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Quantitative Review II

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  1. Quantitative Review II Goods and Service Design (Chapter 6 )Process Selection (Chapter 7 )Lean Manufacturing (Chapter 17 )Facility Location + Facility & Work Design (Chapters 9 & 8)Value Chains and Supply Chain Design (Chapters 2 & 9)

  2. Chapter 6 Goods and Service DesignChapter 7 Process Selection, Design, and AnalysisKey Concepts:1) Taguchi Loss Function2) Reliability Management

  3. Taguchi Loss Function • Taguchi measured quality as the variation from the target value of a design specification and then translated that variation into an economic “loss function” that expresses the cost of variation in monetary terms • Where : • x = is the actual value of the dimension • T = target limits • L(x) = the monetary value of the loss associated with deviating from the target limit “T” • k = the constant that translates the deviation into dollars • Design a manufactured good with a smaller design tolerance = better quality

  4. Sample: A quality engineer has a manufacturing specification (in cm) of 0.200 plus or minus 0.05. Historical data indicates that if the quality characteristic takes on values larger than .25 cm or smaller than .15 cm, the product fails and a cost of $75 is incurred. Determine the Taguchi Loss Function and estimate the loss for a dimension of 0.135 cm. T = 0.2 ± 0.05 (cm) 0.15 ~ 0.25 (cm) L (X) = $75 , if ( x-T ) is outside of the above specified range 75 = k (± 0.05)2 K = 30,000Taguchi Loss Function: L(X) =30,000 (x-T)2

  5. A quality characteristic has a specification (in inches) of 0.200  0.020. If the value of the quality characteristic exceeds 0.200 by the tolerance of 0.020 on either side, the product will require a repair of $150. Develop the appropriate Taguchi loss function (k). The Taguchi Loss Function is: L(x) = k(x - T)2 (x-T)2 = ( 0.020)2 150 = k(0.02)2 or k = 375,000

  6. Suppose that the specifications for a part (in inches) are 7.00 ± 0.25, and that the Taguchi Loss Function is estimated to be L(x) = 8,500(x-T)2. Determine the estimated loss if the quality characteristic under study takes on a value of 7.50 inches. Is the loss greater at 7.50 as opposed to 7.25? T= 7.00 6.75 < specification range < 7.25 L(x) = 8500 (7.50-7.00)2 = $2125 when x=7.5 L(x) = 8500 (7.25 -7)2 = $531.25 when x=7.25

  7. Reliability Management • Reliability is defined as the probability (often expressed as a percentage) that a manufactured good, piece of equipment, or system performs its intended function for a stated period of time under specified operating conditions. • A probability of .97 indicates that on average, 97 out of 100 times the item will perform its function for a given period of time under specified operating conditions • Reliability of a manufactured item depends on the reliability of each component of that system

  8. Reliability Management • Series product components - If one component fails, the entire system fails • Parallel product components – Only if all components fail, the entire system will then fail

  9. Example: Series Product Reliability The manufacturing of compact disks requires four sequential steps. The reliability of each of the steps is 0.96, 0.87, 0.92, and 0.89 respectively. What is the reliability of the process? Rsystem = (p1)(p2)(p3)……(pn) RaRbRcRd = (0.96)(0.87)(0.92)(.89) = 0.6839

  10. Example: Parallel Product Reliability The system reliability for a two-component parallel system is 0.99968. If the reliability of the first component is 0.992, determine the reliability of the second component. 0.99968 = 1 – (1 – 0.992)(1 – p2) 0.99968=1-(.008-.008p2) 0.99968-1= -.008+.008p2 p2 = 0.96

  11. Redundancy .91 B A C B .98 .99 .91 • We have both series and parallel components here in this system • RB = 1- (1-0.91)(1-0.91) = 0.9919 (*parallel Rs is always larger than the each individual R in parallel situation) • Rsystem = (0.98) (0.9919)(0.99) = 0.962 or 96.2% (*series Rs is always smaller than each individual R in series situation)

  12. Given the diagram below, determine the system reliability if the individual component reliabilities are: A = 0.94, B = 0.92, C = 0.97, and D = 0.94. A C B D A and B are parallel components RAB = 1 - (1 - 0.94)(1 - 0.92) =.9952 C and D are also parallel components RCD = 1 - (1 - 0.97)(1 - 0.94) = .9982 A-B and C-D are sequentially arranged Rsystem = RABRCD= (0.9952)(0.9982) = 0.9934

  13. What is the reliability of this system? If you could add one process (must be one of the existing processes) to best improve reliabilty what would be the improved reliability? A B C D 0.987 0.965 0.912 0.988 0.912 The lowest reliability component is C, so we first improve Rc by adding another C in parallel fashion Parallel Reliability = 1- (1-0.912)(1-0.912) = 0.992256 System Reliability = (0.987)(0.965)(0.992256)(0.988) =0.9337 C

  14. Chapter 17Lean Operating Systems & JIT

  15. Key Concepts • Lean Operating Systems/Just In Time Systems The manufacturing and service operations that uses approaches of focusing on the elimination of waste in all forms, and smooth, efficient flow of materials and information throughout the value chain to obtain faster customer response (JIT), higher quality, and lower costs. Lean concepts were initially developed and implemented by the Toyota Motor Corporation.

