1. Tata Letak Fasilitas Manufaktur D0394 Perancangan Sistem Manufaktur Kuliah Ke XXI - XXII

2. QAP Formulation Define, r ij = rate of item movement between departments i and j d rs = distance between locations r and s x ir = 1, if department i is assigned to location r; 0, otherwise 12 3 4 56 1 du d 11 = 0 d 12 = 1 d 13 = 2 d 14 = 1 d 15 = 2 d 16 = 3

3. QAP Formulation Define, r ij = rate of item movement between departments i and j d rs = distance between locations r and s x ir = 1, if department i is assigned to location r; 0, otherwise z = total distance items move objective has quadratic form constraints are assignment contraints –every dept. to one location –every location one dept. Quadratic Assignment Problem

4. Solution Representation Represent a solution to the facility layout problem as a permutation vector a –a = (a(1), a(2), …, a(n)) Element a(i) represents the location to which department i is assigned –a(3) = 5 implies that department 3 is assigned to location 5

5. Solution Representation Represent a solution as a permutation vector a –Element a(i) represents the location to which department i is assigned –Example: a = (2, 4, 5, 3, 1, 6) 12 3 4 56 location siteslayout design

6. Solution Representation Represent a solution as a permutation vector a –Element a(i) represents the location to which department i is assigned –Example: a = (2, 4, 5, 3, 1, 6) 12 3 4 56 location sites 51 4 2 36 layout design

7. Solution Evaluation Assume that the direction of flow is unimportant –So weight between departments i and j is w ij = r ij + r ji Assume distance matrix is symmetric Total flow cost is

8. Solution Evaluation Assume that the direction of flow is unimportant –So weight between departments i and j is w ij = r ij + r ji Assume distance matrix is symmetric Total flow cost is a = (2, 4, 5, 3, 1, 6); C(a) = 114 51 4 2 36 Flow matrix (r ij )

9. Solution Evaluation Given a, the total cost for department k is given by What is the cost if the locations of departments u and v are exchanged? (a represents the new layout)

10. Pairwise Exchange

11. a = (2, 4, 5, 3, 1, 6) 51 4 2 36 C(a) = 114 a = (2, 3, 5, 4, 1, 6) 51 2 4 36 Exchange departments 2 and 4 C(a) = 104  C 24 (a) = 10

12. Pairwise Exchange If a least total cost assignment, a *, is found, then if any two departments are exchanged  C uv (a * )  0. –Necessary condition for a least total cost assignment –Not sufficient, in general, since k-way interchanges (k > 2) may improve the solution

13. Solution Generation Construction Heuristics –Begin with the basic problem data and build up a solution in an iterative manner General Procedure –Let, a(i) = 0 if department i has not been assigned to a location –Let, a(F) be the set of locations assigned to departments in set F 0. While  F  < n 1. select i  FA specification implementation requires 2. select r  a(F)particular rules for performing these steps 3. a(i)  r 4. F  F  {i} 5. End

14. Construction Heuristics Many reasonable rules are possible for steps 1 and 2. Consider, –Random department selection in step 1 –Minimize additional total cost for partial solution in step 2 Partial solution is (F, a(F)) with cost C(a(F)) If we augment the partial solution by assigning department k to location r, we obtain an increase in cost as follows

15. Construction Heuristics Specific Procedure 1. Randomly select i  {1,2,…,n} 2. a(i)  1 3. While  F  < n 4. Randomly select i  F 5. p i (a(F)  k}) = min {p i (a(F)  r})  r  a(F)} 6. a(i)  k 7. F  F  {i} 8. End Could repeat several times and pick best solution Many variations on this basic procedure

16. Construction Heuristics Example –Randomly select department 3 –Assign to location 1; a(3) = 1 3

17. Construction Heuristics Example –Randomly select department 3 –Assign to location 1; a(3) = 1 –Randomly select department 4 3

18. Construction Heuristics Example –Randomly select department 3 –Assign to location 1; a(3) = 1 –Randomly select department 4 w 43 d 21 = (2)(1) = 2 w 43 d 31 = (2)(2) = 4 w 43 d 41 = (2)(1) = 2 w 43 d 51 = (2)(2) = 4 w 43 d 61 = (2)(3) = 6 –Assign to location 2; a(4) = 2 3 4

19. Construction Heuristics Example –Assign 3 to location 1; a(3) = 1 –Assign 4 to location 2; a(4) = 2 –Randomly select department 2 w 42 d 32 + w 32 d 31 = 2+8 = 10 w 42 d 42 + w 32 d 41 = 4+4 = 8 w 42 d 52 + w 32 d 51 = 2+8 = 10 w 42 d 62 + w 32 d 61 = 4+12 = 14 –Assign to location 4; a(2) = 4 3 4 2

