## Preview Extract

This document contains Chapters 20 to 22 Chapter 20 QUALITY PROGRAMS FOR MANUFACTURING REVIEW QUESTIONS 20.1 What are the two aspects of quality in a manufactured product? List some of the product characteristics in each category. Answer: The two aspects of quality identified in the text are (1) product features and (2) freedom from deficiencies. Product features include design configuration, size of product, function and performance, aesthetic appeal, ease of use, reliability, and serviceability. Freedom from deficiencies includes absence of defects, conformance to specifications, components within tolerance, and no missing parts. 20.2 Discuss the differences between the traditional view of quality control and the modern view. Answer: As discussed in the text, traditional QC focused on inspection. In many factories, the only department responsible for quality control was the inspection department. Much attention was given to sampling and statistical methods. The modern view of quality control emphasizes management involvement in quality, focus on customer satisfaction, continuous improvement, and encouraging the involvement of all employees. 20.3 What are the three main objectives of Total Quality Management? Answer: As identified in the text, the three main objectives of Total Quality Management are (1) achieving customer satisfaction, and (2) encouraging involvement of the entire work force, and (3) continuous improvement. 20.4 What do the terms external customer and internal customer mean? Answer: External customers are those who buy the final product or service. Internal customers are those within the same organization; for example, the assembly department is the customer of the parts production department within the same company. 20.5 Manufacturing process variations can be divided into two types: (1) random and (2) assignable. Distinguish between these two types. Answer: Random variations result from intrinsic variability in the process. All processes are characterized by these kinds of variations. Random variations cannot be avoided; they are caused by factors such as inherent human variability and minor variations in raw materials. Random variations typically form a normal statistical distribution. This kind of variability continues so long as the process is operating normally. Assignable variations indicate an exception from normal operating conditions. Something has occurred in the process that is not accounted for by random variations. Reasons for assignable variations include operator mistakes, defective raw materials, tool failures, and equipment malfunctions. 20.6 What is meant by the term process capability? Answer: Process capability equals ±3 standard deviations about the mean output value (a total range of 6 standard deviations), under the assumptions that (1) the output is normally distributed, and (2) steady state operation has been achieved and the process is in statistical control. 20.7 What is a control chart? Answer: A control chart is a graphical technique in which statistics computed from measured values of a certain process output characteristic are plotted over time to determine if the process remains in statistical control. The chart consists of three horizontal lines that remain constant over time: a center line (CL), a lower control limit (LCL), and an upper control limit (UCL). The center line is usually set at the nominal design value, and the upper and lower control limits are generally set at ±3 standard deviations of the sample means. 20.8 What are the two basic types of control charts? Answer: The two basic types of control charts are (1) control charts for variables, and (2) control charts for attributes. Control charts for variables require measurements of the quality characteristic of interest. Control charts for attributes simply require a determination of either the fraction of defects in the sample or the number of defects in the sample. 20.9 What is a histogram? Answer: A histogram is a statistical graph consisting of bars representing different values or ranges of values, in which the length of each bar is proportional to the frequency or relative frequency of the value or range. 20.10 What is a Pareto chart? Answer: A Pareto chart is a special form of histogram, in which attribute data are arranged according to some criteria such as cost or value. It provides a graphical display of the tendency for a small proportion of a given population to be more valuable than the much larger majority. 20.11 What is a defect concentration diagram? Answer: A defect concentration diagram is a drawing of the product (or other item of interest), with all relevant views displayed, onto which are sketched the various types of defects or other problems of interest at the locations where they each occurred. 20.12 What is a scatter diagram? Answer: A scatter diagram is an x-y plot of the data taken of two variables of interest. 20.13 What is a cause and effect diagram? Answer: The cause and effect diagram, also known as a fishbone diagram, is a graphical-tabular chart used to list and analyze the potential causes of a given problem. The diagram consists of a central stem leading to the effect (the problem), with multiple branches coming off the stem listing the various groups of possible causes of the problem. 20.14 What is Six Sigma? Answer: As defined in the text, Six Sigma is a quality-focused program that utilizes worker teams to accomplish projects aimed at improving an organization’s operational performance. The name Six Sigma is derived from the Normal statistical distribution, in which the Greek letter sigma (σ) is the standard deviation or measure of dispersion in a normal population. Six sigma implies near perfection in a process, and that is the goal of a Six Sigma program. There is a strong emphasis on the customer and customer satisfaction in Six Sigma. 20.15 What are the general goals of Six Sigma? Answer: As listed in the text, the general goals of Six Sigma are (1) better customer satisfaction, (2) high quality products and services, (3) reduced defects, (4) improved process capability through reduction in process variations, (5) continuous improvement, and (6) cost reduction through more effective and efficient processes. 20.16 Why does 6σ in Six Sigma really mean 4.5σ? Answer: When the Motorola engineers devised the six sigma standard, they considered processes that operate over the long run. And over the long run processes tend to shift from the original process mean to the right or left. To compensate for these likely shifts, Motorola selected to use 1.5σ as the magnitude of the shift, while leaving the original ±6σ limits in place for the process. Also, the sigma level in Six Sigma includes only one tail of the distribution, instead of two tails usually associated with the normal distribution. Thus, when 6σ is used in Six Sigma, it really refers to 4.5σ and only one tail in the normal probability tables. 20.17 What does DMAIC stand for? Answer: DMAIC stands for define, measure, analyze, improve, and control. 20.18 Why is defects per million (DPM) not necessarily the same as defects per million opportunities (DPMO)? Answer: Defects per million (DPM) refers to the total number of defects per million parts, thus allowing that a given defective part may contain more than one type of defect. Defects per million opportunities (DPMO) also acknowledges this fact that there may be more than one type of defect that occurs in each unit and takes into account the complexity of the product or service so that entirely different types of products and services can be compared on the same sigma scale. 20.19 What is a robust design in Taguchi’s quality engineering? Answer: A robust design is one in which the function and performance of the product or process are relatively insensitive to variations that are difficult or impossible to control. In product design, robustness means that the product can maintain consistent performance with minimal disturbance due to variations in its operating environment. In process design, robustness means that the process continues to produce good product in spite of uncontrollable variations in its operating environment. 