15 Lot-by-Lot Acceptance Sampling for Attributes Chapter 15 Dr. Shokri Selim, KFUPM

The acceptance sampling problem An old method used in the 1930’s and 40’s. The purpose of AS is to inspect received lots of products and decide whether to accept or reject the lot; lot disposition, or lot sentencing. Accepted lots are put into production Rejected lots may be returned to supplier or subjected to other lot-disposition action Sampling methods may also be used during various stages of production Chapter 15 Dr. Shokri Selim, KFUPM

The purpose of AS is sentencing lots and not estimate lot quality. Note that: The purpose of AS is sentencing lots and not estimate lot quality. 2. It could happen that, lots of same quality be sentenced differently based on the sample. 3. AS is an audit tool 4. Control charts are used to signal departures from quality. Chapter 15 Dr. Shokri Selim, KFUPM

Approaches to lot sentencing Accept without any inspection Supplier process is very good and defectives are rare or there is no economic justification to look for defectives 100% inspection Use if defective components can cause high failure cost or supplier process does not meet specifications Acceptance sampling Chapter 15 Dr. Shokri Selim, KFUPM

When is AS used? Chapter 15 Dr. Shokri Selim, KFUPM

Advantages of sampling Chapter 15 Dr. Shokri Selim, KFUPM

Disadvantages of sampling There is a risk of accepting “bad” lots and rejecting “good” lots. Less information is generated about the product or about the process that manufactured it. Acceptance sampling requires planning and documentation where as 100% inspection does not Chapter 15 Dr. Shokri Selim, KFUPM

Types of sampling plans One major classification is by data type, variables and attributes Another is based on the number of samples required for a decision. Chapter 15 Dr. Shokri Selim, KFUPM

Types based on number of samples Single-sampling plans Select a sample of size n. If the number of defectives ≤ c, accept lot, else, reject the lot. Double-sampling plans Select a sample of size n, then depending on the number of defective accept the lot reject the lot take a second sample and decide on both samples Multiple-sampling plans similar to double but with more than 2 samples Sequential-sampling plans units are selected one at a time and decision to accept or reject or continue is made Chapter 15 Dr. Shokri Selim, KFUPM

Factors to consider include: Administrative efficiency Single-, double-, multiple-, and sequential sampling plans can be designed to produce equivalent results. A lot of the some quality level has the same probability of being accepted by these plans. Factors to consider include: Administrative efficiency Type of information produced by the plan Average amount of inspection required by plan Impact of the procedure on manufacturing flow Chapter 15 Dr. Shokri Selim, KFUPM

Lot formation There are a number of important considerations in forming lots for inspection, including: Lots should be homogeneous. Same machine, same raw material, same operator Larger lots are preferred over smaller ones. More economic Lots should be conformable to materials-handling systems used in both supplier and consumer facilities. Lot packaging minimizes risk of damage Selection of sample is easier Chapter 15 Dr. Shokri Selim, KFUPM

Random sampling Assign a number to each item and select n random numbers to form a sample or Randomly select the length, depth and width in the container or Stratify the lot into layers, then cubes. Chapter 15 Dr. Shokri Selim, KFUPM

Guidelines for using acceptance sampling The selection of a sampling plan depends on the objective and the history of the supplier. Non-static nature of AS plans If supplier is known for quality, start with sampling plan for attributes. If quality is proven, may use skip-a-lot policy. If capability is high may stop sampling. If supplier quality is not known, use an attribute plan. If quality is good may use a variable plan, and help them in SPC Companies start with AS and shift to SPC as the quality improves. Chapter 15 Dr. Shokri Selim, KFUPM

Single sampling plans for attributes Definition of a single sampling plan The plan is defined by the sample size n and the acceptance number c. If the number of defectives, d ≤ c accept the lot, else reject the lot. Chapter 15 Dr. Shokri Selim, KFUPM

Types of Lot Size Type A: Lot size is finite Type B: Lot size in infinite Chapter 15 Dr. Shokri Selim, KFUPM

Type B OC curves The OC curve gives the probability of accepting the lot given the lot fraction defective Main assumption: lot size is very large Let p = probability a unit is defective d = number of defectives in a sample of size n Probability of accepting a lot: Chapter 15 Dr. Shokri Selim, KFUPM

