MP4001 / MP4001P

QUALITY ASSURANCE AND MANAGEMENT

STATISTICAL QUALITY CONTROL (SQC)

Dr Wu ZhangMAE NTUMAE, NTU

Office: N3 2 02 14 Tel: 67904445 email:Office: N3.2-02-14, Tel: 67904445, email: mzwu@ntu.edu.sg

TEXTMontgomery, D. C., (2009), Introduction to Statistical

1

g y, , ( ),Quality Control, John Wiley & Sons.

Statistical Quality Control (SQC) concerns the use ofStatistical Quality Control (SQC) concerns the use ofstatistical methods and other problem-solving techniques toimprove the quality of the products in the manufacturingp q y f p f gindustry and service section.

CONTENTS1 QC seven tools

2 Control charts

3 Process capability

4 Gage repeatability and reproducibility

25 Acceptance sampling plan

1 QC Seven Tools

1.1 Histogram (Bar Chart) 1-2

1.2 Check Sheet 1-12

1.3 Pareto Chart 1-14

1.4 Cause and Effect Diagram 1-16

1.5 Defect Concentration Diagram 1-18

31.6 Scatter Diagram 1-20

2 Control Charts

2.1 Introduction 2-1

2 2 Control Charts for Variables 2 252.2 Control Charts for Variables 2-25

2.2.1 & R Charts 2-26x2.2.2 Rational Subgroups 2-69

2 2 3 Cusum (Cumulative Sum) Chart 2-722.2.3 Cusum (Cumulative Sum) Chart 2-72

2.2.4 Runs Rules 2-88

2.3 Control Charts for Attribute 2-94

2 3 1 np Chart 2-942.3.1 np Chart 2 94

2.3.2 c Chart 2-115

4

3 Process Capability

3.1 Process Capability Ratio Cp 3-11

3 2 P C bili R i C 3 343.2 Process Capability Ratio Cpk 3-34

3 3 Process Capabilit Ratio C 3 473.3 Process Capability Ratio Ckm 3-47

4 Gage Repeatability and Reproducibility4 Gage Repeatability and Reproducibility

4.1 Gage (Measurement) Error 4-14.1 Gage (Measurement) Error 4 1

4.2 The Gage Capability Cgage 4-195

g p y gage

5 Acceptance Sampling Plan

5.1 Single Sampling Plan 5-7

5.2 Double Sampling Plan 5-29

5.3 Rectifying Inspection 5-51

5.4 Sampling Plan Standard 5-65

6

1 QC (Quality Control) SEVEN TOOLS

“QC seven tools” is a powerful collection of problemQC seven tools is a powerful collection of problem-solving tools useful in achieving product qualityimprovement. Most of them (except the control chart) areimprovement. Most of them (except the control chart) arevery easy to use, but quite effective for many applications.

(1) Histogram(1) Histogram(2) Check sheet(3) Pareto chart(3) Pareto chart(4) Cause and effect diagram(5) Defect concentration diagram(5) Defect concentration diagram(6) Scatter diagram(7) Control chart

1

( )

1.1 Histogram (Bar Chart)g ( )

It is a very useful graphical technique for summarizing and presenting data.

Table 1-1 presents the observations of the inside diameter Φ for the piston-ring (individual observations

diff t f h th )are different from each other).

Number of samples: 25Number of samples: 25

Number of observations per sample (sample size): 5Number of observations per sample (sample size): 5

Total number of observations: 25 × 5 = 1252

o a u be o obse va o s: 5 5 5

Table 1-1 Forged piston-ring inside diameter (mm)g p g ( )

3

Figure 1-1 is the corresponding histogram, which consists of 13 bars of equal width, between the minimum (73.967) and maximum (74.030) values of the diameter.

Usually, the number of bars in a histogram is between 4 and 20.

The height of a bar is equal to the number of the diameter values inside the interval under that bar. For example,values inside the interval under that bar. For example,

1st bar: 73.965 73.970, one diameter value2nd bar:73 970 73 975 zero diameter value

2nd bar:73.970 73.975, zero diameter value

4th bar: 73.980 73.985, eight diameter values

Th hi h h b h lik l h Φ l ill f ll

The higher the bar, the more likely that a Φ value will fall in the corresponding interval.

4

5Figure 1-1 Histogram for piston-ring diameter data (F2-4 p39)

The diameter Φ is a random variable. Three properties of the probability distribution of Φ is critical to the quality of the product:

1.Shape of the distribution (normal or uniform ?)

2. Location (or average, mean value (μ))

3. Scatter (or spread, variability, standard deviation (σ))

Such information cannot be easily perceived from Table 1-1, but can be clearly illustrated by the histogram in Fig , y y g g1-1. Because the histogram is similar to the probability distribution curve.

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1. The histogram shows that the distribution of ring diameter i hl t i d i d l ( i l i l )is roughly symmetric and unimodal (a single maximum value), similar to a normal distribution.

2. The average is close to 74 mm.

3. The variability is quite high, as the min. diameter value is y q gas small as 73.967 mm, and the max. value is as large as 74.030 mm. The range is (74.030 – 73.967) = 0.063.

From the viewpoint of product quality, the small the variability (or the range), the better the quality is.y ( g ), q yThis information is very useful for determining whether the manufactured diameter can satisfy the design specificationsmanufactured diameter can satisfy the design specifications and whether a process improvement program should be carried out.

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Example 1-1p

The specification for the piston-ring diameter is: 74 ± 0.020 mm

Upper specification limit: USL = 74 + 0.020 = 74.020

Lower specification limit: LSL = 74 - 0.020 = 73.980

From the histogram in Fig 1-2, we can easily find that

One piston-ring is defective due to excessively small Φ.

Three piston-rings are defective due to excessively large Φ.

8The defective rate is (1 + 3) / 125 = 3.2%

9Figure 1-2 Histogram with specification limits

If we can improve the process by

1 Training operators for better skill, or

2 Purchasing machines with higher precision

The variation of Φ will be reduced (Fig 1-3), that is, the observed diameter values cluster more closely to the ynominal value of 74mm.

As a result, the defective rate may be reduced to zero.

10

11Figure 1-3 Histogram for improved process

1.2 Check SheetCheck sheet is also very useful in collecting either historical or current operating data about a process. Figure 1-4 is a check h t f i ti ti th i t f d f t th t dsheet for investigating the various types of defects that occurred

on a tank.

The time-oriented summary at the bottom is particularly valuable in looking for trends or other patterns of defects.

For example, extraordinarily more defects are found in June and July of 1988 (also in Jan and Feb of 1989). People are alarmed to conduct investigation, and find out and remove the roots of the problems in these months.

Also, the category-oriented summary at the right-hand-side can tell which types of defects occur more frequently, or more

12dominant.

13Figure 1-4 A check sheet to record defects on a tank (f4-15 p151)

1.3 Pareto Chart

Pareto chart is a graph of the frequency distribution ofPareto chart is a graph of the frequency distribution ofdefects arranged by category. The defect type occurringmost frequently is plotted at the most left-hand sidemost frequently is plotted at the most left-hand side.

Through this chart the user can quickly and visuallyid tif th t f tl i t f d f tidentify the most frequently occurring types of defects.Thus the causes of these defect types will be identified andattacked firstattacked first.

In Figure 1-5, 129 out of the 166 defects (78%) areattributable to the five most dominant types.

(36 + 34 + 29 + 17 + 13) / 166 129 / 166 78%(36 + 34 + 29 + 17 + 13) / 166 = 129 / 166 = 78%

So if the causes of these 5 types of defects can be found14

So, if the causes of these 5 types of defects can be foundand removed, majority of the defects are eliminated.

15Figure 1-5 Pareto chart of the tank defect data (f4-16 p153)

1.4 Cause and Effect Diagram

Th d ff di i i hThe cause and effect diagram is a systematic approach to discover the potential causes of the defects.

Figure 1-6 shows the cause and effect diagram for the tank defect problem. The diagram consists of the defect box, the center line and multi-level potential cause categories.

Usually, a team is formed to build the cause and effect y,diagram through brainstorming. The team members include engineers and experts from all of the relevant departments.

After the cause and effect diagram has been built, we may have to rank the causes in order to identify those that arehave to rank the causes in order to identify those that are most likely to impact the problem, and then take the corrective actions correspondingly.

16

p g y

17Figure 1-6 Cause and effect diagram for the tank defect problem (f4-18 p156)

1.5 Defect Concentration Diagram

A defect concentration diagram is the picture of a unit, showingall relevant views. Then the various types of defects are drawn onthe picture, and the diagram is analyzed to determine thepotential causes of the defects from the location of the defects.

Figure 1-7 presents a defect concentration diagram of arefrigerator. Surface-finish defects are identified by the darkshaded areas on the refrigerator. It is found that all the defectsappear at the middle of the refrigerator. A follow-up investigationdiscloses that some problem during the transportation of the unitis responsible for the defects. Because, when moving the unit, ab lt i d d th iddl f th f i t d thi b lt ibelt is used around the middle of the refrigerator, and this belt ismade of abrasive material!

18Using other belts with proper materials will solve this problem.

19Figure 1-7 Surface finish defects on a refrigerator (f4-19 p156)

1.6 Scatter Diagram

The scatter diagram is a useful plot for identifying a potential relationship between two variables Y and X. Usually, X indicates the cause of a quality problem; and Y is the quality result.

Data are collected in pairs on the two variables,say (yi, xi), for i p , y (yi, i),= 1, 2, …, n. Then, each yi is plotted against the corresponding xiin the scatter diagram.

Figure 1-8 shows a scatter diagram relating defective rate Yagainst the temperature X in a process. The scatter diagramagainst the temperature X in a process. The scatter diagram indicates a strong positive correlation between Y and X; that is, as X is increased, Y also increases.,

In a quality improvement program, X should be decreased as much as possible in order to cut down the defective rate

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much as possible in order to cut down the defective rate.

21Figure 1-8 A scatter diagram (f4-21 p158)

2 CONTROL CHARTS

2.1 Introduction

Natural variability

Any production process, regardless of how well designed orhow carefully maintained, will generate a certain amount ofi h i i i hi f i biliinherent or natural variability. This type of variability maybe reduced, but cannot be completely eliminated.

The natural variability is the cumulative effect of many small,unknown and essentially unavoidable causes i e the chanceunknown, and essentially unavoidable causes, i.e., the chancecauses. Examples: ambient temperature, humidity, dust,sunshine, vibration of a nearby machine, nearby high way,

1

sunshine, vibration of a nearby machine, nearby high way,oscillation of electrical network.

A process that is operating with only the unavoidable natural variability is said to be in control. An in-control status isvariability is said to be in control. An in control status is considered as a normal and acceptable status.

