Rule Based Layout Planning and Its Multiple Objectives

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1 Rule ased Layout Planning and Its Multiple Objectives Mohammad Komaki, Shaya Sheikh, ehnam Malakooti Case Western Reserve University Systems Engineering Abstract In this paper, we present a novel constructive approach for solving facility layout problems considering objectives of total flow and qualitative closeness. This approach, called Head-Tail, utilizes a set of rules to solve single and multiple row layout problems. We apply pair-wise exchange method to improve the partially constructed layouts. The computational efficiency of Head-Tail approach is examined by comparing it with Systematic Layout Planning. The results show that Head-Tail method can find efficient solution with promising quality. A bi-criteria numerical example that minimizes total flow and total qualitative closeness is also presented. Key words: Rule-based layout planning, multi-objective optimization, Head-Tail Method. Introduction Effective layout designs are necessary in many different businesses such as manufacturing facilities, and service industries. iminishing physical space availability, high cost of re-layouts, safety issues, and substantial long-term investments are good reasons for getting the design right the first time. ifferent criteria can be considered in facility layout design including operational costs, installation costs, utilization of space and equipment, work-force management, materials handling, inventory work in process, inventory of incoming materials and inventory of outgoing final products, ease of future expansion, and safety factors. Most of these objectives are difficult to quantify. In this paper, new methods are introduced for solving multiple objective layout problems. Researchers have used many criteria to come up with the optimal layout design. Raman et al. [] developed three factors of flexibility, productive area utilization and closeness gap to measure the layout effectiveness. Rosenblatt [] discussed a facility layout problem with two criteria of minimizing costs and maximizing closeness rating. ortenberry and Cox [] discussed a multicriteria facility layout problem that combines quantitative with qualitative criteria and minimizes total weighted work-flow. Malakooti [,,6] presented heuristic methods for solving multiobjective facility problems.. Malakooti and Tsurushima [] proposed an expert system for solving multi-crtieria manufacturing problems. Islier [8] developed a genetic algorithm for solving multi objective facility layout by minimizing transportation load, maximizing compactness of departments and minimizing the difference between requested and available area. Pelinescu and Wang [9] presented a hierarchical multi criteria optimal design of fixture layout. Yang and Kuo [0] proposed a hierarchical AHP and data envelopment analysis approach for solving layout design problem. Chen and Sha [,] developed a five-phase heuristic method using paired comparison matrix for solving multi-objective facility layout problem. Sahin and Ramazan [] and Singh & Singh [] have used heuristic method for solving multi-criteria layout problems. In this paper, we develop a new rule based constructive approach called Head-Tail method. We use this method for solving bi-criteria layout optimization problems. Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA.

