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COMPUTERIZED LAYOUT_II. APRIL 2013. Layout Levels. From: http://www.strategosinc.com/facility_plan_levels.htm. Algorithmic Approaches ( i). • SLP is “ informal ” • Does not provide a formal procedure or algorithm for critical steps

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COMPUTERIZED LAYOUT_IIAPRIL 2013Layout LevelsFrom: http://www.strategosinc.com/facility_plan_levels.htmPrepared by: Asst.Prof.Dr. Nevra AKBILEKAlgorithmicApproaches (i)• SLP is “informal”• Does not provide a formal procedure oralgorithmforcriticalsteps• Algorithm: a precise rule (or set of rules)specifying how to solve some problem– Has an objectivefunction– AmenabletocomputerimplementationPrepared by: Asst.Prof.Dr. Nevra AKBILEKAlgorithmicApproachesTypes of input– Qualitative “flow” data (i.e. relationship chart)– Quantitative data (i.e. from-to chart, flow-between)– BothClassificationof algorithms– Based on objectivefunction• Distancebased• Adjacencybased– Format of layoutrepresentation• Discreterepresentation• Continuousrepresentation– Primaryfunction• Improvement• ConstructionPrepared by: Asst.Prof.Dr. Nevra AKBILEKMeasuringFlowQuantitativeAppropriate when large volumes of material/people move between departmentsFrom-tochartFlow-betweenchartQualitativeHow important is adjacency to two departments?Often applied to the layout of office environmentsActivity relationshipdiagram(orchart)Prepared by: Asst.Prof.Dr. Nevra AKBILEKQualitativea. From-to- chartb. FlowbetweenchartIf total flow is neededanddirection is not important.Prepared by: Asst.Prof.Dr. Nevra AKBILEKQuantitaiveClassification of Algorithms-LayoutRepresentation Format– Discrete• The area of each department is rounded off to the nearest integer number of grids.• A smaller grid size yields a finer resolution and gives more flexibility in departmentshapes, but• Results in a larger number of grids which complicates computations. Why?– Continuous• Does not use a grid• More flexible but more difficult to use• Usually limited to rectangular building and departmentsPrepared by: Asst.Prof.Dr. Nevra AKBILEKClassification of AlgorithmsConstruction-based– Develop ‘from scratch' and progressively build layoutConstruction algorithmImprovementalgorithmALDEPCORELAPPLANET(Graph-basedmethods)CRAFTMCCRAFTMULTIPLE(Two-pairmethods)BLOCPLANLOGICMixInteger ProgrammingImprovement- based– Start with an initial layout and try to improve it throughincrementalchangesConstruction-based– Develop ‘from scratch' andprogressively build layoutPrepared by: Asst.Prof.Dr. Nevra AKBILEKComputerizedLayoutAlgorithmsAutomated Layout Design Program (ALDEP) ALDEP is a construction type algorithm. This algorithm uses basic data on facilities and builds a design by successively placing the department in the layout. Computerized Relationship Layout Planning (CORELAP) This algorithm is based on Muthers’s procedure given in systematic layout planning. This computer algorithm was developed by R.C. Lee. Prepared by: Asst.Prof.Dr. Nevra AKBILEKComputerizedLayoutAlgorithmsComputerized Relative Allocation of Facilities technique (CRAFT) A number of computerized layout programs have been developed since the 1970s to help devise good process layouts. Of these, the most widely applied is the Computerized Relative Allocation of Facilities Technique (CRAFT). The main objective of the CRAFT is finding the optimal layout by interchanging the department pair wise accordingly total transportation cost is minimized. Prepared by: Asst.Prof.Dr. Nevra AKBILEKGenerating Layout Alternatives(Procedures)ProceduresConstruction procedures“Greenfield” layout, the layout of a new facilityImprovement proceduresChanges/ improvements to existing facilitiesPrepared by: Asst.Prof.Dr. Nevra AKBILEKComputer-Aided Layout (cont.)Adistance-based objective functioncan be expressed as:An adjacency based objectivefunctioncan be expressed as:Usefrom-tochart as input dataUserelationshipchart as input data Prepared by: Asst.Prof.Dr. Nevra AKBILEKCRAFT-(ComputerizedRelativeAllocation of FacilitiesTechnique)First computer-aided layout algorithm (1963) The input data is represented in the form of aFrom-To chart, or qualitative data. The main