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Facilities Design

Explore multiple-facility and location-allocation models to optimize distribution networks, customer assignments, and capacity planning in industrial engineering.

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Facilities Design

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  1. Facilities Design S.S. Heragu Industrial Engineering Department University of Louisville

  2. Chapter 12:Advanced Location Models

  3. Chapter 12: Advanced Location Models 12.1 INTRODUCTION 12.2 LOCATION MODELS 12.2.1 Multiple-Facility Problems with Rectilinear Distances 12.2.2 Multiple-Facility Problems with Euclidean Distances 12.3 ALLOCATION MODELS 12.3.1 Network Flow Model 12.3.2 Two-Stage Transportation Model 12.3.3 Vehicle Routing Problem

  4. Chapter 12: Advanced Location Models 12.4 LOCATION-ALLOCATION MODELS 12.4.1 Set Covering Model 12.4.2 Uncapacitated Location-Allocation Model 12.4.3 Comprehensive Location- Allocation Model 12.5 SUMMARY

  5. Introduction • How many new facilities are to be located in the distribution network consisting of previously established facilities and customers? • Where should they be located? • How large should each new facility be? In other words, what is the capacity of the new facility?

  6. Introduction • How should customers be assigned to the new and existing facilities? More specifically, which facilities should be serving each customer? • Can more than one facility serve a customer?

  7. 12.2LOCATIONMODELS

  8. 12.2.1Multiple-Facility Problemswith Rectilinear Distances

  9. Model 1:

  10. Model 1:

  11. Model 1: A similar definition of y+ij, y-ij, xa+ij, xa-ij, yb+ij, and yb-ij yields |yi - yj| = y+ij + y-ij yi - yj = y+ij - y-ij |xi - aj| = xa+ij + xa-ij xi - aj = xa+ij - xa-ij |yi - bj| = yb+ij + yb-ij yi - bj = yb+ij - yb-ij

  12. Model 1: Thus, the transformed linear model is:

  13. Model 1: (xi - xj) = x+ij - x-ij Subject to: yi - yj = y+ij - y-ij xi - aj = xa+ij - xa-ij yi - bj = yb+ij - yb-ij x+ij, x-ij, y+ij , y-ij> 0, i, j = 1, 2, ...,n xa+ij, xa-ij, yb+ij , yb-ij> 0, i = 1, 2, ...,n, j =1,2,...,m xi, yi unrestricted in sign, i = 1,2,...,n

  14. Model 1: • xij+ or xij-, but not both, can be greater than 0. (If both are, then the values of xij+ and xij- do not satisfy their definition in (2) and (3)). Similarly, only one of (i) yij+, yij-, (ii) xij+, xaij-, (iii) ybij+, ybij-, must be greater than 0. • Model 1 can be simplified by noting that: • xi can be substituted as ai + xij+ - xaij- • yi may also be substituted similarly, resulting in a model with 2n fewer constraints and variables than model 1.

  15. Example 1: • Tires and Brakes, Inc., is an automobile service company specializing in tire and brake replacement. It has four service centers in a metropolitan area and a warehouse that supplies tires, brakes and other components to the service centers. The company manager has determined that he needs to add two more warehouses so as to improve component delivery service.

  16. Example 1: He wants to ensure that the location of the two new warehouses is such that the cost of delivery components from the new warehouse to the existing facilities (four service centers and existing warehouse) as well as between the new warehouses is minimal. The four service centers and warehouse are located at the following coordinate locations: (8, 20), (8, 10), (10, 20), (16, 30), and (35, 20) It is anticipated that there will be one trip per day between the new warehouses.

  17. Example 1: In addition, the number of trips between the new warehouses and four service centers as well as the existing warehouse is provided below. SC1 SC2 SC3 SC4 W1 W2 7 7 5 4 2 W3 3 2 4 5 2 Develop a mode similar to the transformed Model 1 to minimize distribution cost and solve it using LINGO, LINDO or the LP Solver in Excel

