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Flow Time and Theoretical Flow Time

Flow Time and Theoretical Flow Time. 1: Flow Time 2: Paths on a Network of Activities and the Critical Path 3: Flow Time Levers 4. Critical Path Method and Slacks. 1. Competitive Advantages of a shorter flow time. Shorter response time ( both in production and product design ).

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Flow Time and Theoretical Flow Time

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  1. Flow Time and Theoretical Flow Time 1: Flow Time 2: Paths on a Network of Activities and the Critical Path 3: Flow Time Levers 4. Critical Path Method and Slacks Ardavan Asef-Vaziri

  2. 1 Ardavan Asef-Vaziri

  3. Competitive Advantages of a shorter flow time • Shorter response time (both in production and product design). • Delayed differentiation (Postponement). Moving from the extreme of MTS towards MTO. • Lower inventory costs (due to Little’s Law). Ardavan Asef-Vaziri

  4. Lean Operations: The Real Cost of Inventory Inventory adversely affects all competing edges (P/Q/V/T) • Has cost: • Physical carrying costs • Financial costs • Causes obsolescence: • Due to market changes • Due to technology changes • Leads to poor quality: • Feedback loop is long • Hides problems: • Unreliable suppliers. • High defect rate. • Long tools changeover times. • Frequent machine breakdowns. • Causes long flow time Ardavan Asef-Vaziri

  5. Direct Methods of Measuring Flow Time • Randomly sample flow units over an extended period of time. • Measure the flow time for each flow unit from entry to exit. • Compute the average of flow times. • During a given month, a sample of 50 applications was taken • The average flow time = 20.85 working days. Ardavan Asef-Vaziri

  6. Indirect Methods of Measuring Flow Time • Count the number of units produced over an extended period of time. • R = number of units produced / duration of time period . • Count the number of units of inventory at random points during the time period. Compute the average inventory (I). • Compute flow time T =I/R 200 applications processed during 20 days; average throughput of R=200/20 = 10 applications per day. The number of applications were counted at 4 random points during these days, the average inventory (I) was 860/4 =215. T = I/R  T = 215/10  flow time is 21.5 days Ardavan Asef-Vaziri

  7. Flow Time Example The “Steal a Deal” gift shop specializes in heavily discounted merchandise for the holiday season. The store prices are so appealing that lines of people queue up in the early AM hours in front of the store doors in order to secure a place in the line. On the morning after Thanksgiving, one of the most busy days of the year, the store opened its doors at 8:00AM. From 7:45 to 8:00, 120 customers arrive and are already waiting in line. From 8:00AM to 10:00AM, new customers arrive at the rate of 5 per minute. After 10:00AM, the rate reduces to 2 per minute. The store admits customers at the rate of 4 per minute. Ardavan Asef-Vaziri

  8. Schematic Representation of the Flow Dynamics Rp<Ra=5/min Rp= 4/min Ra-Rp=1/min 120 Rp-Ra=2/min 7:15 8:00 10:00 12:00 9:00 Ardavan Asef-Vaziri

  9. Dynamics of Inventory and Waiting Time a) How many customers are in the waiting line at 10 AM? 120 at 8:00 Buffer increases at rate of 1/min. At 10:00 thee are 120+1(120) = 240 b) Jacob plans to arrive to the store at 9:00AM. How long should he expect to wait? 120 at 8:00 Buffer increases at rate of 1/min. At 9:00 thee are 120+1(60) = 180 Ardavan Asef-Vaziri

  10. Dynamics of Inventory and Waiting Time Buffer is served at rate of 4/min 180/4 = 45 min. It takes Jacob 45 min to get in 9:00 + 45 min = 9:45 c) Rachel does not want to wait more than 15 minutes. When should she show up (in terms of the number of customers in the waiting line)? Buffer is served at rate of 4/min In 15 min they can serve 15(4) = 60. When there are 60 customers in line. Ardavan Asef-Vaziri

  11. Dynamics of Inventory and Waiting Time d) Rachel does not want to wait more than 15 minutes. When should she show up (in terms of time)? At 10:00 there are 240 customers in the waiting line. Rachel should arrive when there are no more than 60 customers in line because those in front of her are served at rate of 4/ min. 2 new customers per minutes arrive after 10:00 AM. Customers are reduced at rate of (4-2) = 2/min (240-60)/2 = 90 min 10:00 + 90 min = 11:30 Ardavan Asef-Vaziri