It means that when a process is in control, there is still variability, but very small.

It is inappropriate to interfere or adjust an in-control process with only natural variability. Because it may lead to over-correction and introduce even larger variability and, thus, make thithings worse.

We should let the in control process continue to run2

We should let the in-control process continue to run.

System variability

Another kind of variability the system variability is generallyAnother kind of variability, the system variability, is generally large in magnitude, and is incurred by the assignable causes

It usually arises from three sources: (1) Operator errors: Using incorrect cutter, setting wrong speed.

(2) Defective raw material: erroneous or low quality material

(3) I dj f hi Adj i i l(3) Improper adjustment for machines: Adjusting an in-controlprocess when there is only natural variability. It will turn the small natural variability to large system variabilitynatural variability to large system variability.

System variability represents an abnormal and unacceptable out-of-control status. System variability should be signaled, identified and removed as soon as possible. Then, the process goes back to

i l i h l i l i bili3

an in-control status with only minor natural variability.

f ( h f i i iμ mean of x (or the average of x in engineering terms). x is a random variable or a quality characteristic (e g a dimension) Each randomcharacteristic (e.g. a dimension). Each random variable has a probability distribution as in Fig 2-1.

σ standard deviation of x (approximately equal to R/6, where R is the range of the probability distribution

f R/6) i f th i bilit fof x, σ = R/6). σ is a measure of the variability of x.

μ0 the value of μ when the process is in control (usually, the nominal or target value of x).

σ0 the value of σ when the process is in control (σ0 = R0 /6, where R0 is the value of R when the process

4

0 , 0 pis in control).

5Figure 2-1 In-control and out-of-control processes (f5-1 p180)

Wh d h i bili f i ll i iWhen μ = μ0 and σ = σ0, the variability of x is small, it is the natural variability. The process is considered in controlcontrol.

When μ ≠ μ and/or σ ≠ σ the variability of x is large itWhen μ ≠ μ0 and/or σ ≠ σ0, the variability of x is large, it is the system variability. The process is considered out of control.control.

Usually, a processes will operate in the in-control status y p pfor most or a long time period.

Eventually, however, assignable causes will occur, seemingly at random, resulting in a "shift" to an out-of-

l ( ≠ d/ ≠ b h)6

control status (μ ≠ μ0 and/or σ ≠ σ0, or both).

7

In Figure 2-2, a machine tool is cutting a shaft.

Until time t1, the process is in control. There is only small natural variability. As a result, both the mean and standard deviation of y ,the process are at in-control levels, μ0 and σ0.

After time t the process is out of controlAfter time t1, the process is out of control.

At time t1, assignable cause 1 occurs (e.g. the breakage of th tt f th hi t l) it ff t i t hift ththe cutter of the machine tool), its effect is to shift the process mean to a new valueμ1 > μ0.

At time t2, assignable cause 2 occurs, resulting in a process standard deviation change, σ1> σ0 (even though,μ = μ0).

At time t3, assignable cause 3 presents, resulting in μ2 < μ0and σ1 > σ0.

8

1 0

9Figure 2-2 Chance and assignable causes of variation (f4-1 p131)

It is desired to detect the out-of-control status at the firsttime in order to minimize the number of defectives. Here,,the control chart comes to play.

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A control chart has the following three usages:

1. On-line process control. Distinguishing between the in-control and out-of-control status.

(1)If the process is in control, avoid over-interference,leave the process aloneleave the process alone.

(2)If the process is out of control, a prompt signalis given and the process is shut down immediatelyis given and the process is shut down immediately before many defects are manufactured. Then, theinvestigation and corrective action may be undertakeng yto resume the process to an in-control status.

2.Estimating the parameters (e.g., μ0 and σ0) of the process.

3 D i i h bili11

3.Determining the process capability.

A control chart is a graphical display of the sample points versus the sample number or time.

A sample is a group of n observed values x1, x2, …, xn of a quality characteristic x (e.g., a diameter, weight, velocity).( g , , g , y)

A statistic is a function in terms of the n observed xi (i = 1, 2, …, n)n).

A widely used statistic is the sample mean, denoted by . Sample i th f th l f i l It i

xmean is the average of the n values of xi in a sample. It is a measure of the central tendency of x, or the estimate of the mean value μ of xvalue μ of x.

nxxxxxn

n /21

x (2-1)nxn

xi

i /1

In a control chart each sample point carries the value of the

x (2 1)

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In a control chart, each sample point carries the value of thestatistic (e.g. ) of a corresponding sample.x

Table 2-1 Diameter of a shaft (mm) (n = 5)

SampleNo. x1 x2 x3 x4 x5 x

1 74.030 74.002 74.019 73.992 74.008 74.0102 73 995 73 992 74 001 74 011 74 004 74 0012 73.995 73.992 74.001 74.011 74.004 74.0013 73.988 74.024 74.021 74.005 74.002 74.008. . . . . . .. . . . . . .

A control chart that is used to inspect the sample mean is calledxthe chart. On a chart, each sample point carries or indicatesthe value of a sample.x

x x

Since, , this control chart is actually to monitor the mean μ(central tendency of the process distribution) or to insure μ =

x

13

μ(central tendency of the process distribution), or to insure μ = μ0. 14Figure 2-3 A control chart (f5-3 p188)

Three Elements of a Control Chart

(1)Sample size (n ) the number of observations in a sample

(2)S li i t l (h) h i i l b l(2)Sampling interval (h) the time interval between two samples

(3)Control limits:

Central line (CL): target value of the statistic when a process is in control (e g the statisticprocess is in control (e.g., the statistic of a chart is , and the target of is μ0when a process is in control. So, for x

xx xp ,

chart, CL = μ0).

Upper control limit (UCL) upper limit of the sample pointsUpper control limit (UCL) upper limit of the sample points

Lower control limit (LCL) lower limit of the sample points

15

Basic Operational Rules for a control chart:Basic Operational Rules for a control chart:

(1) If all sample points fall within the control limits LCL( ) p pand UCL, the process is thought in control (there is onlysmall, natural variability), see Fig. 2-3.

(2) If a sample point falls beyond the control limits(below LCL or above UCL), the process is considered asout of control (there is large, system variability), see Fig2 42-4.

It is how a control chart distinguishes between the inIt is how a control chart distinguishes between the in-control and out-of-control status --- depending on the positions of the sample points relative to the control

16

positions of the sample points relative to the control limits LCL and UCL.

17Figure 2-4 An out-of-control process detected by a control chartx

The values of the control limits LCL and UCL are very iti l t th ti h t i ti f th t lcritical to the operating characteristics of the control

chart.

If the control limits LCL and UCL are too tight (Fig 2-5), some sample points may fall beyond the control limitssome sample points may fall beyond the control limits even when the process is in control. It means that the control chart will produce signals indicating process out p g g pof control, when process is actually in control.

These signals are called False alarms. Because they mislead the operators by telling them the process is out of control when the process is in fact in control. False alarms waste time and effort to search for a problem (an

i bl ) th t d t i t F l l l18

assignable cause) that does not exist. False alarms also make the operator lose confidence on the control chart.

19Figure 2-5 Too tight control limits result in false alarms

On the other hand, if the control limits LCL and UCL are ,too wide (Fig 2-6), the sample points will not fall beyond the control limits even when the process is already out of control.

Or in other words, the control chart tells the operators that the process is still in control when the process is in f t t f t l Th f il d l f d t ti tfact out of control. The failure or delay of detecting out-of-control status will result in a great number of defective productsdefective products.

20

21Figure 2-6 Too tight control limits result in false alarms

3-sigma Control Limits3-sigma control limits are the best and most widely used trade-off. They are neither too tight nor too wide.

For a chart CLUCL 3

x

(2-2)xxCLxUCL 3

xxCLxLCL 3 xxxThe control limits and are below or above the central line by 3 times of (see Fig 2-3)

xLCL xUCL

CL xcentral line by 3 times of (see Fig. 2 3).

Since it is for a chart, sigma means the standard deviation of It is noted that of is different from the standard

x x

x

xx

of . It is noted that of is different from the standard deviation σx (or simply σ) of x. The calculation of will be discussed later

xx

xx

22

discussed later.

By using the 3-sigma control limits,

(1) When a process is in control, the possibility of false alarm is very low or almost impossible (see Fig 2-3)is very low, or almost impossible (see Fig 2 3).

(2) When a process is out of control, the capability of detecting ( ) p , p y gthe out-of-control status is quite high. A signal is likely to be produced very soon after the problem occurs (see Fig 2-4).

The general idea of the 3-sigma control limits can be applied to the chart, as well as many different types of control charts. x

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Two main categories of control chartsTwo main categories of control charts

(1) Control charts for variables( )

To control the quality characteristic that can be expressed in terms of a continuous numerical measurementterms of a continuous, numerical measurement (e.g.,dimension, weight, volume).

E l th di t ( 74 0 02 ) f h ftExample: the diameter x (= 74 0.02mm) of a shaft.

(2) Control charts for attributes

To control the quality characteristics that can only take integer valuesg

Example: the number d of defectives in a box of integrated

24circuits.

2.2 Control Charts for Variables

T l h d i i blTwo control charts are used to monitor a variable x:

chart monitoring the mean al e of (ens re )x chart, monitoring the mean value of x (ensure μ = μ0)

R chart monitoring the standard deviation of x

x

R chart, monitoring the standard deviation of x (ensure σ = σ0)

The two &R charts should be used together in order to monitor both μ and σ (ensure both μ = μ0 and σ = σ0).

x

25

μ ( μ μ0 0)

2.2.1 & R chartsx

Suppose that x is a quality characteristic (e.g., a diameter). If x1, x2, … , xn are the observations of a sample of size n, then the x2, , n p ,sample mean is also a random variable.

nxxxx / (2 3) nxxxx n /21 (2-3)

x is the value of individual product. Its probability distribution can be built based on the data under columns 1 to

bl5 on Table 2-2.

i th l f d t Th b bilitx is the average value of n products. The probability distribution of is built based on the data under column 6 on Table 2 2

xx

26

Table 2-2.

Table 2-2 Preliminary samples collected in phase I

27

If x has a normal distribution with mean μx and standard deviation σx, then the sample mean also follows a normal distributed with

x

(2-4)nxxxx / (2-4)

The mean of is equal to the mean of x, but the standard deviation of is smaller than σx of x by a factor of .

x x xx x n

Conventionally, μx and σxare denoted as μ and σ, respectively, for simplicity thusfor simplicity, thus

nxx / (2-5)

For example, if μ = 74, σ = 0.01, and the sample size n = 5. Then

28

Then 004.05/01.0/74 nxx

29Figure 2-7 How the control chart works (f4-4 p135)

According to the Central Limits Theorem, even if the g ,distribution of x is non-normal, the distribution of is approximately normal as long as n 3 or 4. Moreover, eq (2-5) is still valid.

