IBM Implementation of Fluctuation Smoothing Morrison, Campbell, Dews and LaFreniere Implementation of a Fluctuation Smoothing Production Control Policy in IBM’s 200mm Wafer Fab James R. Morrison Central Michigan University Department of Engineering and Technology Brian Campbell Elizabeth Dews John LaFreniere IBM, Systems and Technology Group CDC – ECC 2005 – 1 –
IBM Implementation of Fluctuation Smoothing Morrison, Campbell, Dews and LaFreniere Presentation Overview • IBM’s 200mm wafer fabricator • Competing control methodologies • The basic FSVCT policy and extensions • Estimation of cycle times • Implementation challenges • Performance evaluation • Concluding remarks CDC – ECC 2005 – 2 –
IBM Implementation of Fluctuation Smoothing Morrison, Campbell, Dews and LaFreniere IBM’s 200mm Semiconductor Wafer Fabricator Lots enter • Reentrant process flow – 30 or more visits to certain tool groups Tool Group 1 • Products from over 30 different semiconductor Tool Group 5 Tool Group 2 technologies Tool Group 6 Tool Group 3 • Dozens of products within each technology Tool Group 4 • Up to 400 stages of processing for a single product • Over 200 distinct tool groups Tool Group 7 • Production capacity on the order of 1000 wafers/day Tool Group 8 • Cycle times generally ranging from 40 to 60 days Tool Group 12 Tool Group 9 with actual production time around 20 days Tool Group 10 • Historical focus on technology excellence rather than manufacturing efficiency Tool Group 11 CDC – ECC 2005 – 3 –
IBM Implementation of Fluctuation Smoothing Morrison, Campbell, Dews and LaFreniere Competing Control Methodologies • Objective: Choose next lot to enter into production (lot sequencing or dispatching) to minimize overall cycle time and meet due date targets • Variant of continuous flow manufacturing (CFM) – Existing policy – Produce a fixed minimum number of lots of each kind per day – Not sensitive to changes in loading – Prioritization scheme employed to implement preference • Critical ratio (CR) – Due date based policy – Considered easy to implement • KANBAN – Enforce maximum queue lengths at each stage of production • Fluctuation smoothing for the variation of cycle time (FSVCT) – Reduce variation of cycle time by driving all lots to the same average CT – Successful in many simulation studies • Fluid limit inspired policies – Theory suggests potential for improved cycle time performance – Requires optimization and fab model • Deterministic finite – horizon mathematical and constraint programming – ILOG software product (own CPLEX optimization engine) – Requires detailed data and setup CDC – ECC 2005 – 4 –
IBM Implementation of Fluctuation Smoothing Morrison, Campbell, Dews and LaFreniere Toward the Basic FSVCT Policy • First “law” of queueing D/D/1 queue at 95% loading – Normalized CT = 1 – M/D/1 queue at 95% loading Normalized CT = 10.5 – Variation leads to cycle time • Reducing the variation of lateness – Let d( l ) = due date (time) of lot l Let h ( l ) = – expected remaining cycle time of lot l – expected lateness = – [ h ( l ) – ( d( l ) – Now ) ] – Slack( l ) = – Select lot with least slack (greatest expected lateness) drive all lots to the same lateness • Example: – Now = 0 – Let d( l ) = due date (time) of lot l = 10.8 days Let h ( l ) = – expected remaining cycle time of lot l = 12.3 days – expected lateness = – [ h ( l ) – ( d( l ) – Now ) ] – Slack( l ) = – [ 12.3 – ( 10.8 – 0 ) ] = – 1.5 days = – Give priority to lots with greater expected lateness CDC – ECC 2005 – 5 –
IBM Implementation of Fluctuation Smoothing Morrison, Campbell, Dews and LaFreniere Basic FSVCT Policy • Reducing the variation of lateness – Let e( l ) = exit date of lot l – Above policy should reduce the variation of lateness( l ) = e( l ) – d( l ) = exit date – due date • Reducing the variation of cycle time – If set d( l ) = arrival date (time) = a( l ) – Expect a reduction in the variation of e( l ) – a( l ) = cycle time Slack( l ) = – expected cycle time = – [ h ( l ) – ( a( l ) – Now ) ] – • Example: – Now = 0 – Let a( l ) = arrival date (time) of lot l = -15.5 days Let h ( l ) = – expected remaining cycle time of lot l = 14.5 days = – expected cycle time = – [ h (l) – d(l) ] – Slack( l ) = – [ 14.5 – ( – 15.5 – 0 ) ] = – 30 days – Give priority to lots with greater expected total cycle time • Expect a reduction in overall cycle time if we reduce the variation of the cycle times CDC – ECC 2005 – 6 –
