lecture17 - MIT 2.852 Manufacturing Systems Analysis...

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Unformatted text preview: MIT 2.852 Manufacturing Systems Analysis Lecture 17 Long Lines Stanley B. Gershwin Spnng,2004 Copyright @2004 Stanley B. Gershwin. All rights reserved. ®E<§wa®w®mr Long Lines 0 Difficulty: *No simple formula for calculating production rate or inventory levels. * State space is too large for exact numerical solution. a: It all butter sizes are N and the length of the line is k, the number of states is S = 2’“(N + 1)’“—1. *ifN = 10 and k: = 20, S = 6.41 x 1025. a: Decomposition seems to work successfully. Copyright @2004 Stanley B. Gershwin. All rights reserved. 2 @E Long Lines Copyright @2004 Stanley B. Gershwin. All rights reserved. Okfls w If: Decomposition M B @k @mb [email protected]@lfi* Decomposition Long Lines - Consider an observer in Buffer Bi. * Imagine the material flow that the observer sees entering and leaving the buffer. c We construct a two-machine line (ie, we find r1, 131, T2, 332, and N) such that an observer in its buffer will see almost the same thing. 0 The parameters are chosen as functions of the behaviors of the other two-machine lines. Copyright @2004 Stanley B. Gershwin. All rights reserved. 4 em ®E<§wa®r Long Lines Examples Three-machine line — production rate. .8— : 135 .5— (EL; '5‘ '25 p1 = .05 4\ N1=N2=5 T @323 .65 .l .2 .3 .4 .5 .6 .7 Copyright @2004 Stanley B. Gershwin. All rights reserved. ®E<§wa®r Long Lines Examples Three-machine line — total average inventory r1=r2=r3—.2 p1=.05 N1=N2=5 Copyright @2004 Stanley B. Gershwin. All rights reserved. @iE Long Lines 50 Machines; r=0.‘i', p=0.01', mu=1.0', N=20.0 15 Average Buffer LemI el 0 5 Buh‘er Nu mber Copyright @2004 Stanley B. Gershwin. All rights reserved. 10 - — 0 _ _ 10 15 23 35 30 35 40 45 5 0 e Hg @t Examples Distribution of em material in a line with identical machines and butte rs. Explain the shape. @E Long Lines 15 10 Average Buffer Level 50 Machines; r=0.1', p=0.0‘l', mu =10; N=20.0 EXCEPT pl10)=0.03?5 | 0 5 10 15 23 25 30 35 40 45 Bulfer Nu mber Copyright @2004 Stanley B. Gershwin. All rights reserved. 50 .MS. Hg @. Examples Effect of a _ bottleneck. Identical - machines and buffers, except for _ M10. mmeemem Long Lines Examples \‘gg: Continuous material model. 20— . Eight-machine, seven-buffer line. 0 For each machine, 7' = .075, p = .009, p, = 1.2. o For each buffer (except Buffer 6), N = 30. Copyright @2004 Stanley B. Gershwin. All rights reserved. 9 mmeemem Long Lines 25 I I I I I I I I | H1 H3 H5 20 - “T 15— 10- Copyright @2004 Stanley B. Gershwin. All rights reserved. Examples o Which m are decreasing and which are increas- ing? 0 Why? 10 mmeemem Long Lines Examples Which has a higher production rate? . 9—Maohine line with two buffering options: *8 buffers equally sized; and *2 buffers equally sized. *Mfiwiwh 9 WWW Copyright @2004 Stanley B. Gershwin. All rights reserved. 11 mmeemem Long Lines Examples | 8 buffers 2 buffers 0.95 0 Continuous model; all machines have 7' = .019, p = .001, p, = 1. o What are the asymptotes? 0.9 0.85 0.8 0.?5 0.7 o Is 8 buffers always faster? 0.55 0 L000 2000 3000 4000 5000 5000 TI‘000 8 000 9000 100 Total Buffer Space Copyright @2004 Stanley B. Gershwin. All rights reserved. 12 mmeemem Long Lines Examples _ o Is 8 buffers always faster? — 0 Perhaps not, but difference is not significant in systems with very small buffers. 10 100 1003 100 Copyright @2004 Stanley B. Gershwin. All rights reserved. 13 mmeemem Long Lines Optimal buffer space distribution. 0 Design the buffers for a 20-machine production line. 0 The machines have been selected, and the only decision remaining is the amount of space to allocate for in-process inventory. 0 The goal is to determine the smallest amount of in-process inventory space so that the line meets a production rate target. Copyright @2004 Stanley B. Gershwin. All rights reserved. 14 mmoomom Long Lines Optimal buffer space distribution. 0 The common operation time is one operation per minute. 0 The target production rate is .88 parts per minute. Copyright @2004 Stanley B. Gershwin. All rights reserved. 15 mmeeeem Long Lines Optimal buffer space distribution. o Case 1 MTTF= 200 minutes and MTTR = 10.5 minutes for all machines (P = .95 parts per minute). Copyright @2004 Stanley B. Gershwin. All rights reserved. 15 emeeeem Long Lines Optimal buffer space distribution. 0 Case 1 MTTF= 200 minutes and MTTR = 10.5 minutes for all machines (P = .95 parts per minute). 0 Case 2 Like Case 1 except Machine 5. For Machine 5, MTTF = 100 and MTTR = 10.5 minutes (P = .905 parts per minute). Copyright @2004 Stanley B. Gershwin. All rights reserved. 17 emeeeee Long Lines Optimal buffer space distribution. o Case 1 MTTF= 200 minutes and MTTR = 10.5 minutes for all machines (P = .95 parts per minute). . Case 2 Like Case 1 except Machine 5. For Machine 5, MTTF = 100 and MTTR = 10.5 minutes (P = .905 parts per minute). 0 Case 3 Like Case 1 except Machine 5. For Machine 5, MTTF = 200 and MTTR = 21 minutes (P = .905 parts per minute). Copyright @2004 Stanley B. Gershwin. All rights reserved. 18 mmeemem Long Lines Optimal buffer space distribution. Are buffers really needed? Line Production rate with no buffers, parts per minute Case 1 .487 Case 2 .475 Case 3 .475 Yes. How were these numbers calculated? Copyright @2004 Stanley B. Gershwin. All rights reserved. 19 mmoomom Long Lines Optimal buffer space distribution. Solution lCase‘l ——AII Machines Identical lCase2—— MachineS Boltleneck— I'u'I'TF =1CO EICaseS—— MachineS Bottleneck — I'u'I'TFl = 21 Line Space Case 1 430 Buffer Size Case 2 485 Case 3 523 I I I I I I I" 3 9 10 11 12 13 14 15 1B 1? 13 19 Buffer I I I 1 2 3 4 5 8 Copyright @2004 Stanley B. Gershwin. All rights reserved. 20 mmeemem Long Lines Optimal buffer space distribution. 0 Observation from studying buffer space allocation problems: * Buffer space is needed most where buffer level variability is greatest! Copyright @2004 Stanley B. Gershwin. All rights reserved. 21 mmeemem Long Lines Profit as a function of buffer sizes mo --— am --—-- 310 —-—-- 222:: oThree-maohine, continuous as: ----— Profit 332 " m- material line. on — .1, p3- — .0141:- = 1. OH = N2) —(fi1 + 172,2). Copyright @2004 Stanley B. Gershwin. All rights reserved. 22 ...
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This note was uploaded on 02/24/2012 for the course ENGINEERIN 2.852 taught by Professor Stanleyb.gershwin during the Spring '04 term at MIT.

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lecture17 - MIT 2.852 Manufacturing Systems Analysis...

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