ThroughputFlowTimeAndLittlesLaw 4xgs

ThroughputFlowTimeAndLittlesLaw 4xgs - Overview Throughput...

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Unformatted text preview: Overview Throughput Concepts, Flow Time, and Little's Law Peter L. Jackson Professor School of O.R. and I.E. 1/31/2011 Throughput Concepts What is it Worth to Increase Throughput? g p Flow time and Lead time Little's Law What is it Worth to Reduce Flow Time? Industrial Data and Systems Analysis 1 1/31/2011 Industrial Data and Systems Analysis 2 What is Throughput? What is Cycle Time? Activity of a repetitive process Output = input ( throughput), unless... Measure of interest: throughput rate Flights per hour out-patients per day hour, day, Definition of cycle time is not consistent: even within the same company! Takt time Inverse of Flow time Time spent in system by single unit (includes queue time) gallons per second, messages per hour, etc. Industrial Data and Systems Analysis 3 1/31/2011 throughput rate Units of time per unit of output Time between successive outputs T Toyota S t t System 1/31/2011 Industrial Data and Systems Analysis 4 Throughput Detractors Unscheduled time Starved / blocked by production control system The M.F.D. THRUPUT Game Speed loss Quality loss Scrap, rework Breakdowns Setup time Planned downtime partial shifts maintenance meetings training meetings, 1/31/2011 Industrial Data and Systems Analysis 5 1/31/2011 Industrial Data and Systems Analysis 6 Throughput Targets DGR (daily going rate): units per day required to Will the Process Achieve Target? T t? Begin with throughput rate achievable support final assembly schedule (level schedule) Cycle time (Takt time): time allowed between successive units of output to achieve DGR under ideal operating conditions Adjust for breakdowns setups, quality breakdowns, setups loss, speed loss Ex Adjust for speed loss: Ex. Equipment runs at 100y% speed Throughput rate adjusted for efficiency = g p j y Takt time = effective daily operating time / DGR Stated as requirement: process must be unadjusted rate * y designed and operated to achieve these targets g p g Industrial Data and Systems Analysis 7 Adjust for planned downtime 1/31/2011 1/31/2011 Industrial Data and Systems Analysis 8 Adjust for Repair Cycles repair TTR operational TTF repair Adjust for Setup Cycles setup Setup time run Run time setup MTTF: mean time to fail (when running) MTTR: mean time to repair Mean repair cycle length = MTTR+MTTF Fraction of time equipment is operational = MTTF / (MTTR+MTTF) Throughput rate adjusted for repair = Unadjusted rate * MTTF / (MTTR MTTF) (MTTR+MTTF) 1/31/2011 R ti Run time = average l t size / production lot i d ti rate Setup cycle length = Setup time + run time Fraction of time equipment is running = Run time / (Setup cycle length) ( p y g ) Throughput rate adjusted for setups = unadjusted rate * run time / cycle length 9 1/31/2011 Industrial Data and Systems Analysis Industrial Data and Systems Analysis 10 Adjust for Scrap I 100(1-x)% 100x% O Adjust for Rework Cycles I 100(1-x)% 100x% O Output rate is 100(1-x)% of input rate Throughput rate adjusted for scrap = Suppose reject rate applies to reworked parts as unadjusted throughput rate * (1-x) well Expected number of work and rework cycles per part = x n=1/(1-x) n= 0 Throughput rate adjusted for rework = unadjusted rate * (1-x) j ( ) 1/31/2011 Industrial Data and Systems Analysis 11 1/31/2011 Industrial Data and Systems Analysis 12 Aside: Rework analysis Let N= number of work + rework cycles y Aside: Rework Analysis (cont d) (cont'd) Unadjusted time per piece = 1/unadjusted E[ N ] = P{N > n} n =0 0 since N is nonnegative r.v. P{N > 0} = 1 since every part has at least one cycle P{N > n} = x n , n 1 i.e. part must fail n times in a row E[ N ] = 1 + x + x + x + = x n = 2 3 n =0 1 1- x 13 throughput rate Adjusted time per piece = j p p E[N]/unadjusted throughput rate Adjusted throughput rate = 1/adjusted j g p j time per piece = unadjusted throughput rate / E[N] Adjusted throughput rate = unadjusted throughput rate * (1-x) 1/31/2011 1/31/2011 Industrial Data and Systems Analysis Industrial Data and Systems Analysis 14 Adjust for Planned Downtime Begin with 24 hour clock g Subtract planned downtime to get Multi Stage Multi-Stage Throughput 1 Blocked operation 2 Bottleneck operation 3 Starved operation available hours Multiply throughput per hour by available hours to get achievable throughput per day (nominal t put) t'put) Compare to DGR If DGR > nominal t'put then design is t put In a series of operations the operation operations, inadequate with the lowest throughput rate determines the throughput rate of the system