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1 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Made-to-stock (MTS) operations. Product is manufactured and stocked in advance. Safety.

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Presentation on theme: "1 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Made-to-stock (MTS) operations. Product is manufactured and stocked in advance. Safety."— Presentation transcript:

1 1 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Made-to-stock (MTS) operations. Product is manufactured and stocked in advance. Safety inventory protects against stockouts due to variability of arrival time and processing time. Inventory also permits economies of scale. Make-to-order (MTO) operations. Each order is specific, cannot be stored in advance. Ex. banks, restaurants, retail checkout counters, airline reservation, hospitals, repair shops, call centres. Production systems also try to follow Dell Computer model. We needs to maintain sufficient capacity to deal with uncertainty in both arrival and processing time. Safety Capacity vs. Safety Inventory. Make to Stock (MTS) vs. Make to Order (MTO)

2 2 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 A Call Centre Sales Reps Processing Calls (Service Process) Incoming Calls (Customer Arrivals) Calls on Hold (Service Inventory) Answered Calls (Customer Departures) Blocked Calls (Due to busy signal) Abandoned Calls (Due to long waits) Calls In Process (Due to long waits) The Call Centre Process

3 3 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 The time of the arrival of an order is not known ahead of time. It is a random variable with estimated mean and standard deviation. The time of the next telephone call is not known. The service time is not known (precisely) ahead of time. It is a random variable with estimated mean and standard deviation. The time a customers spends on the web page of amazon.com is not precisely known. The time a customer spends speaking with the teller in the bank is unknown Characteristics of Waiting Lines (Queuing Systems)

4 4 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Article: The Psychology of Waiting Lines Unoccupied time feels longer than occupied time. Pre-process waits feels longer than in-process waits. Anxiety makes waits seem longer. Uncertain waits are longer than known, finite waits. Unexplained waits are longer than explained waits. Unfair waits are longer than equitable waits. The more valuable the service, the longer I will wait. Solo waiting feels longer than group waiting.

5 5 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Variability in arrival time and service time leads to Idleness of resources Waiting time of customers (orders) to be processed We are interested in evaluating two measures: Average waiting time of flow units. Average waiting time in the waiting line and in the system (Waiting line + Processor). Average number of flow units. The average number of orders (customers) waiting in the waiting line (to be then processed). Let us first look at the Servers or Processors Characteristics of Queuing Systems

6 6 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Tp : Processing time. Tp units of time. Ex. on average it takes 5 minutes to serve a customer. Rp : processing rate. Rp flow units are handled per unit of time. If Tp is 5 minutes. Compute Rp. Rp= 1/5 per minute, or 60/5 = 12 per hour. AVERAGE Processing Time Tp AVERAGE Processing Rate Rp

7 7 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Tp : processing time. Rp : processing rate. What is the relationship between Rp and Tp? If we have one resource Rp = 1/Tp What is the relationship between Rp and Tp when we have more than one resource; We have c recourses Rp = c/Tp Each customer always spends Tp unites of time with the server More than One Server; c Servers

8 8 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Tp = 5 minutes. Processing time is 5 minute. Each customer on average is with the server for 5 minutes. c = 3, we have three servers. Processing rate of each server is 1/5 customers per minute, or 12 customer per hour. Rp is the processing rate of all three servers. Rp = c/Tp Rp = 3/5 customers/minute, or 36 customers/hour. Average Processing Rate of c Servers

9 9 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Inter-arrival Time (Ta) and Arrival Rate (Ra) Ta : customer inter-arrival time. On average each 10 minutes one customer arrives. Ra: customer arrival (inflow) rate. What is the relationship between Ta and Ra Ta = every ten minutes one customer arrives How many customers in a minute? 1/10; Ra= 1/Ta= 1/10 Ra = 1/10 customers per min; 6 customers per hour Ra= 1/Ta

10 10 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Throughput = Min (Ri,Rp) Ra MUST ALWAYS <= Rp. We will show later that even Ra=Rp is not possible. Incoming rate must be less than processing rate. Throughput = Flow Rate R = Min (Ra, Rp). Stable Process = Ra< Rp R = Ra Safety Capacity Rs = Rp – Ra

