Design _ Layout and Location of by chenshu

VIEWS: 30 PAGES: 111

									      Facilities Design

           S.S. Heragu
Decision Sciences and Engineering
       Systems Department
 Rensselaer Polytechnic Institute
       Troy NY 12180-3590



                                    1
  Chapter 13

  Basic Models
     for the
Location Problem



                   2
Outline
•   13.1 Introduction
•   13.2       Important Factors in Location
    Decisions
•   13.3       Techniques for Discrete Space
    Location Problems
     - 13.3.1 Qualitative Analysis
     - 13.3.2 Quantitative Analysis
     - 13.3.3 Hybrid Analysis

                                               3
Outline Cont...
•   13.4       Techniques for Continuous Space
    Location Problems
     - 13.4.1 Median Method
     - 13.4.2 Contour Line Method
     - 13.4.3 Gravity Method
     - 13.4.4 Weiszfeld Method
•   13.5       Facility Location Case Study
•   13.6       Summary
•   13.7       Review Questions and Exercises
•   13.8       References
                                              4
McDonald’s
•   QSCV Philosophy
•   11,000 restaurants (7,000 in USA, remaining
    in 50 countries)
•   700 seat McDonald’s in Pushkin Square,
    Moscow
•   $60 million food plant combining a bakery,
    lettuce plant, meat plant, chicken plant, fish
    plant and a distribution center, each owned
    and operated independently at same location


                                                5
McDonald’s cont...
•   Food taste must be the same at any
    McDonald, yet food must be secured locally
•   Strong logistical chain, with no weak links
    between
•   Close monitoring for logistical performance
•   300 in Australia
•   Central distribution since 1974 with the help
    of F.J. Walker Foods in Sydney
•   Then distribution centers opened in several
    cities
                                                6
McDonald’s cont...
•   2000 ingredients, from 48 food plants,
    shipment of 200 finished products from
    suppliers to DC’s, 6 million cases of food and
    paper products plus 500 operating items to
    restaurants across Australia
•   Delivery of frozen, dry and chilled foods
    twice a week to each of the 300 restaurants
    98% of the time within 15 minutes of
    promised delivery time, 99.8% within 2 days
    of order placement
•   No stockouts, but less inventory
                                                 7
Introduction
•   Logistics management can be defined as the
    management of transportation and
    distribution of goods.

    - facility location
    - transportation
    - goods handling and storage.


                                             8
Introduction Cont...
Some of the objectives in facility location
 decisions:
(1) It must first be close as possible to raw
 material sources and customers;
(2) Skilled labor must be readily available in the
 vicinity of a facility’s location;
(3) Taxes, property insurance, construction and
 land prices must not be too “high;”
(4) Utilities must be readily available at a
 “reasonable” price;


                                                9
 Introduction Cont...
• (5) Local , state and other government
   regulations must be conducive to business;
   and
(6) Business climate must be favorable and the
   community must have adequate support
   services and facilities such as schools,
   hospitals and libraries, which are important
   to employees and their families.




                                              10
Introduction Cont...
Logistics management problems can be
  classified as:

  (1) location problems;

  (2) allocation problems; and

  (3) location-allocation problems.


                                       11
List of Factors Affecting
Location Decisions
• Proximity to raw materials sources
• Cost and availability of energy/utilities
• Cost, availability, skill and productivity of
    labor
•   Government regulations at the federal, state,
    country and local levels
•   Taxes at the federal, state, county and local
    levels
•   Insurance
•   Construction costs, land price
                                                  12
List of Factors Affecting
Location Decisions Cont...
• Government and political stability
• Exchange rate fluctuation
• Export, import regulations, duties, and tariffs
• Transportation system
• Technical expertise
• Environmental regulations at the federal,
    state, county and local levels
•   Support services


                                               13
List of Factors Affecting
Location Decisions Cont...
• Community services, i.e. schools, hospitals,
    recreation, etc.
•   Weather
•   Proximity to customers
•   Business climate
•   Competition-related factors




                                             14
13.2
Important Factors in Location
Decisions
•   International
•   National
•   State-wide
•   Community-wide




                                15
13.3.1
Qualitative Analysis
Step 1: List all the factors that are important,
  i.e. have an impact on the location decision.
Step 2: Assign appropriate weights (typically
  between 0 and 1) to each factor based on the
  relative importance of each.
Step 3: Assign a score (typically between 0 and
  100) for each location with respect to each
  factor identified in Step 1.



