Report

C Waiting Lines For Operations Management, 9e by Krajewski/Ritzman/Malhotra © 2010 Pearson Education Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C–1 Two ways to address waiting lines Queuing theory Certain types of lines can be described mathematically Requires that assumptions are valid for your situation Systems with multiple lines that feed each other are too complex for queuing theory Simulation Building mathematical models that attempt to act like real operating systems Real-world situations can be studied without imposing on the actual system C-2 C–2 Why do we have to wait? Why do services (and many manufacturing situations) have queues? Arrival time variance Processing time variance of capacity – can we afford to always have more people/servers than customers? Costs C-3 C–3 Waiting Costs and Service Costs Cost Total expected cost Cost of providing service Minimum total cost Cost of waiting time Low level of service C-4 Optimal service level High level of service C–4 Costs of Queues Cost of capacity Annoyed customers Wasted capacity Lost customers Space Too Little Queue Too Much Queue Possible opportunity: e.g. wait in the bar for a restaurant table C-5 C–5 Waiting Line Examples Situation Arrivals Servers Service Process Bank Customers Teller Deposit, etc. Doctor’s office Patient Doctor Treatment Traffic intersection Cars Light Controlled passage Assembly line Parts Workers Assembly 1–800 software support User call-ins Tech support personnel Technical support C-6 C–6 Waiting Line Terminology Queue: Waiting line Arrival: 1 person, machine, part, etc. that arrives and demands service Queue discipline: Priority rules for determining the order that arrivals receive service Channel: Number of waiting lines Phase: Number of steps in service C-7 C–7 Assumptions Arrivals At random (Poisson, exponential distributions) Fixed (appointments, service intervals) Service times Variable Fixed (exponential, normal distributions) (constant service time) Other Size of arrival population, priority rules, balking, reneging C-8 C–8 Structure of Waiting-Line Problems Customer population Service system Waiting line Priority rule Service facilities Served customers Figure C.1 – Basic Elements of Waiting-Line Models Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C–9 Structure of Waiting-Line Problems 1. An input, or customer population, that generates potential customers 2. A waiting line of customers 3. The service facility, consisting of a person (or crew), a machine (or group of machines), or both necessary to perform the service for the customer 4. A priority rule, which selects the next customer to be served by the service facility Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 10 Customer Population The source of input Finite or infinite source Customers from a finite source reduce the chance of new arrivals Customers from an infinite source do not affect the probability of another arrival Customers are patient or impatient Patient customers wait until served Impatient customer can either balk (not even enter a long line) or join the line and later renege (leave before being served) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 11 The Service System Number of lines A single-line keeps servers uniformly busy and levels waiting times among customers A multiple-line arrangement is favored when servers provide a limited set of services Arrangement of service facilities Single-channel, single-phase Single-channel, multiple-phase Multiple-channel, single-phase Multiple-channel, multiple-phase Mixed arrangement Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 12 The Service System Service facilities Service facilities (a) Single line Figure C.2 – Waiting-Line Arrangements (b) Multiple lines Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 13 The Service System Service facility (a) Single channel, single phase Service facility 1 Service facility 2 (b) Single channel, multiple phase Figure C.3 – Examples of Service Facility Arrangements Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 14 The Service System Service facility 1 Service facility 2 (d) Multiple channel, multiple phase (e.g., Laundromat) (c) Multiple channel, single phase Figure C.3 – Examples of Service Facility Arrangements Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Service facility 1 Service facility 3 Service facility 2 Service facility 4 C – 15 The Service System Routing for Routing for Routing for (e) Mixed arrangement Service