2. Facilities Layout and Location The three most important criteria in locating a factory: Location! Location! Location!

3. Supply Chain Homework Formulation X i,j = amount of raw natural gas sent from field i to plant j 10 6 ft 3, i = A,B,C,D,F; j = F,G,H,I Y jkl = 10 6 ft 3 finished gas product k produced at plant j and sent to customer l; l = J,K,L,M,N,O P k production cost of gas k; $/10 6 ft 3 T i,j = delivery cost by pipeline from field i to plant j; $/10 6 ft 3 T jl = delivery cost by truck from plant j to customer l; $/10 6 D ij = distance of pipeline from field i to plant j; miles D jl = distance of road from plant j to customer l; miles A jk = amount of product k produced at plant j as a fraction of one unit of raw natural gas

4. Supply Chain Homework Formulation

5. Facilities Layout Techniques Applied to: Hospitals Warehouses Schools Offices Workstations e.g. offices cubicles or manufacturing cells Banks Shopping centers Airports Factories

6. Objectives of a Facility Layout Problem Minimize investment in new equipment Minimize production time Utilize space most efficiently Provide worker convenience and safety Maintain a flexible arrangement Minimize material handling costs Facilitate manufacture Facilitate organizational structure

7. Types of Layouts Fixed Position Layouts – suitable for large items such as airplanes. Product Layouts – work centers are organized around the operations needed to produce a product. Process Layouts – grouping similar machines that have similar functions. Group Technology Layouts – layouts based on the needs of part families.

8. Fixed Position Layout

9. Product layout

10. Process layout

11. Group Technology Layout

12. Computerized Layout Techniques CRAFT. An improvement technique that requires the user to specify an initial layout. Improves materials handling costs by considering pair-wise interchange of departments. COFAD. Similar to CRAFT, but also includes consideration of the type of materials handling system. ALDEP. Construction routine (does not require user to specify an initial layout). Uses REL chart information. CORELAP. Similar to ALDEP, but uses more careful selection criteria for initial choosing the initial department PLANET. Construction routine that utilizes user specified priority ratings.

13. Activity Relationship Chart The desirability of locating pairs of operations near each other: A – absolutely necessary E – especially important I – important O – ordinary importance U – unimportant X - undesirable Reason code: 1 - Flow of material 2 - Ease of supervision 3 - Common personnel 4 - Contact necessary 5 - Convenience

14. Offices Floor manager Conference room Post office Parts shipment Repair & Servicing Receiving Inspection E/4 I/5 I/1 U U U U U U E/1 U A/1 E E/1 O U O/5 U U I/2 U U I/1 E U U U Activity Relationship Chart

15. Assembly Layout n assembly areas to be located on the factory floor in m possible locations (m  n) Minimize material handling requirements Material handling is from the receiving (raw material) to each assembly area and from each assembly area to shipping The Factory Floor Receiving Shipping assembly locations

16. The Cost Coefficients e i,1 = trips per day from assembly area i to receiving e i,2 = trips per day from assembly area i to shipping d j,1 = distance in feet from location j to receiving d j,2 = distance in feet from location j to shipping c i, j = cost per day for assembly area i to be located in location j c i,j = e i,1 d j,1 + e i,2 d j,2

17. An Optimization Model Let X i,j = 1 if assembly area i is at location j; 0 otherwise

18. Discrete Location Assignment Problem Location AssemblyABCD 194136271 262198496 375881880 411M8121 materials handling costs 1 – B 2 – A 3 – C 4 – D costs = $114

19. Locate M new facilities among N potential sites with k existing facilities A (k x N)B (k x M) C (N x M) C = A t B Facility i located at site F: C 11 = (1)(4) + (2)(3) + (4)(6) + (3)(2) + (5)(7) = 75

20. From-to Chart (distances in feet) To From sawsmillingpunch press drillslathessanders saws 1840306524 milling 18 38751630 punch press 4038223812 drills 3075225046 lathes 6516385060 sanders 2430124660

21. From-to Chart - trips per day To From sawsmillingpunch press drillslathessanders saws 43261440 milling 7560 23 punch press 4516 drills 22 28 lathes 45 3060 sanders 12

22. From-to Chart - Cost per day ($) To From sawsmillingpunch press drillslathessanders saws 154.820884520 milling 570900 138 punch press 34238.4 drills 330 280 lathes 144 300720 sanders 72 Cost per day = cost per foot x distance in feet x trips per day

23. Quadratic Assignment Problem Problem: Place each of n facilities in one of n locations where work-in-process moves among the facilities n! possible solutions. Plant Layout Facilities Lathe shop drilling sanding rough polishing finishing inspection galvanizing paint shop packaging quality control raw material storage finished goods storage assembly cell 1 cell 2 cell 3

