facilities layout and location the three most important criteria in locating a factory: location!...
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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