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This document contains Chapters 13 to 15 Chapter 13 OVERVIEW OF MANUFACTURING SYSTEMS REVIEW QUESTIONS 13.1 What is a manufacturing system? Answer: The definition given in the text is the following: A manufacturing system is a collection of integrated equipment and human resources, whose function is to perform one or more processing and/or assembly operations on a starting raw material, part, or set of parts. 13.2 Name the four components of a manufacturing system. Answer: As listed in the text, the four components are (1) production machines plus tools, fixtures, and other related hardware, (2) a material handling system and/or work positioning system, (3) a computer system to coordinate and/or control the preceding components, and (4) human workers to operate and manage the system. 13.3 What are the three classifications of production machines, in terms of worker participation? Answer: In terms of worker participation, the machines can be classified as (1) manually operated, (2) semi-automated, and (3) fully automated. 13.4 What are the five material handling functions that must be provided in a manufacturing system? Answer: The five material handling functions that must be provided in a manufacturing system are (1) loading work units at each station, (2) positioning the work units at the station, (3) unloading the work units from the station, (4) transporting work units between stations in manufacturing systems comprised of multiple workstations, and (5) temporary storage of work units to prevent starving of workstations. 13.5 What is the difference between fixed routing and variable routing in manufacturing systems consisting of multiple workstations? Answer: In fixed routing, the work units always flow through the same sequence of workstations. This means that the work units are identical or similar enough that the processing sequence is the same. In variable routing, work units are transported through a variety of different station sequences. This means that the manufacturing system processes or assembles different styles of work units. 13.6 What is a pallet fixture in work transport in a manufacturing system? Answer: A pallet fixture is a work holder that is designed to be transported by the material handling system. The part is accurately attached to the fixture on the upper face of the pallet, and the under portion of the pallet is designed to be moved, located, and clamped in position at each workstation in the system. 13.7 A computer system is an integral component in a modern manufacturing system. Name four of the eight functions of the computer system listed in the text. Answer: The eight functions of the computer system identified in the text are (1) communicate instructions to workers, (2) download part programs to computer-controlled machines, (3) control the material handling system, (4) schedule production, (5) failure diagnosis, (6) safety monitoring, (7) quality control, and (8) operations management. 13.8 What are the four factors that can be used to distinguish manufacturing systems in the classification scheme proposed in the chapter? Answer: The four factors identified in the text are (1) types of operations performed, (2) number of workstations and system layout, (3) level of automation, and (4) system flexibility. 13.9 Why is manning level inversely correlated with automation level in a manufacturing system? Answer: Manning level is inversely correlated with automation level in a manufacturing system because the number of workers required to operate the system tends to be reduced as the level of automation increases. 13.10 What is flexibility in a manufacturing system? Answer: Flexibility is the capability that allows a manufacturing system to cope with a certain level of variation in part or product style without interruptions in production for changeovers between models. 13.11 What are the three capabilities that a manufacturing system must possess in order to be flexible? Answer: As identified in the text, the three capabilities are (1) identification of the different work units, (2) quick changeover of operating instructions, and (3) quick changeover of the physical setup. 13.12 Name the basic categories of manufacturing systems, as they are identified in the text. Answer: As identified in the text, the three categories are (1) single-station cells, (2) multistation systems with fixed routing, and (3) multi-station systems with variable routing. 13.13 What is a production line? Answer: As defined in the text, a production line consists of a series of workstations laid out so that the work unit moves from one station to the next, and a portion of the total work content is performed on it at each station. Chapter 14 SINGLE-STATION MANUFACTURING CELLS REVIEW QUESTIONS 14.1 Name three reasons why single-station manned cells are so widely used in industry. Answer: The reasons given in the text are the following: (1) It requires the shortest amount of time to implement. (2) It usually requires the least capital. (3) Technologically, it is the easiest system to install and operate. Its maintenance requirements are usually minimal. (4) For many situations, particularly for low quantities, it results in the lowest cost per unit produced. (5) In general, it is the most flexible manufacturing system with regard to changeovers from one part or product style to the next. 14.2 What does the term semi-automated station mean? Answer: As defined in the text, a semi-automated station is a machine that is controlled by some form of program during a portion of the work cycle, and the worker’s function is simply to load and unload the machine each cycle, and periodically change cutting tools. 14.3 What is a single-station automated cell? Answer: A single station automated cell is a fully-automated machine capable of unattended operation for a time period longer than one machine cycle. A worker is not required to be at the machine except periodically to load and unload parts or otherwise tend it. 14.4 What are the five enablers that are required for unattended operation of automated cells designed to produce identical parts or products? Answer: As given in the text, the five enablers of unattended operation of automated cells designed to produce identical parts or products are the following: (1) a programmed cycle that allows the machine to perform every step of the processing or assembly cycle automatically; (2) a parts storage system and a supply of parts that permit continuous operation beyond one machine cycle; (3) automatic transfer of work parts between the storage system and the machine (automatic unloading of finished parts from the machine and loading of raw work parts to the machine); (4) periodic attention of a worker who performs the necessary machine tending functions (e.g., parts loading and unloading of the storage subsystem) for the particular processing or assembly operation; and (5) built-in safeguards that protect the system against operating under conditions that may be unsafe for humans or self-destructive or destructive to the work units being processed or assembled. 14.5 What are the additional three enablers that are required for unattended operation of automated cells designed for product variety? Answer: As given in the text, the three enablers of unattended operation of automated cells designed for product variety are (1) a work identification system that can distinguish between the different starting work units entering the station, so that the correct processing sequence can be used for that part or product style; (2) a program downloading capability to transfer the machine cycle program corresponding to the identified part or product style; and (3) a quick setup changeover capability so that the necessary workholding devices and other tools for each part are available on demand. 14.6 What is an automatic pallet changer? Answer: An automatic pallet changer is a part handling subsystem that is used to exchange pallet fixtures between the machine tool worktable and the load/unload position or part storage system. 14.7 What is a machining center? Answer: As defined in the text, a machining center is a machine tool capable of performing multiple machining operations on a workpart in one setup under NC program control. Operations performed on a machining center are those that use a rotating cutting tool, such as milling, drilling, reaming, and tapping. 14.8 What is the difference between a horizontal machining center and a vertical machining center? Answer: A horizontal machining center has a horizontal spindle (to hold rotating cutting tools) while a vertical machining center has a vertical spindle. 14.9 What are some of the features on a NC machining center used to reduce nonproductive time in the work cycle? Answer: The text lists the following features: (1) automatic tool-changer; (2) automatic work part positioner; and (3) automatic pallet changer. 