20.20 What is ISO 9000? Answer: ISO 9000 is a set of international standards on quality developed by the International Organization for Standardization (ISO). It is not a standard for the products or services. Instead, ISO 9000 establishes standards for the systems and procedures used by a facility that affect the quality of the products and services produced by the facility. ISO 9000 includes a glossary of quality terms, guidelines for selecting and using the various standards, models for quality systems, and guidelines for auditing quality systems. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Process Capability (Note: Problems 20.2, 20.5, 20.25, 20.26, 20.27, and 20.28 require the use of standard normal distribution tables not included in this book.) 20.1 (A) A turning process is in statistical control and the output is normally distributed, producing parts with a mean diameter = 45.025 mm and a standard deviation = 0.035 mm. Determine the process capability. Answer: Process capability PC = µ ± 3σ = 45.025 ± 3(0.035) = 45.025 ± 0.105 mm The upper and lower limits of the process capability range are: 44.920 to 45.130 mm 20.2 In Problem 20.1, the design specification on the part is that the diameter = 45.000 ± 0.150 mm. (a) What proportion of parts fall outside the tolerance limits? (b) If the process is adjusted so that its mean diameter = 45.000 mm and the standard deviation remains the same, what proportion of parts fall outside the tolerance limits? Answer: (a) Given process mean µ = 45.025 mm, σ = 0.035 mm, tolerance limits = 44.850 to 45.150. On the lower side of the tolerance limit, using the standard normal distribution, z = (44.850 - 45.025)/0.035 = −5.0. Conclusion: there are virtually no defects on the lower side of the tolerance. On the upper side of the tolerance limit, z = (45.150 - 45.025)/0.035 = + 3.57 Using tables of the standard normal distribution, Pr(z > 3.57) = 0.0002 The proportion of defects with the current process mean = 0.0002 = 0.02% (b) Resetting the process mean to µ = 45.000 mm with σ = 0.035 mm (same as before) and with tolerance limits remaining at 44.850 to 45.150. On the lower side of the tolerance limit, using the standard normal distribution, z = (44.850 − 45.000)/0.035 = −4.29. Using tables of the standard normal distribution, Pr(z 4.29) ≈ 0.0000 The proportion of defects with new process mean ≈ 0.00% 20.3 An automated tube-bending operation produces parts with an included angle = 91.2°. The process is in statistical control and the values of included angle are normally distributed with a standard deviation = 0.55°. The design specification on the angle = 90.0° ± 2.0°. (a) Determine the process capability. (b) If the process is adjusted so that its mean = 90.0°, determine the process capability index. Answer: (a) PC = 91.2 ± 3(0.55) = 91.2° ± 1.65° The upper and lower limits of the process capability range are 89.55° to 92.85° (b) If µ = 90° UTL - LTL = 92° - 88° = 4° PCI = 4°/(6 × 0.55°) = 1.212 20.4 A plastic extrusion process is in statistical control and the output is normally distributed. The extrudate is subsequently cut into individual parts, and the extruded parts have a critical cross-sectional dimension = 13.65 mm with standard deviation = 0.27 mm. Determine the process capability. Answer: Process capability PC = µ ± 3σ = 13.65 ± 3(0.27) = 13.65 ± 0.81 mm The upper and lower limits of the process capability range are 12.84 to 14.46 mm 20.5 In Problem 20.4, the design specification on the part is that the critical cross-sectional dimension = 13.5 ± 1.0 mm. (a) What proportion of parts fall outside the tolerance limits? (b) If the process were adjusted so that its mean diameter = 13.5 mm and the standard deviation remained the same, what proportion of parts would fall outside the tolerance limits? (c) With the adjusted mean at 13.5 mm, determine the process capability index. Answer: (a) Given the process mean µ = 13.65 mm and σ = 0.27 mm (same as before) and tolerance limits = 12.5 to 14.5 mm. On the lower side of the limit, using the standard normal distribution, z = (12.5 − 13.65)/0.37 = −3.11. Using tables of the standard normal distribution, Pr(z 2.30) = 0.0107 Proportion of defects with the current process mean = 0.0009 + 0.0107 = 0.0116 = 1.16% (b) Given process mean µ = 13.5 mm and σ = 0.37 mm (same as before) and tolerance limits 12.5 to 14.5 mm. On the lower side of the tolerance limit, z = (12.5 − 13.5)/0.37 = −2.70. Using tables of the standard normal distribution, Pr(z 2.70) = 0.0035 Proportion of defects with new process mean = 0.0035 + 0.0035 = 0.0070 = 0.70% (c) Process capability index PCI = 2.0/(6 × 0.37) = 0.901 Control Charts 20.6 (A) Seven samples of 5 parts each have been collected from an extrusion process which is in statistical control, and the diameter of the extrudate has been measured for each part. (a) Determine the values of the center line, LCL, and UCL for and R charts. The calculated values of x and R for each sample are given below (measured values are in inches). (b) Construct the control charts and plot the sample data on the charts. Answer: 20.7 Ten samples of size n = 8 have been collected from a process in statistical control, and the dimension of interest has been measured for each part. (a) Determine the values of the center line, LCL, and UCL for the and R charts. The calculated values of and R for each sample are given below (measured values are in mm). (b) Construct the control charts and plot the sample data on the charts. Answer: 20.8 In 10 samples of size n = 8 for a process that is in statistical control, the average value of the sample means is in for the dimension of interest, and the mean of the ranges of the samples is in. Determine (a) lower and upper control limits for the chart and (b) lower and upper control limits for the R chart. Answer: 20.9 In 20 samples each of size n = 6 for a process that is in statistical control, the grand mean of the samples is for the characteristic of interest, and the mean of the ranges of the samples is . Determine (a) lower and upper control limits for the chart and (b) lower and upper control limits for the R chart. Answer: 20.10 A p chart is to be constructed for a process that is in statistical control. Eight samples of 20 parts each have been collected, and the average number of defects per sample = 2.4. Determine the center line, LCL and UCL for the p chart. Answer: 20.11 Twelve samples of equal size have been taken to prepare a p chart for a process that is in statistical control. The total number of parts in these 12 samples was 600 and the total number of defects counted was 96. Determine the center line, LCL and UCL for the p chart. Answer: 20.12 The yield of good chips during a certain step in silicon processing of integrated circuits averages 89%. This processing step is considered to be in statistical control. The number of chips per wafer is 156. Determine the center line, LCL, and UCL for the p chart that would be used for this processing step. Answer: 20.13 (A) The upper and lower control limits for a p chart are: LCL = 0.10 and UCL = 0.30. Determine the sample size n that is used with this control chart. Answer: 20.14 The lower and upper control limits for a p chart are: LCL = 0 and UCL = 0.25. The center line of the p chart is at 0.11. Determine the sample size n that is used with this control chart. Answer: Because p is not centered between the given LCL and UCL values, it is inferred that the actual calculated value of LCL was less than zero and was rounded up to zero. Accordingly, the UCL and center line should be used to calculate the sample size n. The given LCL value should not be used in the calculations. 20.15 Twelve cars were inspected after final assembly. The number of defects ranged between 87 and 139 defects per car with an average of 116. Assuming that the assembly process was in statistical control, determine the center line and upper and lower control limits for the c chart that might be used in this situation. Answer: 20.16 For each of the three control charts in Figure P20.16, identify whether or not there is evidence that the process is out of control. Answer: (a) With reference to indicators in Section 20.4.1, samples 6 through 13 are all above the center line. (b) With reference to indicators in Section 20. 4.1, samples 11 and 13 are beyond + 2σ. (c) With reference to indicators in Section 20. 