1. Facility Location

2. BMW In the late 1980’s fluctuating exchange rates and rising costs convinced BMW that it was time to consider operating a new production facility outside the European borders. A “blank page” approach was used to compile a list of 250 potential worldwide sites. Analysis pared the list down to 10 options; a location in the United States was preferred due to its proximity to a large market segment for BMW’s automobiles. BMW spent 3 1/2 years considered the labor climate, port and road access, geographical requirements and constraints, airport access, and its relations with the governments. The plant was located in Spartanburg, SC, and now employs approximately 4,700 workers who produce more than 500 vehicles a day.

3. LI & FUNG European retailer order 10,000 garments Buy the best Japanese zipper & button from Chaina Weave & dye in Taiwan Manufacture garments in Thailand Buy yarn from a Korean mfgr. 5 weeks later 10,000 garments reach Europe

4. Ellora Times In 2001, Ellora Time Pvt. Ltd. (Ellora), a company based in Gujarat, India, was the world's largest manufacturer of clocks. It also manufactured calculators, telephones, timepieces and educational toys. Ajanta and Orpat were closely held Ellora companies with a combined investment of Rs 2 billion. Almost all their products, marketed through a countrywide network of 25,000 dealers and 180 service stations, were leaders in their respective categories. For the year 1999-00, the group recorded a combined turnover of over Rs 2.50 billion. Both Ajanta and Orpat received awards by the Government of India for superior exports performance throughout the 1990s. Ajanta, an ISO 9002 certified company, had even received the ‘Best Electronics Industry'award many times.

5. Ellora Times In early 2001, Ellora shocked the corporate world by announcing its decision to shift its manufacturing activities to China.

6. Mitshubishi at Haldia Hero Hona at Uttranchal Wipro in Himachal Tata in Singur Bio-con in Andhrapradesh

7. Factors Affecting the Location Decision Economic – Site acquisition, preparation and construction costs – Labor costs, skills and availability – Utilities costs and availability – Transportation costs – Taxes

8. Factors Affecting the Location Decision Non-economic – Labor attitudes and traditions – Training and employment services – Community’s attitude – Schools and hospitals – Recreation and cultural attractions – Amount and type of housing available

9. Facility Types and Their Dominant Locational Factors Heavy Manufacturing – Near their raw material sources – Abundant supply of utilities – Land and construction costs are inexpensive Light Manufacturing – Availability and cost of labor Warehousing – Proximity to transportation facilities – Incoming and outgoing transportation costs

10. Facility Types and Their Dominant Locational Factors R&D and High-Tech Manufacturing – Ability to recruit/retain scientists, engineers, etc. – Near companies with similar technology interests Retailing and For-Profit Services – Near concentrations of target customers Government and Health/Emergency Services – Near concentrations of constituents

11. Some Reasons the Facility Location Decision Arises Changes in the market – Expansion – Contraction – Geographic shift Changes in inputs – Labor skills and/or costs – Materials costs and/or availability – Utility costs

12. Some Reasons the Facility Location Decision Arises Changes in the environment – Regulations and laws – Attitude of the community Changes in technology

13. Analyzing Industrial Facility Locations Locating a Single Facility – A simple way to analyze alternative locations is conventional cost analysis Locating Multiple Facilities – More sophisticated techniques are often used: Linear programming, computer simulation, network analysis, and others

14. Facility Location Models

15. Location Analysis Methods Factor rating method Load-distance model Center of gravity approach Break-even analysis Transportation method Dimensional Analysis

16. Location Rating Factor  Identify important factors  Subjectively score each factor (0 - 100)  Weight factors (0.00 - 1.00)  Sum weighted scores

17. Location Factor Rating: Example Labor pool and climate Proximity to suppliers Wage rates Community environment Proximity to customers Shipping modes Air service LOCATION FACTOR Rating 90604545301515 Normalized SCORES (0 TO 100)

