This document contains Chapters 18 to 19 Chapter 18 GROUP TECHNOLOGY AND CELLULAR MANUFACTURING REVIEW QUESTIONS 18.1 What is group technology? Answer: As defined in the text, group technology is a manufacturing philosophy in which similar parts are identified and grouped together to take advantage of their similarities in design and production. 18.2 What is cellular manufacturing? Answer: As defined in the text, cellular manufacturing is an application of group technology in which dissimilar machines or processes have been aggregated into cells, each of which is dedicated to the production of a part or product family or a limited group of families. 18.3 What are the production conditions under which group technology and cellular manufacturing are most applicable? Answer: The conditions identified in the text are (1) the plant currently uses traditional batch production and a process type layout, which results in much material handling effort, high in process inventory, and long manufacturing lead times; and (2) it is possible to group the parts into part families. 18.4 What are the two major tasks that a company must undertake when it implements group technology? Answer: The two major tasks are (1) identifying the part families and (2) rearranging production machines into machine cells. 18.5 What is a part family? Answer: As defined in the text, a part family is a collection of parts that are similar either because of geometric shape and size or because similar processing steps are required in their manufacture. 18.6 What are the three methods for solving the problem of grouping parts into part families? Answer: The three methods are (1) intuitive grouping, a.k.a. visual inspection method, (2) parts classification and coding, and (3) production flow analysis. 18.7 What is the difference between a hierarchical structure and a chain-type structure in a classification and coding scheme? Answer: In a hierarchical structure, also known as a monocode, the interpretation of each successive symbol depends on the values of the preceding symbols. In a chain-type structure, also known as a polycode, the interpretation of each symbol in the sequence is always the same; it does not depend on the values of preceding symbols. 18.8 What is production flow analysis? Answer: As defined in the text, production flow analysis is a method for identifying part families and associated machine groupings that uses the information contained on production route sheets rather than part drawings. Work parts with identical or similar routings are grouped into part families. 18.9 What are the typical objectives when implementing cellular manufacturing? Answer: As enumerated in the text, the objectives are to (1) shorten manufacturing lead times, by reducing setup, work part handling, waiting times, and batch sizes; (2) reduce work-in-process inventory; (3) improve quality; (4) simplify production scheduling; and (5) reduce setup times. 18.10 What is the composite part concept, as the term is applied in group technology? Answer: The composite part concept is based on part families. It conceives of a hypothetical part for a given family that includes all of the design and manufacturing attributes of the family. In general, an individual part in the family will have some of the features that characterize the family, but not all of them. The composite part possesses all of the features. 18.11 What are the four common GT cell configurations, as identified in the text? Answer: The four GT cell configurations listed in the text are (1) single-machine cell, (2) group-machine cell with manual handling, (3) group-machine cell with semi-integrated handling, and (4) flexible manufacturing cell or flexible manufacturing system. 18.12 What is the key machine concept in cellular manufacturing? Answer: The key machine concept acknowledges that there is typically a certain machine in a cell that is more expensive to operate than the other machines or that performs certain critical operations. This machine is referred to as the key machine. It is important that the utilization of this key machine be high, even if it means that the other machines in the cell have relatively low utilizations. The other machines are referred to as supporting machines, and they should be organized in the cell to keep the key machine busy. 18.13 What is the difference between a virtual machine cell and a formal machine cell? Answer: Virtual machine cells involve the creation of part families and the dedication of equipment to the manufacture of these part families, but without the physical rearrangement of machines into formal cells. The machines in the virtual cell remain in their original locations in the factory. Formal machine cells represent the conventional GT approach in which a group of dissimilar machines are physically relocated into a cell that is dedicated to the production of one or a limited set of part families. 18.14 What is the principal application of group technology in product design? Answer: As indicated in the text, the principal application of GT in design is to implement a design retrieval system that reduces part proliferation. 18.15 What is the application of rank order clustering? Answer: The application of rank order clustering in GT is grouping machines into cells based on the part-machine incidence matrix, which in turn is based on route sheets. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Rank Order Clustering 18.1 (A) Apply the rank order clustering technique to the part-machine incidence matrix in the following table to identify logical part families and machine groups. Parts are identified by letters, and machines are identified numerically. Answer: 18.2 Apply the rank order clustering technique to the part-machine incidence matrix in the following table to identify logical part families and machine groups. Parts are identified by letters, and machines are identified numerically. Answer: 18.3 Apply the rank order clustering technique to the part-machine incidence matrix in the following table to identify logical part families and machine groups. Parts are identified by letters, and machines are identified numerically. Answer: 18.4 Apply the rank order clustering technique to the part-machine incidence matrix in the following table to identify logical part families and machine groups. Parts are identified by letters, and machines are identified numerically. Answer: Note: packs G, C, and F in group I require processing in machine group II (machines 2, 7, and 8), and packs D and H require processing in machine group I (machines 1 and 3). 