2 . Head-Tail Layout Head-Tail approach starts with the pair of departments that have the highest flow as the initial partial layout. More departments are progressively added to the existing partial layout until all departments are assigned. The problem is to find the order of departments that minimizes the total flow as measured in Equation (). The selection of the location of each department is based on qualitative or quantitative rules and other pertinent information. We call sp = (i j) as a candidate pair of departments and sq = ( ) as a partially constructed sequence. Generic Steps and Rules of Head-Tail Method. Consider the ranking of all pairs (i, j). Set the highest ranked pair as sq.. Consider the highest ranked unassigned pair, sp = (i, j), such that a. oth i and j are not in sq b. Either i or j is in sq.. Use relevant Head-Tail Rules to combine sp and sq.. Repeat steps and until all stations are assigned to sq.. Try to improve upon the obtained solution by use of the Pair-Wise Exchange method. Temporarily iscarding Unassigned Pairs: If a pair sp = (i,j) does not have any common element with sq = (... ), then sp = (i,j) is discarded temporarily only in the current iteration, but it is reconsidered in the next iteration. Permanently iscarding Assigned Pairs: If a pair sp = (i, j) has two common elements with sq = ( i.., j ), then sp = (i,j) is discarded permanently (because both i and j are already assigned). Consider a distance matrix,, consisting of elements dij where dij is the distance between pairs of departments i and j. Also consider a flow matrix, W, where each element (wij) represents the flow from department i to department j. or every pair of departments (i, j) where i<j, create a new wij equal to (wij + wji). These values are used in calculating the total flow. or a problem with n departments, there will be n locations. The distances can be presented by a symmetric flow matrix; that is, the distance from department i to department j, dij, is equal to the distance from department j to location i, dji. The total flow for a given sequence solution is: T = n n wd () ij ij i= j= Rules for Single Row Layout Order all pairs (i, j) in descending order of wij in a flow Matrix. Label the ordered vectors of wij by wo. Consider sp = (i j), sq = ( ), and wo = Vector of wij in descending order of wij. Conflict Resolution Rule: If there is a common department in sequences sp and sq, i.e. sp = (i j) and sq = ( n i m ), then place the new department, j, next to the common department, i, in sq on the side that has a higher ranking to its adjacent department, i.e. compare j n and j m flows and insert j on the side that has higher flow. Linear Mirror Property: The mirror image of sequence sq, i.e. its reverse sequence, is equivalent to sq. oth sequences will have the same total flow. In the following, we show how to solve a single-row layout using Head-Tail. Example : Single-Row Layout Consider the information given below. rectangle. Suppose the layout should be in the form of a x6 Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA.

3 istance 6 Locations lows V Ranking of lows V A C E A C E A A C C E - 0 E igure : Initial ata for Example. sp = (A ); then set sq = (A ). sp = (C ); The mirror image is sp = ( C). Therefore, sq = (A C).. iscard ( ) temporarily as it does not have a common element with sq.. sp = ( ); there is conflict for location of, use the Conflict Resolution Rule to find the location for. This is shown by the following graphical representation.?? Since the flow of (,A) = 0 is more than the flow of (,C) = 0, locate between A and ; therefore, sq = (A C). Reconsider ( ), the location of should be decided. (A C)?? The flow of (, ) = 0 is more than the flow of (,A) = 0; locate on the right side of. Therefore, sq = (A C).. Consider ( C), discard it because it is a duplicate of elements in the sequence. 6. Consider ( E), the location of E should be determined. (A C)?? E The flow of (E,A) = 0 is equal to the flow of (E,) = 0, therefore E can be put on either side of ; arbitrarily, locate it on the right side of. Therefore, sq = (A E C). The final layout is (A E C). Using the above given distances and Equation, the total flow of this layout is Total low = (0*) + (0*) + (0*) + (0*) + (00*) + (0*) + (60*) + (0*) + (0*) + (0*) + (0*) + (80*) + (0*) + (60*) + (0*) = 60 Rules for Multi-Row Layout We can extend the single row Head-Tail method to solve multiple row layout problems. The single row approach expands horizontally. The multi-row layout approach uses all rules of the single row Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA.

