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Facility Location Logistics Management Factors that Affect Location Decisions Distance Measures Classification of Planar Facility Location Problems 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 Planar Multi-Facility Location Problems –Minisum Location Problem with Rectilinear Distances

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Logistics Management Logistics Management can be defined as the management of the transportation and distribution of goods. The term goods includes raw materials or subassemblies obtained from suppliers as well as finished goods shipped from plants to warehouses or customers. Logistics management problems can be classified into three categories: –Location Problems: involve determining the location of one or more new facilities in one or more of several potential sites. The cost of locating each new facility at each of the potential sites is assumed to be known. It is the fixed cost of locating a new facility at a particular site plus the operating and transportation cost of serving customers from this facility-site combination. –Allocation Problems: assume that the number and location of facilities are known a priori and attempt to determine how each customer is to be served. In other words, given the demand for goods at each customer center, the production or supply capacities at each facility, and the cost of serving each customer from each facility, the allocation problem determines how much each facility is to supply to each customer center. –Location-Allocation Problems:involve determining not only how much each customer is to receive from each facility but also the number of facilities along with their locations and capacities.

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Factors that Affect Location Decisions Proximity to source of raw materials. Cost and availability of energy and utilities. Cost, availability, skill, and productivity of labor. Government regulations at the federal, state, county, and local levels. Taxes at the federal, state, county, and local levels. Insurance. Construction costs and land price. Government and political stability. Exchange rate fluctuation. Export and import regulations, duties, and tariffs. Transportation system. Technical expertise. Environmental regulations at the federal, state, county and local levels. Support services. Community services - schools, hospitals, recreation, and so on. Weather. Proximity to customers. Business climate. Competition-related factors.

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Distance Measures Rectilinear distance (L 1 norm) –d(X, P i ) = |x - a i | + |y - b i | Straight line or Euclidean distance (L 2 norm) –d(X, P i ) = Tchebyshev distance (L norm) –d(X, P i ) = max{|x - a i |, |y - b i |} X = (x, y) P i = (a i, b i ) X = (x, y)

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Classification of Planar Facility Location Problems Facility Location Single- Facility Multi- Facility Minisum Minimax Rectilinear Euclidean Tchebyshev Rectilinear Euclidean Tchebyshev Rectilinear Euclidean Tchebyshev Rectilinear Euclidean Tchebyshev Minisum Minimax # of facilitiesObjectives Distance measures

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Planar Single-Facility Location Formulations Minisum Formulation : Min f(x) = where X = (x, y) : location of the new facility P i = (a i, b i ) : location of the i-th existing facility, i = 1, …, m w i : weight associated to the i-th existing facility For example, w i =, where c i : cost per hour of travel,t i : number of trips per month, v i : average velocity. Minimax Formulation : Min f(x) = Max {w i d(X, P i )} Min z s. t. w i d(X, P i ) z, i = 1, …, m i = 1, …, m

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Insights for the Minisum Problem with Euclidean Distance Majority Theorem : When one weight constitutes a majority of the total weight, an optimal new facility location coincides with the existing facility which has the majority weight. w5w5 w1w1 w2w2 w4w4 w3w3 P1P1 P2P2 P3P3 P4P4 P5P5 Weight proportional to w i String Hole Horizontal pegboard

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Minisum Location Problem with Rectilinear Distances Min f(x, y) = Note thatf(x, y) = f 1 (x) + f 2 (y) where f 1 (x) = f 2 (y) = The cost of movement in the x direction is independent of the cost of movement in the y direction, and viceversa. Now, we look at the x direction. f 1 (x) is convex a local min is a global min.