14.10 What is the difference between a turning center and a mill-turn center? Answer: A turning center performs turning and related fixed tooling operations, while a mill-turn center is also capable of performing operations that use rotating cutting tools. 14.11 What is a machine cluster? Answer: As defined in the text, a machine cluster is a collection of two or more machines producing parts or products with identical cycle times and serviced (usually loaded and unloaded) by one worker. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Parts Storage and Multitasking Machines 14.1 (A) A CNC machining center has a programmed cycle time of 25.0 min for a certain part. The time to unload the finished part and load a starting work unit = 5.0 min. (a) If loading and unloading are done directly onto the machine tool table and no automatic storage capacity exists at the machine, what are the cycle time and hourly production rate? (b) If the machine tool has an automatic pallet changer so that unloading and loading can be accomplished while the machine is cutting another part, and the repositioning time = 30 sec, what are the total cycle time and hourly production rate? (c) If the machine tool has an automatic pallet changer that interfaces with a parts storage unit whose capacity is 12 parts, and the repositioning time = 30 sec, what are the total cycle time and hourly production rate? Also, how long does it take to perform the loading and unloading of the 12 parts by the human worker, and what is the time the machine can operate unattended between parts changes? Answer: (a) Tc = 25.0 + 5.0 = 30.0 min/pc Rc = 60/30 = 2.0 pc/hr (b) Tc = Max(25.0, 5.0) + 0.5 = 25.5 min/pc Rc = 60/25.5 = 2.35 pc/hr (c) Tc = Max(25.0, 5.0) + 0.5 = 25.5 min/pc Rc = 60/25.5 = 2.35 pc/hr Time to load/unload = 12(5.0) = 60 min UT = 12(25.5) - 60 = 306 - 60 = 246.0 min = 4.1 hr 14.2 A machine shop operates one 8-hr shift, five days per week. Each part processed on a CNC machine tool of interest has a programmed cycle time of 37 min. At the end of each programmed cycle, a worker unloads and loads the machine which takes 5 min. Thus the total work cycle time is 42 min, and the worker is idle most of that cycle. The plant manager needs to increase output and is considering three alternatives: (1) install and automatic pallet changer (APC) that would have repositioning time of 30 sec, which would increase output slightly, (2) purchase a second machine, with no APC, which would double output, or (3) purchase an automated pallet storage and handling system with a storage capacity that is sufficient to allow the machine to operate overnight (two shifts), which would approximately triple the output. The repositioning time between the storage system and the CNC machine is 30 sec. The worker would unload and load the storage system at the beginning of the day shift and then be assigned other work until later in the shift when it would be time to load the storage system for overnight operation. Unloading and loading of the APC in (1) and the storage system in (3) take the same 6.0 min per cycle because both involve pallets. Determine (a) the output per week for each of the three alternatives, (b) the storage capacity of the storage system in alternative (3) that would achieve overnight operation, and (c) the amount of time the worker would be kept busy each day unloading and loading the storage system. Answer: (a) The work cycle time with the APC in (1) and the storage system in (3) is the same: Tc = 37 + 0.5 = 37.5 min. For alternative (1), weekly output Rpw = 40(60)/37.5 = 64 pc For alternative (2), weekly output is twice the current output, for which the cycle time Tc = 42 min. Thus weekly output = 2(40)(60)/42 = 114.3 rounded down to 114 pc For alternative (3), weekly output Rpw = 3(40)(60)/37.5 = 192 pc (b) The capacity of the storage system would have to be sufficient to operate for 16 hr (overnight operation) plus an allowance for unloading the last part produced during the night and loading at least one starting work unit. With a cycle time of 37.5 min, the minimum capacity would be np = UT/Tc = 16(60)/37.5 = 25.6. With an allowance of one part, np = 25.6 + 1 = 26.6 rounded to 27 pc (c) At the beginning of the shift, the worker would load enough parts for the 8-hr day shift, approximately half of the storage system capacity, say 14 parts. Later in the day, the worker would unload the parts finished during the shift and load enough starting parts to keep the machine busy overnight. The number of parts completed during the shift = 8(60)/37.5 = 12.8 rounded down to 12 parts with one part still on the machine tool. Thus, 27 – 2 = 25 parts to be loaded near the end of the shift. Time to load/unload in the morning = 14(6.0) = 84 min = 1.4 hr Time to load/unload in the afternoon = 25(6.0) = 150 min = 2.5 hr That’s a total of 3.9 hr or almost 50% of the 8-hr day. 14.3 A batch of 35 parts is ordered by a customer about every six months. The parts are currently processed sequentially through five conventional machines, listed in the table below with setup times and work cycle times per piece. These machines all require an operator to be in attendance during the work cycle. There is a delay of 10 hr/machine due to transport between machines and waiting in queues of other parts processed by these machines. A recommendation has been made to process the parts on a new multitasking machine that would complete the batch in one setup, which would involve a simple fixture and a setup time of 2.0 hr. The work cycle on the multitasking machine consists of the same operations that are accomplished by the five machines, but the cycle time would be less than the sum of the five cycle times by one-half because of tool path improvements and reduced part handling. Determine the manufacturing lead time (how long it takes to complete the batch of 35 parts, including delays) for processing on (a) the five conventional machines and (b) the multitasking machine. Assume that a delay of 10 hr would occur for the multitasking machine due to other work on that machine. Answer: (a) The sum of the setup times on the five conventional machines TTsu = 6.5 hr The sum of the work cycle times = QΣTc = 35(70) = 2450 min = 40.833 hr The delay (nonoperation) times TTno = 5(10) = 50 hr MLT for the five conventional machines = 6.5 + 40.833 + 50 = 97.333 hr (b) For the multitasking machine MLT = 2.0 + 35(70)/2 + 10 = 32.42 hr Determining Workstation Requirements 14.4 (A) A total of 9000 stampings must be produced in the press department during the next three days. Manually operated presses (one operator per press) will be used to complete the job and the cycle time is 24 sec. Each press must be set up with a punch-and-die set before production starts. Setup time is 2.0 hr, and availability is assumed to be 100%. How many presses and operators must be devoted to this production during the three days, if there are 7.5 hr of available time per machine per day? Answer: The workload consists of 8000 stampings at 24 sec per piece WL = 9000(24/60 min) + 2(60)n = 3600 + 120 n (min) = 60.0 + 2n (hr) Time available per press during the three days AT = 3(7.5) = 22.5 hr n = 60 + 2n/22.5 22.5n = 60 + 2n 20.5n = 60 n = 60/20.5 = 2.93 rounded up to 3 presses and operators 14.5 A stamping plant must supply an automotive final assembly plant with sheet metal stampings. The plant operates one 8-hr shift for 250 days/yr and must produce 5,000,000 good quality stampings annually. Batch size = 8,000 good stampings produced per batch. Scrap rate = 3%. On average it takes 4.0 sec to produce each stamping when the presses are running. Before each batch, the press must be set up, and that takes 2.5 hr per setup. Availability of the presses is 96% during production and 100% during setup. (a) How many stamping presses will be required to accomplish the specified production? (b) What is the proportion of time spent in setup for each batch? Answer: (a) Production: WL = 5,000,000(4/3600)/1 – 0.03 = 5,727.4 hr/yr AT = 250(8)(0.96) = 1920 hr/yr per press Setup: number batches/yr = 5,000,000/8,000 = 625 batches = 625 setups WL = 625(2.5) = 1562.5 hr/yr AT = 250(8) = 2000 hr/yr per press. n = 5727.4/1920 + 1562.5/2000 = 2.98 + 0.78 = 3.76 rounded up to 4 presses (b) Run time for each batch = 8000(4/3600)/(1 - .03)(0.96) = 9.545 hr Setup time for each batch = 2.5 hr Proportion of time spent in setup for each batch = 2.5/(2.5 + 9.545) = 0.207 = 20.7% Note that this is the same proportion obtained by using press fractions: Proportion of press fractions = 0.78/(0.78 + 2.98) = 0.207 14.6 A new forging plant must supply parts to a construction equipment manufacturer. Forging is a hot operation, so the plant will operate 24 hr/day, five days/wk, 50 wk/yr. Total output from the plant must be 800,000 