3.Process AnalysisNew diagram a (1).pptx

process analysis diagram and detail description

1. Process Analysis

2. 2 Overview Process: Introduction Process Types : Project, Job, Batch, Line and Continuous Process Analysis Processes Flow Chart Process and Flows – Important Concepts Cycle Time Throughput WIP Lead Time Process Capacity Bottleneck Batch Process Little’s Formula Rework, Parallel Path Effect of Variability on Performance of Process

3. What is a process? The basic building block of operations is the process. Operations as a Transformation Process The Transformation process Input transformed resources Material Information Customers Input transforming resources Facilities Staff Environment Environment OUTPUT INPUT Goods Services A process is a series or set of activities that interact to produce a result; it may occur once-only or be recurrent or periodic (Source: Wikipedia). Example: Let’s say you want to get a book from your book shelf. There are many small tasks involved such as getting up from your desk, walking to the shelf, opening the shelf door, identifying the book, picking it up, closing the shelf and going back to your desk. Each step described above is called a task. Processes exist to create value for the customers.

4. Processes in Value Chain A value chain stage such as, manufacturing is comprises of processes designed to add value by transforming relevant inputs into useful outputs. Processes involve tasks and associated resources. Resource Sets R1.1 R1.2 P1 P2 P3 Processes Supplier Mfg DC Value Chain R1.3 T1.1 T1.2 T1.3 Task Set, T1

5. Process Types Examples . Project Construction, Consulting Job Shop Machine Shop, Beauty Shop Batch Bakery, Classroom Line Flow Assembly Line, Cafeteria Line Continuous Flow Paper mill, Central heating Generic Process Types Processes are generically classified based on the product variety and product volume The classification is not so much based on technical attributes of the processes/ resources but from the “business dimensions” it has.

6. 6 Project Process Production of a large-scale, complex product. If a project produces physical goods then the product typically cannot be moved . Examples : construction projects such as a dam/building, software projects Features - Unique products built on site (products like s/w are exception) Coordination of large scale resources Processes are highly flexible Resources dedicated for the project and released at the completion of the project Firm sells capability and experience

7. 7 Jobbing Process A process designed to produce a series of small, single unit products. The same product may not be required in future or orders for the product are infrequent. Examples: Optometric glass making, Specialty machine shops Features - - Makes unique products like project but if involves physical goods then they are small and can be moved - Skilled worker is responsible - Wide product range, flexible process -Usually low investment in automated plant equipment ( mfg example) -material usually purchased only when order is received

8. 8 Examples: Jobbing Bank Machine Shop Hospital

9. 9 Batch Process A process where small groups of similar items are produced together. The product comes out from the process in groups at discrete point of times. Examples: most products manufactured ( 80% of the world's manufacturing ) Features - Process large batch at station before sending to next station Similar items on a repeat basis Wide range of volumes

10. 10 Examples: Batch Process Batch processing of chemical Garment

11. 11 (Connected) Line Processes A high-volume, high investment form of manufacturing designed to produce similar discrete products cheaply . Examples: Car production, Pharmaceuticals and Computers . Features - Dedicated to one product or family Standard products, high volumes Units processed & transferred in synchronized manner Processing can be broken into discrete steps and the stages balanced to synchronize flow

12. 12 Line Process Examples Car Production (Assembly Line) Mass Covid Vaccination (Line) Fast Food McDonald’s over 95 billion served

13. 13 Continuous (Flow) Processes A never-ending form of production Examples: Process industries- Paper, Chemicals, Petroleum. Features Material continuous & processing continuous (no discrete steps) Large volumes of uniform products Narrow range of standard products Highly interconnected Basic material processed in stages into one or more products

14. 14 Continuous Process Examples Paper (Continuous) Glass Making (Continuous)

15. 15 Exercise: Identify the process type used Example Process type Assembly of appliances and cars Running a political event Clothing Mfg. Toys Mfg. Developing a new product Specialist tool maker Furniture restores Processing student applications Installing an ERP package Manufacturer of Component parts Beer bottling plant Electricity utilities Steel making

16. 16 Continuum of Process Structures in Mfg. Type Stage connectivity Product flow Variety of products Product volume Process/ transfer unit Line Process Batch Process Jobbing Process Project Continuous process Unique One Unique One Tightly connected Fixed pattern Few Large Continuous repetitive Loosely connected or disconnected Moderate Moderate Batch Flexible pattern Many Small Process

17. 17 Continuum of Process Structures in Mfg. Type Stage connectivity Product flow Variety of products Product volume Process/ transfer unit Line Process Batch Process Jobbing Process Project Continuous process Unique One Unique One Tightly connected Fixed pattern Few Large Continuous repetitive Loosely connected or disconnected Moderate Moderate Batch Flexible pattern Many Small Process CAKE BREAD PASTRY KING SOOPER BAKERY SHOP

18. -Project -Job Shop -Batch -Line Flow -Continuous Flow Summary: Classification of Processes by Process Architecture These process structures differ in several respects such as: •Volume - ranging from a quantity of one to large scale mass production. •Number of products - ranging from the capability of producing a multitude of different products to producing only one specific product. •Flow - ranging from a large number of possible sequences of activities to only one possible sequence. •Flexibility - Changes may be changes in production volume or changes in the product mix. •Capital investment - ranging from using lower cost general purpose equipment to expensive specialized equipment. •Variable cost - ranging from a high unit cost to a low unit cost. • Labor content and skill - ranging from high labor content with high skill to low content and low skill.