4 approach, but allows both horizontal and vertical expansions. The following rules show how to progressively augment sq by combining existing sq = ( i ), the partially built layout, with a new sp = (i j); i.e. find a location for department j. y use of the following rules, all departments will be assigned and a complete solution will be obtained. Expanding Phase Rule (Attaching Sequences, Horizontal and Vertical) a) Horizontal: If there is one common department in the pair of sequences sp & sq, such that the common department is either at the head or the tail of sequence sp and sq, then form a new sequence by attaching sp to sq by their common department. or example, given sq = (C E) and sp = (A C), new sequence is sq = (A C E). b) Vertical: A sequence can be attached in the middle of an existing sequence at a 90 angle if the new sequence has one common department with the existing sequence. or example, adding the sequence ( ) to the sequence (A C ) in igure a results in the sequence in igure b. Rule (Attaching Sequences, iagonal): If there is a common department in sequences sp and sq, e.g. sp = (i j) and sq = ( i ), and Rule fails, i.e. department j cannot be placed next to i in sq, then insert it at a angle on the side which is available and has the highest ranking priority with respect to its adjacent departments. Rule (Extending): Keep the existing sequence as open (extended) as possible by stretching all connected lines as straight as possible. That is, add the new sequence to the old sequence in the form of a straight line if possible. or example, adding (E ) to the sequence in igure b results in the sequence in igure c. Reshaping Phase Rule (Rotation): To fit the partial constructed layout tree into the given layout locations, if possible, use vertical, horizontal, or diagonal rotation of branches while satisfying closeness preferences. In Example, in igure c, E is rotated around resulting in igure d to allow to fold down to C in igure e. Rule (olding): To fit the partially constructed layout tree into the given layout locations, if possible, bend the branches of the layout tree while satisfying closeness preferences. or example, suppose the given layout is x where the given constructed sequence is shown in igure c. Consider folding branch --E. It can be folded to the right or left. ecause the ranking of (E,C) is higher than (E,A), fold the branch to the right connecting E to C as shown in igure d. igure d uses the folding rule to fit into the given layout space. To do this, bend A and upwards and connect them as shown in igure e. This is the final layout which now fits in the x space. A C igure a: Expanding A C igure b: Expanding A C E igure c: Expanding Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA. A C E igure d: olding A C E igure e: olding itting Phase Rule 6 (Conflict Resolution for Multi-Row): If the generated solution by using the reshaping rules does not fit into the given layout space, cut low priority (ranking) departments and paste them

5 into the closest empty spaces based on closeness preferences. Suppose that igure a shows the sequence obtained from Phase I where the layout space is x. In Phase II folding results in igure b. The sequence given in igure b cannot be folded to fit the layout s shape of x. Suppose that the departments (A,) have the lowest flow of all attached departments in the layout; then A is the best choice to be cut. Apply rule 6 by cutting department A and pasting it to the open space. The result is shown in igure c. A C E A C igure a:given Solution igure b:olding igure c:cut and Paste A Example : Head-Tail Multi-Row Layout Consider the information given below. Suppose the layout should be in the form of a x rectangle. istance 6 Locations lows V Ranking of lows V A C E A C E A A C C E - 0 E igure : Initial ata for Example Expanding Steps. sp = (A ). sq = (A ).sp = (C ) or sp = ( C). sq = (A C). sp = ( ). Ignore sp temporarily.. sp = ( ). sp can be attached to sq, see igure a.. sp = ( ). can be attached to, see igure b. 6. sp = ( E). E can be attached to. Since (E,C) has a higher flow than (E,A), E is attached to the right side of. See igure c All departments are assigned; reshape to fit the layout to the x block requirement. Reshaping Steps. Consider igure c. (E,C) has a flow of 0 compared to a flow of 0 for (E,A). (,C) has a flow of 0 compared to a flow of 0 for (,A). Therefore, the advantage of having next to C (an improvement of 0) outweighs the advantage from having E next to C (an improvement of 0). Therefore, in igure c, E can be rotated from left to the right side. This rotation will allow to fold towards (closer) to C instead of A. The result of this rotation rule is shown in igure d. 8. In igure e, is folded down toward C. The final solution layout is shown in igure f. The total flow for this solution is Total low = Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA. E C A E