                                              16
13.3.1
Qualitative Analysis
Step 4: Compute the weighted score for each
  factor for each location by multiplying its
  weight with the corresponding score (which
  were assigned Steps 2 and 3, respectively)
Step 5: Compute the sum of the weighted
  scores for each location and choose a
  location based on these scores.




                                            17
Example 1:
•A payroll processing company has recently
won several major contracts in the midwest
region of the U.S. and central Canada and
wants to open a new, large facility to serve
these areas. Since customer service is of
utmost importance, the company wants to be
as near it’s “customers” as possible.
Preliminary investigation has shown that
Minneapolis, Winnipeg, and Springfield, Ill.,
would be the three most desirable locations
and the payroll company has to select one of
these three.
                                                18
Example 1: Cont...
A subsequent thorough investigation of each
location with respect to eight important factors
has generated the raw scores and weights
listed in table 2. Using the location scoring
method, determine the best location for the new
payroll processing facility.




                                             19
Solution:
Steps 1, 2, and 3 have already been completed
for us. We now need to compute the weighted
score for each location-factor pair (Step 4), and
these weighted scores and determine the
location based on these scores (Step 5).




                                               20
Table 2. Factors and Weights for
Three Locations
Wt. Factors                      Location
                               Minn.Winn.Spring.
.25   Proximity to customers     95    90   65
.15   Land/construction prices   60    60   90
.15   Wage rates                 70    45   60
.10   Property taxes             70    90   70
.10   Business taxes             80    90   85
.10   Commercial travel          80    65   75

                                             21
Table 2. Cont...

Wt. Factors                 Location
                      Minn. Winn. Spring.
.08 Insurance costs   70    95    60
.07 Office services   90    90    80




                                     22
Solution: Cont...
From the analysis in Table 3, it is clear that
Minneapolis would be the best location based
on the subjective information.




                                                 23
Table 3. Weighted Scores for the
Three Locations
in Table 2
Weighted Score                   Location
                           Minn. Winn. Spring.
Proximity to customers     23.75 22.5 16.25
Land/construction prices   9     9      13.5
Wage rates                 10.5 6.75 9
Property taxes             7     9      8.5
Business taxes             8     9      8.5

                                           24
Table 3. Cont...
Weighted Score            Location
                    Minn. Winn. Spring.
Commercial travel   8     6.5    7.5
Insurance costs     5.6   7.6    4.8
Office services     6.3   6.3    5.6




                                    25
Solution: Cont...
Of course, as mentioned before, objective
measures must be brought into consideration
especially because the weighted scores for
Minneapolis and Winnipeg are close.




                                              26
  13.3.2
Quantitative
 Analysis




               27
General Transportation Model
Parameters
  cij: cost of transporting one unit from
  warehouse i to customer j
  ai: supply capacity at warehouse i
  bi: demand at customer j
Decision Variables
  xij: number of units transported from
  warehouse i to customer j


                                            28
General Transportation Model
                         tion Cost
 Minimize Total Transporta
             m   n
 Z   cij xij
         i 1 j 1
Subject to
  n

 x
 j 1
        ij    ai , i  1,2,...,m (supply restriction at warehou i)
                                                                se
 m

 x
 i 1
        ij    b j , j  1,2,..., n (demand requiremen t at market j)
 xij  0, i, j  1,2,..., n (non - negativity restrictio ns)

                                                               29
Example 2:
Seers Inc. has two manufacturing plants at
Albany and Little Rock supplying Canmore
brand refrigerators to four distribution centers
in Boston, Philadelphia, Galveston and Raleigh.
Due to an increase in demand of this brand of
refrigerators that is expected to last for several
years into the future, Seers Inc., has decided to
build another plant in Atlanta or Pittsburgh.
The expected demand at the three distribution
centers and the maximum capacity at the
Albany and Little Rock plants are given in Table
4.
                                                 30
Example 2: Cont...
Determine which of the two locations, Atlanta
or Pittsburgh, is suitable for the new plant.
Seers Inc., wishes to utilize all of the capacity
available at it’s Albany and Little Rock
Locations




