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Slide 1 Matching Supply with Demand: An Introduction to Operations Management Gérard Cachon ChristianTerwiesch All slides in this file are copyrighted.

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Presentation on theme: "Slide 1 Matching Supply with Demand: An Introduction to Operations Management Gérard Cachon ChristianTerwiesch All slides in this file are copyrighted."— Presentation transcript:

1 Slide 1 Matching Supply with Demand: An Introduction to Operations Management Gérard Cachon ChristianTerwiesch All slides in this file are copyrighted by Gerard Cachon and Christian Terwiesch. Any instructor that adopts Matching Supply with Demand: An Introduction to Operations Management as a required text for their course is free to use and modify these slides as desired. All others must obtain explicit written permission from the authors to use these slides.

2 Slide 2 Variability – Waiting Times

3 Slide 3 Blocked calls (busy signal) Abandoned calls (tired of waiting) Calls on Hold Sales reps processing calls Answered Calls Incoming calls Call center Financial consequences Lost throughputHolding cost Lost goodwill Lost throughput (abandoned) $$$ Revenue $$$Cost of capacity Cost per customer  At peak, 80% of calls dialed received a busy signal.  Customers getting through had to wait on average 10 minutes  Extra telephone expense per day for waiting was $25,000. An Example of a Simple Queuing System

4 Slide 4 7:00 7:107:207:307:407:508:00 Patient Arrival Time Service Time 1 2 3 4 5 6 7 8 9 10 11 12 0 5 10 15 20 25 30 35 40 45 50 55 4 4 4 4 4 4 4 4 4 4 4 4 A Somewhat Odd Service Process

5 Slide 5 Patient Arrival Time Service Time 1 2 3 4 5 6 7 8 9 10 11 12 0 7 9 18 22 25 30 36 45 51 55 5 6 7 6 5 2 4 3 4 2 2 3 Time 7:107:207:307:407:508:007:00 Patient 1Patient 3Patient 5Patient 7Patient 9Patient 11 Patient 2Patient 4Patient 6Patient 8Patient 10Patient 12 0 1 2 3 2 min.3 min.4 min.5 min.6 min.7 min. Service times Number of cases A More Realistic Service Process

6 Slide 6 7:007:107:207:307:407:50 Inventory (Patients at lab) 5 4 3 2 1 0 8:00 7:00 7:107:207:307:407:508:00 Wait time Service time Patient Arrival Time Service Time 1 2 3 4 5 6 7 8 9 10 11 12 0 7 9 18 22 25 30 36 45 51 55 5 6 7 6 5 2 4 3 4 2 2 3 Variability Leads to Waiting Time

7 Slide 7 Processing Modeling Variability in Flow Inflow Demand process is “random” Look at the inter-arrival times a: average inter-arrival time CV a = Often Poisson distributed: CV a = 1 Constant hazard rate (no memory) Exponential inter-arrivals Difference between seasonality and variability Buffer St-Dev(inter-arrival times) Average(inter-arrival times) Time IA 1 IA 2 IA 3 IA 4 Processing p: average processing time Same as “activity time” and “service time” CVp = Can have many distributions: CV p depends strongly on standardization Often Beta or LogNormal St-Dev(processing times) Average(processing times) Outflow No loss, waiting only This requires u<100% Outflow=Inflow Flow Rate Minimum{Demand, Capacity} = Demand = 1/a

8 Slide 8 Average flow time T Theoretical Flow Time Utilization100% Increasing Variability Waiting Time Formula Service time factor Utilization factor Variability factor Inflow Outflow Inventory waiting I q Entry to systemDepartureBegin Service Time in queue T q Service Time p Flow Time T=T q +p Flow rate The Waiting Time Formula

9 Slide 9 Waiting Time Formula for Multiple (m) Servers Inflow Outflow Inventory waiting I q Entry to systemDepartureBegin Service Time in queue T q Service Time p Flow Time T=T q +p Inventory in service I p Flow rate Waiting Time Formula for Multiple, Parallel Resources

10 Slide 10 Customers send emails to a help desk of an online retailer every 2 minutes, on average, and the standard deviation of the inter-arrival time is also 2 minutes. The online retailer has three employees answering emails. It takes on average 4 minutes to write a response email. The standard deviation of the service times is 2 minutes. Estimate the average customer wait before being served. Example: Online retailer

