Presentation on theme: "Make to Stock (MTS) vs. Make to Order (MTO)"— Presentation transcript:
1 Make to Stock (MTS) vs. Make to Order (MTO) 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.
2 A Call Centre The Call Centre Process Sales Reps Processing Calls (Service Process)Incoming Calls(Customer Arrivals)on Hold(Service Inventory)Answered Calls(Customer Departures)Blocked Calls(Due to busy signal)Abandoned Calls(Due to long waits)Calls In ProcessThe Call Centre Process
3 Characteristics of Waiting Lines (Queuing Systems) 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
4 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 Characteristics of Queuing Systems Variability in arrival time and service time leads toIdleness of resourcesWaiting time of customers (orders) to be processedWe 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
6 AVERAGE Processing Time Tp AVERAGE Processing Rate Rp 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.
7 More than One Server; c Servers 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
8 Average Processing Rate of c Servers 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.
9 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 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 Buffer (waiting line) and Processors (Servers) Flow time T = Ti TpInventory I = Ii IpTi: waiting time in the inflow bufferIi: number of customers in the inflow bufferWhat is the waiting time in the servers (processors)? Throughput?
12 Utilization is Always Less than 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
13 Given the Utilization, How Many Flow Units are in the Processor(s) Given a single server. And a utilization of U= 0.5 How many flow units are in the server ?U = 0.5 means50% of time there is 1 flow unit in the server50% of time there is 0 flow unit in the server0.5 0 = 0.5Average Inventory in the server is equal to utilizationIp= 1U = U
14 Given the Utilization, How Many Flow Units are in the Processor(s) Given 2 servers. And a utilization of U = 0.5How many flow units are in the servers ?U = 0.5 means50% of time there is 1 flow unit in each server50% of time there is 0 flow unit in each server0.5 0 = 0.5 flow unit in each serverAverage Inventory in the server is equal to utilization timesthe number of servers Ip= 2U = cU
15 What We Have Learned Without Looking for any Formula 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 = 3 0.8 = 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
16 Characteristics of Queuing Systems Variability in arrival time and service time leads toIdleness of resourcesWaiting time of customers (orders) to be processedWe 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).
17 Utilization and Variability 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 toHigh inflow rate RLow processing rate Rp = c / Tp, which may be due to small-scale c and/or slow speed 1 /TpThe higher the variability, the longer the waiting line.
18 Drivers of Process Performance 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 timesCp = coefficient of variation for processing times
19 Now Let’s Look at the Rest of the System; The Little’s Law Applies Everywhere Flow time T Ti TpInventory I = Ii IpI = R TIi = R TiIp = R TpR = I/T = Ii/Ti = Ip/TpWe have already learnedRp = c/TpU= R/Rp = Ip/c
20 Operational Performance Measures Flow time T = Ti TpInventory I = Ii IpTi: waiting time in the inflow buffer = ?Ii: 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 The Queue Length APPROXIMATION Formula Utilization effect U-partVariability effectV-partU= R /Rp,where Rp = c / TpCa and Cp are the Coefficients of VariationStandard Deviation/Mean of the inter-arrival or processing times (assumed independent)
22 Factors affecting Queue Length This part captures the capacity utilization effect. It shows that queue length increases rapidly as U approaches 1.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 Utilization – Variability - Delay Curve AverageTime in System TUtilization (U)U100%TpVariabilityIncreases
24 Lessons LearnedIf 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.
25 Coefficient of Variations for Alternative Distributions
26 Exact Ii for Poisson Arrival Rate, Exponential Service Time
27 Terminology and Classification of Waiting Lines 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
28 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.=AVERAGE () Avg. interarrival time = 6 min. Ra = 1/6 arrivals / min. =STDEV() Std. Deviation = 3.94 Ca = 3.94/6 = 0.66 C2a = (0.66)2 =
29 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.Tp= 5 minutes Rp = 1/5 processes/min. Std. Deviation = 2.83 Cp = 2.83/5 = 0.57 C2p = (0.57)2 =
30 Example: Utilization and Safety Capacity Ra=1/6 < RP =1/5 R = Ra U= R/ RP = (1/6)/(1/5) = 0.83 With c = 1, the average number of passengers in queue is as follows:Ii = [(0.832)/(1-0.83)] ×[( )/2] = 1.56On average 1.56 passengers waiting in line, even though safety capacity is Rs= RP - Ra= 1/5 - 1/6 = 1/30 passenger per minute, or 2 per hour.