11 11 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 What is the waiting time in the servers (processors)? Throughput? Buffer (waiting line) and Processors (Servers) Flow time T = T i + T p Inventory I = I i + I p Ti: waiting time in the inflow buffer Ii: number of customers in the inflow buffer

12 12 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 U = Utilization U =inflow rate / processing rate U = throughout / process capacity U = R/ Rp < 1 Safety Capacity = Rp – R For example, R = 6 per hour, processing time for a single server is 6 min Rp= 12 per hour, U = R/ Rp = 6/12 = 0.5 Safety Capacity = Rp – R = 12-6 = 6 Utilization is Always Less than 1

13 13 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Given a single server. And a utilization of U= 0.5 How many flow units are in the server ? Given the Utilization, How Many Flow Units are in the Processor(s) U = 0.5 means 50% of time there is 1 flow unit in the server 50% of time there is 0 flow unit in the server = 0.5 Average Inventory in the server is equal to utilization Ip= 1U = U

14 14 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Given the Utilization, How Many Flow Units are in the Processor(s) Given 2 servers. And a utilization of U = 0.5 How many flow units are in the servers ? U = 0.5 means 50% of time there is 1 flow unit in each server 50% of time there is 0 flow unit in each server = 0.5 flow unit in each server Average Inventory in the server is equal to utilization timesthe number of servers Ip= 2U = cU

15 15 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Processing time: Tp, Ex. Tp = 5 minutes Number of servers: c, Ex. c=3 Tp is also waiting time in the server, no mater one server or c servers. Tp in this example is always 5 min. Processing rate Rp= c/Tp. Ex. Rp =3/5 per min; 36/hr Utilization: U. Ex. U = 0.8 in our example Number of the flow units in all servers, Ip = cU In our example, Ip = = 2.4 Can we compute R? TR = I Tp R = cU R = cU/Tp 5 R = 2.4 R = 0.48 flow units per minute or 28.8 / hr We learned it without looking at any formula What We Have Learned Without Looking for any Formula

16 16 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Variability in arrival time and service time leads to Idleness of resources Waiting time of customers (orders) to be processed We are interested in evaluating two measures: Average waiting time of flow units. Average waiting time in the waiting line and in the system (Waiting line + Processor). Average number of flow units. The average number of orders (customers) waiting in the waiting line (to be then processed). Characteristics of Queuing Systems

17 17 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Two key drivers of process performance are Utilization and Variability. The higher the utilization the longer the waiting line. High capacity utilization U= R/Rp or low safety capacity Rs =R – Rp, due to High inflow rate R Low processing rate Rp = c / Tp, which may be due to small-scale c and/or slow speed 1 /Tp The higher the variability, the longer the waiting line. Utilization and Variability

18 18 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Variability in the interarrival time and processing time can be measured using standard deviation. Higher standard deviation means greater variability. Coefficient of Variation: the ratio of the standard deviation of interarrival time (or processing time) to the mean. Ca = coefficient of variation for interarrival times Cp = coefficient of variation for processing times Drivers of Process Performance

19 19 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Now Lets Look at the Rest of the System; The Littles Law Applies Everywhere Flow time T T i + T p Inventory I = I i + I p R I = R T R = I/T = Ii/Ti = Ip/Tp Ii = R Ti Ip = R Tp We have already learned Rp = c/Tp U= R/R p = Ip/c

20 20 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Operational Performance Measures Flow time T = T i + T p Inventory I = I i + I p T i : waiting time in the inflow buffer = ? I i : number of customers waiting in the inflow buffer =? A set of excel sheet are provided at the end of this lecture. We relay on the approximation formula, given shortly, and our understanding of the waiting line logics. We also provide an exact table for a special case.

21 21 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 U= R /R p,where R p = c / T p C a and C p are the Coefficients of Variation Standard Deviation/Mean of the inter-arrival or processing times (assumed independent) The Queue Length APPROXIMATION Formula Utilization effect U-part Variability effect V-part

22 22 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 This part captures the capacity utilization effect. It shows that queue length increases rapidly as U approaches 1. Factors affecting Queue Length This part captures the variability effect. The queue length increases as the variability in interarrival and processing times increases. Even if the processing capacity is not fully utilized, whenever there is variability in arrival or in processing times, queues will build up and customers will have to wait.