19. 19 Process Types vs Layout Types ‘ The process type does not refer to the same things as the layou t (or physical arrangement)’ type . For example, consider a process comprising of three task sets represented by, T1, T2, T3 carried out using resources R1, R2 and R3 respectively. Is there a unique way of physically placing them in a facility? R1 R2 R3 T1 T2 T3 R1 R2 R3 T1 T2 T3 R1 R2 R3 T1 T2 T3 All these 3 systems are capable of making the same set of products occupying the same volume & variety positions.

20. 20 Variety of products and services How much Flexibility of the process; volume, mix, technology and design What type and degree Volume Expected output Job Shop Batch Repetitive/Line Continuous Questions Before Selecting A Process Architecture

21. 21 Product Variety Product Process Matrix High Low Product Volume Low High Project (Ship building, Movie production) Mass/Line (automobile, most consumer durables) Continuous (steel making, refinery) Jobbing (specialist toolmaker, speciality tailors) Batch (component parts of automobile, most clothing) It is a strategic tool for selection of processes. Correct Processes occupy positions along diagonal of volume variety matrix. Not a correct fit Not a correct fit

22. Process Analysis “ If you cannot describe what you are doing as a process, you do not know what you are doing. ” W.E. Deming

23. Process Analysis An operation is composed of processes designed to add value by transforming relevant inputs into useful outputs. The first step to improving a process is to map and analyse it: To understand the activities & their relationships To identify inefficient tasks to eliminate/improve To assess against the relevant performance measures. Map a process! Assess its elements against the relevant performance measures. Cost Flow Time Inventory Quality Etc.

24. Process Analysis: Broad Steps Define the process boundaries that mark the entry points of the process inputs and the exit points of the process outputs. Construct a process flow diagram that illustrates the various process activities and their interrelationships. Determine the capacity of each step in the process. Calculate other measures of interest. Identify the bottleneck, that is, the step having the lowest capacity. Evaluate further limitations in order to quantify the impact of the bottleneck. Use the analysis to make operating decisions and to improve the process.

25. Process Flow Charts Process flow diagram (or process flowchart ) is a valuable tool for understanding the process using graphic elements to represent tasks, flows, and storage. Rectangles: represent tasks Arrows : Represent flows of material and information. Material flow usually is represented by a solid line and information flow by a dashed line. Inverted triangles : represent storage (inventory). Storage bins commonly are used to represent raw material inventory, work in process inventory, and finished goods inventory. Diamond : Represents decision points.

26. Example of a Two-Stage Serial System A A A A B B B B 5 10 15 20 7 12 Gantt Chart A B 5 min 2 min Raw material Sub assembly Step-1 Step-2 to assembly

27. Simultaneous activities Different products produced Alternate Paths 30% 70% Examples: Non-Serial Tasks

28. Example of a two-stage Non-Serial System A1 B 2 min A2 5 min 5 min RawMat-1 Component-2 Component-1 RawMat-2 Assembled part A1 A1 A1 A1 5 10 15 20 7 12 A2 A2 A2 A2 5 10 15 20 B B B 17 B 22 Gantt Chart

29. How to Draw a Process Flow Diagram Focus on one or two types of flow units. Define the process boundaries and choose an appropriate level of detail. Include only those steps that are likely to affect the process flow or the economics of the process. Sizes and exact locations of arrows, boxes, and triangles do not carry any special meaning. Use different colors for different routes for clarity.

30. Case: Kristen's Cookie Co. (A) Task: Perform a process analysis to answer some questions related to making important business decisions in setting up facility, buying resources/equipment, pricing, order policy etc.

31. Process Capacity: The capacity is defined as the maximum output from the process or resource (e.g., worker, machine, factory, organization), measured in units produced per unit of time. Thus capacity can be found by dividing total available time by the processing time per unit. Example of Capacity of a Single Stage Process : If a welder can do a job in 1 minute, then his/her hourly capacity will be [1hr *60 (minutes/hr) ]/ (1 minute) = 60 units/hr Note : Effective capacity (in long term) refers to the output that the manager is most likely to attend in actual setting and this can be less than the maximum output rate due to factors such as maintenance, scheduling difficulties and worker availability. To account for this in setting long term capacity the time availability may be reduced by a factor. For instance, welder’s weekly capacity in above example (1-8 hr shift x 5 working day) may be say 60 *40* 0.9 =2160 units/week (instead of 60 *40=2400 units/week). Process Performance Measures What do we need to measure in order to understand how the designed process will perform? Task 1 Task 2 Task 3

32. Capacity of multi-stage process: concept of bottleneck Q1. Which task is the bottleneck for line with given task times? 20 minutes 36 minutes 30 minutes The bottleneck is defined as the process step (station) in the flow diagram with the lowest capacity (the “weakest link”). Although the bottleneck is often the process step with the longest processing time, it is important to always look at the capacities for making a judgement (why?) . Q3. What is the capacity of this process? The capacity of the multi-stage process is bottleneck capacity, hence, 2 units/ hr 3 units/ hr 5 units/ hr 2 units/ hr # workers used Q2. What is the capacity of different stage?