  12. Average Inventory e) Compute the average inventory from 7:45AM to 12:00PM. Inventory starts from 0, goes up to 120 in 15 min. Therefore during 0.25 hour the average inventory is (0+120)/2 = 60. Inventory then starts from 120, goes up to 240 in 2 hours. Therefore, during 2 hour the average inventory is (120+240)/2 = 180. Then inventory goes down from 240 to 0 in 240/2 = 120 min = 2 hour; at 12:00. Therefore during 2 hours the average inventory is (240+0)/2 = 120. The average inventory 0.25(60) + 2(180) + 2(120) = 615 Ardavan Asef-Vaziri

  13. Average Inventory We should divide this by the sum of the relative weights; 0.25 + 2 + 2 = 4.25 The average inventory (customers in the line) = 615/4.25 = 144.7 Average inventory is the same, regardless the unit of time used (for hours, minutes and in this case, 4.25 hours). Ardavan Asef-Vaziri

  14. Average Throughput f) Compute the average throughput from 7:45AM to 12:00PM. From 7:45 to 8:00; 0.25 hours 120 customers arrive From 8:00 to 10:00; 2 hours (2×60)(5) = 600 customers arrive From 10:00 to 12:00; 2 hours (2×60)(2) = 240 customers arrive A total of 120+5(60×2)+2(120) = 960 arrive over the time interval of 7:45 to 12:00PM, that is 4 hours and 15 min. or in 4.25 hours. R is 960 customers per 4.25 hours. R = 960/4.25 = 225.882 per hour. R = 225.882/60 = 3.76 per min. Ardavan Asef-Vaziri

  15. Average Flow Time g) Compute the average flow time from 7:45AM to 12:00PM. RT = I, R = 3.76 per min, I = 144.7 T = 144.7/3.76 = 38.44 min. Ardavan Asef-Vaziri

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  17. Process Flow Chart Process : Network of activities performed by resources 1. Flow unit: The unit of analysis– Level of detail 2-3. Network of Activities & Storage/Buffers • Define Process Boundaries • Activities with activity times • Buffers with waiting flow times • Routes:precedence relationships (solid lines) with throughputs 4. Resources & Allocation • Who does what? 5. Information Structure & flow (dashed lines) Key for capacity analysis Ardavan Asef-Vaziri

  18. Decision Event Physical flow Information flow Flowcharting Activity Buffer • See the total process; asystems view • Defineflow units andprocess boundaries • Include only the key steps • Clarify the level of detail needed. Processes can be broken down into sub-activities. On the other hand, cascading allows several activities to be combined in a single sub-process • Depicts resources required to carry out activities • Identify the processes that need attention (weak points)

  19. Wondershed Inc: Narrative Representation • Separate the roof and the base sheets 2. Punch the base 3. Punch the roof 4. Form the base 5. Form the roof 6. Sub-assemble the base 7.Assemble 8.Inspect Ardavan Asef-Vaziri

  20. Activity A Activity B Activity C End Start Wondershed Inc. : Schematic Representation Punch base Form base Sub assemble base Separate Sheet Start Punch roof Form roof End Assemble Inspect Ardavan Asef-Vaziri

  21. Theoretical Flow Time Theoretical Flow Time: The minimum time for processing a flow unit. Activity Time: The time required to complete an activity. An activity cannot be broken down into smaller activities that can be executed in parallel. What does that mean??? Buffer Activity Entry Exit Ardavan Asef-Vaziri

  22. Activity Time 1 Activity Time 2 Activity Time 3 Activity Time 1 Activity Time 2 Activity Time 3 Activity Time 1 Activity Time 4 Activity Time 5 Activity Time 2 Activity Time 3 Activity Time 6 Sequential Activities Sequential Activities Parallel Activities Critical Path Theoretical flow time of each path in the process flowchart = sum of the activity times of all activities on that path Critical Path: the longest path Critical Activities: activities that lie on the critical path Theoretical flow time of the process = Time of the theoretical critical path How many Paths? Ardavan Asef-Vaziri

  23. Wondershed Inc. Theoretical Flow Time Path 1 (roof) Start  1  3  5  7  8  End 10 +20 +10 +10+30 = 80 min Path 2 (base) Start  1  2 4  6  78 End 10 +25+5+ 10 +10+30 = 90 min Theoretical Flow Time = 90 min November 11, 2011 Ardavan Asef-Vaziri 23

  24. Theoretical Flow Time Theoretical Flow Time: The minimum amount of time required for processing a typical flow unit without any waiting. Activity Time: The time required by a typical flow unit to complete the activity. Unless stated otherwise, we assume that an activity cannot be broken down into smaller activities that can be executed in parallel. What does that mean??? Buffer Activity Entry Exit Ardavan Asef-Vaziri

