This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Visual Representation in MS Excel xcel
Each column represents 5 minutes of run duration Cyclic Scheduling
Peter L. Jackson
School of Operations Research and Information Engineering Cornell University
3/28/2011 Color shades indicate precedence: lighter shades must precede darker shades
1 3/28/2011 Industrial Data and Systems Analysis Industrial Data and Systems Analysis 2 Completed Schedule Is Makespan Important? What does makespan tell us? If we divide the number of jobs by the makespan, How long it takes to complete a set of jobs does that tell us the throughput rate of the factory? f t ? Silly almost no factory operates like that Silly, So, no. The #jobs/makespan does NOT equal Only if the factory waits until those jobs are done 86 columns x 5 min. per column = 430 min before starting another set of jobs throughput minutes Makespan = total time required to complete all jobs = 430 minutes
3/28/2011 Makespan does not determine throughput Makespan is important but it is not the only measure of interest Industrial Data and Systems Analysis 3 3/28/2011 Industrial Data and Systems Analysis 4 Toyota Standard Operation Routine in Cell R ti i C ll Cell consists of all machines required to process q p a part Operator manually loads (L), unloads (U), and inspects (I) a part from each machine i t tf h hi Operator walks around cell in cyclic fashion, keeping equipment busy Operator walk cycle determines throughput of cell To increase throughput, either add more operators or more equipment Depends on bottleneck analysis
Industrial Data and Systems Analysis 3/28/2011 5 3/28/2011 Industrial Data of the Factory With a Future. Black, J.T. 1991. The Design and Systems Analysis McGrawHill, Inc. 6 ISBN 0070055505 Observations Required cycle time came from topdown top down Cycle Length, Cycle Time, Throughput Th h t Operator walk cycle takes 110 sec. Two parts are produced per cycle Cycle time per p y p part is therefore 55 sec. Throughput rate is 1/55 sec * 3600 sec./hr, or, analysis (seconds per day/required daily quantity Bottleneck operation is "Finish mill outer and inner ID": 70 sec. Impossible to meet cycle time with just one about 65 pieces per hour So, cycle length = 110 sec. (time to repeat machine for this operation Second BR machine added to cell Routing modified to visit each BR machine once per cycle ( p y (two p parts output p cycle) p per y )
3/28/2011 cycle) C l ti Cycle time = 55 sec. per part t Throughput rate = 65 parts per hour (rate of production) d ti )
7 3/28/2011 Industrial Data and Systems Analysis Industrial Data and Systems Analysis 8 Cycle Time vs Flow Time Different question: how long does it take to make one part? Call this "flow time" I h to b at l It has be least 118 sec. = 2 21 0 sec. 27+21+70 (the sum of the machine processing times) Flow time can be much longer than cycle time Flow time can be much longer than sum of process times Little's Law: inventory in process will be equal to product of average flow time and average th throughput rate h t t
Industrial Data and Systems Analysis
9 3/28/2011 3/28/2011 Industrial Data of the Factory With a Future. Black, J.T. 1991. The Design and Systems Analysis McGrawHill, Inc. 10 ISBN 0070055505 Flow Time (First Part)
First part is removed from cold header at time 0 p It is loaded onto VM by about time 17 It is loaded onto HM by about time 88 It is loaded into BR at about time 50 on the next cycle It finishes on the BR at about time 10 on the next cycle (the third cycle) Cycle is 110 sec. long so that puts us at 230 sec. The operator unloads and deburrs it by time 45 Cycle is 110 sec. long so the part is in the cell for 265 sec. sec = 110 + 110+45 sec (flow time of this part) sec.