33. Throughput or throughput rate or flow rate– It is the actual total volume of production through a facility ( e.g., worker, machine, factory, organization) per unit time or the average rate at which units flow past a specific point in the process. Capacity utilization – It is a measure how intensely a process is being used to produce goods/services. Method 1 (Based on time activation). If a worker is busy 45 minutes every hour has a utilization of 75% (=45/60*100). Utilization =Time worked/ Time Available Method 2 (Based on output quanitity ). Alternatively, utilization can be calculated as the actual throughput divided by capacity. Utilization =Throughput or Flow Rate / Capacity A worker capable of producing 60 units per hr who only produces 30 hrs per hr has utilization = 30/60*100 = 50%. Process Performance Measures Remark: Realised Capacity is throughput, hence at maximum capacity realisation or throughput rate, utilization is 100%.

34. Cycle time (CT) – It is the time gap between successive units that are output from the process. Cycle time can be thought of as the time required for a task to repeat itself. Example of single stage process : Cycle time of a which gives output of 30 units per hour will be 2 minutes/unit. Process Performance Measures 2 minutes per unit 3 minutes per unit 5 minutes per unit 1 minute per unit 4 minutes per unit Multi-step process: A process consists of 5 steps carried out in 5 different work centres with each assigned one dedicated worker A B C D E Cycle Time of each step The cycle time of the process is equal to the cycle time of the bottleneck task T hus , in this Example CT of the process = 5 minutes.

35. Cycle Time & Throughput Relation Cycle Time : Average time for completion of a unit at a production step or process. Measured as time/unit . [Note: different kinds of products require different processing, hence averaging] Throughput Rate : Average number of units processed over a unit time interval. Measured as units/time 1 Cycle Time Throughput rate = Key relationship Capacity = (maximum possible) throughput rate = throughput rate at minimum possible Cycle Time) [Note: 1) computed this way, the maximum possible throughput rate or Minimum possible Cycle Time through a process stage measures its capacity] 2) Takt Time = target cycle time to meet throughput demand of usually a day (say, 480 units production demand in an 8-hr shift will give a takt time = 8*60/480= 1 min)

36. Cycle Time & Capacity of Each Step 2 minutes per unit 3 minutes per unit 5 minutes per unit 1 minute per unit 4 minutes per unit 30 units per hour 20 units per hour 12 units per hour 60 units per hour 15 units per hour Capacities for Each Step (work Centre) of the Process A process consists of 5 steps carried out in 5 different work centres with each assigned one dedicated worker A B C D E Capacity Minimum Cycle Time Note: Realised Capacity is throughput. How much capacity is likely to be realised? For process? At each step?

37. Cycle Time of the Process: Dedicated Operators Transient to Steady Sate 2 min per unit 3 min per unit 5 min per unit 1 min per unit 4 min per unit 30 units per hour 20 units per hour 12 units per hour 60 units per hour 15 units per hour A B C D E CAPACITY THROUGHPUT RATE 20 units per hour 12 units per hour 12 units per hour 12 units per hour 30 units per hour 20 units per hour 12 units per hour 12 units per hour 12 units per hour 20 units per hour 12 units per hour 12 units per hour 12 units per hour 12 units per hour 12 units per hour Transient Steady State (system stablized at bottleneck rate …… …… …… …… …… Time

38. Capacity and cycle time with Multiple Resources/Servers Suppose for a step, the Best (minimum possible) Cycle Time = T time per unit Capacity of a single server = 1/T units per time unit If this step has “m” identical servers in parallel Capacity = m/T units per time unit Output 1 unit every T/m time units Cycle Time = T Cycle Time = T for each booth

39. Flow time or throughput time: the average time that a unit requires to flow through the process from the entry point to the exit point. The flow time is the length of the longest path through the process. Flow time includes both processing time and any time the unit spends between steps. Process time – This is the value added time i.e., the average time that a unit is worked on. Process time is flow time less idle time. Idle time - time when no activity is being performed, for example, when an activity is waiting for work to arrive from the previous activity. The term can be used to describe both machine idle time and worker idle time. Flow time, Process Time and Idle Time A B C D t=10 t=10 t=0 t=20 t=40 t=50 t=30 t=40 t=30 t=20

40. Flow time, Process Time and Idle Time A B 5 min 2 min Raw material Finished product Step-1 Step-2 A Two Step Process Product thru Step A Thru B 5 10 Idle Time 3 min 8 Gantt Chart of Processing of a Product Flow Time (10 min) Process Time or Value Added Time 5 + 2 = 7 min 5 min 2 min Waiting to move

41. 2 min per unit 3 min per unit 5 min per unit 1 min per unit 4 min per unit 30 units per hour 20 units per hour 12 units per hour 60 units per hour 15 units per hour A B C D E Cycle Time & Throughtput : Dedicated Operators CAPACITY THRUPUT RATE 12 units per hour 12 units per hour 12 units per hour 12 units per hour 12 units per hour Steady Sate Q1. What is the Value Added Time? Value added time (Process Time) = 2+3+5+1+4=15 minutes per unit Q2. At Steady State determine, Cycle Time and Capacity of the Entire Process with dedicated Operator? Cycle Time of Process = 5 minutes per unit Process capacity = throughput rate of bottleneck= 12 units per hour Cycle Time A process consists of 5 steps carried out in 5 different work centres with each assigned one dedicated worker