18. Location Factor Rating: Example Labor pool and climate Proximity to suppliers Wage rates Community environment Proximity to customers Shipping modes Air service LOCATION FACTOR.30.20.15.15.10.05.05 Rating 90604545301515 Normalized SCORES (0 TO 100)

19. Location Factor Rating: Example Labor pool and climate Proximity to suppliers Wage rates Community environment Proximity to customers Shipping modes Air service LOCATION FACTOR 801006075658550 Site 1 SCORES (0 TO 100)

20. Location Factor Rating: Example Labor pool and climate Proximity to suppliers Wage rates Community environment Proximity to customers Shipping modes Air service LOCATION FACTOR 801006075658550 Site 1 65919580909265 Site 2 SCORES (0 TO 100)

21. Location Factor Rating: Example Labor pool and climate Proximity to suppliers Wage rates Community environment Proximity to customers Shipping modes Air service LOCATION FACTOR 801006075658550 Site 1 65919580909265 Site 2 90757280956590 Site 3 SCORES (0 TO 100)

22. Location Factor Rating: Example Labor pool and climate Proximity to suppliers Wage rates Community environment Proximity to customers Shipping modes Air service LOCATION FACTOR.30.20.15.15.10.05.05 WEIGHT 801006075658550 Site 1 65919580909265 Site 2 90757280956590 Site 3 SCORES (0 TO 100)

23. Location Factor Rating 24.00 20.00 9.00 11.25 6.50 4.25 2.50 77.50 Site 1 19.50 18.20 14.25 12.00 9.00 4.60 3.25 80.80 Site 2 27.00 15.00 10.80 12.00 9.50 3.25 4.50 82.05 Site 3 WEIGHTED SCORES Site 3 has the highest factor rating

24. Example

25. Nearness to the port Proximity to suppliers Availability of skilled labor Govt. Policies Projected cost of operation Shipping modes Educational Infrastructure LOCATION FACTOR 80709050607040 Score 75605070708060 Site 1 70807045609080 Site 2 90 90858560807080 Site 3 SCORES (0 TO 100)

26. 26 Center of Gravity Method The Center of Gravity Method is a tool that seeks to compute geographic coordinates for a potential single new facility that will minimize costs. The Center of Gravity Method takes many factors into account including:  Markets  Volume of goods shipped  Shipping costs

27.  Locate facility at center of geographic area  Based on weight and distance traveled establish grid-map of area  Identify coordinates and weights shipped for each location Center-of-Gravity Technique

28. Grid-Map Coordinates x i, y i =coordinates of existing facility i W i =annual weight shipped from facility i x1x1x1x1 x2x2x2x2 x3x3x3x3x y2y2y2y2y y1y1y1y1 y3y3y3y3 1 (x 1, y 1 ), W 1 2 (x 2, y 2 ), W 2 3 (x 3, y 3 ), W 3

29. Grid-Map Coordinates where, x, y =coordinates of new facility at center of gravity x i, y i =coordinates of existing facility i W i =annual weight shipped from facility i  n WiWiWiWi i = 1  xiWixiWixiWixiWi n x =  n WiWiWiWi i = 1  yiWiyiWiyiWiyiWi n y = x1x1x1x1 x2x2x2x2 x3x3x3x3x y2y2y2y2y y1y1y1y1 y3y3y3y3 1 (x 1, y 1 ), W 1 2 (x 2, y 2 ), W 2 3 (x 3, y 3 ), W 3

30. Example ABCD x200100250500 y200500600300 Wt7510513560

31. Center-of-Gravity Technique: Example ABCD x200100250500 y200500600300 Wt7510513560 y700 500 600 400 300 200 100 0 x 700500600400300200100 A B C D (135) (105) (75) (60) Miles Miles

32. Center-of-Gravity Technique: Example (cont.) x = = = 238 n  WiWi i = 1  xiWixiWi n  n WiWi  yiWiyiWi n y = = = 444 (200)(75) + (500)(105) + (600)(135) + (300)(60) 75 + 105 + 135 + 60 (200)(75) + (100)(105) + (250)(135) + (500)(60) 75 + 105 + 135 + 60