18.5 The following table lists the weekly quantities and routings of ten parts that are being considered for cellular manufacturing in a machine shop. Parts are identified by letters and machines are identified numerically. For the data given, (a) develop the part-machine incidence matrix, and (b) apply the rank order clustering technique to the part-machine incidence matrix to identify logical part families and machine groups. Answer: Machine Cell Organization and Design 18.6 (A) Four machines used to produce a family of parts are to be arranged into a GT cell. The from-to data for the parts processed by the machines are shown in the table below. (a) Determine the most logical sequence of machines for this data. (b) Construct the network diagram for the data, showing where and how many parts enter and exit the system. (c) Compute the percentages of in-sequence moves, bypassing moves, and backtracking moves in the solution. (d) Develop a feasible layout plan for the cell. Answer: (c) % in-sequence moves = (50 + 40 + 50)/170 = 0.824 = 82.4% % bypassing moves = (20 + 10)/170 = 0.176 = 17.6% % backtracking moves = 0 (d) Layout plan: In-line sequence or U-shaped layout is appropriate for the given flows with no back tracking. 18.7 In Problem 18.5, two logical machine groups are identified by rank order clustering. For each machine group, (a) determine the most logical sequence of machines for this data. (b) Construct the network diagram for the data. (c) Compute the percentages of in-sequence moves, bypassing moves, and backtracking moves in the solution. Answer: (a) Hollier method applied to first machine group (machines 2, 3, 4, 7): 18.8 Five machines will constitute a GT cell. The from-to data for the machines are shown in the table below. (a) Determine the most logical sequence of machines for this data, and construct the network diagram, showing where and how many parts enter and exit the system. (b) Compute the percentages of in-sequence moves, bypassing moves, and backtracking moves in the solution. (c) Develop a feasible layout plan for the cell based on the solution. From: To: 1 2 3 4 5 1 0 10 80 0 0 2 0 0 0 85 0 3 0 0 0 0 0 4 70 0 20 0 0 5 0 75 0 20 0 Answer: (b) % in-sequence moves = (10 + 85 + 20)/360 = 0.319 = 31.9% % bypassing moves = (75 + 20 + 80)/360 = 0.486 = 48.6% % backtracking moves = 70/360 = 0.194 = 19.4% (c) Student exercise. There is no single correct solution for this design problem. 18.9 A GT machine cell contains three machines. Machine 1 feeds machine 2 which is the key machine in the cell. Machine 2 feeds machine 3. The cell is set up to produce a family of five parts (A, B, C, D, and E). The operation times for each part at each machine are given in the table below. The products are to be produced in the ratios 4:3:2:2:1, respectively. (a) If 35 hr/wk are worked, determine how many of each product will be made by the cell. (b) What is the utilization of each machine in the cell? Part Operation time Machine 1 Machine 2 Machine 3 A 4.0 min. 15.0 min. 10.0 min. B 15.0 min. 18.0 min. 7.0 min. C 26.0 min. 20.0 min. 15.0 min. D 15.0 min. 20.0 min. 10.0 min. E 8.0 min. 16.0 min. 10.0 min. Answer: (a) Compute time to produce units in given ratio: Machine 1: T = 4(4) + 3(15) + 2(26) + 2(15) + 1(8) = 151 min Machine 2: T = 4(15) + 3(18) + 2(20) + 2(20) + 1(16) = 210 min Machine 3: T = 4(10) + 3(7) + 2(15) + 2(10) + 1(10) = 121 min Machine 2 is the bottleneck machine that determines cell output Time available = 35(60) = 2100 min. Number of cycles to produce the products in the ratio given = 2100/210 = 10 cycles Thus, output = 10(4 + 3 + 2 + 2 + 1) = 10(12) = 120 pc (b) Machine 1 utilization U1 = 10(151)/2100 = 0.719 = 71.9% Machine 2 utilization U2 = 100% Machine 3 utilization U3 = 10(121)/2100 = 0.576 = 57.6% 18.10 A GT cell will machine the components for a family of parts. The parts come in several different sizes and the cell will be designed to quickly change over from one size to the next. This will be accomplished using fast-change fixtures and downloading the part programs from the plant computer to the CNC machines in the cell. The parts are rotational type, and so the cell must be able to perform turning, boring, facing, drilling, and cylindrical grinding operations. Accordingly, there will be several machine tools in the cell, of types and numbers to be specified by the designer. To transfer parts between machines in the cell, the designer may elect to use a belt or similar conveyor system. Any conveyor equipment of this type will be 0.4 m. wide. The arrangement of the various pieces of equipment in the cell is the principal problem to be considered. The raw work parts will be delivered into the machine cell on a belt conveyor. The finished parts must be deposited onto a conveyor that delivers them to the assembly department. The input and output conveyors are 0.4 m wide, and the designer must specify where they enter and exit the cell. The parts are currently machined by conventional methods in a process-type layout. In the current production method, there are seven machines involved but two of the machines are duplicates. "From-to" data have been collected for the jobs that are relevant to this problem. The from-to data indicate the number of work parts moved between machines during a typical 40 hr week. The two categories "parts in" and parts out" indicate parts entering and exiting the seven machine group. A total of 400 parts on average are processed through the seven machines each week. However, as indicated by the data, not all 400 parts are processed by every machine. Machines 4 and 5 are identical and assignment of parts to these machines is arbitrary. Average production rate capacity on each of the machines for the particular distribution of this parts family is given in the table below. Also given are the floor space dimensions of each machine in meters. Assume that all loading and unloading operations take place in the center of the machine. Operation 6 is currently a manual inspection operation. It is anticipated that this manual station will be replaced by a coordinate measuring machine (CMM). This automated inspection machine will triple throughput rate to 15 parts/hr from 5 parts/hr for the manual method. The floor space dimensions of the CMM are 2.0 m x 1.6 m. All other machines currently listed are to be candidates for inclusion in the new machine cell. (a) Analyze the problem and determine the most appropriate sequence of machines in the cell using the data contained in the From-To chart. (b) Construct the network diagram for the cell, showing where and how many parts enter and exit the cell. (c) Determine the utilization and production capacity of the machines in the cell as you have designed it. (d) Prepare a layout (top view) drawing of the GT cell, showing the machines, the robot(s), and any other pieces of equipment in the cell. (e) Write a one-page (or less) description of the cell, explaining the basis of your design and why the cell is arranged as it is. Answer: (a) Use Hollier method to analyze sequence. (d) and (e) Cell design and one page essay: Student exercises. There is no single correct solution to this design problem. Performance Metrics in Cell Operations 18.11 (A) A family of three parts is processed through a group-technology cell consisting of two machines. The table below lists production quantities (Qj), production times (Tpij), and machine fractions for each family member (fij). Assume the nonoperation times (Tno) are all equal to 20 min per machine. Determine (a) average hourly production