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

Tata Letak Fasilitas Manufaktur

D0394 Perancangan Sistem Manufaktur

Kuliah Ke XXI - XXII

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

1 2 3

4 5 6

1 du

1 du

d11 = 0

d12 = 1

d13 = 2

d14 = 1

d15 = 2

d16 = 3

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

min

: , , ,

, , ,

, ,

z r d x x

st x i n

x r n

x i r

ij rs ir jss

n

r

n

j

n

i

n

irr

n

iri

n

ir

1111

1

1

1 1

1 1

0 1

• 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

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


• 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)

1 2 3

4 5 6

location sites layout design


• 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)

1 2 3

4 5 6

location sites

5 1 4

2 3 6

layout design

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

C w d a i a jijj i

n

i

n

a

,11

1

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

a = (2, 4, 5, 3, 1, 6); C(a) = 114

5 1 4

2 3 6

From/To 1 2 3 4 5 61 0 3 2 2 1 22 1 0 3 1 0 53 4 1 0 1 2 24 0 1 1 0 4 15 3 2 0 2 0 56 2 3 4 1 5 0

C w d a i a jijj i

n

i

n

a

,11

1

Flow matrix (rij)

Solution Evaluation

• Given a, the total cost for department k is given by

p w d a i a k w d a k a j

p w d a k a j w

k iki k

kjk j n

k kjj

n

kk

a

a

, ,

,

1

1

0with

• What is the cost if the locations of departments u and v are exchanged? (a represents the new layout)

C C C

C w d a i a j w d a i a j

uv

uv ijj i

n

i

n

ijj i

n

i

n

a a a

a

, ,11

1

11

1

Pairwise Exchange

C w d a i a u w d a i a v w d a u a v

w d a i a u w d a i a v w d a u a v

C w d a i a u w d a i a v w d a u a v

w d a i a v

uv iu iv uvi

n

i

n

iu iv uvi

n

i

n

uv iui

n

ivi

n

uv

iu

a

a

, , ,

, , ,

, , ,

,

11

11

1 1

i v

ivi u

uv

uv iu ivi

n

uv

w d a i a u w d a u a v

C w w d a i a u d a i a v w d a u a v

, ,

, , , a1

2

Pairwise Exchange

a = (2, 4, 5, 3, 1, 6)

5 1 4

2 3 6

C(a) = 114

a = (2, 3, 5, 4, 1, 6)

5 1 2

4 3 6

Exchange departments 2 and 4

C(a) = 104C24(a) = 10

Pairwise Exchange

• If a least total cost assignment, a*, is found, then if any two departments are exchanged Cuv(a*) 0.

– Necessary condition for a least total cost assignment

– Not sufficient, in general, since k-way interchanges (k > 2) may improve the solution

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 F A specification implementation requires

2. select r a(F) particular rules for performing these steps

3. a(i) r

4. F F {i}

5. End

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

p a F r w d r a jk kjj F

,


• Specific Procedure1. Randomly select i {1,2,…,n}2. a(i) 13. While F< n

4. Randomly select i F

5. pi(a(F) k}) = min {pi(a(F) r}) r a(F)}6. a(i) k7. F F {i}

8. End

• Could repeat several times and pick best solution• Many variations on this basic procedure


• Example– Randomly select department 3

– Assign to location 1; a(3) = 1

From/To 1 2 3 4 5 61 0 3 2 2 1 22 1 0 3 1 0 53 4 1 0 1 2 24 0 1 1 0 4 15 3 2 0 2 0 56 2 3 4 1 5 0

3




– Randomly select department 4

From/To 1 2 3 4 5 61 0 3 2 2 1 22 1 0 3 1 0 53 4 1 0 1 2 24 0 1 1 0 4 15 3 2 0 2 0 56 2 3 4 1 5 0

3




– Randomly select department 4• w43d21 = (2)(1) = 2

• w43d31 = (2)(2) = 4

• w43d41 = (2)(1) = 2

• w43d51 = (2)(2) = 4

• w43d61 = (2)(3) = 6


From/To 1 2 3 4 5 61 0 3 2 2 1 22 1 0 3 1 0 53 4 1 0 1 2 24 0 1 1 0 4 15 3 2 0 2 0 56 2 3 4 1 5 0

3 4


• Example– Assign 3 to location 1; a(3) = 1

– Assign 4 to location 2; a(4) = 2

– Randomly select department 2• w42d32 + w32d31 = 2+8 = 10

• w42d42 + w32d41 = 4+4 = 8

• w42d52 + w32d51 = 2+8 = 10

• w42d62 + w32d61 = 4+12 = 14


From/To 1 2 3 4 5 61 0 3 2 2 1 22 1 0 3 1 0 53 4 1 0 1 2 24 0 1 1 0 4 15 3 2 0 2 0 56 2 3 4 1 5 0

3 4

2





– Randomly select department 5• w25d34 + w45d32 + w35d31 = 16

• w25d54 + w45d52 + w35d51 = 12

• w25d64 + w45d62 + w35d61 = 22


From/To 1 2 3 4 5 61 0 3 2 2 1 22 1 0 3 1 0 53 4 1 0 1 2 24 0 1 1 0 4 15 3 2 0 2 0 56 2 3 4 1 5 0

3 4

2 5






– Randomly select department 1• w51d35 + w21d34 + w41d32 + w31d31 = 34

• w51d65 + w21d64 + w41d62 + w31d61 = 34


From/To 1 2 3 4 5 61 0 3 2 2 1 22 1 0 3 1 0 53 4 1 0 1 2 24 0 1 1 0 4 15 3 2 0 2 0 56 2 3 4 1 5 0

3 4 1

2 5


• 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

From/To 1 2 3 4 5 61 0 3 2 2 1 22 1 0 3 1 0 53 4 1 0 1 2 24 0 1 1 0 4 15 3 2 0 2 0 56 2 3 4 1 5 0

3 4 1

2 5 6


• 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

2 6 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

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

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 Cuv(a)3. Decide whether or not to make the interchange4. Decide whether or not to continue

– A specific implementation requires rules for performing each of the steps

Improvement Heuristics

• Many reasonable rules exist for these steps. Consider,– Enumeration of all pairs in step 1 and 4

– Make exchange if Cuv(a) > 0

– Alternatively, make exchange between u and v such that Cuv(a) is the largest value for a given u.

Steepest Descent Pairwise Interchange

a a0

done falseWhile (not.done)

done truemax 0For i = 1 to n-1

For j = i+1 to nIf (Cij(a) > max) then

max Cij(a) u iv jdone false

EndifEndfor

Endfor

If (max > 0) thentemp a(u)a(u) a(v)a(v) temp

EndifEndwhile

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 Cuv(a) so it is never made

5 4 1

6 2 3

4 5 1

2 6 3

initial solution good solution

SDPI

VNZ Heuristic

• Order departments by TFCi : TFC[1] TFC[2] … TFC[n]

• Phase 1– Set m = M1 = [1] and M2 = [2]– Order list of departments i by non-increasing Cim(a). Proceed

through list making each switch provided Cis(a) > 0 (where a is updated assignment vector as switches are made) Repeat for m = M2

• Phase 2– Evaluate Cij(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.

TFC w d a i a ji ijj i

,

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

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