31 Example: Other Performance Measures 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 = = 2. 39
32 Example: Now suppose we have two servers. 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 = 0.417On average passengers waiting in line.safety capacity is Rs= RP - Ra= 2/5 - 1/6 = 7/30 passenger per minute, or 14 passengers per hour
33 Other Performance Measures for Two Servers Other performance measures: Ti=Ii/R = (0.076)(6) = 0.46 minutes Compute T? T = Ti+Tp Since TP = 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) = 0.91cρRsIiTiTI10.830.031.569.3814.382.420.4170.230.0770.465.460.91
34 Processing Time, Waiting Time; Long Waiting Line Ra =R= 10/hour Tp = 5 minutes Interarrival time Poisson Service time exponentialRa =10/hourIncreasing CapacityPolling
35 More Servers + Specialization Ra/2 =10/hourRa/2 =10/hour
36 Polling: Lower Waiting Time, Longer Processing Time (Perhaps) Ra =10/hour
37 Effect of PoolingServer 1Queue 1Ra/2RaServer 2Queue 2Ra/2Ra =R= 10/hour Tp = 5 minutes Interarrival time Poisson Service time exponentialServer 1QueueServer 2Ra
38 Effect of Pooling : 2M/M/1 Server 1Queue 1Ra/2Ra/2 = R= 5/hourTp = 5 minutesc = 1 Rp = 12 / hourU= 5/12 = 0.417RaServer 2Queue 2Ra/2
39 Comparison of 2M/M/1 with M/M/2 Server 1Queue 1Ra/2RaServer 2Queue 2Ra/2Server 1QueueServer 2Ra
40 Effect of Pooling: M/M/2 Ra=R= 10/hour Tp = 5 minutes c = 2 Rp = 24 /hour U= 10/24 U= AS BEFORE for each processorServer 1QueueServer 2Ra
41 Effect of Pooling Under Design A, Under Design B, We have Ra = 10/2 = 5 per hour, and Tp = 5 minutes, c =1, we arrive at a total flow time of 8.6 minutesUnder Design B,We have Ra =10 per hour, Tp= 5 minutes, c=2, we arrive at a total flow time of 6.2 minutesSo 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 queueDesign B, the waiting time of customer is partially dependent on each preceding customer’s processing timeCombining queues reduces variability and leads to reduce waiting times
42 Performance Measures Financial Performance Measures Sales: Throughput RateCost: Capacity utilization, Number in queue / in systemCustomer service: Waiting Time in queue /in systemPerformance Improvement LeversDecrease variability in customer inter-arrival and processing times.Decrease capacity utilization.Synchronize available capacity with demand.
43 1. Variability Reduction Levers Customers arrival are hard to controlScheduling, reservations, appointments, etc….Variability in processing timeIncreased training and standardization processesLower employee turnover rate more experienced work forceLimit product variety, increase commonality of parts
44 2. Capacity Utilization Levers 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 optionsManage Arrivals: Decrease inflow rate RaManage Capacity: Increase processing rate RpManaging ArrivalsBetter scheduling, price differentials, alternative servicesManaging CapacityIncrease scale of the process (the number of servers)Increase speed of the process (lower processing time)
45 3. Synchronizing Capacity with Demand Capacity Adjustment StrategiesPersonnel shifts, cross training, flexible resourcesWorkforce planning & season variabilitySynchronizing of inputs and outputs, Better scheduling
49 Assignment 1: M/M/1 Performance Evaluation 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?
50 Assignment 2: M/M/1 Performance Evaluation What if the arrival rate is 11 per hour?
51 Assignment 3: M/G/cA 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 ?
52 Assignment 4: M/M/c Example 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 customer’s call is answered)?
53 Assignment 5: M/D/c Example Suppose the service time is a constantWhat is the answer of the previous question?