forgings per year in batches of 1250 parts per batch. Anticipated scrap rate = 3%. Each forging cell will consist of a furnace to heat the parts, a forging press, and a trim press. Parts are placed in the furnace an hour prior to forging; they are then removed, forged, and trimmed one at a time. The complete cycle takes 1.5 min per part. Each time a new batch is started, the forging cell must be changed over, which consists of changing the forging and trimming dies for the next part style. This takes 3.5 hr on average. Each cell is considered to be 96% reliable (availability = 96%) during operation and 100% reliable during changeover. (a) Determine the number of forging cells that would be required in the new plant. (b) What is the proportion of time spent in setup for each batch? Answer: Production: WL = 800,000(1.5/60)/1 – 0.03 = 20,618.6 hr/yr AT = 50(5)(3)(8)(0.96) = 5760 hr/yr Setup: number batches/yr = 800,000/1250 = 640 batches/yr = 640 setups/yr WL = 640(3.5) = 2240 hr/yr per cell AT = 50(5)(3)(8) = 6000 hr/yr n = 20618.6/5760 + 2240/6000 = 3.58 + 0.37 = 3.95 rounded up to 4 forging cells (b) Run time for each batch = 1250(1.5/60)/(1 - .03)(0.96) = 33.56 hr Setup time for each batch = 3.5 hr Proportion of time spent in setup for each batch = 3.5/(3.5 + 33.56) = 0.094 = 9.4% Note that this is the same proportion obtained by using press fractions: Proportion of press fractions = 0.37/(.37 + 4.58) = 0.094 14.7 (A) A plastic injection molding plant will be built to produce 4,000,000 molded parts/yr. The plant will run three 8-hr shifts per day, five days/wk, 52 wk/yr. For planning purposes, the average batch size = 5000 moldings, average changeover time between batches = 5 hr, and average molding cycle time per part = 22 sec. Assume scrap rate = 2%, and availability for each molding machine = 97%, which applies to both run time and changeover time. (a) How many molding machines are required in the new plant? (b) What is the time required to produce each batch? Answer: (a) Production: WL = 4,000,000(22/3600)/1 – 0.02 = 24,943 hr/yr Setup: number batches/yr = 4,000,000/5000 = 800 batches = 800 setups WL = 800(5) = 4000 hr/yr AT = 3(5)(52)(8)(0.97) = 6052.8 hr/yr per machine n = 24943 + 4000/6052.8 = 4.78 rounded up to 5 molding machines (b) Run time for each batch = 5000(22/3600)/(1 - .02)(0.97) = 32.14 hr Time required to produce each batch = 32.14 + 5 = 37.14 hr 14.8 A plastic extrusion plant will be built to produce 30 million meters of plastic extrusions per year. The plant will run three 8-hr shifts per day, 360 days/yr. For planning purposes, the average run length = 7500 meters of extruded plastic. The average changeover time between runs = 3.5 hr, and average extrusion speed = 15 m/min. Assume scrap rate = 1%, and average uptime proportion per extrusion machine = 95% during run time and 100% during changeover. If each extrusion machine requires a floor area of 2 m by 25 m, and there is an allowance of 40% for aisles and office space, what is the total area of the extrusion plant? Answer: Production: WL = 30,000,000/15(60)(1 – 0.01) = 33,670.0 hr/yr AT = 360(3)(8)(0.95) = 8208 hr/yr. Changeover: number runs/yr = 30,000,000/7500 = 4000 runs/yr = 4000 changeovers/yr WL = 4000(3.5) = 14,000 hr/yr AT = 360(3)(8) = 8640 hr/yr per machine n = 33,670/8208 + 14,000/8640 = 4.10 + 1.62 = 5.72 rounded up to 6 machines A = 6(2 × 25)(1 + 40%) = 420 m2 14.9 Future production requirements in a machine shop call for several automatic bar machines to be added to produce three new parts (A, B, and C). Annual quantities and cycle times for the three parts are given in the table below. The machine shop operates one 8-hr shift for 250 days/yr. The machines are expected to be 95% reliable, and the scrap rate is 3%. How many automatic bar machines will be required to meet the specified annual demand for the three new parts? Assume setup times are negligible. Answer: AT = 250(8)(0.95) = 1900 hr/yr per machine WL = 25000(5/60)/1 – 0.03 + 40000(7/60)/1 – 0.03 + 10000(10/60)/1 – 0.03 = 2147.7 + 4811.0 + 1700.7 WL = 8,659.4 hr/yr n = 8,659.4 /1900 = 4.6 rounded up to 5 machines 14.10 A certain type of machine will be used to produce three products: A, B, and C. Sales forecasts for these products are: 52,000, 65,000, and 70,000 units/yr, respectively. Production rates for the three products are, respectively, 12, 15, and 10/hr; and scrap rates are, respectively, 5%, 7%, and 9%. The plant will operate 50 wk/yr, 10 shifts/wk, and 8 hr/shift. It is anticipated that production machines of this type will be down for repairs on average 10 percent of the time. How many machines will be required to meet demand? Assume setup times are negligible. Answer: AT = 50(10)(8)(1 - 0.10) = 3600 hr/yr per machine WL = 52,000/12(1 - 0.05) + 65,000/15(1 - 0.07) + 70,000/10(1 - 0.09) = 4561.4 + 4659.5 + 7692.3 WL = 16,913.2 hr/yr n = 16,913.2/3600 = 4.67 rounded up to 5 machines 14.11 An emergency situation has arisen in the milling department, because the ship carrying a certain quantity of a required part from an overseas supplier sank on Friday evening. A certain number of machines in the department must therefore be dedicated to the production of this part during the next week. A total of 1000 of these parts must be produced, and the production cycle time per part = 16.0 min. Each milling machine used for this rush job must first be set up, which takes 5.0 hr. A scrap rate of 2% can be expected. Assume availability = 100%. (a) If the production week consists of 10 shifts at 8.0 hr/shift, how many machines will be required? (b) It so happens that only two milling machines can be spared for this emergency job, due to other priority jobs in the department. To cope with the emergency situation, plant management has authorized a three-shift operation for six days next week. Can the 1000 replacement parts be completed within these constraints? Answer: (a) WL = 1000(16/60)/(1 – 0.02) = 272.1 hr/wk AT = 10(8) - 5 = 80 - 5 = 75 hr/wk per machine n = 272.1/75 = 3.63 rounded up to 4 milling machines (b) AT = 6(3)(8) - 5 = 139 hr/wk per machine n = 272.1/139 = 1.96 rounded up to 2 milling machines. Yes, the job can be completed. 14.12 A machine shop has one CNC vertical machining center (VMC) to produce two parts (A and B) used in the company’s main product. The VMC is equipped with an automatic pallet changer (APC) and a parts storage system that holds 10 parts. One thousand units of the product are produced/yr, and one of each part is used in the product. Part A has a machining cycle time of 50 min. Part B has a machining cycle time of 80 min. These cycle times include the operation of the APC. No changeover time is lost between parts. The anticipated scrap rate is zero. The machining center is 95% reliable. The machine shop operates 250 days/yr. How many hours must the CNC machining center operate each day to supply parts for the product? Answer: Annual WL = 1000(50 + 80)/60 = 2166.67 hr/yr At 250 days/yr, daily workload WL = 2166.67/250 = 8.67 hr/day Factoring in the reliability, hr/day = 8.67/0.95 = 9.13 hr/day Machine Clusters 14.13 (A) The CNC grinding section has a large number of machines devoted to grinding shafts for the automotive industry. The grinding cycle takes 3.6 min and produces one part. At the end of each cycle an operator must be present to unload and load parts, which takes 40 sec. (a) Determine how many grinding machines the worker can service if it takes 20 sec to walk between the machines and no machine idle time is allowed. (b) How many seconds during the work cycle is the worker idle? (c) What is the hourly production rate of this machine cluster? Answer: (a) n = 3.6(60) + 40/40 + 20 = 256/60 = 4.27 Use n1 = 4 grinding machines (b) Worker idle time IT = 256 - 4(60) = 256 - 240 = 16 sec (c) Tc = 256 sec = 4.267 min Rc = 4(60/4.267) = 56.25 pc/hr 14.14 A worker is currently responsible for tending two machines in a machine cluster. The service time per machine is 0.35 min and the time to walk between machines is 0.15 min. The machine automatic cycle time is 1.90 min. If the worker's hourly rate = $12/hr and the hourly rate for each machine = $18/hr, determine (a) the current hourly rate for the cluster, and (b) the current cost per unit of product, given that two units are produced by each machine during each machine cycle. (c) What is the percent idle time of the worker? (d) What is the optimum number of machines that should be used in the machine cluster, if minimum cost per unit of product is the decision criterion? Answer: (a) Co = $12 + 2($18) = $48.00/hr (b) Tc = Tm + Ts = 1.90 + 0.35 = 2.25 min/cycle Rc = 2(2)(60/2.25) = 106.67 pc/hr Cpc = ($48/hr)/(106.67 pc/hr) = $0.45/pc (c) Worker engagement time/cycle = 2(Ts + Tr) = 2(0.35 + 0.15) = 1.0 min Idle time IT = (2.25 – 1.0)/2.25 = 0.555 = 55.5% (d) n = 1.90 + 0.35/0.35 + 0.15 = 2.25/0.5 = 4.5 machines n1 = 4 machines: Cpc(4) = 