1. G – 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Acceptance Sampling Plans G For Operations Management, 9e by Krajewski/Ritzman/Malhotra © 2010 Pearson Education PowerPoint Slides by Jeff Heyl   AQL LTPD

2. G – 2 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Acceptance Sampling Plan Decisions An inspection procedure used to determine whether to accept or reject a specific quantity of materials  Impact of TQM Basic procedure  Take random sample  Accept or reject, based on results Producer, or seller, is origin of the material or service Consumer, or buyer, is destination of the material or service Sampling plans

3. G – 3 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Quality and Risk Decisions Acceptable quality level (AQL) is the quality level desired by the consumer Producer’s risk (  ) is the probability that a shipment having exactly this level of quality will be rejected  Rejecting a good (AQL) lot is a type I error  Consumers also desire low producer’s risk because sending good materials back to the supplier disrupts the consumer’s production processes  Most often the producer’s risk is set at 0.05, or 5 percent

4. G – 4 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Quality and Risk Decisions Lot tolerance proportion defective (LTPD), the worst level the customer can tolerate Consumer’s risk, (  ) is the probability a shipment having exactly this level of quality (the LTPD) will be accepted  Accepting a bad (LTPD) lot is a type II error  A common value for the consumer’s risk is 0.10, or 10 percent

5. G – 5 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Single-Sampling Plans States the sample size, n, and the acceptable number of defectives, c  The accept-reject decision is based on the results of one sample taken at random from a large lot  If the quality characteristic of the sample passes the test (defects ≤ c ), accept the lot  If the sample fails (defects > c ) there may be complete inspection of the lot or the entire lot is rejected A good lot could be rejected if the sample includes an unusually large number of defects A bad lot could be accepted if the quality in the sample is better than in the lot

6. G – 6 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Double-Sampling Plans Two sample sizes, ( n 1 and n 2 ), and two acceptance numbers ( c 1 and c 2 ) Take a random sample of relatively small size n 1, from a large lot If the sample passes the test (≤ c 1 ), accept the lot If the sample fails (> c 2 ), the entire lot is rejected If the sample is between c 1 and c 2, then take a larger second random sample, n 2 If the combined number of defects ≤ c 2 accept the lot, otherwise reject

7. G – 7 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Sequential Sampling Plans Results of random samples of one unit, tested one-by-one, are compared to sequential-sampling chart Chart guides decision to reject, accept, or continue sampling, based on cumulative results Average number of items inspected (ANI) is generally lower with sequential sampling

8. G – 8 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Reject Continue sampling Accept 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 – 0 – Cumulative sample size ||||||| 10203040506070 Number of defectives Sequential Sampling Chart Figure G.1 – Sequential-Sampling Chart

9. G – 9 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Operating Characteristic Curve Perfect discrimination between good and bad lots requires 100% inspection Select sample size n and acceptance number c to achieve the level of performance specified by the AQL, , LTPD, and  Drawing the OC curve The OC curve shows the probability of accepting a lot P a, as a dependent function of p, the true proportion of defectives in the lot For every possible combination of n and c, there exists a unique operating characteristics curve

10. G – 10 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Operating Characteristic Curve   Ideal OC curve Typical OC curve 1.0 AQL LTPD Probability of acceptance Proportion defective Figure G.2 – Operating Characteristic Curves

11. G – 11 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Constructing an OC Curve EXAMPLE G.1 The Noise King Muffler Shop, a high-volume installer of replacement exhaust muffler systems, just received a shipment of 1,000 mufflers. The sampling plan for inspecting these mufflers calls for a sample size n = 60 and an acceptance number c = 1. The contract with the muffler manufacturer calls for an AQL of 1 defective muffler per 100 and an LTPD of 6 defective mufflers per 100. Calculate the OC curve for this plan, and determine the producer’s risk and the consumer’s risk for the plan.