  16. Four Basic Lean Principles • Elimination of Waste: Eliminate any activities that do not add value in an organization. Includes overproduction, waiting time, processing, inventory, and motion. • Increased Speed and Response: Better process designs allow efficient responses to customers’ needs and the competitive environment. • Improved Quality: Poor quality creates waste, so improving quality is essential to the lean environment. • Reduced Cost: Simplifying processes and improving efficiency translates to reduced costs.

  17. Push Production/Distribution Systems • Apush systemproduces finished goods inventory in advance of customer demand using a forecast of sales. • Parts and subassemblies are “pushed” through the operating system based on a predefined schedule that is independent of actual customer demand. • A traditional automobile factory and distribution system is a good example of a push system.

  18. Just-in-Time Systems (JIT) • In a pull system, employees at a given operation (work station) go to the source of the required parts, such as machining or subassembly, and withdraw the units as they need them • By pulling parts from each preceding workstation, the entire manufacturing process is synchronized to the final-assembly schedule. • Finished goods are made to coincide with the actual rate of customer demand, resulting in minimal inventories and maximum responsiveness.

  19. Just-in-Time Systems (JIT) • JIT systems are sometimes called a Kanban system. • AKanbanis a flag or a piece of paper that contains all relevant information for an order. • Slips, called Kanban cards, are circulated within the system to initiate withdrawal and production items through the production process. • The Kanban cards are simple visual controls.

  20. Just-in-Time Systems (JIT) • The withdraw Kanban authorizes the material handler to transfer empty containers to the storage area. Next, a production Kanban triggers production of parts. Finally, the full container is delivered to the material handler. • See Exhibit 17.3 for how a two-card Kanban System works.

  21. Number of Kanban Cards Required K= d(p + w)(1 + ) C Where: K = the number of Kanban cards in the operating system. d = the average production rate. p = the processing time per container. w = the waiting time of Kanban cards (that is, the time it takes to deliver the container). C = the capacity of a standard container in the proper units of measure (parts, items, etc.). JIT practice is to set the lot size or container size equal to about 5% to 20% of a days demand or between 20 to 90 minutes worth of demand. Also, the container is sized to be able to be moved by the operator without help, if possible.  = safety stock as a %, usually ranging from 0 to 1.

  22. Anna works on an assembly line where it takes her 20 minutes to produce 53 units of a product needed to fill a container. It takes her an additional 5 minutes to transport the container to Josh, who works at the next station. The company uses a safety stock of 15%. The current assembly line uses 5 kanbans between Anna's and Josh's stations. What is the demand for the product? • P=20 min, W= 5 min, C=53,  =.15, K=5 • Calculate d • K= d(p + w)(1 + ) • C • d(p + w)(1+ ) = KC • d(20+5)(1+.15)=5*53 • d= 9.217 /minute

  23. Computing the Number of Kanbans: an aspirin manufacturer has converted to JIT manufacturing using Kanban containers. They wish to determine the number of containers at the bottle filling operation which fills at a rate of 200 per hour. Each container holds 25 bottles, it takes 30 minutes to receive more bottles (processing plus delivery time) and safety stock is set at 10% .

  24. Chapter 9Facility Location

  25. Key Concepts Location analysis method Analysis should follow 3 steps • Identify dominant location factors • Develop location alternatives • Evaluate locations alternatives • Factor rating method (where to locate a production facility) –Choose the one with highest factor rating score • Load-distance model (where to locate a warehouse) –Choose the one with the lowest load-distance score

  26. Factor Rating Example The owners of Speedy Logistics, a company that provides overnight delivery of documents, are considering where to locate their new facility in the Midwest. They have narrowed their search down to two locations and have decided to use factor rating to make their decision. They have listed the factors they consider important and assigned a factor score to each location based on a 5-point scale 1 lowest, 5 highest). The information is shown here. What is the best location, and what is its factor rating score?      

