Group Technology and Facility Layout

Transcription

1 Group Techology ad Facility Layout Chapter 6 Beefits of GT ad Cellular Maufacturig (CM) REDUCTIONS Setup tie Ivetory Material hadlig cost Direct ad idirect labor cost IMPROVEMENTS Quality Material Flow Machie ad operator Utilizatio Space Utilizatio Eployee Morale Process layout Group techology layout VMM BM BM VMM VMM BM BM VMM Saple part-achie processig idicator atrix Rearraged part-achie processig idicator atrix M a c h i e P P [a] = a P3 1 1 M a c h i e M1 M4 M6 M2 M3 M5 M7 P P3 1 1 [a] = a P

2 Rearraged part-achie processig idicator atrix Rearraged part-achie processig idicator atrix M1 M4 M6 M2 M3 M5 M7 P P3 1 1 [a] = a P M a c h i e M1 M4 M6 M2 M3 M5 M7 P P P3 1 2 [a] = a P r P4 2 1 t P5 1 2 P Classificatio ad Codig Schees Classificatio ad Codig Schees Hierarchical No-hierarchical Hybrid Classificatio ad Codig Schees MICLASS Nae of Coutry Characteristics syste Developed TOYODA Japa Te digt code MICLASS The Netherlads Thirty digit code TEKLA Norway Twelve digit code BRISCH Uited Kigdo Based o four to six digit priary code ad a uber of secodary digits DCLASS USA Software-based syste without ay fixed code structure NIASH USSR A hierarchical code of te to fiftee digits ad a serial uber OPITZ West Geray Based o a five digit priary code with a four digit secodary code

3 Advatages of Classificatio ad Codig Systes Maxiize desig efficiecy Maxiize process plaig efficiecy Siplify schedulig Clusterig Approach Rak order clusterig Bod eergy Row ad colu askig Siilarity coefficiet Matheatical Prograig Rak Order Clusterig Algorith Rak Order Clusterig Algorith Step 1: Assig biary weight BW j = 2 -j to each colu j of the partachie processig idicator atrix Step 2: Deterie the decial equivalet DE of the biary value of each row i usig the forula j DEi 2 a j1 Step 3: Rak the rows i decreasig order of their DE values Break ties arbitrarily Rearrage the rows based o this rakig If o rearrageet is ecessary, stop; otherwise go to step 4 Step 4: For each rearraged row of the atrix, assig biary weight BW i = 2 -i Step 5: Deterie the decial equivalet of the biary value of each colu j usig the forula i DE j 2 a i1 Step 6: Rak the colus i decreasig order of their DE values Break ties arbitrarily Rearrage the colus based o this rakig If o rearrageet is ecessary, stop; otherwise go to step 1 Rak Order Clusterig Exaple 1 Rak Order Clusterig Exaple 1 Biary weight Biary value 74 P [a] = P P P Biary value Biary weight 32 P [a] = P P P

4 Rak Order Clusterig Exaple 1 Rak Order Clusterig Exaple 1 M4 M6 M1 M2 M3 M5 M7 Biary weight Biary value 112 P [a] = P P P M4 M6 M1 M2 M3 M5 M7 Biary value Biary weight 32 P [a] = P P P5 1 1 ROC Algorith Solutio Exaple 1 Bod Eergy Algorith M4 M6 M1 M2 M3 M5 M7 Biary value Biary weight 32 P [a] = P P P5 1 1 Step 1: Set i=1 Arbitrarily select ay row ad place it Step 2: Place each of the reaiig -i rows i each of the i+1 positios (ie above ad below the previously placed i rows) ad deterie the row bod eergy for each placeet usig the forula i1 i1 j1 a a a, j i1, j i1 Select the row that icreases the bod eergy the ost ad place it i the correspodig positio Bod Eergy Algorith BEA Exaple 2 Step 3: Set i=i+1 If i <, go to step 2; otherwise go to step 4 Step 4: Set j=1 Arbitrarily select ay colu ad place it Step 5: Place each of the reaiig -j rows i each of the j+1 positios (ie to the left ad right of the previously placed j colus) ad deterie the colu bod eergy for each placeet usig the forula j1 a ( a i1 j1 i, j1 ai, j1) Colu Row Step 6: Set j=j+1 If j <, go to step 5; otherwise stop 4

