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This document contains Chapters 16 to 17 Chapter 16 AUTOMATED PRODUCTION LINES REVIEW QUESTIONS 16.1 Name three of the four conditions under which automated production lines are appropriate. Answer: The four conditions listed in the text are (1) high product demand, (2) stable product design, (3) long product life, and (4) multiple operations are required to produce the product. 16.2 What is an automated production line? Answer: As defined in the text, an automated production line consists of multiple workstations that are automated and linked together by a work handling system that transfers parts from one station to the next. 16.3 What is a pallet fixture, as the term is used in the context of an automated production line? Answer: As defined in the text, a pallet fixture is a work holding device that is designed to (1) fixture the part in a precise location relative to its base and (2) be moved, located, and accurately clamped in position at successive workstations by the transfer system. 16.4 What is a dial-indexing machine? Answer: A dial-indexing machine is an automated system consisting of multiple workstations that process work parts attached to fixtures around the periphery of a circular worktable, and the table is indexed (rotated in fixed angular amounts) to position the parts at the stations. 16.5 Why are continuous work transport systems uncommon on automated production lines? Answer: Continuous work transport systems are uncommon on automated lines due to the difficulty in providing accurate registration between the station work heads and the continuously moving parts. 16.6 Is a Geneva mechanism used to provide linear motion or rotary motion? Answer: A Geneva mechanism provides rotary motion. 16.7 What is a storage buffer as the term is used for an automated production line? Answer: As defined in the text, a storage buffer is a location in a production line where parts can be collected and temporarily stored before proceeding to downstream stations. 16.8 Name three reasons for including a storage buffer in an automated production line? Answer: The text lists the following five reasons: (1) to reduce the effect of station breakdowns, (2) to provide a bank of parts to supply the line, (3) to provide a place to put the output of the line, (4) to allow for curing time or other required delay associated with processing, and (5) to smooth cycle time variations. 16.9 What are the three basic control functions that must be accomplished to operate an automated production line? Answer: The three basic control functions are (1) sequence control to coordinate the sequence of actions of the transfer system and associated workstations, (2) safety monitoring to ensure that the production line does not operate in an unsafe condition, and (3) quality control to monitor certain quality attributes of the work parts produced on the line. 16.10 Name some of the industrial applications of automated production lines. Answer: Applications listed in the text include machining, sheet metal forming and cutting, spot welding of car bodies in final assembly plants, painting and plating operations, and assembly. 16.11 What is the difference between a unitized production line and a link line? Answer: A unitized production line is an automated production line that consists of standard modules and is assembled in an appropriate configuration to satisfy the production requirements of the customer. A link line is a production line that consists of standard machine tools that are connected together by standard or special material handling devices. 16.12 What are the three problem areas that must be considered in the analysis and design of an automated production line? Answer: The three problem areas identified in the text are (1) line balancing – the same basic problem as in manual assembly lines; (2) processing technology – cutting tool technology, speeds and feeds, and so on; and (3) system reliability – due to the complexity of an automated production line. 16.13 As the number of workstation on an automated production line increases, does line efficiency (a) decrease, (b) increase, or (c) remain unaffected? Answer: (a). Line efficiency decreases because each additional station increases the probability of a line stop. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Geneva Mechanism 16.1 (A) A rotary worktable is driven by a Geneva mechanism with five slots. The driver rotates at 24 rev/min. Determine (a) the cycle time, (b) available process time, and (c) indexing time each cycle. Answer: 16.2 A Geneva with six slots is used to operate the worktable of a dial-indexing machine. The slowest workstation on the dial-indexing machine has an operation time of 2.5 sec, so the table must be in a dwell position for this length of time. (a) At what rotational speed must the driven member of the Geneva mechanism be turned to provide this dwell time? (b) What is the indexing time each cycle? Answer: (a) θ = 360/6 = 60° Ts = 180 + 60/360N = 0.667/N = 2.5 sec (given) N = 0667/2.5 = 0.2667 rev/sec = 16 rev/min (b) Tr = 180 – 60/360(0.2667) = 1.25 sec 16.3 Solve the previous problem except that the Geneva has eight slots. Answer: (a) = θ = 360/8 = 45° Ts = 180 + 45/360N = 0.625/N = 2.5 sec(given) N = 0.625/2.5 = 0.25 rev/sec = 15 rev/min (b) Tr = 180 – 45/360(0.25) = 1.50 sec Automated Production Lines (No Internal Storage) 16.4 (A) A 12-station automated production line has an ideal cycle time of 30 sec. Line stops occur on average once every 20 cycles. When a line stop occurs, average downtime is 4.0 min. Cost of each starting work part is $1.55, and the cost to operate the line is $66/hr. Tooling cost is $0.27 per work part. Determine (a) average hourly production rate, (b) line efficiency, and (c) cost of a workpiece produced. Answer: (a) Tp = 0.5 + (1/20)(4) = 0.5 + 0.2 = 0.7 min Rp = 1/0.7 = 1.4286 pc/min = 85.7 pc/hr (b) E = 0.5/0.7 = 0.714 = 71.4% (c) Cpc = 1.55 + 66(0.7)/60 + 0.27 = 1.55 + 0.77 + 0.27 = $2.59/pc 16.5 In the operation of a 10-station transfer line, the ideal cycle time is 1.08 min. Line stops occur due to random mechanical and electrical failures once every 28 cycles on average. When a line stop occurs, average downtime is 6.0 min. In addition to these downtimes, the tools at each workstation on the line must be changed every 100 cycles, which takes a total of 12.0 min for all ten stations. Determine (a) average hourly production rate, (b) line efficiency, and (c) proportion downtime. Answer: There are two causes of downtime: (1) random failures and (2) tool changes. (a) F1Td1 = (1/28)(6.0) = 0.214 min F2Td2 = 12.00/100 = 0.12 min Tp = 1.08 + 0.214 + 0.12 = 1.414 min Rp = 1/1.414 = 0.7072 pc/min = 42.4 pc/hr (b) E = 1.08/1.414 = 0.764 = 76.4%, (c) D = (0.214 + 0.12)/1.414 = 0.236 = 23.6% 16.6 A six-station dial-indexing machine operates at an ideal cycle rate of 10/min and has a breakdown frequency of 0.03 stops/cycle. Average downtime per breakdown is 3.5 min. Determine (a) average hourly production rate and (b) line efficiency. Answer: (a) Tc = 1/Rc = 1/20 = .05 min = 3.0 sec Tp = 0.1 + 0.03(3.5) = 0.1 + 0.105 = 0.205 min Rp = 60/0.205 = 292.7 pc/hr (b) E = 0.1/0.155 = 0.488 = 48.8% 16.7 In the operation of a 15-station automated production line, the ideal cycle time = 0.85 min. Breakdowns occur at a