4.1, each of samples 6 through 11 is lower than its predecessor. Determining Sigma Level in Six Sigma 20.17 (A) A garment manufacturer produces 15 different dress styles, and every year new dress styles are introduced and old styles are discarded. Whatever the style, the final inspection department checks each dress before it leaves the factory for eight features that are considered critical-to-quality for customer satisfaction. The inspection report for last month indicated that a total of 301 deficiencies of the eight features were found among 6,250 dresses produced. Determine (a) defects per million opportunities and (b) sigma level for the manufacturer’s production performance. Answer: (a) Given Nd = 301, Nu = 6,250, and No = 8 DPMO = 1,000,000(301)/(6,250)(8) = 6020 defects/million (b) From Table 20.5, this corresponds to around the 4.0 sigma level. 20.18 A producer of cell phones checks each phone prior to packaging, using seven critical-to-quality characteristics that are deemed important to customers. Last year, out of 205,438 phones produced by the company, a total of 578 phones had at least one defect, and the total number of defects among these 578 phones was 1692. Determine (a) the number of defects per million opportunities and corresponding sigma level, (b) the number of defects per million and corresponding sigma level, and (c) the number of defective units per million and corresponding sigma level. Answer: (a) Given No = 7 CTQ features, Nd = 1692 defects among all units, Ndu = 578 defective phones, Nu = 205,438. DPMO = 1,000,000(1,692)/(205,438)(7) = 1177 defects/million opportunities From Table 20.5, this corresponds to the 4.5 sigma level. (b) DPM = 1,000,000(1,692)/(205,438) = 8236 defects/million From Table 20.5, this corresponds to around the 3.9 sigma level. (c) DUPM = 1,000,000(578)/(205,438) = 2814 defective units/million From Table 20.5, this corresponds to around the 4.3 sigma level. 20.19 The inspection department in an automobile final assembly plant checks cars coming off the line against 85 features that are considered critical-to-quality for customer satisfaction. During a one-month period, a total of 16,578 cars were produced. For those cars, a total of 1,989 defects of various types were found, and the total number of cars that had one or more defects was 512. Determine (a) the number of defects per million opportunities and corresponding sigma level, (b) the number of defects per million and corresponding sigma level, and (c) the number of defective units per million and corresponding sigma level. Answer: (a) Given Nd = 1989 defects among all units, Ndu = 512 defective cars, Nu = 16,578, and No = 85 (a) DPMO = 1,000,000(1,989)/(16,578)(85) = 1412 defects/million opportunities From Table 20.5, this corresponds to the 4.5 sigma level. (b) DPM = 1,000,000(1,989)/(16,578) = 119,978 defects/million From Table 20.5, this corresponds to around the 2.7 sigma level. (c) DUPM = 1,000,000(578)/(16,578) = 34,865 defective units/million From Table 20.5, this corresponds to around the 3.3 sigma level. 20.20 A digital camera maker produces three different models: (1) base model, (2) zoom model, and (3) zoom model with extra memory. Data for the three models are shown in the table below. The three models have been on the market for one year, and the first year’s sales are given in the table. Also given are critical-to-quality (CTQ) characteristics and total defects that have been tabulated for the products sold. Higher model numbers have more CTQ characteristics (opportunities for defects) because they are more complex. The category of total defects refers to the total number of defects of all CTQ characteristics for each model. For each of the three models, determine (a) the number of defects per million opportunities and corresponding sigma level, (b) the number of defects per million and corresponding sigma level, and (c) the number of defective units per million and corresponding sigma level. (d) Does any one model seem to be produced at a higher quality level than the others? (e) Determine aggregate values for DPMO, DPM, and DUPM and their corresponding sigma levels for all models made by the camera maker. Answer: Summarizing the data, for Model 1, Nu = 62,347, No = 16, Ndu = 127, and Nd = 282. For Model 2, Nu = 31,593, No = 23, Ndu = 109, and Nd = 429. For Model 3, Nu = 18,662, No = 2, Ndu = 84, and Nd = 551. (a) Model 1: DPMO = 1,000,000(282)/(62,347)(16) = 283 defects/million opportunities From Table 20.5, this corresponds to the 4.9 sigma level. Model 2: DPMO = 1,000,000(429)/(31,593)(23) = 590 defects/million opportunities From Table 20.5, this corresponds to the 4.7 sigma level. Model 3: DPMO = 1,000,000(551)/(18,662)(29) = 1018 defects/million opportunities From Table 20.5, this corresponds to the 4.6 sigma level. (b) Model 1: DPM = 1,000,000(282)/(62,347) = 4523 defects/million opportunities From Table 20.5, this corresponds to the 4.1 sigma level. Model 2: DPM = 1,000,000(429)/(31,593) = 13,579 defects/million opportunities From Table 20.5, this corresponds to the 3.7 sigma level. Model 3: DPM = 1,000,000(551)/(18,662) = 29,525 defects/million opportunities From Table 20.5, this corresponds to the 3.4 sigma level. (c) Model 1: DUPM = 1,000,000(127)/(62,347) = 2037 defects/million opportunities From Table 20.5, this corresponds to the 4.4 sigma level. Model 2: DUPM = 1,000,000(109)/(31,593) = 3450 defects/million opportunities From Table 20.5, this corresponds to the 4.2 sigma level. Model 3: DUPM = 1,000,000(84)/(18,662) = 4501 defects/million opportunities From Table 20.5, this corresponds to the 4.6 sigma level. (d) No model stands out, but model 1 has the highest quality level (e) Summarizing the data for all models, Nu = 62,347 + 31,593 + 18,662 = 112,602 units, Ndu = 127 + 109 + 84 = 320 defective units, Nd = 282 + 429 + 551 = 1262 defects, and No = ΣNuNo = 62,347(16) + 31,593(23) + 18,662(29) = 2,265,389 opportunities. DPMO = 1,000,000(1,262)/2,265,389 = 557 defects/million opportunities From Table 20.5, this corresponds to the 4.7 sigma level. DPM = 1,000,000(1,262)/112,602 = 11,208 defects/million From Table 20.5, this corresponds to the 3.8 sigma level. DUPM = 1,000,000(320)/112,602 = 2842 defective units/million From Table 20.5, this corresponds to the 4.3 sigma level. Taguchi Loss Function 20.21 A certain part dimension on a power garden tool is specified as 25.50 ± 0.30 mm. Company repair records indicate that if the ±0.30 mm tolerance is exceeded, there is a 75% chance that the product will be returned for replacement. The cost associated with replacing the product, which includes not only the product cost itself but also the additional paperwork and handling associated with replacement, is estimated to be $300. Determine the constant k in the Taguchi loss function for this data. Answer: E{L(x)} = 0.75(300) + 0.25(0) = $225.00 225 = k(0.30)2 = 0.09k k = 225/0.09 = $2500 20.22 (A) The design specification on the resistance for an electronic component is 0.50 ±0.02 ohm. If the tolerance is exceeded and the component is scrapped, the company suffers a $20 cost. (a) What is the implied value of the constant k in the Taguchi quadratic loss function? (b) If the output of the process that sets the resistance is centered on 0.50 ohm, with a standard deviation of 0.005 ohm, what is the expected loss per unit? Answer: (a) 200 = k(0.02)2 = 0.0004k k = 200/0.0004 = $50,000 (b) E{L(x)} = kσ2 = 50,000(0.005)2 = 50,000(0.000025) = $1.25/unit 20.23 The Taguchi quadratic loss function for a particular component in a piece of earth moving equipment is L(x) = 3500(x - N)2 , where x = the actual value of a critical dimension and N is the nominal value. If N = 150.00 mm, determine the value of the loss function for tolerances of (a) ±0.20 mm and (b) ±0.10 mm. Answer: (a) L(x) = 3500(0.20)2 = $140.00 (b) L(x) = 3500(0.10)2 = $35.00 20.24 The Taguchi loss function for a certain component is given by L(x) = 8000(x - N)2 , where x = the actual value of a dimension of critical importance and N is its nominal value. Company management has decided that the maximum loss that can be accepted is $10.00. (a) If the nominal dimension is 30.00 mm, at what value should the tolerance on this dimension be set? (b) Does the value of the nominal dimension have any effect on the tolerance that should be specified? Answer: (a) 10 = 8000(x - N)2 (x - N)2 = 10/8000 = 0.00125 (x - N) = 0.0354 mm (b) N has no effect. It is (x - N) that affects L(x). 