OPM 3000: Fall 2023 Managing Service Systems Professor Yuan-Mao Kao Chapter 16 of the Textbook Managing Service Systems 1 The Framework of the Course Chapters Forecasting Demand: Customers Lean Operations Accuracy measures Fluctuation Trend Seasonality Waste reduction, TPS Management Process Flow Metrics Flow rate, flow time, and inventory Capacity Analysis & Improvement Find the bottleneck resources to improve the process capacity Non-repetitive Variability: Demand/Supply Products Services Project Management Inventory & Supply Chain Service System Management Project duration Critical activities EOQ Newsvendor Reorder point Order-up-to level Supply chain strategies Waiting time Waiting customers Managing systems CPM & PERT Managing Service Systems Quality Management Statistical Process Control Process capability Defective probability Control outputs Quality Improvement 2 What Is a Waiting Line? Managing Service Systems 3 The Impact of Waiting • Which component of the customer utility would be impacted if customers need to wait for a long while? Performance Consumption Utility Fit Utility Price Location Inconvenience Timing • Why are customers still willing to purchase if they can anticipate a long waiting during the holiday sales activity (e.g., the Black Friday)? Managing Service Systems 4 Waiting Line Management • Why do waiting lines (queues) form? • What factors cause long wait for customers? • How can we estimate wait? • How can we reduce wait? Managing Service Systems 5 Case: Pizzas-4-U • The business plan for Pizzas-4-U is to accept pizza orders for pickup over the phone and to compete with superior, fresh, made-to-order deep-dish pizza while providing excellent service. • The advertising campaign will feature a promise: “If your pizza is not ready in 20 minutes, that pizza plus your next order is free!” • An extensive production analysis has revealed that each pizza will require an average of 15 minutes oven time and 2 minutes preparation time. • An extensive market analysis has revealed that demand will average 20 pizzas per hour, and P-4-U has ordered 4 ovens that can each hold one pizza. • P-4-U is looking for an investor – do you want to invest? Managing Service Systems 6 Recap: Process Analysis • We can draw a process flow diagram for Pizzas-4-U: Demand rate: 20/hour Preparing Baking Worker Oven Resource Processing Time (min) Capacity (pizzas/ hour) Number of Resource Worker 2 1 Oven 15 4 Capacity of the Resource Pool (pizzas/ hour) • Process capacity: • Flow rate: • Based on the analysis, do you think Pizzas-4-U can serve their customers well? Managing Service Systems 7 Causes of Waiting: Demand vs. Capacity • We know the demand rate is 20 pizzas/hour, and the capacity is as below: Resource Processing Time (min) Capacity (pizzas/hour) Number of Resource Capacity of the Resource Pool (pizzas/hour) Worker 2 30/hour 1 30/hour Oven 15 4/hour 4 16/hour • Let’s consider a simple scenario: • The order arrives exactly in each 3 minutes (and there are 20 orders in an hour) • Each order has just one pizza requested • The processing times of Pizzas-4-U are constant Order 1 2 3 4 5 6 Managing Service Systems Arrive At 3 6 9 12 15 18 Finish Preparing At 5 8 11 14 17 20 Finish Baking At 20 (1st Oven) 23 (2nd Oven) 26 (3rd Oven) 29 (4th Oven) 35 (1st Oven) 38 (2nd Oven) Flow Time 17 17 17 17 20 20 8 Causes of Waiting: Demand vs. Capacity (cont’d) • If we consider a few more customers: Order Arrive At Finish Preparing At Finish Baking At Flow Time 1 2 3 4 5 6 7 8 9 10 3 6 9 12 15 18 21 24 27 30 5 8 11 14 17 20 23 26 29 32 20 (1st Oven) 23 (2nd Oven) 26 (3rd