33. Center-of-Gravity Technique: Example (cont.) ABCD x200100250500 y200500600300 Wt7510513560 y 700 500 600 400 300 200 100 0 x 700500600400300200100 A B C D (135) (105) (75) (60) Miles Miles Center of gravity (238, 444)

34. Copyright 2006 John Wiley & Sons, Inc. Supplement 7-34 Center of Gravity with Excel

35. Example

36. © 2007 Pearson Education Finding the Center of Gravity for Health Watch

37. Example Existing FacilityAnnul loadCost of moving one unit Coordinate locations W27910(20,30) X47310(70,10) Y35010(50,40) z26610(10,80)

38. Load Distance Method

39. Euclidean or rectilinear distance measure may be used.

40. Distance Measures Rectilinear Distance d AB = |20 – 80| + |10 – 60| = 110 Euclidian Distance d AB = (20 – 80) 2 + (10 – 60) 2 = 78.1 = 78.1 What is the distance between (20,10) and (80,60)?

41. Load-Distance Technique Compute (Load x Distance) for each site Choose site with lowest (Load x Distance) Distance can be actual or straight-line

42. Load-Distance Calculations  l i d i i = 1 n LD = LD = load-distance value l i = load expressed as a weight, number of trips or units being shipped from proposed site and location i d i = distance between proposed site and location i d i = (x i - x) 2 + (y i - y) 2 (x,y) = coordinates of proposed site (x i, y i ) = coordinates of existing facility where, or Ix i - xI + Iy i - yI

43. Load-Distance: Example Potential Sites SiteXY 1360180 2420450 3250400 Suppliers ABCD X200100250500 Y200500600300 Wt7510513560

44. Load-Distance: Example Potential Sites SiteXY 1360180 2420450 3250400 Suppliers ABCD X200100250500 Y200500600300 Wt7510513560 Compute distance from each site to each supplier = (200-360) 2 + (200-180) 2 d A = (x A - x 1 ) 2 + (y A - y 1 ) 2 Site 1 = 161.2 = (100-360) 2 + (500-180) 2 d B = (x B - x 1 ) 2 + (y B - y 1 ) 2 = 412.3 d C = 434.2 d D = 184.4

45. Load-Distance: Example (cont.) Site 2 d A = 333 d C = 226.7 d B = 323.9 d D = 170 Site 3 d A = 206.2 d C = 200 d B = 180.4 d D = 269.3 Compute load-distance i = 1 n  l i d i LD = Site 1 = (75)(161.2) + (105)(412.3) + (135)(434.2) + (60)(434.4) = 125,063 Site 2 = (75)(333) + (105)(323.9) + (135)(226.7) + (60)(170) = 99,791 Site 3 = (75)(206.2) + (105)(180.3) + (135)(200) + (60)(269.3) = 77,555* * Choose site 3

46. Example

47. Example: Matrix Manufacturing is considering where to locate its warehouse in order to service its four Ohio stores located in Cleveland, Cincinnati, Columbus, Dayton. Two sites are being considered; Mansfield and Springfield, Ohio. Use the load-distance model to make the decision.

48. Break-Even Analysis Break-even analysis can help a manager compare location alternatives on the basis of quantitative factors that can be expressed in terms of total cost. 1.Determine the variable costs and fixed costs for each site. 2.Plot the total cost lines—the sum of variable and fixed costs—for all the sites on a single graph 3.Identify the approximate ranges for which each location has the lowest cost. 4.Solve algebraically for the break-even points over the relevant ranges.