Industrial Data and Systems Analysis 3/28/2011 Industrial Data of the Factory With a Future. Black, J.T. 1991. The Design and Systems Analysis McGrawHill, Inc. 11 ISBN 0070055505 3/28/2011 12 3/28/2011 Industrial Data of the Factory With a Future. Black, J.T. 1991. The Design and Systems Analysis McGrawHill, Inc. 13 ISBN 0070055505 3/28/2011 Industrial Data of the Factory With a Future. Black, J.T. 1991. The Design and Systems Analysis McGrawHill, Inc. 14 ISBN 0070055505 Flow Time (Second Part) Second part is removed from cold header at Implication A cyclic schedule can tell us time 55 on first cycle It is deburred by time 100 on the third cycle y y
sec. (same as for first part, a coincidence) Flow time for second part = 11055+110+100 = 265 Average flow time = 265 sec. Flow time is the makespan of a single part Average inventory = 265 sec. *1/55 sec. =4.82 That is, on average, there are 4.82 parts in this cell
3/28/2011 Cycle length of the cyclic pattern Cycle time per p y p part Throughput rate Average flow time per part (= part makespan) Average inventory in the system A cyclic Gantt chart (such as the Toyota standard operations routine) helps us to t d d ti ti ) h l t visualize the pattern
Industrial Data and Systems Analysis
16 Industrial Data and Systems Analysis 15 3/28/2011 Example from Tufte's "The Visual Display f Quantitative I f Di l of Q tit ti Information" ti " Marey train schedule, 1885 ParisLyon Marey's ParisLyon Train Schedule, S h d l 1885 Horizontal line length: stop length Vertical V ti l spacing: di t i distance b t between stations t ti Slope: speed Intersection: time/place of crossing Train schedules are repeated each day A cyclic schedule Marey diagram is a cyclic Gantt chart Note how it wraps around itself
Tufte p. 31. 3/28/2011 http://books.nap.edu/html/hs_math/images/tl_f8.gif Industrial Data and Systems Analysis 17 3/28/2011 Industrial Data and Systems Analysis 18 Using Schedules to Look for "Meet d Greet" P "M t and G t" Possibilities ibiliti Meet and greet: crews swap trains at Re Entrant ReEntrant Flow Semiconductor fabrication Basic sequence of photolithography
Evaporate metal onto silicon Apply and cure photoresist Expose photoresist to circuitry pattern Etch away metal unprotected by exposed photoresist Apply protective dielectric material midpoint and return home Few opportunities Many y opportunities Repeated for each layer of circuitry Jobs cycle through fabrication plant over and g p over again: flow shop but with reentrant flows
Industrial Data and Systems Analysis
20 3/28/2011 Industrial Data and Systems Analysis 19 3/28/2011 Easy as ABC ABC Example Suppose a batch of pp wafers requires the following process times: Operation p Evap Photo Batch Time (minutes) 120 150 220 80 120
22 Suppose each wafer Expose Etch Dielectric requires three layers 3/28/2011 Industrial Data and Systems Analysis 21 3/28/2011 Industrial Data and Systems Analysis Sample 4Hour Cyclic Schedule 4 Hour What Can We Deduce from this Cyclic Schedule? C li S h d l ? Cycle length = 240 minutes = 4 hours Cycles per batch =3 (the number of layers) Note: now we have cycles per batch rather than parts per cycle (multiplication not division) Hour: 1 2 3 4 Cycle time per batch: 4 hours * 3 = 12 hours Throughput rate: 0.083 batches per hour To determine flow time, we need to trace a , batch requiring three layers through the schedule
Industrial Data and Systems Analysis 3/28/2011 Industrial Data and Systems Analysis 23 3/28/2011 24 Tracing a Batch in a Cyclic Schedule S h d l
Layer 1 Layer 2 Layer 3 Computing Flow Time 8 cycles +210 minutes = 2130 minutes Hour: 1 2 3 4 Flow time = 2130 minutes = 35.5 hours Avg. inventory = 35.5 hours * 0.083 Alternatively, roll out the schedule...
3/28/2011 batches per hour = 2.94 batches Industrial Data and Systems Analysis 25 3/28/2011 Industrial Data and Systems Analysis 26 Cyclic Schedules ... Are a form of line flow simulation easily constructed rapidly solved C t Capture steady state b h i t d t t behavior Allow measurement of throughput flow time Ignore uncertainties Encourage understanding timing issues are central 3/28/2011 Industrial Data and Systems Analysis 27 ...
View
Full
Document
This note was uploaded on 03/18/2012 for the course ORIE 3120 taught by Professor Jackson during the Spring '09 term at Cornell.
 Spring '09
 JACKSON

Click to edit the document details