facility 1 : 1–2–4 : 2–4–3 : 3–2–1–4 Service facility 2 Service facility 3 Service facility 4 Figure C.3 – Examples of Service Facility Arrangements Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 16 Priority Rule First-come, first-served (FCFS)—used by most service systems Other rules Earliest due date (EDD) = urgent service Shortest processing time (SPT) = express line Preemptive discipline—allows a higher priority customer to interrupt the service of another customer or be served ahead of another who would have been served first Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 17 Probability CUSTOMERS IN THE SYSTEM ARRIVALS x (Poisson) n Probability that Service Time t Probability SERVICE (Exponential) Time t C – 18 Probability Distributions The sources of variation in waiting-line problems come from the random arrivals of customers and the variation of service times Arrival distribution Customer arrivals can often be described by the Poisson distribution with mean = T and variance also = T Arrival distribution is the probability of n arrivals in T time periods Interarrival times are the times between successive arrivals Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 19 Poisson Distributions for Arrival Rates 0.30 0.30 0.25 0.20 0.20 Probabilit y Probabilit Probabilit yy 0.25 0.15 0.15 0.10 0.10 0.05 0.00 0.05 0 1 2 3 4 5 6 7 8 9 10 11 12 x =2 0.00 0 1 2 3 4 5 6 7 x 8 9 10 11 12 =4 = average arrival rate = number/period C-20 C – 20 Poisson Distribution (T)n -T Pn = e for n = 0, 1, 2,… n! where Pn = Probability of n arrivals in T time periods = Average numbers of customer arrivals per period e = 2.7183 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 21 Probability of Customer Arrivals EXAMPLE C.1 Management is redesigning the customer service process in a large department store. Accommodating four customers is important. Customers arrive at the desk at the rate of two customers per hour. What is the probability that four customers will arrive during any hour? SOLUTION In this case using customers per hour, T = 1 hour, and n = 4 customers. The probability that four customers will arrive in any hour is [2(1)]4 16 –2 –2(1) P4 = = e = 0.090 e 4! 24 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 22 Negative Exponential Distribution Probability that Service Time t 1.0 –-t t , for t > 0 Probability Service is greater Probability that that Service TimeTime is greater thanthan t = et=e for t > 0 0.9 0.8 Average Service Rate () = 3 customers per hour Average Service Rate () = 3 customers per hour Average per Average ServiceTime Time==20 20minutes minutes per customer customerService 0.7 0.6 0.5 Average Service Rate Service () =Average 1 customer per Rate () = hour1 customer per hour 0.4 0.3 0.2 0.1 0.0 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 Time t in Hours C-23 C – 23 Service Time Service time distribution can be described by an exponential distribution with mean = 1/ and variance = (1/ )2 Service time distribution: The probability that the service time will be no more than T time periods can be described by the exponential distribution P(t ≤ T) = 1 – e–T where μ = average number of customers completing service per period t = service time of the customer T = target service time Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 24 Service Time Probability EXAMPLE C.2 The management of the large department store in Example C.1 must determine whether more training is needed for the customer service clerk. The clerk at the customer service desk can serve an average of three customers per hour. What is the probability that a customer will require less than 10 minutes of service? SOLUTION We must have all the data in the same time units. Because = 3 customers per hour, we convert minutes of time to hours, or T = 10 minutes = 10/60 hour = 0.167 hour. Then P(t ≤ T) = 1 – e–T P(t ≤ 0.167 hour) = 1 – e–3(0.167) = 1 – 0.61 = 0.39 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 25 Using Waiting-Line Models Balance costs against benefits Operating characteristics Line length Number of customers in system Waiting time in line Total time in system Service facility utilization Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 26 Single-Server Model Single-server, single line of customers, and only one phase Assumptions are 1. Customer population is infinite and patient 2. Customers arrive according to a Poisson distribution, with a mean arrival rate of 3. Service distribution is exponential with