  27. Choose location 1 as its factor rating score is 450, higher than that of location 2.

  28. Load-Distance Example Aalogistics Co. has just signed a contract to deliver products to three locations, and they are trying to decide where to put their new warehouse. The three delivery locations are Chicago, Kansas City, and Memphis. The two potential sites for the warehouse are Peoria and St. Louis. The x, y coordinates for the delivery locations and warehouses are as follows: The total quantity to be delivered to each destination is: 400to Chicago, 150 to Kansas City, and 100 to Memphis. Where to Locate? • Peoria • St. Louis Sites:

  29. Load Distance Model • Calculate the rectilinear distance: • 2. Multiply by the number of loads = 45 miles

  30. The warehouse should be located at Peoria with load-distance score at 8750

  31. Two potential warehouse locations: Mansfield vs. Springfield need to deliver to four locations: Cleveland, Columbus, Dayton and Cincinnati

  32. Distances to Springfield Location City Distance to Springfield (6,6.5) Cleveland (11,22) I11-6I + I22-6.5I = 20.5 Columbus (10,7) I10-6I + I7-6.5I = 4.5 Cincinnati (4,1) I4-6I + I1-6.5I = 7.5 Dayton (3,6) I3-6I + I6-6.5I = 3.5

  33. Distances to Mansfield Location City Distance to Mansfield (11,14) Cleveland (11,22) I11-11I + I22-14I = 8 Columbus (10,7) I10-11I + I7-14I = 8 Cincinnati (4,1) I4-11I + I1-14I = 20 Dayton(3,6) I3-11I + I6-14I = 16

  34. Identify Loads • City Load Between City & Warehouse • Cleveland 15 • Columbus 10 • Cincinnati 12 • Dayton 4

  35. LD Score for Springfield • City Distance to Springfield • Cleveland 20.5 x 15 =307.5 • Columbus 4.5 x 10 = 45 • Cincinnati 7.5 x 12= 90 • Dayton 3.5 x 4 = 14 • 456.5 • Load Distance Score = 456.5

  36. LD Score for Mansfield • City Distance to Mansfield • Cleveland 8 x 15 = 120 • Columbus 8 x 10 = 80 • Cincinnati 20 x 12 = 240 • Dayton 16 x 4 = 64 • 504 • Load Distance Score = 504

  37. The load-distance score for Mansfield is higher than for Springfield. The warehouse should be located in Springfield

  38. Chapter 8 Facility and Work Design

  39. Key Concepts • Facility Layout - specific arrangement of physical facilities • PURPOSES • to support business strategy, competitive priorities • Minimize delays in materials handling and customer movement • Maintain flexibility, use labor and space effectively • Promote high employee morale and customer satisfaction • Provide for good housekeeping and maintenance • Four Types of Facility Layout • 1. Process layout • 2. Product layout • 3. Hybrid cellular (group technology) layout • 4. Fixed-position layout

  40. Process layout design • – grouping of similar activities or consisting of similar • functional equipments • e.g. Grocery stores, Legal offices, Hospitals • 2. Product layout design • – arrangement based on the sequence of operations that is performed • e.g. automobile assembly lines, winemaking industry

  41. Key Concepts • Study process layout design • Using load-distance score to choose a layout plan • 2. Study product layout • Using Assembly line balancing to design work stations

  42. Process Layout Question The following from-to matrix shows daily customer trips between departments of Fresh Foods Grocery Their proposed layout looks like the following What is the load-distance for Fresh Foods proposed layout?

  43. 1 2 To evaluate multiple layout plans, choose the one with smallest load-distance score- a good proposed layout would require less walking

  44. Product Layout Design Grouping tasks among workstations so that each workstation has –in the ideal case- the same amount of work Step 1: Identify tasks & immediate predecessors Step 2: Calculate the cycle time Step 3: Determine the output rate Step 4: Compute the theoretical minimum number of workstations Step 5: Assign tasks to workstations (balance the line) Step 6: Compute efficiency, idle time & balance delay

  45. Layout Calculations • Step 2:Determine cycle time • Cycle time is determined by the bottleneck or station that takes the longest • Cycle time = Station A (50 seconds) since it is the bottleneck • Step 3:Determine output rate • Step 4: Compute the theoretical minimum number of stations • TM = number of stations needed to achieve 100% efficiency (every second is used) • Round up or down depending on circumstances Need 4 work stations

  46. Layout Calculations • Step 6: Compute efficiency, idle time & balance • delay Balance Delay = 1 - Assembly Line Efficiency or 1-1.00 = “0” idle time

  47. The following table shows the tasks required to assemble an aluminum storm door and the length of time needed to complete each task. Total task time: 232 • Calculate the cycle time • What is the maximum output per hour for this line with an operator at each station? • Calculate the minimum number of work stations required and round up • What is the efficiency of the line if the theoretical minimum number of stations is used? Round down. Use the combination of stations shown.

  48. bottleneck locate near E Total task time: 232 B F A C E H G D • Cycle time – is determined by the bottleneck or station that takes the • longest: 70 sec.

  49. 2. 55 B 70 3. 15 A 12 E F C You need 4 workstations G D 55 20 5 H 32 75 23 47 55 Balance the line: (A,D), (B), (C,E,G), (F,H), assign these tasks to 4 work stations

  50. 4. After the activities are grouped and assigned to 4 work stations, the new cycle time becomes 75 seconds.