  18. SUBJECT TO 2) - XP12 + XN12 + X1 - X2 = 0 3) - XP21 + XN21 - X1 + X2 = 0 4) - YP12 + YN12 + Y1 - Y2 = 0 5) - YP21 + YN21 - Y1 + Y2 = 0 6) - XAP11 + XAN11 + X1 = 8 7) - XAP12 + XAN12 + X1 = 8 8) - XAP13 + XAN13 + X1 = 10 9) - XAP14 + XAN14 + X1 = 16 10) - XAP15 + XAN15 + X1 = 35 11) - XAP21 + XAN21 + X2 = 8 12) - XAP22 + XAN22 + X2 = 8 13) - XAP23 + XAN23 + X2 = 10 14) - XAP24 + XAN24 + X2 = 16 15) - XAP25 + XAN25 + X2 = 35 16) - YBP11 + YBN11 + Y1 = 20 17) - YBP12 + YBN12 + Y1 = 10 18) - YBP13 + YBN13 + Y1 = 20 19) - YBP14 + YBN14 + Y1 = 30 20) - YBP15 + YBN15 + Y1 = 20 21) - YBP21 + YBN21 + Y2 = 20 22) - YBP22 + YBN22 + Y2 = 10 23) - YBP23 + YBN23 + Y2 = 20 24) - YBP24 + YBN24 + Y2 = 30 25) - YBP25 + YBN25 + Y2 = 20 END FREE X1 FREE X2 FREE Y1 FREE Y2 Example 1 (LINDO format): MIN XP12 + XN12 + YP12 + YN12 + XP21 + XN21 + YP21 + YN21 + 7 XAP11 + 7 XAN11 + 7 YBP11 + 7 YBN11 + 7 XAP12 + 7 XAN12 + 7 YBP12 + 7 YBN12 + 5 XAP13 + 5 XAN13 + 5 YBP13 + 5 YBN13 + 4 XAP14 + 4 XAN14 + 4 YBP14 + 4 YBN14 + 2 XAP15 + 2 XAN15 + 2 YBP15 + 2 YBN15 + 3 XAP21 + 3 XAN21 + 3 YBP21 + 3 YBN21 + 2 XAP22 + 2 XAN22 + 2 YBP22 + 2 YBN22 + 4 XAP23 + 4 XAN23 + 4 YBP23 + 4 YBN23 + 5 XAP24 + 5 XAN24 + 5 YBP24 + 5 YBN24 + 2 XAP25 + 2 XAN25 + 2 YBP25 + 2 YBN25

  19. Example 1 (LINGO Format): Data: N = 2; M = 5; Enddata Sets: NewFac/1..N/: X, Y; ExistFac/1..M/: A, B; NewLinks(NewFac,NewFac): F, XP, XN, YP, YN; ExistLinks(NewFac,ExistFac): G, XAP, XAN, YBP, YBN; Endsets Data: F = 0 1 1 0; G = 7 7 5 4 2 3 2 4 5 2; A = 8 8 10 16 35; B = 20 10 20 30 20; Enddata ! Objective function; Min= @SUM(NewLinks(i,j): F(i,j)*(XP(i,j)+XN(i,j)+YP(i,j)+YN(i,j)))+ @SUM(ExistLinks(i,j): G(i,j)*(XAP(i,j)+XAN(i,j)+YBP(i,j)+YBN(i,j))); ! Constraints; @FOR (NewLinks(i,j): X(i)-X(j)=XP(i,j)-XN(i,j); Y(i)-Y(j)=YP(i,j)-YN(i,j)); @FOR (ExistLinks(i,j): X(i)-A(j)=XAP(i,j)-XAN(i,j); Y(i)-B(j)=YBP(i,j)-YBN(i,j)); End

  20. Example 1 (LINGO Solution): Global optimal solution found. Objective value: 370.0000 Total solver iterations: 16 Variable Value X( 1) 8.000000 X( 2) 10.00000 Y( 1) 20.00000 Y( 2) 20.00000 XP( 2, 1) 2.000000 XN( 1, 2) 2.000000 XAP( 2, 1) 2.000000 XAP( 2, 2) 2.000000 XAN( 1, 3) 2.000000 XAN( 1, 4) 8.000000 XAN( 1, 5) 27.00000 XAN( 2, 4 6.000000 XAN( 2, 5) 25.00000 YBP( 1, 2) 10.00000 YBP( 2, 2) 10.00000 YBN( 1, 4) 10.00000 YBN( 2, 4) 10.00000

  21. 12.2.2Multiple-Facility Problemswith Euclidean Distances Consider the following objective for the euclidean distance problem.