a mean service rate of 4. Mean service rate exceeds mean arrival rate 5. Customers are served FCFS 6. The length of the waiting line is unlimited Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 27 Single-Server Model = Average utilization of the system = Rn = Probability that n customers are in the system = (1 – ) n L = Average number of customers in the service system = – Lq = Average number of customers in the waiting line = L W = Average time spent in the system, including service 1 = – Wq = Average waiting time in line = W Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 28 Calculating the Operating Characteristics EXAMPLE C.3 The manager of a grocery store in the retirement community of Sunnyville is interested in providing good service to the senior citizens who shop in her store. Currently, the store has a separate checkout counter for senior citizens. On average, 30 senior citizens per hour arrive at the counter, according to a Poisson distribution, and are served at an average rate of 35 customers per hour, with exponential service times. Find the following operating characteristics: a. Probability of zero customers in the system b. Average utilization of the checkout clerk c. Average number of customers in the system d. Average number of customers in line e. Average time spent in the system f. Average waiting time in line Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 29 Calculating the Operating Characteristics SOLUTION The checkout counter can be modeled as a single-channel, single-phase system. Figure C.4 shows the results from the Waiting-Lines Solver from OM Explorer. Figure C.4 – Waiting-Lines Solver for Single-Channel, Single-Phase System Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 30 Calculating the Operating Characteristics Both the average waiting time in the system (W) and the average time spent waiting in line (Wq) are expressed in hours. To convert the results to minutes, simply multiply by 60 minutes/ hour. For example, W = 0.20(60) minutes, and Wq = 0.1714(60) = 10.28 minutes. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 31 Analyzing Service Rates EXAMPLE C.4 The manager of the Sunnyville grocery in Example C.3 wants answers to the following questions: a. What service rate would be required so that customers average only 8 minutes in the system? b. For that service rate, what is the probability of having more than four customers in the system? c. What service rate would be required to have only a 10 percent chance of exceeding four customers in the system? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 32 Analyzing Service Rates SOLUTION The Waiting-Lines Solver from OM Explorer could be used iteratively to answer the questions. Here we show how to solve the problem manually. a. We use the equation for the average time in the system and solve for W= 1 – 8 minutes = 0.133 hour = 1 – 30 0.133 – 0.133(30) = 1 = 37.52 customers/hour Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 33 Analyzing Service Rates b. The probability of more than four customers in the system equals 1 minus the probability of four or fewer customers in the system. 4 P = 1 – n P =0 n 4 and n = 1 – n (1 – ) =0 = 30 = 0.80 37.52 Then, P = 1 – 0.21(1 + 0.8 + 0.82 + 0.83 + 0.84) = 1 – 0.672 = 0.328 Therefore, there is a nearly 33 percent chance that more than four customers will be in the system. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 34 Analyzing Service Rates c. We use the same logic as in part (b), except that is now a decision variable. The easiest way to proceed is to find the correct average utilization first, and then solve for the service rate. P (n>k) = 1 – (1 – )(1 + + 2 + 3 + 4) = 1 – (1 – )(1 + + 2 + 3 + 4) + (1 + + 2 + 3 + 4) = 1 – 1 – – 2 – 3 – 4 + + 2 + 3 + 4 + 5 = 5 = k+1 or = P1/5 If P = 0.10 = (0.10)1/5 = 0.63 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 35 Analyzing Service Rates Therefore, for a utilization rate of 63 percent, the probability of more than four customers in the system is 10 percent. For = 30, the mean service rate must be 30 = 0.63 = 47.62 customers/hour Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 36 Multiple-Server Model Service system has only one phase, multiple-channels Assumptions (in addition to single-server model) There are s identical servers The service distribution for each server is exponential The mean service time is 1/ s should always exceed Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 37 Multiple-Server