20.25 Two alternative manufacturing processes, A and B, can be used to produce a certain dimension on one of the parts in an assembled product. Both processes can produce parts with an average dimension at the desired nominal value. The tolerance on the dimension is ± 0.15 mm. The output of each process follows a normal distribution. However, the standard deviations are different. For process A, σ = 0.12 mm; and for process B, σ = 0.07 mm. Production costs per piece for A and B are $7.00 and $12.00, respectively. If inspection and sortation is required, the cost is $0.50 per piece. If a part is found to be defective, it must be scrapped at a cost = its production cost. The Taguchi loss function for this component is given by L(x) = 2500(x - N)2 , where x = value of the dimension and N is its nominal value. Determine the average cost per piece for the two processes. Answer: Process A: E{L(x)} = kσ2 = 2500(0.12)2 = $36.00 % scrap: Using the standard normal distribution tables, z = 0.15/0.12 = 1.25 Scrap rate q = 2(0.5 - 0.3944) = 0.2112 Cpc = 7.00 + 0.50 + 0.2112(7.00) + 36.00 = $44.98 Process B: E{L(x)} = kσ2 = 2500(0.07)2 = $12.25 % scrap: Using the standard normal distribution tables, z = 0.15/0.07 = 2.143 Scrap rate q = 2(0.5 - 0.4838) = 0.0324 Cpc = 12.00 + 0.50 + 0.0324(12.00) + 12.25 = $25.14 20.26 Solve Problem 20.25, except that the tolerance on the dimension is ± 0.30 mm rather than ± 0.15 mm. Answer: Process A: E{L(x)} = kσ2 = 2500(0.12)2 = $36.00 % scrap: Using the standard normal distribution tables, z = 0.30/0.12 = 2.5 Scrap rate q = 2(0.5 - 0.4938) = 0.0124 Cpc = 7.00 + 0.50 + 0.0124(7.00) + 36.00 = $43.59 Process B: E{L(x)} = kσ2 = 2500(0.07)2 = $12.25 % scrap: Using the standard normal distribution tables, z = 0.30/0.07 = 4.29 Scrap rate q = 0 100% inspection and sortation are not necessary: Cpc = 12.00 + 12.25 = $24.25 20.27 Solve Problem 20.25, except that the average value of the dimension produced by process B is 0.10 mm greater than the nominal value specified. The average value of the dimension produced by process A remains at the nominal value N. Answer: Process A: E{L(x)} = kσ2 = 2500(0.12)2 = $36.00 % scrap: Using the standard normal distribution tables, z = 0.15/0.12 = 1.25 Scrap rate q = 2(0.5 - 0.3944) = 0.2112 Cpc = 7.00 + 0.50 + 0.2112(7.00) + 36.00 = $44.98 Process B: E{L(x)} = k[(µ - N)2 + σ2] = 2500[0.102 + 0.072] = 2500[0.01 + 0.0049] = $37.25 Scrap: Using the standard normal distribution tables, z1 = (0.15 + 0.10)/0.07 = 3.57 Scrap rate q1 = 0.5 - 0.4998 = 0.0002 z2 = (0.15 - 0.10)/0.07 = 0.71 Scrap rate q2 = 0.5 - 0.2612 = 0.2388 q = q1 + q2 = 0.0002 + 0.2388 = 0.2390 Cpc = 12.00 + 0.50 + 0.239(12.00) + 37.25 = $52.62 20.28 Two different manufacturing processes, A and B, can be used to produce a certain component. The specification on the dimension of interest is 100.00 mm ± 0.20 mm. The output of process A follows the normal distribution, with µ = 100.00 mm and σ = 0.10 mm. The output of process B is a uniform distribution defined by f(x) = 2.0 for 99.75 ≤ x ≤ 100.25 mm. Production costs per piece for processes A and B are each $5.00. Inspection and sortation cost is $0.50 per piece. If a part is found to be defective, it must be scrapped at a cost = twice its production cost. The Taguchi loss function for this component is given by L(x) = 2500(x - N)2 , where x = value of the dimension and N is its nominal value. Determine the average cost per piece for the two processes. Answer: Process A: E{L(x)} = 2500σ2 = 2500(0.10)2 = $25.00 Scrap: Using the standard normal distribution tables, z = 0.20/0.10 = 2.0 Scrap rate q = 2(0.5 - 0.4773) = 0.0454 Cpc = 5.00 + 0.50 + 0.0454(10.00) + 25.00 = $30.95 Process B: For uniform distribution, E{L(x)} = 2500(0.020833) = $52.08 Scrap: q = 0.05/0.25 = 0.20 Cpc = 5.00 + 0.50 + 0.20(10.00) + 52.08 = $59.58 Chapter 21 INSPECTION PRINCIPLES AND PRACTICES REVIEW QUESTIONS 21.1 What is inspection? Answer: As defined in the text, inspection is the activity of examining the product, its components, subassemblies, or materials out of which it is made, to determine whether they conform to design specifications. 21.2 Briefly define the two basic types of inspection. Answer: The two basic types of inspection are (1) inspection for variables, in which one or more quality characteristics of interest are measured using an appropriate measuring instrument or sensor; and (2) inspection for attributes, in which the part or product is inspected to determine whether it conforms to the accepted quality standard. Inspection by attributes can also involve counting the number of defects in a product. 21.3 What are the four steps in a typical inspection procedure? Answer: As defined in the text, the four steps are (1) presentation of the item for examination; (2) examination of the item for nonconforming feature(s); (3) deciding whether the item satisfies the defined quality standards; and (4) action, such as accepting or rejecting the item, or sorting the item into the most appropriate quality grade. 21.4 What are the Type I and Type II errors that can occur in inspection? Answer: A Type I error is when an item of good quality is incorrectly classified as being defective. It is a “false alarm.” A Type II error is when an item of poor quality is erroneously classified as being good. It is a “miss.” 21.5 What is quality control testing as distinguished from inspection? Answer: Whereas inspection is used to assess the quality of the product relative to design specifications, testing refers to the assessment of the functional aspects of the product: Does the product operate the way it is supposed to operate? Thus, QC testing is a procedure in which the item being tested (product, subassembly, part, or material) is observed during actual operation or under conditions that might be present during operation. 21.6 What are the Type I and Type II statistical errors that can occur in acceptance sampling? Answer: A Type I error occurs when a batch of product is rejected when in fact it is equal to or better than the acceptable quality level (AQL). The associated probability of a Type I error is called the producer’s risk. A Type II error occurs when a batch of product is accepted when in fact its quality is worse than the lot tolerance percent defective (LTPD). The probability of this error is called the consumer’s risk. 21.7 Describe what an operating characteristic curve is in acceptance sampling. Answer: An operating characteristic curve for a given sampling plan gives the probability of accepting a batch as a function of the possible fraction defect rates that might actually exist in the batch. 21.8 What are the two problems associated with 100% manual inspection? Answer: The first problem is the expense involved. Instead of dividing the time of inspecting the sample over the number of parts in the production run, the inspection time per piece is applied to every part. The inspection cost sometimes exceeds the cost of making the part. The second problem with 100% manual inspection is the problem of inspection accuracy. There are almost always errors associated with 100% inspection (Type I and II errors), especially when human inspectors perform the inspection procedure. 21.9 What are the three ways in which an inspection procedure can be automated? Answer: As identified in the text, the three ways to implement automated inspection are (1) automated presentation of parts by an automatic handling system with a human operator still performing the examination and decision steps; (2) automated examination and decision by an automatic inspection machine, with manual loading (presentation) of parts into the machine; and (3) completely automated inspection system in which parts presentation, examination, and decision are all performed automatically. 21.10 What is the difference between off-line inspection and on-line inspection? Answer: Off-line inspection is performed away from the manufacturing process, and there is generally a time delay between processing and inspection. It is often accomplished using statistical sampling methods. On-line inspection is when the inspection procedure is performed when the parts are made, either as an integral step in the processing or assembly operation, or immediately afterward. 21.11 Under what circumstances is process monitoring a suitable alternative to actual inspection of the quality characteristic of the part or product? Answer: The use of process monitoring as an alternative to product inspection relies on the assumption of a deterministic cause-and-effect relationship between the process parameter(s) that can be measured and the quality characteristic(s) that must be maintained within tolerance. Accordingly, by controlling the process parameter(s), indirect control of product quality is achieved. 