Layout Planning Procedures Layout Procedures Algorithmic Approach: Relative placement of departments on the basis of their “closeness ratings” or “material flow intensities” • Two main types of algorithms: – Construction-type: new layout “from scratch” – Improvement-type: improve existing layout Four Fundamental Elements of Layout Procedures • Space planning units (i.e., workstations) • Space • Affinities (closeness) • Constraints The layout is a result of arranging the space planning units on the available space with the goal of accommodating the closeness relationships (affinities) without violating any of the relevant constraints. Conceptual Illustration of Layout Planning Space Planning Units Block Layout Affinity Diagram Affinities Space Requirements Constraints (Activity) Relationship Chart Closeness Relationships A = absolutely necessary E = especially important I = important O = ordinary closeness acceptable U = unimportant X = not desirable to be close 1 Receiving (1) Milling (2) 2 E U Press (3) Screw Mch. (4) U O E 6 7 4 5 7 U 1 U 2 U U I 6 U I O U E Shipping (7) 5 O I A Plating (6) 4 I U I Assembly (5) 3 3 Activity Relationships • Multiple relationships influence the layout – Material flow – Personnel flow – Equipment flow – Information flow • Guideline on number of relationships of different ratings: – <5% (A), – <5% (X), – <12% (A or E), – <25% (A or E or I), – <40% (A or E or I or O), – >50% (U) Relationship (Activity) Diagram Purpose: Depict spatially the relationships of the activities Constructing an (Activity) Relationship Diagram • Complicated by the lack of a binary ruling (yes/no) on whether or not a rating is satisfied – If a 25m separation satisfies a closeness rating, does a 50m separation also satisfy the closeness rating? – Judgment and discretion are required • Traditionally, manual “cut and try” process • Try to place A’s and E’s first to minimize length of thick relationship lines. • Try to keep lines of same value the same length • Try to avoid crossing lines, usually indicates change is possible • Make X’s longest lines • Not every relationship (U) needs to be shown • Generate multiple alternatives An Improved Relationship Diagram Space Relationship Diagram Space Relationship Diagram • Combine space requirements with relationship information to develop a space-relationship diagram, • Expand sizes of relationship chart nodes to reflect the approximate shape and size of areas • Generate multiple alternatives • The same rules apply as in the development of the original relationship diagram. Important Note • This is not a precise method. Rather, it is an approach to help find a good layout. Any two people applying the approach, working independently, will probably not obtain the same layout. This problem is a design problem – not an analysis problem. Block Layout (Block Plan) Alternatives Logistics System • Flow of materials into a manufacturing facility – Materials management system • Flow of materials, parts, supplies within a manufacturing facility – Material flow system • Flow of products from a manufacturing facility – Physical distribution system What Flows in a Facility? • Materials – Raw material – Work in process – Scrap – Finished goods – Paperwork – Trash • Equipment • Personnel • Information Flow Planning Flow Principles Aisle Widths Flow Planning 1. Maximize direct flow paths (no intersections, no backtracking) 2. Minimize flow (work simplification) 3. Minimize flow costs. FACTORS AFFECTING AISLE WIDTHS • Use of the aisle: material, personnel, handling equipment, machinery, and other equipment • Frequency of use: volume of traffic (at peak loads) • Speed of travel permitted or desired • One-way traffic or both • Possible future conditions of these points SUGGESTED AISLE WIDTHS 1. 