5 BEA Exaple 2 BEA Exaple 2 Row Selected Where Placed Row Arrageet 1 Above Row Below Row Above Row 2 3 Below Row 2 4 Above Row Below Row Row Bod Maxiize Eergy Eergy 4 Yes 4 Yes BEA Exaple 2 BEA Solutio Exaple 2 Colu Where Placed Colu Selected Arrageet 2 Left of Colu 1 2 Right of Colu 1 3 Left of Colu Right of Colu Left of Colu 1 4 Right of Colu 1 Colu Bod Maxiize Eergy Eergy 4 Yes 4 Yes Row ad Colu Maskig Algorith R&CM Algorith Exaple 3 Step 1: Draw a horizotal lie through the first row Select ay 1 etry i the atrix through which there is oly oe lie Step 2: If the etry has a horizotal lie, go to step 2a If the etry has a vertical lie, go to step 2b Step 2a: Draw a vertical lie through the colu i which this 1 etry appears Go to step 3 Step 2b: Draw a horizotal lie through the row i which this 1 etry appears Go to step 3 Step 3:If there are ay 1 etries with oly oe lie through the, select ay oe ad go to step 2 Repeat util there are o such etries left Idetify the correspodig achie cell ad part faily Go to step 4 Step 4: Select ay row through which there is o lie If there are o such rows, STOP Otherwise draw a horizotal lie through this row, select ay 1 etry i the atrix through which there is oly oe lie ad go to Step 2 M a c h i e P P [a] = a P

6 R&CM Algorith Exaple 3 R&CM Algorith - Solutio M a c h i e P P [a] = a P3 1 1 M a c h i e M1 M4 M6 M2 M3 M5 M7 P P3 1 1 [a] = a P Siilarity Coefficiet (SC) Algorith s where a k 1 ki k 1 aa ki kj a a a a ki kj ki kj 1 if part k requires achie i 0 otherwise, SC Algorith Exaple 4 M a c h i e P P [a] = a P3 1 1 SC Algorith Exaple 4 SC Algorith Exaple 4 Machie Pair SC Value Cobie ito oe cell? {1,2} 0/4= {1,3} 0/4= {1,4} 1/2 No {1,5} 0/3= {1,6} 1/2 No {1,7} 0/3= {2,3} 1/2 No {2,4} 0/5= {2,5} 2/3 Yes {2,6} 0/5= {2,7} 1/4 No {3,4} 0/5= {3,5} 1/4 No {3,6} 0/5= {3,7} 1/4 No {4,5} 0/4= {4,6} 2/2=1 Yes {4,7} 0/4= {5,6} 0/4= {5,7} 1/3 No {6,7} 0/4= Machie/Cell Pair SC Value Cobie ito oe cell? {1, (2,5)} {1, (4,6)} 1/2 Yes {1,3} {1,7} {(2,5), (4,6)} {(2,5), 3} 1/2 Yes {(2,5), 7} 1/3 No {(4,6), 3} {(4,6), 7} {3,7} 1/4 No 6

7 SC Algorith Exaple 4 SC Algorith Solutio Exaple 4 Machie/Cell Pair SC Value Cobie ito oe cell? {(1,4,6) (2,3,5)} {(1,4,6), 7} {(2,3,5), 7} 1/3 Yes Machie/Cell Pair SC Value Cobie ito oe cell? {(1,4,6) (2,3,5,7)} Matheatical Prograig Approach s k 1 where a ik aa ik jk k 1 a a a a ik jk ik jk 1 if part i requires achie k 0 otherwise, Weighted Mikowski etric d w a a k ki kj k1 r is a positive iteger w k is the weight for part k d istead of s to idicate that this is a dissiilarity coefficiet 1/ r r Special case where w k =1, for k=1,2,,, is called the Mikowski etric Easy to see that for the Mikowski etric, whe r=1, above equatio yields a absolute Mikowski etric, ad whe r=2, it yields the Euclidea etric The absolute Mikowski etric easures the dissiilarity betwee part pairs P-Media Model P- Media Model Exaple 5 Miiize Subject to dx i1 j1 j1 x 1 i =1,2,, x jj j1 x x i, j =1,2,, jj P x =0 or 1 i, j =1,2,, Setup LINGO odel for this exaple [d] =

8 Desig & Plaig i CMSs Desig & Plaig i CMSs Machie Capacity Safety ad Techological Costraits Upper boud o uber of cells Upper boud o cell size Iter-cell ad itra-cell aterial hadlig cost iiizatio Machie Utilizatio Machie Cost iiizatio Job schedulig i cells Throughput rate axiizatio M 4 MC 1 M 1 R M 6 M 2 M 3 MC 2 M 5 AGV M 1 M 4 M 7 MC 1 M 6 MC 2 R M 5 M 7 AGV M 1 M 2 M 3 Groupig ad Layout Project Groupig ad Layout Project - Solutio See iput data file for GTLAYPC progra Ru GTLAYPC progra See output data file for GTLAYPC progra 2023 Cell Cell Cell 3 8