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Layout and Design Introduction Fixed-position layout Job shop production I

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Layout and DesignKapitel 4 / 2 (c) Prof. Richard F. Hartl Introduction Strategic decisions (location problems) Concretion and realization on tactical level (Layout and design, configuration) Operational decisions (operational planning)

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Layout and DesignKapitel 4 / 3 (c) Prof. Richard F. Hartl Introduction An efficient layout facilitates and reduces costs of material flow, people, and information between areas. 4 concepts will be introduced here: Fixed-position layout Large, bulky workpieces (ships,…) Job shop production (Process/Function-oriented production) High variety, low volume Cellular manufacturing systems Individual products Flow shop production (Object-oriented production) Low variety, high volume cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 4 (c) Prof. Richard F. Hartl Introduction Example: Production and assembly of 4 parts (A, B, C, D) A: saw -> turn -> mill -> drill B: saw -> mill -> drill -> paint C: grind -> mill -> drill -> paint D: weld -> grind -> turn -> drill Minimum equipment: 1 weld 1 grind 1 saw 1 turn 2 mills 2 drills 1 paint Equipment requirements PartWeldGrindSawTurnMillDrillPaint A B C D cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 5 (c) Prof. Richard F. Hartl Introduction 1. Fixed-postion layout Workpiece TurnDrillMillGrindWarehouseAssemblyStoresWeldPaintSawMillDrill cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 6 (c) Prof. Richard F. Hartl Introduction 2. Job shop layout cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 7 (c) Prof. Richard F. Hartl Introduction 3. Cellular manufacturing system cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 8 (c) Prof. Richard F. Hartl Introduction 4. Flow shop layout cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 9 (c) Prof. Richard F. Hartl Introduction Selection of layout: Characteristic of workpiece Variety of production Volume of production Combinations: Components: job shop and/or Celluars sytems; assembly: flow shop system cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 10 (c) Prof. Richard F. Hartl Fixed-position layout Workpiece too large or cumbersome to be moved trough its processing steps -> processes are brought to the product rather than otherwise. Processes are arranged in the right sequence around the workpiece Right processes at the workpiece at the right location in the right time cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 11 (c) Prof. Richard F. Hartl Fixed-position layout Advantages: Material movement is reduced. Promotes job enlargement by allowing individuals or teams to perform the “whole job”. Highly flexible; can accommodate changes in product design, product mix, and production volume. Independence of production centres allows scheduling to achieve minimum total production time. cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 12 (c) Prof. Richard F. Hartl Fixed-position layout Limitations: Increased movement of personnel and equipment. Equipment duplication may occur. Higher skill requirements for personnel. General supervision required. Cumbersome and costly positioning of material and machinery. Low equipment utilization. cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 13 (c) Prof. Richard F. Hartl Job shop production High variety – low volume Rapid changes in mix or volume „In-between conditions“ Collection of processing departments or cells Each containing a collection of machines processing similiar operations Each product (group) undergoes a different sequence of operations Different products have different material flows and are moved from one department to another in the appropriate sequence. High degree of interdepartmental flow cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 14 (c) Prof. Richard F. Hartl Job shop production Advantages: Better utilization of machines can result -> fewer machines are required. A high degree of flexibility exists relative to equipment or manpower allocation for specific tasks (break down,…) Comparatively low investment in machines The diversity of tasks offers a more interesting and satisfying occupation for the operator. Specialized supervision is possible. cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 15 (c) Prof. Richard F. Hartl Job shop production Limitations: Longer flow lines are needed -> material handling is more expensive. Production planning and control systems are more involved than for other layouts. Usually, total production time is longer than for other layouts. Due to the fact that jobs have to queue before being processed in a machine job comparatively large amounts of in-process inventory occur. Comparatively high degree of (machine) idle time because machines have to wait until the subsequent job is finished with its foregoing process. Space and capital are tied up by work in process. Because of the diversity of the jobs in specialized departments, higher grades of skill are required. cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 9 cf. Francis, R., McGinnis, L., White, J., Facility Layout and Location: An Analytical Approach, Prentice Hall, 1992