23 23 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Utilization – Variability - Delay Curve Variability Increases Average Time in System T Utilization (U) U 100% TpTp

24 24 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 If inter-arrival and processing times are constant, queues will build up if and only if the arrival rate is greater than the processing rate. If there is (unsynchronized) variability in inter-arrival and/or processing times, queues will build up even if the average arrival rate is less than the average processing rate. If variability in interarrival and processing times can be synchronized (correlated), queues and waiting times will be reduced. Lessons Learned

25 25 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Coefficient of Variations for Alternative Distributions

26 26 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Exact Ii for Poisson Arrival Rate, Exponential Service Time

27 27 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Terminology: The characteristics of a queuing system is captured by five parameters; Arrival pattern, Service pattern, Number of server, Restriction on queue capacity, The queue discipline. M/M/1; Poisson arrival rate, Exponential service times, one server, No capacity limit. M/G/12/23; Poisson arrival rate, General service times, 12 servers, Queue capacity is 23 Terminology and Classification of Waiting Lines

28 28 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 =AVERAGE () Avg. interarrival time = 6 min. R a = 1/6 arrivals / min. =STDEV() Std. Deviation = 3.94 C a = 3.94/6 = 0.66 C 2 a = (0.66) 2 = Example; Coefficient of Variation of Interarrival Time A sample of 10 observations on Interarrival times in minutes 10,10,2,10,1,3,7,9, 2, 6 minutes.

29 29 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 T p = 5 minutes R p = 1/5 processes/min. Std. Deviation = 2.83 C p = 2.83/5 = 0.57 C 2 p = (0.57) 2 = Example: Coefficient of Variation of Processing Time A sample of 10 observations on Processing times in minutes 7,1,7 2,8,7,4,8,5, 1 minutes.

30 30 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 R a =1/6 < R P =1/5 R = R a U= R/ R P = (1/6)/(1/5) = 0.83 With c = 1, the average number of passengers in queue is as follows: Example: Utilization and Safety Capacity On average 1.56 passengers waiting in line, even though safety capacity is Rs= R P - Ra= 1/5 - 1/6 = 1/30 passenger per minute, or 2 per hour. I i = [( )/(1-0.83)] ×[( )/2] = 1.56

31 31 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Waiting time in the line? Ti=Ii/R = (1.56)(6) = 9.4 minutes. Waiting time in the system? T = Ti+Tp Since Tp= 5 T = Ti+ Tp= 14.4 minutes Total number of passengers in the process is: I = RT = (1/6) (14.4) = 2.4 Alternatively, 1.56 in the buffer. How many in the process? I = = Example: Other Performance Measures

32 32 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Compute R, Rp and U: Ta= 6 min, Tp = 5 min, c=2 R = Ra= 1/6 per minute Processing rate for one processor 1/5 for 2 processors Rp = 2/5 U = R/Rp = (1/6)/(2/5) = 5/12 = Example: Now suppose we have two servers. On average passengers waiting in line. safety capacity is Rs= R P - R a = 2/5 - 1/6 = 7/30 passenger per minute, or 14 passengers per hour

33 33 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Other performance measures: Ti=Ii/R = (0.076)(6) = 0.46 minutes Compute T? T = Ti+Tp Since T P = 5 T = Ti + Tp= = 5.46 minutes Total number of passengers in the process is: I= 0.08 in the buffer and in the process. I = (0.417) = 0.91 Other Performance Measures for Two Servers cρRsRs IiIi TiTi TI

34 34 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Processing Time, Waiting Time; Long Waiting Line Increasing Capacity Polling Ra =R= 10/hour Tp = 5 minutes Interarrival time Poisson Service time exponential Ra = 10/hour

35 35 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 More Servers + Specialization Ra/2 = 10/hour Ra/2 = 10/hour

36 36 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Polling: Lower Waiting Time, Longer Processing Time (Perhaps) Ra = 10/hour

37 37 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Effect of Pooling RaRa Server 1 Queue Server 2 RaRa Queue 2 R a /2 Server 1 Queue 1 R a /2 Ra =R= 10/hour Tp = 5 minutes Interarrival time Poisson Service time exponential