6 00. E igure a: Expanding igure b: Expanding igure c: Expanding E E E igure d: Rotation of E igure e: olding igure f: Solution Example : Comparing Head-Tail Layout and SLP Consider a facility consisting of seven departments, each with individual space requirements as shown in the table below. Each department can be broken up into units or blocks of 00 square feet. The total allotted area for the new facility is 00 square feet and must be laid within a rectangular space with a dimension of x6 blocks. Each department can fit into a square or a rectangle (e.g. a department can be presented by a x6 or x block shape). epartment 6 Sum Square feet No. of locks 6 The matrix below shows the given desired proximity or closeness for each pair of departments. We first use SLP to find a layout solution to this problem for a. 00x600 facility, and b. 00x00 facility. 6 A O A I A A U E O E X X U U E U O O I I 6 X igure 6 show the initial solution where each department is presented by a node, and the desired closeness for each pair of departments is presented by lines that connect the two departments. igure shows the first iteration of the rearrangement of the relationship diagram to minimize the strain on these bands. 6 igure 6: Initial graph to present the problem 6 igure : irst iteration to improve the total closeness igure 8 shows the next iteration. Re-arrangement of the network should be repeated until no more Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA. 6

7 improvement can be achieved. Once the final graphical relationship is achieved, the space requirements for each department are considered. igure 9 shows the area of each department. 6 igure 8: urther rearrangement to improve the total closeness igure 9: Relationship diagram showing departmental space requirements 6 6 Now we can collapse and rotate the spatial relationship diagram shown in igure 9 to fit into the x6 space. This is shown in igure 0. or the x block facility, the solution is shown in igure. 6 6 igure 0: Relationship diagram for a igure : Layout solution for a x area x6 area The above solution generated by SLP is based on space requirements, availability, closeness preferences, and other relative factors. The A, E, I, O, U, and X ratings can be converted to the numerical values,,,,,, and 0, respectively. Then the ranking of the pairs of departments will be performed. Note that pairs that tie have same ranking. Rankings of Pairs (i,j) X - X X - The construction of the layout by the Head-Tail method is as follows. Expanding Steps Step : sp = ( ). sq = ( ) Step : sp = ( ). sq = ( ). Steps,, and : epartments 6,, and are added; see igure. Step 6: Next consider sp = ( ). epartment can be located in four possible locations diagonally from department as shown in igure, Step 6. ecause pairs (,) and (,6) have the highest ranking, it will be located next to and Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA.

8 Step 6: Step : Connect 6 to Step : Connect to Step : Connect to Connect to 6 and igure : Graphical Representation of layout tree Steps,,, and 6. Since all departments are assigned, the expansion phase is complete. Reshaping Steps Step : Consider the next pair, ( 6). epartments and 6 cannot be folded. Consider the next pair, ( ). epartments and cannot be folded. Consider the next pair, ( ). old closer to (see Step in igure ). itting Steps Step 8: Present the actual size and shape of the departments. Step 9: Move the departments to fit in the layout size of x6 units. 6 6 Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA. 8 6 Step : Connect to Step 8: Insert Spaces Step 9: inal Layout igure : Graphical Representation of Steps, 8, and 9 The final layout is shown in Step 9 of igure. The center of each department is presented by a. The distances between each pair of departments are shown in the following table The total preferential closeness is: Total Closeness = Closeness(i,j)*istance(i,j) n n =,60 i= j= Comparison of Head-Tail and SLP Solutions: The above example was solved by SLP method. The layout and the distances for the SLP solution are shown in igure. The total closeness for the SLP solution is,00. The Rule-based Planning solution (Example ) was,60 which is considerably better than the solution found by the SLP method

