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3214-5-1%20(part%202)

3214-5-1%20(part%202) - 4.4 GROUP TECHNOLOGY(GT SYSTEMS...

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Unformatted text preview: 4.4 GROUP TECHNOLOGY (GT) SYSTEMS Review of GT systems: • GT: make similar products in a similar way - similar products ⇒ similar processes ⇒ family - similar way ⇒ same group GT flowline, cell, or center 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 1 - make each product in one group (ideally) • P/Q ratio: medium (medium variety of products, medium quantity of each) • objective: minimize flowtime for a variety of similar products ISE 5204 Course Notes 4.4 - 2 2006 John P. Shewchuk 4.4.1 Family/Machine Type Group Formation Given: N different part types (products) to be made. M different machine types required. Processing data: 1 if product i needs machine type m aim = 0 o/w 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 11 Find: Part type families. Machine type groups. Assignment of families to groups. Objective: Minimize intergroup flow (part flow between groups). (other objectives possible) Solution: Let’s use Binary Ordering (most common). 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 12 Binary Ordering (6.4.2 text) Express processing requirements in matrix form, e.g., part type, i 1 A machine B type, m C D 1 1 2 1 3 1 1 1 4 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 13 Rearrange rows and columns of matrix to obtain block diagonal structure, e.g., 1 A B1 C B1 A C D 3 2 1 1 1 2 3 1 1 1 4 1 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 14 Determine solution from final structure, e.g., 13 B1 1 C 11 A D • 2 4 mach group types 1 B,C A,D 2 products 1,3 2,4 12 11 note: usually consider specific criteria (keep groups same size, maximum group size, etc.) in forming groups. ISE 5204 Course Notes 4.4 - 15 2006 John P. Shewchuk 1. Order rows (machine types). Find rm 1234↓ A 1 4 qi =“weight” of col. i 10 B1 1 = 2N–i ∀ i 2 C 1 D 1 15 rm =rank of row m qi → 8 4 2 1 N = aim qi ∀ m Rows → B,D,A,C i =1 Sort rows by nonincreasing rm . If rows previously ordered and no change, done. Σ 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 17 2. Order columns (part types). Find qm 1234↓ B1 1 8 qm =“weight” of row m D 1 14 = 2M–m ∀ m A 1 2 C 1 1 ri =rank of col. i M ri → 8 6 9 4 = aim qm ∀ i Cols. → 3,1,2,4 m =1 Sort columns by nonincreasing ri . If no change, done; else repeat from step 1. Σ 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 18 Example 1 A B C D E F 1 1 1 1 1 32 16 8 4 2 ⇒ row order = B,D,E,A,C,F ISE 5204 Course Notes 2 3 1 4 1 5 1 1 6 10 36 10 1 17 1 17 4 1 2006 John P. Shewchuk 4.4 - 19 B D E A C F 32 1 1 16 1 18 1 1 4 1 1 2 1 1 32 24 6 33 6 24 ⇒ column order = 4,1,2,6,3,5 ISE 5204 Course Notes 4.4 - 20 1 1 2 3 4 1 5 6 2006 John P. Shewchuk B D E A C F 4 1 1 1 2 1 1 6 1 1 3 5 48 12 12 13 13 32 1 1 1 1 32 16 8 4 2 ⇒ row order = B,F,D,E,A,C ISE 5204 Course Notes 2006 John P. Shewchuk 4.4 - 21 B F D E A C 32 16 1 1 8 1 1 4 1 12 1 11 48 32 12 12 3 3 no change in column order ⇒ done ISE 5204 Course Notes 4.4 - 22 4 1 1 1 1 2 6 3 5 2006 John P. Shewchuk Grouping: 4 1 B 1 1 F 11 D E A C 2 6 3 5 1 1 1 21 1 1 3 1 1 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 23 Final solution: group 1 2 3 mach types B,F D,E A,C products 1,4 2,6 3,5 How good is this solution? no intergroup flow ⇒ optimal (w.r.t. minimizing intergroup flow). 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 24 What if part type 3 needs machine type D as well as A and C? B F D E A C 4 1 1 1 1 2 6 3 5 1 1 1 1 1 1 1 1 1 D needed in two groups ⇒ potential problem 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 25 One approach for handling this problem: look at machine utilizations. • If ρD high, use one (or more) type D machines each group 2 E D A D 3 C high ρD ⇒ can justify extra machine 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 26 • If ρD low, try and share a single type D machine 2 E D A C 3 e.g., back-to-back U cells. note: routings must allow (arrangement affected) 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 27 How good are these solutions? • Both solutions are optimal. (zero intergroup flow) • In some cases, it may be better to simply allow intergroup flow. (e.g., D in group 2 only: part type 3 travels to group 2 for processing) See also Example 6.2 text. 2006 John P. Shewchuk ISE 5204 Course Notes 4.4 - 28 Pros of binary ordering: • • Simple. Fast and efficient – can handle large industrial problems. Uses limited information. - doesn’t consider setup times, processing times, batch sizes, etc. Doesn’t actually give final solution. - human judgement required. ISE 5204 Course Notes 4.4 - 29 Cons: • • 2006 John P. Shewchuk ...
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