42. Cycle Time & Throughput of the Process: Single Operator 2 minutes per unit 3 minutes per unit 5 minutes per unit 1 minute per unit 4 minutes per unit What are value added time & the Cycle Time of the Entire Process with Single Operator? A process consists of 5 steps carried out in 5 different work centres with just one worker A B C D E Throughput Time = Process Time (15 min) + Waiting Time (0 min)=15 min Process Time or Value Added Time = 2+3+5+1+4=15 minutes per unit Cycle Time = 15 minutes per unit (will repeat after 15 minutes) Process Capacity or Max Throughput Rate = 60/15 = 4 units per hour

43. Flow time or throughput time: the average time that a unit requires to flow through the process from the entry point to the exit point. The flow time is the length of the longest path through the process. Flow time includes both processing time and any time the unit spends between steps. Lead time - Time between initiation of order and receipt of goods. Sometimes this is used for process also like process lead time then it amounts to be the same as flow time. Flow time, Process Time and Idle Time Task 1 Task 2 FGI Task 1 Task 2 Customer Order MTO MTS Customer Order Customer Order Cycle (Lead Time) Customer Order Cycle (Lead Time)

44. Direct Labor Capacity Utilization Idle time - time when no activity is being performed, for example, when an activity is waiting for work to arrive from the previous activity. The term can be used to describe both machine idle time and worker idle time. Direct labor content ( value added time ) - the amount of labor (in units of time) actually contained in the product. Excludes idle time when workers are not working directly on the product. Also excludes time spent maintaining machines, transporting materials, etc. Direct labor utilization - the fraction of labor capacity that actually is utilized as direct labor .

45. 2 min per unit 3 min per unit 5 min per unit 1 min per unit 4 min per unit 30 units per hour 20 units per hour 12 units per hour 60 units per hour 15 units per hour A B C D E Capacity and Direct Labor Utilization CAPACITY: Labour THRUPUT RATE 12 units per hour 12 units per hour 12 units per hour 12 units per hour 12 units per hour Steady Sate Cycle Time DIRECT LABOR UTILIZATION 12/30*100 = 40% 12/20*100 = 60% 12/12*100 = 100% 12/60*100 = 20% 12/15*100 = 80% UTILIZATION = THRUPUT RATE/ LINE CAPACITY

46. 2 min per unit 3 min per unit 5 min per unit 1 min per unit 4 min per unit A B C D E Capacity and Direct Labor Utilization Process Cycle Time Cycle Time (each stage) DIRECT LABOR UTILIZATION 2/5*100 = 40% 3/5*100 = 60% 5/5*100 = 100% 1/5*100 = 20% 4/5*100 = 80% UTILIZATION = STAGE CYCLE TIME/ LINE CYCLE TIME 5 min per unit 5 min per unit 5 min per unit 5 min per unit 5 min per unit

47. Practice Problem Consider the following process. All steps (A, B, C, D, and E) are necessary to create each finished unit. Each step employs a single worker . Task times are shown for each step. 1. What is the fastest a rush order for one unit can go through the process? 10 min + 4 min + 12 min + 5 min = 31 min. 2. Working eight hours a day, what is the daily capacity of the process? (8 hrs/day*60 min/hr)/12 min/unit = 40 units/day.

48. Practice Problem Consider the following process. All steps (A, B, C, D, and E) are necessary to create each finished unit. Each step employs a single worker . Task times are shown for each step. 3. During the day, what is the average % labor utilization of the five workers? - Labor time used ( labor content) = 5 + 10 + 4 + 12 + 5 = 36 min. - Labor time available per cycle =(the cycle time of the process)*(the number of workers) = 12 min*5 workers = 60 min. -Utilization = Labor time used ( labor content) / Labor time available per cycle = 36/60 = 60%.

49. Computing Cycle Times in Batch Process Batch Process - The number of units of the same product type produced before changing to different product type. Batch Size- Total Quantity of the same type produced together is called batch size. A process which has a batch size greater than 1 is a batch process. Set-up time - the time required to prepare the equipment to perform an activity on a batch of units. Run time- time to process one unit of a product one after the other in the batch Product A (Batch size=5, Parts {A1, A2, A3, A4, A5} Product B (Batch size=3, Parts={B1, B2, B3} SETUP for A RUN A1 RUN A2 RUN A3 RUN A4 RUN A5 SETUP for B RUN B1 RUN B2 RUN B3 8 min 2 min 2 min 2 min 2 min 2 min 12 min 1.5 min 1.5 min 1.5 min

50. Computing Cycle Times in Batch Process Processing a Batch of Car Cycle Time = Set-up Time + (Batch size) x (Time per unit) Batch size Example: Producing 100 cars. It takes 50 hours to set up the production line. On average, production takes 5 hours per car. Cycle Time = Avg. Time to Process One Unit of Product Car (Batch size=100 Parts) SETUP RUN P1 RUN P2 RUN P3 RUN P4 RUN P5 ….. RUN P98 RUN P99 RUN P100 50 hr 5 hr 5 hr 5 hr 5 hr 5 hr ….. 5 hr 5 hr 5 hr Cycle Time = (50+100*5)/100 = 550/100= 5.5 hr 50+100*5= 550