Sample OC curve Chapter 15 Dr. Shokri Selim, KFUPM

The ideal OC curve Reject lot if p > 0.025 Can we choose n and c that give similar OCC? Chapter 15 Dr. Shokri Selim, KFUPM

Effect of n Chapter 15 Dr. Shokri Selim, KFUPM

Effect of C Chapter 15 Dr. Shokri Selim, KFUPM

Effect of n and c on OC curves Chapter 15

Type-A and Type-B OC curves If the lot size is vey large, we use the binomial distribution to model the P(d ≤ c) If the lot size is finite we use the hypergeometric distribution. Chapter 15 Dr. Shokri Selim, KFUPM

Constructing Type A OCC Let N be the lot size D be the number of defectives in the lot N be the sample size C be the maximum number of defectives allowed DEMO Chapter 15 Dr. Shokri Selim, KFUPM

Relation between Types A and B If N is large (N ≥ 10n ) both graphs are close. DEMO Type A and Type B OC curves Chapter 15 Dr. Shokri Selim, KFUPM

Behavior of Type B OC curve for c = 0 Plans with c = 0. OC curve far from the ideal OCC. Pa falls sharply as p increases Chapter 15 Dr. Shokri Selim, KFUPM

Behavior of Type A OC curve for c = 0 Plans with c = 0 and N = 10n OC curve far from the ideal OCC. Pa falls sharply as p increases N = 10n but the OCCs behave differently. p = D/N Chapter 15 Dr. Shokri Selim, KFUPM

Specific points on the OC curve Acceptable quality level, AQL = the least quality level for the supplier’s process that a consumer would consider to be acceptable. The consumer assigns high acceptance probability to it. Lot tolerance percent defective, LTPD = Rejectable quality level, RQL = the least quality level, that the consumer is willing to accept with small acceptance probability. We can design a plan that almost satisfies both conditions. Chapter 15 Dr. Shokri Selim, KFUPM

Designing a single-sampling plan with a specified OC curve p1 = AQL p2 = RQL DEMO Chapter 15 Dr. Shokri Selim, KFUPM

N ≥ 10 n Binomial monograph Chapter 15 Dr. Shokri Selim, KFUPM

N ≥ 10 n Binomial monograph Chapter 15 Dr. Shokri Selim, KFUPM

Chapter 15

Rectifying inspection Incoming lots Fraction Defectives P0 Outgoing lots Fraction Defectives P1 < P0 Inspection activity Rejected lot has 0 defectives Accepted lot has P0 fraction defectives No of defectives = P0(N-n) Average number of defective units = pa (N – n)p0 + ( 1 – pa)x0 Average outgoing quality, AOQ = pa(N – n)p0/N Chapter 15 Dr. Shokri Selim, KFUPM

Example N = 1000, n =100, c = 3, p = 0.01 pa = 0.9816 Average outgoing quality = AOQ = pa(N – n)p/N = 0.009 Chapter 15 Dr. Shokri Selim, KFUPM

AOQL is the maximum point on the curve AOQ for N very large AOQ = pa(N – n)p/N = pa(1 – n/N)p ≈ pa p AOQL is the maximum point on the curve AOQ limit = worst AQL For the example; AOQL = 0.019417 Chapter 15 Dr. Shokri Selim, KFUPM

Average total inspection Number of inspected items: If d ≤ c, n units will be inspected. If d > c, N units will be inspected. ATI = pa n + ( 1 – pa )*N = n + ( 1 – pa )*( N – n) If N= 1000, n= 100, c = 3, p= 0.01 pa = 0.9816 and ATI = 116.56 Chapter 15 Dr. Shokri Selim, KFUPM

ATI = n + ( 1- pa )*( N – n) Does the behavior of the graph makes sense? ATI for sampling plan: n = 89, c = 2 for lot sizes of 1000, 5000, 10,000 Chapter 15 Dr. Shokri Selim, KFUPM

Selection of n and c based on AOQL and ATI If AOQL is specified, the solution is not unique We find n and c that minimize ATI given some AOQL value Chapter 15 Dr. Shokri Selim, KFUPM

Double, multiple, and sequential sampling Double Sampling Plans n1 = sample size on the first sample c1 = acceptance number of the first sample n2 = sample size on the second sample c2 = acceptance number of the second sample If d1 in the first sample is ≤ c1 accept the lot If d1 in the first sample is > c2 reject the lot Otherwise take 2nd sample. If d1 + d2 ≤ c2 accept the lot Otherwise, reject the lot. Chapter 15 Dr. Shokri Selim, KFUPM