11 Slide 11 Target Wait Time (TWT) Service Level = Probability{Waiting Time  TWT} Example: Deutsche Bundesbahn Call Center - now (2003): 30% of calls answered within 20 seconds - target: 80% of calls answered within 20 seconds Formulas: see Service Operations course / simulation 0 0.2 0.4 0.6 0.8 1 050100150200 Waiting time [seconds] Fraction of customers who have to wait x seconds or less Waiting times for those customers who do not get served immediately Fraction of customers who get served without waiting at all 90% of calls had to wait 25 seconds or less Service Levels in Waiting Systems

12 Slide 12 Independent Resources 2x(m=1) Pooled Resources (m=2) 0.00 10.00 20.00 30.00 40.00 50.00 60.00 70.00 60%65% m=1 m=2 m=5 m=10 70%75%80%85%90%95% Waiting Time T q Utilization u Implications: + balanced utilization + Shorter waiting time (pooled safety capacity) - Change-overs / set-ups Managerial Responses to Variability: Pooling

13 Slide 13 Service times: A: 9 minutes B: 10 minutes C: 4 minutes D: 8 minutes D A C B 9 min. 19 min. 23 min. Total wait time: 9+19+23=51min B D A C 4 min. 12 min. 21 min. Total wait time: 4+13+21=38 min Flow units are sequenced in the waiting area (triage step) Shortest Processing Time Rule - Minimizes average waiting time - Problem of having “true” processing times First-Come-First-Serve - easy to implement - perceived fairness Sequence based on importance - emergency cases - identifying profitable flow units Managerial Responses to Variability: Priority Rules in Waiting Time Systems

14 Slide 14 0 20 40 60 80 100 120 140 160 0:152:003:455:307:159:00 10:4512:30 14:1516:0017:4519:3021:1523:00 Number of customers Per 15 minutes 0 20 40 60 80 100 120 140 160 0:152:003:455:307:159:00 10:4512:30 14:1516:0017:4519:3021:1523:00 Time Number of customers Per 15 minutes 0 0.2 0.4 0.6 0.8 1 0:00:000:00:090:00:170:00:260:00:350:00:430:00:520:01:000:01:09 Distribution Function Empirical distribution (individual points) Exponential distribution 0 0.2 0.4 0.6 0.8 1 0:00:000:00:090:00:170:00:260:00:350:00:430:00:520:01:000:01:09 Distribution Function Inter-arrival timeInter-arrival time Empirical distribution (individual points) Exponential distribution Seasonality vs. variability Need to slice-up the data Within a “slice”, exponential distribution (CV a =1) See chapter 6 for various data analysis tools Data in Practical Call Center Setting

15 Slide 15 What to Do With Seasonal Data Measure the true demand data Slice the data by the hour (30min, 15min) Apply waiting model in each slice

16 Slide 16 Variability is the norm, not the exception Variability leads to waiting times although utilization<100% Use the Waiting Time Formula to - get a qualitative feeling of the system - analyze specific recommendations / scenarios Managerial response to variability: - understand where it comes from and eliminate what you can - accommodate the rest by holding excess capacity Difference between variability and seasonality - seasonality is addressed by staffing to (expected) demand Managing Waiting Systems: Points to Remember

17 Slide 17 Call Center: Lean Diagnostic Example

18 Slide 18 Large UK Retail bank- operates each week day from 8.30am to 6pm Handles three types of calls (agents are dedicated to call type) All subsequent information is about type 1 calls (similar numbers for type 2 and 3 calls) Type 1 calls are direct payments for mortgages / loans that customers do using checks or credit cards Agents make 30 pounds per hour loaded salary Typical time commitments from agents: Taking calls / “wrap time” Six weeks of paid vacation Average absenteeism rate of 6% Other time commitments for the agents, such as breaks, training, special duties: 15% Call Center Overview Picture from Switzer & CO, illustrative only