as a whole Downstream operations will be starved Upstream operations will be blocked If DGR < nominal t'put then process will be idle (unscheduled time) 1/31/2011 Industrial Data and Systems Analysis 15 1/31/2011 Industrial Data and Systems Analysis 16 Sample Problem 1 Speed and Quality Losses What is effective throughput rate of the The engineering specifications indicate that an automated lathe can turn a certain metal part at a rate of 100 pieces per hour. However, in practice the lathe is run at only 95% of that speed. The scrap rate from the machine is only about 1 part in 10,000. Rework occurs at about the same rate. On average the machine is out of service for repairs 30 minutes for every 20 hours of operation. y p Furthermore, jobs arrive in batches of 250 units and there is a setup time required of 15 minutes for each job. lathe considering speed loss, scrap, and rework? (pieces per hour) Assume rework % expressed in terms of non- scrapped units i apply rework adjustment after scrap i.e. l k dj t t ft adjustment 1/31/2011 Industrial Data and Systems Analysis 17 1/31/2011 Industrial Data and Systems Analysis 18 Breakdown Losses Down for repairs Transitions (Setups) In service In service In service In service ... In service In service What fraction of time in service is spent in What fraction of potential operational production (rather than in setup)? time is the equipment available for service? 1/31/2011 Industrial Data and Systems Analysis 19 1/31/2011 Industrial Data and Systems Analysis 20 Net Effect What is the effective throughput rate of Sample Problem 2 the lathe considering all effects of speed, quality, reliability, and transition? Automobile assembly p y plant Production rate: 40 jobs per hour Sales rate per week: 3,040 vehicles Plant Pl t operates 5 d t days per week, t k two shifts per hift day + overtime Daily going rate: 608 vehicles Hours per shift: 8 Planned downtime per shift: 20 minutes Compute: Operating hours per day: 15.333 Takt time: 1.513 minutes per vehicle 1/31/2011 Industrial Data and Systems Analysis 21 1/31/2011 Industrial Data and Systems Analysis 22 Actual Performance There are quality problems if we run the Performance Against Target Effective throughput: 34.1 vehicles per hour g p p Takt time: 1.513 minutes per vehicle Effective cycle time: 1.759 minutes per y p line too fast We have slowed the line speed by 10% p y % There are lots of little interruptions on the line that cause it to slow down Duration of interruption: 20 seconds Frequency of interruption: 10 per hour Uninterrupted hours per hour of operation: 0.947 p Industrial Data and Systems Analysis 23 vehicle "Measure" : a quantity describing performance (ex. effective cycle ti ) f ( ff ti l time) "Metric": a measure used as a standard (ex. takt time) Our cycle times in the plant are 16% too high, compared to our metric g , p 1/31/2011 1/31/2011 Industrial Data and Systems Analysis 24 What is it Worth to Increase Throughput? Th h t? Approach: pp assume production target is fixed As throughput rate increases, time required What Does Overtime Cost? decreases Time is money: compute wages saved A il bl regular ti Available l time h hours per week: k 76.67 Available regular time production per week: 2614.7 Overtime hours required: 12.47 12 47 1/31/2011 Workers per line per shift: 50 Salaries+benefits per year: $70,000 Weeks per year: 50 Salary+benefits per worker-week: $1,400 Benefit rate: 25% Salary per hour: $28 Overtime per hour: $42 (no benefits) Total cost per overtime hour: $2,100 Industrial Data and Systems Analysis 26 Industrial Data and Systems Analysis 25 1/31/2011 What is it Worth to Increase Throughput? Th h t? Current predicted overtime cost: $26,185 What are Other Ways to Measure Benefit of I B fit f Increasing Th i Throughput? h t? per week Annualized cost: $1,309,259 $ , , What is impact of cutting speed loss in half (without hurting q ( g quality)? $492,593 y) $ , per year What is impact of cutting interruption frequency (or duration) in half? $246,296 per year 1/31/2011 Industrial Data and Systems Analysis 27 1/31/2011 Industrial Data and Systems Analysis 28 How Can You Increase Throughput? Th h t? Continuous improvement Measure, Analyze, Act Observe factory, interview workers and staff, y Overview Throughput Concepts What is it Worth to Increase Throughput? g p collect data on problems Brainstorm solutions, implement and test Recitation this week Use SQL to analyze instances of throughput Flow time and Lead time Little's Law What is it Worth to Reduce Flow Time? problems 1/31/2011 Industrial Data and Systems Analysis 29 1/31/2011 Industrial Data and Systems Analysis 30 What is Flow