Oven) 29 (4th Oven) 35 (1st Oven) 38 (2nd Oven) 41 (3rd Oven) 44 (4th Oven) 50 (1st Oven) 53 (2nd Oven) 17 17 17 17 20 20 20 20 23 23 Flow Time – Processing Time 0 0 0 0 3 3 3 3 6 6 • Because of the insufficient capacity, the waiting time increases when more and more customers arrive! • Also, more and more customers are in the waiting line (queue)! Managing Service Systems 9 Causes of Waiting: Demand vs. Capacity (cont’d) • To improve the performance, Pizzas-4-U decides to add one more oven. Now, it has 5 ovens in total: Resource Processing Time (min) Capacity (pizzas/hour) Number of Resource Capacity of the Resource Pool (pizzas/hour) Worker 2 30/hour 1 30/hour Oven 15 4/hour 5 20/hour • Process capacity: 20 pizzas/hour • Flow rate = min demand rate, process capacity = 20, 20 = 20/hour • Do you think all customers can get their pizzas within 20 minutes after they make the order? • What if the orders do not arrive in each 3 minutes? • What if some orders request multiple pizzas? • What if the processing times are not exactly 2 and 15 minutes? Managing Service Systems 10 Causes of Waiting: Variability • The process analysis is based on a long-run performance. • Even if the process capacity is sufficient to handle the demand, customers may still need to wait due to the variability of demand or supply • Predictable variability • Seasonality (by month, week, day, hour) • Manage resources to deal with known fluctuations in demand • Unpredictable variability • • • • Inherent randomness When demand arrives is uncertain How long processing/service takes is uncertain Need probabilities to characterize uncertainty Managing Service Systems 11 Causes of Waiting: Variability (cont’d) • If demand rate < process capacity: • Sometimes customers need to wait when the arrival of customers is faster than the service • In general, the arrival is slower than the service, so the system can easily deal with the customers in the waiting line (queue) • If demand rate = process capability: • It is more likely that the arrival is faster than the service, and customers need to wait due the variability • Because, in general, the arrival is as fast as the service, once a customer is in the waiting line (queue), it is difficult to deal with that customer • To avoid customers’ waiting, Pizzas-4-U decides to add more capacity as a buffer. Now, it has 6 ovens • Process capacity = min 30, 4 × 6 = 24 > 20 = demand rate • With this extra capacity, what is the expected customer waiting time? What is the expected number of the customers in the queue? Managing Service Systems 12 Simulation: Demand Rate < Process Capacity https://cloud.anylogic.com/assets/embed/?modelId=dba4abd8-4b7b-41a6-b62a-bef250dc1754 Managing Service Systems 13 Simulation: Demand Rate > Process Capacity https://cloud.anylogic.com/assets/embed/?modelId=dba4abd8-4b7b-41a6-b62a-bef250dc1754 Managing Service Systems 14 Short Summary: Waiting Line (Queue) No Variability With Variability Demand < Process Capacity No queue Stable system Queue exists Stable system Demand = Process Capacity No queue Stable system Queue builds up Unstable system Demand > Process Capacity Queue builds up Unstable system Queue builds up Unstable system • We can analyze the performance of the waiting line in a service system where the demand rate < process capacity • If the demand rate ≥ process capacity and there exists variability, the system is not stable, and the first goal is to increase the capacity Managing Service Systems 15 Short Summary: Key Drivers of Waiting Time Waiting Time Variability Managing Service Systems Low capacity in relation to