21.12 What is the difference between distributed inspection and final inspection in quality control? Answer: Distributed inspection is when inspection stations are located along the line of flow of the part or product. In the most extreme case, there is an inspection station following every operation. Final inspection is when one comprehensive inspection is performed on the part or product after all operations have been completed. PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Inspection Accuracy 21.1 (A) For Example 21.1 in the text, develop a table of outcomes similar in format to Table 21.3 in the text. The entries will be the probabilities of the various possible outcomes in the inspection operation. Answer: From Example 21.1, p1 = 0.9535, p2 = 0.5714, and actual q = 14/100 = 0.14 Conforming items: Pr(accept) = p1 (1 − q) = 0.9535(1 − 0.14) = 0.820 Pr(reject) = (1 − p1)(1 − q) = (1 − 0.9535)(1 − 0.14) = 0.040 Nonconforming items: Pr(accept) = (1 − p2)q = (1 − 0.5714)(0.14) = 0.060 Pr(reject) = p2 q = (0.5714)(0.14) = 0.080 21.2 An inspector reports a total of 16 defects out of a total batch size of 200 parts. On closer examination, it is determined that 3 of these reported defects were in fact good pieces, while a total of 7 defective units were undetected by the inspector. What is the inspector’s accuracy in this instance? Specifically, what are the values of p1 and p2? (b) What was the true fraction defect rate? Answer: (a) Actual quantity of defects = (16 − 3) + 7 = 20 defects Actual quantity of good units = 200 − 20 = 180 good units Inspector correctly found 200 − 16 − 7 = 177 good units p1 = 177/180 = 0.9833 Inspector found 16 − 3 = 13 defects p2 = 13/20 = 0.650 (b) True defect rate q = 20/200 = 0.100 21.3 For the preceding problem, develop a table of outcomes similar in format to Table 21.3 in the text. The entries in the table should represent the probabilities of the various possible outcomes in the inspection operation. Answer: From Problem 21.2, p1 = 0.9833, p2 = 0.650, and q = 0.100 Conforming items: Pr(accept) = p1 (1 − q) = 0.9833(1 − 0.10) = 0.885 Pr(reject) = (1 − p1)(1 − q) = (1 − 0.9833)(1 − 0.10) = 0.015 Nonconforming items: Pr(accept) = (1 − p2)q = (1 − 0.65)(0.10) = 0.035 Pr(reject) = p2 q = (0.65)(0.10) = 0.065 21.4 An inspector’s accuracy has been assessed as follows: p1 = 0.96 and p2 = 0.60. The inspector is given the task of inspecting a batch of 500 parts and sorting out the defects from good units. If the actual defect rate in the batch is 0.05, determine (a) the expected number of Type I and (b) Type II errors the inspector will make. (c) Also, what is the expected fraction defect rate that the inspector will report at the end of the inspection task? Answer: (a) E(Type I error): Probability according to Table 21.3 in text is (1 − p1)(1 − q) (1 − p1)(1 − q) = (1 − 0.96)(1 − 0.05) = (0.04)(0.95) = 0.038 In 500 pc, E(number of Type I errors) = 0.038(500) = 19 pc (b) E(Type II error): Probability according to Table 21.3 in text is (1 − p2)q (1 - p2)q = (1 − 0.60)(0.05) = 0.4(0.05) = 0.02 In 500 pc, E(number of Type II errors) = 0.02(500) = 10 pc (c) E(number of defects reported). Probability of rejection in Table 21.3 is 1 − p1 − q(1 − p1 − p2) 1 − p1 − q(1 − p1 − p2) = 1 − 0.96 − 0.05(1 − 0.96 − 0.60) = 0.04 − 0.05(−0.56) = 0.068 In 500 pc, E(number of defects reported) = 0.068(500) = 34 pc Fraction defect rate reported = 34/500 = 0.068 21.5 An inspector must inspect a production batch of 500 parts using a gaging method. If the actual fraction defect rate in the batch is 0.02, and the inspector’s accuracy is given by p1 = 0.96 and p2 = 0.84, determine (a) the number of defects the inspector can be expected to report and (b) the expected number of Type I and Type II errors the inspector will make. Answer: (a) Probability of rejection in Table 21.3 = 1 − p1 − q(1 − p1 − p2) E(q) = (1 − p1 − q(1 − p1 − p2))Q = (1 − 0.96 − 0.02(1 − 0.96 − 0.84))(500) = (0.04 − 0.02(−0.8))(500) = 0.056(500) = 28 pc (b) Probability of Type I errors according to Table 21.3 in text is (1 − p1)(1 − q) E(Type I errors) = (1−p1)(1−q)Q = (1 − 0.96)(1 − 0.02)(500) = 0.04(0.98)(500) = 19.6 pc Probability of Type II errors according to Table 21.3 in text is (1 − p2)q E(Type II errors) = (1 − p2)q Q = (1 − 0.84)(0.02)(500) = 1.6 pc Effect of Fraction Defect Rate 21.6 (A) A batch of 2000 raw work units is processed through 12 operations, each of which has a fraction defect rate of 0.02. How many defect-free units and how many defects are in the final batch? Answer: Qf = Qo(1 − qi)n = Qo (1 − 0.02)12 = 2000(0.98)12 = 2000(0.7847) Qf = 1569 defect-free units D = 2000 − 1569 = 431 defects 21.7 A silicon wafer has a total of 200 integrated circuits (ICs) at the beginning of its fabrication sequence. A total of 60 operations are used to complete the integrated circuits, each of which inflicts damages on 1.0% of the ICs. The damages compound, meaning that an IC that is already damaged has the same probability of being damaged by a subsequent process as a previously undamaged IC. How many defect-free ICs remain at the end of the fabrication sequence? Answer: Qf = Qo(1 − qi)n = Qo (1 − 0.01)60 = 200(0.99)60 = 200(0.54716) Qf = 109 defect-free ICs 21.8 A batch of work parts is processed through a sequence of nine processing operations, which have fraction defect rates of 0.03, 0.05, 0.02, 0.04, 0.06, 0.01, 0.03, 0.04, and 0.07, respectively. A total of 13,974 completed parts are produced by the sequence. What was the starting batch quantity? Answer: Qf = Qo (1 − q1)(1 − q2) . . . (1 − q9) 5000 = Qo (0.97)(0.95)(0.98)(0.96)(0.94)(0.99)(0.97)(0.96)(0.93) = Qo (0.6987) Qo = = 20,000 pc 21.9 (A) A production line consists of six workstations, as shown in Figure P21.9. The six stations are as follows: (1) first manufacturing process, scrap rate is q1 = 0.10; (2) inspection for first process, separates all defects from first process; (3) second manufacturing process, scrap rate is q3 = 0.20; (4) inspection for second process, separates all defects from second process; (5) rework, repairs defects from second process, recovering 70% of the defects from the preceding operation and leaving 30% of the defects as still defective; (6) third manufacturing process, scrap rate q6 = zero. If the output from the production line is to be 100,000 defect-free units, what quantity of raw material units must be launched onto the front of the line? Answer: Qf = Qo (1 − q1)(1 − q3 + q3(1 − q5))(1 − q6) 100,000 = Qo (1−0.1)(1−0.2 + 0.2 × 0.70)(1−0) = Qo (0.9)(0.8 + 0.14)(1.0) = 0.846 Qo Qo = = 118,203 pc 21.10 A certain industrial process can be depicted as in Figure P21.10. Operation 1 is a disassembly process in which each unit of raw material is separated into one unit each of parts A and B. These parts are then processed separately in operations 2 and 3, respectively, which have scrap rates of q2 = 0.05 and q3 = 0.10. Inspection stations 4 and 5 sort good units from bad for the two parts. Then the parts are assembled back together in operation 6, which has a fraction defect rate q6 = 0.15. Final inspection station 7 sorts good units from bad. The desired final output quantity is 100,000 units. (a) What is the required starting quantity (into operation 1) to achieve this output? (b) Will there be any leftover units of parts A or B, and if so, how many? Answer: (a) Let Qi = output quantity of process i and Qi-1 = input quantity to process i. Critical path through diagram = 1 - 3 - 5 - 6 - 7 (path B) since fraction defect rate q3 (process 3) is greater than q2. Q7 = Qf = 100,000 pc. Q7 = Q5(1 − 0.15) Q5 = Q3 Q3 = Q1(1 − 0.10) Q1 = Qo(1 − 0) = Qo 100,000 = Qo(1 − 0)(1 − 0.10)(1 − 0.15) = 0.765Qo Qo = = 130,719 pc If alternative path 1 - 2 - 4 - 6 - 7 (path A) had been taken, then the calculations are as follows: 100,000 = Qo(1 − 0)(1 − 0.05)(1 − 0.15) = 0.8075Qo Qo = = 123,839 pc, which would not be enough units to supply path B. (b) Leftover units of A = 130,719 − 123,839 = 6880 pc 21.11 A certain component is produced in three sequential operations. Operation 1 produces defects at a rate q1 = 5%. Operation 2 produces defects at a rate q2 = 8%. Operation 3 produces defects at a rate q3 = 10%. Operations 2 and 3 can be performed on units that are already defective. If 10,000 starting parts are processed through the sequence, (a) how many units are expected to be defect-free, (b) how many units are expected to have exactly one defect, and (c) how many units are expected to have all