Facilities Layout and Location

The three most important criteria in locating a factory: Location! Location! Location!

Supply Chain Homework Formulation

Xi,j = amount of raw natural gas sent from field i to plant j 106 ft3 , i =

A,B,C,D,F; j = F,G,H,I

 Yjkl = 106 ft3 finished gas product k produced at plant j and sent to customer

l; l = J,K,L,M,N,O

 Pk production cost of gas k; $/106 ft3

 Ti,j = delivery cost by pipeline from field i to plant j; $/106 ft3

 Tjl = delivery cost by truck from plant j to customer l; $/106

 Dij = distance of pipeline from field i to plant j; miles

 Djl = distance of road from plant j to customer l; miles

 Ajk = amount of product k produced at plant j as a fraction of one unit of

raw natural gas

Supply Chain Homework Formulation

min

subject to:

(field capacity)

(plant capacity)

(for each k)

(customer demands for gas )

ij ij jl jklki j j k l

ij ij

ij ji

jk ij jkli j j l

jkl klj

T X T P Y

X S

X R

A X Y

Y D l k

Facilities Layout Techniques Applied to:

Hospitals Warehouses Schools Offices Workstations

e.g. offices cubicles or manufacturing cells Banks Shopping centers Airports Factories

Objectives of a Facility Layout Problem

Minimize investment in new equipment Minimize production time Utilize space most efficiently Provide worker convenience and safety Maintain a flexible arrangement Minimize material handling costs Facilitate manufacture Facilitate organizational structure

Types of Layouts

Fixed Position Layouts – suitable for large items such as airplanes.

Product Layouts – work centers are organized around the operations needed to produce a product.

Process Layouts – grouping similar machines that have similar functions.

Group Technology Layouts – layouts based on the needs of part families.

Fixed Position Layout

Product layout

Process layout

Group Technology Layout

Computerized Layout Techniques

CRAFT. An improvement technique that requires the user to specify an initial layout. Improves materials handling costs by considering pair-wise interchange of departments.

COFAD. Similar to CRAFT, but also includes consideration of the type of materials handling system.

ALDEP. Construction routine (does not require user to specify an initial layout). Uses REL chart information.

CORELAP. Similar to ALDEP, but uses more careful selection criteria for initial choosing the initial department

PLANET. Construction routine that utilizes user specified priority ratings.

Activity Relationship Chart

The desirability of locating pairs of operations near each other:

A – absolutely necessaryE – especially importantI – importantO – ordinary importanceU – unimportantX - undesirable

Reason code:1 - Flow of material2 - Ease of supervision3 - Common personnel4 - Contact necessary5 - Convenience

Offices

Floor manager

Conference room

Post office

Parts shipment

Repair & Servicing

Receiving

Inspection

E/4I/5

I/1U

U

UU

UU

E/1U

A/1

E

E/1

O

U

O/5

U

UI/2

I/2U

UI/1

E

U

U

U

Activity Relationship Chart

Assembly Layout n assembly areas to be located on the

factory floor in m possible locations (m n) Minimize material handling requirements

Material handling is from the receiving (raw material) to each assembly area

and from each assembly area to shipping

The Factory Floor

Rec

eivi

ng

Sh

ipp

ing

assembly locations

The Cost Coefficients

ei,1 = trips per day from assembly area i to receivingei,2 = trips per day from assembly area i to shippingdj,1 = distance in feet from location j to receivingdj,2 = distance in feet from location j to shipping

ci,j = cost per day for assembly area i to be located in location jci,j = ei,1dj,1 + ei,2dj,2

An Optimization Model

1 1

1

1

. . 1 1,2,...,

1 1,2,...,

n n

ij iji j

n

iji

n

ijj

Min z c x

s t x j n

x i m

Let Xi,j = 1 if assembly area i is at location j; 0 otherwise

Discrete Location Assignment Problem

LocationAssembly A B C D

1 94 13 62 712 62 19 84 963 75 88 18 804 11 M 81 21

materials handling costs 1 – B2 – A3 – C4 – Dcosts = $114

Locate M new facilities among N potential sites with k existing facilities

new sitesF G H

A 1 3 5existing B 2 3 4facilties C 4 5 6

D 3 1 2E 5 2 3

distance in miles

new facilitiesI j k

A 4 2 2existing B 3 2 1facilties C 6 4 7

D 2 1 4E 7 8 9

trips per day

A (k x N) B (k x M)