rate for the cell, (b) utilization of each machine and average utilization of the cell, (c) manufacturing lead time, and (d) work-in-process. A spreadsheet calculator is recommended for this problem. Machine 1 Machine 2 Part Qj Tp1 (min) f1j Tp2 (min) f2j A 1 4.00 0.400 3.00 0.300 B 1 3.00 0.200 6.00 0.400 C 1 5.00 0.250 6.00 0.300 Answer: A spreadsheet calculator was used to perform the calculations. Hourly production rates for each machine and family member were computed using Equation (18.2). The quantities of each family member produced in one hour were calculated as Qij = fijRpij. The total for each column represents the hourly output of all parts from each machine (13.0 pc/hr). Finally, the MLT values were obtained from MLTj = Tp1j + Tp2j + 2Tno. These were averaged to obtain 99.7 min = 1.661 hr. Summary: (a) Hourly production rate for each machine and for the cell Rp = 13.0 pc/hr Note that the two machines have the same production rate, which means that the workload in the cell is balanced. (b) Utilization for each machine is given by Σfij for each machine i. Thus, U1 = 0.85, and U2 = 1.0. Average cell utilization U = 0.925. (c) Average manufacturing lead time MLT = 49 min = 0.817 hr (d) Average work-in-process WIP = (13.0 pc/hr)(0.8167 hr) = 10.6 pc 18.12 A group-technology cell has three machines and is used to process a family of four parts. The table below lists production quantities (Qj), production times (Tpij), and machine fractions for each family member (fij). Assume the nonoperation times (Tno) are all the same at 40 min per machine. Determine (a) average hourly production rate for the cell, (b) utilization of each machine and average utilization of the cell, (c) manufacturing lead time, and (d) work-in-process. A spreadsheet calculator is recommended for this problem. Machine 1 Machine 2 Machine 3 Part Qj Tp1 (min) f1j Tp2 (min) f2j Tp3 (min) f3j A 1 4.0 0.4 3.0 0.3 2.0 0.2 B 1 2.0 0.2 4.0 0.4 1.8 0.18 C 1 5.0 0.1 6.0 0.12 2.5 0.05 D 1 4.0 0.3 2.0 0.15 3.33 0.25 Answer: A spreadsheet calculator was used to perform the calculations. Hourly production rates for each machine and family member were computed using Equation (18.2). The quantities of each family member produced in one hour were calculated as Qij = fijRpij. The total for each column represents the hourly output of all parts from each machine (17.7 pc/hr). Finally, the MLT values were obtained from MLTj = Tp1j + Tp2j + Tp3j + 3Tno. These were averaged to obtain 129.9 min = 2.165 hr. Summary: (a) Hourly production rate for each machine and for the cell Rp = 17.7 parts/hr Note that all three machines have the same production rate, which means that the workload in the cell is balanced. (b) Utilization for each machine is given by Σfij for each machine i. Thus, U1 = 1.0, U2 = 0.97, and U3 = 0.68. Average cell utilization U = 0.883. (c) Average manufacturing lead time MLT = 129.9 min = 2.165 hr (d) Average work-in-process WIP = (17.7 parts/hr)(2.165 hr) = 38.3 parts 18.13 Three machines are used in a group-technology cell to process a family of five parts. The table below lists production quantities (Qj), production times (Tpij), and machine fractions for each family member (fij). Part E requires a changeover time of 5 min on each machine so it is produced in batches of 5 parts. Nonoperation times (Tno) are the same for all parts and machines: 25 min per machine. Determine (a) average hourly production rate for the cell, (b) utilization of each machine and average utilization of the cell, (c) manufacturing lead time, and (d) work-in-process. A spreadsheet calculator is recommended for this problem. Machine 1 Machine 2 Machine 3 Part Qj Tp1 (min) f1j Tp2 (min) f2j Tp3 (min) f3j A 1 1.50 0.200 2.25 0.300 1.50 0.200 B 1 2.50 0.200 4.00 0.320 2.25 0.180 C 1 4.0 0.100 2.00 0.050 1.00 0.025 D 1 1.75 0.100 2.10 0.120 3.50 0.200 E 5 3.00 0.250 2.52 0.210 2.40 0.200 Answer: A spreadsheet calculator was used to perform the calculations. Hourly production rates for each machine and family member were computed using Equation (18.2). The quantities of each family member produced in one hour were calculated as Qij = fijRpij. The total for each column represents the hourly output of all parts from each machine (22.73 pc/hr). Finally, the MLT values were obtained from MLTj = Tp1j + Tp2j + Tp3j + 3Tno. These were averaged to obtain 82.25 min = 1.371 hr. Summary: (a) Hourly production rate for each machine and for the cell Rp = 22.73 parts/hr Note that all three machines have the same production rate, which means that the workload in the cell is balanced. (b) Utilization for each machine is given by Σfij for each machine i. Thus, U1 = 0.85, U2 = 1.00, and U3 = 0.805. Average cell utilization U = 0.885. (c) Average manufacturing lead time MLT = 82.25 min = 1.371 hr (d) Average work-in-process WIP = (22.73 parts/hr)(1.371 hr) = 31.2 parts 18.14 A group-technology cell consists of four machines and processes a family of six part styles in equal quantities. During eight hr, a total of 180 parts are produced, and each part spends 32 min in the cell on average, either being processed by one of the machines or waiting to be processed. All six part styles are processed through all four machines in the same order. One machine is the key machine which is utilized 100%. The other three machines have utilizations ranging between 60% and 85%. Five of the parts in the family require no changeover time on any machine, but the sixth part requires a changeover time of 4 min on the key machine. The six parts are currently produced consecutively, so a disproportionate amount of the time of the key machine is spent processing the sixth part. (a) What is the average work-in-process (number of parts) in the cell at any moment? A proposal has been made to process the sixth part in batches, so the 4-min changeover time would be spread over the number of units in the batch. How many hours would be required to process the same 180 parts if the sixth part were processed in batch sizes of (b) 10 parts and (c) 30 parts? Answer: (a) If 180 parts are processed in 8 hr, the hourly rate Rp = 180/8 = 22.5 pc/hr. Given that each part spends 32 min in the cell, MLT = 32 min. WIP = Rp(MLT) = 22.5(32/60) = 12 parts in the cell at any moment. (b) A total of 180 parts are produced of a family of 6 part styles in equal quantities. That means 180/6 = 30 of each part style. Let Tc = the average cycle time to process one part. For five of the parts, Tp = Tsu + Tc = 0 + Tc. For the sixth part Tp = 4 + Tc. Total time TT = 8 hr = 480 min. TT = 480 = 30(5)Tc + 30(4 + Tc) = 150Tc + 120 + 30Tc = 180Tc + 120 180Tc = 480 – 120 = 360 Tc = 360/180 = 2.0 min If the sixth part were processed in a batch size of 10 parts, TT = 30(5)(2.0) + 3(4 + 10 × 2.0) = 300 + 12 + 60 = 372 min = 6.2 hr (c) If the sixth part were processed in a batch size of 30 parts, TT = 30(5)(2.0) + (4 + 30 × 2.0) = 300 + 4 + 60 = 364 min = 6.067 hr 18.15 A GT assembly cell consisting of five workstations and five workers produces eight similar subassemblies at an average rate of 16 units/hr. All subassemblies go