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Minisum Location Problem with Rectilinear Distances (cont.) The coordinates of the existing facilities are sorted so that a 1 a 2 a 3 …. Now, we consider the case of m = 3. Case x a 1 : f 1 (x) = w 1 |a 1 - x| + w 2 |a 2 - x| + w 3 |a 3 - x| = - (w 1 + w 2 + w 3 )x + w 1 a 1 + w 2 a 2 + w 3 a 3 = - W x + w 1 a 1 + w 2 a 2 + w 3 a 3,where W = w 1 + w 2 + w 3 Case a 1 x a 2 : f 1 (x) = w 1 |a 1 - x| + w 2 |a 2 - x| + w 3 |a 3 - x| = (w 1 - w 2 - w 3 )x - w 1 a 1 + w 2 a 2 + w 3 a 3 = (- W + 2 w 1 ) x - w 1 a 1 + w 2 a 2 + w 3 a 3 …

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Objective Function f 1 (x) a3a3 a2a2 a1a1 w3w3 w2w2 w1w1 w 1 + w 2 + w 3 w 1 + w 2 - w 3 w 1 - w 2 - w 3 - w 1 - w 2 - w 3 The slope changes sign x f 1 (x)

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Minisum Location Problem with Rectilinear Distances (cont.) Slopes of f 1 (x) : M 0 = - (w 1 + w 2 + w 3 ) = - W M 1 = 2 w 1 + M 0 M 2 = 2 w 2 + M 1 M 3 = 2 w 3 + M 2 = w 1 + w 2 + w 3 = W Median conditions : f 1 (x) is minimized at the point where the slope changes from nonpositive to nonnegative. M 1 = w 1 - w 2 - w 3 < 0 2 w 1 < (w 1 + w 2 + w 3 ) = W w 1 < W/2 M 2 = w 1 + w 2 - w 3 0 2 (w 1 + w 2 ) (w 1 + w 2 + w 3 ) = W (w 1 + w 2 ) W/2

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Example 1 m = 3 a 1 = 10a 2 = 20a 3 = 40 w 1 = 5w 2 = 6w 3 = 4 W = w 1 + w 2 + w 3 = 15 W/2 = 7.5 w 1 = 5 < 7.5 w 1 + w 2 = 11 > 7.5 Minimizing point : a 2 = 20 Problem Data : Solution :

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Linear Programming Formulation Min f 1 (x) = w 1 |a 1 - x| + w 2 |a 2 - x| + w 3 |a 3 - x| Min z = w 1 (r 1 + s 1 ) + w 2 (r 2 + s 2 ) + w 3 (r 3 + s 3 ), Dual variables s. t. x - r 1 + s 1 = a 1, : y 1 x - r 2 + s 2 = a 2, : y 2 x - r 3 + s 3 = a 3, : y 3 r j, s j 0, j = 1, 2, 3. Relationships among variables : a j - x = r j - s j, |a j - x| = r j + s j, r j, s j 0. If both r j, s j > 0, we can reduce each by j = min {r j, s j }. This maintains feasibility and reduces z In an optimal solution, at least one of the r j and s j is 0, i. e., r j s j = 0.

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Linear Programming Formulation (cont.) Dual Problem : Max g = - a 1 y 1 - a 2 y 2 - a 3 y 3 + (w 1 a 1 + w 2 a 2 + w 3 a 3 ) s. t. y 1 + y 2 + y 3 = w 1 + w 2 + w 3 = W 0 y j 2 w j, j = 1, 2, 3 Min a 1 y 1 + a 2 y 2 + a 3 y 3 s. t. y 1 + y 2 + y 3 = W 0 y j 2 w j, j = 1, 2, 3 Complementary slackness conditions : 0 < y j * < 2 w j x* = a j 1 2 0 y 1 2 w 1 a 1 0 y 2 2 w 2 a 2 a 3 0 y 3 2 w 3 W W a 1 a 2 a 3

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Example 1 : Dual Solution f 1 (x) = 5 |x - 10| + 6 |x - 20| + 4 |x - 40| W = 15 y 1 * = 10 y 2 * = 5 y 3 * = 0 0 < y 2 * < 12 x* = a 2 = 20 1 2 0 y 1 10 10 0 y 2 12 20 40 0 y 3 8 15

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Minisum Location Problem with Euclidean Distances Min f(x, y) = Colinear case : all the points are in a line The problem reduces to minimizing f 1 (x), which is the rectilinear distance problem. (a i, b i ) The optimum location is always in the convex hull of {(a 1, b 1 ), …, (a m, b m )} (a i, b i )

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Non-colinear Case The graph of is a cone (strictly convex function). f(x, y) = is strictly convex unless the convex hull is a line segment. (a i, b i ) contours (a i, b i, 0) y x y x