                                                    31
Table 4. Costs, Demand and
Supply Information

              Bost. Phil. Galv. Rale.   Supply
                                        Capacity
Albany        10    15    22    20      250
Little Rock   19    15    10    9       300
Atlanta       21    11    13    6       No limit
Pittsburgh    17    8     18    12      No limit
Demand        200   100   300   280


                                              32
Table 5. Transportation Model
with Plant at Atlanta

              Bost. Phil. Galv. Rale.   Supply
                                        Capacity
Albany        10    15    22    20      250
Little Rock   19    15    10    9       300
Atlanta       21    11    13    6       330
Demand        200   100   300   280     880



                                              33
Table 6. Transportation Model
with Plant at Pittsburgh

              Bost. Phil. Galv. Rale.   Supply
                                        Capacity
Albany        10    15    22    20      250
Little Rock   19    15    10    9       300
Pittsburgh    17    8     18    12      330
Demand        200   100   300   280     880




                                              34
Min/Max Location Problem:

             Location
           d11 d12   d1n
           d21 d22   d2n
    Site

           dm1 dm2   dmn




                            35
13.3.3
Hybrid Analysis
•   Critical
•   Objective
•   Subjective




                  36
Hybrid Analysis Cont...
CFij = 1 if location i satisfies critical factor j,
       0 otherwise
OFij = cost of objective factor j at location i
SFij = numerical value assigned
       (on scale of 0-1)
       to subjective factor j for location i
wj = weight assigned to subjective factor
       (0< w < 1)


                                                      37
Hybrid Analysis Cont...
                                      p
CFM i  CFi1CFi 2    CFip   CFij ,
                                     j 1

i  1,2,...,m
                   q       q
         maxi  OFij    OFij
OFMi              j 1    j 1       , i  1,2,...,m
              q
                                q
                                      
       maxi  OFij   min i  OFij 
             j 1             j 1 
          r
SFM i   w j SFij , i  1,2,...,m
         j 1




                                                          38
Hybrid Analysis Cont...
The location measure LMi for each location is
then calculated as:

      LMi = CFMi [  OFMi + (1- ) SFMi ]

     Where  is the weight assigned to the
     objective factor.

We then choose the location with the highest
location measure LMi
                                                39
Example 3:
Mole-Sun Brewing company is evaluating six
candidate locations-Montreal, Plattsburgh,
Ottawa, Albany, Rochester and Kingston, for
constructing a new brewery. There are two
critical, three objective and four subjective
factors that management wishes to incorporate
in its decision-making. These factors are
summarized in Table 7. The weights of the
subjective factors are also provided in the
table. Determine the best location if the
subjective factors are to be weighted 50 percent
more than the objective factors.
                                              40
             Table 7:
Critical, Subjective and Objective
Factor Ratings for six locations for
Mole-Sun Brewing Company, Inc.



                                       41
Table 7. Cont...
 Location                   Factors
                 Critical
             Water     Tax
             Supply Incentives
 Albany        0         1
 Kingston      1         1
 Montreal      1         1
 Ottawa        1         0
 Plattsburgh   1         1
 Rochester     1         1

                                      42
Table 7. Cont...
Location                         Factors
              Critical             Objective
                         Revenue    Labor      Energy
                                    Cost        Cost
Albany                     185         80        10
Kingston                   150        100        15
Montreal                   170         90        13
Ottawa                     200        100        15
Plattsburgh                140         75         8
Rochester                  150         75        11

                                                   43
Table 7. Cont...
 Location                  Factors
                         Subjective
               Community    Ease of
                Attitude Transportation
                  0.3         0.4
 Albany           0.5         0.9
 Kingston         0.6         0.7
 Montreal         0.4         0.8
 Ottawa           0.5         0.4
 Plattsburgh      0.9         0.9
 Rochester        0.7         0.65
                                          44
Table 7. Cont...
 Location                   Factors
                          Subjective
                  Labor         Support
               Unionization     Services
                   0.25            0.05
 Albany            0.6             0.7
 Kingston          0.7             0.75
 Montreal          0.2             0.8
 Ottawa            0.4             0.8
 Plattsburgh       0.9             0.55
 Rochester         0.4             0.8
                                           45
Table 8. Location Analysis of
Mole-Sun Brewing Company,
 Inc., Using Hybrid Method



