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Facility Location
Professor Dah-Chuan Gong
October 2014
Course Focus
Single FacilityLocation Problems
Rectilinear and
Product Design-- QFD
Flow Systemsand Clustering
Methodology (GT)
Systematic Layout Planning
Facility Facility Material Handling Equipment
Process Design CRAFT andBLOCPLAN
Location Layout WarehouseOperations Automated Storage and
Retrieval System
Material Handling(Facility Planning)
MaterialHandlingWarehouse Sizing
and Layout
Order Pickin Anal sis
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Conveyor Systemand AGVS
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Introduction (Location-Allocation)
Location Problems: involve determining the location of one or morenew ac es n one or more o severa po en a s es. e cos olocating each new facility at each of the potential sites is assumed to beknown. It is the fixed cost of locatin a new facilit at a articular siteplus the operating and transportation cost of serving customers from thisfacility-site combination.Allocation Problems: assume that the number and location of facilitiesare known a priori and attempt to determine how each customer is to be
. ,center, the production or supply capacities at each facility, and the costof serving each customer from each facility, the allocation problem
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determines how much each facility is to supply to each customer center.
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Introduction (Location-Allocation)
-much each customer is to receive from each facility but also thenumber of facilities along with their locations and capacities.
Location Allocation
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Risdy Furnitures Logistics Problem
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RisdiyonoRisdiyono ID: 107988ID: 1079885
Risdy Furnitures Logistics Problem
Summary result of the total cost for five alternatives
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Planar Single-Facility Location Problems
Minisum Location Problem with Rectilinear Distances
Minisum Location Problem with Euclidean Distances
Minimax Location Problem with Rectilinear Distances
Minimax Location Problem with Euclidean Distances
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Example: Tyler Emergency Medical Services (EMS)
,are shown on the following map.
The o ulation densit for each of the cit s tracts is also shown.The darker red areas have up to 5,000 people per square mile.
The southeast part of Tyler, census tract 18.03, has experiencedrapid growth, with its population almost doubling in the lasttwelve years.
e res en s o s rac ave comp a ne a a es oo ongfor the EMS vehicles to reach them.
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Population Density of Tyler,Texas
Areas of rapid growth.
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Example: Tyler Emergency Medical Services (EMS)
A general guideline for locating EMS facilities in urban areas
is that an EMS vehicle should be able to answer 95 percentof its calls within 10 minutes in tracts that have a populationdensit of 1 000 eo le er s uare mile.
Census tract 7, on the west side of the city with apopulation density of 967 people per square mile, shouldbe included in the study as well.
Thus, the census tracts that are as dark as or darker than
census rac , s ou e w n a -m nu e r ve mezone of an EMS facility.
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coverage goals for Tyler?
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Census Track 7
10 minuteresponse
zones
With MapPoint, it is easy to calculate ar ve me zone y us seec ng e
pushpin and going under Tools on the
menu bar to select drive time zone interms of the number of minutes of drive
Some areas not incovera e zone.
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time..
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Three EMS locationswere chosen through atrial and error approachand evaluation using
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.
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Minisum Examples
, ,station, or library in a metropolitan area
Locate a dock in a warehouse for purpose of loading
Locate a copy machine in a library
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Minisum location problem
.machine (or facility) receives parts from existing machines (orfacilities) and supplies parts to existing machines. There is ao a cos o mov ng par s o an rom e new mac ne. o a
cost depends on the location of the new machine.
cost.
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Notation
m = the number of existing machines i i, i
X = the location of new machine = (x, y)t = the number of trips per month between P and Xd(X, P i) = the distance between X and P itid(X, P i) = total distance items travel per month to and from X and P ivi= the average velocity of items traveling to and from the two machines (Xand P i)t/v d X, P = the total travel time er month between the two machines
ci = the cost per unit time (hour) of travel between the two machines
(citi/vi)d(X, P i) = the cost per month involving the two machines
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= wid(X, P i), where w i = c i(ti/vi)
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The total cost of movement between the new machine and all theexisting machines
= f(X) A minisumlocation problem
==m
=i i i m m
1
,,...,
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==++= i i i m m P X d w P X d w P X d w X f 111 ),(),(...),()(
P 3P
P 1
X
P 2 P 5
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Majority Theorem
When one weight constitutes a majority of the total weight, anop ma new ac y oca on co nc es w e ex s ng ac ywhich has the majority weight.