38 38 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Effect of Pooling : 2M/M/1 Ra Server 2 Queue 2 R a /2 Server 1 Queue 1 R a /2 Ra/2 = R= 5/hour Tp = 5 minutes c = 1 Rp = 12 / hour U= 5/12 = 0.417

39 39 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Comparison of 2M/M/1 with M/M/2 Ra Server 2 Queue 2 Ra/2 Server 1 Queue 1 Ra/2 Server 1 Queue Server 2 Ra

40 40 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Effect of Pooling: M/M/2 Server 1 Queue Server 2 Ra Ra=R= 10/hour Tp = 5 minutes c = 2 Rp = 24 /hour U= 10/24 U= AS BEFORE for each processor

41 41 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Under Design A, We have Ra = 10/2 = 5 per hour, and Tp = 5 minutes, c =1, we arrive at a total flow time of 8.6 minutes Under Design B, We have Ra =10 per hour, Tp= 5 minutes, c=2, we arrive at a total flow time of 6.2 minutes So why is Design B better than A? Design A the waiting time of customer is dependent on the processing time of those ahead in the queue Design B, the waiting time of customer is partially dependent on each preceding customers processing time Combining queues reduces variability and leads to reduce waiting times Effect of Pooling

42 42 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Financial Performance Measures Sales: Throughput Rate Cost: Capacity utilization, Number in queue / in system Customer service: Waiting Time in queue /in system Performance Improvement Levers Decrease variability in customer inter-arrival and processing times. Decrease capacity utilization. Synchronize available capacity with demand. Performance Measures

43 43 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Customers arrival are hard to control Scheduling, reservations, appointments, etc…. Variability in processing time Increased training and standardization processes Lower employee turnover rate more experienced work force Limit product variety, increase commonality of parts 1. Variability Reduction Levers

44 44 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 If the capacity utilization can be decreased, there will also be a decrease in delays and queues. Since U=R/Rp, to decrease capacity utilization there are two options Manage Arrivals: Decrease inflow rate Ra Manage Capacity: Increase processing rate Rp Managing Arrivals Better scheduling, price differentials, alternative services Managing Capacity Increase scale of the process (the number of servers) Increase speed of the process (lower processing time) 2. Capacity Utilization Levers

45 45 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Capacity Adjustment Strategies Personnel shifts, cross training, flexible resources Workforce planning & season variability Synchronizing of inputs and outputs, Better scheduling 3. Synchronizing Capacity with Demand

46 46 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 The M/M/c Model EXACT Formulas

47 47 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 The M/G/1 Model EXACT Formulas

48 48 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 The M/G/1 Model EXACT Formulas

49 49 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Example: The arrival rate to a GAP store is 6 customers per hour and has Poisson distribution. The service time is 5 min per customer and has Exponential distribution. On average how many customers are in the waiting line? How long a customer stays in the line? How long a customer stays in the processor (with the server)? On average how many customers are with the server? On average how many customers are in the system? On average how long a customer stay in the system? Assignment 1: M/M/1 Performance Evaluation

50 50 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Assignment 2: M/M/1 Performance Evaluation What if the arrival rate is 11 per hour?

51 51 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 A local GAP store on average has 10 customers per hour for the checkout line. The inter-arrival time follows the exponential distribution. The store has two cashiers. The service time for checkout follows a normal distribution with mean equal to 5 minutes and a standard deviation of 1 minute. On average how many customers are in the waiting line? How long a customer stays in the line? How long a customer stays in the processors (with the servers)? On average how many customers are with the servers? On average how many customers are in the system ? On average how long a customer stay in the system ? Assignment 3: M/G/c

52 52 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 A call center has 11 operators. The arrival rate of calls is 200 calls per hour. Each of the operators can serve 20 customers per hour. Assume interarrival time and processing time follow Poisson and Exponential, respectively. What is the average waiting time (time before a customers call is answered)? Assignment 4: M/M/c Example

53 53 Ardavan Asef-Vaziri Jan 2011Operations Management: Waiting Lines 1 Suppose the service time is a constant What is the answer of the previous question? Assignment 5: M/D/c Example


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