12. G – 12 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Constructing an OC Curve SOLUTION Let p = 0.01. Then multiply n by p to get 60(0.01) = 0.60. Locate 0.60 in Table G.1. Move to the right until you reach the column for c = 1. Read the probability of acceptance: 0.878. Repeat this process for a range of p values. The following table contains the remaining values for the OC curve. Values for the Operating Characteristic Curve with n = 60 and c = 1 Proportion Defective ( p ) np Probability of c or Less Defects ( P a )Comments 0.01 (AQL)0.60.878  = 1.000 – 0.878 = 0.122 0.021.20.663 0.031.80.463 0.042.40.308 0.053.00.199 0.06 (LTPD)3.60.126  = 0.126 0.074.20.078 0.084.80.048 0.095.40.029 0.106.00.017

13. G – 13 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Constructing an OC Curve 0.878 0.663 0.463 0.308 0.199 0.126 0.078 0.048 0.029 0.017 (AQL) (LTPD) 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| 12345678910 Proportion defective (hundredths) Probability of acceptance  = 0.126  = 0.122 Figure G.3 – The OC Curve for Single-Sampling Plan with n = 60 and c = 1

14. G – 14 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application G.1 A sampling plan is being evaluated where c = 10 and n = 193. If AQL = 0.03 and LTPD = 0.08. What are the producer’s risk and consumer’s risk for the plan? Draw the OC curve. SOLUTION Finding  (probability of rejecting AQL quality) p = np = P a =  = 0.03 5.79 0.965 0.035 (or 1.0 – 0.965) Finding  (probability of accepting LTPD quality) p = np = P a =  = 0.08 15.44 0.10

15. G – 15 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application G.1 1.0 – 0.8 – 0.6 – 0.4 – 0.2 – 0.0 – ||||||||||| 0246810 Probability of acceptance Percentage defective  = 0.035  = 0.10

16. G – 16 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Explaining Changes in the OC Curve Sample size effect  Increasing n while holding c constant increases the producer’s risk and reduces the consumer’s risk n Producer’s Risk ( p = AQL) Consumer’s Risk ( p = LTPD) 600.1220.126 800.1910.048 1000.2640.017 1200.3320.006

17. G – 17 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Explaining Changes in the OC Curve 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| 12345678910 (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance n = 60, c = 1 n = 80, c = 1 n = 100, c = 1 n = 120, c = 1 Figure G.4 –Effects of Increasing Sample Size While Holding Acceptance Number Constant

18. G – 18 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Explaining Changes in the OC Curve Acceptance level effect  Increasing c while holding n constant decreases the producer’s risk and increases the consumer’s risk c Producer’s Risk ( p = AQL) Consumer’s Risk ( p = LTPD) 10.1220.126 20.0230.303 30.0030.515 40.0000.706

19. G – 19 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Explaining Changes in the OC Curve 1.0 – 0.9 – 0.8 – 0.7 – 0.6 – 0.5 – 0.4 – 0.3 – 0.2 – 0.1 – 0.0 – |||||||||| 12345678910 (AQL) (LTPD) Proportion defective (hundredths) Probability of acceptance n = 60, c = 1 n = 60, c = 2 n = 60, c = 3 n = 60, c = 4 Figure G.5 –Effects of Increasing Acceptance Number While Holding Sample Size Constant

20. G – 20 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Acceptance Sampling Plan Data AQL BasedLTPD Based Acceptance Number Expected Defectives Sample Size Expected Defectives Sample Size 00.050952.299638 10.3552363.887565 20.8112815.321789 31.36751376.6697111 41.96801977.9894133 52.62562639.2647154 63.283832810.5139175 73.979439811.7726196 84.693646912.9903217 95.423754214.2042237 106.163561615.4036257

21. G – 21 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Average Outgoing Quality AOQ is the expected (or Average) proportion of defects that a particular sampling plan would allow to pass through (Outgoing Quality) inspection Rectified inspection – defects found during the sampling process are removed and reworked or replaced with conforming material where p =true proportion defective of the lot P a =probability of accepting the lot N =lot size n =sample size

22. G – 22 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Average Outgoing Quality Rejected lots are subjected to 100% inspection AOQL is the maximum value of the average outgoing quality over all possible values of the proportion defective Different sampling plans have different AOQs and AOQLs

23. G – 23 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Calculating the AOQL EXAMPLE G.2 Suppose that Noise King is using rectified inspection for its single-sampling plan. Calculate the average outgoing quality limit for a plan with n = 110, c = 3, and N = 1,000. Use Table G.1 (pp. G.9–G.11) to estimate the probabilities of acceptance for values of the proportion defective from 0.01 to 0.08 in steps of 0.01. SOLUTION Use the following steps to estimate the AOQL for this sampling plan: Step 1: Determine the probabilities of acceptance for the desired values of p. These are shown in the following table. However, the values for p = 0.03, 0.05, and 0.07 had to be interpolated because the table does not have them. For example, P a for p = 0.03 was estimated by averaging the P a values for np = 3.2 and np = 3.4, (or 0.603 + 0.558)/2 = 0.580.