three defects? Answer: (a) Quantity of defect-free pieces Qf = Qo(1 − q1)(1 − q2)(1 − q3) Qf = 10,000(1 − 0.05)(1 − 0.08)(1 − 0.10) = 10,000(0.95)(0.92)(0.90) = 7866 pc (b) Quantity of pieces with exactly one defect: D1 = 10,000(0.05)(0.92)(0.90) = 414 pc D2 = 10,000(0.95)(0.08)(0.90) = 684 pc D3 = 10,000().95)(0.92)(0.10) = 874 pc Total pieces with exactly one defect = 414 + 684 + 874 = 1972 pc (c) Quantity of pieces with all three defects D1,2,3 = 10,000(0.05)(0.08)(0.10) = 4 pc 21.12 An industrial process can be depicted as in Figure P21.12. Two components are made, respectively, by operations 1 and 2, and then assembled together in operation 3. Scrap rates are as follows: q1 = 0.20, q2 = 0.10, and q3 = 0. Input quantities of raw components at operations 1 and 2 are 25,000 and 20,000, respectively. One of each component is required in the assembly operation. Trouble is that defective components can be assembled just as easily as good components, so inspection and sortation is required in operation 4. Determine (a) how many defect-free assemblies will be produced and (b) how many assemblies will be made with one or more defective components. (c) Will there be any leftover units of either component, and if so, how many? Answer: (a) Determine the limiting operation, 1 or 2. Let Qi = output quantity of good units from operation i. Q1 = 25,000(1 − q1) = 25,000(0.80) = 20,000 pc Q2 = 20,000(1 − q2) = 20,000(0.90) = 18,000 pc Conclusion: operation 2 is the limiting operation. Quantity of defect-free assemblies = Qf Qf = 20,000(1 − q1)(1 − q2)(1 − q3) = 20,000(0.8)(0.9)(1.0) = 20,000(0.72) = 14,400 pc (b) Quantity of pieces with one or more defects Df = Qo − Qf = 20,000 − 14,400 = 5,600 pc Check: Df = 20,000(0.8 × 0.1 + 0.2 × 0.9 + 0.2 × 0.1) = 1600 + 3600 + 400 = 5,600 pc (c) Leftover units from operation 1 = 25,000 − 20,000 = 5000 pc Inspection Costs 21.13 (A) Two inspection alternatives are to be compared for a processing sequence consisting of 20 operations performed on a batch of 100 starting parts: (1) one final inspection and sortation operation following the last operation, and (2) distributed inspection with an inspection and sortation after each operation. The cost of each processing operation Cpr = $1.00 per unit processed. The fraction defect rate at each operation = 0.03. The cost of the single final inspection and sortation in alternative (1) is Csf = $2.00 per unit. The cost of each inspection and sortation in alternative (2) is Cs = $0.10 per unit. Compare total processing and inspection costs per batch for the two cases. Answer: Alternative (1): Final inspection: Cb = Qo(nCpr + Csf) = 100(20 × 1.00 + 2.00) = 100(22.00) = $2200.00 Alternative (2): Distributed inspection: Cb = Qo(1 + (1 − q) + (1 − q)2 + . . + (1 − q)n-1)(Cpr + Cs) = 100(1 + (0.97) + (0.97)2 + . . . + 0.97)19 (1.00 + 0.10) = 100(15.2068)(1.10) = $1672.75 21.14 In the preceding problem, instead of inspecting and sorting after every operation, the 20 operations will be divided into groups of five, with inspections after operations 5, 10, 15, and 20. Following the logic of Equation (21.11), the cost of each inspection will be five times the cost of inspecting for one defect feature; that is, Cs5 = Cs10 = Cs15 = Cs20 = 5($0.10) = $0.50 per unit inspected. Processing cost per unit for each operation remains the same as before at Cpr = $1.00, and Qo = 100 parts. What is the total processing and inspection cost per batch for this partially distributed inspection system? Answer: Cb = Qo(5Cpr + Cs5) + Qo(1−q)5(5Cpr + Cs10) + Qo(1−q)10(5Cpr + Cs15) + Qo (1−q)15(5Cpr + Cs20) = 100(5.00 + 0.50) + 100(1−0.03)5(5.00 + 0.50) + 100(1−0.03)10(5.00 + 0.50) + 100(1−0.03)15(5.00 + 0.50) = 100(5.50)(1 + (0.97)5 + (0.97)10 + (0.97)15) = 550(1 + 0.8587 + 0.7374 + 0.6333) = 550(3.2294) = $1776.17 21.15 A processing sequence consists of 10 operations, each of which is followed by an inspection and sortation operation to detect and remove defects generated in the operation. Defects in each process occur at a rate of 0.04. Each processing operation costs $1.00 per unit processed, and the inspection/sortation operation costs $0.30 per unit. (a) Determine the total processing and inspection costs for this distributed inspection system. (b) A proposal is being considered to combine all of the inspections into one final inspection and sortation station following the last processing operation. Determine the cost per unit of this final inspection and sortation station that would make the total cost of this system equal to that of the distributed inspection system. Answer: (a) Distributed inspection Equation (21.11): Let Qo = 1, thus Cb = cost/unit. Cb = Qo(1 − (1 − q) + (1 − q)2 + . . . + (1 − q)9)(Cpr + Cs) = 1(1 + (0.96) + (0.96)2 + . . . + (0.96)9)(1.00 + 0.20) = (1 + 0.96 + 0.9216 + . . . + 0.6925)(1.30) = 8.3791(1.30) = $10.893/pc (b) Final inspection Equation (21.9): Again, let Qo = 1, thus Cb = cost/unit. Cb = Qo(nCpr + Csf) = 1(10 × 1.00 + Csf) = 10.00 + Csf Set Cb for distributed inspection = Cb for final inspection: 10.893 = 10.00 + Csf Csf = $0.893/pc 21.16 This problem is intended to show the merits of a partially distributed inspection systems in which inspections are placed after processing steps that generate a high fraction defect rate. The processing sequence consists of eight operations with fraction defect rates for each operation as follows: Three alternatives are to be compared: (1) fully distributed inspection, with an inspection after every operation; (2) partially distributed inspection, with inspections following operations 4 and 8 only; and (3) one final inspection station after operation 8. All inspections include sortations. In alternative (2), the inspection procedures are each designed to detect all of the defects for the preceding four operations. The cost of processing is Cpr = $1.00 for each of operations 1 through 8. Inspection/sortation costs for each alternative are given in the table below. Compare total processing and inspection costs for the three cases. Answer: Alternative (1): Fully distributed inspection Equation (21.10): Qo = 1, thus Cb = cost/unit. Cb = 1.10 + 0.99(1.10) + (0.99)2(1.10) + (0.99)3(1.10) + (0.99)3(0.89)(1.10) + (0.99)3(0.89)(0.99)(1.10) + (0.99)3(0.89)(0.99)2(1.10) + (0.99)3(0.89)(0.99)3(1.10) = 1.10(1 + 0.99 + 0.9801 + 0.9703 + 0.8636 + 0.8549 + 0.8464 + 0.8379) = 1.10(7.3432) = $8.078/pc Alternative (2): Partially distributed inspection Equation (21.14) is adapted for the data in this problem. Again, Qo = 1, thus Cb = cost/unit Cb = (4Cpr + Cs4) + (1 − q1)(1 − q2)(1 − q3)(1 − q4)(4Cpr + Cs8) = (4.40) + (0.99)3(0.89)(4.40) = 4.40(1 + 0.8636) = $8.20/pc Alternative (3): One final inspection Equation (21.8): again, Qo = 1, thus Cb = cost/unit Cb = (8Cpr + Csf) = 8.00 = 0.80 = $8.80/pc Conclusion: The partially distributed inspection achieves 83% of the savings of a fully distributed inspection system with only 25% of the inspection stations. Inspection or No Inspection 21.17 (A) A batch of 1000 parts has been produced and a decision is needed whether or not to 100% inspect the batch. Past history with this part suggests that the fraction defect rate is around 0.02. Inspection cost per part is $0.15. If the batch is passed on for subsequent processing, the damage cost for each defective unit in the batch is $9.00. Determine (a) batch cost for 100% inspection and (b) batch cost if no inspection is performed. (c) What is the critical fraction defect value for deciding whether to inspect. Answer: (a) Cb for 100% inspection = QCs = 1000(0.15) = $150 (b) Cb for no inspection = QqCd = 1000(0.02)(9.00) = $180 (c) Since the average q is greater than 0.01667, the decision should be to inspect 100%. 21.18 Given the data from the preceding problem, sampling inspection is being considered as an alternative to 100% inspection. The sampling plan calls for a sample of 50 parts to be drawn at random from the batch. Based on the operating characteristic curve for this sampling plan, the probability of accepting the batch is 95% at the given defect rate of q = 0.015. Determine the batch cost for sampling inspection. Answer: Cb = Cs Qs + (Q − Qs)qCdPa + (Q − Qs)Cs(1 − Pa) = 0.15(50) + (1000 − 50)(0.02)(9.00)(0.95) + (1000 − 50)(0.15)(0.05) = $177.08 Chapter 22 INSPECTION TECHNOLOGIES REVIEW QUESTIONS 22.1 Define the term measurement. Answer: Measurement is a procedure in which an unknown quantity is compared to a known standard, using an accepted and consistent system of units. 