rate of once every 35 cycles on average, and the average downtime per breakdown is 9.2 min. The production line is located in a plant that works an 8-hr day, 5 days/wk. Determine (a) line efficiency and (b) number of parts the production line produces in a 40-hr week? Answer: (a) Tp = 0.85 + 9.2/35 = 0.85 + 0.263 = 1.113 min E = 0.85/1.113 = 0.764 = 76.4% (b) Rp = 60/1.113 = 53.9 pc/hr Weekly production Rpw = 40(53.9) = 2156 pc/wk 16.8 A 17-station in-line transfer machine has an ideal cycle time of 1.35 min. Station breakdowns occur with a probability of 0.01. Average downtime is 8.0 min per line stop. The starting work part is a casting that costs $3.20. Operating cost of the transfer line is $108/hr, and tooling cost is $0.07 per piece per station. Determine (a) ideal production rate, (b) frequency of line stops, (c) average actual production rate, (d) line efficiency and (e) cost per completed part. Answer: (a) Rc = 1/Tc = 1/1.35 = 0.741 pc/min = 44.44 pc/hr (b) F = np = 17(0.01) = 0.17 (c) Tp = 1.35 + 0.17(8) = 1.35 + 1.36 = 2.71 min Rp = 60/2.71 = 22.14 pc/hr (d) E = 1.35/2.71 = 0.498 = 49.8% (e) Cpc = 3.20 + (108/60)(2.71) + 17(0.07) = 3.20 + 4.88 + 1.19 = $9.27/pc 16.9 A ten-station rotary-indexing machine performs machining operations at nine workstations, and the tenth station is used for unloading and loading parts. The longest process time on the line is 1.75 min and the loading/unloading operation requires less time than this. It takes 9.0 sec to index the machine between workstations. Stations break down with a frequency of 0.006, which is considered equal for all ten stations. When breakdowns occur, it takes an average of 8.0 min to diagnose the problem and make repairs. The starting work part costs $2.50 per unit. Operating cost of the indexing machine is $96/hr, and tooling cost is $0.38 per piece. Determine (a) line efficiency, (b) average hourly production rate, and (c) completed part cost. Answer: (a) F = np = 10(0.006) = 0.06 Tc = 1.75 + 9/60 = 1.90 min Tp = 1.90 + 0.06(8) = 1.90 + 0.48 = 2.38 min/pc E = 1.90/2.38 = 0.798 = 79.8% (b) Rp = 60/2.38 = 25.2 pc/hr (c) Cpc = 2.50 + (96/60)(2.38) + 0.38 = 2.50 + 3.81 + 0.38 = $6.69/pc 16.10 (A) A transfer line has six stations that function as listed in the table below. Transfer time = 0.18 min. Average downtime per occurrence = 8.0 min. A total of 20,000 parts must be processed through the transfer machine. Determine (a) proportion downtime, (b) average hourly production rate, and (c) how many hours of operation are required to produce the 20,000 parts. Station Operation Process time pi 1 Load part 0.78 min 0 2 Drill three holes 1.25 min 0.02 3 Ream two holes 0.90 min 0.01 4 Tap two holes 0.85 min 0.04 5 Mill flats 1.32 min 0.01 6 Unload parts 0.45 min 0 Answer: (a) Tc = 1.32 + 0.18 = 1.50 min F = 0.02 + 0.01 + 0.04 + 0.01 = 0.08 Tp = 1.50 + 0.08(8.0) = 1.50 + 0.64 = 2.14 min D = 0.64/2.14 = 0.299 = 29.9% (b) Rp = 60/2.14 = 28.04 pc/hr (c) H = 20,000(2.14/60) = 713.3 hr 16.11 The cost to operate a 20-station transfer line is $144/hr. The line operates with an ideal cycle time of 0.90 min. Downtime occurrences happen on average once per 34 cycles. Average downtime per occurrence is 10.0 min. It is proposed that a new computer system and associated sensors be installed to monitor the line and diagnose downtime occurrences when they happen. This new system is expected to reduce downtime per occurrence from 10 min to 7.5 min. (a) If the cost of purchasing and installing the new system is $12,000, how many parts must the system produce for the savings to pay for the computer system? (b) How many hours of operation will be required to produce this number of parts? Answer: (a) Current operation: Tp = 0.90 + 10/34 = 0.90 + 0.294 = 1.194 min/pc Cpc = ($144/60)(1.194) = $2.8656/pc With computer system: Tp = 0.90 + 7.5/34 = 0.90 + 0.221 = 1.121 min/pc Cpc = ($144/60)(1.121) = $2.6904/pc 12,000 = (2.8656 – 2.6904)Q = 0.1752 Q Q = 12,000/0.1752 = 68,493 pc (b) With the system installed, Tp = 1.121 min and Rp = 60/1.121 = 53.52 pc/hr H = 68,493/53.52 = 1279.8 hr 16.12 The operation of a 16-station transfer line has been logged for five days (40 hr). During this time there were a total of 127 downtime occurrences on the line for a total downtime of 682 min. Of the total occurrences, 105 were station failures and 22 were transfer mechanism failures. The line performs a sequence of machining operations, the longest of which takes 0.52 min. The transfer mechanism takes 0.08 min to move parts from one station to the next each cycle. Determine the following based on the five-day observation period: (a) number of parts produced, (b) line efficiency, (c) production rate, (d) average downtime per line stop, and (e) frequency rate associated with the transfer mechanism failures. Answer: (a) Tc = 0.52 + .08 = 0.60 min QTp = QTc + QFTd QTp = 40(60) = 2400 min, QFTd = 682 min, and QTc = 0.60Q 2400 min = 0.60Q + 682 0.60Q = 2400 - 682 = 1718 min Q = 1718/0.60 = 2863 pc (b) E = 1718/2400 = 0.716 = 71.6% (c) Rp = 2863 pc/40 hr = 71.6 pc/hr (d) Td = 682/127 = 5.37 min (e) Transfer mechanism breakdown frequency: p = 22/2863 = 0.0077 breakdowns/pc 16.13 An eight-station rotary indexing machine performs the machining operations shown in the table below, with processing times and breakdown frequencies for each station. Transfer time is 0.15 min. A study of the system was undertaken, during which time 2000 parts were completed. The study also revealed that when breakdowns occur, the average downtime is 7.0 min. For the study period, determine (a) average hourly production rate, (b) line uptime efficiency, and (c) how many hours were required to produce the 2000 parts. Station Process Process time Breakdowns 1 Load part 0.50 min 0 2 Mill top 0.85 min 22 3 Mill sides 1.10 min 31 4 Drill two holes 0.60 min 47 5 Ream two holes 0.43 min 8 6 Drill six holes 0.92 min 58 7 Tap six holes 0.75 min 84 8 Unload part 0.40 min 0 Answer: (a) Tc = 1.10 + 0.15 = 1.25 min/cycle Number of breakdowns = 0 + 22 + 31 + 47 + 8 + 58 + 84 + 0 = 250 F = 250/2000 = 0.125 Tp = 1.25 + 0.125(7.0) = 2.125 min/pc Rp = 60/2.125 = 28.2353 pc/hr (b) E = 1.25/2.125 = 0.588 = 58.8% (c) Total time H = 2000(2.125)/60 = 70.83 hr 16.14 A 14-station transfer line has been observed for 50 hours to identify type of downtime occurrence, how many occurrences, and time lost. The results: 68 line stops were due to tool changes for a total downtime of 329 min, 45 line stops were random station failures for 242 min, and 20 line stops resulted from transfer mechanism failures for 98 min. The ideal cycle time for the line is 0.60 min, which includes transfer time. Determine (a) how many parts were produced during the 50 hours, (b) line efficiency, (c) average production rate per hour, and (d) frequency associated with transfer mechanism failures. Of the three reasons for downtime occurrences, which one has the longest average downtime per occurrence? Answer: Total downtime = 329 + 242 + 98 = 669 min Number of downtime occurrences = 68 + 45 + 20 = 133 occurrences (a) Total uptime = 50(60) - 669 = 2331 min Number of parts = (2331 min)/(0.60 min/pc) = 3885 pc (b) E = 2331/3000 = 0.777 = 77.7% (c) Rp = 3885 pc/50 hr = 77.7 pc/hr (d) Transfer mechanism failures: p = 20/3885 = 0.00515 (e) Tool change Td = 329/68 = 4.84 min Station failure Td = 242/45 = 5.38 min Transfer mechanism Td = 98/20 = 4.90 min Station failures are associated with the longest average downtime per occurrence. 