51. Setup time: 15 min A B Question: What is the cycle time between points A and B of the process, if we work in batches of 10? And for batches of 30? Production Time: 2 min/unit Computing Cycle Times in Batch Process Cycle Time for Batch of 10 = (15+10*2)/10 = 35/10= 3.5 min Cycle Time for Batch of 30 = (15+30*2)/30 = 75/30= 2.5 min

52. How do we analyze a complex process… Look at the process step by step Determine maximum throughput rate (i.e. capacity) of each step Identify the process bottleneck (smallest capacity). The capacity of the process is equal to the capacity of the bottleneck

53. Example : hammer production process Description Work begins at the machining center. Here two lines form the heads of the hammers and place them in a buffer. Handles are attached at the assembly step. Finished hammers are sent to the next stage, where they are packed and shipped. assembly pack and ship machining machining WIP WIP

54. Process Data: Machining (each line) : Set up 80 min. 4 min per unit processing. Batch size 200 . Two identical lines. Assembly : Manual by team of two workers ( NO set up ). Each hammer requires 40 min processing. 34 workers available. Pack and ship : 30 min set up, 2 min per unit processing. Lot sizes of 100 . assembly pack and ship machining machining WIP WIP Let’s analyze the hammer process…

55. Step 1: Machining Look at any one line. 200 units require: 80 + 200  4 = 880 minutes per 200 units The throughput rate is: 200 / 880 = 0.227 units/minute = 13.63 units/hour But we have two identical lines , so for the machining step capacity is 2  13.63 = 27.26 units/hour.

56. Step 2: Assembly 1 unit requires 40 min processing time, so the throughput rate is per team of two worker: 1 unit / 40 min = (1/40)*60 units/ hr per team = 1.5 units/ hr 34 workers available, but 2 workers are required for each unit, so they can form 17 teams and thus assembly capacity is: 17  1.5 = 25.5 units/ hr

57. Step 3: Pack and ship Similar to machining: 30 + 100  2 = 230 min per 100 units Pack & ship capacity is: 100 / 230 = 0.43 units/min = 26.09 units / hr

58. Hammer process: what is the capacity? Process Step Capacity (units/hr) Machining 27.26 Assembly 25.50 Pack & Ship 26.09 Assembly is the bottleneck!

59. Case: Circuit Board Fabricators, Inc.

60. Throughput, Throughput time and Work in Process Inventory: Little’s Law Throughput Time (THT) : Average time that a unit takes to go through the entire (single/multi-stage) process including waiting time . Measured as time. This measure concerns the product delivery time promising. Work in Process (WIP) Inventory : Average number of units in system (or part of the system) over a time interval. Measured as units. This measure concerns planning of waiting area and inventory budget. Throughput /flow rate (TH)– Actual output through a facility per unit time or the average rate at which units flow past a specific point in the process. This measure concerns the product delivery quantity promising. Key relationship Little’s Law W.I.P. Inventory  =  Throughput Rate  x  Throughput Time

61. Little’s Law TH TH Queue Processing Job Throughput Time, THT Number of jobs in the system, WIP =TH x THT Input Output Input Rate= Input Rate = Throughput, TH W.I.P. Inventory  =  Throughput Rate  x  Flow Time WIP = TH x THT or L = λ x W (in queuing notation) This relation is known as Little's Law , named after John D.C. Little who proved it mathematically in 1961.

62. Little’s Law TH TH Queue Processing Job Queue Time Job Time Throughput Time, THT Number of jobs in the system, WIP =TH x THT Input Output Input Rate= Input Rate = Throughput, TH WIP queue = TH x Queue Time WIP job = TH x Job Time

63. Assumption: A Stable Process In a Stable Process Average inflow rate is the same as average outflow rate, say TH . For the above to happen, process capacity must be greater than average inflow rate, while average inventory is not zero. Ideally, TH should be equal to the customer demand . Average inflow rate is 600 passengers/ hr , or 10 passengers/min Process Capacity Scanner can handle 12 passengers per minute It can handle inflow Is the Security Checkpoint a stable Process? Can we have Process Capacity 9 instead of 12 in a stable process? If we have Process Capacity 20 instead of 12, could we still have people waiting in the line?

