ch05-Flows.ppt
- 1. 1 Managing Process Flows Chapter 5 Business Process Modeling, Simulation and Design
- 2. 2 Overview • Processes and Flows – Important Concepts – Throughput – WIP – Cycle Time – Little’s Formula • Cycle Time Analysis • Capacity Analysis • Managing Cycle Time and Capacity – Cycle time reduction – Increasing Process Capacity • Theory of Constraints
- 3. 3 • A process = A set of activities that transforms inputs to outputs • Two main methods for processing jobs 1. Discrete – Identifiable products or services Examples: Cars, cell phones, clothes etc. 2. Continuous – Products and services not in identifiable distinct units Examples: Gasoline, electricity, paper etc. • Three main types of flow structures 1. Divergent – Several outputs derived from one input Example: Dairy and oil products 2. Convergent – Several inputs put together to one output Example: Car manufacturing, general assembly lines 3. Linear – One input gives one output Example: Hospital treatment Processes and Flows – Concepts
- 4. 4 • Inflow and Outflow rates typically vary over time – IN(t) = Arrival/Inflow rate of jobs at time t – OUT(t) = Departure/Outflow rate of finished jobs at time t – IN = Average inflow rate over time – OUT = Average outflow rate over time • A stable system must have IN=OUT= – = the process flow rate – = process throughput Process Throughput
- 5. 5 0 2 4 6 8 10 12 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 t Jobs IN(t) OUT(t) Process Inflow and Outflow vary over time
- 6. 6 • All jobs that have entered the process but not yet left it • A long lasting trend in manufacturing has been to lower WIP by reducing batch sizes – The JIT philosophy – Forces reduction in set up times and set up costs • WIP = Average work in process over time • WIP(t) = Work in process at time t – WIP(t) increases when IN(t)>OUT(t) – WIP(t) decreases when IN(t)<OUT(t) Work-In-Process
- 7. 7 The WIP Level Varies With Process Inflow and Outflow t1 t2 t3 WIP(t) WIP OUT(t) > IN(t) IN(t) > OUT(t) OUT(t) = IN(t)
- 8. 8 • The difference between a job’s departure time and its arrival time = cycle time – One of the most important attributes of a process – Also referred to as throughput time • The cycle time includes both value adding and non-value adding activity times – Processing time – Inspection time – Transportation time – Storage time – Waiting time • Cycle time is a powerful tool for identifying process improvement potential Process Cycle Time
- 9. 9 • States a fundamental and very general relationship between the average: WIP, Throughput (= ) and Cycle time (CT) – The cycle time refers to the time the job spends in the system or process • Implications, everything else equal – Shorter cycle time lower WIP – If increases to keep WIP at current levels CT must be reduced • A related measure is (inventory) turnover ratio – Indicates how often the WIP is entirely replaced by a new set of jobs Little’s Formula (Due to J.D.C. Little (1961)) Little’s Formula: WIP = ·CT Turnover ratio = 1/CT
- 10. 10 • The task of calculating the average cycle time for an entire process or process segment – Assumes 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 Cycle Time Analysis
- 11. 11 • 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 = percentage of jobs requiring rework (rejection rate) • Assuming a job is never reworked more than once • Assuming a reworked job is no different than a regular job Rework CT = (1+r)T CT = T/(1-r)
- 12. 12 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? a) If no jobs are rejected and sent for rework. b) If 25% of the jobs need rework but never more than once. c) 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)
- 13. 13 • 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 – pi = The probability that a job is routed to path i – Ti = The time to go down path i Multiple Paths CT = p1T1+p2T2+…+pmTm= m 1 i i iT p
- 14. 14 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)
- 15. 15 • 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 – Ti = Average process time for process segment i to be completed Processes with Parallel Activities CTparallel = Max{T1, T2,…, TM}
- 16. 16 • 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)
- 17. 17 • Measured as the percentage of the total cycle time spent on value adding activities. • Theoretical Cycle Time = the cycle time which we would have if only value adding activities were performed – That is if the activity times, which include waiting times, are replaced by the processing times • See example – Cycle time analysis Exercise 9 & 10, Laguna & Marklund Chapter 5 Cycle Time Efficiency Cycle Time Efficiency = CT Time Cycle l Theoretica
- 18. 18 • Focus on assessing the capacity needs and resource utilization in the process 1. Determine the number of jobs flowing through different process segments 2. Determine capacity requirements and utilization based on the flows obtained in 1. • The capacity requirements are directly affected by the process configuration Flowcharts are valuable tools Special features to watch out for Rework Multiple Paths Parallel Activities • Complements the cycle time analysis! Capacity Analysis
- 19. 19 • A rework loop implies an increase of the flow rate for that process segment • Definitions – N = Number of jobs flowing through the rework loop – n = Number of jobs arriving to the rework loop from other parts of the process – r = Probability that a job needs rework • Assuming a job is never reworked more than once • Assuming a reworked job is no different than a regular job The Effect of Rework on Process Flows N = (1+r)n N = n/(1-r)