Since n is usually equal to or larger than 3, therefore, when designing a chart, we always assume that the distribution

f i l dl f th di t ib ti fx

of is normal regardless of the distribution of x.

Since engineers are usually very familiar with the normal

x

Since engineers are usually very familiar with the normal distribution, the design of the chart becomes easier.x

30

31Figure 2-8 Central limits theorem

Sample range R

The range of a random variable x is the difference between the maximum and minimum of all the possible values of x.maximum and minimum of all the possible values of x.

Usually, the range is approximately equal to 6σ of the y, g pp y qprobability distribution of x. Or in other words, the range is different from σ only by a constant of six. As a result, the range can be used as a measure of σ of the process distribution of x.

The statistic used by the R chart is the sample range R. The l i ti t f th l l xsample range is an estimate of the real range, as sample mean

is an estimate of μ. The sample range R is equal to the difference between the maximum and the minimum of the n

x

32

difference between the maximum and the minimum of the n values of xi in a sample.

For example, if the five observed diameters in a sample of size 5 p , pare

x1 = 2, x2 = 4, x3 = 1, x4 = -2, x5 = 0

Then, the sample range for this sample is

4 ( 2) 6 (2 6)R = x2 - x4 = 4 – (-2) = 6 (2-6)

The value of R is always equal to or larger than zero

minmaxxxR

The value of R is always equal to or larger than zero.

A R chart is used to monitor the sample range A sample point onA R chart is used to monitor the sample range. A sample point on the R chart carries the value of the sample range R of a sample. The R chart displays these sample points versus the sample

33

p y p p pnumber or time (see Figure 2-9). 34Figure 2-9 A R chart

Since the calculation of sample range R is much simpler than the calculation of sample standard deviation, the R chart is usedthe calculation of sample standard deviation, the R chart is used widely in industry to ensure σ = σ0.

When process is in control: σ = σ0, R = R0 = 6σ0.

When process is out of control: σ = σ1 ≠ σ0, R = R1 ≠ 6σ0.

Conversely,

If R = R0 = 6σ0, it means σ = σ0, the process may be in control (we still have to check if μ = μ0).

If R ≠ R0, it means σ ≠ σ0, the process must be out of control.

35

A l d i d f h &R hActual design procedure of the &R chart

(1) Sample size n decided by the available manpower and

x

(1) Sample size n --- decided by the available manpower and instrument. n is usually equal to 4, 5 or 6, for the &R charts

xcharts.

(2) Sampling interval h --- decided by the production shift(2) Sampling interval h decided by the production shift and rate.

(3) Central line CL and control limits LCL, UCL.

36

Design of the chartx

3-sigma control limits of the chartx

CLUCL 033

CLn

CLUCL

0

00 33

x

xxx

(2-7)

nCLLCL 0

0

0

33 xxx

x

n/, xxNote,

When process is in control, n/, 00 xx

Here, μ0 and σ0 are the mean and standard deviation of the individual observation x when the process is in control.

37

In Eq (2-7), both LCL and UCL depend on μ0 and σ0. However, in practice, μ0 and σ0are usually unknown and must be estimated from the observed data.

Phase I (Build the chart): In this phase, m (at least 20 to 25) preliminary samples are taken (see Table 2-2). Each ) p y p ( )preliminary sample also contains n observations of the quality characteristic x.

The purpose of taking the preliminary samples in phase I is not to monitor the process, but to collect sufficient data, so p , ,that, at the end of phase I, μ0 and σ0 can be estimated from these data, and the control charts can be built.

Phase II (Use the chart): The control charts (built at the end of phase I) now can be used to monitor or control the

38

p )forthcoming production.

39Figure 2-10 Phase I and phase II (m = 25)

Table 2-2 shows the 25 preliminary samples available at the end f h I l Th lit h t i ti i th di tof a phase I example. The quality characteristic x is the diameter

of a piston ring.

We first calculate the sample mean and sample range for each of the preliminary samples as listed at the right most 2 columns inthe preliminary samples as listed at the right most 2 columns in Table 2-2. For example,

010.745

008.74992.73019.74002.74030.741

x

038.0992.73030.745

1 R0047401174001749927399573

019.0992.73011.74

001.745

004.74011.74001.74992.73995.73

2

2

R

x

40

019.0992.73011.742R(2-8)......

Estimation of μ0

The estimate of μ0 is the grand average (average of averages)0

x

mxxxx m 21

0 (2-9)

In Table 2-2,

00174028.1850998.73001.74010.74 x(2-10)

001.742525

x

x is an even more accurate and reliable estimate of μ0,x is an even more accurate and reliable estimate of μ0, compared to each individual .x

41

Estimation of σ0

First, calculate average range (average of sample ranges)RRRR (2 11)

mRRRR m 21 (2-11)

In Table 2-2,

5810035001900380 (2-12)023.0

25581.0

25035.0019.0038.0

R

(2 12)

h h i f i l l d bˆThen, the estimate of σ0 is calculated by0

/ˆ dR (2 13)20 / dR (2-13)

The constant d2 is dependent only on the sample size n and can 42

2 p y pbe found from Table 2-3. For example, if n = 5, d2 = 2.326.

Table 2 3 Factors for constructing variable control chartsTable 2-3 Factors for constructing variable control charts

43

After μ0 and σ0 have been estimated, we can build the &R xcharts at the end of phase I. For the chart, the basic formulae are (see Eq (2-7))

x

RxdRxUCL 200

00

3/3ˆ3ˆ3 x

xCLndnnn

00

200

ˆ

x

x

(2-14)

ndRx

ndRx

nnLCL

2

200

00

3/3ˆ3ˆ3 x

44

The simplified formulae:The simplified formulae:

RAxnd

RxUCL2

3

x

xCLnd

2

x

(2-15)

RAxnd

RxLCL2

2

3

x

The constantnd

A2

3 (2-16)

nd2

A2 also only depends on n and can be directly found in Table 2-3 For example when n = 5 A2 = 0 577Table 2 3. For example, when n 5, A2 0.577

Eq (2-15) makes the computation slightly simpler, but 45

q ( ) p g y p ,loses some underlying information.

A sample point in a chart carries the value of the samplexA sample point in a chart carries the value of the sample mean of a sample.

x

If all sample points fall around the central line CL of the chart (Fig 2-11)

xc a t ( g )

≈ μ0 → μ ≈ μ0x μ0 μ μ0

It means that the mean value or the average of the quality characteristic x is equal or close to μ0. The process is likely to be in control (we still have to check if σ= σ0).

46

Figure 2-11 An in-control process monitored by a chartx47

Figure 2 11 An in control process monitored by a chartx

If a sample point falls above the upper limit UCL of the chart (Fig 2-12 (a))

> μ0 → μ > μ0x

It means that the mean value or the average of x has increased and process is out of controlincreased, and process is out of control.

If a sample point falls below the lower limit LCL of the xIf a sample point falls below the lower limit LCL of the chart (Fig 2-12 (b))

x

< μ0 → μ < μ0x

48It means that the mean value or the average of x has decreased, and process is out of control.

49Figure 2-12 How chart detecting μ ≠ μ0x

Design of the R chart

The R chart plots the points of sample range R versus the sample number or time (see Figure 2-9).( g )

The sample size n and sampling interval h of the R chart are the same as for the chart. The n observations of x in each sample are used to calculate both sample mean and sample range R.x

x

The 3-sigma control limits for the R chart:

RRRR

RCLRCLUCL 33

RRRR

R

RCLLCLRCL

33

(2-17)

50

(1) is the average value of the sample ranges obtained Rfrom the m preliminary samples. is taken as the target CLRof R.

R

(2) The standard deviation σR of R is calculated as

Rdd

dRdddR

330303 ˆ (2-18)ddR

2230303 (2 18)

The constant d is also dependent on the sample size n andThe constant d3 is also dependent on the sample size n and can be found from Table 2-3

(3) If the calculated value of LCLR is smaller than zero, LCLR is pushed up to zero. Because the sample range will

51never be smaller than zero.

The basic formulae for calculating the 3-sigma control limits for the R chart (see Eq (2-17)):

33 333d

RdRRddRRUCL RR

22

RCLdd

R (2-19)

2

3

2

3 333d

RdRRddRRLCL RR

52

The simplified formulae for the R chart:p

RDRdd

dRdRUCLR 4

33 313

RCLdd

R

22

(2-20)

RDRdd

dRdRLCLR 3

2

3

2

3 313

where,

dd (2-21)

2

3

4

2

3

331,31

ddD

ddD

(2-21)

D3 and D4 depend on sample size n and can be found in Table 2-3 For example when n = 5 D3 = 0 D4 = 2 115

53

2 3. For example, when n 5, D3 0, D4 2.115

A sample point in a R chart carries the value of a sample Rrange R.

If all sample points fall around the central line CL of the RIf all sample points fall around the central line CL of the Rchart (Fig 2-13 (a))

R ≈ R0 → 6σ ≈ 6σ0, → σ ≈ σ0

It means that the standard deviation or the spread of the quality characteristic x is equal or close to σ0. The process is likely to be in control (we still have to check if μ = μ0).

54

55Figure 2-13 How R chart works

If a sample point falls above the upper limit UCL of the R chart (Fig 2-13 (b))

R > R0 → 6σ > 6σ0, → σ > σ0

It means that the standard deviation or the spread of x has increased. The process is out of control and the product quality p p q ydeteriorates.

If a sample point falls below the lower limit LCL of the R chartIf a sample point falls below the lower limit LCL of the R chart (Fig 2-10 (c))

R < R0 → 6σ < 6σ0, → σ < σ0

It means that the standard deviation or the spread of x hasIt means that the standard deviation or the spread of x has decreased. It indicates that the product quality is improved. However the process is still considered as out of control

56

However, the process is still considered as out of control because of σ ≠ σ0

Retrospective Check

The control limits of the and R control charts depend on the xvalues of the in-control μ0 and σ0 that are estimated from the data contained in the m preliminary samples in phase I (Fig 2-10) It i li th t d i h I i d t t l10). It implies that, during phase I, in order to accurately estimate μ0 and σ0, the process should be in an in-control status How can we ensure it?status. How can we ensure it?

Customarily the control limits of the and R control chartsxCustomarily, the control limits of the and R control charts determined based on the data of the preliminary samples are treated only as the trial control limits. They allow us to

x

y ycheck whether the process was in control when the mpreliminary samples were taken.