C (N x M)

C = At B new faciltiesi j k

F 75 67 89new sites G 67 49 66

H 93 68 91Facility i located at site F:C11 = (1)(4) + (2)(3) + (4)(6) + (3)(2) + (5)(7) = 75

From-to Chart (distances in feet)

To

From

saws milling punch

press

drills lathes sanders

saws 18 40 30 65 24

milling 18 38 75 16 30

punch

press40 38 22 38 12

drills 30 75 22 50 46

lathes 65 16 38 50 60

sanders 24 30 12 46 60

From-to Chart - trips per day

To

From

saws milling punch

press

drills lathes sanders

saws 43 26 14 40

milling 75 60 23

punch

press 45 16

drills 22 28

lathes 45 30 60

sanders 12

From-to Chart - Cost per day ($)

To

From

saws milling punch

press

drills lathes sanders

saws 154.8 208 84 520

milling 570 900 138

punch

press 342 38.4

drills 330 280

lathes 144 300 720

sanders 72

Cost per day = cost per foot x distance in feet x trips per day

Quadratic Assignment Problem

Problem: Place each of n facilities in one of n locations wherework-in-process moves among the facilitiesn! possible solutions.

Plant LayoutFacilitiesLathe shopdrillingsandingrough polishingfinishinginspectiongalvanizingpaint shoppackagingquality controlraw material storagefinished goods storageassemblycell 1cell 2cell 3

Quadratic Assignment Problem

let ci,j,k,l = the “cost” of placing machine i in location k and machine j in location l ci,j,k,l = fi,j dk,l

where fi,j = the number of trips per day between machine i to machine j dk,l = the distance in feet between location k and location l

let xi,k = 1 if facility i is placed in location k; 0 otherwiselet xj,l = 1 if facility j is placed in location l; 0 otherwise

Min z c x x

subj to x k n

x i n

i j k l i kljki

j l

i ki

i kk

, , , , ,

,

,

: , , ,...,

, , ,...,

1 1 2

1 1 2

E F G HA 3 1 5 EB 20 4 6 FC 10 14 7 GD 12 8 9 H A B C D

dk,lfi,j

locations

facilities LocationE F G H CostA B C D

A B C DA: 20(3) + 10(1) + 12(5) = 130B: 14(4) + 8(6) = 104C: 9(7) = 63

total = 297

297

E F G HA 3 1 5 EB 20 4 6 FC 10 14 7 GD 12 8 9 H A B C D

dk,lfi,j

locations

facilities

Best pairwise exchange: From baselinesolution, compute all possible interchanges.Select the best one; then repeat until nofurther improvement is obtained.

LocationE F G H CostA B C D 297

B A C D 289A B D C 301D B C A 339A C B D 276C B A D 309A D C B 328

LocationE F G H CostA C B D 276

C A B D 297D C B A 351A C D B 318A D B C 290B C A D 280

n2( ) paired

exchanges

Any Improvement Heuristic: From baselinesolution, interchange pairs until an improve-ment is obtained. Then repeat until all pairshave been found with no further improvement.

LocationE F G H CostA B C D 297

B A C D 289B A D C 309B C A D 280D C A B 343 etc.