objective behind CRAFT is to minimize total transportationcost: Improvement-typelayoutalgorithm Prepared by: Asst.Prof.Dr. Nevra AKBILEKDistanceDistanceCalculations Rectilinear distance from centroid to centroid Euclidean distance from centroid to centroid Prepared by: Asst.Prof.Dr. Nevra AKBILEKCentroidsRectilinear distance from A to B: D (AB) = 1.5 + 1 = 2.5 Rectilinear distance from B to C: D (BC) = (5-1.5) + (1+1.5) = 3.5 + 2.5 = 6 DistanceCalculationsPrepared by: Asst.Prof.Dr. Nevra AKBILEKStepsin CRAFT1. Start with an initial layout with all departments made up of individual square grids (Note: each grid represents the same amount of space)Calculate centroid of each department and rectilineardistance between pairs of departments centroids (stored in adistancematrix).Departments i and j exchange New centroid i = centroid j New centroid j = centroid i Only consider exchanging adjacent departments 2. Find the cost of the initial layout by multiplying the– From-To (flow) chart,– unitcostmatrix, and– From-To (distance) matrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKSteps in CRAFT3. Improve the layout by performing all-possible two or threewayexchanges – At each iteration, CRAFT selects the interchange that results in themaximum reduction in transportation costs– These interchanges are continued until no further reduction is possible(If the estimated cost of the best exchange in (2) is higher than the best cost found so far, stop Else, go to step1)Prepared by: Asst.Prof.Dr. Nevra AKBILEKCRAFT Properties• CRAFT only exchanges departments that are– Adjacent (share at least one common edge)– HaveequalareasThe actual size of cost reduction can beoverestimated or underestimated • Adjacency is a necessary but not sufficient criteria forswappingdepartmentsQuality of final solution depends on the initial layout Final solution may be locally optimal, not globally optimal Prepared by: Asst.Prof.Dr. Nevra AKBILEKCRAFT PropertiesDepartment representation Discrete grids No shape restrictions Objective Distance based Prepared by: Asst.Prof.Dr. Nevra AKBILEKIn-class ExercisesExchange departments 2 and 4 in the layout shown below. All departments are fixed except 2 & 4.If the flow from A to B is 4, A to C is 3, and B to C is 9, and all move costs are 1, what is the layout cost?Prepared by: Asst.Prof.Dr. Nevra AKBILEKWeakness of CRAFTIt is not always possible to exchange two unequal size, adjacent departments without splitting the larger one. (adjacentandequalareas) Centroid Initialsolution_based Discrete Rectangularshape Prepared by: Asst.Prof.Dr. Nevra AKBILEKExamplePrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleA - RowmaterialstoreB - FabricinspectiondepartmentC - CuttingdepartmentD -SawingdepartmentE - FinishingdepartmentF - WashingdivisionG - IroningdivisionH - PackagingdivisionI - FinishgoodstorePrepared by: Asst.Prof.Dr. Nevra AKBILEKCostmatrixSTEP 1 INPUTS Initiallayout Costmatrix (cij ) Flowmatrix Area of eachdepartment Transportation cost is among departments is Zero. So we got cost matrix based on that assumption Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleTo compute the flow matrix we use no of movement between the departments Numbers of movements between the departments are given bellow. FlowmatrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleSTEP 1Compute the centroids of the department in present layout Calculationof centroids A 26.59 54.77 B 17.5 40 C 10 17.5 D 45.68 19.54 E 60 50 F 77.5 52.5 G 77.5 40 H 77.5 17.5 I 92.5 30 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleSTEP 1PREPARATION OF DISTANCE MATRIX Initialsolution Distancematrix DistancematrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleSTEP 2CALCULATE THE PRESENT LAYOUT COST Initial cost matrix :Initial cost 119.3 + 108.64 +38.18 +68.68+ 181.36 +23.86 + 150 +377.2 + 537.36 +80 + 50 +180+110 = 2024.58 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleSTEP 3FIND THE POSIBLE INTERCHANGESA – B CommonboarderA – C Not possibleA –D CommonboarderA – E CommonboarderA – F Not possibleA – G Not possibleA – H Not possibleA – I Not possibleB – C CommonboarderB – D CommonboarderB – E Not possibleB – F Not possibleB – G Not possibleB – H