9 6 igure : SLP Layout Solution. I-CRITERIA LAYOUT PLANNING In bi-criteria facility layout problem, the two objectives are: Minimize Total low f = Minimize Total Closeness f = n n i= j= n n i= j= wd ij wp or example, suppose that the flow between departments A and is very high and from the total flow point of view should be located very close to each other. However, from a safety point of view, they should be very far from each other. The above bi-criteria problem can be used to handle such conflicting objectives. Suppose that an additive utility function can be used to rank multi-criteria layout alternatives. Let w and w be the weights of importance of the two objective functions, respectively. irst find the normalized values of the two objective functions. To normalize the flow information, find the largest flow (dij,max) and smallest flow (dij,min), and then use the formula shown below to normalize all flow values presented by dij: they will be between 0 and. Similarly, closeness information can be normalized between 0 and using the following formula where pij,max is the largest and pij,min is the smallest pij values. (wij - w ij,min ) (pij -p ij,min ) vij = pij = (wij,max - w ij,min ) (pij,max -p ij,min ) Then for each pair of departments (i,j), calculate zij = wdij + wpij The matrix of zij values is called the composite matrix, Z. The additive utility function (U) is defined as the weighted sum of the flow and the closeness values using weights of importance of w and w for the flow and closeness, respectively. In order to minimize the additive utility function U = wf + wf, n n Minimize U = d ij(wd ij' + wp ij') = n n dz () ij ij i- j= ij ij ij i= j= The value of zij = wdij + wpij is constant for i, j, and given weights. Therefore, they only need to be calculated once for each pair of departments. Therefore, for the generated Z matrix, the bicriteria layout problem can be solved by Head-Tail method. The alternative with the smallest U value is the best layout. Example : i-criteria Layout istances for the SLP Solution Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA. 9

10 or a four department facility layout problem presented in igure, find the best bi-criteria layout solution. Suppose that the decision maker s weights of importance for flow and closeness objectives are w = 0. and w = 0., respectively. Consider using 0 to numerical ratings for X to A qualitative ratings. Location low Matrix A C A Assignment A C C Qualitative Matrix Converted Qualitative Matrix A C A C A - X/0 E/ A/ A U/ E/ - C - X/0 C igure : Initial data for the bi-criteria layout problem irst convert qualitative closeness data to quantitative data as presented in igure. Then normalize flow matrix and closeness matrix values as shown in igure 6. Normalized low Matrix Normalized Qualitative Matrix (v ij ) = (v ij-0)/(00-0) (p ij ) = (p ij-0)/(-0) A C A C A C A A A C C - 0 C igure 6: Normalized flow and qualitative closeness values A composite matrix, Z, can be generated by using zij = wvij + wpij = 0.vij + 0.pij. Solving the i-criteria Problem by the HEA-TAIL Method: irst rank all pairs in descending order of zij values: they are (A,), (A,C), (,), (,C), (A,), and (C,). The steps of HEA-TAIL for assigning departments are: ) (A ); ) (C A ); ) (C A ) This is the final solution, and its mirror is ( A C). The objective values for this layout are: f = *00 + *0 + *0 + *60 + *0 + *90 = 60 and f =*0+*+*+*+*+*0 =6 No improved solution can be found by applying the Pairwise Exchange Method to the final solution of this example. Using the additive utility as the performance measure for each facility, the first iteration of the Pair- Wise Exchange method produces the results shown below. Coefficient acility Pairs Initial distance Pair-Wise Exchange new distance A AC A C C 0. A 0.66 AC 0. A Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA. 0

11 0.0 C C U = d ij z ij i j The lowest score is.68 for swapping A and C. Improved Layout: C A The swap is made and the initial layout for the second iteration is as follows: Coefficient acility Initial Pair-Wise Exchange new distance pairs distance A AC A C C 0. A 0.66 AC 0. A 0.0 C C U= d ij z ij i j The lowest score is.0 for swapping A and. The new layout is. Improved Layout: C A Next iteration is: Coefficient acility Initial Pair-Wise Exchange new distance pairs distance A AC A C C 0. A 0.66 AC 0. A 0.0 C C U= d ij z ij i j The lowest score is.0 for swapping and. The new layout is shown below. Improved Layout: C A Continuing with the next iteration: Coefficient acility Initial Pair-Wise Exchange new distance pairs distance A AC A C C 0. A 0.66 AC 0. A 0.0 C C U= d ij z ij i j Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA.