the demand Demand Variability High Demand Service Variability Low Capacity 16 Example: Queue with a Single Server • Let’s start with a simple setting: a system with one activity and one resource Queue Server • First, we need to know the demand information: how frequent do customers arrive the system? • Interarrival time: the time between the arrivals of two consecutive customers to a system • For example, if customer 1 arrives at 10:00 am, and customer 2 arrives at 10:08 am, the Interarrival time is 8 minutes • We define the average interarrival time as ππ 1 ππ • The reciprocal of the average interarrival time is the demand rate • For example, if the average interarrival time is 3 minutes, then on average, there are 1 20 customers arrive in an hour: demand rate = 20/hours = 3 / minutes Managing Service Systems 17 Example: Queue with a Single Server (cont’d) • A service system with one activity and one resource: Queue Server • In addition to the average interarrival time ππ, we also want to know the variability of the interarrival time • We only need the relative variability, and therefore we consider the coefficient of variation πΆπΆππππ , where Standard deviation of interarrival time πΆπΆππππ = average interarrival time • For example, if the interarrival time has mean = 10 minutes and standard 2 deviation = 2 minutes ⇒ ππ = 10, πΆπΆππππ = = 0.2 Managing Service Systems 10 18 Example: Queue with a Single Server (cont’d) • A service system with one activity and one resource: Queue Server • Next, we also want to know the information about the processing time. • We define the average processing time as ππ 1 ππ • The reciprocal of the processing time is the capacity • We also consider the coefficient of variation of the processing time πΆπΆππππ , where Standard deviation of processing time πΆπΆππππ = average processing time • For example, if the processing time has mean = 5 minutes and standard deviation = 2 2 minutes ⇒ ππ = 5, πΆπΆππππ = = 0.4 5 • We consider the process where demand rate < capacity ⇒ Managing Service Systems 1 ππ < 1 ππ ⇒ ππ > ππ 19 The Analysis Framework of a Waiting Line Flow rate = demand rate Managing Service Systems 20 Estimate Average Waiting Time π»π»ππ • We know the following information: • Interarrival time (demand): average value ππ, coefficient of variation πΆπΆππππ • Processing time (supply): average value ππ, coefficient of variation πΆπΆππππ • First, we can calculate the utilization. Let’s define the service utilization as π’π’, and Flow Rate Demand rate 1⁄ππ ππ ⇒ π’π’ = = = <1 π’π’ = Capacity Capacity 1⁄ππ ππ • The average waiting time ππππ is πΆπΆππππ2 + πΆπΆππππ2 π’π’ × ππππ = ππ × 1 − π’π’ 2 Managing Service Systems 21 Practice: Average Waiting Time π»π»ππ • A server operates with 75% utilization. The average processing time is 2 minutes and the standard deviation of processing time is 1 minute. The coefficient of variation of the arrival process is 1. What is the average time in the queue for customers? • π’π’ = 0.75 • ππ = 2 1 2 • πΆπΆππππ = 1, πΆπΆππππ = = 0.5 • The average waiting time is πΆπΆππππ2 + πΆπΆππππ2 π’π’ × ππππ = ππ × 1 − π’π’ 2 2 0.75 1 + 0.52 1.25 × = 3.75 =2× =2×3× 1 − 0.75 2 2 • On average, it takes 3.5 minutes for waiting. Managing Service Systems 22 Estimate Average Waiting Time (cont’d) • ππππ = ππ × π’π’ 1−π’π’ × πΆπΆππππ2 +πΆπΆππππ2 2 • We will not talk about how to derive this formula. It is based on some advanced knowledge in probability • You need to know the high-level ideas about each component in the formula: • The first term ππ: • If a customer needs to wait, how long he/she should wait depends