16.15 An automated production line operates with an ideal cycle time of 35 sec. Line stops are characterized by a mean time between failures of 70 min and a mean time to repair of 8.0 min. What is the average hourly production rate? Answer: Given Tc = 35 sec = 0.5833 min/pc Line efficiency E = Availability A = MTBF MTTR/ MTBF = 70 – 8/70 = 0.886 E = Tc/Tp, therefore Tp = Tc/E = 0.5833/0.886 = 0.658 min/pc Rp = 60/0.658 = 91.1 pc/hr 16.16 A machine shop is negotiating with a potential customer on a job that would consist of producing 120,000 parts in the first year for a contracted price of $450,000. It is not known whether there would be a continuation of the job after that, so the shop must break even on the work during that first year. The engineering department has proposed an automated production line that would operate with an ideal cycle time of 1.50 min. It is anticipated that the line efficiency would be 80% and that average down time per line stop would be 7.0 min. Material cost will be $1.10 per starting work part and tooling cost is estimated at $0.25 per part. A 3% scrap rate must be planned for this job, so more than 120,000 parts must be processed to achieve the contracted quantity. Finally, to separate the good parts from the defectives, an inspection cost of $0.06 per part must be factored in. (a) How many hours would the line have to operate to produce the 120,000 parts? (b) During that time, how many breakdowns would occur? (c) Ignoring overhead costs, what is the maximum installed cost of the line for the shop to break even on the job. Answer: (a) Given Tc = 1.5 min/pc E = Tc/Tp, therefore Tp = Tc/E = 1.5/0.80 = 1.875 min/pc To produce 120,000 parts, the starting quantity Qo = 120,000/(1-0.03) = 123,711 H = 123,711(1.875/60) = 3866 hr (b) Given Td = 7.0 min, QTp = QTc + QFTd 3866 = 123,711(1.5/60) + 123,711F(7.0/60) 3866 = 3093 + 14,433F F = (3866 – 3093)/14,433 = 0.05356 Number of breakdowns = QF = 123,711(0.05356) = 6626 breakdowns (c) QCpc = QCm + QCoTp + QCt 450,000 = 123,711(1.10) + 123,711(1.875/60)Co + 123,711(0.25) + 123,711(0.06) 450,000 = 136,082 + 3866Co + 30,928 + 7423 3866Co = 450,000 – 136,082 – 30,928 + 7423 = 275,567 Co = 275,567/3866 = $71.28/hr Maximum installed cost of production line = ($71.28/hr)(3866 hr) = $275,567 16.17 (A) A part is to be produced on an automated production line. Total work content time to make the part is 36 min, and this work will be divided evenly among the workstations, so that the processing time at each station is 36/n, where n = the number of stations. In addition, the time required to transfer parts between workstations is 6 sec. Thus, the cycle time = (0.1 + 36/n) min. In addition, it is anticipated that the station breakdown frequency will be 0.005, and that the average downtime per breakdown will be 8.0 min. (a) How many workstations should be included in the line to maximize production rate? Also, what are (b) the hourly production rate and (c) line efficiency for this number of stations? Answer: (a) Tc = 6/60 + 36/n = 0.1 + 36/n FTd = pnTd = 0.005n(8.0) = 0.04n Tp = 0.1 + 36/n + 0.04n dTdnp/dn = −36/n2 + 0.04 = 0, rearranging, 36/n2 = 0.04 n2 = 36/0.04 = 900 n = 30 stations (b) Tp = 0.1 + 36/30 + 0.04(30) = 0.1 + 1.2 + 1.2 = 2.5 min/pc Rp = 60/2.5 = 24 pc/hr (c) E = (0.1 + 1.2)/2.5 = 1.3/2.5 = 0.52 = 52% Automated Production Lines with Storage Buffers (Appendix A16) 16.18 (A) A 30-station transfer line has an ideal cycle time of 0.75 min, an average downtime of 6.0 min per line stop occurrence, and a station failure frequency of 0.01 for all stations. A proposal has been submitted to locate a storage buffer between stations 15 and 16 to improve line efficiency. Determine (a) the current line efficiency and production rate, and (b) the maximum possible line efficiency and production rate that would result from installing the storage buffer. Answer: (a) Tp = 0.75 + 30(0.01)(6.0) = 0.75 + 1.8 = 2.55 min/pc E = 0.75/2.55 = 0.2941 = 29.41% Rp = 1/2.55 = 0.392 pc/min = 23.53 pc/hr (b) Tp1 = Tp2 = 0.75 + 15(0.01)(6.0) = 0.75 + 0.90 = 1.65 min/pc E∞ = 0.75/1.65 = 0.4545 = 45.45% Rp = 1/1.65 = 0.6061 pc/min = 36.36 pc/hr 16.19 Given the data in Problem 16.18, solve the problem except that (a) the proposal is to divide the line into three stages, that is, with two storage buffers located between stations 10 and 11, and between stations 20 and 21, respectively; and (b) the proposal is to use an asynchronous line with large storage buffers between every pair of stations on the line; that is a total of 29 storage buffers. Answer: (a) Tp1 = Tp2 = Tp3 = 0.75 + 10(0.01)(6.0) = 0.75 + 0.60 = 1.35 min/pc For each stage, E = 0.75/1.35 = 0.5555 = 55.55% Rp = 1/1.35 = 0.7407 pc/min = 44.44 pc/hr (b) Tp1 = Tp2 = . . . = Tp29 = 0.75 + 0.01(6.0) = 0.75 + 0.06 = 0.81 min/pc For each stage, E = 0.75/0.81 = 0.926 = 92.6% Rp = 1/0.81 = 1.235 pc/min = 74.1 pc/hr 16.20 In Problem 16.18, if the capacity of the proposed storage buffer is to be 20 parts, determine (a) line efficiency, and (b) production rate of the line. Assume that the downtime (Td = 6.0 min) is a constant. Answer: 16.21 Solve Problem 16.20 but assume that the downtime (Td = 6.0 min) follows the geometric repair distribution. Answer: 16.22 In the transfer line of Problems 16.18 and 16.20, suppose it is more technically feasible to locate the storage buffer between stations 11 and 12, rather than between stations 15 and 16. Determine (a) the maximum possible line efficiency and production rate that would result from installing the storage buffer, and (b) the line efficiency and production rate for a storage buffer with a capacity of 20 parts. Assume that the downtime per line stop (Td = 6.0 min) is a constant. Answer: 16.23 A proposed synchronous transfer line will have 20 stations and will operate with an ideal cycle time of 0.5 min. All stations are expected to have an equal probability of breakdown, p = 0.01. The average downtime per breakdown is expected to be 5.0 min. An option under consideration is to divide the line into two stages, each stage having 10 stations, with a buffer storage zone between the stages. It has been decided that the storage capacity should be 20 units. The cost to operate the line is $96.00/hr. Installing the storage buffer would increase the line operating cost by $12.00/hr. Ignoring material and tooling costs, determine (a) line efficiency, production rate, and unit cost for the one-stage configuration, and (b) line efficiency, production rate, and unit cost for the optional two-stage configuration (assume a constant repair time). Answer: (a) For the current line operation: Tp = 0.5 + 20(.01)(5) = 1.5 min E = 0.5/1.5 = 0.333 = 33.3% Rp = 60/1.5 = 40 pc/hr Cpc = ($96/60)(1.5) = $2.40/pc (b) For the proposed two-stage line: Td/Tc = 10, b = 20: B = 2, L = 0 F1 = F2 = 10(0.01) = 0.10, r = 0.1/0.1 = 1.0 Tp = 0.5 + 0.1(5) = 1.0 min h(20) = 2/3 = 0.6667 Eo = 0.5/1.5 = 0.3333, E2 = 0.5/1.0 = 0.5, and D1 = 0.01(10)(5)/1.5 = 0.3333. E = .3333 + .3333(.6667)(.5) = 0.4444 Tp = 0.5/0.4444 = 1.125 min, Rp = 60/1.125 = 53.33 pc/hr Co = $96 + $12 = $108/hr Cpc = ($108/60)(1.125) = $2.025/pc 16.24 A two-week study has been performed on a 12-station transfer line that is used to machine engine heads for an automotive company. During 