Model = Average utilization of the system = s P0 = Probability that zero customers are in the system = s1 n0 / n / 1 s n! s! 1 1 Pn = Probability that n customers are in the system / P0 0 n s = n! n / P n s 0 s! s n s n Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 38 Multiple-Server Model Lq = Average number of customers in the waiting line P0 / s = s! 1 2 Wq = Average waiting time of customers in line = Lq W = Average time spent in the system, including service = Wq 1 L = Average number of customers in the service system = W Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 39 Estimating Idle Time and Costs EXAMPLE C.5 The management of the American Parcel Service terminal in Verona, Wisconsin, is concerned about the amount of time the company’s trucks are idle (not delivering on the road), which the company defines as waiting to be unloaded and being unloaded at the terminal. The terminal operates with four unloading bays. Each bay requires a crew of two employees, and each crew costs $30 per hour. The estimated cost of an idle truck is $50 per hour. Trucks arrive at an average rate of three per hour, according to a Poisson distribution. On average, a crew can unload a semitrailer rig in one hour, with exponential service times. What is the total hourly cost of operating the system? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 40 Estimating Idle Time and Costs SOLUTION The multiple-server model is appropriate. To find the total cost of labor and idle trucks, we must calculate the average number of trucks in the system. Figure C.5 shows the results for the American Parcel Service problem using the Waiting-Lines Solver from OM Explorer. Manual calculations using the equations for the multiple-server model are demonstrated in Solved Problem 2 at the end of this supplement. The results show that the four-bay design will be utilized 75 percent of the time and that the average number of trucks either being serviced or waiting in line is 4.53 trucks. That is, on average at any point in time, we have 4.53 idle trucks. We can now calculate the hourly costs of labor and idle trucks: Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 41 Estimating Idle Time and Costs Labor cost: $30(s) = $30(4) = $120.00 Idle truck cost: $50(L) = $50(4.53) = 226.50 Total hourly cost = $346.50 Figure C.5 – Waiting-Lines Solver for Multiple-Server Model Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 42 Little’s Law Relates the number of customers in a waiting line system to the waiting time of customers Using the notation from the single-server and multiple-server models it is expressed as L = W or Lq = Wq Holds for a wide variety of arrival processes, service time distributions, and numbers of servers Only need to know two of the parameters Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 43 Little’s Law Ls = Ws DMV Count arrivals () Periodically count # customers in line (Lq) Lq = Wq Estimate average time in line (Wq) Widget manufacturing Determine average process time (Ws) Determine average arrival rate () C-44 Estimate average work-in-process (Ls) C – 44 Little’s Law Service Estimate W L customers Average time = W = in the facility customer/hour 30 = = 0.75 hours or 45 minutes 40 Manufacturing Estimate the average work-in-process L Work-in-process = L = W = 5 gear cases/hour (3 hours) = 15 gear cases Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 45 Finite-Source Model Assumptions Follows the assumption of the single-server, except that the customer population is finite Having only N potential customers If N > 30, then the single-server model with the assumption of infinite customer population can be used for a rough approximation Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 46 Finite-Source Model P0 = Probability that zero customers are in the system = N n 0 N n ! N! n 1 = Average utilization of the server = 1 – P0 Lq = Average number of customers in the waiting line = N 1 P 0 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 47 Finite-Source Model L = Average number of customers in the service system = N 1 P0 Wq = Average waiting time in line = L q N L 1 Ws = Average time spent in the system, including service = L N L Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 1 C – 48 Analyzing Maintenance Costs EXAMPLE C.6 The Worthington Gear Company installed a bank of 10 robots about 3 years ago. The robots greatly increased the firm’s labor productivity, but recently attention has focused on