on the service time of the previous customers → the waiting time depend on service time • The second term π’π’⁄ 1 − π’π’ : • If the server is much busier, the customer needs to wait for a longer time • If the server is almost totally idle (π’π’ → 0), the waiting time is close to 0 as well • The third term πΆπΆππππ2 + πΆπΆππππ2 ⁄2: • If the relative variability of demand or supply increases, the customer is expected to wait for a longer time • If there is no variation (πΆπΆππππ = πΆπΆππππ = 0), there is no waiting time either (ππππ = 0) Managing Service Systems 23 Average Waiting Time vs. Utilization • Please note that the waiting time is not linearly increasing in utilization! ππππ = ππ × πΆπΆππππ2 + πΆπΆππππ2 π’π’ × 1 − π’π’ 2 π’π’ 1 − π’π’ • That’s why having almost 100% utilization could be costly! Managing Service Systems 24 Flow Time and Average Number of Customers • By definition, flow time = the time between the start and finish • Flow time = waiting time + processing time • Average flow time = ππππ + ππ • How can we get the average number of customers in the queue/system? • Use Little’s Law! • The average number of customers in the queue (πΌπΌππ ): ππππ 1 πΌπΌππ = π π × ππππ = × ππππ = ππ ππ • The average number of customers in service (πΌπΌππ ): 1 ππ πΌπΌππ = π π × ππ = × ππ = ππ ππ • The average number of customers in the system: πΌπΌππ + πΌπΌππ Managing Service Systems 25 Practice (cont’d) • A server operates with 75% utilization. The average processing time is 2 minutes and the standard deviation of processing time is 1 minute. The coefficient of variation of the arrival process is 1. What is the average time in the queue for customers? 1 2 • π’π’ = 0.75, ππ = 2, πΆπΆππππ = 1, πΆπΆππππ = = 0.5 • ππππ = ππ × • π’π’ = ππ ππ = • Flow time: 2 ππ π’π’ 1−π’π’ × πΆπΆππππ2 +πΆπΆππππ2 2 = 3.75 = 0.75 ⇒ ππ = 2.6667 • ππππ + ππ = 3.75 + 2 = 5.75 • Average number of customers: • In queue: πΌπΌππ = • In service: ππππ ππ ππ πΌπΌππ = ππ = 3.75 2.6667 = 0.75 = 1.4063 • In system: πΌπΌππ + πΌπΌππ = 2.1563 Managing Service Systems 26 Practice: Average Waiting Time • Average waiting time: ππππ = ππ × π’π’ 1−π’π’ × πΆπΆππππ2 +πΆπΆππππ2 2 • You are attending an outdoor event, where there is only one portable toilet offered to all customers. On average, a person goes to toilet every 5 minutes, with a standard deviation of 5 minutes. It takes 2 minutes for each person using the toilet with a standard deviation of 3 minutes. How long would you expect to wait when you want to use the toilet? • Information: 5 5 • Interarrival: ππ = 5, πΆπΆππππ = = 1 3 2 • Service: ππ = 2, πΆπΆππππ = = 1.5 • Utilization: π’π’ = ππ ππ 2 5 = = 0.4 • Average waiting time: ππππ = ππ × π’π’ 1−π’π’ Managing Service Systems × πΆπΆππππ2 +πΆπΆππππ2 2 =2 0.4 × 0.6 × 12 +1.52 2 2 3 =2× × 3.25 2 = 2.167 27 Flow Time and Average Number of Customers • By definition, flow time = the time between the start and finish • Flow time = waiting time + processing time • Average flow time = ππππ + ππ • How can we get the average number of customers in the queue/system? • Use Little’s Law! • The average number of customers in the queue (πΌπΌππ ): ππππ 1 πΌπΌππ = π π × ππππ = × ππππ = ππ ππ • The average number of customers in service (πΌπΌππ ): 1 ππ πΌπΌππ = π π × ππ = × ππ = ππ ππ • The average number of customers in the system: πΌπΌππ + πΌπΌππ Managing Service Systems 28 Practice (cont’d) • You are attending an outdoor event, where there is only one portable toilet offered to all customers. On average, a person goes to toilet every 5 minutes, with a standard deviation of 5 minutes. It takes 2 minutes for each person using the toilet with a standard deviation of 3 minutes. • How long would you expect to wait when you want to use the toilet? • ππππ = ππ × π’π’ 1−π’π’ × πΆπΆππππ2 +πΆπΆππππ2 2 =2 0.4 × 0.6 × 12 +1.52 2 = 2.167 • How long would you expect to spend on waiting and using the toilet? • ππππ + ππ = 2.167 + 2 = 4.167 • How many people are expected to wait for the toilet? 