,general class of distances (Lp) including rectilinear distances.
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Major Types of Distances
Rectilinear distance (Manhattan distance, Right-angle distance,ec angu ar s ance
Euclidean distanceChebyshev distance (or Tchebychev distance)
Given 2 points , A (x1, y1) and B (x2, y2) in the space
Lp = ( x1 x2 p y1 y2 p )1/p
When p=1, Lp Rectilinear distance
When p=2, Lp Euclidean distance
When p= , Lp = max( x1 x2 , y1 y2 ) , Chebyshev
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distance
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Minisum Location Problem with Rectilinear Distance
X = (x, y) and Pi = (ai , bi )
)()()(),()( y f x f b y a x w y x f X f m
i i i i 21
1
+=+== =
where and=
=m
i i i a x w x f
11 )()(
=
=m
i i i b y w y f
12 )()(
We can minimize the total cost of movement by solving
of movement in the x direction and minimizing the cost
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.
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Example 1
Suppose that a conveyor line is being planned to run into aware ouse. e ne w eg n a e po n , a coor na es arein units of tens of feet) and run parallel to the x-axis into thewarehouse. Items entering the warehouse on the line are picked upat the end of the line and transported directly to one of the truckdocks at the points P1=(7,10), P2=(15,7), P3=(15,3), and P4=(12,0).
e o a annua cos per or ranspor ng ems e ween eend of line and points P1 through P4 will be $160, $40, $60, and$140, respectively. The equivalent annual cost per 10 ft of conveyoris $180.Find the length of the conveyor so as to minimize the total cost.
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Example 1(continued)
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Example 1: Data and Cost Function
wi 160 40 60 140 180
Pi (7,10) (15,7) (15,3) (12,0) (0,5)
1510012140716001801 +++=)( x x x x x f 1700516024001802605140522 =++++== )()()()()()()( f y f
Method 1-- Apply the majority theorem (median condition)
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Example 1: Method 2-- Check on the slope changes
The coefficient of x in any The minimizing point is x=7,
points is the sum of the
weights of the points to the leftof x minus the sum of the
1 =
weights of the points to theright of x.
,but the first is the sum of theslope of the previous interval(from left to right) and twicethe weight of the point
common to both intervals.-220=-580+2(180),
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100= -220+2(160)
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Example 1: MathCAD Graph(different data set)
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Example 2: Contour Line Method
Minimize the function f(X) for which there arem =4 existing
facilities, with locations P1=(4,2), P2=(8,5), P3=(11,8), and P4=(13,2),with weights 1, 2, 2, and 1, respectively.
We normalize the weights by dividing by their sum (6) to obtain1/6, 1/3, 1/3, and 1/6, respectively.
wi 1/6 1/3 1/3 1/6
i, , , ,
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Example 2: Contour Line Method(continued)
Contour line (or called levelne : every po n on e
contour line has the samevalue of the function f.Contour set (level set): eachcontour set whoseoun ary s a con our ne,
is the set of all pointshaving values of f(X) nolonger than those of the
points on the contour line.
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852
131184111
61
31
31
61
1
++=
+++=
f
x x x x x f )(
The slope of every contour line passing through a box B is the negativeratio of the bottom margin x coefficient to the left margin y coefficient.
3
2
=
3
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13
=
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Optimum location
The contourThe contourline passesP1=(4,2)line passesP1=(4,2)
P1
What is the value ofthis contour line?
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Contour Lines for Three Simple Rectilinear Location
Observations?
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Contour Line Construction Procedure
1. Pass a horizontal line and a vertical line through each existing.
leftmost and beyond the rightmost existing facility location.
Similarly, each vertical line should extend beyond thebottommost and beyond the topmost existing facility location.
2. For each vertical line, total the weights of the existing facilities.
3. For each horizontal line, total the weights of the existingfacilities l in on the line and write the total at the left of theline.
4. The vertical lines partition the plane into columns. For each
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column, compute the coefficient of x and write the coefficient inthe bottom margin.
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Contour Line Construction Procedure (continued)
5. The horizontal lines partition the plane into rows. For each row,
margin.
6. The slope for every contour line passing through a given box is thenegative ratio of the number in the bottom margin to the number inthe left margin.