80 hours of observation, the line was down 42 hr, and 1689 parts were completed. The accompanying table lists the machining operation performed at each station, the process times, and the downtime occurrences for each station. Transfer time between stations is 6 sec. To address the downtime problem, it has been proposed to divide the line into two stages, each consisting of six stations. The storage buffer between the stages would have a storage capacity of 20 parts. Determine (a) line efficiency and production rate of the current one-stage configuration and (b) line efficiency and production rate of the proposed two-stage configuration. (c) Given that the line is to be divided into two stages, should each stage consist of six stations as proposed, or is there a better division of stations into Answer: 16.25 In Problem 16.24, the current line has an operating cost of $66.00/hr. The starting workpart is a casting that costs $4.50 per piece. Disposable tooling costs $1.25 per piece. The proposed storage buffer will add $6.00/hr to the operating cost of the line. Does the improvement in production rate justify this $20 increase? Answer: Current line: Cpc = 4.50 + 66(2.842/60) + 1.25 = $8.88/pc Two stage line in which division into stages is between stations 6 and 7: Cpc = 4.50 + 72/25.31 + 1.25 = $8.59/pc Two stage line in which division into stages is between stations 7 and 8. See solution of Problem 16.24(c): Cpc = 4.50 + 72/27.23 + 1.25 = $8.39/pc Improvement is justified for either division. 16.26 A 16-station transfer line can be divided into two stages by installing a storage buffer between stations 8 and 9. The probability of failure at any station is 0.01. The ideal cycle time is 1.0 min and the downtime per line stop is 10.0 min. These values are applicable for both the one-stage and two-stage configurations. The downtime should be a considered constant value. The cost of installing the storage buffer is a function of its capacity. This cost function is Cb = $0.60b/hr = $0.01b/min, where b = the buffer capacity. However, the buffer can only be constructed to store increments of 10 (in other words, b can take on values of 10, 20, 30, etc.). The cost to operate the line itself is $120/hr. Ignore material and tooling costs. Based on cost per unit of product, determine the buffer capacity b that will minimize unit product cost. Answer: With no buffer storage (b = 0): Tp = 1.0 + 16(0.01)(10) = 2.6 min/pc Cpc = 120(2.6/60) = 2.00(2.6) = $5.20/pc, Eo = 1/2.6 = 0.3846 With two stages, each stage would operate the same way, with n = 8 stations Tp = 1.0 + 8(0.01)(10) = 1.8 min/pc E1 = E2 = 1/1.8 = 0.5555 D1' = (8 × 0.01 × 10)/(1 + 16 × 0.01 × 10) = 0.3077 h(b) = B/B + 1 , b = 10, 20, 30, 40, etc. → B = 1, 2, 3, 4, etc. Buffer cost = 0.1B/min, CL = (2.0 + 0.1B)/min B = 0 Cpc = (2.0 + .1 × 0)(0 + 1)/.3846 = $5.20/pc B = 1 Cpc = (2.0 + .1 × 1)(1 + 1)/.9401 = $4.468/pc B = 2 Cpc = (2.0 + .1 × 2)(2 + 1)/1.4956 = $4.413/pc B = 3 Cpc = (2.0 + .1 × 3)(3 + 1)/2.0511 = $4.485/pc B = 4 Cpc = (2.0 + .1 × 4)(4 + 1)/2.6066 = $4.604/pc B = 5 Cpc = (2.0 + .1 × 5)(5 + 1)/3.1621 = $4.744/pc Lowest cost is at B = 2 or b = 20 unit capacity 16.27 (A) The uptime efficiency of a 20 station automated production line is only 40%. The ideal cycle time is 48 sec, and the average downtime per line stop occurrence is 3.0 min. Assume the frequency of breakdowns for all stations is equal (pi = p for all stations) and that the downtime is constant. To improve uptime efficiency, it is proposed to install a storage buffer with a 15-part capacity for $14,000. The present production cost is $4.00 per unit, ignoring material and tooling costs. How many units would have to be produced for the $14,000 investment to pay for itself? Answer: F = np = 20p Tp = Tc + 20pTd Tp = Tc/E = 0.8/0.4 = 2.0 min/pc 2.0 = 0.8 + 20p(3) = 0.8 + 60p 60p = 2.0 - 0.8 = 1.2 p = 1.2/60 = 0.02 Cpc = CoTp = $4.00 = Co(2.0), Co = $2.00/min Two stage line each with 10 stations: Eo = 0.4 Tp = 0.8 + 10(0.02)(3.0) = 1.4 min E2 = 0.8/1.4 = 0.5714 D1' = 0.6/2.0 = 0.3 With constant repair time, b = 15 and Td/Tc = 3.75 so B = 4, L = 0 h(b) = h(15) = 4/(4 + 1) = 0.8 E = 0.4 + 0.3(0.8)(0.5714) = .5371 Tp = Tc/E = 0.8/0.5371 = 1.4895 min Cpc = 2.00(1.4895) = 2.979/pc Break-even point Q: 4.00Q = 14,000 + 2.979Q (4.00 - 2.979)Q = 14,000 1.021Q = 14,000 Q = 13,712 pc 16.28 A transfer line is divided into two stages with a storage buffer between them. Each stage consists of 9 stations. The ideal cycle time of each stage = 1.0 min, and frequency of failure for each station is 0.01. Average downtime per stop is 8.0 min, and a constant downtime distribution is assumed. Determine the required capacity of the storage buffer such that the improvement in line efficiency compared to a zero buffer capacity would be 80% of the improvement yielded by a buffer with infinite capacity. Answer: For b = 0, F = 18(0.01) = 0.18 Tp = 1.0 + 0.18(8.0) = 1.0 + 1.44 = 2.44 min/pc Eo = 1.0/2.44 = 0.4098 For b = ∞, F = 9(0.01) = 0.09 Tp = 1.0 + 0.09(8.0) = 1.0 + 0.72 = 1.72 min/pc E∞ = 1.0/1.72 = 0.5814 E1 = E2 = 0.5814 ΔE = 0.5814 - 0.4098 = 0.1716 80% of ΔE = 0.80(0.1716) = 0.1373 E = 0.4098 + 0.1373 = 0.5471 = E at 80% improvement Eb = Eo + D1' h(b)E2, so D1' h(b)E2 = 0.1373 D1' = 0.72/2.44 = 0.2951 0.1373 = 0.2951(h(b)) (0.5814) h(b) = = 0.80 For ease of calculations, ignore the second term. If B turns out to be an integer, the second term does not apply. If B is not an integer, a trial and error solution will be required to find B and L. h(b) = 0.80 = B/B + 1 0.80(B + 1) = B 0.80 B + 0.80 = B 0.80 = 0.20B B = 4 → b = B Td/Tc = 4(8/1) = 32 = buffer capacity 16.29 A 20-station transfer line presently operates with a line efficiency E = 1/3. The ideal cycle time = 1.0 min. The repair distribution is geometric with an average downtime per occurrence of 8.0 min, and each station has an equal probability of failure. It is possible to divide the line into two stages with 10 stations each, separating the stages by a storage buffer of capacity b. With the information given, determine the required value of b that will increase the efficiency from E = 1/3 to E = 2/5. Answer Chapter 17 AUTOMATED ASSEMBLY SYSTEMS REVIEW QUESTIONS 17.1 Name three of the four conditions under which automated assembly technology should be considered. Answer: The four conditions named in the text are (1) high product demand, so that millions of units are produced, (2) stable product design, because design changes require tooling changes in the assembly system, (3) the assembly consists of no more than a limited number of components, say around a dozen or fewer, and (4) the product is designed for automated assembly. 17.2 What are the four automated assembly system configurations listed in the text? Answer: The four configurations are (a) in-line assembly machine, (b) dial-type assembly machine, (c) carousel assembly system, and (d) single-station assembly machine. 17.3 Name the typical hardware components of a workstation parts delivery system. Answer: The typical components are (1) hopper, (2) parts feeder, (3) selector and/or orientor, (4) feed track, and (5) escapement and placement device. 