24. Quadratic Assignment Problem let c i,j,k,l = the “cost” of placing machine i in location k and machine j in location l c i,j,k,l = f i,j d k,l where f i,j = the number of trips per day between machine i to machine j d k,l = the distance in feet between location k and location l let x i,k = 1 if facility i is placed in location k; 0 otherwise let x j,l = 1 if facility j is placed in location l; 0 otherwise

25. E F G H A 3 1 5E B 20 4 6F C 10 14 7 G D 12 8 9H A B C D d k,l f i,j locations facilities Location E F G H Cost A B C D A: 20(3) + 10(1) + 12(5) = 130 B: 14(4) + 8(6) = 104 C: 9(7) = 63 total = 297 297

26. E F G H A 3 1 5E B 20 4 6F C 10 14 7 G D 12 8 9H A B C D d k,l f i,j locations facilities Best pairwise exchange: From baseline solution, compute all possible interchanges. Select the best one; then repeat until no further improvement is obtained. Location E F G H Cost A B C D 297 B A C D 289 A B D C 301 D B C A 339 A C B D 276 C B A D 309 A D C B 328 Location E F G H Cost A C B D 276 C A B D 297 D C B A 351 A C D B 318 A D B C 290 B C A D 280 n2n2 ( ) paired exchanges

27. Any Improvement Heuristic: From baseline solution, interchange pairs until an improve- ment is obtained. Then repeat until all pairs have been found with no further improvement. Location E F G H Cost A B C D 297 B A C D 289 B A D C 309 B C A D 280 D C A B 343 etc.

28. Facility Location Goal is to find the optimal location of one or more new facilities. Optimality depends on the objective used. In many systems, the objective is to minimize some measure of distance. Two common distance measures: Straight line distance (Euclidean distance). Rectilinear Distance (as might be measured following roads on city streets).

29. Facility Location Analysis Locating on a plane sphere 3-dimensional space network discrete location Minimize costs distances weighted distances travel time Single versus multiple facilities Straight-line (Euclidean) versus Rectilinear Distances

30. Locating on the Plane-Euclidean Distances x y (a,b) (x,y) (x – a) (y – b) h h 2 = (x – a) 2 + (y – b) 2

31. The Centroid Problem let x = the x-coordinate y = the y-coordinate (a i,b i ) = coordinate of i th facility w i = weight placed on the i th facility Euclidean distance squared

32. The Solution convex function

33. The Euclidean Distance Problem

35. If at least one half of the cumulative weight is associated with an existing facility, the optimum location for the new facility will coincide with the existing facility. I call this the Majority Theorem. It is also true that the optimum location will always fall within the convex hull formed from the existing points.

36. A Simple Proof using an Analog Model The convex hull

37. Centroid Problem - Example x y 1 2 3 4 5 x,y (2,2) (6,3) (9,1) (8,5) (3,6) 5 10 6 5 2

38. Centroid Problem Location w i a i b i w i x i w i y i P15221010 P210363060 P326312 6 P46854830 P559145 5 total28145111 x* = 145/28 = 5.2y* = 111/28 = 4.0

39. Euclidean Distance - Example x y central repair facility w = vehicles serviced per day 1 3 2 (1100,400) (500,800) (1400,1200) w 2 = 8 w 1 = 4 w 3 = 8

40. Euclidean Distance - Example x y central repair facility w = vehicles serviced per day 1 3 2 (1100,400) (500,800) (1400,1200) w 2 = 8 w 1 = 4 w 3 = 8 Centroid solution: x 0 = [4(500)+8(1100)+8(1400)]/20 = 1100 y 0 = [4(800)+8(400)+8(1200)]/20 = 800 OPTIMAL SOLUTION: x* = 1098; y* = 649

41. Locating on the Plane-Rectilinear Distances x y (a,b) (x,y) (x – a) (y – b)

42. Rectilinear Distance Properties of the optimum solution: 1. can solve for x, y independently 2. x coordinate will be the same as one of the a i 3.y coordinate will be the same as one of the b i 4.Optimum x (y) coordinate is at a median location with respect to the weights; i.e. no more than half the movement is to the left (up) or right (down) of the location.

43. x 1 x 2 x 3 = x* x 4 x 5 An Intuitive Proof – weights = 1 79610 x 1 x 2 x* x 3 x 4 x 5 7 6 3610 distance = (7+9) + 9 + 6 + (6+10) = 47 distance = (7+6) + 6 +3 + (3 + 6) + (3 + 6+10) = 50

44. w i w i + w 1 w 3 + w j w j x i x 1 x 2 = x* x 3 x j Let x 2 be a median location. That is w 2 + w 3 + w j  w i + w 1 w 3 + w j  w i + w 1 + w 2 w i w i + w 1 w 3 + w j w j x i x 1 x 2 x 3 x j x* w 2 + w 3 + w j but More of an Intuitive Proof