Not possibleB – I Not possibleC – D CommonboarderC – E Not possibleC – F Not possibleC – G Not possibleC – H Not possibleC – I Not possibleD – E CommonboarderD – F Not possibleD – G CommonboarderD – H CommonboarderD – I Not possibleE – F CommonboarderE – G CommonboarderE – H Not possibleE – I Not possibleF – G CommonboarderF – H Not possibleF – I CommonboarderG –H CommonboarderG – I CommonboarderH – I Common board Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleSTEP 3FIND THE COST OF INTERCHAGING DEPARTMENT Interchange A & B DistancematrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleCostmatrixTotal cost = 119.3 + 97.28 + 52.5 + 60 + 170 +23.86 + 269.3 + 377.2 + 537.36 + 44.78 + 20 + 80+ 50 + 180 + 110 = 2191.50 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleInterchange A & D DistancematrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleCostmatrixTotal cost 243.2 + 108.64 + 44.68 + 52.28 + 67.72 +48.64+ 150 + 538.6 + 458.16 + 80 + 20 + 50 + 180 +110 Total costis 2151.92Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleInterchange A & E DistancematrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleCostmatrixTotal cost 262.5 + 89.48 + 38.18 + 27.5 + 100 +52.5 +150 +377.5 +430 +54.32 +212 +20 +50 + 180 +110 Total costis 2153.98 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleInterchange B & C DistancematrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleCostmatrixTotal cost 219.3 + 108.64 + 38.18 + 68.67 + 180.36 + 43.86 + 150 + 486.4 + 537.36 + 44.78 +20 + 80 +50 +180 + 110 Total cost2317.55 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleInterchange B & D DistancematrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleCostmatrixTotal cost 119.3 + 53.44 + 38.18 + 68.67 + 181.36 + 54.12 + 188.5 + 525 + 618 + 52.5 +80 +20 +50 + 180 + 110 Total cost2339.07 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleInterchange C & D DistancematrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleCostmatrixTotal cost 119.3 + 107.72 + 38.18 + 68.67 + 181.36 + 23.86 + 243.2 + 377.2 + 990 + 82.5 + 80 + 180 + 110 + 50 Total cost2651.99 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleInterchange D & E DistancematrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleCostmatrixTotal cost 119.3 + 108.6 + 38.18 + 68.67 +181.36 + 23.86 +150 +725 +537 + 44.75 + 240.88 + 50 + 180 +110 Total cost is 2577.6 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleInterchange D & G DistancematrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleCostmatrixTotal cost 119.3 + 108.6 + 38.18 + 54.32 + 181.36 + 23.8 + 150 + 900 + 330 + 27.5 + 80 +20 + 259.12 + 270.88 + 110 Total costis 2673 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleInterchange D & H DistancematrixPrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleCostmatrixTotal cost 119.3 + 176.4 + 38.18 + 68.67 + 108.64 + 23.86 + 150 + 675 + 600 + 52.5 + 80 +20 + 50 + 418.24 + 229.12 Total cost is 2809.91 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleSummary of approximate total cast due to pair wise interchanges Minimum valuePrepared by: Asst.Prof.Dr. Nevra AKBILEKExampleSTEP 3COMPAIRING INITIAL COST AND INTERCHANGES Cost of the initial layout is 2024.58when comparing it with cost of interchanging departments, it is less than every interchanging cost. In the other words all interchanging costs are higher than the initial cost. Therefore, present layout can be considered as the optimal layout. Prepared by: Asst.Prof.Dr. Nevra AKBILEKCORELAPCORELAP: Computerized Relationship Layout Planning Developed for main frame computers Construction type Adjacency-basedmethod CORELAP uses A=4, E=3, I=2, O=1, U=0 and X=-1 values Selection of the departments to enter the layout is based on Total Closeness Rating Total Closeness Rating (TCR) for a department is the sum of the numerical values assigned to the closeness relationships between the department and all other departments. Prepared by: Asst.Prof.Dr. Nevra AKBILEKCORELAP(Departmentselection)1. The first department placed in the layout is the one with the greatest TCR value. If there is a tie, then choose the one with more A’s (E’s, etc.). 