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Layout and DesignKapitel 4 / 16 (c) Prof. Richard F. Hartl Job shop production Eliminiation of negative effects Production planning (operational level) Optimized machine allocation (tactical level) -> Assignment problems: LAP (linear assignment problem) QAP (quadratic assignment problem)

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Layout and DesignKapitel 4 / 17 (c) Prof. Richard F. Hartl Linear Assignment Problem Simplest optimization problem in intra-company location planning Gegeben: n machines (activities, workers) n potential locations (periods, projects) c ij... cost of running maschine i on location j Any machine can be assigned to any location It is required to use all locations by assigning exactly one machine to each location total cost of the assignments are to be minimized.

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Layout and DesignKapitel 4 / 18 (c) Prof. Richard F. Hartl Linear Assignment Problem 3 machines, 4 locations and the following costs c ij locationj = i \ j1234 machine i =215 Machine 2 cannot be assigned to location 2 -> cost

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Layout and DesignKapitel 4 / 19 (c) Prof. Richard F. Hartl Linear Assignment Problem If number of locations ≠ number of machines: Add dummymachines (-rows) or dummylocations (-columns) with costs 0 (this is always possible) locationsj = i \ j1234 machine i =215 dummy40000 Location with dummy machine -> location stays empty

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Layout and DesignKapitel 4 / 20 (c) Prof. Richard F. Hartl Linear Assignment Problem Formulation as TP: Each LAP can be interpreted as special case of a TP with each supplier (=machine) having a capacity of 1 and each customer (= location) having a demand of 1. Since, for the TP it is guaranteed (due to the special problem structure) that all decision variables are integer, we end up with a feasible solution for the LAP (n variables with value 1 and all others 0 (LAP: m=n basis variables; TP: m+n-1 basis variables) TP:

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Layout and DesignKapitel 4 / 21 (c) Prof. Richard F. Hartl Linear Assignment Problem 1 if maschine i is assigned to location j 0 otherwise x ij = Cost: Constraints: = 1 für i = 1,...,n... each machine is assigned exactly once = 1 für j = 1,...,n... each location is allocated with 1 machine = 0 oder 1 für i = 1,...,n und j = 1,...,n

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Layout and DesignKapitel 4 / 22 (c) Prof. Richard F. Hartl Linear Assignment Problem In order to solve LAPs exactly we are going to make use of the following important problem characteristic: it is always possible to reduce (or increase) all entries of any column or row by a certain value without changing the optimal solution (only the absolute costs change, the relation stays the same). We use this characteristic to generate the maximum number of 0 entries. Example: Optimal solution (column minimum method): A-I B-II C-III Cost: 1+2+3=6 IIIIII A1815 B6210 C793

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Layout and DesignKapitel 4 / 23 (c) Prof. Richard F. Hartl Linear Assignment Problem Cost reduction (Columns -> Rows): The relation of assignment costs for each machine/locations does not change. Optimal solution (column minimum method): A-I B-II C-III (Reduced) cost: 0 Reduced cost + reduction values: = 6 (= Total assignment cost without cost reduction) IIIIII A0612 B507 C

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Layout and DesignKapitel 4 / 24 (c) Prof. Richard F. Hartl Kuhn´s Algorithm Kuhn´s algorithm finds the exact solution is based on adding/subtracting values to/from given cost factors in order to find the lowest opportunity cost (i.e. not-obtained profits) 3 steps to be followed: 1. Cost reduction -> Generation of 0 elements 2. Try to determine the optimal assignment. If this is not possible draw the minimum number of lines necessary to cover all zeros in the matrix. 3. Adapt the cost factors in the matrix and return to step 2. cf. Heizer, J., Render, B., Operations Management, Prentice Hall, 2006, Chapter 15

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Layout and DesignKapitel 4 / 25 (c) Prof. Richard F. Hartl Kuhn´s Algorithm Step 1: Cost reduction Subtract the smallest number in each column from every number in that column Subtract the smallest number in each row from every number in that row. We obtain a matrix with a series of zeros, meaning zero opportunity costs (at least one zero in each column and each row) No cost reduction in columns already including zero elements