57

If all m points in phase I plot inside the trial control limitsIf all m points in phase I plot inside the trial control limits (Fig. 2-10), then we conclude that the process was in control when the m preliminary samples were taken, and the trial p y p ,control limits can be used as the final control limits in phase II for controlling future production.

If one or more of the m sample points plot out of the trial control limits,

(1) E i i h f th t f t l i t l ki f(1) Examining each of the out-of-control points, looking for the possible problems (assignable causes). If found, the assignable causes must be removed and the out of controlassignable causes must be removed, and the out-of-control points are discarded. The trial control limits are recalculated, using only the remaining sample points

58

using only the remaining sample points.

(2) In some cases, it may not be possible to find out a responsible assignable cause for the out-of-control p gpoints, then we may either

(i) Eliminate the out-of-control points, or

(ii) i h f l i(ii) Retain the out-of-control points

If i l d th l t t fIf m is large, and there are only one or two out-of-control points, either treatment (i) or (ii) will not distort the control limits significantly The controldistort the control limits significantly. The control chart will work satisfactorily.

59

Steps for setting up the & R chartsx

(1) Decide the sample size n (based on the available resources) and the sampling interval h (based on the working shifts).

(2) In phase I, take m preliminary samples, and calculate the sample mean and sample range R for eachthe sample mean and sample range Ri for each sample (eq (2-8)).

ix

(3) Calculate the grand average (eq (2-9)) and the average of sample ranges (eq (2-10)).

xRg p g ( q ( ))

(4) Find out the constants d2 and d3 from Table 2-3 based

60

2 3on sample size n.

(5) Calculate the central line CL and the trial control limits ( )LCL a nd UCL for both the and R charts (eqs (2-14) and (2-19)).

x

(Sometimes, the simplified formulae eqs (2-15) and (2-20) also can be used to calculate the control limits).

(6) Pl t th t h t(6) Plot the two charts.

(7) Carry out retrospective check If it is passed the trial(7) Carry out retrospective check. If it is passed, the trial control limits are launched as the final control limits.

(8) Start phase II, using the two charts to monitor the process.

61

p

EXAMPLE 2-1: Inside diameter x of the piston rings

(1) Decide the sample size n and the sampling interval h.

n = 5, h = 1 hour

(2) In phase I, take 25 preliminary samples (Table 2-2), and calculate the sample mean and sample range R forxcalculate the sample mean and sample range Ri for each sample.

ix

62

(3) Calculate the grand average and the average of x Rsample ranges.

25

x001.74

25028.1850

25998.73001.74010.74

251 i

ixx

5810

25

iR023.0

25581.0

251 i

i

R

(4) Find out the constants d2 and d3 based on sample size n = 5.

From Table 2-3, d2 = 2.326, d3 = 0.864.

63

(5) Calculate the central line CL and the control limits LCL and UCL for both the and R charts.x

chart:x

014.745326.2

023.03001.7432

nd

RxUCLx

98873023.03001743001.74

RLCL

xCLx

988.735326.2

001.742

nd

xLCLx

64

R chartR chart

049.03262

023.0864.03023.033

dRdRUCL

R

0230864033023.0

326.22

RdRCL

d

R

0026.0326.2

023.0864.03023.032

3

d

RdRLCLR

(6) Plot the and R charts (Figure 2-14).x( ) ( g )

(7) Carry out retrospective check.

65 66Figure 2-14 and R chartsx

(8) After the retrospective check has been passed, we can start phase II monitoring the processphase II, monitoring the process.

[1] chart is to monitor the central location of x (ensure μ= μ0);[1] chart is to monitor the central location of x (ensure μ μ0); and the R chart is to monitor the spread of x (ensure σ = σ0). The two charts are running side by side.0) g y

[2] Take a sample of n (five) observations for every one hour (h = 1 hr), calculate the sample mean and sample range R. For example, the five observations for a sample are

x

x1=74.030, x2=74.002, x3=74.019, x4=73.992, x5=74.008,

010.745

008.74992.73019.74002.74030.74 x

67038.0992.73030.74 R

[3] Plot sample mean on the chart; and plot samplingx x[3] Plot sample mean on the chart; and plot sampling range R on the R chart.

x x

[4] If (the sample point falls inside the control limits of the chart) and (the sample point R falls inside the control limits of

xx

the R chart), the process is in control (μ = μ0, σ= σ0).

[5] If (the sample point falls beyond the control limits of the chart) and/or (the sample point R falls beyond the control li it f th R h t) th i t f t l ( ≠

xx

limits of the R chart), the process is out of control (μ ≠ μ0,and/or σ ≠ σ0).

68

2.2.2 Rational Subgroups

The rational subgroup concept means that samples (also ll d b ) h ld b l t d i ti dcalled subgroups) should be selected in a time order

according to the working shift of production, so that the control chart is more effective in detecting the out of controlcontrol chart is more effective in detecting the out-of-control status. Time order subgrouping also helps to identify the source of the problemsource of the problem.

69

Concentrated sampling

When the primary purpose of using the control chart is to detect out-of-control status, each sample should consist of , punits that are produced at the same time (or as closely as possible). Concentrated samples are usually taken at the end of each sampling interval.

iRandom sampling

Wh th t l h t i l d t k d i i b tWhen the control chart is employed to make decision about the overall quality level of a product, samples should consist of units that are randomly picked up over the wholeconsist of units that are randomly picked up over the whole sampling interval h and, therefore, are representative of all units.

70

units.

Concentrated sampling is used most widely.

71Figure 2-15 Concentrated and random sampling

2.2.3 Cusum (Cumulative Sum) Chart

In this section, we only consider the process mean and assume that σ is always equal to σ So as long as the process is that σ is always equal to σ0. So, as long as , the process is in control.

Th h t i it ff ti f d t ti l l

0

The chart is quite effective for detecting large scale mean shiftsδ. However, it may become ineffective to small and medium mean shifts

x

medium mean shifts.

For the chart, the sample mean shift isx

0

(2-22)

(2 23)0 x (2-23)

When the process is in control , δ = 0.0

72When the process is out of control , δ ≠ 0.0

Figure 2-16 Mean shift δ73

g

So, the cusum chart comes to play, because it is much more effective than the chart to detect small and medium mean xshifts.

But, the design and operation of the cusum chart are more , g pcomplicated, and its fundamental algorithm is more difficult to understand.

In this section, in order to simplify the discussion, we use the x chart, that is a special chart with sample size n = 1. x, p pIn each sample of the x chart, there is a single observation x. Thus, = x/1 = x. From Eq (2-23),x

x

0ˆ x (2-24)

In this section, xi represents the single observation of the ith sample. The estimate of the mean shift for this sample is

iˆ

ˆ740

ˆ ii x (2-25)

Fig 2-17 displays a x chart for monitoring the diameter x of a g p y gshaft. The in-control parameters are μ0 = 10 and σ0 = 1, and the control limits are

LCL = 7, CL = 10, UCL = 13 (2-26)

In Table 2-4, the values of xi of the first 20 samples are b d h th i i t l All f th 20observed when the process is in control. All of these 20 xi

values fall within the control limits.

75 76Figure 2-17 An x control chart

Table 2-4: Data for the cusum chart example (μ0 = 10)

77 78

Then, the cutter of the machine tool is broken suddenly after the 20th sample. It increases the process mean to μ = 11. Thethe 20th sample. It increases the process mean to μ 11. The process becomes out of control.

δ = μ - μ0 = 11 – 10 = 1 > 0 (2-27)

The subsequent observations x (i = 21 22 ) are taken underThe subsequent observations xi (i = 21. 22, …..) are taken under the out-of-control status. But, the corresponding sample points still fall within the control limits of the x chart. It means thatstill fall within the control limits of the x chart. It means that the x chart fails to detect the out-of-control status, because it is not very effective for detecting the mean shift of small y gmagnitude (for example, δ = 1).

The reason is that the x chart (or chart) makes decisions only based on the data of the current (or last) sample point without

id i h i f h i l i h

x

79considering the series of the previous sample points or the operational history of the process.

Telling a story: In a company, employees are required to arrive and start working a 8:00am in the morning. However, a lateness of no more than 5 minutes will not be punished.

An operator is later every day, but always before 8:05am (i δ < 5) Th h b ht(i.e., δi < 5) Then he can never be caught.

80

Later the company changes the rule. It will penalize any employee whose cumulated lateness time is more than 10 minutes. Since the sum C4 of the first four δi (i.e., the sum of th l t ti i th fi t f d )the lateness time in the first four days).

C = δ + δ + δ + δ = 11 > 10 (2 28)C4 = δ1 + δ2 + δ3 + δ4 = 11 > 10 (2-28)

the operator is caught in just four days with the new rulethe operator is caught in just four days with the new rule being effective.

Thi l ill t t th f th l ti

This example illustrate the power of the cumulative sum, or the capability of the CUSUM chart. This chart incorporates all the information in the sequence of the sample points byall the information in the sequence of the sample points by monitoring the cumulative sums Ci of the mean shifts .

I i h h CUSUM h i ff i h h h81

It is why the CUSUM chart is more effective than the x chart (or chart) for detecting small process shifts.x

Ci is the sum of the values from the first sample to theith (current) sample.

C1 = δ1

C δ + δC2 = δ1 + δ2

C = δ + δ + δC3 = δ1 + δ2 + δ3

C 1211

ii

C

iiiC

121 (2-29)

82

iiiiiCC

1121)( Or, (2-30)iiiii 1121)( ( )

It means:

C1 = δ1

C2 = δ1 + δ2 = C1 + δ2

C3 = δ1 + δ2 + δ3 = C2 + δ3 (2-31)

………

83

For example, in Table 2-4,

C1 = δ1 = -0.55

C2 = C1 + δ2 = -0.55 + (-2.01) = -2.56

C3 = C2 + δ3 = -2.56 + (-0.71) = -3.27

C4 = C3 + δ4 = -3.27 + 1.66 = -1.61 (2-32)

A CUSUM h t l t th l i t f C C CA CUSUM chart plots the sample points of C1, C2, C3, …. That is, the ith sample point of a CUSUM chart carries the value of C

84

value of Ci.

85Figure 2-18 Plot of the cumulative sum

If μ= μ0 (process in control),

The x values of some samples are larger than μ0 (see Table 2-4)4),

x > μ → = (x μ ) > 0 → C > C → C ↑ˆxi > μ0 → = (xi - μ0) > 0, → Ci > Ci-1 → Ci ↑

(Note: Ci = Ci 1 + )

i

(Note: Ci Ci-1 + )

But, the x values of some other samples are smaller than μ0,

i

, p μ0,

xi < μ0 → = (xi - μ0) < 0, → Ci < Ci-1 → Ci ↓i

As a result, Ci sometimes increases, and sometimes decreases,

i

86always wanders around zero.