64. 7 A fast food restaurant processes on the average 5000 lbs. of hamburger per week. The observed inventory level of raw meat, over a long period of time, averages 2500 lbs. Determine inventory turnover ratio. Average inventory level, WIP = 2500 lbs Throughput, TH = 5000 lbs /week - Average time spent by a pound of meat in inventory, THT=WIP/TH= 2500/5000 = 1/2 week; -1/THT= 2 times per week = 2*52/ yr =104 / yr is the inventory turnover ratio Little’s Law Example: Inventory Management

65. 7 Little’s Law Example: Services Management A restaurant processes on average 1500 customers per day (=15 hours). On average, there are 50 customers waiting to place an order, waiting for an order to arrive or eating. TH=1500 customers/day=100 customers/hour; WIP= 50 customers; Counter Eating Time in Counter Eating Time Time in Restaurant, THT Number of customers in the system, (WIP =TH x THT=50 customers) Input Output Input Rate= Input Rate = Throughput, TH=150/day= 100/ hr WIP c = TH x Queue Time WIP e = TH x Job Time b) Out of the 50 customers, 40 customers on the average are eating. Find avg. wait time at counter. Find average time spent in the restaurant. THT=WIP/TH= 50/100 = 1/2 hours, average time in the restaurant. TH= 100, WIPc = 50−40 = 10 customers at the service counter; THTc = WIPc /TH= 10/100 hours = 6 minutes average wait at the counter

66. 7 Little’s Law Example: Workforce Management Thus, the company hires an average of 194.6 + 144 = 338.6 new employees per month, TH=338.6×12 = 4063.2 new employees / year. OR 4063.2/36,000×100% = 11.28% labor turnover during a year A certain Japanese company has 36,000 employees, 20% of whom are women. The average term of employment for a woman is 37 months, whereas for men it is 200 months. Assume that the total employment level and the mix of men and women are stable over time. Determine employee turnover. WIPw = average no of women in system = 36,000×0.2 = 7,200 woman THw = 7200/37= 194.6 women/ month is the average number of new women employees hired per month. WIPm = average number of men in system = 36,000×0.8 = 28,800 men, THm =28,800/200= 144 men/month .

67. Little’s Law Example: Monetary Flow. For the new euro introduction in 2002, Wim Duisenberg had to decide how many new Euro coins to stamp by 2002. Euroland’s central banks’ cash-in-coins handling was estimated at €3 00 billion per year. The average cash-in-coins holding time by consumers and businesses was estimated at 2 months. How many Euro coins were to be made? TH= €3 00 billion per year ; THT = 2 months WIP = THxTHT = 300/ yr * 2mo = 300/ yr * 1/6 yr = 50 Euro’s: NYT 2001: “As of Jan. 1, some 50 billion new coins and 14.5 billion euro notes are to be pumped into circulation.” 9

68. Little’s Law -- Caveats Applies to the long run average of a stable system In any given time period (sample) the average may be different (especially for small samples) In an unstable, or dynamic system, the average may not be very useful In systems with variance, we often need to know about more than the average

69. Practice Problem: Process of Car Repair in a Garage Car Checks in Inspection and Estimates Work Order Preparation Repair Process Wait Wait 4 cars per hr 5 min 2 cars 25% quit 75% stay 5 min 12 min 1 car 30 min Question1. What is the average flow time for: (a) customers quitting the process? (b) customers staying & getting car repaired? (c) an average customer Question2. Show Little’s Law can be applied to parts of the system as well as at the whole system level. Question1. Car Checks in Inspection and Estimates Work Order Preparation Repair Process Wait Wait TH= 4 cars per hr 5 min 2 cars 25% quit 75% stay 5 min 12 min 1 car 30 min TH= 4 cars/ hr WIP= 2 cars THT= WIP /TH = 2/4 =0.5 hrs = 30 min TH= 3 cars/ hr WIP= 1 car THT= WIP/TH = 1/3 =0.33 hrs = 20 min THquit = 1 car/ hr THstay = 3 car/ hr THTquit = 5 + 30 + 12 = 47 minutes THTstay = 47+5+20+30 =103 minutes THTavg = 0.25 *47+ 0.75*103 = 88.25 minutes

70. Practice Problem: Process of Car Repair in a Garage Question1. What is the average flow time for: (a) customers quitting the process? (b) customers staying & getting car repaired? (c) an average customer Question2. Show Little’s Law can be applied at component level (for parts) as well as at the systems level (whole). Question2. Car Checks in Inspection and Estimates Work Order Preparation Repair Process Wait Wait TH= 4 cars per hr 5 min 2 cars 25% quit 75% stay 5 min 12 min 1 car 30 min TH= 4 cars/ hrs THT= 5/60 hrs WIP= THT*TH = 1/3 cars THq = 1 car/ hr THs= 3 car/ hr TH= 4 cars/ hrs THT= 12/60 hrs WIP= THT*TH = 4/5 cars TH= 3 cars/ hrs THT= 5/60 hrs WIP= THT*TH = 1/4 cars TH= 3 cars/ hrs THT= 30/60 hrs WIP= THT*TH = 3/2 cars Method 2: TH= 4 cars/ hrs (inbound rate) = 1 cars/ hrs + 3 cars/ hrs (outbound rate) THTaverage = 88.25 minutes (from previous slide) = 1.4708 hrs WIP =THT x TH ? Method 1): WIP= 1/3 + 2 + 4/5 + 1/4+ 1 + 3/2 = 5.8833 WIP = THTaverage x TH = 1.4708 hrs * 4 cars / hr = 5.8833 Cars

71. Make-to-order (MTO) vs. make-to-stock (MTS) Task 1 Task 2 FGI Task 1 Task 2 Customer Order Make-to-Stock: Demand is satisfied by FGI (finished goods inventory). Highly standardized products made for finished goods inventory. Important metric for customer service? Make-to-Order : Process for making customized products to meet individual customer requirements. Important metric for customer service? MTO MTS Customer Order Other Process Terminology Customer Order Cycle Customer Order Cycle