Time? Accounting for Lead Time Lead time: the time Cycle time Inverse of Flow time Time spent in system by single unit (includes queue time) Cookie flow time: 32 min. + queue time throughput rate Ex. Oven cycle time: 32 min. Two trays of 8 cookies per cycle Cookie cycle time: 2 min./cookie 1/31/2011 from when you trigger an action until you see the result Typically, lead time is duration of time between order placement and order fulfillment (delivery) 31 1/31/2011 Industrial Data and Systems Analysis Industrial Data and Systems Analysis 32 Lead Time Analysis Order Sent Credit Approved Schedule Released Material Picked / Produced Shipments Consolidated Long Haul Transport Drop Material Received / Inspected / Order Verified Queue Factor Order lead time is flow time of order through the g Order Lead Time Material Packed / Shipped Short Haul Delivery entire system (order processing, factory scheduling, production, packing, shipping, and receiving) g) Distinguish between value-added time and nonvalue added time customer Value-added time: material transformation + transport to Non-value-added time: queue time+internal transport time Order Scheduled Order Received Production Lead Time Pick Materials Operation 1 Operation 2 Operation 3 Queue factor: (time in queue+value added time) / Operation Lead Time Production Transport Queue Setup Run Transport Queue value added time If measured rigorously, queue time is many orders of magnitude larger than value added time value-added Industrial Data and Systems Analysis 1/31/2011 Industrial Data and Systems Analysis 33 1/31/2011 34 How to Reduce Lead Times Continuous improvement Measure, Analyze, Act , y , Overview Throughput Concepts What is it Worth to Increase Throughput? g p How to reduce value added time: Increase speed of operations Flow time and Lead time Little's Law What is it Worth to Reduce Flow Time? How to reduce queue time Increase frequency of operations Reduce variability 1/31/2011 Industrial Data and Systems Analysis 35 1/31/2011 Industrial Data and Systems Analysis 36 Cumulative Orders Cumulative C l ti Units Additions to "On Order On Order" Cumulative C l ti Units r: Order input rate r rT: Units ordered during order lead time Time 0 Time 0 T: Order lead time 1/31/2011 Industrial Data and Systems Analysis 37 1/31/2011 Industrial Data and Systems Analysis 38 Delivery of "On-Order" On Order Cumulative C l ti Units Integral of On Order Process Cumulative C l ti Units r r rT r rT 0 Time 0 Time T T T T Area = 2 * T*(rT) = rT2 T (rT) 1/31/2011 Industrial Data and Systems Analysis 39 1/31/2011 Industrial Data and Systems Analysis 40 Integral of On Order Process Cumulative C l ti Units Average Units on Order Average Units on Order = Integral of On r rT 0 Time Order Process / T Average Units on Order = rT g Average Units on Order = Throughput rate * Order Lead Time Little's Law: L = W Applicable in wide variety of systems When is it not applicable? pp T Area = 2 * T*(rT) = rT2 T (rT) 1/31/2011 Industrial Data and Systems Analysis 41 1/31/2011 Industrial Data and Systems Analysis 42 Little s Little's Law 600 students admitted annually to four- How Much Money is Tied Up in Inventory? I t ? Automotive example continued: p Sales: 3,040 vehicles per week Production and transport time p vehicle: 8 p per 43 year degree program 2400 students in program, on average 15 days for ship to cross Pacific Ocean, 700 vehicles per ship, one ship per week 1500 vehicles i oceanic t hi l in i transit, on average it days Pipeline inventory: 3,474.3 vehicles Vehicle cost: $12,000 Pipeline inventory investment: $41,691,429 Interest rate to finance i I fi investment: 12% Annual pipeline inventory interest charge: $5,002,971 $5 002 971 Industrial Data and Systems Analysis 1/31/2011 Industrial Data and Systems Analysis 1/31/2011 44 What is it Worth to Reduce Flow Time? Fl Ti ? Reducing material flow time reduces What are Other Benefits of Reducing Flow Ti Fl Times and L d Ti d Lead Times? ? inventory ("pipeline inventory") Reducing production and transportation flow time by 1 day would reduce pipeline inventory by 12.5 % 12 5 Interest savings per year: $625,371 1/31/2011 Industrial Data and Systems Analysis 45 1/31/2011 Industrial Data and Systems Analysis 46 Overview Throughput Concepts What is it Worth to Increase Throughput? g p Flow time and Lead time Little's Law What is it Worth to Reduce Flow Time? 1/31/2011 Industrial Data and Systems Analysis 47 ...
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This note was uploaded on 03/18/2012 for the course ORIE 3120 taught by Professor Jackson during the Spring '09 term at Cornell.

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