If μ> μ0 (process is out of control with increasing mean shift δ > 0)

The x values of almost all samples are larger than μThe x values of almost all samples are larger than μ0,

x > μ0 → = (x - μ0) > 0 → C > C 1 → C ↑ˆxi > μ0 → (xi - μ0) > 0 → Ci > Ci-1 → Ci ↑

Most of the time, Ci increases point by point until going

i

Most of the time, Ci increases point by point until going beyond an upper limit UCLcusum.

In Fig 2 18 the 26th sample point goes above UCLIn Fig 2-18, the 26th sample point goes above UCLCUSUM, and the CUSUM chart signals the out-of-control status.

The control limit UCLcusum of the CUSUM chart is calculated by Markov chain method or Monte Carlo simulation.

A similar CUSUM chart can be used to detect the out-of-control status with decreasing mean shift (μ < μ0 or δ < 0).

87

2 2 4 Runs Rules2.2.4 Runs Rules

In this section, again, we only consider the process mean and , g , y passume that σ is always equal to σ0.

A control chart can signal an out-of-control condition by

(1)Basic rule when one sample point falls beyond the control limit LCL or UCL;

(2)Runs rules when several sample points exhibit some nonrandom patternnonrandom pattern.

88

A control chart may presents some nonrandom patterns of y p pthe sample points when the process is out of control.

Such nonrandom patterns are seldom found when the process is in control. Therefore, their occurrence is a strong indication that the process is out of control (even if all of the sample points are within the control limits).

When the basic rule is used together with the runs rules, the capability of the control charts will be enhanced Howevercapability of the control charts will be enhanced. However, using the runs rules increases the difficulty of operation.

89 90Figure 2-19 Runs rule 1

(2)Four out of five consecutive points plot at a distance of 1-σ or beyond from the center line on one side (points 16,1 σ or beyond from the center line on one side (points 16, 17, 18, 19, 20 on Fig 2-20).

91Figure 2-20 Runs rule 2

(3)Eight consecutive points plot on one side of the center line (points 13 14 15 16 17 18 19 20 in Fig 2 21)line. (points 13, 14, 15, 16, 17, 18, 19, 20 in Fig 2-21).

92Figure 2-21 Runs rule 3

The out-of-control status in Fig 2-17 can be detected if the 3rd runs rule is applied to the last eight sample pointsruns rule is applied to the last eight sample points.

93

2.3 Control Charts for Attribute

To control the quality characteristics (e.g., d and C) which can only take integer valuesonly take integer values.

2.3.1 np Chart

The np control chart is used to monitor the number d of nonconforming items found in a sample of size n. A g pnonconforming item is a defective product, which fails to meet the quality requirement on the product.

For example, if five defectives are found in a sample i i 100 i d i i b d ( 100) h d 5containing 100 integrated circuit boards (n = 100), then d = 5.

94

Fraction nonconforming p is the ratio of the number D of nonconforming items in a population to the total number N of items in that population.

p = D / N (2-33)

p can be estimated by(2-34)ndp /ˆ (2 34)ndp /

95

Example 2-2

Last year, a company produces 10000 TVs, among which 295 are nonconforming. A sample of 100 TVs are inspected and 3 TVs in the sample are found nonconforming. Therefore

N = 10000, D = 295, p = D / N = 0.0295

n = 100, d = 3, = d / n = 0.03p

The sample size n is usually much smaller than N. Therefore, estimating p by d/n (inspecting only n items in a sample) is g p y ( p g y p )much easier than calculating p by D/N (inspecting the whole population of N items).

96

d is a random number and may have any integral value i

(0 ≤ i ≤ n). Moreover, d follows a binomial distribution, which is determined by two parameters n and p.

inid pp

in

iPidob

)1()(Pr

ini ppii

ni

)1()!(!

! (2-35)ppini

)()!(!

)(21)!(21!21! iiii )(21)!(,21!,21! ininiinn

Mean and standard deviation of d areMean and standard deviation of d are

)1(, pnpnp (2-36)97

)1(, pnpnpdd

(2 36)

For example, if the cardboards are inspected with sample p , p psize of four, and 10% of the cardboards are nonconforming (n = 4 and p = 0.1), d may be equal to 0, 1, 2, 3, or 4

6561.0)1.01(1.0)!04(!0

!4)0( 040

dP

!4

2916.0)1.01(1.0)!14(!1

!4)1( 141

dP

00360)101(10!4)3(

0486.0)1.01(1.0)!24(!2

!4)2(

343

242

d

P

P

0001.0)1.01(1.0)!44(!4

!4)4(

0036.0)1.01(1.0)!34(!3

)3(

444

d

d

P

P

)!44(!4

98

Note: Pd(0) + Pd(1) + Pd(2) + Pd(3) + Pd(4) = 1 = 100%

It means that d cannot take any value smaller than zero for larger than fourlarger than four.

4.01.04 npd

(2-38)pd

(2-38)

It means that, on average, there will be 0.4 nonconforming items in each sample.

60)101(104)1( pnp (2 39)6.0)1.01(1.04)1( pnpd

(2-39)

99

Cumulative Binomial probabilitiesCumulative Binomial probabilities

p

c0.05 0.10 0.20 0.30 0.40 0.50

n = 20 0 0.358 0.122 0.012 0.001 0.000 0.000

1 0.736 0.392 0.069 0.008 0.001 0.000

2 0.925 0.677 0.206 0.035 0.004 0.000

3 0.984 0.867 0.411 0.107 0.016 0.001

100

np Control Chart: monitoring the number d of f i it i l f i Th f thnonconforming items in a sample of size n. The name of the

np chart reflects the fact that the mean of d is equal to np. Each sample point on the np chart carries the value of d of aEach sample point on the np chart carries the value of d of a sample.

When using the np chart, the process is considered as in control if p = p0, where p0 is the in-control (or acceptable, or p p0, p0 ( p ,nominal) value of fraction nonconforming p. The value of p0is usually very small, but cannot be zero (p0 = 0 is just a dream). When process is in control,

p = p0, )1(,000

pnpnpdd

(2-40)

101

The fraction nonconforming p is estimated by d/n, or approximatelyapproximately,

p = d / n, or d = np (2-41)

Or in other words, d is different from p only by a constant of n. As a result, d can be used as a measure of p of the product, and the np chart can be used to ensure p = p0.

If d = np0 p = np0 / n = p0 (2-42)If d np0, p np0 / n p0, (2 42)

Then, the process is in control. As such, np0 is the target of d.

In practice, if d is equal or close to np0, p will be equal or close to p0, and the process is thought in control.

On the other hand, if d significantly deviates from np0, p must differ from p0, and the process will be signalled as out of

102

differ from p0, and the process will be signalled as out of control.

3-sigma np chart

)1(33000

pnpnpCLUCLddd

3 sigma np chart

)(

0

000

npCLppp

d

ddd

(2-43)

)1(33000

pnpnpCLLCLddd

Sh ld h l l i i ld i C C 0Should the calculation yield a negative LCLd, set LCLd = 0. Because d cannot be a negative number.

103

In Eq (2-43), both LCLd and UCLd depend on p0. However, in practice, p0 is usually unknown. To estimate p0, we take m preliminary samples. For the ith sample (see Table 2-5)

d mindp i

i,,2,1ˆ (2-44)

Even though each is an estimate of p0, the average of is an even more accurate and reliable estimate of p0.

ip p ipp0

pm

i ˆ

mpp i 1

0(2-45)

For example, from Table 2-5,

12.016.030.024.0

1042313.0

3012.016.030.024.0

0

pp

Table 2-5 Data of Cardboard (n = 50)

105

Then, p0 in equation (2-43) will be replaced by .p

)1(3CL

ppnpnUCLd (2 46)

)1(3 ppnpnLCLpnCL

d

d

(2-46)

Sample size of the np chart is determined by the production &Sample size of the np chart is determined by the production & inspection rate and the p0 value. If p0 is very small, sufficiently large n should be chosen. An empirical formula is as follows: g p

n = c / p0

where, c is a constant between 0.5 and 1.5. For example, if p00 01 h ld b b 50 d 150

106= 0.01, n should be set between 50 and 150.

EXAMPLE 2-3 Design an np Chart to Monitor the g pnumber of Defective Cardboards (Table 2-5)

(1) Decide the sample size n and the sampling interval h.

n = 50, h = 0.5 hr

(2) I h I t k 30 li i l d l l t(2) In phase I, take 30 preliminary samples, and calculate the fraction nonconforming pi for each sample (eq (2-44)).

(3) Calculate the average fraction nonconforming (eq (2-45))

p45)).

2313012.016.030.024.01

pp

m

ii

107

2313.030

m

p

(4)Calculate the central line CLd and the control limits ( ) dLCLd and UCLd for the np chart (eq (2-46)).

1020)231301(2313003231300)1(3 ppnpnUCLd

510.20)2313.01(2313.05032313.050

565112313050CL 565.112313.050 pnCLd

6202)231301(231305032313050)1(3

ppnpnLCLd

620.2)2313.01(2313.05032313.050

108

(5) Round off the control limits(5) Round off the control limits

Fractional limits:Fractional limits:

If 2.620 d 20.510, the process is in control..6 0 d 0.5 0, t e p ocess s co t o .

If d < 2.620 or d > 20.510, the process is out of control.

p

But, since d is always an integer, the operational rules of this np chart is

(i) l i i i l if 3 d 20(i) claiming in-control status if 3 d 20;

(ii) signalling out of control status if d < 3 or d > 20

109

(ii) signalling out-of-control status if d < 3 or d > 20.

Or in other words, the fractional limits are exactly equivalent to the following integral limits:to the following integral limits:

If 3 d 20, the process is in control. , p

If d < 3 or d > 20, the process is out of control.

The round-off rule: the fractional LCL is rounded up, and the fractional UCL is rounded down.

Si th i t l li it i t h dl th th f ti lSince the integral limits are easier to handle than the fractional limits, the former will be used in the implementation of the np chart

110

chart.

Figure 2 23 Round off of the control limits111

Figure 2-23 Round off of the control limits

(6) Plot the np chart

112Figure 2-24 An np control chart

(7) Start phase II using the np chart to monitor d of(7) Start phase II, using the np chart to monitor d of the process.

As long as the sample points remain within the control limits, d is equal or close to CL (=np0), or p is equal or ts, d s equa o c ose to C ( np0), o p s equa oclose to p0, the process is thought in control. That is, the defective rate of the Cardboard is kept around the pacceptable level of p0.