0.5(12/4 + 18(2.25/60) = $0.394/pc n2 = 5 machines: Cpc(5) = 0.5(12 + 18 × 5)(0.50/60) = $0.425/pc Use n1 = 4 machines 14.15 In a machine cluster, the appropriate number of production machines to assign to the worker is to be determined. Let n = the number of machines. Each production machine is identical and has an automatic processing time Tm = 4.0 min. The servicing time Ts = 12 sec for each machine. The full cycle time for each machine in the cell is Tc = Ts + Tm. The repositioning time for the worker is given by Tr = 5 + 3n, where Tr is in sec. Tr increases with n because the distance between machines increases with more machines. (a) Determine the maximum number of machines in the cell if no machine idle time is allowed. For your answer, compute (b) the cycle time and (c) the worker idle time expressed as a percent of the cycle time? Answer 14.16 An industrial robot will service n production machines in a machine cluster. All production machines are identical and have the same processing time of 130 sec. The robot servicing and repositioning time for each machine is given by the equation (Ts + Tr) = 15 + 4n, where Ts is the servicing time (sec), Tr is the repositioning time (sec), and n = number of machines that the robot services. (Ts + Tr) increases with n because more time is needed to reposition the robot as n increases. The full cycle time for each machine in the cell is Tc = Ts + Tm. (a) Determine the maximum number of machines in the cell such that machines are not kept waiting. For your answer, (b) what is the machine cycle time, and (c) what is the robot idle time expressed as a percent of the cycle time Tc? Answer: (a) Given Tm = 130 sec, Tc = Tm + Ts = 130 + Ts n = 130 +Ts/15 + 4n We can deduce that if there were only one machine (n = 1), then repositioning time would be zero (Tr = 0). Thus, for n = 1, (Ts + Tr) = 15 + 4(1) = 19 sec and (Tm + Ts) = 130 + 19 Substituting this value into the equation, n = 130 +19/15 + 4n = 149/15 + 4n 15n + 4n2 = 149 4n2 + 15n - 149 = 0 Use quadratic equation to find n. Use n = 4.51 (ignore n = -8.26) → For no machine waiting, n1 = 4 machines (b) Tc = 130 + 19 = 149 sec (c) Robot work time = n(Ts + Tr) = 4(15 + 4 x 4) = 4(31) = 124 sec Robot idle time % = (149 – 124)/149 = 0.168 = 16.8% 14.17 A factory production department consists of a large number of work cells. Each cell consists of one human worker performing electronics assembly tasks. The cells are organized into sections within the department, and one foreman supervises each section. It is desired to know how many work cells should be assigned to each foreman. The foreman’s job consists of two tasks: (1) provide each cell with a sufficient supply of parts that it can work for 4.0 hr before it needs to be resupplied and (2) prepare production reports for each work cell. Task (1) takes 18.0 min on average per work cell and must be done twice per day. The foreman must schedule the resupply of parts to every cell so that no idle time occurs in any cell. Task (2) takes 9.0 min per work cell and must be done once per day. Neither the workers nor the foreman are allowed to work more than 8.0 hr/day. Each day, the cells continue production from where they stopped the day before. (a) What is the maximum number of work cells that should be assigned to a foreman, on the condition that the work cells must never be idle? (b) With the number of work cells from part (a), how many idle minutes does the foreman have each day? Answer: Since the foreman must resupply each cell twice in an 8.0 hr day and also prepare a report on each cell once each day, the foreman’s day can be modeled as follows: 2n(18) + 1n (9) = 8(60) 36n + 9n = 480 45n = 480 n = 10.67. Use n1 = 10 cells (b) Foreman’s idle time each day = 480 - 45n = 480 - 450 = 30 min Chapter 15 MANUAL ASSEMBLY LINES REVIEW QUESTIONS 15.1 What are the four factors that favor the use of manual assembly lines? Answer: The four factors listed in the text are: (1) Demand for the product is high or medium. (2) The products are identical or similar. (3) The total work to assemble the product can be divided into small work elements. (4) It is technologically impossible or economically infeasible to automate the assembly operations. 15.2 What are the four reasons given in the text that explain why manual assembly lines are so productive compared to alternative methods in which multiple workers each perform all of the tasks to assemble the product? Answer: The four reasons are the following: (1) Specialization of labor, which asserts that when a large job is divided into small tasks and each task is assigned to one worker, the worker becomes highly proficient at performing the single task. (2) Interchangeable parts, in which each component is manufactured to sufficiently close tolerances that any part of a certain type can be selected for assembly with its mating component. (3) Work flow principle, which involves moving the work to the worker rather than vice versa. (4) Line pacing, in which workers on an assembly line must complete their assigned tasks within a certain cycle time. 15.3 What is a manual assembly line? Answer: The definition given in the text is the following: A manual assembly line is a production line that consists of a sequence of workstations where assembly tasks are performed by human workers. Products are assembled as they move along the line. At each station, a portion of the total work is performed on each unit. 15.4 What do the terms starving and blocking mean? Answer: Starving is the situation in which the assembly operator has completed the assigned task on the current work unit, but the next unit has not yet arrived at the station. The worker is thus starved for work. Blocking means that the operator has completed the assigned task on the current work unit but cannot pass the unit to the downstream station because that worker is not yet ready to receive it. The operator is therefore blocked from working. 15.5 Identify and briefly describe the three major categories of mechanized work transport systems used in production lines? Answer: The three major categories of mechanized work transport systems are (1) continuous transport, which uses a continuously moving conveyor that operates at constant speed to transport work units; (2) synchronous transport, also known as intermittent transport, in which all work units are moved simultaneously between stations with a quick discontinuous motion; and (3) asynchronous transport, in which work units are moved independently rather than synchronously between stations. 15.6 To cope with product variety, three types of assembly line are described in the text. Name the three types and explain the differences between them. Answer: The three types of assembly line are (1) single model line, in which only one product is made and there is no variation in the product; (2) batch-model line, in which each type of product is produced in batches and the line setup is changed over between each batch; and (3) mixed-model line, in which different models are produced simultaneously on the same line but no changeover is required between models. 15.7 What does the term line efficiency mean in production line terminology? Answer: Line efficiency is the proportion uptime on the production line. It is the ratio of the theoretical cycle time Tc to the actual average production time Tp when downtime on the line is averaged in. 15.8 The theoretical minimum number of workers on an assembly line w* is the minimum integer that is greater than the ratio of the work content time Twc divided by the cycle time Tc. Two factors are identified in the text that make it difficult to achieve this minimum value in practice. Name the two factors. Answer: The two factors are (1) repositioning losses, in which either the worker or the work unit must be positioned at the start of each work cycle, and (2) the line balancing problem, which refers to the difficulty encountered in allocating all of the work elements in the work content time equally among workers on the line. 15.9 What is the difference between the cycle time Tc and the service time Ts? Answer: The cycle time Tc is the time interval between arrivals of work units at a station, while service time Ts is the actual amount of time available at the station to perform the assigned task. The difference is the repositioning time Tr. That is, Ts = Tc – Tr. 15.10 What is a minimum rational work element? Answer: As defined in the text, a minimum rational work element is a small amount of work that has a specific limited objective, such as adding a component to the base part or joining two components. A minimum rational work element cannot be subdivided any further without loss of practicality. 15.11 What is meant by the term precedence constraint? Answer: Precedence constraints are restrictions on the order in which the work elements can be performed. Some elements must be done before others. For example, to create a threaded hole, the hole must be drilled before it can be tapped. 