49. Break-Even Analysis An operations manager has narrowed the search for a new facility location to four communities. The annual fixed costs (land, property taxes, insurance, equipment, and buildings) and the variable costs (labor, materials, transportation, and variable overhead) are shown below. Total costs are for 20,000 units. Fixed CostsVariable CostsTotal Costs Communityper Yearper Unit(Fixed + Variable) A$150,000$62$1,390,000 B$300,000$38$1,060,000 C$500,000$24$ 980,000 D$600,000$30$1,200,000

50. Example

51. The operations manager for Mile-High Beer has narrowed the search for a new facility location to seven communities. Annual fixed costs (land, property taxes, insurance, equipment, and buildings) and variable costs (labor, materials, transportation, and variable overhead) are shown below. Mile-High Beer

52. Example Santro Electronics is considering 2 locations for the audio equipment factory Ahmedabad & Chennai. At Ahmedabad fixed cost is estimated at Rs.1 million and the variable cost at Rs.1,200 per audio equipment. At Chennai fixed cost is Rs. 1.2 million and variable cost is Rs. 1100 per audio equipment. The selling price of the equipment will be Rs. 3000 per unit irrespective of the location. Decide which location is the best.

53. The Transportation Method The transportation method is a quantitative approach that can help solve multiple-facility location problems. The transportation method does not solve all facets of the multiple-facility location problem. It utilizes linear programming to minimize the cost of shipping products from two or more plants, or sources of supply, to two or more warehouses, or destinations.

54. The Transportation Method The Sunbelt Pool Company has a plant in Phoenix and three warehouses. It is considering building a new 500-unit plant because business is booming. One possible location is Atlanta. The cost to ship one unit from Atlanta to San Antonio. Initial Tableau

55. Example

56. Dimensional Analysis  Considers both tangible and intangible costs  Intangibles could include (lack of ) facilities e.g. for education, shopping, recreation, social life.  Intangibles could be quantified on a scale.  Weightages could be assigned to each ‘cost.’  A pair of sites is compared by a ratio.

57. Formula for Dimensional Analysis  If ‘C’ are costs, M & N are the two sites, and ‘w’ are weightages, the relative demerit of site M to N is: ( C 1 M / C 1 N ) X ( C 2 M / C 2 N ) X ….. X ( C z M / C z N ) where, C z M is the cost ‘z’ for site M.  If the above is >1, site N is superior. W1W1 W2W2 WzWz

58. Example .. Costs Site LaborPowerEducational Facilities for children's (Score) Recreational Facilities (Score) MRs.1.50.000Rs.40,00,00022 NRs.1.00.000Rs.25,0000064 Weightage1122

59. FACILITY LAYOUT PROBLEM

60. Once a firm has decided where a facility will be located, the next important decision is the Arrangement of people and Equipment within the facility.

61. FACILITY LAYOUT PROBLEM Facility Layout problem involves the location of departments (or sections) within the facility AND the arrangement of people and equipment within each department..

62. FACILITY LAYOUT PROBLEM The layout decision will certainly affect the Flow of materials In-plant Transportation cost Equipment utilization General productivity and effectiveness of the business.

63. FACILITY LAYOUT PROBLEM Usually the layout is planned to minimize a particular criterion: Minimizing total traveling time, total cost, total delays, etc. There are also situations in which the layout may be designed to maximize a criterion: Maximize quality, flexibility, or space utilization.

64. Costs associated with a plant layout  Costs of customer dissatisfaction due to poor service (delivery, responsiveness, quality, flexibility)  Costs of movement of materials  Costs of space  Costs of spoilage of materials  Costs of employee dissatisfaction  Costs of changes required with operational changes

65. Basic Production Layout Formats Process Layout (also called job-shop or functional layout) Product Layout (also called flow-shop layout) Group Technology (Cellular) Layout Fixed-Position Layout

66. Process Layout Similar pieces of equipment that perform similar functions are grouped together. For example; all drill machines are grouped and placed together.

67. Process Layout An example LLLL MM MM DD DD DD GGG LLLL Product A Product C Product B

68. Product Layout The pieces of equipment required to make a Particular product are grouped together, as in an Automobile assembly line.

69. Product Layout L LDL DMG Product A Product B Product C Step 1 Step 2 Step 3 Step 4 7-14 G LDMLG Step 5

70. Fixed Layout The equipment is brought to the object being processed, and the object does not move. Example; house construction.