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Non-colinear Case (cont.) First order optimality conditions : Any point where the partial derivatives are zero is optimal. Let and (x 0, y 0 ) is optimal

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Non-colinear Case (cont.) = 0

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No-colinear Case (cont.) If the optimal solution is in an exiting facility (a i, b i ), then. A simple way to avoid the problem of division by zero is to “perturb” the problem as follows : where > 0 and small. f(x,y) is flat near the optimum. x y f(x,y) (x*,y*)

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Weiszfeld’s Algorithm Initialization : Iterative step (k = 1, 2, …) : Terminating conditions : or

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Example 2 Problem Data : m = 4 P 1 = (0, 0)w 1 = 1 P 2 = (0, 10)w 2 = 1 P 3 = (5, 0)w 3 = 1P 4 = (12, 6)w 4 = 1 Solution : x 0 = (5+12)/4 = 4.25y 0 = (10+6)/4 = 4 k 1 2 5 10 Optimum (x, y) 4.023, 3.116 3.949, 2.647 3.958, 2.124 3.995, 2.011 4.000, 2.000 f(x, y) 24.808 24.665 24.600 24.597

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Minimax Location Problem with Rectilinear Distances Possible example : locating an ambulance with the existing facilities being the locations of possible accidents. X P3P3 P4P4 P2P2 P1P1 h3h3 h4h4 h2h2 h1h1 Hospital Hospital : Poss. Accident : Ambulance :

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Minimax Location Problem with Rectilinear Distances (cont.) Notation EF (existing facilities) locations : P i = (a i, b i ), i = 1, …, m NF (new facility = ambulance) location : X = (x, y) Travel distance from EF i to the nearest hospital = h i, i = 1, …, m Travel distance from NF to EF i = |x - a i | + |y - b i | Formulation : Min g(x, y) whereg(x, y) = max {|x - a i | + |y - b i | + h i : i = 1, …, m} Min z s. t. |x - a i | + |y - b i | + h i z, i = 1, …, m

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Minimax Location Problem with Rectilinear Distances (cont.) We want to make the inequalities linear |x - a i | + |y - b i | z - h i x - a i + y - b i z - h i (1) a i - x + b i - y z - h i (2) a i - x + y - b i z - h i (3) x - a i + b i - y z - h i (4) Make the intersection as small as possible with the largest diamond as small as possible. (2) (4) (3) (1)

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Minimax Location Problem with Rectilinear Distances (cont.) Minz s. t. x + y - z a i + b i - h i, i = 1,..., m x + y + z a i + b i + h i, i = 1,..., m - x + y - z - a i + b i - h i, i = 1,..., m - x + y + z - a i + b i + h i, i = 1,..., m Minz s. t. x + y - z c 1 wherec 1 = Min {a i + b i - h i } x + y + z c 2 c 2 = Max {a i + b i + h i } - x + y - z c 3 c 3 = Min {- a i + b i - h i } - x + y + z c 4 c 4 = Max {- a i + b i + h i } i = 1,..., m

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Minimax Location Problem with Rectilinear Distances (cont.) Minz s. t.- x - y + z - c 1 x + y + z c 2 x - y + z - c 3 - x + y + z c 4 c 5 = Max {c 2 - c 1, c 4 - c 3 }z = c 5 /2 Optimal solution : (x 1, y 1, z 1 ) = 1/2 (c 1 - c 3, c 1 + c 3 + c 5, c 5 ) (x 2, y 2, z 2 ) = 1/2 (c 2 - c 4, c 2 + c 4 - c 5, c 5 ) The line segment joining (x 1, y 1 ) and (x 2, y 2 ) is the set of optimal NF locations z (c 2 - c 1 )/2 (lower bound) z (c 4 - c 3 )/2 (lower bound)

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Minimax Location Problem with Euclidean Distances Examples : helicopter in an emergency unit, radio transmitter EF : (a i, b i ), i = 1, …, m NF : (x, y) min g(x, y) whereg(x, y) = max {[(x - a i ) 2 + (y - b i ) 2 ] 1/2, i = 1, …, m} min z s. t. [(x - a i ) 2 + (y - b i ) 2 ] 1/2 z, i = 1, …, m min z s. t. (x - a i ) 2 + (y - b i ) 2 z, i = 1, …, m (a i, b i ) (x, y)