113

If a d sample point plots outside of the control limits, p has most likely shifted to a new level and the process is thought out of control.

(d > UCL) indicates that d > np0 (or p > p0), the defective rate becomes higher than p We must shut down the process thenbecomes higher than p0. We must shut down the process, then search and remove the problem.

(d < LCL) means that d < np0 (or p < p0). Even though defective rate gets lower, the process is still thought as out of control a e ge s owe , e p ocess s s oug as ou o co o(because p ≠ p0).

We also have to look into the process to find out the reason for improvement. The np chart may have to be redesigned, because

114the value of p0 is changed and the control limits depends on p0.

2.3.2 c chart

h i i h b C f f i i ic chart is to monitor the number C of nonconformities in a unit of n items. It is mainly used to handle some minor (nonfatal) defects in the products(nonfatal) defects in the products.

A nonconformity is a minor defect in common languageA nonconformity is a minor defect in common language.

A unit is an entity for which it is convenient to keepA unit is an entity for which it is convenient to keep records. It could be a single item or a group of n items.

115

Depending on the nature and severity of the nonconformitiesDepending on the nature and severity of the nonconformities, it is quite possible that a unit containing several nonconformities is acceptable and not classified asnonconformities is acceptable, and not classified as nonconforming.

A “nonconforming” means that a unit as a whole is unsatisfactory and must be rejected.y j

A “nonconformity” means a minor defect on one of the nitems in a unit. The unit may, or may not, be rejected.

116

117Figure 2-25 A unit of three cars

Suppose ci is the number of nonconformities in the ith item within a unit. And C is the total number of nonconformities in all of the nitems in a unit of n items.

n

(2 47)

i

in ccccC1

21 (2-47)

In the example in Fig 2 25 a unit is comprised of three carsIn the example in Fig 2-25, a unit is comprised of three cars shipped by a container (n = 3). A nonconformity is a tiny spot on the surface of a car due to some painting problemthe surface of a car due to some painting problem.

So, ci is the number of tiny spots on the ith car in a unit, and C isSo, ci is the number of tiny spots on the ith car in a unit, and C is the total number of the tiny spots in all three cars of a unit. In Fig 2-25,,

c1 = 2, c2 = 1, c3 = 0

3012C (2 48)118

3012321 cccC (2-48)

For this example, based on the quality requirement, the p , q y q ,Quality Engineer specifies that a unit is nonconforming when C is larger than 4. That is,

If C 4, the whole unit (three cars) is conforming. Even though there may be a few nonconformities (between 1 and 4) on some items in this unit, the unit

h l ill b t das a whole will be accepted.

But if C > 4 the whole unit (three cars) isBut, if C > 4, the whole unit (three cars) is nonconforming. Even though the impact of each individual nonconformity is not severe the number ofindividual nonconformity is not severe, the number of nonconformities in a unit is intolerably high, the unit as a whole will be rejected.

119

j

In unit 3, the third car contains no nonconformity (c3 = 0). However, since unit 3 as a whole is nonconforming, this car will b j t d t th ith th i thi it

120be rejected together with other cars in this unit.

C is a random number and follows a Poisson distribution.

!718.2

!)()(Pr

iieiPiCob

ii

C

(2-49)

The Poisson distribution is uniquely determined by a single parameter λ (> 0) which is the mean value μ of C Or inparameter λ (> 0) which is the mean value μC of C. Or in engineering terms, λ is the average of the number of nonconformities in each unit In the example in Table 2-6nonconformities in each unit. In the example in Table 2 6,

λ ≈ (3 + 1 + 5) / 3 = 3 (2-50)

The standard deviation σC of C is equal to the square root of λ.

(2-51)μC = λ, C

121

Suppose, the average number of tiny spots (nonconformities) in a unit is three That isin a unit is three. That is,

λ = 3, μC = 3, 732.13 C

0498.0!03)0(

03

ePC

14936.0!13)1(

!013

ePC

22404.0!23)2(

!123

ePC

(2-52)

22404.0!33)3(

!233

ePC

!3

122

123Figure 2-26 Poisson distribution

3-sigma c chart:g

33 CCC

CLCLUCL

(2-53)

33

CCC

C

CLLCLCL (2-53)

In Eq (2-53), both LCL and UCL depend on λ. However, in practice λ is usually unknown But λ can be estimated frompractice, λ is usually unknown. But, λ can be estimated from the observed values of C in m preliminary units.

124

For example (Table 2-7), m = 26, n = 100. Cj is the total number of nonconformities in the jth unit containing 100number of nonconformities in the jth unit containing 100 printed circuit boards.

C1 = 21, C2 = 24, …., C26 = 15

8519152421ˆ 2621 CCC (2 54)85.192626

2621 (2-54)

Now, replacing λ by ,p g y

ˆ

ˆ3ˆ C

CL

UCL

(2 55)

ˆ3ˆ

C

C

LCL

CL (2-55)

Should the calculation yields a negative value for LCLC, set LCL 0 Beca se the n mber of nonconformities cannot be

125

LCLC = 0. Because the number of nonconformities cannot be negative.

Table 2-7 Number of nonconformities in units of 100 printed circuit boardscircuit boards

(m = 26, n = 100)

126

EXAMPLE 2-4 Design of a c chart to monitor the number of nonconformities in units of n printed circuit boards (Table 2-7)

(1)Decide the number n of items per unit and the sampling i t l hinterval h.

n = 100 h = 0 5 hrn = 100, h = 0.5 hr

(2)In phase I take 26 preliminary units and calculate the(2)In phase I, take 26 preliminary units, and calculate the number Cj of nonconformities for each unit.

C1 = 21, C2 = 24, …., C26 = 15

127

(3)Estimate the mean λ (eq (2-54)).( ) ( q ( ))

85.1926

15242126

ˆ 2621 CCC2626

(4)Calculate the central line CL and the control(4)Calculate the central line CLC and the control limits LCLC and UCLC for the c chart (eq (2-55)).

ˆ22.3385.19385.19ˆ3ˆ CUCL

48.685.19385.19ˆ3ˆ85.19

C

C

LCL

CL

128

(5)Round off the control limits

Since C is always an integer, the control limits of the c chart should be rounded off. Particularly, LCLC should be rounded up

d UCL h ld b d d d f h h Th iand UCLC should be rounded down, as for the np chart. Thus, in this example, the final control limits are: LCLC = 7 and UCLC = 3333.

(6)Plot the c chart.

(7)Start phase II, using the c chart to monitor C of the process.

If 7 C 33 th i th ht i t l (th b

If 7 C 33, the process is thought in control (the number of nonconformities in each unit is close to the average value of )

85.19ˆ average value of ).

If C < 7 or C > 33, the process is thought out of control

129

3 PROCESS CAPABILITY

Process capability refers to the capability that a manufactured quality characteristic x conforms to the design specifications q y g pon the tolerance. The higher the process capability is, the lower the defective rate.

Specification limits: decided by the design engineer based on the functional requirements. They are fixed values in a process q y pcapability study.

Lower specification limit: LSLLower specification limit: LSL

Upper specification limit: USL

Design tolerance = USL - LSL (3 -1)

An individual part is classified as a defective, if x is smaller 1

d v dua pa t s c ass ed as a de ect ve, x s s a ethan LSL or larger than USL.

For example, the design specifications of the diameter x of a shaft is decided by the design engineer as 74 ± 0.05mm. Then, the nominal or target value, T, of the diameter is 74mm, and

LSL 74 0 05 73 95LSL = 74 – 0.05 = 73.95mm

USL = 74 + 0 05 = 74 05mmUSL = 74 + 0.05 = 74.05mm

Design tolerance = USL – LSL = 74 05 – 73 95 = 0 1mmDesign tolerance USL LSL 74.05 73.95 0.1mm

An individual shaft is classified as a defective, if the ,diameter x is smaller than 73.95mm or larger than 74.05mm.

2

3Figure 3-1 Target and specifications

Natural tolerance limits: acquired from the actual probability distribution of x (see Fig 3-1) varying along with the changes of the manufacturing status.

L LNTL ( h i i l f )Lower: LNTL (the minimum value of x)Upper: UNTL (the maximum value of x)

The probability distribution of x, as well as LNTL and UNTL can be changed by using different machinesUNTL, can be changed by using different machines, processes, materials and so on.

4

x is usually assumed to have a normal distribution. Therefore, x may be infinitely small or infinitely large from a theoreticalx may be infinitely small or infinitely large from a theoretical viewpoint.

But from an engineering viewpoint, it is almost impossible (with a probability of only 0.0027) that the x value will fall ( p y y )below (μ - 3σ) or above (μ + 3σ). Thus, in engineering practice, we can take (μ - 3σ) as the minimum possible value of x; and take (μ + 3σ) as the maximum possible value of x.

LNTL ≈ μ - 3σLNTL μ 3σUNTL≈ μ + 3σ (3-2)

Distribution range R= UNTL LNTLDistribution range, R= UNTL - LNTL = (μ + 3σ) – (μ - 3σ) = 6σ (3-3)

5

Figure 3 2 Lower and upper natural control limits6

Figure 3-2 Lower and upper natural control limitsin the normal distribution (f9-1 p432)

μ and σ can be estimated from the measured data (e.g., the data bt i d d i i t l h t)obtained during running a control chart).

Suppose four observations are available x = 2 0 x = 2 1 x =Suppose four observations are available, x1 = 2.0, x2 = 2.1, x3 = 1.9, x4 = 2.2, n = 4

05.24

2.29.11.20.2ˆ 4321 n

xxxx (3-4)

ˆ 2 xn

i

1ˆ 1

n

xi

i

14)05.22.2()05.29.1()05.21.2()05.20.2( 2222

7129.0 (3-5)

14609.12.21/ˆ RRROr (3 6)146.0059.2222

ddd

Or, (3-6)

d2 = 2 059 for n = 4d2 2.059, for n 4.

Then, if using ,129.0ˆ

437.2129.0305.2ˆ3ˆ663.1129.0305.2ˆ3ˆ

UNTLLNTL

(3-7)

8

In order to achieve a high process capability so that a g p p ymanufactured quality characteristic x conforms to the design specification on the tolerance (or the value of x falls between LSL and USL) with a high probability, we may

(1)make the standard deviation σ of x small. Because it makes the distribution range R (= 6σ) of x likely to fall within

ifi ti li it LSL d USLspecification limits LSL and USL.

Compare Fig 3 1 (a) and (c)Compare Fig 3-1 (a) and (c)

(2)make the mean μ of x close to the target T (usually T is set(2)make the mean μ of x close to the target T (usually, T is set at the center between LSL and USL).