                                46
Table 7. Cont...
Location                     Factors
              Critical Objective Subjective LMi
                       Sum of       SFMi
                     Obj. Factors
Albany                   -95        0.7     0
Kingston                 -35        0.67    0.4
Montreal                 -67        0.53    0.53
Ottawa                   -85        0.45    0
Plattsburgh              -57        0.88    0.68
Rochester                -64        0.61    0.56

                                              47
               13.4
         Techniques For
Continuous Space Location Problems




                                48
13.4.1 Model for Rectilinear
Metric Problem
Consider the following notation:
fi = Traffic flow between new facility and
   existing facility i
ci = Cost of transportation between new facility
   and existing facility i per unit
xi, yi = Coordinate points of existing facility i




                                                49
Model for Rectilinear Metric
Problem (Cont)
 The median location model is then to minimize:

          m
   TC     c f [| x  x |  | y
          i 1
                 i   i   i         i    y |]


Where TC is the total distribution cost




                                                50
Model for Rectilinear Metric
Problem (Cont)
Since the cifi product is known for each facility,
it can be thought of as a weight wi
corresponding to facility i.
                  m                m
 Minimize TC     w [ | x  x |] w [| y
                  i 1
                         i   i
                                  i 1
                                         i   i    y |]




                                                      51
Median Method:
Step 1: List the existing facilities in non-
  decreasing order of the x coordinates.
Step 2: Find the jth x coordinate in the list at
  which the cumulative weight equals or
  exceeds half the total weight for the first
  time, i.e.,
   j 1       m                 j        m
               wi                         wi
    wi   2 and
   i 1   i 1
                               wi   2
                              i 1   i 1


                                                   52
Median Method (Cont)
Step 3: List the existing facilities in non-
  decreasing order of the y coordinates.
Step 4: Find the kth y coordinate in the list
  (created in Step 3) at which the cumulative
  weight equals or exceeds half the total
  weight for the first time, i.e.,
   k 1      m               k        m
              wi                        wi
   wi   2 and
  i 1   i 1
                             wi   2
                            i 1   i 1


                                                53
Median Method (Cont)
Step 4: Cont... The optimal location of the new
  facility is given by the jth x coordinate and the
  kth y coordinate identified in Steps 2 and 4,
  respectively.




                                                 54
Notes
1. It can be shown that any other x or y
   coordinate will not be that of the optimal
   location’s coordinates
2. The algorithm determines the x and y
   coordinates of the facility’s optimal location
   separately
3. These coordinates could coincide with the x
   and y coordinates of two different existing
   facilities or possibly one existing facility


                                                55
Example 4:
Two high speed copiers are to be located in the
fifth floor of an office complex which houses
four departments of the Social Security
Administration. Coordinates of the centroid of
each department as well as the average number
of trips made per day between each department
and the copiers’ yet-to-be-determined location
are known and given in Table 9 below. Assume
that travel originates and ends at the centroid
of each department. Determine the optimal
location, i.e., x, y coordinates, for the copiers.
                                                56
 Table 9. Centroid Coordinates
and Average Number of Trips to
            Copiers




                                 57
Table 9.
Dept.   Coordinates    Average number of
  #     x    y        daily trips to copiers
  1     10   2                 6
  2     10   10                10
  3     8    6                 8
  4     12   5                 4




                                               58
 Solution:
Using the median method, we obtain the
following solution:
Step 1:

Dept. x coordinates in   Weights Cumulative
  # non-decreasing order          Weights
 3            8              8           8
 1            10             6           14
 2            10             10          24
 4            12             4           28
                                              59
Solution:
Step 2: Since the second x coordinate, namely
  10, in the above list is where the cumulative
  weight equals half the total weight of 28/2 =
  14, the optimal x coordinate is 10.




                                              60
 Solution:
Step 3:

Dept. x coordinates in   Weights Cumulative
  # non-decreasing order          Weights
  1          2             6        6
  4          5             4        10
  3          6             8        18
  2          10            10       28



                                        61
Solution:
Step 4: Since the third y coordinates in the
  above list is where the cumulative weight
  exceeds half the total weight of 28/2 = 14, the
  optimal coordinate is 6. Thus, the optimal
  coordinates of the new facility are (10, 6).