through all five stations. The number of parts in the cell at any moment is twice the number of workers. The lead worker in the cell claims that the time a given work unit spends in the cell is less than 20 min. His reasoning is that with five workers each producing at 16 units/hr, the average time at each station = 60/16 = 3.75 min. With five stations, the total time is 5 × 3.75 = 18.75 min. Is he correct or is the time a given work unit spends in the cell more than what he claims? Answer: Rp = 16 pc/hr and WIP = 2(5) = 10 pc, so MLT = 10/16 = 0.625 hr = 37.5 min The lead assembly worker is not correct. He probably never heard of Little’s formula. Chapter 19 FLEXIBLE MANUFACTURING CELLS AND SYSTEMS REVIEW QUESTIONS 19.1 Name three production situations in which FMS technology can be applied? Answer: The situations are similar to those identified for cellular manufacturing. They are the following: (1) Presently the plant either produces parts in batches or uses manned GT cells and management wants to automate. (2) It must be possible to group a portion of the parts made in the plant into part families, whose similarities permit them to be processed on the machines in the flexible manufacturing system. (3) The parts or products made by the facility are in the mid-volume, mid-variety production range. The appropriate production volume range is 5000 to 75,000 parts per year. 19.2 What is a flexible manufacturing system? Answer: The definition provided in the text is the following: A flexible manufacturing system (FMS) is a highly automated GT machine cell, consisting of a group of processing workstations (usually CNC machine tools), interconnected by an automated material handling and storage system, and controlled by a distributed computer system. The reason the FMS is called flexible is that it is capable of processing a variety of different part styles simultaneously at the various workstations, and the mix of part styles and quantities of production can be adjusted in response to changing demand patterns. 19.3 What are the three capabilities that a manufacturing system must possess in order to be flexible? Answer: The three capabilities that a manufacturing system must possess in order to be flexible are (1) the ability to identify and distinguish among the different incoming part or product styles processed by the system, (2) quick changeover of operating instructions, and (3) quick changeover of physical setup. 19.4 Name the four tests of flexibility that a manufacturing system must satisfy in order to be classified as flexible. Answer: The four tests of flexibility that a manufacturing system must satisfy in order to be classified as flexible are the following, as identified in the text: (1) Part variety test. Can the system process different part styles in a non-batch mode? (2) Schedule change test. Can the system readily accept changes in production schedule: changes in either part mix or production quantities? (3) Error recovery test. Can the system recover gracefully from equipment malfunctions and breakdowns, so that production is not completely disrupted? (4) New part test. Can new part designs be introduced into the existing product mix with relative ease? 19.5 What is the dividing line between a flexible manufacturing cell and a flexible manufacturing system, in terms of the number of processing stations in the system? Answer: The dividing line is between 3 and 4 processing stations. A flexible manufacturing cell has 2 or 3 processing stations, while a flexible manufacturing system has 4 or more processing stations. 19.6 What is the difference between a dedicated FMS and a random-order FMS? Answer: As defined in the text, a dedicated FMS is designed to produce a limited variety of part styles, and the complete universe of parts to be made on the system is known in advance. By comparison, a random-order FMS is more flexible and is more appropriate when the part family is large, there are substantial variations in part configurations, there will be new part designs introduced into the system and engineering changes in parts currently produced, and the production schedule is subject to change from day to day. 19.7 What are the three basic components of a flexible manufacturing system? Answer: The text lists the following as the three basic components: (1) workstations, (2) material handling and storage system, and (3) computer control system. People are required to manage and operate the system. 19.8 Name the seven functions performed by human resources in an FMS. Answer: The seven functions listed in the text are (1) loading raw work parts into the system, (2) unloading finished parts (or assemblies) from the system, (3) changing and setting tools, (4) performing equipment maintenance and repair, (5) performing NC part programming, (6) programming and operating the computer system, and (7) managing the system. 19.9 What are the five functions of the material handling and storage system in a flexible manufacturing system? Answer: The five functions of the material handling and storage system identified in the text are the following: (1) random independent movement of work parts between stations; (2) handling of a variety of work part configurations; (3) temporary storage; (4) convenient access for loading and unloading work parts; and (5) compatibility with computer control. 19.10 What is the difference between the primary and secondary handling systems that are common in flexible manufacturing systems? Answer: The primary handling system establishes the basic layout of the FMS and is responsible for moving parts between stations in the system, whereas the secondary handling system consists of transfer devices, automatic pallet changers, and similar mechanisms located at the workstations in the FMS. The function of the secondary handling system is to transfer work from the primary system to the workstation and to position parts with sufficient accuracy and repeatability to perform the processing or assembly operation. 19.11 The text lists four categories of layout configurations that are found in flexible manufacturing systems. Name them. Answer: The four categories listed in the text are (1) in-line layout, (2) loop layout, (3) open field layout, and (4) robot-centered layout. 19.12 What are the benefits that can be expected from a successful FMS installation? Answer: The text lists the following benefits of a successful FMS installation: (1) increased machine utilization, (2) fewer machines required to accomplish the same amount of work, (3) reduction in factory floor space required, (4) greater responsiveness to change, (5) reduced inventory requirements, (6) lower manufacturing lead times, (7) reduced labor requirements and higher labor productivity, and (8) opportunity for unattended operation. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Bottleneck Model 19.1 (A) A flexible machining cell consists of two machining workstations and a load/unload station. Station 1 is the load/unload station with one server (human worker). Station 2 consists of two identical CNC milling machines. Station 3 has one CNC drill press. The stations are connected by a part-handling system that has two work carriers. The mean transport time is 1.5 min. The FMC produces two parts, A and B. The part mix fractions and process routings for the two parts are presented in the table below. The operation frequency fijk = 1.0 for all operations. Determine (a) maximum production rate of the FMS, (b) corresponding production rates of each product, (c) utilization of each station, and (d) number of busy servers at each station. A spreadsheet calculator is recommended for this problem. Part j Part mix pj Operation k Description Station i Process time Tcijk A 0.4 1 Load 1 4 min 2 Mill 2 30 min 3 Drill 3 10 min 4 Unload 1 2 min B 0.6 1 Load 1 4 min 2 Mill 2 40 min 3 Drill 3 15 min 4 Unload 1 2 min Answer: The computations were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the top row of the table. Station 2 has the highest WL/s ratio so it is the bottleneck station. Equation (19.2) (19.4) (19.7) (19.10) Station # Servers WL WL WL/s U BS 1 (Load/unload) 1 6 6 0.333 0.333 2 (Mill) 3 36 18* 1 2 3 (Drill) 2 13 13 0.722 0.722 4 (Part handling) 2 4.5 2.25 0.125 0.25 (a) To compute the FMS production rate, use Equation (19.5): Rp* = 2/36.0 = 0.05555 pc/min = 3.333 pc/hr (b) For the production rate of each product, multiply Rp* by its respective part mix fraction. RpA* = 3.333(0.4) = 1.333 pc/hr RpB* = 3.333(0.6) = 2.00 pc/hr (c) Utilizations of stations are listed in the sixth column of the spreadsheet table (labeled U). (d) Mean number of busy servers at each station are listed in the seventh column of the spreadsheet table (Labeled BS). 19.2 (A) A flexible manufacturing cell consists of two machining workstations plus a load/unload station. The load/unload station is station 1 with one server (human worker). Station 2 consists of one CNC machining center. Station 3 has one CNC drill press. The three stations are connected by a part-handling system that has one work carrier. The mean transport time is 2.0 min. The FMC produces three parts, A, B, and C. The part mix fractions and process routings for the three parts are presented in the table below. The operation frequency fijk = 1.0 for all operations. Determine (a) maximum production rate of the FMC, (b) corresponding production rates of each product, (c) utilization of each machine in the system, and (d) number of busy servers at each station. A spreadsheet calculator is recommended for this problem. Part j Part mix pj Operation k Description Station i Process time Tcijk A 0.2 1 Load 1 3 min 2 Mill 2 20 min 3 Drill 3 12 min 4 Unload 1 2 min B 0.3 1 Load 1 3 min 2 Mill 2 15 min 3 Drill 3 30 min 4 Unload 1 2 min C 0.5 1 Load 1 3 min 2 Drill 3 14 min 3 Mill 2 22 min 4 Unload 1 2 min Answer: The computations were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the top row of the table. Station 2 has the highest WL/s ratio so it is the bottleneck station. Equation (19.2) (19.4) (19.7) (19.10) Station # Servers WL WL WL/s U BS 1 (Load/unload) 1 5 5 0.256 0.256 2 (Mill) 1 19.5 19.5* 1.0 1.0 3 (Drill) 1 18.4 18.4 0.944 0.944 4 (Part handling) 1 7.5 7.5 0.385 0.385 (a) To compute the FMS production rate, use Equation (19.5): Rp* = 1/19.5 = 0.05128 pc/min = 3.077 pc/hr (b) For the production rate of each product, multiply Rp* by its respective part mix fraction. RpA* = 0.05128(0.2) = 0.01026 pc/min = 0.615 pc/hr RpB* = 0.05128(0.3) = 0.01538 pc/min = 0.923 pc/hr RpC* = 0.05128(0.5) = 0.02564 pc/min = 1.538 pc/hr (c) Station utilizations are listed in the sixth column of the spreadsheet table (labeled U). (d) Mean number of busy servers at each station are listed in the seventh column of the spreadsheet table (labeled BS). 19.3 Solve the previous problem except the number of servers at station 2 (CNC milling machines) = 3 and the number of servers at station 3 (CNC drill presses) = 2. Note that with the increase in the number of machines from two to five, the FMC is now an FMS according to the definitions in Section 19.1.2. A spreadsheet calculator is recommended for this problem. Answer: The computations were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the top row of the table. Station 3 has the highest WL/s ratio so it is the bottleneck station. Equation (19.2) (19.4) (19.7) (19.10) Station # Servers WL WL WL/s U BS 1 (Load/unload) 1 5 5 0.543 0.543 2 (Mill) 3 19.5 6.5 0.707 2.120 3 (Drill) 2 18.4 9.2* 1.0 2.0 4 (Part handling) 1 7.5 7.5 0.815 0.815 (a) To compute the FMS production rate, use Equation (19.5): Rp* = 2/18.4 = 0.1087 pc/min = 6.522 pc/hr (b) For the production rate of each product, multiply Rp* by its respective part mix fraction. RpA = 0.1087(0.2) = 0.02174 pc/min = 1.304 pc/hr RpB = 0.1087(0.3) = 0.03261 pc/min = 1.957 pc/hr RpC = 0.1087(0.5) = 0.05435 pc/min = 3.261 pc/hr (c) Station utilizations are listed in the sixth column of the spreadsheet table (labeled U). (d) Mean number of busy servers at each station are listed in the seventh column of the spreadsheet table (Labeled BS). 19.4 An FMS consists of three stations plus a load/unload station. Station 1 loads and unloads parts using two servers (material handling workers). Station 2 consists of two identical CNC horizontal milling machines. Station 3 consists of three identical CNC vertical milling machines. Station 4 consists of two identical CNC drill presses. The machines are connected by a part-handling system that has two work carriers and a mean transport time = 2.0 min. The FMS produces four parts, A, B, C, and D, whose part mix fractions and process routings are presented in the table below. The operation frequency fijk = 1.0 for all operations. Determine (a) maximum production rate of the FMS, (b) utilization of each machine in the system, and (c) average utilization of the regular stations. A spreadsheet calculator is recommended for this problem. Answer: The computations were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the top row of the table. Station 4 has the highest WL/s ratio so it is the bottleneck station. Equation (19.2) (19.4) (19.7) Station # Servers WL WL WL/s U 1 (Load/unload) 2 7.0 3.5 0.520 2 (H Mill) 2 8.7 4.35 0.647 3 (V Mill) 3 11.3 3.77 0.560 4 (Drill) 2 13.45 6.725* 1.0 5 (Part handling) 2 7.2 3.6 0.535 (a) To compute the FMS production rate, use Equation (19.5): Rp* = 2/13.45 = 0.1487 pc/min = 8.92 pc/hr (b) Station utilizations are listed in the sixth column of the spreadsheet table (labeled U). 19.5 Solve Problem 19.4 except the number of drill presses = 3. A spreadsheet calculator is recommended for this problem. Answer: The computations were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the top row of the table. Station 4 has the highest WL/s ratio so it is the bottleneck station. (a) To compute the FMS production rate, use Equation (19.5): Rp* = 3/13.45 = 0.223 pc/min = 13.38 pc/hr (b) Station utilizations are listed in the sixth column of the spreadsheet table (labeled U). Comment: Station 4 is still the bottleneck, but FMS production rate and utilization have increased significantly compared to Problem 19.4. 