2. 3. 4. 5. 6. 7. 8. For personnel only (2 persons to pass) For two-wheel hand truck (no passing or turning with load) For stock truck (where trucker must pass around it) For stock truck (where other trucks and workers must pass) For hand-operated fork truck, pallet transporter, semilive skid and jack For 2,000-pound fork truck For 4,000-pound fork truck For 6,000-pound fork truck 30" minimum 30" minimum 20" plus width of truck 38" plus 2 times truck width 5 to 8' depending on load size 8' to 10' 10' to 12' 12' to 14' Measuring Flow • Flow among departments is one of the most important factors in the arrangement of departments within a facility. • Evaluating alternative arrangements require both Quantitative and Qualitative measures • Quantitative Flow Measurement From-To-Chart (large volumes moved) • Qualitative Flow Measurement Relationship Chart (low volumes but intensive communication & organizational relationships) From-To-Chart Product A: 1-2-3-4 Product B: 3-4-2-1-2 1 Production volume = 10 Production volume = 20 2 1 2 3 4 3 4 1 2 3 4 - 10+20 = 30 - - 20 - 10 - - - - 10+20 = 30 - 20 - - Assumption Component B is twice as large as Component A & moving 2 units of A is equivalent to moving 1 unit of B 1 2 3 4 1 2 3 4 1 2 3 4 - 10/20 - - - 10+40 = 50 - - 20 - 10 - 40 - 10 - - - - 10/20 - - - 10+40 = 50 - 20 - - - 40 - - 1 2 3 4 Using Component A as the reference product Converting Flows to Closeness Ratings This is not unique! Many different scales could be constructed. Developing a Relationship Diagram From Relationship Diagram to Layout Plans • Relative placement of departments on the basis of their “closeness ratings” or “material flow intensities” – can be reduced to an algorithmic process • Reflects two basic steps of many heuristic design procedures: construction followed by improvement An Intuitive Construction Method for Generating the Activity Relationship Diagram • Two issues must be addressed: – The order of placement of departments, and – Their relative locations Example Total Closeness Rating – TCR Total closeness rating for department i = The sum of the values for the relationships with department i A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Order of Placement Approach 1 (closeness rating): • First department (D1) à Highest TCR – If tie exists, pick one with highest number of A relationships • If a department has an X relationship with D1, it becomes the last placed department – if more than one department has an X relationship with D1, the one with smallest TCR is the last placed department • Second department (D2) à The department having an A relationship with D1 and having greatest TCR • If a department has an X relationship with D2, it becomes – the last placed department if no department had an X relationship with D1 – the next-to-last placed department if a department had an X relationship with D1 • Third department (D3) à The one having greatest TCR among those having an A relationship with one of the placed departments – If no A relationship, then E relationship, etc. • So on… Order of Placement • 10 – 5 – 8 – 9 – 7 – 6 – 2 – 3 – 1 – 4 – – – – – – – – – 10 has the highest TCR 5 has a higher TCR than 8 8 has an A relation with 10 9 due to its E relationship with 10 7 due to its E relationship with 9 6 due to its E relationship with 7 2 due to its I relationships and TCR 3 due to its I relationship with 2 1 due to its E relationship with 3 Order of Placement Another Approach: • Descending order of TCRs 10 – 5 – 8 – 9 – 7 – 6 – 3 – 1 – 2 – 4 Order of Placement (revised TRC method) Yet Another Approach: • Instead of looking for A relationships (if not then E, etc.) with departments already placed, we can look at sum of relationship values of an unplaced department with the placed departments. • 10 – 5 – 8 …..? Order of Placement 1 2 3 4 6 7 9 10 5 8 Total A = 10000 U U U 0 E = 1000 I I O 200 U U U 0 I U U 100 U U U 0 U U U 0 E U U 1000 I = 100 O = 10 U=0 X = - 10000 Order of Placement 1 2 3 4 6 7 10 5 8 9 Total U U U U 0 I I O I 300 U U U U 0 I U U U 100 U U U I 100 U U U E 1000 A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Order of Placement 1 2 3 4 6 10 5 8 9 7 Total U U U U U 0 I I O I I 400 U U U U U 0 I U U U U 100 U U U I E 1000 A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Order of Placement 1 2 3 4 10 5 8 9 7 6 Total U U U U U U 0 I I O I I I 500 U U U U U U 0 I U U U U O 110 A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Order of Placement 10 5 8 9 7 6 2 Total 1 U U U U U U O 10 3 U U U U U U I 100 4 I U U U U O O 120 A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Order of Placement 10 5 8 9 7 6 2 4 Total 1 U U U U U U O O 20 3 U U U U U U I U 100 A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Then, order of placement is 10-5-8-9-7-6-2-4-3-1 Class Exercise Calculate the TCR scores find the placement order for the following closeness rating. Dep 1 2 3 4 5 6 1 -- O A E I U 2 O -- U E O O 3 A U -- I U A 4 E E I -- U O 5 I O U U -- X 6 U O A O X -- A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Exercise Dep 1 2 3 4 5 6 A E I O U X TCR 1 -- O A E I U 1 1 1 1 1 0 11110 2 O -- U E O O 0 1 0 3 1 0 1030 3 A U -- I U A 2 0 1 0 2 0 20100 4 E E I -- U O 0 2 1 1 1 0 2110 5 I O U U -- X 0 0 1 0 2 1 -9890 6 U O A O X -- 1 0 0 2 1 1 20 A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Exercise Dep 1 2 3 4 5 6 A E I O U X TCR 1 -- O A E I U 1 1 1 1 1 0 11110 2 O -- U E O O 0 1 0 3 1 0 1030 3 A U -- I U A 2 0 1 0 2 0 20100 4 E E I -- U O 0 2 1 1 1 0 2110 5 I O U U -- X 0 0 1 0 2 1 -9890 6 U O A O X -- 1 0 0 2 1 1 20 • 3 – 3 has the highest TCR Exercise Dep 1 2 3 4 5 6 A E I O U X TCR 1 -- O A E I U 1 1 1 1 1 0 11110 2 O -- U E O O 0 1 0 3 1 0 1030 3 A U -- I U A 2 0 1 0 2 0 20100 4 E E I -- U O 0 2 1 1 1 0 2110 5 I O U U -- X 0 0 1 0 2 1 -9890 6 U O A O X -- 1 0 0 2 1 1 20 • 3–1 – 3 has the highest TCR – 1 breaking the tie with 6 (higher TCR) Exercise Dep 1 2 3 4 5 6 A E I O U X TCR 1 -- O A E I U 1 1 1 1 1 0 11110 2 O -- U E O O 0 1 0 3 1 0 1030 3 A U -- I U A 2 0 1 0 2 0 20100 4 E E I -- U O 0 2 1 1 1 0 2110 5 I O U U -- X 0 0 1 0 2 1 -9890 6 U O A O X -- 1 0 0 2 1 1 20 • 3–1–6 -5 – 3 has the highest TCR – 1 breaking the tie with 6 (higher TCR) – 6 has a A relation with 3 and 5 has X relation with 6 Exercise Dep 1 2 3 4 5 6 A E I O U X TCR 1 -- O A E I U 1 1 1 1 1 0 11110 2 O -- U E O O 0 1 0 3 1 0 1030 3 A U -- I U A 2 0 1 0 2 0 20100 4 E E I -- U O 0 2 1 1 1 0 2110 5 I O U U -- X 0 0 1 0 2 1 -9890 6 U O A O X -- 1 0 0 2 1 1 20 • 3–1–6–4 -5 – 3 has the highest TCR – 1 breaking the tie with 6 (higher TCR) – 6 has a A relation with 3 – 4 due to its E relationship with 1 Exercise Dep 1 2 3 4 5 6 A E I O U X TCR 1 -- O A E I U 1 1 1 1 1 0 11110 2 O -- U E O O 0 1 0 3 1 0 1030 3 A U -- I U A 2 0 1 0 2 0 20100 4 E E I -- U O 0 2 1 1 1 0 2110 5 I O U U -- X 0 0 1 0 2 1 -9890 6 U O A O X -- 1 0 0 2 1 1 20 • 3–1–6–4–2–5 – 3 has the highest TCR – 1 breaking the tie with 6 (higher TCR) – 6 has a A relation with 3 – 4 due to its E relationship with 1 – 2 (the last item to place) Exercise Dep 1 2 3 4 5 6 A E I O U X TCR 1 -- O A E I U 1 1 1 1 1 0 11110 2 O -- U E O O 0 1 0 3 1 0 1030 3 A U -- I U A 2 0 1 0 2 0 20100 4 E E I -- U O 0 2 1 1 1 0 2110 5 I O U U -- X 0 0 1 0 2 1 9900 6 U O A O X -- 1 0 