                                               62
Equivalent Linear Model for the
Rectilinear Distance, Single-
Facility Location Problem
Parameters
  fi = Traffic flow between new facility and
  existing facility i
  ci = Unit transportation cost between new
  facility and existing facility i
  xi, yi = Coordinate points of existing facility i
Decision Variables
   x, y= Optimal coordinates of the new facility
  TC = Total distribution cost
                                                  63
Equivalent Linear Model for the
Rectilinear Distance, Single-
Facility Location Problem
The median location model is then to

                 m               m
 Minimize TC    w [ | x  x |] w [| y
                 i 1
                        i   i
                                 i 1
                                        i   i    y |]




                                                     64
Equivalent Linear Model for the
Rectilinear Distance, Single-
Facility Location Problem
Since the cifi product is known for each facility,
  it can be thought of as a weight wi
  corresponding to facility i. The previous
  equation can now be rewritten as follows
                   m                m
  Minimize TC     w [ | x  x |] w [| y
                   i 1
                          i   i
                                   i 1
                                          i   i    y |]




                                                      65
Equivalent Linear Model for the
Rectilinear Distance, Single-
Facility Location Problem
Define
    ( xi  x ) if xi  x  0
x i 
     0          otherwise
    ( x  xi ) if xi  x  0
x i 
     0          otherwise
We can observe that, whether( xi  x )  0 or  0,
xi  x  x  i  x  i and
( xi  x )  x  i  x  i                           66
Equivalent Linear Model for the
Rectilinear Distance, Single-
Facility Location Problem
A similar definition of y  i , y  i yields
 yi  y  y  i  y  i and
( yi  y )  y  i  y  i




                                               67
Equivalent Linear Model for the
Rectilinear Distance, Single-
Facility Location Problem
         d
Transforme Linear Model
                  n
Minimize         wi ( x  i  x  i  y  i  y  i )
                 i 1
Subject to
( xi  x )  x  i - x  i , i  1,2,...,n
( yi  y )  y  i - y  i , i  1,2,...,n
x  i , x  i , y  i , y  i  0, i  1,2,...,n
x , y , unrestricted in sign
                                                         68
       13.4.2
Contour Line Method




                      69
Algorithm for Drawing Contour
Lines:
Step 1: Draw a vertical line through the x
  coordinate and a horizontal line through the y
  coordinate of each facility
Step 2: Label each vertical line Vi, i=1, 2, ..., p
  and horizontal line Hj, j=1, 2, ..., q where Vi=
  the sum of weights of facilities whose x
  coordinates fall on vertical line i and where
  Hj= sum of weights of facilities whose y
  coordinates fall on horizontal line j


                                                70
Algorithm for Drawing Contour
Lines (Cont)          m
Step 3: Set i = j = 1; N0 = D0 =   w   i
                                    i=1
Step 4: Set Ni = Ni-1 + 2Vi and Dj = Dj-1 + 2Hj.
  Increment i = i + 1 and j = j + 1
Step 5: If i < p or j < q, go to Step 4. Otherwise,
  set i = j = 0 and determine Sij, the slope of
  contour lines through the region bounded by
  vertical lines i and i + 1 and horizontal line j
  and j + 1 using the equation Sij = -Ni/Dj.
  Increment i = i + 1 and j = j + 1

                                                 71
Algorithm for Drawing Contour
Lines:
Step 6: If i < p or j < q, go to Step 5. Otherwise
  select any point (x, y) and draw a contour line
  with slope Sij in the region [i, j] in which (x, y)
  appears so that the line touches the
  boundary of this line. From one of the end
  points of this line, draw another contour line
  through the adjacent region with the
  corresponding slope
Step 7: Repeat this until you get a contour line
  ending at point (x, y). We now have a region
  bounded by contour lines with (x, y) on the
  boundary of the region                            72
Notes on Algorithm for Drawing
Contour Lines
1. The number of vertical and horizontal lines
   need not be equal
2. The Ni and Dj as computed in Steps 3 and 4
   correspond to the numerator and
   denominator, respectively of the slope
   equation of any contour line through the
   region bounded by the vertical lines i and i +
   1 and horizontal lines j and j + 1