22.2 What is metrology? Answer: Metrology is the science of measurement. 22.3 What are the seven basic quantities used in metrology upon which all other variables are derived? Answer: The seven basic quantities are length, mass, time, electric current, temperature, luminous intensity, and matter. 22.4 What is the difference between accuracy and precision in measurement? Define these two terms. Answer: Measurement accuracy is the degree to which the measured value agrees with the true value of the quantity of interest. A measurement procedure is accurate when it is absent of systematic errors, which are positive or negative deviations from the true value that are consistent from one measurement to the next. Precision is a measure of repeatability in a measurement process. Good precision means that random errors in the measurement procedure are minimized. 22.5 With respect to measuring instruments, what is calibration? Answer: Calibration is a procedure in which the measuring instrument is checked against a known standard. 22.6 What is meant by the term contact inspection? Answer: Contact inspection involves the use of a mechanical probe or other device that makes contact with the object being inspected. The purpose of the probe is to measure or gage the object in some way. 22.7 What are some of the advantages of noncontact inspection? Answer: The text lists four advantages of noncontact inspection: (1) They avoid damage to the surface that might result from contact inspection. (2) Inspection cycle times are inherently faster. (3) Noncontact methods can often be accomplished on the production line without the need for any additional handling of the parts, whereas special handling and positioning of the parts are usually required in contact inspection. (4) There is increased opportunity for 100% automated inspection. 22.8 What is meant by the term coordinate metrology? Answer: Coordinate metrology is the science concerned with the measurement of the actual shape and dimensions of an object and comparing these with the desired shape and dimensions, as might be specified on a part drawing. It consists of the evaluation of the location, orientation, dimensions, and geometry of the part or object. 22.9 What are the two basic components of a coordinate measuring machine? Answer: The two basic components of a CMM are (1) the probe head and probe that contacts the surface of an object and (2) the mechanical structure that provides motion of the probe in three-dimensional Cartesian axes. 22.10 Name the four categories into which the methods of operating and controlling a CMM can be classified. Answer: The four categories are (1) manual drive, (2) manual drive with computer–assisted data processing, (3) motor drive with computer–assisted data processing, and (4) direct computer control with computer–assisted data processing. 22.11 What does the term reverse engineering mean in the context of coordinate measuring machines? Answer: Reverse engineering means taking an existing physical part and constructing a computer model of the part geometry based on a large number of measurements of its surface by a CMM. 22.12 The text lists six characteristics of potential applications for which CMMs are most appropriate. Name the six characteristics. Answer: The seven characteristics listed in the text are (1) Many inspectors are currently performing repetitive manual inspection operations. (2) The application involves postprocess inspection. (3) Measurement of geometric features requires multiple contact points. (4) Complex part geometry. (5) A wide variety of parts must be inspected. (6) Repeat orders are common. 22.13 What are some of the arguments and objections to the use of inspection probes mounted in toolholders on machine tools? Answer: The two main arguments and objections are the following: (1) Errors that are inherent in the cutting operation will also be manifested in the measuring operation. For example, if there is misalignment between the machine tool axes, thus producing out-of-square parts, this condition will not be identified by the machine-mounted probe because the movement of the probe is affected by the same axis misalignment. (2) Machine–mounted inspection probes take time above and beyond the regular machining cycle. Time is required to program the inspection routines, and time is lost during the cutting sequence for the probe to perform its measurement function. 22.14 What is the most common method used to measure surfaces of a part? Answer: The most common surface measuring instruments are stylus–type instruments which are electronic devices that utilize a cone–shaped diamond stylus with very small point radius to traverse across the test surface at a constant slow speed. The stylus moves vertically to trace the deviations from the nominal flat surface of the part. 22.15 What is machine vision? Answer: Machine vision can be defined as the acquisition of image data, followed by the processing and interpretation of these data by computer for some useful application. 22.16 The operation of a machine vision system can be divided into three functions. Name and briefly describe them. Answer: The three functions are the following: (1) Image acquisition and digitization. This involves the use of a digital camera and a digitizing system to store the image data. The image is obtained by dividing the viewing area into a matrix of pixels, in which each pixel has a value proportional to the light intensity of that portion of the scene. An analog-to-digital converter (ADC) converts the intensity value of each pixel into its equivalent digital value. (2) Image processing and analysis. This is designed to separate regions of interest in the image (using techniques such as edge detection) and to characterize the object in the image by means of its features (such as length and area). (3) Interpretation. This is concerned with recognizing the object, usually by comparing it with predefined models or standard values. 22.17 What are the two types of image sensors used in machine vision cameras? Answer: The two types of image sensors used in machine vision cameras are (1) charge-coupled device (CCD) and (2) complementary metal-oxide semiconductor (CMOS). 22.18 What is the largest application of machine vision in industry? Answer: The largest application of machine vision in industry is quality control inspection. 22.19 What is an optical comparator? Answer: An optical comparator is an optical device that projects the shadow of an object (e.g., a work-part) against a large screen in front of an operator. The object can be moved in the x-y axes, permitting the operator to obtain dimensional data using cross hairs on the screen. 22.20 The word laser is an acronym for what? Answer: The word laser stands for light amplification by stimulated emission of radiation). PROBLEMS Answers to problems labeled (A) are listed in the Appendix at the back of the book. Inspection Metrology 22.1 A measuring instrument is being designed to inspect two part dimensions that are considered key characteristics (Section 21.1.2). The two dimensions with tolerances are 205.5 ± 0.25 mm and 57.0 ± 0.20 mm. To what level of precision should the instrument be designed to measure these part dimensions? Answer: Based on the rule of 10, the precision of the instrument should be ± 0.020 mm 22.2 (A) A digital measuring device has a full-scale range of 250 mm, and its storage register has 12 bits for each measurement. What is the measurement resolution of this device? Answer: 22.3 A digital scale has a range of 30 kg, and its storage register capacity is 10 bits. It is used in a packaging line, on which the net weight of each product is specified as 20 ± 0.40 kg. (a) What is the measurement resolution of this scale? (b) Is this resolution sufficient for this application based on the rule of 10? Answer: (b) The rule of 10 states that the measurement precision, which corresponds to the measurement resolution in this problem, should be one-tenth of the tolerance. With a resolution of 0.029 kg and a tolerance of 0.40 kg, the resolution of the scale is sufficient for this application. 