Chapter 15 Dr. Shokri Selim, KFUPM

Chapter 15 Dr. Shokri Selim, KFUPM

Advantage of double sampling over single sampling plans HW: read the advantages and disadvantages Chapter 15 Dr. Shokri Selim, KFUPM

Constructing the OC curve for double sampling plans Chapter 15 Dr. Shokri Selim, KFUPM

Supplementary OC curve Primary OC curve Supplementary OC curve Chapter 15 Dr. Shokri Selim, KFUPM

Example n1 = 50, C1 = 3, n2 = 150, C2 = 6 Chapter 15 Dr. Shokri Selim, KFUPM

The average sample number It is the average number of inspected units ASN = n1 + n2 Pr( c1 < d1 ≤ c2 ) What is the assumption here ? The assumption We complete the inspection of the second sample even after the total number of defectives has exceeded c2 Chapter 15 Dr. Shokri Selim, KFUPM

Curtailment It is stoppage of sampling when the rejection condition is satisfied Do not do it with the first sample Why ? To be able to estimate the fraction defective. However, can do on the second sample. Chapter 15 Dr. Shokri Selim, KFUPM

Rectifying Inspection with double sampling If all defective items are discovered, either in sampling or 100% inspection, and are replaced with good ones: Chapter 15 Dr. Shokri Selim, KFUPM

Multiple Sampling Plans Example: Cumulative sample size Acceptance number Rejection 20 3 40 1 4 60 5 80 7 100 8 9 Chapter 15 Dr. Shokri Selim, KFUPM

At the completion of stage i: Cumulative sample size Acceptance number Rejection 20 3 40 1 4 60 5 80 7 100 8 9 At the completion of stage i: If d1 + d2 + … + di ≤ acceptance number → Accept lot If d1 + d2 + … + di ≥ rejection number → Reject lot Otherwise take the next sample Usually, the first sample is inspected 100%. Usually, subsequent samples are subject to curtailment Chapter 15 Dr. Shokri Selim, KFUPM

Take 2nd sample, what value of d2 will result in accepting lot? Cumulative sample size Acceptance number Rejection 20 3 40 1 4 60 5 80 7 100 8 9 Suppose d1 = 1 Take 2nd sample, what value of d2 will result in accepting lot? what value of d2 will result in rejecting lot? suppose d2 = 1 Take 3nd sample, what value of d3 will result in accepting lot? what value of d3 will result in rejecting lot? Chapter 15 Dr. Shokri Selim, KFUPM

The values of the d’s that lead to lot acceptance Cumulative sample size Acceptance number Rejection 20 3 40 1 4 60 5 80 7 100 8 9 3 1 2 d1 4   d2 5 d3 7 d4 8 9 d5 Chapter 15 Dr. Shokri Selim, KFUPM

Item by Item Sequential Sampling Plans The x-axis shows cumulative number of items inspected. The y-axis shows cumulative number of defectives If the point is between the two lines, take one more item If the point falls on or above the top line, reject lot If the point falls on or below the bottom line, accept lot Chapter 15 Dr. Shokri Selim, KFUPM

The acceptance and rejection lines Given p1 and 1-α, p2 and β. Suppose p1 = 0.01, α = 0.05, p2 = 0.06, β = 0.1 Chapter 15 Dr. Shokri Selim, KFUPM

Truncation of sampling Sequential sampling could be truncated if the number of units inspected reaches three times the sample size of the equivalent single sample plan For the previous example, the equivalent single sample plan has n = 89. If sentencing does not takes place after 267 units, stop sampling and accept lot Chapter 15 Dr. Shokri Selim, KFUPM

Homework Project 1: Suppose the fraction defective is 0.05, find the ASN for double sampling plan defined by (n1 = 50, c1 = 1, n2 = 50, c2 = 2), in case of curtailment at the second sample Project 2: A single sample plan has n = 50; find c that will result in least ATI and AOQL ≤ 0.01 Project 3: Consider a double sampling plan with n1=50, c1=1, c2 =2. Find the smallest n2 that will result in AQL = 0.01 with α ≤ 0.05, and RQL = 0.1 with β ≤ 0.2. Chapter 15 Chapter 15 Dr. Shokri Selim, KFUPM Dr. Shokri Selim, KFUPM 55