  22. Multiple-Facility Problemswith Euclidean Distances Taking the partial derivatives, we get

  23. Multiple-Facility Problemswith Euclidean Distances

  24. Multiple-Facility Problemswith Euclidean Distances To make sure the denominator is never 0, we add  to it. We then get:

  25. Multiple-Facility Problemswith Euclidean Distances And …

  26. Example 2 Consider Example 1. Assuming the Euclidean metric is more appropriate and that Tire and Brakes, Inc. does not currently have a warehouse, determine where the two new warehouses are to be located.

  27. 12.3ALLOCATIONMODEL

  28. 12.3.1 Network Flow Model

  29. Network Flow Model: Model 2 Consider this notation: cij cost of sending one unit of flow on arc (i, j) Uij upper bound on the flow that can be sent on arc (i, j), i.e., capacity of arc (i,j) Lij lower bound on the flow that can be sent on arc (i, j) Di net flow generated at node i xij number of units of flow on arc (i, j)

  30. Model 2:

  31. Network Simplex Algorithm: Step 1: Construct a spanning tree for the n nodes. The variables xij corresponding to the arcs (i,j) in the spanning tree are basic variables and the remaining are nonbasic. Find a basic feasible solution to the problem so that: (1) the basic variables satisfy Lij < xij < Uij , and (2) the nonbasic variables take on a value of Lij or Uij to satisfy constraint (22). Step 2: Set u1 =0 and find uj, j=2,...,n using the formula ui - uj = cij for all basic variables. Step 3: If ui - uj - cij< 0 for all nonbasic variables xij with a value of Lij , and ui - uj - cij> 0 for all nonbasic variables xij with a value of Uij , then the current basic feasible solution is optimal; stop. Otherwise, go to step 4.

  32. Network Simplex Algorithm: Step 4: Select the variable xi*j* that violates the optimality condition (in step 3) the most, i.e., the largest of the ui - uj - cij for those nonbasic variables with xij = Lij, and the smallest of the ui - uj - cij for those nonbasic variables with xij = Uij,. Make the arc (i*,j*) a basic variable and add arc (i*,j*) to the spanning tree. Make one of the other basic variables in the loop of basic variables [formed by including arc (i*,j*)], a nonbasic variable such that: (1) xi*j* takes on the largest possible value, (2) constraint (21) is satisfied for all the n nodes, and (3) constraint (22) is satisfied for all the arcs in the loop. Remove the arc corresponding to the nonbasic variable just identified so that we have a spanning tree once again. Go to step 2.

  33. Example 3 The Fast Shipping Company manages the distribution of lawnmowers from a company that has two factories (F1 and F2) in the Northeast to two large customer bases (C1 and C2) in the Southwest. For cost and freight consolidation reasons, Fast Shipping would like to route the shipments via three intermediate nodes (T1 - T3) located in the midwest. The relevant data is provide in Tables 12.3-12.5. Setup a model to determine how the shipment is to take place from the two factories to the two destinations via the three intermediate shipment points.

  34. Supply and Demand

  35. Inbound and Outbound Transportation Costs and Arc Capacities

  36. SUBJECT TO 2) X11 + X12 + X13 = 900 3) X21 + X22 + X23 = 600 4) - Y11 - Y21 - Y31 = - 750 5) - Y12 - Y22 - Y32 = - 750 6) X11 + X21 - Y11 - Y12 = 0 7) X12 + X22 - Y21 - Y22 = 0 8) X13 + X23 - Y31 - Y32 = 0 9) X11 <= 500 10) X12 <= 1500 11) X13 <= 350 12) X21 <= 1200 13) X22 <= 750 14) X23 <= 450 15) Y11 <= 1000 16) Y21 <= 750 17) Y31 <= 1000 18) Y12 <= 150 19) Y22 <= 200 20) Y32 <= 1500 END Example 3 (LINDO Format): MIN 8 X11 + 11 X12 + 5 X13 + 12 X21 + 8 X22 + 5 X23 + 6 Y11 + 12 Y21 + 9 Y31 + 3 Y12 + Y22 + 19 Y32