17.4 Name six typical products that are made by automated assembly. Answer: The products identified in the text are alarm clocks, audio tape cassettes, ball bearings, ball point pens, cigarette lighters, computer disks, electrical plugs and sockets, fuel injectors, gear boxes, light bulbs, locks, mechanical pens and pencils, printed circuit board assemblies, pumps for household appliances, small electric motors, spark plugs, video tape cassettes, and wrist watches. 17.5 Considering the assembly machine as a game of chance, what are the three possible events that might occur when the feed mechanism attempts to feed the next component to the assembly work head at a given workstation in a multi-station system? Answer: The three possible events are (1) the component is defective and causes a station jam, (2) the component is defective but does not cause a station jam, and (3) the component is not defective. 17.6 Name some of the important performance measures for an automated assembly system. Answer: The performance measures considered in the text include yield (proportion of assemblies produced with no defective components), production rate, proportion uptime (a.k.a. line efficiency), and unit cost per assembly. 17.7 Why is the production rate inherently lower on a single-station assembly system than on a multi-station assembly system? Answer: Because all of the work elements are performed sequentially at one station in the single-station system, whereas the elements are performed simultaneously at multiple workstations in a multi-station system. 17.8 What are two reasons for the existence of partially automated production lines? Answer: The two reasons given in the text are (1) automation is introduced gradually on an existing manual line and (2) certain manual operations are too difficult or too costly to automate. 17.9 What are the effects of poor quality parts, as represented by the fraction defect rate, on the performance of an automated assembly system? Answer: The two effects given in the text are (1) jams at stations that stop the entire assembly system to adversely affect production rate, uptime proportion, and cost per unit produced; or (2) assembly of defective parts in the product to adversely affect yield of good assemblies and product cost. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Parts Feeding 17.1 (A) The feeder-selector device at one of the workstations of an assembly machine has a feed rate of 56 components/min and provides a throughput of one part in five. The ideal cycle time of the assembly machine is 6 sec. The low-level sensor on the feed track is set at 10 components, and the high-level sensor is set at 25 components. (a) How long will it take for the supply of components to be depleted from the high-level sensor to the low-level sensor once the feeder-selector device is turned off? (b) How long will it take for the components to be resupplied from the low-level sensor to the high-level sensor, on average, after the feeder-selector device is turned on? (c) What are the time proportions that the feeder-selector device is turned on and turned off? Answer: (a) Time to deplete from nf2 to nf1 Rate of depletion = cycle rate Rc = 60/Tc = 60/6 = 10 parts/min Time to deplete Tde = (25 – 10)/10 = 15/10 = 1.5 min (b) Time to resupply from nf1 to nf2 Rate of resupply = fθ - Rc = 56(1/5) – 10 = 11.2 – 10 = 1.2 parts/min Time to resupply = Tre = (25 – 10)/1.2 = 15/1.2 = 12.5 min (c) Total cycle of depletion and resupply = 1.5 + 12.5 = 14.0 min Proportion of time feeder-selector is on = 12.5/14 = 0.893 Proportion of time feeder-selector is off = 1.5/14 = 0.107 17.2 Solve Problem 17.1 but use a feed rate of 50 parts/min. How does the reduced feed rate of the feeder-selector affect the operation of the assembly machine? Answer: (a) Time to deplete from nf2 to nf1 Rate of depletion = cycle rate Rc = 60/Tc = 60/6 = 10 parts/min Time to deplete Tde = (25 - 10)/10 = 1.5 min (b) Time to resupply from nf1 to nf2 Rate of resupply = 50(1/5) - 10 = 10 - 10 = 0 parts/min Time to resupply = Tre = (20 - 10)/0 = 10/0 = ∞ (c) Total cycle time = 1.5 + ∞ = ∞ min Proportion of time feeder-selector turned on = 100% Proportion of time feeder-selector turned off = 0 Answer to question: A feed rate of 50 components per min is probably not sufficient because the feeder-selector will be on all of the time and there will be periods when the feeder-selector will not be able to keep up with the cycle rate. 17.3 The ideal cycle time of an assembly machine is 5 sec. The parts feeder at one of the workstations has a feed rate of 75 components/min and the probability that the components will pass through the selector is 18%. The active length of the feed track (where the highlevel sensor is located) is 350 mm. The low-level sensor on the feed track is located 100 mm from the station work head. The components have a length of 12.5 mm in the feed track direction, and there is no overlapping of parts. (a) How long will it take for the supply of parts to be depleted from the high-level sensor to the low-level sensor once the feeder-selector device is turned off? (b) How long will it take for the parts to be resupplied from the low-level sensor to the high-level sensor, on average, after the feeder-selector device is turned on? (c) What are the time proportions that the feeder-selector device is turned on and turned off? Answer: (a) Number of components in feed track at nf2 is 350/12.5 = 28 components Number of components in feed track at nf1is 100/12.5 = 8 components Rate of depletion = cycle rate Rc = 60/Tc = 60/5 = 12 parts/min Time to deplete from nf2 to nf1 = Tde = (28 – 8)/12 = 20/12 = 1.667 min (b) Time to resupply from nf1 to nf2 Rate of resupply = fθ - Rc = 75(0.18) – 12 = 13.5 – 12 = 1.5 parts/min Time to resupply from nf1 to nf2 = Tre = (28 – 8)/1.5 = 20/1.2 = 16.667 min (c) Total cycle of depletion and resupply = 1.667 + 16.667 = 18.333 min Proportion of time feeder-selector is on = 16.667/18.333 = 0.909 Proportion of time feeder-selector is off = 1.667/18.333 = 0.091 17.4 An assembly machine has eight stations and must produce at an average rate of 500 completed assemblies/hr. Average downtime per breakdown is 2.5 min. When a breakdown occurs, all subsystems (including the feeder) stop. The frequency of breakdowns of the assembly machine is once every 55 parts. Average downtime per breakdown is 2.0 min. One of the stations is an automatic assembly operation that uses a feeder-selector. Components fed into the selector have a 20% probability of passing through. Parts rejected by the selector are fed back into the hopper. What minimum rate must the feeder deliver components to the selector during system uptime in order to keep up with the assembly machine? Answer: Tp = 60/500 = 0.12 min/asby Tp = Tc + FTd = Tc + (1/55)(2.0) = Tc + 0.03636 Tc = 0.12 - 0.03636 = 0.08364 min/asby Rc = 1/Tc = 1/0.08364 = 11.95 asby/min Min fθ = 0.20 f = 11.95 asby/min Feeder rate f = 11.95/0.20 = 59.8 parts/min Multi-Station Assembly Systems 17.5 (A) A ten-station assembly machine has an ideal cycle time of 6 sec. The fraction defect rate at each station is 0.005 and a defect always jams the affected station. When a breakdown occurs, it takes 1.2 min, on average, for the system to be put back into operation. Determine (a) the hourly production rate for the assembly machine, (b) yield of good product (final assemblies containing no defective components), and (c) proportion uptime of the system. Answer: (a) If defects always jam the affected station, then m = 1.0. Tp = 6/60 + 10(1.0)(0.005)(1.2) = 0.16 min/asby Production rate Rp = 60/0.16 = 375 asby/hr (b) Yield Pap = (1 - 0.005 + 1 × 0.005)8 = 1.0 (c) Proportion uptime E = 0.1/0.16 = 0.625 = 62.5% 17.6 Solve Problem 17.5 but assume that defects never jam the workstations. Other data are the same. Answer: If defects never jam, then m = 0. Tp = 6/60 + 10(0)(0.005)(1.2) = 0.10 min/asby Rp = 60/0.10 = 600 asby/hr (b) Yield Pap = (1 - 0.005 + 0 × 0.005)8 = (0.995)8 = 0.9607 (c) Proportion uptime E = 0.1/0.1 = 100% 17.7 Solve Problem 17.5 but assume that m = 0.5 for all stations. Other data are the same. Answer: Tp = 0.1 + 10(0.5)(0.005)(1.2) = 0.13 min/asby Rp = 60/0.13 = 461.5 asby/hr (b) Yield Pap = (1 - 0.005 + 0.5 × 0.005)8 = (0.9975)8 = 0.9802 (c) Proportion uptime E = 0.1/0.13 = 0.769 = 76.9% 17.8 A six-station dial-indexing machine assembles components to a base part. The operations, element times, q and m values for components added are given in the table below (NA means q and m are not applicable to the operation). The indexing time is 2 sec. When a jam occurs, it requires 1.5 min to release the jam and put the machine back in operation. Determine (a) hourly production rate for the assembly machine, (b) yield of good product (final assemblies containing no defective components), and (c) proportion uptime of the system. Answer: (a) Σ(mq) = 0.6(0.015) + 0.8(0.01) + 1(0.02) + .5(0.01) = .042 Tp = 0.1333 + 0.042(1.5) = 0.19633 min/asby Rp = 60/0.19633 = 305.6 asby/hr (b) Pap = (1-0.015 + 0.6 × 0.015)(1-0.01 + 0.8 × 0.01)(1-0.02 + 1 × 0.02)(1-0.01 + 0.5 × 0.01) = (0.994)(0.998)(1.0)(0.995) = 0.98705 (c) E = 0.1333/0.19633 = 0.679 = 67.9% 17.9 A six-station automatic assembly line runs 4000 hr/yr and has an ideal cycle time of 10 sec. Downtime occurs for two reasons. First, mechanical and electrical failures cause line stops that occur with a frequency of once per 120 cycles. Average downtime for these causes is 3.0 min. Second, defective components also result in downtime. The fraction defect rate of each of the six components added to the base part at the six stations is 1.0%, and the probability that a defective component will cause a station jam is 0.5 for all stations. Downtime per occurrence for defective parts is 2.0 min. Determine (a) total number of assemblies produced in one year, (b) number of assemblies with at least one defective component, and (c) number of assemblies with all six defect components. Answer: (a) Tp = 10/60 + (1/120)(3) + 6(0.5)(0.01)(2) = 0.25167 min Rp = 60/0.25167 = 238.41 asby/hr In 2000 hr, Q = 4000(238.41) = 953,642 asby (b) Yield Pap = (1 - 0.01 + 0.5 × 0.01)6 = (0.995)6 = 0.97037 Proportion of assemblies with at least one defect Pqp = 1 - 0.97037 = 0.02963 Number of assemblies with at least one defect = 0.02963(953,642) = 28,254 (c) Proportion of assemblies with all six defects = (0.5 × 0.01)6 = 1.5625(10)-14 Number of assemblies with all six defects = 1.5625(10)-14(953,642) = 1.49(10)-8 The number of products having all six added components defective is virtually zero. 17.10 (A) An eight-station automatic assembly machine has an ideal cycle time of 6 sec. Downtime is caused by defective parts jamming at the individual assembly stations. The average downtime per occurrence is 2.5 min. Fraction defect rate is 0.2% and the probability that a defective part will jam at a given station is 0.6 for all stations. The cost to operate the assembly machine is $95.00/hr and the cost of components being assembled is $0.73 per unit assembly. Ignore other costs. Determine (a) yield of good assemblies, (b) average hourly production rate of good assemblies, (c) proportion of assemblies with at least one defective component, and (d) unit cost of the assembled product. Answer: (a) Pap = (1 - 0.002 + 0.6 × 0.002)8 = (0.9992)8 = 0.9936 (b) Tp = 6/60 + 8(0.6)(0.002)(2.5) = 0.124 min/asby Rp = 60/0.124 = 483.9 asby/hr Rap = (0.9936)(483.9) = 480.8 good asby/hr (c) Pqp = 1 - 0.9936 = 0.0064 (d) Cpc = [0.73 + (95/60)(0.124)]/0.9936 = 0.99263/0.9936 = $0.9323/asby 17.11 An automated assembly machine has four stations. The first station presents the base part, and the other three stations add components to the base. The ideal cycle time for the machine is 3 sec, and the average downtime when a jam results from a defective component is 1.5 min. The fraction defective rates (q) and probabilities that a defective component will jam the station (m) are given in the table below. Quantities of 100,000 for each of the bases, brackets, pins, and retainers are used to stock the assembly line for operation. Determine (a) proportion of good product to total product coming off the line, (b) hourly production rate of good product coming off the line, (c) total number of final assemblies produced, given the starting component quantities. Of the total, how many are good product, and how many are products that contain at least one defective component? (d) Of the number of defective assemblies determined in part (c), how many will have defective base parts? How many will have defective brackets? How many will have defective pins? How many will have defective retainers? Answer: (a) Pap = (1-0.01 + 1 × 0.01)(1-0.02 + 1 × 0.02)(1-0.03 + 1 × 0.03)(1-0.04 + 0.5 × 0.04) = (1.0)(1.0)(1.0)(0.98) = 0.98 (b) Tp = 3/60 + (0.01 + 0.02 + 0.03 + 0.04 × 0.5)(1.5) = 0.17 min/cycle Rap = 0.98(60/0.17) = 345.9 good asby/hr (c) The diagram below shows quantities of components at the four workstations in the assembly machine: Total number of assemblies produced = 95,060 + 1,940 = 97,000 Number of units of good product = 95,060 Number of units containing at least one defect = 1,940 (d) Number of products containing defective base parts = 0 Number of products containing defective brackets = 0 Number of products containing defective pins = 0 Number of products containing defective retainers = 1,940 17.12 A six-station automatic assembly machine has an ideal cycle time of 6 sec. At stations 2 through 6, parts feeders deliver components to be assembled to a base part that is added at the first station. Each of stations 2 through 6 is identical and the five components are identical. That is, the completed product consists of the base part plus five identical components. The base parts have zero defects, but the other components are defective at a rate q. When an attempt is made to assemble a defective component to the base part, the machine stops (m = 1.0). It takes an average of 2.0 min to make repairs and start the machine up after each stoppage. Since all components are identical, they are purchased from a supplier who can control the fraction defect rate very closely. However, the supplier charges a premium for better quality. The cost per component is determined by the following equation: Cost per component = 0.1 + 0.0012/q, where q = fraction defect rate. Cost of the base part is 20 cents. Accordingly, the total cost of the base part and the five components is: Product Material cost = 0.70 + 0.006/q. The cost to operate the automatic assembly machine is $150.00/hr. The problem facing the production manager is this: As the component quality decreases (q increases), downtime increases which drives production costs up. As the quality improves (q decreases), material cost increases because of the price formula used by the supplier. To minimize total cost, the optimum value of q must be