  25. Wondershed Inc. Flow Time Path 1 (roof) Start  1  3  5  7  8  End 30 +110 + 85+55 + 105 = 385 min Path 2 (base) Start  1  2 4  6  78 End 30 +70 +40 + 70 + 55+105 = 370 min Flow Time = 385 min November 11, 2011 Ardavan Asef-Vaziri 25

  26. Flow Time Efficiencies in White Collar Processes Flow Time Efficiency = Theoretical Flow Time / Flow Time Flow Time Efficiency = 90/385 = 23.4% November 11, 2011 Ardavan Asef-Vaziri 26

  27. 6 4 A3 A1 4 3 S E A4 A6 3 2 A2 A5 Flow Time Example: Activity Times What is the Theoretical Flow Time Ardavan Asef-Vaziri

  28. A3 A1 S E A4 A6 A2 A5 Critical Path Method: Paths How many paths? 6 4 4 3 3 2 10 11 8 Critical Path is the longest Path ArdavanAsef-Vaziri

  29. Critical Path Example of Flow Time a) The Critical Path is A1-A4-A6. The theoretical flow time of the process is 4+4+3= 11. b) What will happen if activity A5 is increased from 2 to 4? A5 is not on critical path. Increasing its time by 2 increases the length of the path A2-A5-A6 from 8 to 10. It does not become a critical path. The flow time is still 11. c) What will happen if activity A5 is increased from 2 to 5? ArdavanAsef-Vaziri

  30. Critical Path Example of Flow Time Increasing A5 time by 3 increases the length of the path A2-A5-A6 from 8 to 11. Now both paths of A1-A4-A6 and A2-A5-A6 are critical path. The flow time is still 11. d) What will happen if activity A1 is increased from 4 to 5? Path A1-A4-A6 is still critical and the flow time increases to 12. e) What will happen if activity A3 is increased from 4 to 8? Now path A1-A3 becomes critical and the flow time increases to 4+8 =12. ArdavanAsef-Vaziri

  31. W8 W9 W5 W7 W6 W3 W4 W2 W1 E S Theoretical Critical Path vs. Critical Path The time of the critical path differs from the time of the theoretical critical path. Why? 6 4 A3 A1 4 3 A4 A6 3 2 A2 A5 The critical path itself also may differ from the theoretical critical path. Why? Ardavan Asef-Vaziri

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  33. Levers for Managing Flow Time • To reduce the flow time we must shorten the length of every critical path. • Flow time is the sum of two components—Waiting time and Activity time. • These two components have different natures and the levers available for managing each are distinct. • The main levers for reducing waiting times are • Managing congestion (Utilization and Variability) – Ch3 and Ch8. • Reducing Batch sizes and Safety stock – Ch6 and Ch7. • Synchronization – Ch10. November 11, 2011 Ardavan Asef-Vaziri 33

  34. Reducing Theoretical Flow Time: Shorten the Length of Every Critical Path • Re-Structure - Moving work off the critical path • Move work off the critical path to a noncritical activity • Move work off the critical path to the outer loop (pre-processing or post-processing). • Eliminate - Reducing the work content of critical activities • Work smarter. Reduce non-value-adding part of the activity; Business Process Re-engineering. • Work Faster. Increase the speed of the activity. Better Methods, Training, Advanced Technology, Better Management. • Reduce the amount of extra work. Work right the first time; implement a robust quality management system. • Modify the product sequencing • Priority to the product that requires less processing time - Given market condition. November 11, 2011 Ardavan Asef-Vaziri 34

  35. Reducing Theoretical Flow Time: Shorten the Length of Every Critical Path Whatever approach we take, it must be directed towards the critical path. Reducing the work content of noncritical activities does not reduce the theoretical flow time. Such reduction may still be useful for decreasing total processing costs, increasing process capacity, and reducing the potential for errors and defects. November 11, 2011 Ardavan Asef-Vaziri 35

  36. 1. Restructure the Critical Path Move work off the critical path to a noncritical activity. So that critical activities are performed in parallel rather than sequentially. A :5 days B:5 days C:5 days A 1:3 days A 2:2 days C:5 days B:5 day Moving activities to the outer loop: perform them before or after the process. Instead of assembling a complete hamburger in MTO (make to order), shift towards MTS (made to stock), precook beef patties and keep them ready. November 11, 2011 Ardavan Asef-Vaziri 36