24. G – 24 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Calculating the AOQL Proportion Defective ( p ) np Probability of Acceptance ( P a ) 0.011.100.974 0.022.200.819 0.033.300.581= (0.603 + 0.558)/2 0.044.400.359 0.055.500.202= (0.213 + 0.191)/2 0.066.600.105 0.077.700.052= (0.055 + 0.048)/2 0.088.800.024

25. G – 25 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Calculating the AOQL Step 2: Calculate the AOQ for each value of p. For p = 0.01:0.01(0.974)(1000 – 110)/1000 = 0.0087 The plot of the AOQ values is shown in Figure G.6. For p = 0.02:0.02(0.819)(1000 – 110)/1000 = 0.0146 For p = 0.03:0.03(0.581)(1000 – 110)/1000 = 0.0155 For p = 0.04:0.04(0.359)(1000 – 110)/1000 = 0.0128 For p = 0.05:0.05(0.202)(1000 – 110)/1000 = 0.0090 For p = 0.06:0.06(0.105)(1000 – 110)/1000 = 0.0056 For p = 0.07:0.07(0.052)(1000 – 110)/1000 = 0.0032 For p = 0.08:0.08(0.024)(1000 – 110)/1000 = 0.0017

26. G – 26 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Calculating the AOQL Step 3: Identify the largest AOQ value, which is the estimate of the AOQL. In this example, the AOQL is 0.0155 at p = 0.03. AOQL 1.6 – 1.2 – 0.8 – 0.4 – 0 – ||||||||12345678||||||||12345678 Defectives in lot (percent) Average outgoing quality (percent) Figure G.5 –Average Outgoing Quality Curve for the Noise King Muffler Service

27. G – 27 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Application G.2 Demonstrate the model for computing AOQ Management has selected the following parameters: AQL= 0.01  = 0.05 LTPD= 0.06  = 0.10 n = 100 c = 3 What is the AOQ if p = 0.05 and N = 3000? p = np = P a = AOQ= 0.05 1000(0.05) = 5 0.265 = 0.0128 SOLUTION

28. G – 28 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. An inspection station has been installed between two production processes. The feeder process, when operating correctly, has an acceptable quality level of 3 percent. The consuming process, which is expensive, has a specified lot tolerance proportion defective of 8 percent. The feeding process produces in batch sizes; if a batch is rejected by the inspector, the entire batch must be checked and the defective items reworked. Consequently, management wants no more than a 5 percent producer’s risk and, because of the expensive process that follows, no more than a 10 percent chance of accepting a lot with 8 percent defectives or worse. Solved Problem a.Determine the appropriate sample size, n, and the acceptable number of defective items in the sample, c. b.Calculate values and draw the OC curve for this inspection station. c.What is the probability that a lot with 5 percent defectives will be rejected?

29. G – 29 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem SOLUTION a.For AQL = 3 percent, LTPD = 8 percent,  = 5 percent, and  = 10 percent, use Table G.1 and trial and error to arrive at a sampling plan. If n = 180 and c = 9, np = 180(0.03) = 5.4  = 0.049 180(0.08) = 14.4  = 0.092 Sampling plans that would also work are n = 200, c = 10; n = 220, c = 10; and n = 240, c = 12.

30. G – 30 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem b.The following table contains the data for the OC curve. Table G.1 was used to estimate the probability of acceptance. Figure G.7 shows the OC curve. Proportion Defective ( p ) np Probability of c or Less Defectives ( P a )Comments 0.011.81.000 0.023.60.996 0.03 (AQL)5.40.951  = 1 – 0.951 = 0.049 0.047.20.810 0.059.00.587 0.0610.80.363 0.0712.60.194 0.08 (LTPD)14.40.092  = 0.092 0.0916.20.039 0.1018.00.015

31. G – 31 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Solved Problem c.According to the table, the probability of accepting a lot with 5 percent defectives is 0.587. Therefore, the probability that a lot with 5 percent defects will be rejected is 0.413, or 1 – 0.587  = 0.092  = 0.049 1.0 — 0.9 — 0.8 — 0.7 — 0.6 — 0.5 — 0.4 — 0.3 — 0.2 — 0.1 — 0 — |||||||||| 12345678910 Proportion defective (hundredths)( p ) Probability of acceptance ( P a ) (AQL)(LTPD) 0.996 0.951 0.810 0.587 0.363 0.194 0.092 0.039 0.015 1.000 Figure G.7

32. G – 32 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.