15.12 What is meant by the term balance efficiency? Answer: The balance efficiency Eb is the work content time divided by the total available service time on the line. It can be computed as Eb = Twc/wTs, where Twc = work content time, min; Ts = the maximum available service time on the line (Max{Tsi}), min/cycle; and w = number of workers. 15.13 What is the difference between how the largest candidate rule works and how the Kilbridge and Wester method works? Answer: In the largest candidate rule, the algorithm begins with the work elements listed in descending order of their time values, whereas in the Kilbridge and Wester method, the algorithm operates on the work elements listed according to their precedence order in the precedence diagram. 15.14 What is meant by the term manning level in the context of a manual assembly line? Answer: The manning level is the number of workers per station for a single station. For the entire line, it is the total number of workers on the line divided by the number of stations. 15.15 What are storage buffers and why are they sometimes used on a manual assembly line? Answer: A storage buffer is a location in the production line where work units are temporarily stored. As identified in the text, the reasons to include one or more storage buffers in a production line include: (1) to accumulate work units between two stages of the line when their production rates are different; (2) to smooth production between stations with large task time variations; and (3) to permit continued operation of certain sections of the line when other sections are temporarily down for service or repair. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Analysis of Single-Model Assembly Lines 15.1 (A) Annual demand for a new assembled product is expected to be 75,000 units. Work content time for the product is 25.8 min. The assembly line works 50 wk/yr, 40 hr/wk. Expected line efficiency is 95%. With one worker per station, determine (a) average hourly production rate, (b) cycle time, and (c) ideal minimum number of workers. Answer: (a) Rp = 75,000/(50 × 40) = 37.5 units/hr (b) Tc = 60(0.95)/37.5 = 1.52 min (c) w* = minimum integer ≥ 25.8/1.52 = 16.97 rounded to 17 workers 15.2 A new product is to be assembled on a manual assembly line. The line is to be designed to produce 100,000 units/yr. Work content time for the product is 39.5 min. The assembly line will operate 52 wk/yr, 5 shifts/wk, 8 hr/shift. Expected line efficiency is 96%. There will be one worker per station. Determine (a) average hourly production rate, (b) cycle time, and (c) ideal minimum number of workstations. Answer: (a) Rp = 100,000/(52 × 5 × 8) = 48.08 units/hr (b) Tc = 60(0.96)/48.08 = 1.198 min (c) n* = w* = minimum integer ≥ 39.5/1.198 = 32.97 rounded to 33 workstations 15.3 A product whose work content time = 52.0 min is to be assembled on a manual production line. Required production rate is 30 units/hr. From previous experience, it is estimated that proportion uptime = 0.95, balance efficiency = 93%, and repositioning time = 6 sec. Determine (a) cycle time, (b) ideal minimum number of workers required on the line, (c) more realistic estimate of the number of workers on the line when balance efficiency and repositioning time are accounted for. Answer: (a) Tc = 60(0.95)/30 = 1.9 min (b) w* = Minimum Integer ≥ 52/1.9 = 27.4 rounded to 28 workers (c) Er = (1.9 – 6/60)/1.9 = 0.947 w = Minimum Integer ≥ 52/(1.9 × 0.95 × 0.93 × 0.947) = 32.7 rounded to 33 workers 15.4 A manual assembly line has 20 workstations and produces 24 units/hr with one operator per station. Work content time to assemble the product = 41.8 min. Proportion uptime = 0.94, and repositioning time = 9 sec. Determine the balance delay. Answer: Tc = 60(0.94)/24 = 2.35 min, Ts = 2.35 – 9/60 = 2.20 min w = n = 20 workers and 20 stations Eb = 41.8/(20 × 2.2) = 0.950, d = 1 - 0.95 = 0.05 = 5.0% 15.5 A manual assembly line must be designed for a product with annual demand = 95,000 units. Work content time = 45.0 min. The line will have one worker per station and operate 50 wk/yr, 5 shifts/wk, and 7.5 hr/shift. Work units will be attached to a continuously moving conveyor. Estimated line efficiency = 0.97, balancing efficiency = 0.92, and repositioning efficiency = 0.95. Determine (a) hourly production rate to meet demand and (b) number of workers required. Answer: (a) Rp = 95,000/(50 × 5 × 7.5) = 50.67 units/hr (b) Tc = 60(0.97)/50.67 = 1.149 min w = Minimum Integer ≥ 45.0/(1.149 × 0.92 × 0.95) = 44.8 rounded to 45 workers 15.6 A single model assembly line is being planned to produce a consumer appliance at the rate of 150,000 units/yr. The line will be operated 8 hr/shift, two shifts per day, five days/wk, 52 wk/yr. Work content time = 35.8 min. For planning purposes, it is anticipated that the proportion uptime on the line will be 95%. Determine (a) average hourly production rate, (b) cycle time, and (c) theoretical minimum number of workers required on the line. (d) If the balance efficiency is 0.93 and the repositioning time = 6 sec, how many workers will actually be required? (e) What is the assembly line labor efficiency? Answer: (a) Rp = 150,000/(10 × 8 × 52) = 37.5 pc/hr (b) Tc = 60(0.95)/37.5 = 1.52 min (c) w* = Minimum Integer ≥ 35.8/1.52 = 23.5 rounded to 24 workers (d) Er = (1.52 – 6/60)/1.52 = 0.934 w = Minimum Integer ≥ 35.8/(1.52 × 0.93 × 0.934) = 27.1 rounded to 28 workers (e) Assembly line labor efficiency = EEbEr = 0.95 × 0.93 × 0.934 = 0.825 = 82.5% 15.7 Required production rate for a new product is 45 units/hr and its assembly work content time is 1.25 hr. It will be produced on an assembly line that includes four automated workstations. Because of the automated stations, the line will have an expected uptime efficiency = 90%. The remaining manual stations will each have one worker. It is anticipated that 5% of the cycle time will be lost due to repositioning at the bottleneck station. If the balance delay is expected to be 0.07, determine (a) the cycle time, (b) number of workers, and (c) assembly line labor efficiency on the line. Answer: (a) Tc = 60(0.90)/45 = 1.20 min (b) Er = 1 – 0.05 = 0.95, Eb = 1 - d = 1 - 0.07 = 0.93 w = Minimum Integer ≥ 1.25(60)/(1.2 × 0.95 × 0.93) = 70.7 rounded to 71 workers (c) Assembly line labor efficiency = EEbEr = 0.90(0.93)(0.95) = 0.795 = 79.5% 15.8 A final assembly plant for a new automobile model is to have a capacity of 180,000 units annually. The plant will operate 50 weeks/yr, two shifts/day, 5 days/week, and 7.5 hr/shift. It will be divided into three departments: (1) body shop, (2) paint shop, (3) general assembly department. The body shop welds the car bodies using robots, and the paint shop coats the bodies. Both of these departments are highly automated. General assembly has no automation. There are 15.0 hr of work content time on each car in this third department, where cars are moved by a continuous conveyor. In addition to the workers assembling the product on the line in general assembly, there are 25 utility workers. Determine (a) hourly production rate of the plant, (b) number of workers and workstations in general assembly, given that its manning level is 2.4, balancing efficiency = 94%, proportion uptime = 95%, and a repositioning time of 0.15 min is allowed for each worker. (c) What is the assembly line labor efficiency in general assembly? Answer: (a) Rp = 180,000/(50 × 10 × 7.5) = 48.0 cars/hr (b) Tc = 60(0.95)/48 = 1.1875 min, Er = (1.1875 – 0.15)/1.1875 = 0.874 w = Minimum Integer ≥ 15(60)/(1.1875 × 0.94 × 0.874) = 923 workers In addition, there are 25 utility workers, so total number of workers w = 948 workers Number of workstations n = 948/2.4 = 395 workstations (c) Assembly line labor efficiency = EEbEr = 0.95(0.94)(0.874) = 0.780 = 78.0% 15.9 (A) Production rate for a certain assembled product is 48 units/hr. The assembly work content time = 35 min. The line operates at 92% uptime. Ten workstations have two workers on opposite sides of the line so that both sides of the product can be worked on simultaneously. The remaining stations have one worker. Repositioning time lost by each worker is 0.15 min/cycle. It is known that the number of workers on the line is two more than the number required for perfect balance. Determine (a) number of workers, (b) number of workstations, (c) balance efficiency, and (d) average manning level (ignore utility workers). Answer: (a) Tc = 60(0.92)/48 = 1.15 min, Er = (1.15 – 1.0)/1.15 = 0.8696 For perfect balance, Eb = 1.0. w = Minimum Integer ≥ 35.0/(1.15 × 0.8696) = 35 workers With 2 more workers than required for perfect balance, w = 35 + 2 = 37 workers (b) n = 20/2 + (37 - 20) = 10 + 17 = 27 stations (c) Eb = 35.0/(37 × 1.15) = 0.946 = 94.6% (d) M = 37/27 = 1.37 15.10 The