                                                73
Notes on Algorithm for Drawing
Contour Lines (Cont)
Consider the objective function when the new facility
is located at some point (x, y), i.e., x  x, y  y
       m              m
TC   wi xi  x   wi yi  y
      i 1            i 1




                                                      74
Notes on Algorithm for Drawing
Contour Lines (Cont)
By noting that the Vi’s and Hj’s calculated in
Step 2 of the algorithm correspond to the sum
of the weights of facilities whose x, y
coordinates are equal to the x, y coordinates,
respectively of the ith, jth distinct lines and that
we have p, q such coordinates or lines (p < m, q
< m), the previous equation can be written as
follows
            p            q
     TC   Vi xi  x   H i yi  y
           i 1         i 1

                                                  75
     Notes on Algorithm for Drawing
     Contour Lines (Cont)
     Suppose that x is between the sth and s+1th
     (distinct) x coordinates or vertical lines (since
     we have drawn vertical lines through these
     coordinates in Step 1). Similarly, let y be
     between the tth and t+1th vertical lines. Then
       s                   p                t                q
TC   Vi (x  xi )     V ( x  x)   H ( y  y )   H ( y
                                   i   i          i   i              i   i    y)
      i 1              i  s 1           i 1           i  t 1




                                                                         76
Notes on Algorithm for Drawing
Contour Lines (Cont)
Rearranging the variable and constant terms in
the above equation, we get
       s        p
                        t          q
                                          
 TC   Vi   Vi  x   H i   H i  y
       i 1 i  s 1    i 1   i t 1 
            s             p              t               q
           Vi xi    V x   H y   H y
                                  i i          i   i             i   i
           i 1        i  s 1         i 1           i t 1




                                                                         77
Notes on Algorithm for Drawing
Contour Lines (Cont)
The last four terms in the previous equation
can be substituted by another constant term
c and the coefficients of x can be rewritten
as follows
       s          p            s                 s
 TC   Vi    V  V  V
                          i          i                 i
      i 1     i  s 1       i 1              i 1


Notice that we have only added and
subtracted this term   s

                      Vi                i 1

                                                           78
Notes on Algorithm for Drawing
Contour Lines (Cont)
                                           s       m
Since it is clear from Step 2 that    V   w ,
                                       i 1
                                               i
                                                   i 1
                                                          i


 the coefficient of x can be rewritten as
       s
             s         p
                                s    p
    2 Vi   Vi   Vi  2 Vi   Vi
     i 1    i 1  i  s 1  i 1 i 1
               s     m
            2 Vi   wi
              i 1   i 1
                                       t           m
                                     2 H i   wi
Similarly, the coefficient of y is
                                      i 1         i 1
                                                              79
Notes on Algorithm for Drawing
Contour Lines (Cont)
            s      m
                           t             m
                                               
Thus, TC  2Vi   wi  x  2 H i   wi  y  c
            i 1  i 1    i 1         i 1 
• The Ni computation in Step 4 is in fact
  calculation of the coefficient of x as shown
  above. Note that Ni=Ni-1+2Vi. Making the
  substitution for Ni-1, we get Ni=Ni-2+2Vi-1+2Vi
• Repeating the same procedure of making
  substitutions for Ni-2, Ni-3, ..., we get
• Ni=N0+2V1+2V2+...+2Vi-1+2V1=  m w  2 i V
                                        i  k
                                      i 1    k 1

                                                     80
Notes on Algorithm for Drawing
Contour Lines (Cont)
                                           m        i

Similarly, it can be verified that Di   wi  2 H k
                                          i 1     k 1

            s          m
                              t         m
                                              
Thus, TC  2 Vi   wi  x  2 H i   wi  y  c
            i 1      i 1   i 1     i 1 
         N s x  Dt y  c
which can be rewritten as
    Ns
y    x  (TC  c)
    Dt

                                                    81
Notes on Algorithm for Drawing
Contour Lines (Cont)
The above expression for the total cost function
at x, y or in fact, any other point in the region [s,
t] has the form y= mx + c, where the slope
m = -Ns/Dt. This is exactly how the slopes are
computed in Step 5 of the algorithm




                                                 82
Notes on Algorithm for Drawing
Contour Lines (Cont)
3. The lines V0, Vp+1 and H0, Hq+1 are required for
   defining the “exterior” regions [0, j], [p, j], j =
   1, 2, ..., p, respectively)
4. Once we have determined the slopes of all
   regions, the user may choose any point (x, y)
   other than a point which minimizes the
   objective function and draw a series of
   contour lines in order to get a region which
   contains points, i.e. facility locations,
   yielding as good or better objective function
   values than (x, y)
                                                     83
Example 5:
Consider Example 4. Suppose that the weight
of facility 2 is not 10, but 20. Applying the
median method, it can be verified that the
optimal location is (10, 10) - the centroid of
department 2, where immovable structures
exist. It is now desired to find a feasible and
“near-optimal” location using the contour line
method.