determined. Determine the value of q that minimizes the total cost per assembly. Also, determine the associated cost per assembly and hourly production rate. (Ignore other costs). Answer: Product material cost Cm = 0.20 + 5(0.1 + 0.0012/q) = 0.70 + 0.006/q Tp = 0.1 + 5(1)(q)(2.0) = 0.1 + 10q Cpc = Cm + CLTp = 0.70 + 0.006/q + 2.50(0.1 + 10q) Cpc = 0.70 + 0.006/q + 0.25 + 25q = 0.95 + 0.006/q + 25q Taking the derivative of the cost equation with respect to q: dCpc/dq = -0.006/q2 + 25 = 0 0.006/q2 = 25 q2 = 0.006/25 = 0.00024 q = 0.0155 Using this value in the preceding cost equation, Cpc = 0.70 + 0.006/0.0155 + 2.50(0.1 + 10 × 0.0155) = $1.725/asby Tp = 0.1 + 5(10)(0.0155)(2.0) = 0.255 min/asby Rp = 60/0.255 = 235.3 asby/hr 17.13 A six-station dial-indexing machine is designed to perform four assembly operations at stations 2 through 5 after a base part has been manually loaded at station 1. Station 6 is the unload station. Each assembly operation involves the attachment of a component to the existing base. At each of the four assembly stations, a hopper-feeder is used to deliver components to a selector device that separates components that are improperly oriented and drops them back into the hopper. The system was designed with the operating parameters for stations 2 through 5 as given in the table below. It takes 2 sec to index the dial from one station to the next. When a component jams, it takes an average of 2 min to release the jam and restart the system. Line stops due to mechanical and electrical failures of the assembly machine are not significant and can be neglected. The foreman says the system was designed to produce at a certain hourly rate, which takes into account the jams resulting from defective components. However, the actual delivery of finished assemblies is far below that designed production rate. Analyze the problem and determine the following: (a) The designed average hourly production rate that the foreman alluded to. (b) What is the proportion of assemblies coming off the system that contain one or more defective components? (c) What seems to be the problem that limits the assembly system from achieving the expected production rate? (d) What is the hourly production rate that the system is actually achieving? State any assumptions that you make in determining your answer. Station Assembly time Feed rate f Selector θ q m 2 4 sec 32/min 0.25 0.01 1.0 3 7 sec 20/min 0.50 0.005 0.6 4 5 sec 20/min 0.20 0.02 1.0 5 3 sec 15/min 1.0 0.01 0.7 Answer: Td = 2 min, Tc = 7 + 2 = 9 sec = 0.15 min Cycle rate Rc = 1/0.15 = 6.667 cycles/min (a) Σ(mq) = 1(0.01) + 0.6(0.005) + 1(0.02) + 0.7(0.01) = 0.04 Tp = 0.15 + 0.04(2) = 0.23 min Rp = 60/0.23 = 260.9 asby/hr (b) Pap = (1-0.01 + 1 × 0.01)(1-0.005 + 0.6 × 0.005)(1-0.02 + 1 × 0.02)(1-0.01 + 0.7 × 0.01) Pap = (1)(0.998)(1)(0.997) = 0.995 Pqp = 1 - 0.995 = 0.005 Rap = 260.9(0.995) = 259.6 good asby/hr (c) Station 2: fθ = 32(.25) = 8 components/min Station 3: fθ = 20(.50) = 10 components/min Station 4: fθ = 20(.20) = 4 components/min Station 5: fθ = 15(1.0) = 15 components/min The problem is that the feeder for station 4 is slower than the machine's cycle rate of 6.667 cycles/min (d) If the machine operates at the cycle rate that is consistent with the feed rate of Station 4, then Tc = 15 sec = 0.25 min Tp = 0.25 + 0.04(2) = 0.33 min Rp = 60/0.33 = 181.8 asby/hr Rap = 181.8(0.995) = 180.9 good asby/hr 17.14 For Example 17.4 in the text, which deals with a single-station assembly system, suppose that the sequence of assembly elements were to be accomplished on a seven-station assembly system with synchronous parts transfer. Each element is performed at a separate station (stations 2 through 6) and the assembly time at each respective station is the same as the element time given in Example 17.4. Assume that the handling time is divided evenly (3.5 sec each) between a load station (station 1) and an unload station (station 7). The transfer time is 2 sec, and the average downtime per downtime occurrence is 2.0 min. Determine (a) hourly production rate of all completed units, (b) yield, (c) production rate of good quality completed units, and (d) uptime efficiency. Answer: (a) Tc = 7 + 2 = 9.0 sec = 0.15 min F = 0.02(1.0) + 0.01(0.6) + 0.015(0.8) + 0.02(1.0) + 0.012 = 0.070 Tp = 0.15 + 0.070(2.0) = 0.15 + 0.14 = 0.29 min Rp = 60/0.29 = 206.9 asby/hr (b) Pap = (1.0)(0.996)(0.997)(1.0) = 0.993 (c) Rap = 206.9(0.993) = 205.5 good asby/hr (d) E = 0.15/0.29 = 0.5172 = 51.7% Comment: Comparing the values with those in Example 17.4, production rate is more than double for the multi-station system, yield is the same, and line efficiency is greatly reduced because of the much faster cycle time. Single-station Assembly Systems 17.15 (A) A single-station assembly machine is to be considered as an alternative to the dialindexing machine in Problem 17.8. Use the data given in that problem to determine (a) hourly production rate, (b) yield of good product (final assemblies containing no defective components), and (c) proportion uptime of the system. Handling time to load the base part and unload the finished assembly is 7 sec and the downtime averages 1.5 min every time a component jams. Why is the proportion uptime so much higher than in the case of the dialindexing machine in Problem 17.8? Answer: (a) Tc = 7 + (4 + 3 + 5 + 4 + 3 + 6) = 7 + 25 = 32 sec = 0.5333 min Σ(mq) = 0.6(0.015) + 0.8(0.01) + 1(0.02) + 0.5(0.01) = 0.042 (same as for Problem 17.4) Tp = 0.5333 + 0.042(1.5) = 0.59633 min Rp = 60/0.59633 = 100.6 asby/hr (b) Pap = (1-0.015 + 0.6x0.015)(1-0.01 + 0.8x0.01)(1-0.02 + 1x0.02)(1-0.01 + 0.5x0.01) = 0.98705 (same as for Problem 17.4) (c) E = .5333/.59633 = 0.8943 = 89.43% Comment: Proportion of uptime E is so much higher than in Problem 17.8 because the cycle time is much longer for the single-station machine than for the six-station dial-indexing machine (32 sec versus 6 sec). The average downtime per cycle is the same for both machines, but it is a much lower proportion of the longer cycle time in the single-station case. 17.16 A single-station robotic assembly system performs a series of five assembly elements, each of which adds a different component to a base part. Each element takes 3.5 sec. In addition, the handling time needed to move the base part into and out of position is 4.5 sec. The fraction defect rate is 0.003 for all components, and the probability of a jam by a defective component is 0.7. Average downtime per occurrence is 2.5 min. Determine (a) hourly production rate, (b) yield of good product in the output, and (c) uptime efficiency. (d) What proportion of the output contains a defective component from the third of the five elements performed in the work cycle? Answer: (a) Tp = Tc + nmqTd Tc = 4.5 + 5(3.5) = 22 sec = 0.3667 min Tp = 0.3667 + 5(0.7)(0.003)(2.5) = 0.393 min Rp = 60/0.393 = 152.7 asby/hr (b) Pap = (1 - 0.003 + 0.7(0.003))5 = (0.9991)5 = 0.9955 (c) E = 0.3667/0.393 = 0.933 = 93.3% (d) Pqp3 = 1 - (1 - 0.003 + 0.7(0.003)) = 0.003 - 0.7(0.003) = 0.3(0.003) = 0.0009 17.17 A single-station assembly cell uses an industrial robot to perform a series of assembly operations. The base part and parts 2 and 3 are delivered by vibratory bowl feeders that use selectors to insure that only properly oriented parts are delivered to the robot for assembly. The robot cell performs the elements