work content time is 50 min for a product assembled on a manual production line. The line uses an overhead conveyor that moves at 1.2 m/min. There are 27 workstations on the line, one-third of which have two workers; the rest have one worker each. Repositioning time per worker is 9 sec, and uptime efficiency is 95%. (a) What would be the hourly production rate if the line were perfectly balanced? (b) If the actual production rate is 35 units/hr, what is the balance efficiency of the line? (c) To compensate for service time variability, it has been decided that the tolerance time should be 50% greater than the cycle time. To achieve this, what station length should be used on the line? (d) Determine the elapsed time that each base part spends on the production line? Answer: (a) w = 2(27/3) + 1(27 x 2/3) = 2(9) + 1(18) = 36 workers For perfect balance, Eb = 1.0. Ts = 50.0/(36 × 1.0) = 1.389 min, Tc = 1.389 + 0.15 = 1.539 min Rp = 60(0.95)/1.539 = 37.04 pc/hr (at E = 0.95 and Eb = 1.0) (b) Actual Rp = 35 pc/hr Tc = 60(0.95)/35 = 1.629 min, Ts = 1.629 - 0.15 = 1.479 min Eb = 50.0/(36 × 1.479) = 0.939 = 93.9% (c) Tolerance time Tt = station length Ls divided by conveyor speed. Thus Ls = Ttvc Tt = 1.5Tc = 1.5(1.629) = 2.44 min, Ls = 2.44(1.2) = 2.93 m (d) Elapsed time ET = nTt = 27(2.44) = 65.9 min 15.11 The production rate of a manual assembly line is 38 units/hr. The work content time of the product made on the line is 24.8 min. Work units are attached to a moving conveyor whose speed = 1.5 m/min. Repositioning time per worker is 8 sec, line efficiency is 96%, and manning level is 1.25 (ignoring utility workers). Owing to imperfect line balancing, the number of workers needed on the line is about 10% more than the number required for perfect balance. If the workstations are arranged in a line, and the length of each station is 3.5 m, (a) how long is the entire production line, and (b) what is the elapsed time a base part spends on the line? Answer: (a) Tc = 60(0.96)/38 = 1.516 min Ts = 1.516 - 0.133 = 1.383 min For perfect balance, Eb = 1.0. w = Minimum Integer ≥ 24.8/(1.383 × 1.0) = 17.9 With 10% more workers, w = 1.1(17.9) = 19.7 rounded up to 20 workers Given the manning level M = 1.25, n = 20/1.25 = 16 stations L = (16 stations)(3.5 m/station) = 56 m (b) Given L = 56 m and vc = 1.5 m/min, ET = 56/1.5 = 37.33 min Line Balancing (Single-Model Lines) 15.12 Show that the two statements of the objective function in single model line balancing in Eq. (15.20) are equivalent. Answer: nTs in the first expression = the second expression, since Ts is a constant (Ts = Max{Tsi} in Eq. (15.11)) And Twc in the first expression = in the second expression, according to Eq. (15.15). 15.13 The table below defines the precedence relationships and element times for a new model toy. (a) Construct the precedence diagram for this job. (b) If the ideal cycle time = 1.1 min. repositioning time = 0.1 min, and uptime proportion is assumed to be 1.0, what is the theoretical minimum number of workstations required to minimize the balance delay under the assumption that there will be one worker per station? (c) Use the largest candidate rule to assign work elements to stations. (d) Compute the balance delay for your solution. Work element Te (min) Immediate predecessors 1 0.5 - 2 0.3 1 3 0.8 1 4 0.2 2 5 0.1 2 6 0.6 3 7 0.4 4,5 8 0.5 3,5 9 0.3 7,8 10 0.6 6,9 Answer: (a) Precedence diagram: (b) Ts = Tc - Tr = 1.1 - 0.1 = 1.0 min With M = 1.0, n = w = Minimum Integer ≥ Twc/Ts = 4.3/1.0 = 4.3, Use n = 5 stations (c) Line balancing solution using the largest candidate rule. 15.14 (A) Solve the previous problem using the Kilbridge and Wester method in part (c). Answer: (a) Precedence diagram same as in Problem 15.11. (b) Same as in Problem 15.11: n = 5 stations (c) Line balancing solution using the Kilbridge & Wester method: (d) Same as in Problem 15.11: d = 0.14 = 14% 15.15 Solve the previous problem using the ranked positional weights method in part (c). Answer: (a) Precedence diagram same as in Problem 15.11. (b) Same as in Problem 15.11: n = 5 stations (c) Line balancing solution using the Kilbridge & Wester method: Same as in Problem 15.11: d = 0.14 = 14% 15.16 A manual assembly line is to be designed to make a small consumer product. The work elements, their times, and precedence constraints are given in the table below. The workers will operate the line for 400 min per day and must produce 300 products per day. A mechanized belt, moving at a speed of 1.25 m/min, will transport the products between stations. Because of the variability in the time required to perform the assembly operations, it has been determined that the tolerance time should be 1.5 times the cycle time of the line. (a) Determine the ideal minimum number of workers on the line. (b) Use the Kilbridge and Wester method to balance the line. (c) Compute the balance delay for your solution in part (b). Element Time Te Preceded by: Element Time Te Preceded by: 1 0.4 min - 6 0.2 min 3 2 0.7 min 1 7 0.3 min 4 3 0.5 min 1 8 0.9 min 4, 9 4 0.8 min 2 9 0.3 min 5, 6 5 1.0 min 2, 3 10 0.5 min 7, 8 Answer : Note: In the above solution, the Kilbridge & Wester method was followed very precisely, so that the solution consisted of a total of six stations. A better solution, but one that violates the K&W algorithm, would be to combine elements 9 and 8 at station 4 and elements 7 and 10 at station 5, for a total of only 5 stations. (b) Use cycle time of 1.3 min (station 2) rather than 1.333 min. Using the six stations in the K&W solution, balance delay d = 6(1.3) – 5.6/6(1.3) = 0.282 = 28.2% 15.17 Solve the previous problem using the ranked positional weights method in part (b). Answer: (a) Same solution as in Problem 15.14: (Tc = 1.333 min, Tr = 0, Twc = 5.6 min) w = 5 workers. (b) Line balancing solution using ranked positional weights method. Assume M = 1.0. (c) Use cycle time of 1.3 min (station 2) rather than 1.333 min. Using the five stations in the RPW solution, balance delay d = 5(1.3) – 5.6/5(1.3) = 0.138 = 13.8% 15.18 (A) A manual assembly line operates with a mechanized conveyor. The conveyor moves at a speed of 5 ft/min, and the spacing between base parts launched onto the line is 4 ft. It has been determined that the line operates best when there is one worker per station and each station is 6 ft long. There are 14 work elements that must be accomplished to complete the assembly, and the element times and precedence requirements are listed in the table below. Determine (a) feed rate and corresponding cycle time, (b) tolerance time for each worker, and (c) ideal minimum number of workers on the line. (d) Draw the precedence diagram for the problem. (e) Determine an efficient line balancing solution. (f) For your solution, determine the balance delay. Element Te Preceded by: Element Te Preceded by: 1 0.2 min - 8 0.2 min 5 2 0.5 min - 9 0.4 min 5 3 0.2 min 1 10 0.3 min 6, 7 4 0.6 min 1 11 0.1 min 9 5 0.1 min 2 12 0.2 min 8, 10 6 0.2 min 3, 4 13 0.1 min 11 7 0.3 min 4 14 0.3 min 12, 13 Answer: (a) Assume Tr = 0. fp = (5 ft/min)/(4 ft/asby) = 1.25 asby/min Tc = 0.8 min/asby (b) Tt = (6 ft/station)/(5 ft/min) = 1.2 min/station (c) Twc = Σ Tek = 0.2 + 0.5 + . . + 0.3 = 3.7 min w* = Minimum Integer ≥ (3.7 min)/(0.8 min) = 4.625 → 5 workers (d) Precedence diagram (e) Line balancing solution using the Kilbridge and Wester method Balance delay d = 5(0.8) – 3.7/5(0.8) = 0.075 = 7.5% 15.19 A new small electrical appliance is to be assembled on a single model assembly line. The line will be operated 250 days/yr, 15 hr/day. The work content has been divided into work elements as defined in the table below. Also given are the element times and precedence requirements. Annual production is to be 200,000 units. It is anticipated that the line efficiency will be 0.96. Repositioning time for each worker is 0.08 min. Determine (a) average hourly production rate, (b) cycle time, and (c) theoretical minimum number of workers required to meet annual production requirements. (d) Use one of the line balancing algorithms to balance the line. For your solution, determine (e) balance efficiency and (f) overall labor efficiency on the line. Answer: Workstation Details 15.20 (A) An overhead continuous conveyor is used to carry dishwasher base parts along a manual assembly line while components are being added to them. The spacing between appliances = 2.2 m and the speed of the conveyor = 1.2 m/min. The length of each workstation is 3.5 m. There are a total of 25 stations, four of which have two workers. There are also 3 utility workers assigned to the line. Determine (a) cycle time and feed rate, (b) tolerance time, (c) manning level, and (d) elapsed time a dishwasher base part spends on the line. Answer: (a) Cycle time Tc = sp/vc = (2.2 m)/(1.2 m/min) = 1.833 min Feed rate fp = 1/Tc = 1/1.833 = 0.545 parts/min (b) Tolerance time Tt = (3.5 m/station)/(1.2 