                                              84
Solution:
The contour line method is illustrated using
Figure 1
Step 1: The vertical and horizontal lines V1, V2,
V2 and H1, H2, H2, H4 are drawn as shown. In
addition to these lines, we also draw line V0, V4
and H0, H5 so that the “exterior regions can be
identified
Step 2: The weights V1, V2, V2, H1, H2, H2, H4 are
calculated by adding the weights of the points
that fall on the respective lines. Note that for
this example, p=3, and q=4
                                                 85
Solution:
                 4

Step 3: Since   w
                i 1
                       i    38

  set N0 = D0 = -38

  Step 4: Set
  N1 = -38 + 2(8) = -22;      D1 = -38 + 2(6) = -26;
  N2 = -22 + 2(26) = 30;      D2 = -26 + 2(4) = -18;
  N3 = 30 + 2(4) = 38;        D3 = -18 + 2(8) = -2;
                              D4 = -2 + 2(20) = 38;
  (These values are entered at the bottom of each
           column and left of each row in figure 1)

                                                  86
Solution:
Step 5: Compute the slope of each region.
S00 = -(-38/-38) = -1;    S14 = -(-22/38) = 0.58;
S01 = -(-38/-26) = -1.46; S20 = -(30/-38) = 0.79;
S02 = -(-38/-18) = -2.11; S21 = -(30/-26) = 1.15;
S03 = -(-38/-2) = -19;    S22 = -(30/-18) = 1.67;
S04 = -(-38/38) = 1;      S23 = -(30/-2) = 15;
S10 = -(-22/-38) = -0.58; S24 = -(30/38) = -0.79;
S11 = -(-22/-26) = -0.85; S30 = -(38/-38) = 1;
S12 = -(-22/-18) = -1.22; S31 = -(38/-26) = 1.46;
S13 = -(-22/-2) = -11;    S32 = -(38/-18) = 2.11;
                                                 87
Solution:
Step 5: Compute the slope of each region.

S33 = -(38/-2) = 19;
S34 = -(38/38) = -1;

(The above slope values are shown inside each
  region.)




                                            88
Solution:
    Step 6: When we draw contour lines
          through point (9, 10), we get the
          region shown in figure 1.

    Since the copiers cannot be placed at the
    (10, 10) location, we drew contour lines
    through another nearby point (9, 10).
    Locating anywhere possible within this
    region give us a feasible, near-optimal
    solution.


                                              89
                 13.4.3
Single-facility Location Problem with
    Squared Euclidean Distances




                                   90
La Quinta Motor Inns
Moderately priced, oriented towards business
travelers
Headquartered in San Antonio Texas
Site selection - an important decision
Regression Model based on location
characteristics classified as:
   - Competitive, Demand Generators,
      Demographic, Market Awareness, and
      Physical

                                           91
La Quinta Motor Inns (Cont)
Major Profitability Factors - Market awareness,
hotel space, local population, low
unemployment, accessibility to downtown office
space, traffic count, college students, presence
of military base, median income, competitive
rates




                                             92
 Gravity Method:
The cost function is

                                                          
                     m
  Minimize TC       ci f i ( xi  x ) 2  ( yi  y ) 2
                    i 1

As before, we substitute wi = fi ci, i = 1, 2, ..., m
and rewrite the objective function as
                    m                    m
 Minimize TC       wi ( xi  x ) 2   wi ( yi  y ) 2
                   i 1                  i 1




                                                               93
Gravity Method (Cont)
Since the objective function can be shown to
be convex, partially differentiating TC with
respect to x and y, setting the resulting two
equations to 0 and solving for x, y provides the
optimal location of the new facility
TC     m         m
     2 wi x  2 wi xi  0
 x    i 1      i 1
        m          m
 x   wi xi     w      i
        i 1       i 1
                                              94
Gravity Method (Cont)
 Similarly,
 TC     m         m
      2 wi y  2 wi yi  0
  y    i 1      i 1
         m         m
  y   wi yi    w      i
         i 1      i 1

 Thus, the optimal locations x and y are simply
 the weighted averages of the x and y coordinates
 of the existing facilities
                                            95
Example 6:
Consider Example 4. Suppose the distance
metric to be used is squared Euclidean.
Determine the optimal location of the new
facility using the gravity method.