IBM Implementation of Fluctuation Smoothing Morrison, Campbell, Dews and LaFreniere Incorporating Diverse Cycle Times • A myriad of cycle time expectations – Technologies have very different processing requirements (e.g., 8 days vs. 30 days) – Within technologies, some customers may purchase preferential treatment • G arrival processes to the system – Route through the fab is common (may be Bernoulli) – Expected remaining cycle time at each stage of production is common • Multiple cycle time FSVCT slack Let h g ( l ) = expected remaining cycle time for lot l from arrival process g – – Let CT g = total expected cycle time for lots from arrival process g – Let CT NOR = arbitrary normalization constant > 0 – For a lot l from arrival process g define CT h NOR Slack ( l ) : [ Now a ( l ) ( l )] g CT g • Alternate approaches – Scaled lateness (from the expected cycle time) h Slack ( l ) : [ Now a ( l ) ( l ) CT ] K g g g – For scaling constants K g (capturing the relative importance of lateness for each arrival process g) CDC – ECC 2005 – 7 –
IBM Implementation of Fluctuation Smoothing Morrison, Campbell, Dews and LaFreniere Estimation of Expected Cycle Times Lots enter • Fundamental parameters for multiple cycle Tool Group 1 time FSVCT: – CT g = total expected cycle time for lots Tool Group 5 Tool Group 2 from arrival process g h g ( l ) = expected remaining cycle time for – Tool Group 6 Tool Group 3 lot l from arrival process g Tool Group 4 • Absence of plant model with routes, capacities and cycle times Tool Group 7 • Tool Group 8 How to determine cycle times? – Simulation is out – no model, costly to Tool Group 12 Tool Group 9 create – Measure and use existing CT (update as Tool Group 10 needed) – Set some CTs (preferred customers) Tool Group 11 determine CT imposed on remaining lots CDC – ECC 2005 – 8 –
IBM Implementation of Fluctuation Smoothing Morrison, Campbell, Dews and LaFreniere Estimation of Expected Cycle Times • Resort to a Corollary of Little’s Law: – For each arrival process g: • Denote by P g the number of stages of production • Denote by l g the throughput rate for lots from g – Let L g = aggregate rate at which stages of production are completed = l g P g – Little’s Law: l g = N g / CT g • N g is mean number of g lots • CT g is mean cycle time for a g lot – Aggregate rate of completion of stages of production L g = P g N g / CT g CDC – ECC 2005 – 9 –
IBM Implementation of Fluctuation Smoothing Morrison, Campbell, Dews and LaFreniere Estimation of Expected Cycle Times • • Measure the historical rate of completion EXAMPLE: of stages of production L H • = P g N g / CT g Suppose for two arrival processes g – N 1 = 100 lots, P 1 = 300 stages – N 2 = 400 lots, P 2 = 300 stages • Assume (independent of control policy) • that Assume (independent of control policy) – that There exist collections of arrival processes L c with – = 3000 lot * stages/day P N L L c H g g : • If CT 1 = 40 days g g c g c CT L H g – = P 1 N 1 /CT 1 1 = (300 stages)(100 lots)/(40 days) • We can employ historical throughput rates = 750 lot * stages/day to predict cycle times – L c = L H 1 + L H Fix N g • 2 – Fix P g – Choose/find CT g satisfying assumption L H 2 = L c – L H 1 = 2250 = (300*400)/CT 2 CT 2 = 53.3 days CDC – ECC 2005 – 10 –
IBM Implementation of Fluctuation Smoothing Morrison, Campbell, Dews and LaFreniere Estimation of Expected Cycle Times • Employ this approach for – CT g by application to the fabricator h g ( l ) by application to each toolset – • Determine the consequences of prioritization – P 1 = P 2 = 300 stages – N 1 = 100 lots, N 2 = 400 lots – Cycle time targets: CT T 1 = 20 days, CT T 2 = 50 days – Lots from arrival process 1 are for preferred customers – To ensure CT for preferred customer lots: • Devote L 1 = (300 stages)(100 lots)/(20 days) = 1500 lot * stages/day – The remaining L c – L 1 aggregate throughput is devoted group 2 lots • Let CT 2 = e * CT T 2 • Then T P N / CT e 2 2 2 L T P N / CT c 1 1 1 CDC – ECC 2005 – 11 –
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