Tata Letak Fasilitas Manufaktur

1
Tata Letak Fasilitas Manufaktur

• D0394 Perancangan Sistem Manufaktur
• Kuliah Ke XXI - XXII

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QAP Formulation

• Define,
• rij rate of item movement between departments i
  and j
• drs distance between locations r and s
• xir 1, if department i is assigned to location
  r 0, otherwise

d11 0 d12 1 d13 2 d14 1 d15 2 d16 3
1 du
1 du
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QAP Formulation

• Define,
• rij rate of item movement between departments i
  and j
• drs distance between locations r and s
• xir 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

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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

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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)
location sites
layout design
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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)
5
1
4
2
3
6
location sites
layout design
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Solution Evaluation

• Assume that the direction of flow is unimportant
• So weight between departments i and j is wij
  rij rji
• Assume distance matrix is symmetric
• Total flow cost is

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Solution Evaluation

• Assume that the direction of flow is unimportant
• So weight between departments i and j is wij
  rij rji
• Assume distance matrix is symmetric
• Total flow cost is

Flow matrix (rij)
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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)

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Pairwise Exchange
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Pairwise Exchange
Exchange departments 2 and 4
C(a) 104
DC24(a) 10
C(a) 114
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Pairwise Exchange

• If a least total cost assignment, a, is found,
  then if any two departments are exchanged
  DCuv(a) 0.
• Necessary condition for a least total cost
  assignment
• Not sufficient, in general, since k-way
  interchanges (k gt 2) may improve the solution

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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 ½F½lt n
• 1. select i Ï F A specification implementation
  requires
• 2. select r Ï a(F) particular rules for
  performing these steps
• 3. a(i) r
• 4. F F È i
• 5. End

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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

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Construction Heuristics

• Specific Procedure
• 1. Randomly select i Î1,2,,n
• 2. a(i) 1
• 3. While ½F½lt n
• 4. Randomly select i Ï F
• 5. pi(a(F) Èk) min pi(a(F) Èr) ½ r Ï
  a(F)
• 6. a(i) k
• 7. F F È i
• 8. End
• Could repeat several times and pick best solution
• Many variations on this basic procedure

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Construction Heuristics

• Example
• Randomly select department 3
• Assign to location 1 a(3) 1

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Construction Heuristics

• Example
• Randomly select department 3
• Assign to location 1 a(3) 1
• Randomly select department 4

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Construction Heuristics

• Example
• Randomly select department 3
• Assign to location 1 a(3) 1
• Randomly select department 4
• w43d21 (2)(1) 2
• w43d31 (2)(2) 4
• w43d41 (2)(1) 2
• w43d51 (2)(2) 4
• w43d61 (2)(3) 6
• Assign to location 2 a(4) 2

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Construction Heuristics

• Example
• Assign 3 to location 1 a(3) 1
• Assign 4 to location 2 a(4) 2
• Randomly select department 2
• w42d32 w32d31 28 10
• w42d42 w32d41 44 8
• w42d52 w32d51 28 10
• w42d62 w32d61 412 14
• Assign to location 4 a(2) 4

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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
• w25d34 w45d32 w35d31 16
• w25d54 w45d52 w35d51 12
• w25d64 w45d62 w35d61 22
• Assign to location 5 a(5) 5

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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
• w51d35 w21d34 w41d32 w31d31 34
• w51d65 w21d64 w41d62 w31d61 34
• Assign to location 3 a(1) 3

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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

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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

• Example
• 5 6 largest weights 2 3 close to 6 1 close
  to 3
• a (3, 4, 6, 1, 2, 5)
• C(a) 92

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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

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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 a0
• 1. Select a pair of facilities (u, v)
• 2. Evaluate DCuv(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

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Improvement Heuristics

• Many reasonable rules exist for these steps.
  Consider,
• Enumeration of all pairs in step 1 and 4
• Make exchange if DCuv(a) gt 0
• Alternatively, make exchange between u and v such
  that DCuv(a) is the largest value for a given u.

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Steepest Descent Pairwise Interchange

• a a0
• done false
• While (not.done)
• done true
• max 0
• For i 1 to n-1
• For j i1 to n
• If (DCij(a) gt max) then
• max DCij(a)
• u i
• v j
• done false
• Endif
• Endfor
• Endfor

• If (max gt 0) then
• temp a(u)
• a(u) a(v)
• a(v) temp
• Endif
• Endwhile

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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 DCuv(a) so it is never made

SDPI
initial solution
good solution
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VNZ Heuristic

• Order departments by TFCi TFC1 ³ TFC2 ³ ³
  TFCn
• Phase 1
• Set m M1 1 and M2 2
• Order list of departments i by non-increasing
  DCim(a). Proceed through list making each switch
  provided DCis(a) gt 0 (where a is updated
  assignment vector as switches are made) Repeat
  for m M2
• Phase 2
• Evaluate DCij(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.

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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