Facility Location Goal is to find the optimal location of one or more

new facilities. Optimality depends on the objective used. In many systems, the objective is to minimize some measure of distance. Two common distance measures:

Straight line distance (Euclidean distance). Rectilinear Distance (as might be measured following roads

on city streets).

x a y b

Facility Location Analysis Locating on a

plane sphere 3-dimensional space network discrete location

Minimize costs distances weighted distances travel time

Single versus multiple facilities Straight-line (Euclidean) versus Rectilinear

Distances

Locating on the Plane-Euclidean Distances

x

y

(a,b)

(x,y)

(x – a)

(y – b) h

h2 = (x – a)2 + (y – b)2

2 2h x a y b

The Centroid Problemlet x = the x-coordinate y = the y-coordinate(ai,bi) = coordinate of ith facility wi = weight placed on the ith facility

2 2

1

min ( , )n

i i ii

f x y w x a y b

1

1

2 ( ) 0

2 ( ) 0

n

i ii

n

i ii

fw x a

x

fw y b

y

=

=

¶= - =

¶

¶= - =

¶

å

å

Euclideandistancesquared

The Solution

1

1

2 ( ) 0

2 ( ) 0

n

i ii

n

i ii

fw x a

x

fw y b

y

=

=

¶= - =

¶

¶= - =

¶

å

å

1 1

1 1

1

1

0

*

n n

i i ii i

n n

i i ii i

n

i ii

n

ii

w x w a

x w w a

w ax

w

= =

= =

=

=

- =

=

=

å å

å å

å

å1

1

*

n

i ii

n

ii

wby

w

=

=

=å

å

2

21

2 2

21

2 0

2 0 0

n

ii

n

ii

fw

x

f fw

y x y

=

=

¶= >

¶

¶ ¶= > =

¶ ¶ ¶

å

å

convex function

The Euclidean Distance Problem

2 2

1

min ( , )n

i i ii

f x y w x a y b

2 21

2 21

2 ( )10

2 ( ) ( )

2 ( )10

2 ( ) ( )

ni i

i i i

ni i

i i i

w x af

x x a y b

w y bf

y x a y b

=

=

-¶= =

¶ - + -

-¶= =

¶ - + -

å

å

2 2let ( , )

( ) ( )i

i

i i

wg x y

x a y b

1

1

1

1

( , )Then

( , )

( , )and

( , )

n

i ii

n

ii

n

i ii

n

ii

a g x yx

g x y

b g x yy

g x y

If at least one half of the cumulative weight is associated

with an existing facility, the optimum location for the new facility will coincide with the existing facility. I call this the Majority Theorem. It is also true that the optimum location

will always fall within the convex hull formed from the

existing points.

A Simple Proof using an Analog Model

The convex hull

Centroid Problem - Example

x

y

1

2

3

4

5

x,y

(2,2)

(6,3)

(9,1)

(8,5)

(3,6)

5

10

6

5

2

Centroid Problem

Location wi ai bi wixi wiyi

P1 5 2 2 10 10P2 10 3 6 30 60P3 2 6 3 12 6P4 6 8 5 48 30P5 5 9 1 45 5 total 28 145 111

x* = 145/28 = 5.2 y* = 111/28 = 4.0

Euclidean Distance - Example

x

y

central repair facilityw = vehicles serviced per day

1

3

2

(1100,400)

(500,800)

(1400,1200)

w2 = 8

w1 = 4

w3 = 8

2 2 2 2 2 2min ( , ) 4 ( 500) ( 800) 8 ( 1100) ( 400) 8 ( 1400) ( 1200)f x y x y x y x y

Euclidean Distance - Example

x

y

central repair facilityw = vehicles serviced per day

1

3

2

(1100,400)

(500,800)

(1400,1200)

w2 = 8

w1 = 4

w3 = 8

Centroid solution: x0 = [4(500)+8(1100)+8(1400)]/20 = 1100y0 = [4(800)+8(400)+8(1200)]/20 = 800

OPTIMAL SOLUTION: x* = 1098; y* = 649

Locating on the Plane-Rectilinear Distances

x

y

(a,b)

(x,y)

(x – a)

(y – b) h x a y b

Rectilinear Distance

1

1 1

min ( , )

min min

n

i i ii

n n

i i i ii i

f x y w x a y b

w x a w y b

Properties of the optimum solution:1. can solve for x, y independently2. x coordinate will be the same as one of the ai

3. y coordinate will be the same as one of the bi

4. Optimum x (y) coordinate is at a median location with respect to the weights;

i.e. no more than half the movementis to the left (up) or right (down) of the location.

