Lecture Eight: Facility location

8/10/2019 Lecture Eight: Facility location

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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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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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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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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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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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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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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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Lecture Eight: Facility location