9Compare Fig 3-1 (a) and (b)

Applications of the process capability studies:

(1)Predicting how well the process will hold the tolerance.

(2)Assisting product developers/designers in selecting or modifying a processmodifying a process.

(3)Selecting between competing vendors(3)Selecting between competing vendors.

The process capability is usually measured by one of the three process capability ratios: Cp, Cpk and Ckm.p p

For any of Cp, Cpk and Ckm, the greater its value is, the higher

10the process capability, and the lower the defective rate.

3.1 Process Capability Ratio Cp

Cp is a measure of the distribution range R with respect to design tolerance (USL – LSL).g ( )

6LSLUSL

LNTLUNTLLSLUSL

RLSLUSLCp

(3-8)

Example 3-1: x =74±0.05mm is the diameter of a shaft. A

6LNTLUNTLR

Also suppose mm

USL 74 05 LSL 73 95

0099.0ˆ

USL = 74.05 mm, LSL = 73.95 mm,

68195.7305.74

LSLUSLC 68.10099.06ˆ6

Cp

11

If Cp is great, the distribution range R (UNTL – LNTL) of pthe quality characteristic x is relatively small compared to the design tolerance (USL – LSL), or x is less likely to fall beyond the specification limits and is more likely to conform to the design tolerance requirement.

The greater the Cp, the better the product quality. The minimum requirement is C = 1 otherwise a substantialminimum requirement is Cp = 1, otherwise, a substantial amount of products will fail to meet the design specifications (indicated by the darkened area in Fig 3-3 (c))(indicated by the darkened area in Fig 3 3 (c)).

12

13Figure 3-3 Different values of Cp

Table 3-1 Recommended values of the CTable 3 1 Recommended values of the Cp

14

One-sided specifications

U ifi tiUpper specification

In some applications, x is constrained only by an USL, but has no LSL, i.e.,

x USL (3-9)x USL (3 9)

For example, x is the deflection of a cantilever beam under t i l d F Th d i i ifi th t t b

certain load F. The design engineer specifies that x must be no larger than 0.01mm. i.e.,

x 0.01

USL = 0.01. LSL is not needed, because, we hope the deflection 15

, , pto be as small as possible.

For upper specification, we calculate upper Cp ---- CPUSuppose there is a virtual LSL that is symmetrical with USL about μSuppose there is a virtual LSL that is symmetrical with USL about μ (Fig 3-4 (a)).

3650)(5.0

6

USLLSLUSLLSLUSLCP

(3-10)

365.06

This Cp is taken as CPU (Note, it is independent of LSL), sop ( p )

USLCPU (3 11)

3CPU (3-11)

16

17Figure 3-4 One-sided specifications

Lower specification

In some other applications, x is constrained only by a LSL, but h USL ihas no USL, i.e.,

x LSL (3 12)x LSL (3-12)

For example x is the power of an engine The design engineer

For example, x is the power of an engine. The design engineer specifies that x must be no smaller than 1000 watt, i.e.,

x 1000

LSL = 1000. USL is not needed, because, we hope the power of the engine to be as high as possible.

18

For lower specification, we calculate lower Cp ---- CPL

Suppose there is a virtual USL that is symmetrical with LSLabout μ (Fig 3-4 (b)).about μ (Fig 3 4 (b)).

)(5.0 LSLLSLUSLLSLUSLCP

(3-13)

365.06P (3 3)

This C is taken as CPL (Note it is independent of USL) soThis Cp is taken as CPL (Note, it is independent of USL), so

LSL

3

LSLCPL (3-14)

19

Another quality index that is directly related to the process capability is PPM. It is the number of defective (nonconforming)capability is PPM. It is the number of defective (nonconforming)Parts Per Million. PPM is actually a measure of the defective rate p (or the fraction nonconforming).p ( g)

PPM = 1,000,000 × p (3-15)

p = Prob (x < LSL or x > USL) (3-16)

Table 3-2 shows several PPM values along with the associated l f th C H th tivalues of the Cp. Here, the assumptions are:

(1)The quality characteristic x has a normal distribution(1)The quality characteristic x has a normal distribution.

(2)The process mean is centered between the upper and lower20

(2)The process mean is centered between the upper and lower specification limits (μ = T).

Table 3-2 Cp versus PPM

21

F l if C 1 00 ( Fi 3 3) h b biliFor example, if Cp = 1.00 (see Fig 3-3), the probability pthat x falls beyond LSL or USL is 0.0027. Then,

PPM = 1,000,000 0.0027 = 2700

It is obvious that, the larger the Cp, the smaller the defective rate p and the smaller the PPM.defective rate p and the smaller the PPM.

22

If the quality characteristic x has a normal distribution, but the q y ,process mean is not centered between the upper and lower specification limits (i.e., μ ≠ T), PPM can be calculated easily.

Suppose the mean and standard deviation of the random variable x are μ and σ, respectively. Then, the probability that xvariable x are μ and σ, respectively. Then, the probability that xis smaller than a fixed value X is equal to

X

zXXxob

)(Pr

X

(3-17)

Xzwhere,

The random variable z follows a standard normalThe random variable z follows a standard normal distribution with μ = 0 and σ = 1. Φ( ) is the cumulative probability function of z

23

probability function of z.24

Table 3 325

Table 3-3

F h bl di l h Φ l l fFurthermore, many tables display the Φ values only for z > 0. However, if z < 0,

Φ(z) = 1 - Φ(-z) (3-19)

For example,

Φ(-0.54) = 1 - Φ(0.54) = 1- 0.7054 = 0.2946

26

At the lower end, the number, PPML, of defective Parts Per Million is calculated as follows (Fig 3-5).

PPML = 1,000,000 prob(x < LSL)

1 000 000

LSL= 1,000,000

= 1 000 000

z= 1,000,000

where (3-21)

Lz

LSLzwhere, (3 21)

zL

27

Figure 3 5 Calculation of PPM28

Figure 3-5 Calculation of PPM

At the upper end, the number, PPMU, of defective Parts Per Milli i l l t d f llMillion is calculated as follows.

PPMU= 1,000,000 prob(x > USL)

= 1,000,000 [ 1 - prob(x < USL) ]

1 000 000

USL1= 1,000,000

= 1,000,000

1

Uz1

where, (3-22)

USLzwhere, (3 22)

The overall PPM

zU

PPM = PPML + PPMU (3-23)

29

If the process mean is centered between the upper and p pplowerspecification limits (μ = T), then

prob(x < LSL) = prob(x > USL)

PPML = PPMU = 0.5 PPM (3-20)

30

Example 3-2 (Figure 3-5)

The weight x of a product follows a normal distribution with μ= 10g and σ = 2g. It is required that x be maintained between 5g and 14g, i.e.,

LSL = 5, USL = 14, T = 0.5 (5 + 14) = 9.5

Here, μ ≠ T, and Table 3-2 cannot be used.

31

At the lower end,105LSL 5.2

2105

LSLzL

Φ( L) Φ( 2 5) 1 Φ(2 5) 0 00621Φ(zL) = Φ(-2.5) = 1- Φ(2.5) = 0.00621

PPML = 1,000,000 = 1,000,000 0.00621 = 6210 Lz,000,000 ,000,000 0.006 6 0 L

At the upper end, 1014USL 2

21014

USLzU

Φ(zU) = Φ(2) = 0.97725

PPMU = 1 000 000 z1PPMU = 1,000,000

= 1,000,000 = 22750

Uz1

97725.0132

1,000,000 22750 97725.01

The overall PPMThe overall PPM

PPM = PPML + PPMU = 6210 + 22750 = 28960PPM PPML + PPMU 6210 + 22750 28960

It can be seen that most of the defectives occur at the upper end, as the process mean μ has shifted to the pp p μright side.

33

3.2 Process Capability Ratio Cpk

In Fig 3-6, LSL and USL are the same for all six processes. The standard deviation σ of all processes is equal to 2.p q

Therefore, all of the six processes have the same Cp (Eq (3-8)).

3862LSLUSL263862

6

LSLUSLCp = 2 (3-26)

That is, if judged by Cp, the six processes have the same process capability It is somewhat misleading

34

process capability. It is somewhat misleading.

35Figure 3-6 Relationship of Cp and Cpk

However, obviously, the process capability degenerates from top to bottom because more and more defective units will beto bottom, because more and more defective units will be produced (PPM becomes larger and larger from the top to the bottom).bottom).

Why can’t Cp tell this difference of the process capabilities of the y p p psix processes? Because, Cp only considers the ratio between (USL – LSL) and R (= 6σ), and does not take into account the centering of process distribution between the specification limits.

As a conclusion, Cp is a valid measure of process capability only when μ = T.

When μ ≠ T, other more appropriate measures should be used for the process capability

36

the process capability.

C is another process capability ratio that takes into accountCpk is another process capability ratio that takes into account not only the ratio between (USL – LSL) and R (6σ), but also the location of the process mean μ relative to thethe location of the process mean μ relative to the specification center T.

CPLCPUifCPU (3 27)

CPLCPUifCPLCPLCPUifCPU

CPLCPUCpk ,min (3-27)

Since (Eqs (3-11) and (3-14))

USL

3

USLCPU

LSL

3LSLCPL

ii LSLUSLCCC (3 28)

37

3,

3min,min LSLUSLCPLCPUCpk

(3-28)

If μ > T (Fig 3-7(a)),

33LSLUSL

33

Cpk = CPU =

3USL (3-29)3

When μ > T, the upper side is more critical, most of the d f ti ill b d d t th id Th f CPUdefectives will be produced at the upper side. Therefore, CPUis smaller than CPL. Cpk is equal to CPU.

The more μ moves to the upper side, the smaller the (USL - μ) and the smaller the C kand the smaller the Cpk.

38

39Figure 3-7 Cpk for different cases

If μ < T (Fig 3-7(b)),

33LSLUSL

Cpk = CPL =

3

LSL(3-30)

3

When μ < T, the lower side is more critical, most of the μdefectives will be produced at the lower side. Therefore, CPL is smaller than CPU. Cpk is equal to CPL.

The more μ moves to the lower side, the smaller the (μ - LSL) d h ll h Cand the smaller the Cpk.

40

If μ = T (Fig 3-7(b)),

33LSLUSL

or CPU = CPL

Cpk = CPU =

3USL (3-31)

Or Cpk = CPL = LSL (3-32)Or Cpk CPL 3

(3 32)

From Eq (3-31),From Eq (3 31),

CLSLUSLUSLUSL

)(2C = (3-33)

pC

6323

Cpk = ( )

It indicates that if μ = T the two process capability ratios C41

It indicates that, if μ = T, the two process capability ratios, Cpkand Cp, are equivalent.