1

1

2

n

ii

median w

x1 x2 x3 = x* x4 x5

An Intuitive Proof – weights = 1

7 9 6 10

x1 x2 x* x3 x4 x5

7 6 3 6 10

distance = (7+9) + 9 + 6 + (6+10) = 47

distance = (7+6) + 6 +3 + (3 + 6) + (3 + 6+10) = 50

wi wi + w1 w3 + wj wj

xi x1 x2 = x* x3 xj

Let x2 be a median location. That is w2 + w3 + wj wi + w1

w3 + wj wi + w1 + w2

wi wi + w1 w3 + wj wj

xi x1 x2 x3 xjx*

w2 + w3 + wjbut

More of an Intuitive Proof

wi wi + w1 w3 + wj wj

xi x1 x2 = x* x3 xj

Let x2 be a median location. That is w2 + w3 + wj wi + w1

w3 + wj wi + w1 + w2

wi wi + w1 w3 + wj wj

xi x1 x2 x3 xjx*

wi + w1 + w2but

More of More of an Intuitive Proof

Rectilinear Distance - Example

Warehouses: A B C D E totallocation: (0,0) (3,16) (18,2) (8,18) (20,2)weight: 5 22 41 60 34 162

warehouse x-coord weight cumulative wgtA 0 5 5B 3 22 27D 8 60 87C 18 41 128E 20 34 162

x* = 8

median = 81

Rectilinear Distance - Example

Warehouses: A B C D E totallocation: (0,0) (3,16) (18,2) (8,18) (20,2)weight: 5 22 41 60 34 162

warehouse y-coord weight cumulative wgtA 0 5 5C,E 2 41+34 80B 16 22 102D 18 60 162

y* = 16

median = 81

Rectilinear Distance - Example

x

y

1

(6,6)

(4,20)

(20,22)

(14,4)

(22,8)

4

3

2

5

retail outlet centers

factory(x,y)

w1 = 15

w4 = 9

w2 = 3

w3 = 12

w5 = 6

Rectilinear Distance - Example

retail cumulative retail cumulativeoutlet x-coord weight outlet y-coord weight

#4 4 9 #2 4 3#1 6 24 #1 6 18#2 14 27 #3 8 30#5 20 33 #4 20 39#3 22 45 #5 22 45median = 45/2 = 22.5

(x*,y*) = (6,8)

Rectilinear Distance - Example

x

y

1

(6,6)

(4,20) (20,22)

(14,4)

(22,8)

4

3

2

5

retail outlet centers

factory(6,8)

w1 = 15

w4 = 9

w2 = 3

w3 = 12

w5 = 6

MiniMax Criterion

Locate a facility to minimize the maximum distance traveled from any of n existing facilities

Applications locating a rural health clinic placement of fire stations and other emergency

equipment location of a parking lot

Useful if no continuous travel between all existing facilities; travel will not be performed over all routes

What about a maximin criterion?

An Rectilinear Formulation

i i

i i

i i

i i

i i

Min z

subj. to: x - a y - b ; 1,2,...,

.....

Min z

y - b x - a

y - b x - a

y - b x - a

y - b x - a

z i n

z

z

z

z

Equivalent to finding the center of the smallest diamond that contains all of the (ai, bi)

3 variables4n constraints

Rectilinear Minimax Solution

1

,

( , ) max

( *, *) min ( , )

i ii n

x y

f x y x a y b

f x y f x y

c1 = min (ai + bi)c2 = max (ai + bi)c3 = min (-ai + bi)c4 = max (-ai + bi)c5 = max(c2 – c1, c4 – c3)

x1 = (c1 – c3 )/2y1 = (c1 + c3 + c5)/2

x2 = (c2 – c4 )/2y2 = (c2 + c4 - c5)/20 1

x* = x1 + (1- )x2

y* = y1 + (1- )y2

f(x*,y*) = c5 /2

The Minimax Company wishes to locate a parking lotrelative to its four factories so that the maximum distanceany employee must walk is minimized.