. se e s ng e con our ne cons ruc on me o o con uc acontour line through each point of interest.
. reduced. Then, the optimum location is obtained.
9. Alternativel to determine those oints that minimize f X either
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identify the points where the margin numbers change from negativeto nonnegative or else, equivalently, use the median conditions..
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Example 3(p521)
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Example 3( continued) p521
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Example 3( continued) p521
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Example 3( continued) p521
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Minisum Location Problem with Squared EuclideanDistance =++=
m
P X d w P X d w P X d w X f ...
Ob ective Function
=i 1
( ) ( )[ ]2i2im
i byaxw)y,x(f Minimize += =
( ) ( )[ ] byaxw)y,x(f 2i2im
1ii
+= =
( ) ( ) bywaxw 2im
1ii
2i
m
1ii +=
==
yx 21 +=2
m
= 2
m
bwf =
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i1i
i1= 1i=
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Minisum Location Problem with Squared EuclideanDistance continued
Since f x and f are se arable the o timal solution of
f 1(x) is independent of the optimal solution of f 2(y).Note both f x and f are convex functions. Thus theoptimal x* and y* that minimize f 1(x) and f 2(y) can beobtained by setting the first derivatives equal to zero.
m m
0axw2dx i1ii
1
== = ( )2
1 2 0i i
i w y bdy == =
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Minisum Location Problem with Squared EuclideanDistance continued
* *f 1(x) and f 2(y) can be obtained as following:
m
i iw a m
i iw bm
i
xw
=
m
i
yw
==
It is also called acentroid problem .
= =
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Example 4
,problem (i.e., a minisum location problem with
wi 1 1 1 1
.
Pi (0,0) (0,10) (5,0) (12,6)
(x*, y*) = (4.25, 4.0)
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Minisum Location Problem with Euclidean Distance
2 2( , )
m
Minimize f x y w x a y b= + 1
1/22 2
i
m
i i iw x a y b
=
= +
Since f 1(x) and f 2(y) are not separable, the optimal solution
1i=
of f 1(x) is dependent of the optimal solution of f 2(y).
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Minisum Location Problem with Euclidean Distance
The graph of is a cone (strictly convex[(x a ) (y b )i 2 i 2 + ]12
function).
contours[(x a ) (y b )i
2i
2 + ]12 y
(ai, bi)
a, b, 0
x
y
m2 2
12
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,
convex hull is a line segment.
i i ii=1
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Minisum Location Problem with Euclidean Distance
First order o timalit conditions :
* *f x f x, ,x
=y
=
Any point where the partial derivatives are zero is optimal.Let
( ) ( )2 2 1/2
( , )
[ ]
ii
i i
wx y
x a y b
= +
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Minisum Location Problem with Euclidean Distance
1/2m wf
1
22
i i ii
m
x a y b x ax =
= +
1
,i ii
x y x a =
= =
1
( , )m
i ii
a x y ==
1
( , )m
ii
x y =
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Minisum Location Problem with Euclidean Distance
1/22 2m wf
1 2
i i ii
m
x a y yy =
= +
1
,i ii
x y y =
= =
1
( , )m
i ii
b x y ==
1
( , )m
ii
x y =
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Minisum Location Problem with Euclidean Distance
If we ut the two artial derivatives into a two-tu le we et
the gradient of f evaluated. ==> a general formm=
1i=
.( )
( ) ( ) 0m m
ii i i
XX X X P X P
= =
= =
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Minisum Location Problem with Euclidean Distance--
(0)
.
Step 1: Define and let n=01
( )( )
( )
mi
ii
XWF X P
X
=
=
Step 2: Solve WF (X (n )) and letX (n+1 )= WF (X (n ))
Step 3: If d(WF (X (n )), WF (X (n+1 ))) , then STOP
X (n ) is the solution
otherwise let n = n+1 and o to Ste 2.
Alternatively, replaced( WF (X (n )), WF (X (n+1 ))) by( 1) ( 1) ( ) ( )n n n n+ +
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, ,
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Minisum Location Problem with Euclidean Distance--
this algorithm is to perturb the problem as follows:replace
( ) ( )1/22 2
( , )i i id X P x a y b = + by
( ) ( )1/22 2
( , )i i id X P x a y b = + +
Weiszfelds Algorithm Example (R1, p208)
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