71. Cellular Manufacturing (CM) Layouts Cellular manufacturing is a type of layout in which machines are grouped into what is referred to as a cell. Groupings are determined by the operations needed to perform work for a set of similar items, or part families that require similar processing.

72. Process Layout Example Frontec Company wants to arrange Four of its departments in a Row so that the Total Distance Traveled between Departments is minimized. This part of the building will contain four departments arranged in a row. Frontec wishes to minimize the total daily inter- departmental distance traveled. The number of daily communications between each pair of department is shown below:

73. Example Assume that adjacent departments are 20 feet apart.

74. Example We will use a trial-and-error approach to this problem. Assume that we selected the following configuration for the departments: A-B-C-D. For this configuration, Total communication cost (based on distance) is as follows:

75. Example

77. In terms of total daily communication distance, (B-A-C-D) is the preferred alternative. But the firm has to consider all of the 24 (4! = 4x3x2x1) possible configurations before it knows if this is the optimal configuration.

78. Example This trial-and-error approach becomes time- consuming as the number of departments increases AND It also becomes complex when the cost of communications vary between departments.

79. Product Layout

80. LINE BALANCING Essentially,the layout design seeks to identify minimum number of resources required to meet a targeted production rate and the order in which this sequences are to be used. In the process it seeks to establish a balance among the resources Line Balancing is a method by which tasks are optimally combined without violating the precedence constraint and a certain number of workstation is designed to complete the task.

81. Station 1 Minutes per Unit 6 Station 2 7 Station 3 3 Assembly Lines Balancing Concepts Question: Suppose you load work into the three work stations below such that each will take the corresponding number of minutes as shown. What is the cycle time of this line? Determine the interval between producton of two successive components?

82. Station 1 Minutes per Unit 6 Station 2 7 Station 3 3 Assembly Lines Balancing Concepts Question: Suppose you load work into the three work stations below such that each will take the corresponding number of minutes as shown. What is the cycle time of this line? Determine the interval between producton of two successive components? Answer: One component will come out of the system only every 7 minutes. This measure is known as cycle time.

83. Assembly Line Balancing 1.Determine cycle time: Cycle time could be actual or desires 1.Determine required workstations (theoretical minimum)

84. Assembly Line Balancing 5.Evaluate line efficiency:

85. Balance delay (%) is the amount by which the line falls short of 100%

86. Example A factory working two shifts each of eight hour produces 24000 electric bulbs using a set of workstations. Compute the actual cycle time of the plant operation. There are 8 tasks required to manufacture the bulbs. The sum of all task time is equal to 12 seconds. How many workstations are required to maintain the level of production.

87. Example of Line Balancing You’ve just been assigned the job a setting up an electric fan assembly line with the following tasks:

88. Example of Line Balancing: Structuring the Precedence Diagram ABG H CDEF Task Predecessors ANone BABA CNone DA, C Task Predecessors EDED FEFE GBGB HF, G

89. Example of Line Balancing: Precedence Diagram Question: Which process step defines the maximum rate of production? A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 Answer: Task C is the cycle time of the line and therefore, the maximum rate of production.

90. Example of Line Balancing: The Bottleneck

91. Example of Line Balancing : Determine Cycle Time Question: Suppose we want to assemble 100 fans per day. What would our cycle time have to be?

92. Example of Line Balancing : Determine Cycle Time Question: Suppose we want to assemble 100 fans per day. What would our cycle time have to be? Answer:

93. Example of Line Balancing: Determine Theoretical Minimum Number of Workstations Question: What is the theoretical minimum number of workstations for this problem? Answer:

94. Example of Line Balancing: Determine Theoretical Minimum Number of Workstations Question: What is the theoretical minimum number of workstations for this problem?