(4,5)

(2,8)

(8,6)

(10,12)

c1 = min(ai + bi) = min(10,9,14,22) = 9c2 = max(ai + bi) = max (10,9,14,22) = 22c3 = min (-ai + bi) = min (6,1,-2,2) = -2c4 = max (-ai + bi) = max (6,1,-2,2) = 6c5 = max(c2 – c1, c4 – c3) = max(13,8) = 13

x1 = (c1 – c3 )/2 = 5.5y1 = (c1 + c3 + c5)/2 = 10

x2 = (c2 – c4 )/2 = 8y2 = (c2 + c4 - c5)/2 = 7.5

x* = 5.5 + 8(1- ) y* = 10 + 2.5(1- )f(x*,y*) = c5 /2 = 6.5

85.5

2

10

(8,7.5)

(5.5,10)

2 2

i i

Min z

subj. to: x - a y - b ; 1,2,...,z i n

Equivalent to finding the center of the circle having the smallest radius that contains all of the (xi, yi)

A Euclidean Distance Formulation

Mini-max Euclidean Distance

(4,5)

(2,8)

(8,6)

(10,12)

85.5

2

10

Multifacility Location

1 2

11 1 1

21 1 1

( ) ( )

( )

( )

m n

jk j k ij j ij k m j i

m n

jk j k ij j ij k m j i

Min f x f x

f x v x x w x a

f y v y y w y b

•Locate m facilities relative to n existing facilities•(Weighted) travel occurs among the new facilities

11 1 1

( )

:

m n

ik jk jk ij ij ijj k m j i

j k jk jk

j i ij ij

Min f v c d w e f

subject to

x x c d

x a e f

x

The Mathematical Formulation

I get it. The “c” and “d” variables and the “e” and “f”

variables cannot be in the basis at the same time. At

least one must always be zero (nonbasic).

Example Problem

A company is to locate two factories relative to its three majoroutlet stores. The following data pertains:

Store 1 Store 2 Store 3coordinates (5,18) (12,14) (18,5)weight1 (wi1) 21 18 6weight2 (wi2) 4 23 36interaction v12 = 9note: weights are train carloads per weektravel by rail which is mostly rectilinear

Where in the World are these stores?

0 5 10 15 20 25

20

15

10

5

0

The Formulation – x

1 11 11 12 12

21 21 22 22 31 31 32 32

1 2

1 11 11

1 21 21

1 31 31

2 12 12

2 22 22

2 32 32

( ) 9 21 4

18 23 6 36

:

5

12

18

5

12

18

Min f x c d e f e f

e f e f e f e f

subject to

x x c d

x e f

x e f

x e f

x e f

x e f

x e f

The Formulation – y

2 11 11 12 12

21 21 22 22 31 31 32 32

1 2

1 11 11

1 21 21

1 31 31

2 12 12

2 22 22

2 32 32

( ) 9 21 4

18 23 6 36

:

18

14

15

18

14

15

Min f y c d e f e f

e f e f e f e f

subject to

y y c d

y e f

y e f

y e f

y e f

y e f

y e f

Solver Solution

(x1*,y1*) = (12,14)(x2*,y2*) = (18,14)

coord c-d e1 e2 e3 f1 f2 f3 sumprodx1 12 0 7 0 0 0 0 6 183x2 18 6 13 6 0 0 0 0 190

v wi1 21 18 6 21 18 6 4279 wi2 4 23 36 4 23 36

constr -2.14E-12 5 12 18 5 12 18RHS 0 ai 5 12 18 5 12 18

coord c-d e1 e2 e3 f1 f2 f3 sumprody1 14 0 0 0 9 4 0 0 138y2 14 0 0 0 9 4 0 0 340

v wi1 21 18 6 21 18 6 4789 wi2 4 23 36 4 23 36

constr 0 18 14 5 18 14 5RHS 0 bi 18 14 5 18 14 5

Where are the factories?

(x1*,y1*) = (12,14) ; (x2*,y2*) = (18,14)

0 5 10 15 20 25

20

15

10

5

0