19 Slide 19 Demand information (Type 1 calls) Call volume changes over the course of the day Typical day is shown on the right Calls come in randomly, given this seasonal pattern (CVa=1) Call center operates for 252 days a year (closed week-ends, holidays) Call handling information Type 1 calls take on average 5 minutes (with a CVp=1) This includes actual call time as well as wrap time Actual value varies by agent / customer Number of staffed seats for type 1 calls In each time slot, 9 agents are currently on duty from 8.30am to 6.00pm Service level / targeted wait time The average wait time is targeted to be 2 minutes or less Present wait time is significantly longer for 11-12 time slot At other times, agents have plenty of capacity (e.g., at 8.30am) Cost: about 0.04 pounds per minute for wait or talk time Demand, Call Handling, and Service Level

20 Slide 20 Variation exists across agents handling calls – the following information is about type 1 calls A sample of calls across three agents has been collected (Matthew, Daniel, and Anna) For the purpose of this exercise, consider the following calls: 1b, 1c, 2a, 2b, 2d, 3a, 3b You can assume that this data (and its variation in customer and agent behavior) is representative Audio files have been made available to you Sample of Calls Picture from Livehacker blog

21 Slide 21 You are on site in the call center and have the information provided above The goal of the project is to conduct an initial diagnosis of the call center operation What productivity improvement will you aim for? What is your pitch to the client’s CEO? Detailed assignments: 1. Excel model of status quo (with Tq) 2. Compute current OPE / OEE 3. Detailed, micro-level analysis of calls; estimate improvement potential 4. Evaluate improvements (shorter calls, staffing plan, pooling, other) Assignment

22 Slide 22 What generates queues?  An arrival rate that predictably exceeds the service rate (i.e., capacity) of a process.  e.g., toll booth congestion on the NJ Turnpike during the Thanksgiving Day rush.  Variation in the arrival and service rates in a process where the average service rate is more than adequate to process the average arrival rate.  e.g., calls to a brokerage are unusually high during a particular hour relative to the same hour in other weeks (i.e., calls are high by random chance).

23 Slide 23 Predicable and unpredictable variation at An-ser

24 Slide 24  An inter-arrival time is the amount of time between two arrivals to a process.  An arrival process is stationary over a period of time if the number of arrivals in any sub-interval depends only on the length of the interval and not on when the interval starts.  For example, if the process is stationary over the course of a day, then the expected number of arrivals within any three hour interval is about the same no matter which three hour window is chosen (or six hour window, or one hour window, etc).  Processes tend to be non-stationary (or seasonal) over long time periods (e.g., over a day or several hours) but stationary over short periods of time (say one hour, or 15 minutes). Defining inter-arrival times and a stationary process

25 Slide 25 Arrivals to An-ser over the day are non-stationary The number of arrivals in a 3-hour interval clearly depends on which 3-hour interval your choose … … and the peaks and troughs are predictable (they occur roughly at the same time each day.

26 Slide 26 How to describe (or model) inter-arrival times  We will use two parameters to describe inter-arrival times to a process:  The average inter-arrival time.  The standard deviation of the inter-arrival times.  What is the standard deviation?  Roughly speaking, the standard deviation is a measure of how variable the inter-arrival times are.  Two arrival processes can have the same average inter-arrival time (say 1 minute) but one can have more variation about that average, i.e., a higher standard deviation.

27 Slide 27 Relative and absolute variability  The standard deviation is an absolute measure of variability.  Two processes can have the same standard deviation but one can seem much more variable than the other:  Below are random samples from two processes that have the same standard deviation. The left one seems more variable.

28 Slide 28 Relative and absolute variability (cont)  The previous slide plotted the processes on two different axes.  Here, the two are plotted relative to their average and with the same axes.  Relative to their average, the one on the left is clearly more variable (sometime more than 400% above the average or less than 25% of the average)

29 Slide 29  The coefficient of variation is a measure of the relative variability of a process – it is the standard deviation divided by the average.  The coefficient of variation of the arrival process: Coefficient of variation CV a = 10 /10 = 1 CV a = 10 /100 = 0.1

30 Slide 30  The coefficient of variation of the service time process: Service time variability 1000 900800700 600 500400300200 100 0 800 600 400 200 0 Call durations (seconds) Frequency  Standard deviation of activity times = 150 seconds.  Average call time (i.e., activity time) = 120 seconds.  CV p = 150 / 120 = 1.25

31 Slide 31 A multi-server queue  a = average interarrival time  Flow rate = 1 / a  p = average activity time  Capacity of each server = 1 / p  m = number of servers  Capacity = m x 1 / p  Utilization = Flow rate / Capacity = (1 / a) / (m x 1 / p) = p / (a x m) Multiple servers Customers waiting for service Arriving customers Departing customers  Assumptions:  All servers are equally skilled, i.e., they all take p time to process each customer.  Each customer is served by only one server.  Customers wait until their service is completed.  There is sufficient capacity to serve all demand.