m/min) = 2.917 min/station (c) Manning level for the line M = (4 × 2 + 21 × 1 + 3)/25 = 32/25 = 1.28 (d) Line length L = 25(3.5 m) = 87.5 m Elapsed time ET = (87.5 m)/(1.2 m/min) = 72.9 min 15.21 A moving belt line is used to assemble a product whose work content = 22 min. Production rate = 35 units/hr, and the proportion uptime = 0.96. The length of each station = 2.0 m and station manning level = 1.0 for all stations. The belt speed can be set at any value between 0.6 and 3.0 m/min. It is expected that the balance delay will be about 0.08 or slightly higher. Time lost for repositioning each cycle is 6 sec. (a) Determine the number of stations needed on the line. (b) Using a tolerance time that is 50% greater than the cycle time, what would be an appropriate belt speed and spacing between parts? Answer: (a) Tc = 60(0.096)/35 = 1.646 min Ts = 1.646 - 0.1 = 1.546 min and Eb = 1 - 0.08 = 0.92 Since M = 1.0, n = w = Minimum Integer ≥ 22.0/0.92(1.546) = 15.5 → 16 workstations (b) Use Tt = 1.5Tc = 1.5(1.646) = 2.47 min/station Tt = Ls/vc; rearranging, vc = Ls/Tt = (2 m)/(2.47 min) = 0.81 m/min Base part spacing sp = vcTc = 0.81(2.47) = 2.0 m/part 15.22 In the general assembly department of an automobile final assembly plant, there are 495 workstations, and the cycle time = 0.93 min. If each workstation is 6.2 m long, and the tolerance time = the cycle time, determine the following: (a) speed of the conveyor, (b) center-to-center spacing between units on the line, (c) total length of the general assembly line, assuming no vacant space between stations, and (d) elapsed time a work unit spends in the general assembly department. Answer: (a) Tt = Tc = 0.95 min Conveyor speed vc = Ls/Tt = (6.2 m)/(0.93 min) = 6.667 m/min (b) sp = vcTc = 6.667(0.93) = 6.2 m/car (c) L = 495(6.2) = 3069 m (d) ET = L/vc = 3069/6.667 = 460.3 min Alternative method: ET = nTt = 495(0.93) = 460.3 min 15.23 Total work content for a product assembled on a manual production line is 33.0 min. Production rate of the line must be 47 units/hr. Base parts are attached to a moving conveyor whose speed = 2.2 m/min. Repositioning time per worker is 6 sec, and line efficiency is 94%. Due to imperfect line balancing, the number of workers needed on the line must be two more workers than the number required for perfect balance. Manning level = 1.6, excluding utility workers. Determine (a) the number of workers and (b) the number of workstations on the line. (c) What is the balance efficiency for this line? (d) If the workstations are arranged in a line, and the length of each station is 3.3 m, what is the tolerance time in each station? (e) What is the elapsed time a work unit spends on the line? Answer: (a) Tc = 60(0.94)/47 = 1.2 min, Ts = 1.2 - 0.1 = 1.1 min For perfect balance, Eb = 1.0. w = Minimum Integer ≥ 33.0/1.0(1.1) = 30 workers. With 4 more workers than required for perfect balance, w = 30 + 2 = 32 workers (b) n = w/M = 32/1.6 = 20 stations (c) Eb = 33/32(1.1) = 0.9375 = 93.75% (d) Tt = Ls/vc = (3.3 m)/(2.2 m/min) = 1.5 min (e) ET = 20(1.5) = 30.0 min Alternative method: L = (20 stations)(3.3/station) = 66 m, ET = 66/2.2 = 30 min 15.24 A manual assembly line is to be designed for a certain major appliance whose assembly work content time = 2.0 hours. The line will be designed for an annual production rate of 100,000 units. The plant will operate one 8-hr shift per day, 250 days/yr. A continuous conveyor system will be used and it will operate at a speed = 1.6 m/min. The line must be designed under the following assumptions: balance delay = 6%, uptime efficiency = 96%, repositioning time = 6 sec for each worker, and average manning level = 1.25, not counting utility workers. (a) How many workers will be required to operate the assembly line? If each station is 2.0 m long, (b) how long will the production line be, and (c) what is the elapsed time a work unit spends on the line? Answer: (a) Rp = 100,000/250(8) = 50pc/hr Tc = 60(0.96)/50 = 1.152 min, Ts = 1.152 - 0.1 = 1.052 min w = Minimum Integer ≥ 2(60)/(1 – 0.06)(1.052) = 121.2 → 122 workers (b) n = 122/1.25 = 97.6 rounded up to 98 stations L = (98 stations)(2 m/station) = 196 m (c) ET = 196/1.6 = 122.5 min min Batch-Model and Mixed-Model Assembly Lines (Appendix A15) 15.25 Two products are to be produced on a batch-model assembly line that will operate 40 hr/wk and 50 wk/yr, but some of that time must be devoted to product changeovers between batches. Annual demand (Daj), batch quantities (Qj), work content time (), and changeover time (Tsuj) for each product are listed in the table below. The anticipated line efficiency = 0.97, balance efficiency = 0.95, and repositioning time = 0.2 min. Determine (a) number of workers on the line, (b) production rates for the two products, and (c) the production schedule for the year. (d) Could the number of workers be reduced if the batch sizes were doubled? If so, by how many workers? Product Daj Qj Twcj Tsuj A 16,000 800 48 min 6 hr B 22,000 1100 39 min 5 hr Answer: (a) Based on the annual demands for the two products and the batch quantities, the number of batches of product A would be 16,000/800 = 20 changeovers. For product B, 22,000/1100 = 20 changeovers. The total downtime for these changeovers = 20(6.0) + 20(5.0) = 220 hours. Rp = (16000 + 22000/50(40) - 220) = 21.35 pc/hr Tc = 60(0.97)/21.35 = 2.726 min and Ts = 2.726 – 0.2 = 2.526 min Using Eq. (A15.2), w = Minimum Integer ≥ (16000(48) + 22000(39)/(16000 + 22000)/2.526(0.95)) 17.83 rounded up to 18 workers (b) For product A, TsA = 48/(18 × 0.95) = 2.81 min, TcA = 2.81 + 0.2 = 3.01 min, and RpA = 60(0.97)/3.01 = 19.35 pc/hr For product B, TsB = 39/(18 × 0.95) = 2.28 min, TcB = 2.28 + 0.2 = 2.48 min, and RpB = 60(0.97)/2.48 = 23.46 pc/hr (c) Batch quantities are specified in the table: QA = 800 and QB = 1100. The production schedule will be to set up the line for product A (6 hr) and produce the batch (800/19.35 = 41.34 hr), and then change over to product B (5 hr) and produce the batch (1100/23.46 = 46.89 hr). This cycle is repeated 20 times during the year. Total time = 20(6 + 41.34 + 5 + 46.89) = 1984.4 hr. This is slightly less than 2000 hours because the number of workers was rounded up from 17.83 to 18. (d) If batch quantities were double to QA = 1600 and QB = 2200, this would reduce the number of hours lost due to changeovers. The number of changeovers would be reduced from 20 to 10 for each product. Total downtime is reduced from 200 hr to 110 hr. Thus, production rate could be reduced: Rp = (16000 + 22000)/50(40) - 120 = 20.11 pc/hr Tc = 60(0.97)/20.11 = 2.895 min and Ts = 2.895 – 0.2 = 2.695 min Using Eq. (A15.2), w = Minimum Integer = 16.71 rounded up to 17 workers. The number of workers required would be reduced by one. 15.26 (A) A batch-model assembly line is being planned to produce three portable power tools: a drill, a sander, and a jig saw. The three models will be produced in batches because annual demand for each power tool is not sufficient to employ a line full time for each. Time to change over the line between production runs is 4 hours. The line will operate 40 hr/wk, 50 wk/yr, one worker per station. Data on work content time, cycle time, and annual demand are presented in the table below. Estimated line efficiency = 0.95, and balance efficiency = 0.93. Repositioning time per worker = 9 sec. Batches of each power tool will be produced 10 times/yr, with batch quantities proportional to annual demand for each product. Determine (a) the number of workers on the line, (b) the production rates for the three products, and (c) the production schedule for the year. Answer: Given that there are (50 wks/yr)(40 hr/wk) = 2000 hr/yr. Also, batches of each of the three products are produced 10 times/yr, so a total of 30 batches are produced/yr. Each batch requires a changeover time of 4 hr. Thus a total of 30(4 hr) = 120 hr are lost each year due to changeovers. That leaves 2000 – 120 = 1880 hr/yr of run time. For each product, the batch quantity is proportional to annual demand. (a) Total demand for all three products = 30,000 + 18,000 + 12,000 = 60,000 units. Rp = 60,000/1880 = 31.91 pc/hr Tc = 60(0.95)/31.91 = 1.786 min/pc Ts = 1.786 – 0.15 = 1.636 min/pc To determine the number of workers, use Eq. (A15.2). Weighted average Twc = (30,000 × 26 + 18,000 × 19 + 12000 × 28)/60,000 = 24.3 min and balance efficiency = 0.93. w = 24.3/(0.93 × 1.636) = 15.97 rounded to 16 workers (b) To determine production rates, first compute the service time Ts for each product by rearranging Eq. (15.17), Ts = Twc/(wEb). For the drill, TsA = 26/(16 × 0.93) = 1.747 min and TcA = TsA + Tr = 1.747 + 0.15 = 1.897 min; for the sander, TsB = 19/(16 × 0.93) = 1.277 min and TcB = 1.277 + 0.15 = 1.427 min; and for the jig saw, TsC = 28/(16 × 0.93) = 1.882 min and TcC = 1.882 + 0.15 = 2.032 min. Rearranging Eq. (15.2), Rp = 60E/Tc. For the drill, RpA = 60(0.95)/1.897 = 