45. w i w i + w 1 w 3 + w j w j x i x 1 x 2 = x* x 3 x j Let x 2 be a median location. That is w 2 + w 3 + w j  w i + w 1 w 3 + w j  w i + w 1 + w 2 w i w i + w 1 w 3 + w j w j x i x 1 x 2 x 3 x j x* w i + w 1 + w 2 but More of More of an Intuitive Proof

46. Rectilinear Distance - Example Warehouses: ABCDEtotal location:(0,0)(3,16)(18,2)(8,18)(20,2) weight:522416034162 warehousex-coordweightcumulative wgt A055 B32227 D86087 C1841128 E2034162 x* = 8 median = 81

47. Rectilinear Distance - Example Warehouses: ABCDEtotal location:(0,0)(3,16)(18,2)(8,18)(20,2) weight:522416034162 warehousey-coordweightcumulative wgt A055 C,E241+3480 B1622102 D1860162 y* = 16 median = 81

48. Rectilinear Distance - Example x y 1 (6,6) (4,20) (20,22) (14,4) (22,8) 4 3 2 5 retail outlet centers factory (x,y) w 1 = 15 w 4 = 9 w 2 = 3 w 3 = 12 w 5 = 6

49. Rectilinear Distance - Example retail cumulative outletx-coord weight outlety-coord weight #449#24 3 #1624#1618 #21427#3830 #52033#42039 #32245#52245 median = 45/2 = 22.5 (x*,y*) = (6,8)

50. Rectilinear Distance - Example x y 1 (6,6) (4,20) (20,22) (14,4) (22,8) 4 3 2 5 retail outlet centers factory (6,8) w 1 = 15 w 4 = 9 w 2 = 3 w 3 = 12 w 5 = 6

51. MiniMax Criterion Locate a facility to minimize the maximum distance traveled from any of n existing facilities Applications locating a rural health clinic placement of fire stations and other emergency equipment location of a parking lot Useful if no continuous travel between all existing facilities; travel will not be performed over all routes What about a maximin criterion?

52. An Rectilinear Formulation Equivalent to finding the center of the smallest diamond that contains all of the (a i, b i ) 3 variables 4n constraints

53. Rectilinear Minimax Solution c 1 = min (a i + b i ) c 2 = max (a i + b i ) c 3 = min (-a i + b i ) c 4 = max (-a i + b i ) c 5 = max(c 2 – c 1, c 4 – c 3 ) x 1 = (c 1 – c 3 )/2 y 1 = (c 1 + c 3 + c 5 )/2 x 2 = (c 2 – c 4 )/2 y 2 = (c 2 + c 4 - c 5 )/2 0   1 x* = x 1 + (1- )x 2 y* = y 1 + (1- )y 2 f(x*,y*) = c 5 /2

54. The Minimax Company wishes to locate a parking lot relative to its four factories so that the maximum distance any employee must walk is minimized. (4,5) (2,8) (8,6) (10,12) c 1 = min(a i + b i ) = min(10,9,14,22) = 9 c 2 = max(a i + b i ) = max (10,9,14,22) = 22 c 3 = min (-a i + b i ) = min (6,1,-2,2) = -2 c 4 = max (-a i + b i ) = max (6,1,-2,2) = 6 c 5 = max(c 2 – c 1, c 4 – c 3 ) = max(13,8) = 13 x 1 = (c 1 – c 3 )/2 = 5.5 y 1 = (c 1 + c 3 + c 5 )/2 = 10 x 2 = (c 2 – c 4 )/2 = 8 y 2 = (c 2 + c 4 - c 5 )/2 = 7.5 x* = 5.5 + 8(1- ) y* = 10 + 2.5(1- ) f(x*,y*) = c 5 /2 = 6.5 85.5 2 10 (8,7.5) (5.5,10)

55. Equivalent to finding the center of the circle having the smallest radius that contains all of the (x i, y i ) A Euclidean Distance Formulation

56. Mini-max Euclidean Distance (4,5) (2,8) (8,6) (10,12) 85.5 2 10

57. Multifacility Location Locate m facilities relative to n existing facilities (Weighted) travel occurs among the new facilities

58. The Mathematical Formulation I get it. The “c” and “d” variables and the “e” and “f” variables cannot be in the basis at the same time. At least one must always be zero (nonbasic).

59. Example Problem A company is to locate two factories relative to its three major outlet stores. The following data pertains: Store 1Store 2Store 3 coordinates(5,18)(12,14)(18,5) weight1 (w i1 ) 21 18 6 weight2 (w i2 ) 4 23 36 interaction v 12 = 9 note: weights are train carloads per week travel by rail which is mostly rectilinear

60. Where in the World are these stores? 0 5 10 15 20 25 20 15 10 5 0

61. The Formulation – x

62. The Formulation – y

63. Solver Solution (x 1 *,y 1 *) = (12,14) (x 2 *,y 2 *) = (18,14)

64. Where are the factories? (x 1 *,y 1 *) = (12,14) ; (x 2 *,y 2 *) = (18,14) 0 5 10 15 20 25 20 15 10 5 0