maintenance. The firm does no preventive maintenance on the robots because of the variability in the breakdown distribution. Each machine has an exponential breakdown (or interarrival) distribution with an average time between failures of 200 hours. Each machine hour lost to downtime costs $30, which means that the firm has to react quickly to machine failure. The firm employs one maintenance person, who needs 10 hours on average to fix a robot. Actual maintenance times are exponentially distributed. The wage rate is $10 per hour for the maintenance person, who can be put to work productively elsewhere when not fixing robots. Determine the daily cost of labor and robot downtime. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 49 Analyzing Maintenance Costs SOLUTION The finite-source model is appropriate for this analysis because the customer population consists of only 10 machines and the other assumptions are satisfied. Here, = 1/200, or 0.005 breakdown per hour, and = 1/10 = 0.10 robot per hour. To calculate the cost of labor and robot downtime, we need to estimate the average utilization of the maintenance person and L, the average number of robots in the maintenance system at any time. Figure C.6 shows the results for the Worthington Gear Problem using the Waiting-Lines Solver from OM Explorer. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 50 Analyzing Maintenance Costs Manual computations using the equations for the finite-source model are demonstrated in Solved Problem 3 at the end of this supplement. The results show that the maintenance person is utilized only 46.2 percent of the time, and the average number of robots waiting in line or being repaired is 0.76 robot. However, a failed robot will spend an average of 16.43 hours in the repair system, of which 6.43 hours of that time is spent waiting for service. While an individual robot may spend more than two days with the maintenance person, the maintenance person has a lot of idle time with a utilization rate of only 42.6 percent. That is why there is only an average of 0.76 robot being maintained at any point of time. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 51 Analyzing Maintenance Costs Figure C.5 – Waiting-Lines Solver for Finite-Source Model The daily cost of labor and robot downtime is Labor cost: ($19/hour)(8 hours/day)(0.462 utilization) = $36.96 Idle robot cost: (0.76 robot)($30/robot hour)(8 hours/day) = 182.40 Total daily cost = $219.36 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 52 Decision Areas for Management 1. Arrival rates 2. Number of service facilities 3. Number of phases 4. Number of servers per facility 5. Server efficiency 6. Priority rule 7. Line arrangement Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 53 Using Waiting Line Methods to Analyze Operations Line Length Number of Customers in System – customers in line and being served Waiting Time in Line Total Time in System – time in line and being served Service Facility Utilization – percentage of time servers are busy or idle C-54 C – 54 Remember: and are Rates = Mean number of arrivals per time period If average service time is 15 minutes, then μ is 4 customers/hour e.g., 3 units/hour = Mean number of people or items served per time period e.g., 4 units/hour 1/ = 15 minutes/unit © 1984-1994 T/Maker Co. C – 55 Multi-Channel, Single Phase System (Classic Setup) Service system Queue Arrivals Service facility Served units Service facility Example: Bank customers pick shortest line for one of several tellers. C-56 C – 56 Multi-Channel, Single Phase System (Modern Approach) Service system Queue Arrivals Service facility Served units Service facility Example: Bank customers wait in single line for one of several tellers. C-57 C – 57 Multi-Channel, Single Phase System (Classic versus Modern Approach) Performance measures are the same if customers in Classic approach are allowed to switch lines if their line is moving slower. So why the modern approach? More uniform customer experience (variance around average is smaller) Customers are kept moving more often (psychological benefit) More privacy at service window Easier to manage lines C-58 C – 58 Solved Problem 1 A photographer takes passport pictures at an average rate of 20 pictures per hour. The photographer must wait until the customer smiles, so the time to take a picture is exponentially distributed. Customers arrive at a Poisson-distributed average rate of 19 customers per hour. a. What is the utilization of