32 Slide 32 A multi-server queue – some data and analysis  Suppose:  a = 35 seconds per customer  Flow rate = 1 / 35 customers per second  p = 120 seconds per customer  Capacity of each server = 1 / 120 customers per second  m = number of servers = 4  Then  Capacity = m x 1 / p = 4 x 1 / 120 = 1 / 30 customers per second  Utilization = Flow rate / Capacity = p / (a x m) = 120 / (35 x 4) = 85.7%

33 Slide 33 The implications of utilization  A 85.7% utilization means:  At any given moment, there is a 85.7% chance a server is busy serving a customer and a 1-0.857 = 14.3% chance the server is idle.  At any given moment, on average 85.7% of the servers are busy serving customers and 1 – 0.857 = 14.3% are idle.  If utilization < 100% then:  The system is stable which means that the size of the queue does not keep growing over time. (Capacity is sufficient to serve all demand)  If utilization >= 100% then:  The system is unstable, which means that the size of the queue will continue to grow over time.

34 Slide 34 Stable and unstable queues a = 35 p = 120 m = 4 Utilization = 86% a = 35 p = 120 m = 3 Utilization = 114%

35 Slide 35 Time spent in the two stages of the system T q = Time in queue p = Time in service T = T q + p = Flow time  The above Time in queue equation works only for a stable system, i.e., a system with utilization less than 100% Recall: p = Activity time m = number of servers

36 Slide 36 What determines time in queue in a stable system? The capacity factor: Average processing time of the system = (p/m). Demand does not influence this factor. The utilization factor: Average demand influences this factor because utilization is the ratio of demand to capacity. The variability factor: This is how variability influences time in queue – the more variability (holding average demand and capacity constant) the more time in queue.

37 Slide 37  Suppose:  a = 35 seconds, p = 120 seconds, m = 4  Utilization = p / (a x m) = 85.7%  CV a = 1, CV p = 1  So Flow time = 150 + 120 = 270 seconds.  In other words, the average customer will spend 270 seconds in the system (waiting for service plus time in service) Evaluating Time in queue

38 Slide 38 How many customers are in the system?  Use Little’s Law, I = R x T  R = Flow Rate = (1/a)  The flow rate through the system equals the demand rate because we are demand constrained (utilization is less than 100%)  I q = (1/a) x T q = T q / a  I p = (1/a) x p = p / a  Note, time in service does not depend on the number of servers because when a customer is in service they are processed by only one server no matter how many servers are in the system. I q = Inventory in queueI p = Inventory in service I = I q + I p = Inventory in the system

39 Slide 39 Utilization and system performance  Time in queue increases dramatically as the utilization approaches 100% Activity time = 1 m = 1 CV a = 1 CV p = 1

40 Slide 40 Which system is more effective? Pooled system: One queue, four servers Partially pooled system: Two queues, two servers with each queue. Separate queue system: Four queues, one server with each queue. a = 35 m = 4 For each of the 2 queues: a = 35 x 2 = 70 m = 2 For each of the 4 queues: a = 35 x 4 = 140 m = 1 Across these three types of systems: Variability is the same: CV a = 1, CV p = 1 Total demand is the same: 1/35 customers per second. Activity time is the same: p = 120 Utilization is the same: p / a x m = 85.7% The probability a server is busy is the same = 0.857

41 Slide 41 Pooling can reduce Time in queue considerably Pooled system Partially pooled system Separate queue system a = 35, p = 120, m = 4, CV a = 1, CV p = 1 T q = 150 Flow Time = 270 For each of the two queues: a = 70, p = 120, m = 2, CV a = 1, CV p = 1 T q = 336 Flow Time = 456 For each of the four queues: a = 140, p = 120, m = 1, CV a = 1, CV p = 1 T q = 720 Flow Time = 840