                                            96
Solution - Table 10
Department i   xi     yi   wi   wixi    wiyi
      1        10     2    6    60      12
      2        10     10   10   100     100
      3        8      6    8    64      48
      4        12     5    4    48      20
    Total                  28   272     180

From table 10, we conclude that
x  272 28  9.7 and y  180 28  6.4
                                         97
Example 6. Cont...
If this location is not feasible, we only need to
find another point which has the nearest
Euclidean distance to (9.7, 6.4) and is a feasible
location for the new facility and locate the
copiers there




                                                98
 13.4.4
Weiszfeld
Method




            99
Weiszfeld Method:
 The objective function for the single facility
 location problem with Euclidean distance can
 be written as:
                    m
 Minimize TC  ci f i (x i  x) 2  (y i  y) 2
                   i 1

As before, substituting wi=cifi and taking the
derivative of TC with respect to x and y yields




                                                  100
Weiszfeld Method:
TC 1 m        w i 2(xi  x)
     
 x  2 i 1 (xi  x) 2  (yi  y) 2
         m
                         wixi
                                        
         i 1   (xi  x) 2  (yi  y) 2
                m
                                wix
                                                0
                i 1   (xi  x) 2  (yi  y) 2

                                                  101
Weiszfeld Method:
       m
                     wixi
       
       i 1   (xi  x) 2  (yi  y) 2
x    m
                       wi
       
       i 1   (xi  x) 2  (yi  y) 2




                                        102
Weiszfeld Method:
TC 1 m        w i 2(yi  y)
     
 y  2 i 1 (xi  x) 2  (yi  y) 2
         m
                         w i yi
                                        
         i 1   (xi  x) 2  (yi  y) 2
                m
                                  wiy
                                                0
                i 1   (xi  x) 2  (yi  y) 2

                                                  103
Weiszfeld Method:
       m
                      w i yi
       
       i 1   (xi  x) 2  (yi  y) 2
y    m
                       wi
       
       i 1   (xi  x) 2  (yi  y) 2




                                        104
Weiszfeld Method:
Step 0: Set iteration counter k = 1;
                m                              m

                w x    i       i              w y    i       i
         xk    i 1
                   m
                                    ;   yk    i 1
                                                  m

                w
                 i 1
                            i                  w
                                                i 1
                                                           i




                                                                   105
Weiszfeld Method:
Step 1: Set
                         m
                                          wixi
                         
                         i 1   (xi  x k ) 2  (yi  y k ) 2
              k 1
       x               m
                                            wi
                         
                         i 1   (xi  x k ) 2  (yi  y k ) 2



                                                          106
Weiszfeld Method:
                   m
                                     wixi
                   
                   i 1   (xi  x )  (yi  y ) 2
                                 k   2        k
       x k 1    m
                                   wi
                   
                   i 1   (xi  x )  (yi  y ) 2
                                 k   2        k




•   Step 2: If xk+1 = xk and yk+1 = yk, Stop.
    Otherwise, set k = k + 1 and go to Step 1

                                                    107
Example 7:
Consider Example 5. Assuming the distance
metric to be used is Euclidean, determine the
optimal location of the new facility using the
Weiszfeld method. Data for this problem is
shown in Table 11.




                                                 108
         Table 11.
Coordinates and weights for
      4 departments




                              109
Table 11:
  Departments #   xi   yi   wi
       1          10   2    6
       2          10   10   20
       3          8    6    8
       4          12   5    4




                                 110
Solution:
Using the gravity method, the initial seed can
be shown to be (9.8, 7.4). With this as the
starting solution, we can apply Step 1 of the
Weiszfeld method repeatedly until we find that
two consecutive x, y values are equal.




                                             111

								
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