19.6 A semi-automated flexible manufacturing cell is used to produce three products. The products are made by two automated processing stations followed by an assembly station. There is also a load/unload station. Material handling between stations is accomplished by mechanized carts that move tote bins containing the particular components to be processed and then assembled into a given product. The carts are kept busy while the tote bins are queued in front of the workstations. Each tote bin remains with the product throughout processing and assembly. The details of the FMC can be summarized as follows: Station Description Number of servers 1 Load and unload 2 human workers 2 Process X 1 automated server 3 Process Y 1 automated server 4 Assembly 2 human workers 5 Transport Number of carriers to be determined The product mix fractions and station processing times for the parts are presented in the table below. The same station sequence is followed by all products: 1 → 2 → 3 → 4 → 1. Product j Product mix pj Station 1 Station 2 Station 3 Station 4 Station 1 A 0.25 3 min 10 min 13 min 5 min 2 min B 0.45 3 min 5 min 8 min 10 min 2 min C 0.30 3 min 7 min 6 min 8 min 2 min The average cart transfer time between stations is 2.0 min. (a) What is the bottleneck station in the FMC, assuming that the material handling system is not the bottleneck? (b) At full capacity, what is the overall production rate of the system and the rate for each product? (c) What is the minimum number of carts in the material handling system required to keep up with the production workstations? (d) Compute the average utilization of the regular stations in the FMC. (e) What recommendations would you make to improve the efficiency and/or reduce the cost of operating the FMC? A spreadsheet calculator is recommended for this problem. Answer: The computations were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the top row of the table. Station 3 has the highest WL/s ratio so it is the bottleneck station. Equation (19.2) (19.4) (19.7) Station # Servers WL WL WL/s U 1 (Load/unload) 2 5.0 2.5 0.289 2 (Process X) 1 5.4 5.35 0.618 3 (Process Y) 1 8.65 8.650* 1.0 4 (Assembly) 2 8.15 4.075 0.471 5 (Part handling) ? 8.0 ? ? (a) To compute the FMS production rate, use Equation (19.5): Rp* = 1/8.65 = 0.1156 pc/min = 6.94 pc/hr (b) Bottleneck is station 3: Rp* = 1/6.85 = 0.146 pc/min = 8.76 pc/hr RpA = 0.25(6.94) = 1.73 pc/hr RpB = 0.45(6.94) = 3.12 pc/hr RpC = 0.3(6.94) = 2.08 pc/hr (c) Minimum number of carts required in system: 8.0/nc = 8.65 Rearranging, nc = 8/8.65 = 0.92 → Use 1 cart Comment: With one cart, the utilization of the part-handling system U5 = 0.925. Perhaps two carts should be used instead of only one. It would be a shame if the transport system became the bottleneck station at times. (e) Recommendations: (1) reduce number of servers at station 1 to 1 worker, and (2) reduce number of servers at station 4 to 1 server. 19.7 (A) An FMS consists of four stations. Station 1 is a load/unload station with one server. Station 2 consists of three identical CNC milling machines. Station 3 consists of two identical CNC drill presses. Station 4 is an inspection station with one server that performs inspections on a sampling of the parts. The stations are connected by a part-handling system that has two work carriers and whose mean transport time = 2.0 min. The FMS produces four parts, A, B, C, and D. The part mix fractions and process routings for the four parts are presented in the table below. Note that the operation frequency at the inspection station (f4jk) is less than 1.0 to account for the fact that only a fraction of the parts are inspected. Determine (a) maximum production rate of the FMS, (b) corresponding production rate of each part, (c) utilization of each station in the system, and (d) average utilization of the regular stations (excluding the part-handling system). A spreadsheet calculator is recommended for this problem. Answer: The computations were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the top row of the table. Station 3 has the highest WL/s ratio so it is the bottleneck station. Equation (19.2) (19.4) (19.7) Station # Servers WL WL WL/s U 1 (Load/unload) 1 4.85 4.85 0.674 2 (Mill) 3 19.0 6.33 0.880 3 (Drill) 2 14.4 7.20* 1.000 4 (Inspect) 1 4.0 4.00 0.556 5 (Part handling) 2 5.75 2.87 0.399 (a) To compute the FMS production rate, use Equation (19.5): Rp* = 2/14.4 = 0.1389 pc/min = 8.33 pc/hr (b) For the production rate of each product, multiply Rp* by its respective part mix fraction. RpA = 0.1389 (0.1) = 0.01389 pc/min = 0.833 pc/hr RpB = 0.1389 (0.2) = 0.02778 pc/min = 1.667 pc/hr RpC = 0.1389 (0.3) = 0.04167 pc/min = 2.50 pc/hr RpD = 0.1389 (0.4) = 0.05556 pc/min = 3.333 pc/hr (c) Station utilizations are listed in the sixth column of the spreadsheet table (labeled U). 19.8 A flexible manufacturing system produces a family of 15 parts and consists of six station types, including the part-handling system. The number of servers and workloads for each station are given in the table below. Determine (a) hourly production rate of the system and (b) utilization of each station. (c) If the number of machines at station 3 were increased from 2 to 3, what effect would this have on hourly production rate of the system? (d) If the number of carts at station 6 were reduced from 2 to 1, what effect would this have on hourly production rate of the system? (e) If the number of workers at station 1 were reduced from 2 to 1, what effect would this have on hourly production rate of the system? A spreadsheet calculator is recommended for parts (a) and (b) of this problem. Station Description Number of servers Workload (min) 1 Load and unload 2 human workers 17.8 2 Mill 3 machines 34.4 3 Drill 2 machines 15.0 4 Turn 1 machine 10.5 5 Inspection 1 human worker 8.1 6 Part handling 2 carts 8.3 Answer: The computations were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the top row of the table. Station 2 has the highest WL/s ratio so it is the bottleneck station. Equation (19.2) (19.4) (19.7) Station # Servers WL WL WL/s U 1 (Load/unload) 2 17.8 8.9 0.776 2 (Mill) 3 34.4 11.47 * 1.000 3 (Drill) 2 15.0 7.5 0.654 4 (Turn) 1 10.5 10.5 0.916 5 (Inspection) 1 8.1 8.1 0.706 6 (Part handling) 2 8.3 4.15 0.362 (a) Rp* = 3/34.4 = 0.0872 pc/min = 5.23 pc/hr (b) Station utilizations are given in the sixth column of the table. (c) Because station 2 is the bottleneck station, increasing the number of servers at station 3 would have no effect on hourly production rate. (d) Utilization at station 6 is only 0.362, so reducing the number of carts from 2 to one would seem to have no effect on hourly production rate, according to the bottleneck model. However, the bottleneck model does not consider possible queuing effects that might result in getting parts to stations; thus, starving of stations becomes a significant risk with only one cart. (e) Reducing the number of workers from 2 to 1 at station 1 would result in that station becoming the bottleneck station. Production rate would be reduced to 1/17.8 = 0.0562 