95. A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 Station 1Station 2Station 3 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4

96. A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 Station 1Station 2Station 3 A (4.2-2=2.2) TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4

97. A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1=.2) Idle=.2 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 Station 1Station 2Station 3

98. A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1=.2) Idle=.2 Station 1Station 2Station 3

99. C (4.2-3.25)=.95 Idle =.95 A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1=.2) Idle=.2 Station 1Station 2Station 3

100. A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 Idle =.95 D (4.2-1.2)=3 E (3-.5)=2.5 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1=.2) Idle=.2 Station 1Station 2Station 3

101. A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 Idle =.95 D (4.2-1.2)=3 E (3-.5)=2.5 F (2.5-1)=1.5 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1=.2) Idle=.2 Station 1Station 2Station 3

102. A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 Idle =.95 D (4.2-1.2)=3 E (3-.5)=2.5 F (2.5-1)=1.5 H (1.5-1.4)=.1 Idle =.1 TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1=.2) Idle=.2 Station 1Station 2Station 3

103. A C B DEF G H 2 3.25 1 1.2.5 1 1.4 1 C (4.2-3.25)=.95 Idle =.95 Efficiency=77% D (4.2-1.2)=3 E (3-.5)=2.5 F (2.5-1)=1.5 H (1.5-1.4)=.1 Idle =.1 Efficiency=98% TaskFollowersTime (Mins) A62 C43.25 D31.2 B2 1 E20.5 F11 G11 H01.4 A (4.2-2=2.2) B (2.2-1=1.2) G (1.2-1=.2) Idle=.2 Efficiency=95% Station 1Station 2Station 3

104. Example of Line Balancing: Determine the Efficiency of the Assembly Line

105. Step 1: Identify Tasks & Immediate Predecessors

107. Layout Calculations Step 2: Determine output rate – Vicki needs to produce 60 pizzas per hour Step 3: Determine cycle time – The amount of time each workstation is allowed to complete its tasks – Limited by the bottleneck task (the longest task in a process):

108. Layout Calculations (continued) Step 4: Compute the theoretical minimum number of stations – TM = number of stations needed to achieve 100% efficiency (every second is used) – Always round up (no partial workstations) – Serves as a lower bound for our analysis

109. Last Layout Calculation Step 6: Compute efficiency and balance delay – Efficiency (%) is the ratio of total productive time divided by total time – Balance delay (%) is the amount by which the line falls short of 100%

110. Problem – Draw precedence diagram – Determine cycle time—demand = 50 units/hr – Theoretical minimum no. of work stations – Assign tasks to workstations using cycle time – Efficiency and balance delay of line? – Bottleneck? – Maximum output? TaskImm. predecessorTask time (sec) ANone55 BA30 CA22 DB35 EB, C50 FC15 GF5 HG10 TOTAL222

111. Group Technology: Transition from Process Layout 1. Grouping parts into families that follow a common sequence of steps 2. Identifying dominant flow patterns of parts families as a basis for location or relocation of processes 3. Physically grouping machines and processes into cells

112. Product Layout An example LD L MD G L L M D LD L MD G L L D G M

113. Machine – Component Incident Matrix(MCIM) Before Grouping Machines Components

114. Machine – Component Incident Matrix After Grouping Machines Components

115. Rank Order Clustering(ROC) Machines Components

116. Rank Order Clustering(ROC) 1.Read each row of the MCIM as binary number. Rank the rows in descending order. 2.If there is no change stop. Otherwise go to next step. 3.Rearrange the rows based on ranking. 4.Read each column of the MCIM as binary number. Rank the rows in descending order. 5.If there is no change stop. Otherwise go to next step. 6.Rearrange the rows based on ranking. Go to step 1.

117. Rank Order Clustering(ROC) Value of the binary numberRank ROW1105 ROW2254 ROW3362 ROW4263 ROW5371 Rows as binary numbers

118. Rank Order Clustering(ROC) Machines Components

119. Rank Order Clustering(ROC) Value of the binary numberRank Column1241 Column2125 Column3134 Column4241 Column556 Column6183 Columns as binary numbers

120. Rank Order Clustering(ROC) Machines Components

121. Example) Machines Components 12345678910 A11111 B111 C11111 D111 E111 F11 G111 1.Use ROC to rank families and machine groups 2.What will happen if we did column sorting first and then row?