1 ππ • πΌπΌππ = π π × ππππ = × ππππ = 0.433 • How many people are expected to use the toilet? 1 ππ 1 5 • πΌπΌππ = π π × ππ = × ππ = × 2 = 0.4 Managing Service Systems 29 Practice (cont’d) • A server operates with 75% utilization. The average processing time is 2 minutes and the standard deviation of processing time is 1 minute. The coefficient of variation of the arrival process is 1. • What is the average time in the queue for customers? 1 2 • π’π’ = 0.75, ππ = 2, πΆπΆππππ = 1, πΆπΆππππ = = 0.5 • ππππ = ππ × • π’π’ = ππ ππ = • Flow time: 2 ππ π’π’ 1−π’π’ × πΆπΆππππ2 +πΆπΆππππ2 2 =2 = 0.75 ⇒ ππ = 2.6667 0.75 1+0.52 × × 2 0.25 = 3.75 • ππππ + ππ = 3.75 + 2 = 5.75 • Average number of customers: • In queue: πΌπΌππ = • In service: ππππ ππ ππ πΌπΌππ = ππ = 3.75 2.6667 = 0.75 = 1.4063 • In system: πΌπΌππ + πΌπΌππ = 2.1563 Managing Service Systems 30 Obtaining Demand/Supply Information (*) • Demand: interarrival time Order 1 2 3 4 5 Arrival Time 8:10 AM 8:15 AM 8:23 AM 8:26 AM 8:30 AM Interarrival Time 5 minutes 8 minutes 3 minutes 4 minutes Interarrival time: • Average: ππ = 5 • Standard deviation: 2.16 2.16 = 0.43 • πΆπΆππππ = 5 • Supply: processing time (service time) Order 1 2 3 4 5 Service Start Time 8:10 AM 8:15 AM 8:23 AM 8:26 AM 8:31 AM Service End Time 8:13 AM 8:20 AM 8:26 AM 8:31 AM 8:35 AM Processing Time 3 minutes 5 minutes 3 minutes 5 minutes 4 minutes Processing time: • Average: ππ = 4 • Standard deviation: 1 1 • πΆπΆππππ = = 0.25 4 • What if the detailed data are unavailable? Managing Service Systems 31 Obtaining Demand/Supply Information (cont’d*) • Suppose the process capacity is sufficient (i.e., demand-constrained process): • Demand rate = Flow rate = Flow units/time (e.g., 30 customers/hour) • Average interarrival time = 1/Demand rate (e.g., 1/30 hours = 2 minutes) • Given the utilization of the resource: • Capacity of the resource pool = Flow rate/Utilization • Capacity of a resource = Capacity of the resource pool/number of resource • Average processing time = 1/Capacity of a resource • What about the variability? • Average waiting time: ππππ = ππ × π’π’ 1−π’π’ × πΆπΆππππ2 +πΆπΆππππ2 2 • If πΆπΆππππ = πΆπΆππππ = 1, the formula becomes simpler: ππππ = ππ × • What does πΆπΆππππ = πΆπΆππππ = 1 means? π’π’ 1−π’π’ • The coefficient of variation = 1 implies that average = standard deviation Managing Service Systems 32 A Common Setting for Variability (*) • Exponential distribution has average = standard deviation: • Probability density function: ππ π₯π₯ = ππππ −πππ₯π₯ • Cumulative distribution function: ππ π₯π₯ ≤ π‘π‘ = 1 − ππ −πππ‘π‘ • Mean = standard deviation = 1⁄ππ • Suppose the processing time (service time) follows an exponential distribution: • Service rate = ππππ = capacity (e.g., ππππ = 20 customers/hour) • Average processing time = 1⁄ππππ (e.g., 1/20 hours = 3 minutes) • The standard deviation of the processing time is also 1⁄ππππ ⇒ πΆπΆππππ = 1 • Suppose the interarrival time follows an exponential distribution: • • • • Arrival rate = ππππ = demand rate (e.g., ππππ = 15 customers/hour) Average interarrival time = 1⁄ππππ (e.g., 1/15 hours = 4 minutes) The standard deviation of the interarrival time is also 1⁄ππππ ⇒ πΆπΆππππ = 1 The exponentially distributed interarrival