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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 2Vi 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