 15.10, where the processing times were (b) exponentially distributed? Now assume that at each work station there is a 1% chance of equipment failure while processing a unit. What is the value of preventive maintenance if it reduces equipment failures to this level? Suppose Larry Lujack has another order to promise (see Section 15.10). This order is also for 20 units, but the average processing time is 6 hours per unit at Work Station 1 and 4 hours per unit at Work Station 2. (a) (b) (c) (d) (e) C H A P T E R 1 5 Queuing CD15-31 Using the basic model, estimate the average time needed to complete the order. Assuming that processing times are exponentially distributed, use a spreadsheet simulation to compute the average time. How does it compare with your answer to part (a)? When both arrival and service times are exponentially distributed, then in the basic model waiting plus processing time is also exponentially distributed. Use this fact to estimate without simulation the 99th percentile of the time to complete all 20 units. Use Crystal Ball to find the 99th percentile of the time it takes to complete all 20 units. Compare with your answer to part (c). Suppose that the times were reversed; that is, that processing took an average of 4 hours per unit at Work Station 1 and 6 hours per unit at Work Station 2. Can you use the basic model to estimate the mean and 99% fractile? Why or why not? Use simulation and compare with your answers to parts (b) and (d). Case Study How Many Operators? L. L. Bean realizes the importance of people to its success. The emphasis on training and employee relations tells us that L. L. Bean understands a basic fact of management: People play the major role in determining how well most systems operate. It is obvious that a concern for people in an organization is reflected in the way individuals are treated on a day-to-day basis. It may not be quite as obvious that this same concern plays a fundamental role in designing business systems. Mail order is the heart of L. L. Bean s business. This business is based on orders received over the phone. Bean s lines are open 24 hours a day, 365 days a year. They have an average rate of 78,000 calls per day. A moment s reflection suggests that these calls do not arrive at a uniform rate. There clearly are seasonal effects, as well as variability during each day. To meet its need for phone operators, Bean offers three types of work arrangements: full time, permanent part time, and temporary. This strategy allows for great flexibility in adjusting the number of operators on duty at any moment. It also provides flexibility for employees, who can structure an arrangement that fits with other demands on their time. However, the nagging question remains: How many operators does Bean need, and when? it seems clear that the company wants to balance customer service against staffing expense. Its approach is to consider each of the 168 hours in a week as a period to be staffed. For each hour, the system is modeled as an M/M/s queue that is, a multiserver queue with exponential arrival and service times and s servers (operators). The arrival rate and service rate are estimated from historical data. Cost balancing is done in an intuitive manner: A service standard that management believes is appropriate is the basis of the design. In particular, the Bean system is designed so that no more than 15% of calls wait more than twenty seconds before reaching an operator.

32 CD15-32 C D C H A P T E R S Let s look at a simplified version of this model. Suppose there is a single phone line that answers those who call on the phone number in the period of 1 A.M. 5 A.M. (clearly a dramatically slow time). The company has gathered some data on this line (see Exhibit 1). We see that the first call arrives at 0.45 minutes and it takes 3.66 minutes to process the order for a total amount of $ When the next caller phones at minutes, the operator is still servicing the first caller, and so he or she gets a busy signal. The spreadsheet (BEAN.XLS) has data for the first 100 phone calls. EXHIBIT 1 Phone Order Data Questions 1. Analyze the data to determine what the average length of a call is. What is the average interarrival time? What is the average order amount? 2. Plot a histogram of this data and decide whether you think an exponential distribution is the right one to use for the arrival rate? For the service rate? (Hint: Crystal Ball has a built-in feature to fit data like this to the best distribution.) 3. What percentage of the time is the operator busy? 4. What percentage of the calls are lost due to the busy signal? The current model gives a busy signal if someone else is being helped, but L. L. Bean could certainly add more lines and place the customers on hold until its only operator were able to get to them. After these extra lines were full, additional callers would then get the busy signal. Assume the fixed cost of each extra line is $35 per month and that the average completed call on the number costs the company an additional $2 on average per call. L. L. Bean wonders how to analyze the trade-off if it added the ability for callers to be placed on hold. Assume the profit margin is 45% on all orders placed. More Questions 5. Use the finiteq template to analyze the increased number of calls that could be handled if L. L. Bean added 1, 2, 3, 4, or 5 extra lines. What extra revenue and profit would each of these scenarios generate? 6. Is the extra profit worth the increased cost? What is the optimal number of lines to add? References Edwin Landauer and Linda Becker, Reducing Waiting Time at Security Checkpoints, Interfaces, 19, no. 5 (1989): Richard Larson, Michael Cahn, and Martin Shell, Improving the New York City Arrest-to- Arraignment System, Interfaces, 23, no. 1 (1993):