  37. 2. Work Smarter; Reduce Non-Value-Adding Activities Value-adding activities increase the economic value of a flow unit from the perspective of the customer. Performing surgery, flying airplane, serving meals, manufacturing products, and dispensing loans are value-adding. Non-value-adding activities do not directly increase the value of a flow unit. Moving work or workers, setting up machines, scheduling activities or personnel, sorting, storing, counting, filling out forms, participating in meetings, obtaining approvals or maintaining equipment are non- value -adding. November 11, 2011 Ardavan Asef-Vaziri 37

  38. 2. Reduce Non-Value-Adding Activities Non value adding activities come in two types; necessary or non-necessary. The second type should be eliminated outright. The first type activities can also be eliminated . A process with high fraction of defectives may require a sorting station to separate the defective from the good units. If the process capability is increased so that no defectives are produced, the sorting activity could be eliminated. The primary value-adding activity the accounts-payable (A/P) process is paying the bills in an accurate and timely fashion. The A/P department spends much of its time reconciling contradictory information, verifying, matching documents, and investigating discrepancies. Such activities do not add value but are still necessary. They can be eliminated if the process is modified. November 11, 2011 Ardavan Asef-Vaziri 38

  39. 3. Work Faster; Increase the Speed of Operations • Increase the speed of operation - work faster. Technology, Method, Training, Incentive. The speed of a manual checkout counter can be increased by using bar codes with a scanner, adding a second worker to bag products, or proper incentives, coupled with training and better equipment so that checkout personnel work faster without increasing error rates or jeopardizing service quality. November 11, 2011 Ardavan Asef-Vaziri 39

  40. 4. Do It Right in the First Time; Reduce the Amount of Rework Decreasing the amount of repeat work can often be achieved by process-improvement techniques such as statistical process control, method improvement and training. In data-processing, the goal is to input each piece of data just once; to avoid additional time (as well as cost and errors). Work content = Activity time × Average number of visits Activity time 10 min, rework 10%  number of visits 1.1 Work content = 1.1(10) = 11 min. Theoretical flow time is computed based on work content not based on Activity time November 11, 2011 Ardavan Asef-Vaziri 40

  41. 5. Modifying the Product Sequencing • Modify the product sequencing - do the quickest thing first. Most processes involve a mix of products, characterized by different flow times for the various units of flow. If we give priority to flows unit that move through the process faster, the overall flow time of the process will decrease. • Product Blue needs two days, Product Red needs one day. • (1+3)/2 = 2.0 • (2+3)/2 = 2.5 • Of course, product mix is often dictated by the market, and even when the organization has some control over it, there may be other relevant factors, such as profitability, resource-utilization issues, and market considerations. November 11, 2011 Ardavan Asef-Vaziri 41

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  43. A3 A1 S E A4 A6 A2 A5 Critical Path Method: Paths How many paths? 6 4 4 3 3 2 10 11 8 Critical Path is the longest Path ArdavanAsef-Vaziri

  44. 6 4 A3 A1 0 0 4 3 0 E S A4 A6 0 3 2 0 A2 A5 0 Critical Path Methos: Forward Path; Earliest Starts 4 4 10 10 11 11 4 4 4 10 11 8 8 4 5 11 4 8 8 5 3 3 3 3 5 Ardavan Asef-Vaziri

  45. Forward Path; Earliest Starts Max = 30 35 10 35 35 30 20 5 Ardavan Asef-Vaziri

  46. 6 4 4 A3 4 A1 10 10 11 11 4 0 0 4 4 10 4 3 0 S E A4 A6 8 8 4 0 5 11 4 8 8 3 2 0 A2 A5 5 3 3 0 3 3 5 Backward Path; Latest Starts 5 11 0 4 5 5 11 11 11 4 0 11 4 8 8 11 11 0 4 8 8 11 8 3 3 3 6 6 8 8 6 6 Ardavan Asef-Vaziri

  47. 30 30 30 Backward Path; Latest Starts Min = 35 30 35 45 5

  48. Activity Slack Slack, or float: The amount of time a noncritical task can be delayed without delaying the project Slack—LFT – EFT or LST – EST EST—Earliest Start Time; Largest EFT of all predecessors EFT—Earliest Finish Time; EST + duration for this task LFT—Latest Finish Time; Smallest LST of following tasks LST—Latest Start Time; LFT – duration for this task November 11, 2011 Ardavan Asef-Vaziri 48

  49. A3 A1 E S A4 A6 A2 A5 Critical Path, Slacks 6 4 5 11 0 4 11 11 0 4 4 10 4 8 4 8 11 3 11 4 8 8 3 3 6 6 2 8 3 0 3 5 Ardavan Asef-Vaziri

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