72. Buffering : Keep some inventory between stages 1 1/2 Starving : Stoppage of activity because of lack of material Blocking : Stoppage of flow because there is no storage place 1 0/2 1 2/2 1 Other Process Terminology: (relevant for unbalanced flow) Current/Maximum Resource working (1) /not working (0) Resource working (1) /not working (0)

73. Examples.. CT = 3s CT = 1s FGI Task 1 Task 2 Let’s study this make-to-stock system. What is the capacity of the process? What is the throughput time? What is the average WIP? Is any task starved or blocked? Note: No buffer space between stations, so upstream station has to wait if downstream station is busy Other Process Terminology: (relevant for unbalanced flow)

74. Examples.. CT = 3s CT = 1s FGI Task 1 Task 2 Task 2 starved for 2s. each time. Throughput rate = 20 units/min at Task 1, 60 units/min at Task 2 Capacity (throughput rate) of process = 20 units/min Throughput time = 4 seconds (3s Task 1+ 0 wait + 1s Task 2) = 1/15 min WIP = Throughput rate x Throughput time = 20 units/min x 1/15 min = 1.33 units Other Process Terminology : (relevant for unbalanced flow)

75. Examples.. CT = 1s CT = 3s FGI Task 1 Task 2 What is the capacity of the process? What is the throughput time? What is the average WIP? Is any task starved or blocked? Let’s study this make-to-stock system: Note: No buffer space between stations, so upstream station has to wait if downstream station is busy Other Process Terminology: (relevant for unbalanced flow)

76. Examples.. CT = 1s CT = 3s FGI Task 1 Task 2 Task 1 blocked for 2s. each time. Throughput rate = 60 units/min at Task 1, 20units/min at Task 2 Capacity of process = 20 units/min Throughput time = 6 seconds (1s Task 1+ 2s wait + 3s Task 2) = 0.1 min WIP = Throughput rate x Throughput time = 20 units/min x 0.1 min = 2 units Other Process Terminology: (relevant for unbalanced flow)

77. Examples.. CT = 3s CT = 3s FGI Task 1 Task 2 What is the capacity of the process? Is any task starved or blocked? Let’s study this make-to-stock assembly system: Note: No buffer space between stations CT = 4s Task 3 CT = 2s Task 4 Other Process Terminology: (relevant for unbalanced flow)

78. Example.. CT = 3s¸ CT = 3s FGI Task 1 Task 2 CT = 4s Task 3 CT = 2s Task 4 Tasks 1 and 2 are blocked by Task 3 for 1 second per product . Task 4 is starved for 2 seconds per product. The capacity of the process is 15 units/hour (limited by Task 3). Other Process Terminology: (relevant for unbalanced flow)

79. 79 The task is to calculate the average cycle time for an entire process or process segment Assume that the average activity times for all involved activities are available In the simplest case a process consists of a sequence of activities on a single path The average cycle time is just the sum of the average activity times involved … but in general we must be able to account for Rework Multiple paths Parallel activities Rework, Multiple Paths

80. 80 Many processes include control or inspection points where if the job does not conform it will be sent back for rework The rework will directly affect the average cycle time! Definitions T = sum of activity times in the rework loop r = fraction of jobs requiring rework (rejection rate) Assuming a job is never reworked more than once (one pass) Assuming a reworked job is no different than a regular job (infinite pass) Rework CT = (1+r)T CT = T/(1-r) Hint: In infinite geometric series with first term 1, Sum = 1/(1-r) 1-r = 0.8 r=0.2 A T=10 B (20) 1 unit

81. 81 Example – Rework effects on the average cycle time Consider a process consisting of Three activities, A, B & C taking on average 10 min. each One inspection activity (I) taking 4 minutes to complete. X% of the jobs are rejected at inspection and sent for rework What is the average cycle time? If no jobs are rejected and sent for rework. If 25% of the jobs need rework but never more than once. If 25% of the jobs need rework but reworked jobs are no different in quality than ordinary jobs. 0.75 0.25 A (10) B (10) C (10) I (4)

82. 82 Multiple Paths (alternatives) M1 (2 min/part) M2a (5 min/part) M2b (4 mi/part) M1 (12 mins) M2a (12 mins) M2b (12 mins) Processing individual parts (consider 12 minutes time interval) 2.4 parts 3 parts 6 parts CT =? CT=12/5.4 =2.22 min/part (if both are evenly used)

83. 83 It is common that there are alternative routes through the process For example: jobs can be split in “fast track”and normal jobs Assume that m different paths originate from a decision point p i = The probability that a job is routed to path i T i = The time to go down path i Multiple Paths CT = p 1 T 1 +p 2 T 2 +…+ p m T m =

84. 84 Example – Processes with Multiple Paths Consider a process segment consisting of 3 activities A, B & C with activity times 10,15 & 20 minutes respectively On average 20% of the jobs are routed via B and 80% go straight to activity C. What is the average cycle time? 0.8 0.2 A (10) B (15) C (20)