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Elzinga-Hearn Algorithm (1971) Step 1. Choose any two points and go to Step 2. Step 2.Find the minimum covering circle for the chosen points*. Discard from the set of chosen points those points not defining the minimum covering circle, and go to Step 3. Step 3. If the constructed circle contains all the points, then the center of the circle is a minimax location, so stop. Otherwise, choose some point outside the circle, add it to the set of points defining the circle, and go to Step 2. * Find the minimum covering circle for the chosen points : A.If there are two points, let the two points define the diameter of the circle. B.If there are three points defining a right or obtuse triangle, let the two points opposite to the right or obtuse angle define the diameter of the circle. Otherwise, construct a circle through the three points (see Figure 1). C.If there are four points, construct a circle using as defining points those indicated in Figure 2.

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Elzinga-Hearn Algorithm (cont.) Defining points : BD Defining points : ABD Defining points : ABD Defining points : AD A B Defining points : BCD Defining points : ACD Defining points : ABD Defining points : AD A B A Defining points : CD Defining points : BD Figure 1. Alternative B Figure 2. Alternative C B C C

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Planar Multi-Facility Location Problems Old Facility : New Facility : X2X2 X1X1 P4P4 P3P3 P2P2 P1P1 v 12 w 24 w 23 w 12 w 11

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Minisum Multi-Facility Location Problem with Rectilinear Distances Location of new facilities: X j = (x j, y j ), j = 1, …, n. Location of existing facilities: P i = (a i, b i ), i = 1, …, m. Weight between new facilities j and k: v jk, where k > j. Weight between new facility j and existing facility i: w ji. Problem formulation: Min f((x 1,y 1 ), …, (x n, y n )) = f 1 (x 1, …, x n ) + f 2 (y 1, …, y n ) where f 1 (x 1, …, x n ) = f 2 (y 1, …, y n ) =

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Example 3 Problem Data : n = 2 (NF)m = 3 (EF) v = [v jk ] = w = [w ji ] = x = [x j ] = (x 1, x 2 ) a = [a j ] = (10, 20, 40) Min f 1 (x 1, x 2 ) = 2 |x 1 - x 2 | + 2 |x 1 - 10| + |x 1 - 20| + 4 |x 2 - 20| + 5 |x 2 - 40| Min f 1 (x 1, x 2 ) = 2 (p 12 + q 12 ) + 2 (r 11 + s 11 ) + (r 12 + s 12 ) + 4 (r 21 + s 21 ) + 5 (r 23 + s 23 ) s. t. x 1 - x 2 - p 12 + q 12 = 0 x 1 - r 11 + s 11 = 10 x 1 - r 12 + s 12 = 20 x 2 - r 21 + s 21 = 10 x 2 - r 23 + s 23 = 40 Relationships among variables : x 1 - x 2 = p 12 - q 12,|x 1 - x 2 | = p 12 + q 12, p 12, q 12 0 x i - a j = r ij - s ij,|x i - a j | = r ij + s ij, r ij, s ij 0

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Example 3 (Dual Problem) Max (- 10 u 11 - 20 u 12 - 10 u 21 - 40 u 23 ) + (10 2 + 20 1 + 10 4 + 40 5) Min 10 u 11 + 20 u 12 + 10 u 21 + 40 u 23 s. t. z 12 + u 11 + u 12 = 5 - z 12 + u 21 + u 23 = 7 0 z 12 4 0 u 11 4 0 u 12 2 0 u 21 8 0 u 23 10

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Equivalent Network Flow Problem After drawing the network, the solution can be usually obtained by inspection. N1N1 N2N2 E3E3 E1E1 E2E2 N3N3 1 4 8 12 Cap = u 11 4 u 12 2 z 12 4 u 21 8 u 23 10 (0) 12 (20) (10) (40) 5 7

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Equivalent Network Flow Problem (cont.) Complementary slackness conditions : 1. 0 < z jk * x k * x j * z jk * < 2 v jk x j * x k * In particular, 0 < z jk * < 2 v jk x j * = x k * 2. 0 < u ji * a i x j * u ji * < 2 w ji x j * a i In particular, 0 < u ji * < 2 w ji x j * = a i In Example 3, 0 < z 12 * < 2 v 12 x 1 * = x 2 *, 0 < u 12 = 2 w 12 x 2 * = a 1 = 10. If the network is not connected, then the problem decomposes into independent problems, one for each component.