time implies Poisson arrivals Managing Service Systems 33 A Common Setting for Variability (cont’d*) • The pattern of an exponential distribution with service rate = 5 customers/hour: • Average processing time: 12 minutes 5 4.5 Probability that the processing time < 6 minutes 4 3.5 3 Probability that the processing time is 6 - 12 minutes 2.5 2 1.5 1 0.5 0 6 12 18 24 30 36 Minutes 42 48 54 60 66 72 • Think about whether the demand/service follows such a pattern! Managing Service Systems 34 Service System with Multiple Servers • In practice, a system can have multiple servers and a common queue • Call center • Self checkouts in grocery stores/supermarkets • Restroom • How to measure the performance of this service system? Managing Service Systems 35 Analysis of Multiple-Server Service Systems • The concept of process analysis can be applied to the analysis! • If each server has identical average processing time ππ: • What is the capacity of each server? • What is the total capacity of ππ servers? • The demand does not change! • The arrival of customers is not affected by the number of servers • The average interarrival time and the coefficient of variation keep the same • What is the utilization of all servers? Managing Service Systems 36 Analysis of Multiple-Server Service Systems (cont’d) • Average waiting time with a common queue and ππ servers: πΆπΆππππ2 + πΆπΆππππ2 ππ π’π’ 2 ππ+1 −1 × ππππ = × ππ 1 − π’π’ 2 • Remember that utilization π’π’ = πΉπΉπΉπΉπΉπΉπΉπΉ ππππππππ πΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆπΆ = 1 ππ 1 ×ππ ππ = ππ ππ×ππ • The utilization should be smaller than 1: ππ < ππ × ππ • If there is only one server, we can get: πΆπΆππππ2 + πΆπΆππππ2 πΆπΆππππ2 + πΆπΆππππ2 ππ π’π’ 2 1+1 −1 π’π’ × × = ππ × ππππ = × 1 1 − π’π’ 1 − π’π’ 2 2 • The formula in the single-server service system is the special case when ππ = 1! • We can use Little’s Law to get the average number of waiting customers: πΏπΏππ = π π × ππππ Managing Service Systems 37 Practice: Checkout Services • There are 4 servers in the checkout area. The average interarrival time of customers is 2 minutes. The average processing time is 5 minutes. Assume Poisson arrivals and exponential service time (i.e., πΆπΆππππ = πΆπΆππππ = 1). • What is the average time in the queue for a customer? • Utilization: • ππππ = ππ ππ 5 4 × = × ππ ππ×ππ = 5 2×4 π’π’ 2 ππ+1 −1 1−π’π’ = 0.625 × 0.625 2× 4+1 −1 0.375 πΆπΆππππ2 +πΆπΆππππ2 × 2 12 +12 2 5 4 = × 0.625 10−1 0.375 5 4 = × 0.6252.1623 0.375 ≅ 1.21 • What is the average (total) time in process for a customer? • ππππ + ππ = 1.21 + 5 = 6.21 • What is the average number of waiting customers? 1 2 • πΏπΏππ = π π × ππππ = × 1.21 = 0.605 Managing Service Systems 38 Practice: Hospital • An emergency room of a local hospital employs 3 nurse practitioners who attend to patients arriving at the emergency room. Emergency patients arrive randomly at an average rate of 3 per hour. The amount of time that a nurse practitioner spends with a patient is also random, with a mean of 30 min. Assume that the arrival follows a Poisson process, and the service time is exponentially distributed. • What is the average patient waiting time? • Average interarrival time ππ = 20 minutes • Utilization: • ππππ = = ππ ππ × 30 3 Managing Service Systems × ππ ππ×ππ = 30 20×3 π’π’ 2 ππ+1 −1 1−π’π’ = 0.5 × 0.5 2× 3+1 −1 0.5 πΆπΆππππ2 +πΆπΆππππ2 × 2 12 +12 2 = 30 3 × 0.5 8−1 0.5 = 30 0.51.8284 × 0.5 3 ≅ 5.63 39 Practice: Hospital (cont’d) • An emergency room of a local hospital employs 3 nurse practitioners who attend to patients arriving at the emergency room. Emergency