- 20. 20 N = (1+r)n = (1+0.25)100 = 125 Example – Capacity Analysis with Rework 0.75 0.25 A B C I 100 jobs 125 jobs 125 jobs 125 jobs
- 21. 21 Multiple Paths and process flows • The flow along a certain path depends on – The number of jobs entering the process as a whole (n) – The probability for a job to go along a certain path • Defining – Ni = number of jobs taking path i – pi = Probability that a job goes along path i Parallel Activities and process flows • All jobs still have to go through all activities – if they are in parallel or sequential does not affect the number of jobs flowing through a particular activity Multiple Paths and Parallel Activities Ni = n·pi
- 22. 22 Need to know – Processing times for all activities – The type of resource required to perform the activity – The number of jobs flowing through each activity – The number of available resources of each type Step 1 – Calculate unit load for each resource • The total resource time required to process one job – Ni = Number of jobs flowing through activity i for every new job entering the process – Ti = The processing time for activity i in the current resource – M = Total number of activities using the resource Analyzing Capacity Needs and Utilization (I) Unit load for resource j = M 1 i i i T N
- 23. 23 Step 2 – Calculate the unit capacity • The number of jobs per time unit that can be processed Step 3 – Determine the resource pool capacity • A resource pool is a set of identical resources available for use • Pool capacity is the number of jobs per time unit that can be processed – Let M = Number of resources in the pool Analyzing Capacity Needs and Utilization (II) Unit capacity for resource j = 1/Unit load for resource j Pool capacity = MUnit capacity = M/unit load
- 24. 24 Capacity is related to resources not to activities! • The process capacity is determined by the bottleneck – The bottleneck is the resource or resource pool with the smallest capacity (the slowest resource in terms of jobs/time unit) – The slowest resource will limit the process throughput Capacity Utilization • The theoretical process capacity is obtained by focusing on processing times as opposed to activity times – Delays and waiting times are disregarded The actual process throughput The theoretical capacity! Analyzing Capacity Needs and Utilization (III) Capacity Utilization = Capacity ocess Pr l Theoretica Throughput Actual
- 25. 25 Cycle time and capacity analysis provide valuable information about process performance – Helps identify problems – Increases process understanding – Useful for assessing the effect of design changes • Ways of reducing cycle times through process redesign 1. Eliminate activities 2. Reduce waiting and processing time 3. Eliminate rework 4. Perform activities in parallel 5. Move processing time to activities not on the critical path 6. Reduce setup times and enable batch size reduction Cycle time Reduction
- 26. 26 • Consider a process with three sequences or paths By moving 2 minutes of activity time from path 2 to path 1 the cycle time is reduced by 2 minutes to CT=45 minutes Example – Critical Activity Reduction A B C D E 12 15 18 20 14 Sequence (Path) Time required (minutes) 1. AB E 12+14+15 = 41 2. AC E 12+20+15 = 47 = CT 3. A D E 12+18+15 = 45 Critical path
- 27. 27 • Two fundamental ways of increasing process capacity 1. Add resource capacity at the bottleneck – Additional equipment, labor or overtime – Automation 2. Reduce bottleneck workload – Process redesign Shifting activities from the bottleneck to other resources Reducing activity time for bottleneck jobs • When the goal is to reduce cycle time and increase capacity careful attention must be given to – The resource availability – The assignment of activities to resources • See also example 5.15 in Laguna & Marklund Increasing Process Capacity
- 28. 28 • An approach for identifying and managing bottlenecks – To increase process flow and thereby process efficiency • TOC is focusing on improving the bottom line through – Increasing throughput – Reducing inventory – Reducing operating costs Need operating policies that move the variables in the right directions without violating the given constraints • Three broad constraint categories 1. Resource constraints 2. Market constraints 3. Policy constraints Theory of Constraints (TOC) (I)
- 29. 29 • TOC Methodology 1. Identify the system’s constraints 2. Determine how to exploit the constraints – Choose decision/ranking rules for processing jobs in bottleneck 3. Subordinate everything to the decisions in step 2 4. Elevate the constraints to improve performance – For example, increasing bottleneck capacity through investments in new equipment or labor 5. If the current constraints are eliminated return to step 1 – Don’t loose inertia, continuous improvement is necessary! • See example 5.18 , Chapter 5 in Laguna & Marklund Theory of Constraints (TOC) (II)
- 30. 30 • Consider a process with 9 activities and three resource types. Activities 1, 2 & 3 require 10 minutes of processing and the other activities 5 minutes each. • There are 3 jobs, following different paths being processed • Activities 1, 2 & 3 utilize resource X, activities 4, 5, & 6 resource Y and activities 7, 8 & 9 resource Z. Each resource have 2400 minutes of weekly processing time available Example – Applying the TOC Methodology Job Routing Demand (Units/week) Profit Margin A 4, 8, and 9 50 20 B 1, 2, 3, 5, 6, 7, and 8 100 75 C 2, 3, 4, 5, 6, 7, 8, and 9 60 60