1. L ECTURE 10 C – M IDTERM 2 R EVIEW B US 385

2. A VERAGE O UTGOING Q UALITY (AOQ) Underlying Assumptions Acceptance sampling has reduced the proportion of defective items accepted Rejected lots are improved (e.g. through 100% inspection and the replacement of defects) Definition – AOQ is the average percentage defectives in outgoing goods assuming that rejected lots of incoming goods are 100-percent inspected and defective items in those lots are replaced with good items AOQ = P ac x p P ac = Probability of accepting lot p= Fraction defective

3. A VERAGE O UTGOING Q UALITY (AOQ) ( CONT ’ D ) 3 AOQ (Outgoing proportion defective) (Incoming proportion defective).00.05.06.04.08 0.00.200.40 Approximate AOQL=.082 0.300.10

4. A VERAGE O UTGOING Q UALITY L IMIT (AOQL) Definition – the worst quality of outgoing items This is information that is crucial to buyers so that they know the quality range of product that that they are receiving! AOQL = 0.4(1.25 c + 1) n n – sample size c - number of defects in a specific sample size

5. A VERAGE O UTGOING Q UALITY S AMPLE Q UESTION Construct an AOQ curve for N = 500, n = 10 & c = 1 for known defect levels that range in 0.05 increments from.05 to.40 Determine the AOQ Limit (AOQL)

6. AOQ S AMPLE Q UESTION A manufacturer of electronic devices makes a component in large. After completing process A, the batch is checked using an acceptance sampling plan with the following characteristics: Sample size such that probability of acceptance is 0.80, and average outgoing quality level is 0.05. Following inspection, "rectification" occurs (ie., all defective items discovered during sampling are replaced by items known to be good) and the batch continues to process B. At the end of process B, an acceptance sampling plan with n=4 and c=1 is used to determine whether to pass the batch to the next stage. What is the average defective rate of items produced by process A? (2 marks) Assuming that process B does not cause any new defects in the components, what is the AOQ from process B? (4 marks)

7. C UMULATIVE B INOMIAL T ABLE

8. 8 D ETERMINE A CCEPTANCE S AMPLING P LAN Zypercom, a manufacturer of video interfaces, purchases printed wiring boards from an outside vender, Procard. Procard has set an acceptable quality level of 1% and accepts a 5% risk of rejecting lots at or below this level. Zypercom considers lots with 3% defectives to be unacceptable and will assume a 10% risk of accepting a defective lot. Develop a sampling plan for Zypercom. Zypercom, a manufacturer of video interfaces, purchases printed wiring boards from an outside vender, Procard. Procard has set an acceptable quality level of 1% and accepts a 5% risk of rejecting lots at or below this level. Zypercom considers lots with 3% defectives to be unacceptable and will assume a 10% risk of accepting a defective lot. Develop a sampling plan for Zypercom.

9. 9 S TEP 1 - D ETERMINE “ C ” First divide LTPD by AQL. Then find the value for “c” by selecting the value in the Table below from “LTPD/AQL” column that is equal to or just greater than the ratio above. cLTPD/AQLn AQLcLTPD/AQLn AQL 044.8900.05253.5492.613 110.9460.35563.2063.286 26.5090.81872.9573.981 34.8901.36682.7684.695 44.0571.97092.6185.426 So, c = 6.

10. 10 D ETERMINE S AMPLE S IZE c = 6, from Table n (AQL) = 3.286, from Table AQL =.01, given in problem c = 6, from Table n (AQL) = 3.286, from Table AQL =.01, given in problem Sampling Plan Statement: Take a random sample of 329 units from a lot. Reject the lot if more than 6 units are defective. Sampling Plan Statement: Take a random sample of 329 units from a lot. Reject the lot if more than 6 units are defective. Now given the information below, compute the sample size in units to generate your sampling plan n = 3.286/AQL= 3.286/.01 = 328.6, or 329 (always round up)

11. Given an Acceptance Plan with: Sample size n = 25; Acceptance number c = 2 Determine: 1.AQL 2. LTPD Calculating AQL and LTPD 11

12. 12 E XAMPLE : C ALCULATING AQL AND LTPD ( CONT ’ D ) cLTPD/AQLn AQLcLTPD/AQLn AQL 044.8900.05253.5492.613 110.9460.3553.2063.286 26.5090.81872.9573.981 34.8901.36682.7684.695 44.0571.97092.6185.426 6 Given Acceptance Plan: Sample size n =25 Acceptance number c = 2 Find AQL and LTPD Solution: From Table nAQL =25(AQL)=0.818 AQL = 0.818/25=0.0327 LTPD/AQL=LTPD/0.0327=6.509 LTPD = 6.509X0.0327= 0.213