0 2 1 1 20 Decreasing TCR gives the ordering: • 3–1 –4–2–6 –5 Relative locations No space consideration à All departments have unit size Weighted placement value (WP) = sum of numerical values of all pairs of adjacent departments Order: 10 – 5 – 8 – 9 – 7 – 6 – 2 – 3 – 1 – 4 Locations 1, 3, 5, 7 are “fully adjacent” è full weight Locations 2, 4, 6, 8 are “partially adjacent” è half weight Relative locations Order: 10 – 5 – 8 – 9 – 7 – 6 – 2 – 3 – 1 – 4 Department 5 will be placed next (A relationship with 10) WP1 = WP3 = WP5 = WP7 = 10000 WP2 = WP4 = WP6 = WP8 = 5000 Relative locations Order: 10 – 5 – 8 – 9 – 7 – 6 – 2 – 3 – 1 – 4 Department 8 will be placed next (A relationship with 10, U relationship with 5) WP1 = WP2 = WP10 = 0, WP3 = WP5 = WP7 = WP9 = 5000, WP4 = WP6 = WP8 = 10000 Relative locations Order: 10 – 5 – 8 – 9 – 7 – 6 – 2 – 3 – 1 – 4 Relative locations Order: 10 – 5 – 8 – 9 – 7 – 6 – 2 – 3 – 1 – 4 Relative locations Order: 10 – 5 – 8 – 9 – 7 – 6 – 2 – 3 – 1 – 4 Relative locations Order: 10 – 5 – 8 – 9 – 7 – 6 – 2 – 3 – 1 – 4 Relative locations Order: 10 – 5 – 8 – 9 – 7 – 6 – 2 – 3 – 1 – 4 Relative locations Order: 10 – 5 – 8 – 9 – 7 – 6 – 2 – 3 – 1 – 4 Relative locations Order: 10 – 5 – 8 – 9 – 7 – 6 – 2 – 3 – 1 – 4 Activity Relationship Diagram Exercise Find the relative TCR ordering and build the activity relationship diagram on a 2x3 grid. Dep 1 2 3 4 5 6 1 -- O A E I U 2 O -- U E O O 3 A U -- I U A 4 E E I -- U O 5 I O U U -- X 6 U O A O X -- A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Exercise (revised TCR) Dep 1 2 3 4 5 6 A E I O U X TCR 1 -- O A E I U 1 1 1 1 1 0 11110 2 O -- U E O O 0 1 0 3 1 0 1030 3 A U -- I U A 2 0 1 0 2 0 20100 4 E E I -- U O 0 2 1 1 1 0 2110 5 I O U U -- X 0 0 1 0 2 1 -9900 6 U O A O X -- 1 0 0 2 1 1 20 Revised TRC starts with 3 and then 1 (due to A relation with 3) • 3–1… Exercise (revised TCR) Dep 1 2 3 4 5 6 3 1 TCR 2 O -- U E O O U O 10 4 E E I -- U O I E 1100 5 I O U U -- X U I 100 6 U O A O X -- A U 10000 A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Next comes 6 with the highest r-TCR • 3–1–6 Exercise (revised TCR) Dep 1 2 3 4 5 6 3 1 6 TCR 2 O -- U E O O U O O 20 4 E E I -- U O I E O 1110 5 I O U U -- X U I X -9900 A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Next comes 4 with the highest r-TCR • 3 – 1 –6 – 4 Exercise (revised TCR) Dep 1 2 3 4 5 6 3 1 6 4 TCR 2 O -- U E O O U O O E 1020 5 I O U U -- X U I X U -9900 A = 10000, E = 1000, I = 100, O = 10, U = 0, X = - 10000 Next comes 2 with the highest r-TCR • 3 – 1 –6 – 4 – 2 - 5 Exercise: Activity Relationship Diagram • 3 – 1 – 6 – 4 – 2 – 5 is the final ordering. • Use a 2x3 grid 1 2 3 4 5 6 1 -- O A E I U 2 O -- U E O O 3 A U -- I U A 4 E E I -- U O 5 I O U U -- X 6 U O A O X -- 3 Activity Relationship Diagram • 3 – 1 – 6 – 4 – 2 – 5 is the final ordering. 1 2 3 4 5 6 1 -- O A E I U 2 O -- U E O O 3 A U -- I U A 4 E E I -- U O 5 I O U U -- X 6 U O A O X -- 3 1 Activity Relationship Diagram • 3 – 1 – 6 – 4 – 2 – 5 is the final ordering. 1 2 3 4 5 6 1 -- O A E I U 2 O -- U E O O 3 A U -- I U A 4 E E I -- U O 5 I O U U -- X 6 U O A O X -- 6 3 1 Activity Relationship Diagram • 3 – 1 – 6 – 4 – 2 – 5 is the final ordering. 1 2 3 4 5 6 1 -- O A E I U 2 O -- U E O O 3 A U -- I U A 4 E E I -- U O 5 I O U U -- X 6 U O A O X -- 6 4 3 1 Activity Relationship Diagram • 3 – 1 – 6 – 4 – 2 – 5 is the final ordering. 1 2 3 4 5 6 1 -- O A E I U 2 O -- U E O O 3 A U -- I U A 4 E E I -- U O 5 I O U U -- X 6 U O A O X -- 6 4 3 1 2 Activity Relationship Diagram • 3 – 1 – 6 – 4 – 2 – 5 is the final ordering. 1 2 3 4 5 6 1 -- O A E I U 2 O -- U E O O 3 A U -- I U A 4 E E I -- U O 5 I O U U -- X 6 U O A O X -- 6 4 2 3 1 5
