types of facility layout algorithm
- 1. Types of layout Algorithms Construction algorithms Improvement algorithms Graph-based method ALDEP CORELAP PLANET Pairwise exchange method CRAFT MCCRAFT MULTIPLE 1
- 2. AUTOMATED LAYOUT DESIGN PROGRAM (ALDEP) • ALDEP is basically a construction algorithm, but it can also be used to evaluate two layouts • It uses basic data on facilities & builds a layout by successively placing the layout using relationship information between the departments • Data requirements • Area of Departments • REL Chart with A,E,I,O,U,X rankings • Sweep width • Minimum Department Preferences (MDP) • Ranking values: • A = 64, Absolutely necessary • E = 16, Especially important • I = 4, Important • O = 1, Ordinarily important • U = 0, Unimportant • X = -1024, Undesirable 2
- 3. ALDEP Facility Representation Facility Length FacilityWidth Discrete Representation Sweep Method Facility Length FacilityWidth
- 4. ALDEP Facility Representation A A A A A A A A Facility Length FacilityWidth Dept. size = 8 squares Facility width = 6 squares Sweep width = 1 square C C C C C C C C C C C C C C Facility Length FacilityWidth Dept. size = 14 squares Facility width = 6 squares Sweep width = 2 squares
- 5. Example Dept. Area(sq.m) 1 12,000 2 8000 3 6000 4 12,000 5 8000 6 12,000 7 12,000 A Timber Processing facility need to be located in a plot who’s Length is 360 meters Width is 200 meters. Create a layout using ALDEP to get a optimal facility Layout with sweep value is 2 & MDP=> E(16)
- 6. Example Dept. Area 1 12,000 2 8000 3 6000 4 12,000 5 8000 6 12,000 7 12,000 From the above table we try to scale it down by considering common area value of 400 sq.m. We get number of squares. Departments 1,4,6,7 have same number of squares. Select any one Randomly Number of Squares 30 20 15 30 20 30 30
- 8. Dept. Area 1 12,000 2 8000 3 6000 4 12,000 5 8000 6 12,000 7 12,000 Number of Squares 30 20 15 30 20 30 30 175 Facility Length=18 FacilityWidth=10 • One square is 400 sq.m • Length of square =20m • Width of square= 20m • Length of plot is 360m • Width of plot is 200m Length of layout matrix = 360/20=18 widthoflayoutmatrix 200/20=10
- 9. • We randomly select Dept. 4 as first dept. in the layout matrix. • If a dept. is said to be fixed then start with that dept. or office. • Now sweep Value is 2. So fill the layout matrix accordingly. • Now compare Dept. 4 with our dept. to find which is nearer. • Dept. 4 ->1,2,3,5,6,7 • Dept. Rank ->I,E,U,I,U,U So E (Dept. 2) is selected
- 10. • Now dept. 2 is entered then its now compared with other Depts. • To find which is nearer to the Dept. 2 • Now compare Dept. 2 with other Depts. • Dept. 2 ->1,3,5,6,7 • Dept. Rank->E,U,I,I,U So E (Dept. 1) is selected
- 11. The sequence of selection is 4-2-1-6-5-7-3
- 12. Scoring the layout Neighbouring Departments Rel Grade Rel Value 4-2 E 16 4-1 I 4 2-1 E 16 1-6 U 0 6-5 A 64 6-7 E 16 5-7 I 4 7-3 U 0 120 Total scoring = 120*2=240 Since both the ways we can have neighbour pairs
- 13. In class Exercise Dept. Area(sq.m) 1 12,000 2 8000 3 6000 4 12,000 5 8000 6 12,000 7 12,000 A Timber Processing facility need to be located in a plot who’s Length is 360 meters Width is 200 meters. Create a layout using ALDEP with Dept. 2 is fixed to get a optimal facility Layout with sweep value is 2 & MDP=> I(4)
- 14. In class Exercise-Solution Steps Depts. Area(sq.m) Total scoring = 104*2=208 Since both the ways we can have neighbour pairs 1 2 Fixed 3 4 “I” with 1 2 1 “E” with 2 4 5 “I” with 1 5 6 “A” with 5 6 7 “E” with 6 7 3 Remaining
- 15. CORELAP
- 16. CORELAP • Computerized relationship layout planning. • Computer algorithm was Developed by R.C.Lee. • Scale of layout is limited to a maximum dimension 40 X 40. • Input requirements of CORELAP: • no. of departments and their area • closeness relationship as given by REL chart. • weighted ratings for REL chart entries. • Optional input formation : • scale of output • building length to width ratio • department preassigned.
- 17. Problem: • Create a layout using CORELAP.
- 18. Step 1: Area of Department and Scale Number of Squares is calculated using scale information If input for scale is not given corelap assumes 1 Squares = 6000 Sq. ft. For the first run usually scale is to be assumed by corelap. For improvement one can specify the scale suitably. Number of squares for each department is fixed by dividing the area by scale value and result is kept as integer. Care Should be taken to see that the total number of squares multiplied by the scale value equals approximately the total area.