the photographer? b. How much time will the average customer spend with the photographer? SOLUTION a. The assumptions in the problem statement are consistent with a single-server model. Utilization is = = 19 = 0.95 20 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 59 Solved Problem 1 b. The average customer time spent with the photographer is W = 1 1 = = 1 hour – 20 – 19 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 60 Solved Problem 2 The Mega Multiplex Movie Theater has three concession clerks serving customers on a first come, first-served basis. The service time per customer is exponentially distributed with an average of 2 minutes per customer. Concession customers wait in a single line in a large lobby, and arrivals are Poisson distributed with an average of 81 customers per hour. Previews run for 10 minutes before the start of each show. If the average time in the concession area exceeds 10 minutes, customers become dissatisfied. a. What is the average utilization of the concession clerks? b. What is the average time spent in the concession area? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 61 Solved Problem 2 SOLUTION a. The problem statement is consistent with the multiple-server model, and the average utilization rate is = s = 81 customers/ hour 60 minutes/se rver hour 3 servers 2 minutes/cu stomer = 0.90 The concession clerks are busy 90 percent of the time. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 62 Solved Problem 2 1 b. The average time spent in the system, W, is W = W q Here, Wq = Lq P0 / s Lq = s! 1 2 s1 / / 1 n! s! 1 n0 n and P0 = s 1 We must solve for P0, Lq, and Wq, in that order, before we can solve for W: s1 / / 1 P0 = n! s! 1 n0 n 1 1 = 1 = s 81 / 30 2 . 7 2 1 2 2 .7 1 6 1 0 . 9 1 1 2 . 7 3 . 645 32 . 805 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 3 = 1 40 . 15 = 0.0249 C – 63 Solved Problem 2 P0 / Lq = Wq = s! 1 Lq = 3! 1 0 . 9 7.352 customers = W = Wq 2 0 . 0249 81 / 30 0 . 9 3 s 81 customers/ 1 hour = 0.0908 hours + 2 = 0 . 4411 6 0 . 01 = 7.352 customers = 0.0908 hour 1 30 hour = 0.1241 hour 60 minutes hour = 7.45 minutes With three concession clerks, customers will spend an average of 7.45 minutes in the concession area. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 64 Solved Problem 3 The Severance Coal Mine serves six trains having exponentially distributed interarrival times averaging 30 hours. The time required to fill a train with coal varies with the number of cars, weather-related delays, and equipment breakdowns. The time to fill a train can be approximated by an exponential distribution with a mean of 6 hours 40 minutes. The railroad requires the coal mine to pay large demurrage charges in the event that a train spends more than 24 hours at the mine. What is the average time a train will spend at the mine? SOLUTION The problem statement describes a finite-source model, with N = 6. The average time spent at the mine is W = L[(N – L)]–1, with 1/ = 30 hours/train, = 0.8 train/day, and = 3.6 trains/day. In this case, Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 65 Solved Problem 3 N N! N n ! n 0 n P0 = = = 1 = 1 6 0 .8 6 n ! 3 . 6 n0 6! n 1 6! 0 . 8 6! 0 . 8 6! 0 . 8 6! 0 . 8 6! 0 . 8 6! 0 . 8 6! 0 . 8 6 ! 3 . 6 5 ! 3 . 6 4 ! 3 . 6 3 ! 3 . 6 2 ! 3 . 6 1 ! 3 . 6 0 ! 3 . 6 0 1 2 3 1 = 1 1 . 33 1 . 48 1 . 32 0 . 88 0 . 39 0 . 09 L= N 1 P0 W = L N L 1 3 .6 = 6 0 .8 = 4 1 6 . 49 5 6 = 0.1541 1 0 . 1541 = 2.193 trains 2 . 193 3 . 807 0 . 8 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. = 0.72 day Arriving trains will spend an average of 0.72 day at the coal mine. C – 66 Izzy’s Ice Cream Stand You are opening an ice cream stand that has a single employee (you). You expect to see about 25 customers an hour. It takes you an average of 2 minutes to serve each customer. Customers are served in a FCFS manner. Your research suggests that if there is a line of more than 4 people that some customers will leave without buying anything. In addition, if customers have to wait more than 6 minutes to get their order filled they are not likely to come back. How well will this system do at satisfying customers? C-67 C – 67 Izzy’s Ice Cream Stand (2) = 25 customers/hr ρ = / = .833 = 30 customers/hr L = / ( - ) = 5 customers Pn = probability that n customers are in the system = (1- ρ) ρn P0 P1 P2 P3 P4 = = = = = (1 – .833) x .833 (1 – .833) x .833 (1 – .833) x .833 (1 – .833) x .833 (1 – .833) x .833 = 0 = 1 = = 2 = 3 .167 .139 .116 .097 .080 .167 