in the table below (also given are feeder rates, selector proportion θ, element times, fraction defect rate q, and probability of jam m, and, for the last element, the frequency of downtime incidents p). In addition to the times given in the table, the time required to unload the completed subassembly is 4 sec. When a line stop occurs, it takes an average of 1.8 min to make repairs and restart the cell. Determine (a) yield of good product, (b) average hourly production rate of good product, and (c) uptime efficiency for the cell? State any assumptions you must make about the operation of the cell in order to solve the problem. Element Feed rate f Selector θ Element Time Te q m p 1 15 pc/min 0.30 Load base part 4 sec 0.01 0.6 2 12 pc/min 0.25 Add part 2 3 sec 0.02 0.3 3 25 pc/min 0.10 Add part 3 4 sec 0.03 0.8 4 Fasten 3 sec 0.02 Answer: Assumptions: (1) Feeders continue to operate and deliver parts into the feed track even when a jam occurs during assembly. (2) Low-level quantity nf1 is sufficient to eliminate possibility of a stockout. Tc = Th + Σ Te = 4 + 4 + 3 + 4 + 3 = 18 sec = 0.3 min Tp = Tc + (Σqm + p)Td = 0.3 + (0.01 x 0.6 + 0.02 x 0.3 + 0.03 x 0.8 + 0.02)(1.8) Tp = 0.3 + 0.056(1.8) = 0.4008 min/asby Check feed rates: f1 θ1 = 15(0.3) = 4.5 pc/min or 0.2222 min/pc f2 θ2 = 12(0.25) = 3.0 pc/min or 0.3333 min/pc f3 θ3 = 25(0.1) = 2.5 pc/min or 0.40 min/pc Each of these can be completed within the average cycle time of the robot assembly operation. (a) Pap = (1 - 0.01 + 0.6 × 0.01)(1 - 0.02 + 0.3 × 0.02)(1 - 0.03 + 0.8 × 0.03) = (0.996)(0.986)(0.994) = 0.976 (b) Rp = 1/0.4008 = 2.495 asby/min = 149.7 asby/hr Rap = 149.7(0.9762) = 146.1 good asby/hr (c) E = 0.3/0.4008 = 0.7485 = 74.85% Alternative assumption: (1) Feeder stops when jam occurs. Hence, cycle time is limited by feeder #3. Tc = 0.4 min Tp = 0.4 + (0.056)(1.8) = 0.5008 min (a) Rp = 1/0.5008 = 1.9968 asby/min = 119.8 asby/hr Pap = 0.976 (same as above) (b) Rap = 119.8(0.9762) = 116.9 good asby/hr (c) E = 0.400/0.5008 = 0.7987 = 79.87% Partial Automation 17.18 (A) A partially-automated production line has three mechanized and three manual workstations, a total of six stations. The ideal cycle time is 57 sec, which includes a transfer time of 3 sec. Data on the six stations are listed in the table below. Cost of the transfer mechanism is $0.10/min, cost to run each automated station is $0.12/min, and labor cost to operate each manual station is $0.17/min. It has been proposed to substitute an automated station in place of station 5. The cost of this new station is estimated at $0.25/min and its breakdown rate = 0.02 per cycle, but its process time would be only 30 sec, thus reducing the overall cycle time of the line from 57 sec to 36 sec. Average downtime per breakdown of the current line, as well as for the proposed configuration, is 3.0 min. Determine the following for the current line and the proposed line: (a) hourly production rate, (b) proportion uptime, and (c) cost per unit. Assume that when an automated station stops, the whole line stops, including the manual stations. Also, in computing costs, neglect material and tooling costs. Station Type Process time pi 1 Manual 36 sec 0 2 Automatic 15 sec 0.01 3 Automatic 20 sec 0.02 4 Automatic 25 sec 0.01 5 Manual 54 sec 0 6 Manual 33 sec 0 Answer: For the current line, (a) Tc = 57 sec = 0.95 min, F = 0.01 + 0.02 + 0.01 = 0.04 Tp = 0.95 + 0.04(3.0) = 0.95 + 0.12 = 1.07 min/unit Rp = 60/1.07 = 56.1 asby/hr (b) E = 1.0/1.07 = 0.935 = 93.5% (c) Co = 0.10 + 3(0.12) + 3(0.17) = $0.97/min. Cpc = (0.97)(1.07) = $1.038/asby For the proposed line in which station 5 is automated, (a) Tc = 36 sec = 0.6 min F = 0.01 + 0.02 + 0.01 + 0.02 = 0.06 Tp = 0.6 + 0.06(3.0) = 0.6 + 0.18 = 0.78 min/unit Rp = 60/0.78 = 76.9 units/hr (b) E = 0.6/0.78 = 0.769 = 76.9% (c) Co = 0.10 + 3(0.12) + 0.25 + 2(0.17) = $1.05/min Cpc = (1.05)(0.78) = $0.819/asby 17.19 A manual assembly line has six stations. The service time at each manual station is 60 sec. Parts are transferred by hand from one station to the next, and the lack of discipline in this method adds 12 sec to the cycle time. Hence, the current cycle time is 72 sec. The following two proposals have been made: (1) Install a mechanized transfer system to pace the line; and (2) automate one or more of the manual stations using robots that would perform the same tasks as humans only faster. The second proposal requires the mechanized transfer system of the first proposal and would result in a partially- or fully-automated assembly line. The transfer system would have a transfer time of 6 sec, thus reducing the cycle time on the manual line to 66 sec. Regarding the second proposal, all six stations are candidates for automation. Each automated station would have an assembly time of 30 sec. Thus if all six stations were automated the cycle time for the line would be 36 sec. There are differences in the quality of parts added at the stations; these data are given in the table below for each station (q = fraction defect rate, m = probability that a defect will jam the station). Average downtime per station jam at the automated stations is 3.0 min. Assume that the manual stations do not experience line stops due to defective components. Cost data: Cat = $0.10/min; Cw = $0.20/min; and Cas = $0.15/min. Determine if either or both of the proposals should be accepted. If the second proposal is accepted, how many stations should be automated and which ones? Use cost per piece as the criterion for your decision. Assume for all cases considered that when an automated station stops, the whole line stops, including the manual stations. Station qi mi Station qi mi 1 0.005 1.0 4 0.020 1.0 2 0.010 1.0 5 0.025 1.0 3 0.015 1.0 6 0.030 1.0 Answer: Proposal 1: Current operation: Tc = 1.2 min Co = 6(0.20) = $1.20/min Cpc = 1.20(1.2) = $1.44/asby Proposal: Tc = 1.1 min Co = 0.10 + 6(0.20) = 1.30/min Cpc = 1.30(1.1) = $1.43/asby Conclusion: Accept Proposal 1. Proposal 2: Tc = 36 sec = 0.6 min if all six stations are automated. If fewer stations than six were automated, then the cycle time would still be determined by the one minute manual assembly time. F = 0.005(1.0) + 0.01(1.0) + 0.015(1.0) + 0.02(1.0) + 0.025(1.0) + 0.03(1.0) = 0.105 Tp = 0.6 + 0.105(3.0) = 0.6 + 0.315 = 0.915 min/unit Co = 0.10 + 6(0.15) = 1.00/min Cpc = 1.00(0.915) = $0.915/asby Conclusion: Accept Proposal 2. 17.20 Solve preceding Problem 17.19, except that the probability that a defective part will jam the automated station is m = 0.5 for all stations. Answer: Proposal 1: Current operation: Tc = 1.2 min Co = 6(0.20) = $1.20/min Cpc = 1.20(1.2) = $1.44/asby Proposal: Tc = 1.1 min Co = 0.10 + 6(0.20) = 1.30/min Cpc = 1.30(1.1) = $1.43/asby Conclusion: Accept Proposal 1. Proposal 2: Tc = 36 sec = 0.6 min if all six stations are automated. F = 0.005(0.5) + 0.01(0.5) + 0.015(0.5) + 0.02(0.5) + 0.025(0.5) + 0.03(0.5) = 0.0525 Tp = 0.6 + 0.0525(3.0) = 0.6 + 0.1575 = 0.7575 min/unit Co = 0.10 + 6(0.15) = 1.00/min Pap = (1-0.005 + 0.005x0.5)(0.995)(0.9925)(0.990)(0.9875)(0.985) = 0.9486 Cpc = 1.00(0.7575)/0.9486 = $0.798/asby Conclusion: Accept Proposal 2, assuming the problem of sorting good units from bad can be solved at minimum cost. Solution Manual for Automation, Production Systems, and Computer-Integrated Manufacturing Mikell P. Groover 9780133499612, 9780134605463