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Example 4 Four hospitals located within a city are cooperating to establish a centralized blood-bank facility that will serve the hospitals. The new facility is to be located such that the (total) distance traveled is minimized. The hospitals are located at the following coordinates: P 1 =(5,10), P 2 =(7,6), P 3 =(4,2), and P 4 =(16,3). The number of deliveries to be made per week between the blood-bank facility and each hospital is estimated to be 3, 8, 2, and 10, respectively. Assuming rectilinear travel, determine the optimum location. m = 4P 1 = (5, 10)w 1 = 3P 2 = (7, 6)w 2 = 8 P 3 = (4, 2)w 3 = 2P 4 = (16, 3)w 4 = 10 W = Computation of x*: a 3 = 4w 3 = 2w 3 = 2 a 1 = 5w 1 = 3w 3 + w 1 = 5 a 2 = 7w 2 = 8w 3 + w 1 + w 2 = 13 x* = a 4 = 16w 4 = 10 Computation of y*: b 3 = 2w 3 = 2w 3 = 2 b 4 = 3w 4 = 10w 3 + w 4 = 12 y* = b 2 = 6w 2 = 8 b 1 = 10w 1 = 3

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Example 5 Find the optimal location of an ambulance with respect to four (known) possible accident locations which coordinates are P 1 =(6,11), P 2 =(12,5), P 3 =(14,7), and P 4 =(10,16). The objective is to minimize the maximum distance from the ambulance location to an accident location and from the accident location to its closest hospital. The distances from the accident locations to their closest hospitals are h 1 =10, h 2 =16, h 3 =14, and h 4 =11. Assume that distances are rectilinear. If multiple optima exist, find all optimal solutions. (5, 4) 6 8 10 12 14 (6, 11) (10, 16) (12, 5) (10, 7) (14, 7) (12, 9) h 4 = 11 h 2 = 16 h 3 = 14 h 1 = 10 16 14 12 10 8 6

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Example 5 Solution m = 4P 1 = (6, 11)h 1 = 10P 2 = (12, 5)h 2 = 16 P 3 = (14, 7)h 3 = 14P 4 = (10, 16)h 4 = 11 c 1 = min {a i + b i - h i } = min {6+11-10, 12+5-16, 14+7-14, 10+16-11} = c 2 = max {a i + b i + h i } = max {6+11+10, 12+5+16, 14+7+14, 10+16+11} = c 3 = min {-a i + b i - h i } = min {-6+11-10, -12+5-16, -14+7-14, -10+16-11} = c 4 = max {-a i + b i + h i } = max {-6+11+10, -12+5+16, -14+7+14, -10+16+11} = c 5 = max (c 2 - c 1, c 4 - c 3 }= max { -, - } = Optimal objective value: z* = Set of optimal solutions: line segment defined by the following end points: (x 1 *, y 1 *) = (c 1 - c 3, c 1 + c 3 + c 5 ) = ( -, + + ) = (, ) (x 2 *, y 2 *) = (c 2 - c 4, c 2 + c 4 - c 5 ) = ( -, + - ) = (, )

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Example 6 Given five existing facilities located at points P 1, P 2, P 3, P 4, and P 5 as shown below, determine the optimum location for a new facility which will minimize the maximum distance to the existing facilities. Assume that distances are Euclidean. Elzinga-Hearn algorithm: Figure 1 : Initial set of points = {P 1, P 2, P 3 }; center = C 1. Figure 2 : 2 nd set of points = {P 1, P 2, P 4 }; center = C 2. Figure 3 : 3 rd set of points = {P 2, P 4, P 5 }; center = C 3 (optimal location).

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Figure 1 P3P3 P1P1 P2P2 P4P4 C1C1 P5P5 P2P2 P 3 P 1

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Figure 2 P3P3 P1P1 P2P2 P4P4 P5P5 C2C2 P2P2 P1P1 P4P4

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Figure 3 P3P3 P1P1 P2P2 P4P4 P5P5 C3C3 P4P4 P5P5 P1P1

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