pc/min = 3.37 pc/hr. Extended Bottleneck Model 19.9 (A) Use the extended bottleneck model to solve for the overall production rate, manufacturing lead time, and waiting time for the data in problem 19.2 with the following number of parts in the system: (a) N = 2 parts and (b) N = 4 parts. Answer: From the solution to Problem 19.2, MLT1 = 5.0 + 19.5 + 18.4 + 7.5 = 50.4 min Rp* = 0.05128 pc/min from solution to Problem 19.2. N* = 0.05128(50.4) = 2.58 (a) For N = 2 N* = 2.58, apply case 2 Rp* = 0.05128 pc/min = 3.077 pc/hr MLT2 = 4/0.05128 = 78.0 min, and Tw = 78.0 - 50.4 = 27.6 min 19.10 Use the extended bottleneck model to solve for the overall production rate, manufacturing lead time, and waiting time for the data in problem 19.3 with the following number of parts in the system: (a) N = 3 parts and (b) N = 6 parts. Answer: From the solution to Problem 19.3, MLT1 = 5.0 + 19.5 + 18.4 + 7.5 = 50.4 min Rp* = 0.1087 pc/min from solution to Problem 19.3. N* = 0.1087(50.4) = 5.48 (a) For N = 3 N* = 5.48, apply case 2 Rp* = 0.1087 pc/min = 6.52 pc/hr MLT2 = 6/0.1087 = 55.2 min, and Tw = 55.2 - 50.4 = 4.8 min 19.11 Use the extended bottleneck model to solve for the overall production rate, manufacturing lead time, and waiting time for the data in problem 19.4 with the following number of parts in the system: (a) N = 5 parts, (b) N = 8 parts, and (c) N = 12 parts. (d) Also determine the average utilization of the regular workstations, excluding the part-handling system, for the three cases of N in (a), (b), and (c). Answer: From the solution to Problem 19.4, MLT1 = 7.0 + 8.7 + 11.3 + 13.45 + 7.2 = 47.65 min Rp* = 0.1487 pc/min from solution to Problem 19.4. N* = 0.1487(47.65) = 7.89 (a) For N = 5 N* = 7.09, apply case 2 Rp* = 0.1487 pc/min = 8.92 pc/hr MLT2 = 8/0.1487 = 53.8 min, and Tw = 53.8 - 53.05 = 0.75 min (c) N = 12 > N* = 7.89, apply case 2 Rp* = 0.1487 pc/min = 8.92 pc/hr MLT2 = 12/0.1487 = 80.7 min, and Tw = 80.7 - 53.05 = 27.65 min (d) To obtain the system utilization of the regular stations, the utilizations of the individual stations must be determined. The calculations are summarized in the following table: 19.12 Use the extended bottleneck model to solve for the overall production rate, manufacturing lead time, and waiting time for the data in Problem 19.5 with the following number of parts in the system: (a) N = 8 parts, (b) N = 11 parts, and (c) N = 15 parts. (d) Also determine the overall utilization of the regular workstations, excluding the part-handling system, for the three cases of N in (a), (b), and (c). Answer: From the solution to Problem 19.5, MLT1 = 7.0 + 8.7 + 10.5 + 11.25 + 7.2 = 47.65 min Rp* = 0.223 pc/min from solution to Problem 19.5. N* = 0.223(47.65) = 10.6 (a) For N = 8 N* = 10.6, apply case 2 Rp* = 0.223 pc/min = 13.38 pc/hr MLT2 = 11/0.223 = 49.33 minN , and Tw = 49.33 – 47.65 = 1.68 min (c) N = 15 > N* = 10.6, apply case 2 Rp* = 0.1778 pc/min = 10.67 pc/hr MLT2 = 15/0.223 = 67.26 min, and Tw = 67.26 – 47.65 = 19.61 min (d) To obtain the system utilization of the regular stations, the utilizations of the individual stations must be determined. The calculations are summarized in the following table: 19.13 For the data given in Problem 19.6, use the extended bottleneck model to develop the relationships for production rate Rp and manufacturing lead time MLT each as a function of the number of parts in the system N. Plot the relationships as in Figure 19.10. Answer: See solution for Problem 19.6. MLT1 = 5.0 + 5.4 + 6.85 + 8.15 + 8.0 = 35.15 min Rp* = 0.1156 pc/min = 6.94 pc/hr from Problem 19.6 solution. N* = 0.1156(35.15) = 4.06 Establish point at N = 10 > 4.06 (case 2): MLT2 = 10/0.1156 = 86.5 min Plot 19.14 A flexible manufacturing system is used to produce three products: A, B, and C. The FMS consists of a load/unload station, two automated processing stations, an inspection station, and an automated guided-vehicle system (AGVS) with an individual cart for each product. The conveyor carts remain with the parts during their time in the system, and therefore the mean transport time includes not only the move time, but also the average total processing time per part. The number of servers at each station is given in the following table: Station 1 Load and unload 2 workers Station 2 Process X 3 servers Station 3 Process Y 4 servers Station 4 Inspection 1 server Transport system Conveyor 8 carriers All parts follow either of two routings, which are 1 → 2 → 3 → 4 → 1 or 1 → 2 → 3 → 1, the difference being that inspections at station 4 are performed on only one part in four for each product (f4jk = 0.25). The product mix and process times for the parts are presented in the table below: Product j Part mix pj Station 1 Station 2 Station 3 Station 4 Station 1 A 0.2 5 min 15 min 25 min 20 min 4 min B 0.3 5 min 10 min 30 min 20 min 4 min C 0.5 5 min 20 min 10 min 20 min 4 min The move time between stations is 2 min. (a) Using the bottleneck model, show that the conveyor system is the bottleneck in the present FMS configuration, and determine the overall production rate of the system. (b) Determine how many carts are required to eliminate the AGVS as the bottleneck. (c) With the number of carts determined in (b), use the extended bottleneck model to determine the production rate for the case when N = 8; that is, only eight parts are allowed in the system even though the AGVS has a sufficient number of carriers to handle more than eight. (d) How close are your answers in (a) and (c)? Why? Answer: (a) WL1 = (5 + 4)(0.2 + 0.3 + 0.5) = 9.0 min WL2 = 15(0.2)(1.0) + 10(0.3)(1.0) + 20(0.5)(1.0) = 16.0 min WL3 = 25(0.2)(1.0) + 30(0.3)(1.0) + 10(0.5)(1.0) = 19.0 min WL4 = 20(0.2)(0.5) + 20(0.3)(0.25) + 20(0.5)(0.25) = 5.0 min nt = 1 + 1 + 0.25 + 1 = 3.25, WL5 = 3.25(2) + 9 + 16 + 19 + 5 = 55.5 min Bottleneck is station 5 (transport): Rp* = 8/55.5 = 0.144 pc/min = 8.65 pc/hr (b) Next bottleneck station after transport system = station 2. Find the minimum number of carriers to eliminate the transport system as the bottleneck. 55.5/nc = 5.333, nc = 55.5/5.333 = 10.4 → Use nc = 11 carriers With station 2 as bottleneck, Rp* = 3/16 = 0.1875 pc/min = 11.25 pc/hr (c) N* = 0.1875(55.5) = 10.4 Given N = 8, case 1 applies since 8 N* = 4.03, case 1 applies: Rp* = 0.09107 pc/min = 5.46 pc/hr MLT2 = 8/0.09107 = 87.8 min and Tw = 87.8 – 44.25 = 43.55 min (d) Station utilizations are given in the sixth column of the table. Sizing the FMS 19.17 (A) A flexible manufacturing cell is used to produce four parts. The FMC consists of one load/unload station and two automated processing stations (processes X and Y). The number of servers for each station type is to be determined. The FMS also includes an automated part-handling system with individual carts to transport parts between servers. The carts move the parts from one server to the next, drop them off, and proceed to the next delivery task. Average time required per transfer is anticipated to be 3.5 minutes. The following table summarizes the FMS: Station 1 Load and unload Number of human servers (workers) to be determined Station 2 Process