85. 85 Parallel Operation (work shared by resources) M1 (2 min/part) M2a (5 min/part) M2b (4 mi/part) M1 (10 mins) M2a (10 mins) M2b (12 mins) Batch Processing (Batch size= 5 parts) 2 parts 3 parts 5 parts CT = 12 minutes per batch = 12/5 = 2.4 min /part CT = Max (10, 12) = 12 minutes per batch CT= 10 min/batch

86. 86 If two activities related to the same job are done in parallel the contribution to the cycle time for the job is the maximum of the two activity times. Assuming M process segments in parallel T i = Average process time for process segment i to be completed Processes with Parallel Activities CT parallel = Max{T 1 , T 2 ,…, T M }

87. 87 Consider a process segment with 5 activities A, B, C, D & E with average activity times: 12, 14, 20, 18 & 15 minutes What is the average cycle time for the process segment? Example – Cycle Time Analysis of Parallel Activities A (12) B (14) C (20) D (18) E (15)

88. X-Ray Service Process 1. Patient walks to x-ray lab 2. X-ray request travels to lab by messenger 3. X-ray technician fills out standard form based on info. From physician 4. Receptionist receives insurance information, prepares and signs form, sends to insurer 5. Patient undresses in preparation of x-ray 6. Lab technician takes x-ray 7. Darkroom technician develops x-ray 8. Lab technician checks for clarity-rework if necessary 9. Patient puts on clothes, gets ready to leave lab 10. Patient walks back to physicians office 11. X-rays transferred to physician by messenger

89. Example: X-Ray Service Process 3 2 1 4 7 6 5 11 10 9 start end 25% 75% 7 20 6 5 3 6 12 2 20 3 7 transport support Value added decision Data: Measured actual flow time: 154 minutes 8 1. Patient walks to x-ray lab 2. X-ray request travels to lab by Messenger 3. X-ray lab technician fills out standard form 4. Receptionist processes insurance information 5. Patient undresses in preparation of x-ray in Changing room 6. Lab technician takes x-ray 7. Darkroom technician develops x-ray in dark room 8. Lab technician checks for clarity-rework if necessary 9. Patient puts on clothes, gets ready to leave lab 10. Patient walks back to physicians office 11. X-rays transferred to physician by messenger (6+12+2)*1.25=25 Max(7,20+6)=26 path 2-3 Max(3+7,20)=20 path 11 Path: 2-3-4-5-6-7-8-11: Time: 26+5+3+25+20 =79 minutes

90. Most time inefficiency comes from waiting: E.g.: Flow Times in White Collar Processes Industry Process Average Flow Time Theoretical Flow Time Flow Time Efficiency Life Insurance New Policy Application 72 hrs. 7 min. 0.16% Consumer Packaging New Graphic Design 18 days 2 hrs. 0.14% Commercial Bank Consumer Loan 24 hrs. 34 min. 2.36% Hospital Patient Billing 10 days 3 hrs. 3.75% Automobile Manufacture Financial Closing 11 days 5 hrs 5.60%

91. X-Ray Service Process Resource Pool Resource Time Consumed / Unit Load Load Batch Theoretical Capacity of Res. unit No of units in pool Theoretical capacity of pool Messenger 20+20 min/patient 1 60/40=1.5 patients/hr 6 1.5(6)=9 Patient/hr Receptionist 5 1 60/5=12 1 12 X-ray technician 6+7.5+2.5 1 60/16=3.75 4 15 X-ray lab 6+0.25(6)=7.5 1 60/7.5=8 2 16 Darkroom technician 12+0.25(12)=15 1 60/15=4 3 12 Darkroom 12+0.25(12)=15 1 60/15=4 2 8 Changing room 3+3 1 60/6=10 2 20

92. Utilizations Data: actual observed throughput of 5.5 patients/ hr Resource pool Theoretical capacity Patients/hr Capacity utilization Messenger 9 61.11 Receptionist 12 45.83 X-ray technician 15 36.67 X-ray lab 16 34.38 Darkroom technician 12 45.83 Darkroom 8 68.75 Changing room 20 27.50

93. Effect of Variability on Performance of Process

94. Understand Variability: using Queueing Formula Queueing Parameters l= r a = the rate of arrivals in customers (jobs) per unit time ( t a = 1/ r a = the average time between arrivals). c a = the coefficient of variation (CV) of inter-arrival times. m = the number of machines. r e = the rate of the station in jobs per unit time = m/t e . ( t e = the average processing time). c e = the CV of processing times. u = utilization of station = l a /r e . Note: a station can be described with 5 parameters.

95. Queueing Measures Measures: W q = the expected waiting time spent in queue. W = the expected time spent at the process center, i.e., queue time plus process time. L = the average WIP level (in jobs) at the station. L q = the expected WIP (in jobs) in queue. Relationships: W = W q + t e L = r a  W L q = r a  W q Result: If we know W q , we can compute L, L q , W.

96. The G/G/1 Queue Formula: Observations: Useful model of single machine workstations Separate terms for variability, utilization, process time. W q (and other measures) increase with c a 2 and c e 2 Flow variability, process variability, or both can combine to inflate queue time. Variability causes congestion!

97. The G/G/m Queue Formula: Observations: Useful model of multi-machine workstations Extremely general. Fast and accurate. Easily implemented in a spreadsheet

98. Lead Time vs. Utilization Lead Time ( hrs )

99. Thanks