12 No more improvement is possible, stop. The layout remains with f (total flow) = 60 and f (total closeness) =. inal Layout: C A Efficient rontier y systematically varying weights (w, w) in Equation (), a set of efficient bi-criteria alternatives can be generated. To do this, first generate the Z matrix for each given set of weights and then use head-tail method to solve the problem. Note that for a given set of weights, when PWE method is used, alternatives generated in different iterations of PWE method may also be efficient and should be recorded. That is, each Pairwise Exchange solution in each iteration of PWE method can be efficient. Consider the set of four weights given in the first row of the following table. or each given set of weights, solve the problem by Head-Tail method. The generated solution is recorded under each set of weights. The set of generated four alternatives are all efficient. (w, w ) (0.99, 0.0) (0.6, 0.) (0., 0.) (0.0, 0.99) Layout AC AC AC CA Min. f Min. f Efficient? Yes Yes Yes Yes Conclusion In SLP, the process of improving the initial graph of the layout becomes very difficult as the number of departments increases. Also the evaluation process in SLP is subjective and depends on the layout designer. A different approach for solving layout problem is by use of Head-Tail Layout method which is a powerful tool for solving both structured and ill-structured layout problems. Head-Tail can solve very large layout problems with minimal computational efforts. ecause this method is Head-Tail, it can incorporate complicated constraints while developing the layout solution. urthermore, it can be used in conjunction with Expert systems approaches to solve layout problems. References: [] Raman., S. V. Nagalingam, G. C.I. Lin, Towards measuring the effectiveness of a facilities layout, Robotics and Computer-Integrated Manufacturing, doi:0.06/j.rcim , 008. [] Rosenblatt, M. J., "The facility layout problem: a multi-goal approach." International Journal of Production Research,, 99, pp. -. [] ortenberry, J. C., J.. Cox, Multiple criteria approach to the facilities layout problem, International Journal of Production Research, Vol., Issue, 98, pp.-8. [] Malakooti,., "Multi-Objective acility Layout: A Heuristic Method to Generate All Efficient Alternatives", International Journal of Production Research, Vol., No., 989, pp. -8. [] Malakooti,., Computer Aided acility Layout Selection with Applications to Multiple Criteria Manufacturing Planning Problems. Large Scale Systems: Theory and Applications, Special Issue on Complex Systems Issues in Manufacturing, Vol., 98, pp [6] Malakooti,., A. Tsurushima, "An Expert System Using Priorities for Solving Multiple Criteria acility Layout Problems." Inter. Journal of Production Research, Vol., No., 989, pp [] Yang, Z.,. Malakooti, A Multiple Criteria Neural Network Approach for Layout of Machine -Part Group ormation in Cellular Manufacturing. Third IE Research Conference, 99, pp. 9-. [8] Islier A. A., A genetic algorithm approach for multiple criteria facility layout design, International Journal of Production Research, 998, Vol. 6, No. 6, pp [9] Pelinescu. M., M. Y. Wang, "Multi-objective optimal fixture layout design", Robotics and Computer-Integrated Manufacturing, Vol. 8, 00, pp. 6-. Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA.

13 [0] Yang T., Ch. Kuo, A hierarchical AHP/EA methodology for the facilities layout design problem, European Journal of Operational Research,, 00, pp [] Chen C.-W.,. Y. Sha, Heuristic approach for solving the multi-objective facility layout problem, International Journal of Production Research, Vol., No.,, November 00, pp [] Sha,., C. Chen, "A New Approach to the Multiple Objective acility Layout Problem", Integrated Manufacturing Systems, Vol., No., 00, pp [] Şahin, Ramazan. A Simulated Annealing Algorithm for Solving the i-objective acility Layout Problem. Expert Systems with Applications, Vol. 8, Issue, 0, pp [] Singh S. P., V. K. Singh, Three-Level AHP-ased Heuristic Approach for a Multi-Objective acility Layout Problem. International Journal of Production Research, Vol. 9, Issue, 0, pp. 0-. Proceedings of the 0 ICAM, International Conference on Advanced and Agile Manufacturing, Held at Oakland University, Rochester, MI 809, USA Copyright 0, ISPE and ISAM USA.