X Number of automated servers to be determined Station 3 Process Y Number of automated servers to be determined Station 4 Transport system Number of carts to be determined All parts follow the same routing, which is 1 → 2 → 3 → 1. The product mix and processing times at each station are presented in the table below: Product j Product mix pj Station 1 Station 2 Station 3 Station 1 A 0.1 3 min 23 min 25 min 2 min B 0.3 3 min 16 min 18 min 2 min C 0.4 3 min 19 min 10 min 2 min D 0.2 3 min 35 min 13 min 2 min Required production is 10 parts per hour, distributed according to the product mix indicated. Use the bottleneck model to determine (a) the minimum number of servers at each station and the minimum number of carts in the transport system that are required to satisfy production demand and (b) the utilization of each station for the answers in part (a). A spreadsheet calculator is recommended for this problem. Answer: The computations were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the top row of the table. The workloads (WL) in columns 2 and 3 are computed for Rp = one pc/hr. For Rp = 10 pc/hr, the above workloads would be 10 times the WL values. Column 4 lists the specified Rp multiplied by WL, adjusted for 60 min/hr. (a) Column 5 rounds up the values in column 4 to the nearest integer. (b) Column 6 computes the utilization values for each station. A spreadsheet calculator is recommended for this problem. Equation (19.2) (19.4) (19.22) (19.7) Station WL WL 10WL/60 s U 1 (Load/unload) 5 0.833 1 0.833 2 (Process X) 21.7 3.617 4 0.904 3 (Process Y) 14.5 2.4167 3 0.806 4 (Part handling) 10.5 1.75 2 0.875 19.18 A flexible machining system is being planned that will consist of four workstations plus a part-handling system. Station 1 will be a load/unload station. Station 2 will consist of horizontal machining centers. Station 3 will consist of vertical machining centers. Station 4 will be an inspection station. Four the part mix that will be processed by the FMS, the workloads at the four stations are as follows: WL1 = 7.5 min, WL2 = 22.0 min, WL3 = 18.0 min, and WL4 = 10.2 min. The workload of the part-handling system WL5 = 8.0 min. The FMS will be operated 16 hours per day, 250 days per year. Maintenance will be performed during nonproduction hours, so uptime proportion (availability) is expected to be 97%. Annual production of the system will be 50,000 parts. Determine (a) the number of machines (servers) of each type (station) required to satisfy production requirements and (b) the utilization of each station. (c) What is the maximum possible production rate of the system if the bottleneck station were to operate at 100% utilization? Answer: Rp = 50,000/(4000 × 0.97) = 12.887 pc/hr = 0.2148 pc/min. The remaining computations were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the top row of the table. The workloads (WL) in columns 2 and 3 are computed for Rp = one pc/hr. Column 4 lists the specified Rp (0.2148 pc/min) multiplied by WL. (a) Column 5 rounds up the values in column 4 to the nearest integer. (b) Column 6 computes the utilization values for each station. Equation (19.2) (19.4) (19.22) (19.7) Station WL WL RpWL s U 1 (Load/unload) 7.5 1.61 2 0.805 2 (HMCs) 22 4.72 5 0.945 3 (VMCs) 18 3.87 4 0.967 4 (Inspection) 10.2 2.19 3 0.730 5 (Part handling) 8 1.72 2 0.859 (c) Maximum value of utilization is at station 3. If station 3 were to operate at 100% utilization, Rp* = 12.887/0.967 = 13.33 pc/hr 19.19 Given the part mix, process routings, and processing times for the three parts in Problem 19.2. The FMS planned for this part family will operate 250 days per year and the anticipated availability of the system is 90%. Determine how many servers at each station will be required to achieve an annual production rate of 40,000 parts per year if (a) the FMS operates 8 hours per day, (b) 16 hours per day, and (c) 24 hours per day. (d) Which system configuration is preferred, and why? A spreadsheet calculator is recommended for this problem. Answer: The computations were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the second row of the table. The workloads (WL) in columns 2 and 3 are computed for Rp = one pc/hr from the solution to Problem 2. Columns 4, 5, and 6 list the specified Rp multiplied by WL for each of the three cases, and these values are rounded up to the nearest integer: (a) Annual hours of operation = (8 hr/day)(250 days/yr) = 2000 hr/yr Rp = 40,000/(2000 × 0.90) = 22.22 pc/hr = 0.3704 pc/min (b) Annual hours of operation = (16 hr/day)(250 days/yr) = 4000 hr/yr Rp = 40,000/(4000 × 0.90) = 11.11 pc/hr = 0.1852 pc/min (c) Annual hours of operation = (24 hr/day)(250 days/yr) = 6000 hr/yr Rp = 40,000/(6000 × 0.90) = 7.407 pc/hr = 0.1234 pc/min (a) 8 hr/day (b) 16 hr/day (c) 24 hr/day Equation (19.2) (19.4) (19.22) (19.22) (19.22) Station WL WL RpWL → s RpWL → s RpWL → s 1 (Load/unload) 5 1.85 → 2 0.93 → 1 0.62 → 1 2 (Mill) 19.5 7.22 → 8 3.61 → 4 2.41 → 3 3 (Drill) 18.4 6.82 → 7 3.41 → 4 2.27 → 3 4 (Part handling) 7.5 2.78 → 3 1.39 → 2 0.93 → 1 (d) Configuration (c) is preferable because it requires the minimum capital investment and involves the minimum complexity. Labor cost to operate 24 hr/day would probably be higher even though there are fewer machines that must be managed. 19.20 Given the part mix, process routings, and processing times for the four parts in Problem 19.4. The FMS proposed to machine these parts will operate 20 hours per day, 250 days per year. Assume system availability = 95%. Determine (a) how many servers at each station will be required to achieve an annual production rate of 75,000 parts per year, and (b) the utilization of each workstation. (c) What is the maximum possible annual production rate of the system if the bottleneck station were to operate at 100% utilization? A spreadsheet calculator is recommended for this problem. Answer: (a) H = (20 hr/day)(250 days/yr) = 5000 hr/yr Rp = 75,000/(5000 × 0.95) = 15.789 pc/hr = 0.263 pc/min The remaining computations for parts (a) and (b) were performed using a spreadsheet calculator with the results shown below. Equations used to compute the entries are given in the top row of the table. The workloads (WL) in columns 2 and 3 are computed for Rp = one pc/hr from the solution to Problem 4. Column 4 lists the specified Rp (0.263 pc/min) multiplied by WL, adjusted for 60 min/hr. (a) Column 5 rounds up the values in column 4 to the nearest integer. (b) Column 6 computes the utilization values for each station. Equation (19.2) (19.4) (19.22) (19.7) Station WL WL RpWL s U 1 (Load/unload) 7.0 1.84 2 0.921 2 (H Mill) 8.7 2.29 3 0.763 3 (V Mill) 11.3 2.97 3 0.991 4 (Drill) 13.45 3.54 4 0.884 5 (Part handling) 7.2 1.89 2 0.947 (c) Maximum utilization is at station 3. If station 3 were to operate at 100% utilization, Rp* = 15.789/0.991 = 15.932 pc/hr Annual production = 15.932(20)(250)(0.95) = 75,679 pc/yr Solution Manual for Automation, Production Systems, and Computer-Integrated Manufacturing Mikell P. Groover 9780133499612, 9780134605463
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