- 19. Contd…
- 20. Step 2: Determination of Placement order of departments For the purpose of finding the order of placing departments in the layout the following data is needed. 1. REL Chart 2. Closeness values for entries in the REL chart Closeness values: A- 6 E- 5 I- 4 O- 3 U- 2 X- 1 S – Refuse to diagonal
- 21. Contd… REL CHART: 1 S 2 E U 3 O U S 4 I E U S 5 O I U I S 6 U I O U A S 7 U U U U I E S 1 2 3 4 5 6 7
- 22. Contd… • Using the REL Chart information total closeness rating is computed for all the departments Total closeness rating (TCR) is the sum of the numerical values assigned to the closeness relationship between a department and all other department.
- 23. Contd…
- 24. Contd… • The following steps are carried out to select the placement order of departments • Step 1: Department with largest TCR is selected for first placement. In case of ties the department with the largest area is selected. For the example, department 5 is selected. Dept 5 has the largest TCR value 23
- 25. Contd… Step 2: Review the closeness relationship of dept.5 with other departments to find the department that has high closeness relationship with dept5. closeness of dept. 5 with : 1, 2, 3, 4, 6, 7 O, I, U, I, A, I Dept 6 has the highest closeness relationship with the first selected dept 5. Dept 6 is selected as second in placement order.
- 26. Contd… • Step 3: Scan the REL chart for the closeness values for the unsigned departments with department 5. closeness of dept. 5 with : 1, 2, 3, 4, 6, 7 O, I, U, I, A, I None of the unassigned has an “A” closeness with dept 5 Check the closeness of unassigned department with dept 6 (next in the placement order) closeness of dept. 6 with :1, 2, 3, 4, 7 U, I, O, U, E None of the unassigned has an “A” closeness with dept 6 Check whether any of the unassigned dept has an “E” relationship with the departments already included in the placement order. None of the unassigned has “E” closeness with dept 5. Dept 7 has an “E” closeness with the dept 6. Dept 7 is included third in the order.
- 27. Contd… Step 4: Repeat the procedure of step 3 with the unassigned departments. 1, 2, 3, 4 With Dept 5 : O, I, U, I Dept 6: U, I, O, U Dept 7: U, U, U, U Dept 2 and 4 have an “I” closeness with dept 5. In the case of ties, select based on highest TCR. Dept 2 (TCR=22) is selected and included forth in the placement order Repeat this until all departments are included. Placement order: 5,6,7,2,1,4,3
- 28. Step 3: Placement of department in the layout 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Step 1: the first department in the placement order (dept 5) is placed at the centre of the (40 X 40) layout.
- 29. Contd… • Step 2: the second department from the list is placed adjacent to the first department. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6 6 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
- 30. Contd… • Step 3: Department 7 is to be placed next in the layout. It can be seen that there are many arrangements in which dept 7 can be placed as an adjacent department to the layout. • To aid evaluation of different placing alternatives placing Rating (PR) is calculated. • Placing rating is the sum of the weighted closeness ratings between the entering department and its new neighbours.
- 31. • CORELAP uses the following weighted closeness rating values:
- 32. Contd… • Dept 7 has 2 squares. The following placements are possible 0 0 0 0 0 0 0 0 0 0 0 7 7 0 0 0 0 0 0 6 6 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Dept7with6=E=81 PR=81
- 33. • Option 2: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6 6 7 0 0 0 0 0 0 5 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Dept7with6=E=81 with5=I=27 PR=108
- 34. • Option 3: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6 6 0 0 0 0 0 7 7 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Dept7with6=E=81 with5=I=27 PR=108
- 35. Contd… • Select the Option 3 ( random between 2 and 3 ). Department 7 is placed as shown in figure below. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6 6 0 0 0 0 0 7 7 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
- 36. Contd… Step 4: Now Dept 2 is placed. Dept 2 has 1 square. the best location is to place dept 2 adjacent to two depts (dept 6 and dept 7) so that the PR can be maximized (PR =28). All other arrangement will have PR less than 28. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 6 6 0 0 0 0 0 7 7 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Repeat this procedure until all the departments are selected
- 37. Final Arrangement 0 0 0 0 0 0 0 0 0 4 1 1 3 0 0 0 0 4 2 6 6 0 0 0 0 0 7 7 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
- 38. Step 4: Total Score for the layout Total score for a layout is the sum of the products found by multiplying length of the shortest path between all pairs of department times the numerical closeness for the pair Total score = Summation over all pairs (closeness rating X length of the shortest path) Path is calculated by the rectilinear distance from the border of one department to the border of another department. Therefore if 2 departments are adjacent they will have common border. The path will be zero