patients arrive randomly at an average rate of 3 per hour. The amount of time that a nurse practitioner spends with a patient is also random, with a mean of 30 min. Assume that the arrival follows a Poisson process, and the service time is exponentially distributed. • The hospital is considering reducing the number of nurse practitioners in the emergency room to 2. What’s the effect on the patient waiting time? • Utilization: • ππππ = = ππ ππ × 30 2 × ππ ππ×ππ = 30 20×2 π’π’ 2 ππ+1 −1 1−π’π’ = 0.75 × 0.75 2× 2+1 −1 0.25 πΆπΆππππ2 +πΆπΆππππ2 × 2 12 +12 2 = 30 0.75 6−1 × 0.25 2 = 30 0.751.4495 × 0.25 2 ≅ 39.54 • The waiting time with 3 nurse practitioners: 5.63 • The waiting time increases by 39.54 − 5.63 = 33.91 minutes! Managing Service Systems 40 Practice: Hospital (cont’d) • An emergency room of a local hospital employs 3 nurse practitioners who attend to patients arriving at the emergency room. Emergency patients arrive randomly at an average rate of 3 per hour. The amount of time that a nurse practitioner spends with a patient is also random, with a mean of 30 min. Assume that the arrival follows a Poisson process, and the service time is exponentially distributed. • Can the hospital further decrease the number of nurse practitioners to be 1? 1 ππ • The demand rate: = 3 patients/hour 1 ππ • Capacity of a nurse practitioner: = 1 0.5 hours 1 ππ • A system would be stable only when < ππ ππ = 2 patients/hour ⇒ 3 < 2 × ππ! • The hospital needs at least ππ = 1.5 practitioners! However, we need a positive integer, and therefore the hospital needs at least 2 practitioners. Managing Service Systems 41 Pooled Queue vs. Separate Queue • If the demand rate and service rate are the same in these two systems, which system would have a lower average customer waiting time? Managing Service Systems 42 Pooled Queue vs. Separate Queue (cont’d) • For simplicity, we assume that the systems have a Poisson arrival and exponentially distributed service time • Pooled Queue: • ππ = 25, ππ = 90, ππ = 4 • Utilization: π’π’ = • ππππ = ππ ππ × ππ ππ×ππ = π’π’ 2 ππ+1 −1 1−π’π’ 90 100 = = 0.9 90 0.9 10−1 × 0.1 4 = 179.16 • Separated Queue: • ππ = 25 × 4 = 100, ππ = 90 • Utilization: π’π’ = • ππππ = ππ × π’π’ 1−π’π’ ππ ππ = 90 100 = 90 × = 0.9 0.9 0.1 = 810 • The average waiting time of the separated queue is much higher… why? Managing Service Systems 43 Pooled Queue vs. Separate Queue (cont’d) • Economic of scale: • By combining all the demand and all the supply (servers), the demand and supply can match better, and the customer waiting time is reduced. • Flexibility to serve customers: Managing Service Systems 44 Why Are Separate Queues Still Common? • Layout of the waiting line(s): • Strategic behavior of customers: • The arrival of customers in the separate queue may not be purely random • The customers may observe the waiting line first and decide which line to join • The customers may switch from one queue to another queue • Heterogenous services: • The servers may offer different services (e.g., bank, post office, airport security check, …) Managing Service Systems 45 Summary • Average waiting time: ππππ = ππ ππ × π’π’ 2 ππ+1 −1 1−π’π’ • When the number of server ππ = 1: ππππ = ππ × • Average total time in the system: ππππ + ππ × πΆπΆππππ2 +πΆπΆππππ2 π’π’ 1−π’π’ 2 × πΆπΆππππ2 +πΆπΆππππ2 2 • How to get πΆπΆππππ = πΆπΆππππ = 1? • Poisson arrival • Exponential service time • Average number of customers: • In the waiting line: π π × ππππ = ππππ ⁄ππ • In service: π π × ππ = ππ⁄ππ • The power of pooling: A pooled queue can have a lower average waiting time. Managing Service Systems 46