42 Slide 42 Pooling reduces the number of customers waiting but not the number in service Pooled system Partially pooled system Separate queue system I q = 4.3 I p = 3.4 I q = 4.8 I p = 1.7 I q = 5.1 I p = 0.86 I q = 4.3 I p = 3.4 I q = 9.6 I p = 3.4 I q = 20.4 I p = 3.4 Inventories for each queue within the system Inventories for the total system ( 2 times the inventory in each queue ) ( 4 times the inventory in each queue )

43 Slide 43 Why does pooling work? Pooled system Separate queue system = A customer waiting = A customer in service  Both systems currently have 6 customers  In the pooled system, all servers are busy and only 2 customers are waiting  In the separate queue system, 2 servers are idle and 4 customers are waiting  The separate queue system is inefficient because there can be customers waiting and idle servers at the same time.

44 Slide 44 Some quick service restaurants pool drive through ordering  The person saying “Can I take your order?” may be hundreds (or even thousands) of miles away:  Pooling the order taking process can improve time-in- queue while requiring less labor.  It has been shown that queue length at the drive through influences demand – people don’t stop if the queue is long.  However, this system incurs additional communication and software costs.

45 Slide 45 Limitations to pooling  Pooling may require workers to have a broader set of skills, which may require more training and higher wages:  Imagine a call center that took orders for McDonalds and Wendys … now the order takers need to be experts in two menus.  Suppose cardiac surgeons need to be skilled at kidney transplants as well.  Pooling may disrupt the customer – server relationship:  Patients like to see the same physician.  Pooling may increase the time-in-queue for one customer class at the expense of another:  Removing priority security screening for first-class passengers may decrease the average time-in-queue for all passengers but will likely increase it for first-class passengers.

46 Slide 46 Why the long queues for the women’s restroom?

47 Slide 47 Potty-parity Laws  Women’s Restroom Equity Bill:  Passed NY City Counsel May 2005.  Requires women’s restrooms to have twice the flushing capacity of men’s restrooms (at least a 2:1 ratio of women’s toilets to men’s toilets & urinals)  In the context of Time in queue equation, this law requires m to increase for women, which decreases the capacity factor (p/m), and decreases utilization.  Other solutions:  Unisex restrooms (pools capacity, at least the toilets)  Flexible partitions that can change the size of each restroom.

48 Slide 48 Summary  Even when a process is demand constrained (utilization is less than 100%), waiting time for service can be substantial due to variability in the arrival and/or service process.  Waiting times tend to increase dramatically as the utilization of a process approaches 100%.  Pooling multiple queues can reduce the time-in-queue with the same amount of labor (or use less labor to achieve the same time-in-queue).

49 Slide 49 Recitation Questions (prepare the questions described on the following slides for the R5 recitation)

50 Slide 50 Rent-A-Phone Rent-A-Phone is a new service company that provides European mobile phones to American visitors to Europe. The company currently has 80 phones available at Charles de Gaulle Airport in Paris. There are, on average, 25 customers per day requesting a phone. These requests arrive uniformly throughout the 24 hours the store is open. (Note: This means customers arrive at a faster rate than 1 customer per hour) The corresponding coefficient of variation is 1.

51 Slide 51 Rent-A-Phone (cont) Customers keep their phone for 72 hours on average. The standard deviation of this time is 100 hours. Given that Rent-A-Phone currently does not have a competitor in France providing equally good service, customers are willing to wait for the telephones. Yet, during the waiting period, customers are provided a free calling card. Based on prior experiences, Rent-A-Phone found that the company incurred a cost of $1 per hour per waiting customer, independent of day or night.

52 Slide 52 Rent-A-Phone (Questions) a) What is the average number of telephones the company has in its store? b) How long does a customer, on average, have to wait for the phone? c) What are the total monthly (30 days) expenses for telephone cards? d) Assume that it costs Rent-A-Phone $1000 per unit per month for a new phone. Is it worth it to buy one additional phone? Why?

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