20. Construction Heuristics Example –Assign 3 to location 1; a(3) = 1 –Assign 4 to location 2; a(4) = 2 –Assign 2 to location 4; a(2) = 4 –Randomly select department 5 w 25 d 34 + w 45 d 32 + w 35 d 31 = 16 w 25 d 54 + w 45 d 52 + w 35 d 51 = 12 w 25 d 64 + w 45 d 62 + w 35 d 61 = 22 –Assign to location 5; a(5) = 5 3 4 25

21. Construction Heuristics Example –Assign 3 to location 1; a(3) = 1 –Assign 4 to location 2; a(4) = 2 –Assign 2 to location 4; a(2) = 4 –Assign 5 to location 5; a(5) = 5 –Randomly select department 1 w 51 d 35 + w 21 d 34 + w 41 d 32 + w 31 d 31 = 34 w 51 d 65 + w 21 d 64 + w 41 d 62 + w 31 d 61 = 34 –Assign to location 3; a(1) = 3 3 4 1 25

22. Construction Heuristics Example –Assign 3 to location 1; a(3) = 1 –Assign 4 to location 2; a(4) = 2 –Assign 2 to location 4; a(2) = 4 –Assign 5 to location 5; a(5) = 5 –Assign 1 to location 3; a(1) = 3 –Assign 6 to location 6; a(6) = 6 –a = (3, 4, 1, 2, 5, 6) –C(a) = 108 3 4 1 25 6

23. Construction Heuristics Observations –Many different variations of the construction procedure –Clearly the initial location has an effect as does the department sequence –Intuitively, you want large weights near the center and small weights near the outside Difficult to formalize as a general algorithm 4 5 1 26 3 –Example 5 & 6 largest weights; 2 & 3 close to 6; 1 close to 3 a = (3, 4, 6, 1, 2, 5) C(a) = 92

24. Solution Quality How good is the solution? Lower Bound –Order location pairs by increasing distance, d Preferred locations –Order weights by decreasing flow volume, w Highest activities –“Assign” largest weights to preferred locations –LB = d  w –d = (1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3) –w = (10, 8, 6, 6, 6, 4, 4, 4, 4, 2, 2, 2, 2, 2, 2) –LB = 10(1) + 8(1) + 6(1) + … + 2(3) + 2(3) = 88

25. Improvement Heuristics Modify a given solution so that the total cost is reduced Pairwise interchange –Select two departments and interchange their locations –General Procedure 0. a  a 0 1. Select a pair of facilities (u, v) 2. Evaluate  C uv (a) 3. Decide whether or not to make the interchange 4. Decide whether or not to continue –A specific implementation requires rules for performing each of the steps

26. Improvement Heuristics Many reasonable rules exist for these steps. Consider, –Enumeration of all pairs in step 1 and 4 –Make exchange if  C uv (a) > 0 –Alternatively, make exchange between u and v such that  C uv (a) is the largest value for a given u.

27. Steepest Descent Pairwise Interchange a  a 0 done  false While (not.done) done  true max  0 For i = 1 to n-1 For j = i+1 to n If (  C ij (a) > max) then max  C ij (a) u  i v  j done  false Endif Endfor If (max > 0) then temp  a(u) a(u)  a(v) a(v)  temp Endif Endwhile

28. Improvement Heuristics Pairwise Interchange has several difficulties –May be “trapped” in bad solution Departments 5 and 6 have a large flow between them so if they get trapped on the outside, any exchange that moves one and not the other will have a negative  C uv (a) so it is never made 5 4 1 62 3 4 5 1 26 3 initial solution good solution SDPI

29. VNZ Heuristic Order departments by TFC i : TFC [1]  TFC [2]  …  TFC [n] Phase 1 –Set m = M 1 = [1] and M 2 = [2] –Order list of departments i by non-increasing  C im (a). Proceed through list making each switch provided  C is (a) > 0 (where a is updated assignment vector as switches are made) Repeat for m = M 2 Phase 2 –Evaluate  C ij (a) for each dept. pair 1 and 2, 1 and 3, …, M-1 and M. Exchange i and j if cost is reduced. –Continue until every pair has been examined without making a change or each pair has been examined twice.

30. Improvement Heuristics Initial (starting) solution is important -- try several! Could consider k-wise interchanges –Computational burden increases greatly “Good” starting solution not necessary –In general, more effort should be expended in the improvement phase –Quickly, generate a large variety of starting solutions and then try to improve them