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Layout and DesignKapitel 4 / 26 (c) Prof. Richard F. Hartl Kuhn´s Algorithm Step 2: Optimal solution? Start with a column or row having as few as possible 0 entries Frame one the 0 in this column/row and cross all other 0 in the corresponding column and row Go on with the next column or row having as few as possible not-framed and not-crossed zeros. And so on until all zeros are either framed or crossed. If we are able to make a zero (reduced) cost assignment for all machines we found the optimal solution! Otherwise we have to find the minimum arrangement of lines covering all zeros in the matrix.

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Layout and DesignKapitel 4 / 27 (c) Prof. Richard F. Hartl Kuhn´s Algorithm Example: Step 1:

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Layout and DesignKapitel 4 / 28 (c) Prof. Richard F. Hartl Kuhn´s Algorithm Step 2: Optimal solution!

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Layout and DesignKapitel 4 / 29 (c) Prof. Richard F. Hartl Kuhn´s Algorithm Step 2: If not draw the minimum number of lines covering all zero elements Mark (for example „X“) all rows with no framed 0 Mark all columns having at least 1 crossed 0 in a marked row Mark all rows having a framed 0 in a marked column Repeat until there is no column or row left to be marked Mark each non-marked row and each marked column (shaded) with a continuous line. -> this is the minimum arrangement of lines needed to cover all 0. If the number of lines is equal to the number of rows/columns, an optimal assignment is already possible.

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Layout and DesignKapitel 4 / 30 (c) Prof. Richard F. Hartl Kuhn´s Algorithm Step 3: Adapt the cost factors The smallest not-covered element is the new reduction value (a). Subtract a from all not-covered elements in the matrix. Add a to all elements covered by two lines. Elements covered by 1 line remain unchanged. Go on with step 2.

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Layout and DesignKapitel 4 / 31 (c) Prof. Richard F. Hartl Kuhn´s Algorithm 17,51595, ,510, ,514,5115,5 4,581417,513 9,58,51217,5 -0,5 -4, ,5-5,5 1370,5 6,5 211,58,50 7, ,507,512,5 8,51, ,56,5 11,58, ,5 0 5,5 7 12,5 12 7,50 8,5 0 1,5 7,5 K = 4, , ,5 + 0,5 = 32

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Layout and DesignKapitel 4 / 32 (c) Prof. Richard F. Hartl Kuhn´s Algorithm 12,56, ,58,5205 7, ,512,57,5 8,51,50712 No zero cost assignment! -> Find the minimum arrangement of lines covering all zeros.

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Layout and DesignKapitel 4 / 33 (c) Prof. Richard F. Hartl Kuhn´s Algorithm 12,56, ,58,5205 7, ,512,57,5 8,51, > Adapt the cost elements by adding/subtracting element a (minimum value of all not-covered elements)

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Layout and DesignKapitel 4 / 34 (c) Prof. Richard F. Hartl Kuhn´s Algorithm 12,56, ,58,5205 7, ,512,57,5 8,51,50712 a = 1, , , ,5 3,5 10,5 107 1 additional zero (assignment 5 2) increases the chance the find an assignment with total (reduced) costs of 0.

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Layout and DesignKapitel 4 / 35 (c) Prof. Richard F. Hartl Kuhn´s Algorithm , ,5 7, , ,5 Optimal assignment! Total cost: sum of all reduction values (step 1 and 3): C=C=(4, , ,5 + 0,5) + (1,5) =33,5 Start again with step 2 until a zero cost assignment is possible!

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Layout and DesignKapitel 4 / 36 (c) Prof. Richard F. Hartl Kuhn´s Algorithm Some assignment problems entail, e.g., maximizing profit instead of minimizing cost. To convert a maximization problem to an equivalent minimization problem, we subtract every number in the original matrix from the largest single number in that matrix. ABCDEF I1I I2I I3I I4I I5I I6I Maximize the total profit! [„Cost“ = 20 - Profit]

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