2. If a department has an X relationship with the first one, it is placed last in the layout and not considered. If a tie exists, choose the one with the smallest TCR value. 3. The second department is the one with an A (or E, I, etc.). relationship with the first one. If a tie exists, choose the one with the greatest TCR value. 4. If a department has an X relationship with the second one, it is placed next-to-the-last or last in the layout. If a tie exists, choose the one with the smallest TCR value. 5. The next department is the one with an A (E, I, etc.) relationship with the already placed departments. If a tie exists, choose the one with the greatest TCR value. 6. The procedure continues until all departments have been placed. Placement sequence Prepared by: Asst.Prof.Dr. Nevra AKBILEKCRAFT(Departmentplacement)DepartmentneighborsAdjacent (in position 1, 3, 5 or 7) with department 0 Touching (in position 2, 4, 6 or 8) department 0 Placing rating (PR) is the sum of the weighted closeness ratings between the department to enter the layout and its neighbors. The placement of departments is based on the following steps: 1.The first department selected is placed in the middle. 2. The placement of a department is determined by evaluating PR for all possible locations around the current layout in counterclockwise order beginning at the “western edge”.3. The new department is located based on the greatest PR value. Prepared by: Asst.Prof.Dr. Nevra AKBILEKCORELAP – Example 1 Given the relationship chart and the departmental dimensions below determine thesequence of the placement of the departments in the layout based on the CORELAP algorithm. Place the departments in the layout while evaluating each placement. Prepared by: Asst.Prof.Dr. Nevra AKBILEKExample-1A=4, E=3, I=2, O=1, U=0, X=-1 The first department placed in the layout is the one with the greatest TCR value. If there is a tie, then choose the one with more A’s (E’s, etc.). Theplacementsequence: 5 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExample-1A=4, E=3, I=2, O=1, U=0, X=-1 The second department is the one with an A relationship with the first one (or E, I, etc.). If a tie exists, choose the one with the greatest TCR value. Theplacementsequence: 5-6 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleA=4, E=3, I=2, O=1, U=0, X=-1 The next department is the one with an A (E, I, etc.) relationship with the already placed departments. If a tie exists, choose the one with the greatest TCR value. Theplacementsequence: 5-6-7 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleA=4, E=3, I=2, O=1, U=0, X=-1 Theplacementsequence: 5-6-7-9 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleA=4, E=3, I=2, O=1, U=0, X=-1 Theplacementsequence: 5-6-7-9-3 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleA=4, E=3, I=2, O=1, U=0, X=-1 The placement sequence: 5-6-7-9-3 - 8 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleThe placement sequence: 5-6-7-9-3-8 - 1 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleA=4, E=3, I=2, O=1, U=0, X=-1 The placement sequence: 5-6-7-9-3-8-1 - 2 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleA=4, E=3, I=2, O=1, U=0, X=-1 Theplacementsequence: 5-6-7-9-3-8-1-2-4 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleTheplacementsequence: 5-6-7-9-3-8-1-2-4 Selection is completed!Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleA=4, E=3, I=2, O=1, U=0, X=-1 Both options give the same PR Score PR = A[5,7]+I[6,7] = 4 + 2 = 6 If the location for the department 7 is chosen as shown, the PR would be PR = A [5,7] =4 Theplacementsequence: 5-6-7-9-3-8-1-2-4 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExampleA=4, E=3, I=2, O=1, U=0, X=-1 PR = E[6,9] = 3 PR = E[6,9] +O[5,9]= 3 + 1= 4 Theplacementsequence: 5-6-7-9-3-8-1-2-4 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExamplePR = I[3,5] + O[3,7] + U[3,9] = 2 + 1 + 0 = 3 PR = + E[3,8] + I[8,9] = 3 + 2 = 5 Theplacementsequence: 5-6-7-9-3-8-1-2-4 Prepared by: Asst.Prof.Dr. Nevra AKBILEKExamplePR = I[1,3] + U[1,7] = 2 + 0 = 2 TCR = I[1,2] + I[2,3]= 2 +2 = 4 ContinuewithDepartment 4. Theplacementsequence: 5-6-7-9-3-8-1-2-4 Prepared by: Asst.Prof.Dr. Nevra AKBILEK

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