30.04 pc/hr For the sander, RpB = 60(0.95)/1.427 = 39.95 pc/hr For the jig saw, RpC = 60(0.95)/2.032 = 28.06 pc/hr (c) For the drill, batch quantity QA = 30,000/10 = 3000 units; for the sander, QB = 18,000/10 = 1800 units; for the jig saw, QC = 12,000/10 = 1200 units. The production schedule begins with the setup for the drill (4 hr) followed by the run time (QA/RpA = 3000/30.04 = 99.86 hr). Next is the changeover for the sander (4 hr) followed by the run time (1800/39.95 = 45.06 hr). And then the changeover for the jig saw (4 hr) followed by its run time (1200/28.06 = 42.77 hr). This cycle is repeated ten times during the year. Total time = 10(4 + 99.86 + 4 + 45.06 + 4 + 42.77) = 1997 hr, which is just shy of the scheduled 2000 hr/yr. 15.27 Two models, A and B, are to be produced on a mixed-model assembly line that operates 2000 hr/yr. Annual demand and work content time for model A are 24,000 units and 32.0 min, respectively; and for model B are 40,000 and 21.0 min. Line efficiency = 0.95, balance efficiency = 0.93, repositioning time = 0.10 min, and manning level = 1 for all stations. Determine how many workers must be on the production line in order to produce this workload. Answer: Hourly production rates can be determined using Eq. (A15.6): Rp = (24,000 + 40,000)/2000 = 32 units/hr Tc Ts = 1.78125 - 0.1 = 1.68125 min Alternative method using workload calculations: RpA = 24,000/2000 = 12 units/hr and RpB = 40,000/2000 = 20 units/hr WL = 12(32) + 20(21) = 384 + 420 = 804 min/hr Er = Ts/Tc = 1.68125/1.78125 = 0.94386 AT = 60EEbEr = 60(0.95)(0.93)(0.94386) = 50.03 min/hr/ worker w = 804/50.03 = 16.07 → 17 workers 15.28 (A) Three models, A, B, and C, will be produced on a mixed-model assembly line. Hourly production rate and work content time for model A are 10 units/hr and 45.0 min; for model B are 20 units/hr and 35.0 min; and for model C are 30 units/hr and 25.0 min. Line efficiency is 95%, balance efficiency is 0.94, repositioning efficiency = 0.93, and manning level = 1.3. Determine how many workers and workstations must be on the production line in order to produce this workload. Answer: WL = 10(45) + 20(35) + 30(25) = 450 + 700 + 750 = 1900 min/hr AT = 60(0.95)(0.94)(0.93) = 49.83 min/hr per worker w = 1900/49.83 = 38.1 → 39 workers Since M = 1.3, n = w/M = 39/1.3 = 30 stations 15.29 For Problem 15.27, determine the variable rate launching intervals for models A and B. Answer: For product A, Tcv(A) = = 2.144 min For product B, Tcv(B) = = 1.407 min 15.30 For Problem 15.28, determine the variable rate launching intervals for models A, B, and C. Answer: For product A, Tcv(A) = = 1.320 min For product B, Tcv(B) = = 1.027 min For product C, Tcv(C) = = 0.733 min 15.31 For Problem 15.27, determine (a) the fixed rate launching interval, and (b) the launch sequence of models A and B during one hour of production. Answer: Note that the hourly production rates of the two models (12/hr for A and 20/hr for B) are both divisible by 4 (3 per 15 min for A and 5 per 15 min for B). Thus the sequence repeats every 8 launches. The following table indicates the solution for two cycles (16 launches): 15.32 For Problem 15.28, determine (a) the fixed rate launching interval, and (b) the launch sequence of models A, B, and C during one hour of production. Answer: 15.33 Two models, A and B, are to be assembled on a mixed-model line. Hourly production rates for the two models are: A, 25 units/hr; and B, 18 units/hr. The work elements, element times, and precedence requirements are given in the table below. Elements 6 and 8 are not required for model A, and elements 4 and 7 are not required for model B. Assume E = 1.0, Er = 1.0, and Mi = 1. (a) Construct the precedence diagram for each model and for both models combined into one diagram. (b) Find the theoretical minimum number of workstations required to achieve the required production rate. (c) Use the Kilbridge and Wester method to solve the line balancing problem. (d) Determine the balance efficiency for your solution in (c). Work element k TeAk Preceded by: TeBk Preceded by 1 0.5 min - 0.5 min - 2 0.3 min 1 0.3 min 1 3 0.7 min 1 0.8 min 1 4 0.4 min 2 - - 5 1.2 min 2, 3 1.3 min 2, 3 6 - - 0.4 min 3 7 0.6 min 4, 5 - - 8 - - 0.7 min 5, 6 9 0.5 min 7 0.5 min 8 Twc 4.2 min 4.5 min Answer: (b) Minimum number of workstations to achieve production rates for A and B. Element k RpA TeAk RpB TeBk Σ Rpj Tejk 1 12.5 min 9.0 min 21.5 min 2 7.5 min 5.4 min 12.9 min 3 17.5 min 14.4 min 31.9 min 4 10.0 min - 10.0 min 5 30.0 min 23.4 min 53.4 min 6 - 7.2 min 7.2 min 7 15.0 min - 15.0 min 8 - 12.6 min 12.6 min 9 12.5 min 9.0 min 21.5 min Total = 186.0 min Given M = 1, E = 1, Er = 1, therefore, AT = 60 min, n = w = Min Int ≥186.0/60 = 3.1 → 4 stations. (c) Kilbridge & Wester method to solve line balancing problem (TTs = 60 min): List of elements by column Allocation of elements to workstations Element TTk Column 1 21.5 min I 3 31.9 min II 2 12.9 min II 5 53.4 min III 4 10.0 min III 6 7.2 min III 7 15.0 min IV 8 12.6 min IV 9 21.5 min V 186.0 min Station Element TTk TTsi 1 1 21.5 min 3 31.9 min 53.4 min 2 2 12.9 min 4 10.0 min 6 7.2 min 30.1 min 3 5 53.4 min 53.4 min 4 7 15.0 min 8 12.6 min 9 21.5 min 49.1 min 186.0 min (d) Maximum{TTsi} = 53.4 min Eb = 186.0/4(53.4) = 0.871 = 87.1% 15.34 For the data given in previous Problem 15.33, solve the mixed-model line balancing problem except use the ranked positional weights method to determine the order of entry of work elements. Answer: (a) and (b) Same solution as Problem 15.24 (a) and (b). (c) Ranked positional weights method to solve line balancing problem (TTs = 60 min): Station Element TTk TTsi 1 1 21.5 min 3 31.9 min 53.4 min 2 2 12.9 min 4 10.0 min 6 7.2 min 30.1 min 3 5 53.4 min 53.4 min 4 7 15.0 min 8 12.6 min 9 21.5 min 49.1 min 186.0 min Element RPWA RPWB Σ(RPW) 1 105.0 81.0 3 75.0 66.6 141.6 2 75.0 50.4 125.4 5 57.5 45.0 102.5 4 37.5 - 37.5 6 - 28.8 28.8 7 27.5 - 27.5 8 - 21.6 21.6 9 12.5 9.0 21.5 (d) Maximum{TTsi} = 53.4 min Eb = 186.0/4(53.4) = 0.871 = 87.1% 15.35 Three models A, B, and C are to be assembled on a mixed-model line. Hourly production rates for the three models are: A, 15 units/hr; B, 10 units/hr; and C, 5 units/hr. The work elements, element times, and precedence requirements are given in the table below. Assume E = 1.0, Er = 1.0, and Mi = 1. (a) Construct the precedence diagram for each model and for all three models combined into one diagram. (b) Find the theoretical minimum number of workstations required to achieve the required production rate. (c) Use the Kilbridge and Wester method to solve the line balancing problem. (d) Determine the balance efficiency for the solution in (c). Element TeAk Preceded by TeBk Preceded by TeCk Preceded by 1 0.6 min - 0.6 min - 0.6 min - 2 0.5 min 1 0.5 min 1 0.5 min 1 3 0.9 min 1 0.9 min 1 0.9 min 1 4 - 0.5 min 1 - 5 - - 0.6 min 1 6 0.7 min 2 0.7 min 2 0.7 min 2 7 1.3 min 3 1.3 min 3 1.3 min 3 8 - 0.9 min 4 - 9 - - 1.2 min 5 10 0.8 min 6, 7 0.8 min 6, 7, 8 0.8 min 6, 7, 9 Twc 4.8 min 6.2 min 6.6 min Answer: (a) Precedence diagrams: (b) Given M = 1, E = 1, Er = 1, and therefore, AT = 60 min, n = w = Minimum Integer = 2.78 → 3 workstations and 3 (c) Computation of workload: 15.36 For the data given in previous Problem 15.35, (a) solve the mixed-model line balancing problem except that line efficiency = 0.96 and repositioning efficiency = 0.95. (b) Determine the balance efficiency for your solution. Answer: (a) Available time per station = 60(0.96)(0.95) = 54.72 min. Use the same list of elements by column as in Problem 15.35. (b) Balance efficiency Eb = 167/4(54) = 0.773 = 77.3% 15.37 For Problem 15.35, determine (a) the fixed rate launching interval and (b) the launch sequence of models A, B, and C during one hour of production. Answer: 15.38 Two similar models, A and B, are to be produced on a mixed-model assembly line. There are four workers and four stations on the line (Mi = 1 for i = 1, 2, 3, 4). Hourly production rates for the two models are: for A, 7 units/hr; and for B, 5 units/hr. The work elements, element times, and precedence requirements for the two models are given in the table below. As the table indicates, most elements are common to both models. Element 5 is unique to model A, while elements 8 and 9 are unique to model B. Assume E = 1.0 and Er = 1.0. (a) Develop the mixed-model precedence diagram for the two models and for both models combined. (b) Determine a line balancing solution that allows the two models to be produced on the four stations at the specified rates. (c) Using your solution from (b), solve the fixed rate model launching problem by determining the fixed rate launching interval and constructing a table to show the sequence of model launchings during the hour. Answer: Solution Manual for Automation, Production Systems, and Computer-Integrated Manufacturing Mikell P. Groover 9780133499612, 9780134605463