cumulative .306 .422 .519 .599 Pmore than 4 = 1 – .599 4= .401 Customers in the system C-68 C – 68 Izzy’s Ice Cream Stand (3) = 25 customers/hr = 30 customers/hr ρ = / = .833 L = / ( - ) Lq = ρL = 25 / (30 – 25) = 5 customers = .833 x 5 = 4.17 customers W = 1 / ( - ) Wq = ρW = 1 / (30 – 25) = .2 hr = .833 x 12 min C-69 = 12 min = 10 min Why not L–1? C – 69 Application C.1 Customers arrive at a checkout counter at an average 20 per hour, according to a Poisson distribution. They are served at an average rate of 25 per hour, with exponential service times. Use the single-server model to estimate the operating characteristics of this system. = 20 customer arrival rate per hour = 25 customer service rate per hour SOLUTION 1. Average utilization of system Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. = = 20 = 0.8 25 C – 70 Application C.1 20 =4 25 – 20 2. Average number of customers in the service system L= 3. Average number of customers in the waiting line Lq = L = 0.8(4) = 3.2 4. Average time spent in the system, including service W= 5. Average waiting time in line Wq = W = 0.8(0.2) = 0.16 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. – = 1 1 = = 0.2 – 25 – 20 C – 71 Application C.2 In the checkout counter example, what service rate is required to have customers average only 10 minutes in the system? SOLUTION 1 W= = 0.17 hr (or 10 minutes) – 0.17( – ) = 1, where = 20 customers arrival rate per hour 1 + 0.17(20) = = 25.88 or about 26 customers per hour 0.17 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 72 Application C.3 Suppose the manager of the checkout system decides to add another counter. The arrival rate is still 20 customers per hour, but now each checkout counter will be designed to service customers at the rate of 12.5 per hour. What is the waiting time in line of the new system? s = 2, = 12.5 customers per hour, = 20 customers per hour SOLUTION 1. Average utilization of the system Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. = s = 20 2 12 . 5 = 0.8 C – 73 Application C.3 2. Probability that zero customers are in the system P0 = = Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 1 s 1 1 s! 1 20 1 12 . 5 1 2 20 1 12 . 5 2! 1 0 . 8 = 0.11 C – 74 Application C.3 3. Average number of customers in the waiting line P0 / s Lq = s! 1 2 2 = 4. Average waiting time of customers in line Wq = Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 20 0 . 11 0 .8 12 . 5 2! 1 0 . 8 Lq = 1 . 408 20 2 = 1.408 = 0.0704 hrs (or 4.224 minutes C – 75 Application C.4 DBT Bank has 8 copy machines located in various offices throughout the building. Each machine is used continuously and has an average time between failures of 50 hours. Once failed, it takes 4 hours for the service company to send a repair person to have it fixed. What is the average number of copy machines in repair or waiting to be repaired? = 1/50 = 0.02 copiers per hour = 1/4 = 0.25 copiers per hour Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 76 Application C.4 SOLUTION N N! n 0 N n ! n 1. Probability that zero customers are in the system = P0 = 1 1 8! 8! 8! 0 1 8 0 . 08 0 . 08 0 . 08 8! 7! 0! = 0.44 2. Average utilization of the server = 1 – P0 = 1 – 0.44 = 0.56 3. Average number of customers in the service system L= N Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. = 8 1 0 . 25 0 . 02 P0 1 0 . 44 = 1 C – 77 Application C.5 The Hilltop Produce store is staffed by one checkout clerk. The average checkout time is exponentially distributed around an average of two minutes per customer. An average of 20 customers arrive per hour. What is the average utilization rate? SOLUTION = = 20 = 0.667 30 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 78 Application C.5 What is the probability that three or more customers will be in the checkout area? First calculate 0, 1, and 2 customers will be in the checkout area: Rn = (1 – ) 0 = (0.333)(0.667)0 = 0.333 Rn = (1 – ) 1 = (0.333)(0.667)1 = 0.222 Rn = (1 – ) 2 = (0.333)(0.667)2 = 0.111 Then calculate 3 or more customers will be in the checkout area: 1 – P0 – P1 – P2 = 0.333 – 0.222 – 0.111 = 0.334 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 79 Application C.5 What is the average number of customers in the waiting line? Lq = L = – 20 = 0 . 667 = 1.333 30 20 What is the average time customers spend in the store? W= 1 1 = = 0.1 hr 60 min/hr = 6 minutes – 30 20 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 80 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. C – 81