22.4 A digital measuring instrument has a range of 40.0 in. It is used to measure the lengths of sheet metal parts in an automobile stamping plant. The measurement resolution of the instrument is specified as 0.125 in. How many bits must be in the storage register to achieve this resolution? Answer: Coordinate Metrology (Appendix A22) For ease of computation, numerical values in the following problems are given at a lower level of precision than most CMMs would be capable of. 22.5 (A) Two point locations corresponding to a certain length dimension have been measured by a coordinate measuring machine in the x-y plane. The coordinates of the first point are (12.511, 2.273), and the coordinates of the opposite point are (4.172, 1.985), where the units are inches. The coordinates have been corrected for probe radius. Determine the length dimension that would be computed by the CMM software. Answer: 22.6 The coordinates at the two ends of a certain length dimension have been measured by a CMM. The coordinates of the first end are (120.5, 50.2, 20.2), and the coordinates of the opposite end are (23.1, 11.9, 20.3), where the units are mm. The given coordinates have been corrected for probe radius. Determine the length dimension that would be computed by the CMM software. Answer: 22.7 Three point locations on the surface of a drilled hole have been measured by a CMM in the x-y axes. The three coordinates are: (16.42, 17.17), (20.20, 11.85), and (24.08, 16.54), where the units are mm. These coordinates have been corrected for probe radius. Determine (a) the coordinates of the hole center and (b) the hole diameter, as they would be computed by the CMM software. Answer: 22.8 (A) Three point locations on the surface of a cylinder have been measured by a coordinate measuring machine. The cylinder is positioned so that its axis is perpendicular to the x-y plane. The three coordinates in the x-y axes are: (5.242, 0.124), (0.325, 4.811), and (−4.073, −0.544), where the units are inches. The coordinates have been corrected for probe radius. Determine (a) the coordinates of the cylinder axis and (b) the cylinder diameter, as they would be computed by the CMM software. Answer: 22.9 Two points on a line have been measured by a CMM in the x-y plane. The point locations have the following coordinates: (12.257, 2.550) and (3.341, −10.294), where the units are inches and the coordinates have been corrected for probe radius. Find the equation for the line in the form of Equation (A22.5). Answer: 22.10 Two points on a line are measured by a CMM in the x-y plane. The points have the following coordinates: (100.24, 20.57) and (50.44, 60.46), where the units are mm. The given coordinates have been corrected for probe radius. Determine the equation for the line in the form of Equation (A22.5). Answer: 22.11 The coordinates of the intersection of two lines are to be determined using a CMM to define the equations for the two lines. The two lines are the edges of a machined part, and the intersection represents the corner where the two edges meet. Both lines lie in the x-y plane. Two points are measured on the first line to have coordinates of (5.254, 10.430) and (10.223, 6.052). Two points are measured on the second line to have coordinates of (6.101, 0.657) and (8.970, 3.824). Units are inches. The coordinate values have been corrected for probe radius. (a) Determine the equations for the two lines in the form of Equation (A22.5). (b) What are the coordinates of the intersection of the two lines? (c) The edges represented by the two lines are specified to be perpendicular to each other. Find the angle between the two lines to determine if the edges are perpendicular? Answer: 22.12 Two of the edges of a rectangular part are represented by two lines in the x-y plane on a CMM worktable, as illustrated in Figure P22.12. It is desired to mathematically redefine the coordinate system so that the two edges are used as the x- and y-axes, rather than the regular x-y axes of the CMM. To define the new coordinate system, two parameters must be determined: (a) the origin of the new coordinate system must be located in the existing CMM axis system; and (b) the angle of the x-axis of the new coordinate system must be determined relative to the CMM x-axis. Two points on the first edge (line 1) have been measured by the CMM and the coordinates are (46.21, 22.98) and (90.25, 32.50), where the units are mm. Also, two points on the second edge (line 2) have been measured by the CMM and the coordinates are (26.53, 40.75) and (15.64, 91.12). The coordinates have been corrected for the radius of the probe. Find (a) the coordinates of the new origin relative to the CMM origin and (b) degrees of rotation of the new x-axis relative to the CMM x-axis. (c) Are the two lines (part edges) perpendicular? Answer: 22.13 (A) Three point locations on the flat surface of a part have been measured by a CMM. The three point locations are (225.21, 150.23, 40.17), (14.24, 140.92, 38.29), and (12.56, 22.75, 38.02), where the units are mm. The coordinates have been corrected for probe radius. (a) Determine the equation for the plane in the form of Equation (A22.8). (b) To assess flatness of the surface, a fourth point is measured by the CMM. If its coordinates are (120.22, 75.34, 39.26), what is the vertical deviation of this point from the perfectly flat plane determined in (a)? Answer: Optical Inspection Technologies 22.14 A digital camera that uses a CCD image sensor has a 1024 × 768 pixel matrix. The analog-to-digital converter takes 0.05 µ-sec (0.05 × 10-6 sec) to convert the analog charge signal for each pixel into the corresponding digital signal, including the time to switch between pixels. Determine (a) how much time is required to collect the image data for one frame, and (b) is this time compatible with the processing rate of 30 frames per second? Answer: (a) 1024 × 768 = 786,432 pixels ADC time = 786,432(0.05 × 10-6) = 0.0393 s (b) Comparing this with 1/30 = 0.0333 sec, the analog-to-digital conversion cannot be completed within the time required. 22.15 (A) The pixel count of a digital camera that uses a CCD image sensor is 1600 × 1200. Each pixel is converted from an analog voltage to the corresponding digital signal by an analog-to-digital converter. Each conversion takes 0.015 µ-sec (0.015 × 10-6 sec). (a) Given this time, how long will it take to collect and convert the image data for one frame? (b) Can this be done 30 times/sec? Answer: (a) 1600 × 1200 = 1,920,000 pixels ADC time = 1,920,000(0.015 × 10-6) = 0.0288 sec (b) Comparing this with 1/30 = 0.0333 sec, the analog-to-digital conversion can be completed within the time required. 22.16 A digital camera that uses a CCD image sensor is to have a 1035 × 1320 pixel matrix. An image processing rate of 30 times/sec must be achieved (0.0333 sec per frame). To allow for time lost in other data processing per frame, the total ADC time per frame must be 80% of the 0.0333 sec. In order to be compatible with this speed, in what time period must the analog-to-digital conversion be accomplished per pixel? Answer: 22.17 A scanning laser device, similar to the one shown in Figure 22.12, is to be used to measure the diameters of shafts that are ground in a centerless grinding operation. The part has a diameter of 0.475 in with a tolerance of ±0.002 in. The four-sided mirror of the scanning laser beam device rotates at 250 rev/min. The collimating lens focuses 30° of the sweep of the mirror into a swath that is 1.000 inch wide. It is assumed that the light beam moves at a constant speed across this swath. The photodetector and timing circuitry is capable of resolving time units as fine as 100 nanoseconds (100 × 10-9 sec.). This resolution should be equivalent to no more than 10% of the tolerance band (0.004 in). (a) Determine the interruption time of the scanning laser beam for a part whose diameter is equal to the nominal size. (b) How much of a difference in interruption time is associated with the tolerance of ±0.002 in? (c) Is the resolution of the photodetector and timing circuitry sufficient to achieve the 10% rule on the tolerance band? Answer: 22.18 Triangulation computations are used to determine the distance of parts moving on a conveyor. The setup of the optical measuring apparatus is as illustrated in the text in Figure 22.14. The angle between the beam and the surface of the part is . Suppose for one given part passing on the conveyor, the baseline distance is 7.500 in, as measured by the linear photosensitive detection system. What is the distance of this part from the baseline? Answer: R = L/tan(30) = 7.500/0.5774 = 12.990 in Solution Manual for Automation, Production Systems, and Computer-Integrated Manufacturing Mikell P. Groover 9780133499612, 9780134605463