2 8102.810Manufacturing System ExamplesT. Gutowski1. Push Vs Pull Systems – Video2. Toyota Cell – Video3FMS-Video3.FMS Video4. Aircraft EnginesPush and Pull SystemsPush and Pull SystemsMachinesMachines1234Parts OrdersPush Systems –Order (from centralized decision process) arrives at the front of the system and is produced in batches of size “B”.Q. How long does it take to get one part out of the system?1 2 3 N…..Time = 0Time=T2Time = T1Time = T3Time T2Time = TNPhStPush Systems –123NComment; Of course, thisTime = 0123N…..part can come from inventoryin a much shorter time, but the point is that the push system is not very responsiveTime =Tis not very responsive.= TNIf the process time per part is “t” at each of “N” processes and the batch size is“B”N” processes, and the batch size is B”,it takes time TN=“NBt” to get one part through the systemone part through the system.Pull Systems-Th d i tth d fth li di “ ll d” t fth tThe order arrives at the end of the line and is “pulled” out of the system. WIP between the machines allows quick completion. Q.How long does it take to pull out one part? A.The time to finish the last operation “t”.Comparison between Ph dPllSPush and Pull SystemsPush system characteristics:CentralPush system characteristics:Central decision making, local optimization of equipment utilization leads to largeequipment utilization leads to large batches, large inventories and a sluggish systemsystem.Pull system characteristics: Local decision making emphasis on smooth flow WIPmaking, emphasis on smooth flow,… WIP in system.See HP Video (go to 3:00 minutes)HP VideoHP VideoDots Tacks Tape PackInventory in the system = LTi i h WTime in the system = WLittle’s Law L = λWHP Video ResultsPh t (6)Pll(3)Pll(1)HP Video ResultsPush system (6)Pull (3)Pull (1)Space 2 Tables 2 Tables 1 TableWIP = L 30 12 4“Cycle time” = W 3:17 1:40 0:19Rework Units ≈ WIP26 10 3Q lit P blHiddVi iblVi iblQuality ProblemHiddenVisibleVisibleProduction Rateλ = L / W0.15 0.12 0.21Graphical InterpretationGraphical Interpretation250150200Time [s]50100Number or 00246Inventory, LBatch Size "B"Inventory, LTime in System, WL = λ WL≈ k1BL ≈ k1BW ≈ k2Bλ = L / W = k1/ k2So what are the advantages of the pull systems?•quick responsequick response• low inventoriesbblbl•observable problems (if stopped = problem)• sensitive to state of the factory(if no part = problem) (pp )• possible cooperative problem solvingTPS CellTPS Cell1Operation of cell; Standard work1.Operation of cell; Standard work2. L, λ & W for cell; L=λW3. Increase in production rateMachining CellgOperator moves part from machine to machinefrom machine to machine(including “decouplers”)by making traversearound the cell.Cell FeaturesCell Features•“Synchronized”sequential productionSynchronized, sequential production• Operator decoupled from individual machinesmachines• Operator integrated into all tasks• Goal: single piece Flow• Best with single cycle automatics, but can gy ,be done manually too SVidSee VideoMachining CellWalking segments - 10gsegmentManual(Sec)Walk to(Sec)Machine(Sec)1Raw 35672 Saw 15 3 603 L1103 704567894 L2123 505 HM 12 3 1206VM12037039106VM1203707VM2203608G15360129 F.I. 19 3 10 Finishpart3Totals M+W = 153 490Machining CellParts in the cell ~ 14gManual(Sec)Walk to(Sec)Machine(Sec)R367891011Raw3Saw 15 3 60L1 10 3 704511121314L2 12 3 50HM 12 3 12012313VM1 20 3 70VM2 20 3 60G153601G15360F.I. 19 3 + 3Totals M+W = 153 490Standard Work for CellStandard Work for CellPAJT Bl k C llOtPAJT. Black CellOperators:PROCESS#OPERATIONMan Walk Auto1Raw 32 Saw 15 3 603L1 10 3 70160 18060 80 100 120TIME {secs}20 40 1404L2 12 3 505 HM 12 3 1206VM1 20 3 707VM2 20 3 608G 15 3 509F.I. 196996Cell produces one part every 153 secNote: machine time Max (MTj) < cycle time CTi.e. 120+12 < 153TPS CellTPS Cell1Operation of cell; Standard work1.Operation of cell; Standard work2. L, λ & W for cell; L=λW3. Increase in production rateTPS Cell and Little’s LawTPS Cell and Little s Law•L=λ W we know L andλwhat is W?•L =λ W, we know L and λ, what is W?• Define System Boundaries• Follow part around the cell• Single operator caseMachining CellNumber of round trips; 13g6789101145111213Saw 3+15 + 153#1 decoupler1.5 +15312313decouplerL1 1.5+ 10+153……. …….1Grind 1.5+15+153Manual and walk19+3 outand walk150 153X13=19891989 15021391989 + 150 = 2139By Little’s LawBy Little s LawL = (13 + 1) X (150/153) +13 X (3/153) 13 9813 X (3/153) = 13.98 partsrate, λ = 1/153 parts/secondW = 153 X 13.98 = 2139 secTPS CellTPS Cell1Operation of cell; Standard work1.Operation of cell; Standard work2. L, λ & W for cell; L=λW3. Increase in production rateManual(Sec)Walk to(Sec)Machine(Sec)Raw 3To increase production rate add 2ndworkerSaw 15 3 60L1 10 3+3 70L2 12 3 50HM 12 3 120VM120370WORKER 2VM120370VM2 20 3+3 60G153 60WORKER 1F.I. 19 3 + 3Totals M+W = 159 490Wk180Work 180Work 2 79What is the production rate for this new arrangement?Check max(MTj) < CTCheck max(MTj) < CTWorker 1; 80 = 80Worker 2; 12+120 >79One part every 132 secondsCan we shift work off of the HM to reduce the cycleCan we shift work off of the HM to reduce the cycle time?Manual(Sec)Walk to(Sec)Machine(Sec)Raw 3Saw 15 3 60L1 10 3+3 70L2 12 3 50HM 12 3 12080VM120370WORKER 2VM12037080VM2 20 3+3 6090G15360WORKER 1G15360F.I. 19 3 + 3Totals M+W = 159 490Work 1 80Work 2 79Standard Work for Worker #2Standard Work for Worker #2RdS tOtWk #2Rod SupportOperators:Worker #2ROCESSOPERATIONMan Walk Autofrom decoupler 1.5L2 12 3 50HM12380160 18060 80 100 120TIME {secs}20 40 140HM12380VM1 20 3 80VM2 20 1.5 90Cycle # 1 Cycle # 2+3Operator waiting+3pgOn machineWhat is the new production Rate?Check max(MTj) < CTCheck max(MTj) < CTWorker 1; 80 = 80Worker 2; 110 > 79Hence Worker #2 will be waiting on gVertical Mill #2What is the new production Rate?•The new production rate is;The new production rate is;one part every 110 secP d C W k “idl ” ’t d•Pro and Cons; Worker “idle”, can’t speed up by adding additional worker• Design for flexibility make;Max(MTj) < CT/2(j)Manual(Sec)Walk to(Sec)Machine(Sec)Raw 3A BAlternative solution add 2 HM’sSaw 15 3 60L1 10 3+3 70L2 12 3 50HM 12 3 120VM120370WORKER 2VM120370VM2 20 3+3 60G153 60WORKER 1F.I. 19 3 + 3Totals M+W = 159 490Wk180Work 180Work 2 79Homework #1Homework #1Determine L, λ & WF th 2 t t ith t ll l•For the 2 operator system with two parallel machines
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