  37. Example 3 (LINGO Format): Data: N = 2; M = 3; Enddata Sets: OriginNodes/1..N/: DemandO; DestinationNodes/1..N/: DemandD; IntermediateNodes/1..M/: DemandI; OutboundArcs(OriginNodes,IntermediateNodes): C, X, LO, UO; InboundArcs(IntermediateNodes,DestinationNodes): D, Y, LD, UD; Endsets Data: C = 8 11 5 12 8 5; LO = 0 0 0 0 0 0; UO = 500 500 350 1200 750 450; D = 6 3 12 1 9 19; LD = 0 0 0 0 0 0; UD = 1000 150 750 200 1000 1500; DemandO=900 600; DemandD=750 750; DemandI=0 0 0; Enddata

  38. Example 3 (LINGO Format): ! Objective function; Min= @SUM(OutboundArcs(i,j): C(i,j)*X(i,j))+ @SUM(InboundArcs(i,j): D(i,j)*Y(i,j)); ! Constraints; @FOR (OriginNodes(i): @SUM(IntermediateNodes(j): X(i,j))=DemandO(i)); @FOR (DestinationNodes(j): @SUM(IntermediateNodes(i): -1*Y(i,j))=-DemandD(j)); @FOR (IntermediateNodes(i): @SUM(OriginNodes(j): X(j,i))= @SUM(DestinationNodes(j): Y(i,j))); @FOR (OutboundArcs(i,j): X(i,j)<=UO(i,j)); @FOR (InboundArcs(j,i): Y(j,i)<=UD(j,i)); End

  39. Example 3 (LINGO Solution): Global optimal solution found. Objective value: 23700.00 Total solver iterations: 8 Variable Value X( 1, 1) 500.0000 X( 1, 2) 50.00000 X( 1, 3) 350.0000 X( 2, 2) 150.0000 X( 2, 3) 450.0000 Y( 1, 1) 350.0000 Y( 1, 2) 150.0000 Y( 2, 2) 200.0000 Y( 3, 1) 400.0000 Y( 3, 2) 400.0000

  40. 12.3.2 Two-Stage Transportation Model 1 1 1 2 2 3 2 3 4 q p r

  41. Two-Stage Transportation Model: Parameters Si capacity of supply source i, i = 1, 2, ..., p Pj capacity of plant j, j = 1, 2, ..., q Dk demand at customer k, k = 1, 2, ..., r cij cost of transporting one unit from supply source i to plant j djk cost of transporting one unit from plant j to customer k Decision Variables xij number of units transported from supply source i to plant j yjk number of units transported from plant j to customer k

  42. Model 3:

  43. Four cases arise: • Supply source capacity is unlimited and total plant capacity is more than total demand • Supply source capacity is unlimited and total demand exceeds total plant capacity • Plant capacity is unlimited and total supply source capacity exceeds total demand • Plant capacity is unlimited and total demand exceeds total supply source capacity In the following discussion, the supply sources are assumed to have unlimited capacities and total plant capacity is more than total demand (case (i))

  44. Case 1:

  45. Example 4: 2-Stage Distribution Problem: RIFIN Company has recently developed a new method of manufacturing a type of chemical. It involves refining a certain raw material which can be obtained from four overseas suppliers A, B, C, D who have access to the four ports at Vancouver, Boston, Miami, and San Francisco, respectively. RIFIN wants to determine the location of plants at which the chemical will be refined. The chemical, once refined, will be transported via trucks to five outlets located at Dallas, Phoenix, Portland, Montreal and Orlando.

  46. Example 4: After an initial study, the choice of location for RIFIN’s refineries has narrowed down to Denver, Atlanta and Pittsburgh. Assume that one unit of raw material is required to make one unit of chemical. The amount of raw material from each potential refinery as well as the cost of trucking the chemical to outlets are also provided below. Determine the location of RIFIN’s refining plants, capacities at these plants and distribution pattern for the raw material and processed chemical.

  47. Example 4: Raw Material Source Supply Outlet Demand A 1000 Dallas 900 B 800 Phoenix 800 C 800 Portland 600 D 700 Montreal 500 Orlando 500

  48. Example 4: Raw Material Transportation Cost TO Denver Atlanta Pittsburgh FROM Vancouver 4 13 9 Boston 8 8 5 Miami 12 2 9 San Francisco 11 11 12

  49. Example 4: Chemical Trucking Cost TO Dallas Phoenix Portland Montreal Orlando FROM Den. 28 26 12 30 30 Atla. 10 22 23 29 8 Pitts. 18 21 23 18 21

  50. Figure 12.3: Pictorial representation of RIFIN Example Da V D Ph B Po A M Mo P SF Orl

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