13. S AMPLE P LAN C RITERIA Single Sampling Plan Double Sampling Plan Multiple Sampling Plan

14. C ONTROL L IMIT R EVIEW Group the control limit methods into families Differentiate between the families by associating them with the appropriate given set of data Do preliminary calculations as required Determine control limits Compare finished product dataset to your control limits Comment

15. F AMILY OF C ONTROL C HARTS Sample Mean (Mean Method) & Sample Mean (Range Method) Sample Range Individual Unit & Moving Range p-charts (sample proportion of defects) c-charts (sample number of defects per unit produced)

16. P- CHART & C - CHART Mr. Bubbles, a quality control inspector at the Elora Bicycle Factory is monitoring the quality of their new Super Cycle by inspecting the bicycles for defects. He is taking random samples of 100 bicycles per sample and has taken 10 samples in total. The results of his inspections are as follows: SampleNumber of defective parts 1 12 2 14 3 16 4 17 5 4 6 12 7 7 8 25 9 12 10 21

17. P- CHART & C - CHART The sales & service department of a large chain of sporting goods stores decided that it wanted to track the accuracy of the refund forms for returned goods. Complete forms from a particular store were sampled to check the performance quality for the department. To establish some sort of standard, a total of 30 forms were randomly collected and the number of errors in each was counted, with the following results: Form #Number of Defects per form Form #Number of Defects per Form 1 6161 2 11174 3 4184 4 9192 5 8205 6 3219 7 7222 8 8233 9 10244 10 5256 11 5265 12 7273 13 4281 14 8292 15 2302

18. O PERATING C HARACTERISTICS C URVE C ONSUMER AND P RODUCER R ISK

19. C APACITY P LANNING This material follows that provided for the establishment of control limits – so you can put it in that context Few Definitions Design Specification – a range of acceptable values established by engineering design or customer requirements Process Variability – the actual variability in the process itself Process Capability – the ability of a process to meet the design specification Capability analysis is simply measuring this ability Purpose – to determine if the process output falls within the design specification. If the output is within its design specification, then the process is deemed to be ‘capable’

20. C APACITY P LANNING ( CONT ’ D ) Options for dealing with processes that are n0t capable of producing within the required specification limits: 1. Refine (or redesign) the process to increase controllability and reduce variability 2. Use an alternate process 3. Retain the current system as is and implement 100% inspection 4. Relax design specification without reducing customer satisfaction

21. C APACITY P LANNING ( CONT ’ D ) Process variability is typically measured as + 3 standard deviations from the process mean To determine whether or not a process is capable, compare + 3 standard deviations value (in the process itself) to the acceptable range of variation in the design (specification) of the product An alternate method is to compare 6 standard deviations from either the upper or lower (product) control limit specification So lets recap: We have upper and lower design specifications within which a product must fall The specification width is 6σ – 3σ down from the upper limit and 3σ up from the lower limit If you are given the standard deviations of each machine (i.e. how accurate the process is) then you can determine whether or not they are capable of meeting design specifications by comparing their variability to the design specification spread

22. C APACITY P LANNING ( CONT ’ D ) C P (process capability) = Design Specification Width Process Width = Upper Design Spec – Lower Design Spec 6σ (of the machine – ‘process’)

23. C APACITY P LANNING S AMPLE Q UESTION C P is the capability ratio Therefore the greater the C P (capability ratio) the greater the probability that the output from that machine will fall within the design specification If the previous formula holds, then you obviously need a Cp ratio of > 1.00 for a process to be capable Sample Question – A manager has the option of using one or more of three different machines. The machine capabilities are summarized in the adjoining table. Which machine(s) should he use if the design specification is 101.00 – 101.60? MachineStandard Deviation A.10 B.08 C.13

24. C APACITY P LANNING - C PK Cpk – is used when the process is not centered between the design specification limits Method of Calculation: Calculate the difference between the upper and lower and the process mean and dividing the value by 3σ Upper Design Spec – Process Mean 3σ Process Mean – Lower Design Spec 3σ Cpk is equal to the smaller number

25. C PK – S AMPLE Q UESTION The process’ output has a mean of 9.20 kg and a standard deviation of 0.30 kg. The lower design specification is 8.00 kg and the upper design specification is 10.00 kg. Calculate the Cpk