In Figure 3-6,

For process (a)38505062 22,2min233850,

235062min

pkC

F (b)For process (b)

515251min38535362min

kC 5.15.2,5.1min23

,23

min

pkC

For process (c)For process (c)

13,1min233856,

235662min

pkC2323 p

42

For process (d)

04,0min233862,

236262min

pkC ,23

,23

pk

F ( )For process (e)

505450min38656562min

C 5.05.4,5.0min23

,23

min

pkC

The Cpk value becomes smaller and smaller from process (a) to pk p ( )process (e). It exactly indicates the fact that the process capability degenerates from process (a) to process (e).

43

The value of Cpk depends on the difference (μ - T). It means pthat Cpk takes into account the location of the process mean μrelative to the specification center T.

As a conclusion, when μ ≠ T, Cpk is a more appropriate f bilit th Cmeasure of process capability than Cp.

44

It can be shown that Cpk is always smaller than or equal to Cp.

Cpk < Cp when μ ≠ T

Cpk = Cp only when μ = T. (3-34)

While Cpk is called the actual capability of a process, Cp is called the potential capability. Cp is the maximum possible p p y p pvalue of the process capability that may be achieved for a process when μ = T.

When the process mean μ is far away from the T (μ < T or μ > T), Cpk is very small. Cpk becomes larger and larger when μ is moved towards T from either side. Finally, when μ is

i id t ith T C C45

coincident with T, Cpk = Cp.

Figure 3-8 Increasing Cpk to Cp.

How to improve the actual capability (increasing Cpk)?

(1)Move μ towards T, in order to make the actual capability Cpkcloser to Cp.

46(2)Reduce σ, (σ is the denominator in Eq. (3-28)).

3.3 Process Capability Ratio CkmCkm is defined based on Taguchi’s Loss Function 2

222 )( T (3-35))( ( )where, μ and σ are the mean and standard deviation of a quality characteristic x; and T is the target of x. q y ; g

T = ( LSL + USL)21 (3-36)

Professor Taguchi uses to measure the loss incurred by a d t i t d t lit F E (3 35) th ll th

2deteriorate productquality. From Eq (3-35), the smaller the σvalue or the difference (μ - T), the smaller the value of (or ) Conversely a small (or ) indicates better

2 2 ). Conversely, a small (or ) indicates better

product quality or higher process capability.

2

47

Ckm is calculated by the following formula.LSLUSL

6LSLUSLCkm

(3-37)

C i h i llCkm is greater when is smaller.

Since C depends on and takes into account the location

Since Ckm depends on , and takes into account the location of the process mean μ relative to the specification center T (i.e. μ - T) so like C C can be used to handle the off-center

μ - T), so, like Cpk, Ckm can be used to handle the off-center process.

Same as for Cpk, there are two ways to increase Ckm (or to reduce the loss function ), see Eq (3-35).2 ), q ( )

(1)Reduce |μ - T|, i.e., centering the process mean μ at T.

48(2)Reduce σ.

Example 3-3 (Figure 3-9)

The weights of three products A, B and C follow normal distributions. For all three products, the target value T of the weight is equal to 50g, and LSL = 43g, LSL = 57g.

For product (A), μ = 50, σ = 2

4)5050(2 222 = 24)5050(2

167.1264357

kmC

2

26km

For product (B), μ = 52, σ = 2

8)5052(2 222 = 2.828

49825.0828.264357

kmC 50Figure 3-9 Two processes (f9-10 p445)

For product (C), μ = 50, σ = 3

9)5050(3 222 = 3

778.0364357

kmC

Product (A) has smallest loss function value, or highest process capability ratio Cprocess capability ratio Ckm.

The quality of product A is better than the quality of productThe quality of product A is better than the quality of product B, because the mean of the former is coincident with the target T.ta get .

The quality of product A is also better than the quality of 51

q y p q yproduct C, because the former has smaller standard deviation.

Summary

(1)Cp is the simplest process capability ratio. It is a valid f th bilit h th imeasure of the process capability when the process mean is

coincident with the center between the specification limits. It indicates the potential capability of a processindicates the potential capability of a process.

(2)C k measures process capability based on PPM (Defective(2)Cpk measures process capability based on PPM (DefectiveParts Per Million). It can handle off-center process capability.

(3)Ckm measures process capability based on Taguchi’s loss function. It is also able to handle off-center process capability.

52

4 GAGE REPEATABILITY AND REPRODUCIBILITY

4.1 Gage (Measurement) Error

A i f SQC i l i iAn important aspect of SQC implementation is to ensure adequate accuracy and precision for gage and measuring system.

In any problem involving measurements, some of the observed variability will be due to production itself, and some will be due to measurement or gage error. Gage error includes all the variabilities related to the gage, as well as the measuring system. I b d b h i lf b h i b hIt may be caused by the gage itself, or by the inspectors, or by the measuring conditions and procedures.

Statistical methods can be used to separate the components of variability, as well as to give an assessment of gage capability, so

1that corresponding actions can be taken to reduce the overall variability (errors).

For the following error components (see Figure 4-1):

gageproductiontotal (4-1)

(11) = (10) + (1)

is the deviation of a produced dimension from its target value. isthe measuring deviation from the

production

gageg gproduced dimension. is the total deviation from the target.

gage

total

Both and are random variables and usually follow thenormal distribution. Since is a function in

production gage

totalterms of and . Therefore, is also a normally distributed random variable.

total

productiongage total

2

3Figure 4-1 Error components

4Figure 4-2 Components of variability

The gage error cannot be determined based on an individual d i ti i t d it t b d b th t d ddeviation , instead, it must be measured by the standard deviation of the probabilitydistribution of all values of . Similarly the production error cannot bedetermined based on an

gagegage

gageSimilarly, the production error cannot bedetermined based on an individual deviation , it must be measured by the standard deviation

productiond tideviation .production

standard deviation of . indicates the total error

total totaltotalerror

standard deviation of . indicates the d i

productionproductio productionproduction error.

standard deviation of . indicates the gagegagegage

5gage error.

g ggageg g

Because of Eq (4-1), the variances (squares of the standard deviations σ) satisfy :

222gageproductiontotal (4-2)

Further Decomposition of Gage Error

The gage error can be further decomposed into the error l ti t t bilit d th l ti t d ibilitrelating to repeatability and the error relating to reproducibility.

ilityreproducibityrepeatabilgage (4-3)ypypg g (4 3)

Repeatability regards the inherent error of the gage i lf (i d i i )itself (inadequate precision).

Reproducibility regards the variability due to different6

Reproducibility regards the variability due to different inspectors, or the error made by the inspectors.

Like Eq (4-2),222

ilityreproducibityrepeatabilgage

O ll

(4-4)

Overally,

ilityreproducibityrepeatabilproduction

gageproductiontotal

(4-5)

222

222

222

ilityreproducibityrepeatabilproduction

gageproductiontotal

(4-6)

7

EXAMPLE 4-1 (Table 4-1)

Each of the 20 produced parts has been measured twice b h f th th i t All f thby each of the three inspectors. All of them use a samegage.

8

T bl 4 1 D f E l 4 1Table 4-1 Data for Example 4-1

9

(1)The production error production

The production error is reflected by the differences among the average readings (j = 1 2 20) of the 20 different partsx~average readings (j = 1, 2, …, 20) of the 20 different parts. Because if there is no production error, all of the 20 parts have the exactly same dimension then the difference between the

jx

the exactly same dimension, then the difference between the20 values should be equal or very close to zero.jx~

Each is the average of the three values from the three inspectors for the jth part. For example,

jx~ x

17.203/)0.200.205.20(~1 x

67.233/)5.230.245.23(~2 x

10……

To evaluate σproduction, we set up the following sample based on the 20 values The sample size is (n = 20) and number of samplesx~20 values. The sample size is (n = 20) and number of samples is m = 1.

jx

The sample range R is equal to ( - = 29.67 – 18.17=11.5) and is attributable to the production error σ d ti

15~x 20

~xis attributable to the production error σproduction.

(4-7)5.111/ RRR (4 7)

(4-8)079.37353

5.11

dR

production (4 8)

where, d2 = 3.735 can be found from Table 2-3 for n = 20.

735.32d

11

where, d2 3.735 can be found from Table 2 3 for n 20.

(2)The error relating to repeatability σrepeatability

The error relating to repeatability is reflected by the diff f t di t k b i t fdifference of two readings taken by a same inspector for a same part. Because if the gage is perfectly precise, the two readings will have the exactly same value then thereadings will have the exactly same value, then the difference between them should be equal or very close to zerozero.

To evaluate σrepeatability, we consider the pairs of the two repeatability, preadings from a part taken by an inspector, and use each of the pairs as a sample. There are totally 60 such samples in Table 4-1 (20 samples for each of the three inspectors).

12

13

Each sample contains two readings. The sample size is (n = 2) and number of samples is m = 60.

Th l R f h f th 60 l li t d tThe sample range R of each of these 60 samples are listed at the right hand side. R is attributable to the inherent error of the gage σgage σrepeatibility.

151)212()100()011(

R (4 9)15.1203

R (4-9)

02.1128.115.1

2

dR

ityrepeatabil (4-10)

where, d2 = 1.128 for samples of size n = 2.

14

(3)The error relating to reproducibility σreproducibility

The error relating to reproducibility is reflected by the difference of the three grand average (i = 1, 2, 3) from the three different ix~inspectors. Because if there is no variability among the three inspectors, the three will be equal or very close to each other.ix~

To evaluate σreproducibility, we set up the following sample based on the al es sho n at the bottom of Table 4 1xthe values shown at the bottom of Table 4-1.ix

This single sample contains three . The sample size is (n = 3) and number of samples is m = 1.

ix~

15

p

The sample range R is attributable to the variability relating to the inspectors or the reproducibility σreproducibility.

32.0132.0

R(4-11)

1

19.0693132.0

2

dR

ilityreproducib (4-12)693.12d

where d = 1 693 for sample size (n = 3)where, d2 = 1.693 for sample size (n = 3).

16

Next, the gage error

222 ilityreproducibityrepeatabilgage (4-13)

039.108.119.002.1 22

gage

( )

Here the inherent gage error σ is much larger than theHere, the inherent gage error σrepeatibility is much larger than the error σreproducibility due to the variability of the inspectors.σreproducibility due to the variability of the inspectors.

17

Finally, the total error

222 gageproductiontotal

250.3560.10039.1079.3 22

gage(4-14)

The majority of the total error is attributable to the production error in this example The contribution from theproduction error in this example. The contribution from the gage error is negligible.

Or in other words, the gage error will not significantly influence the total error and the gage is considered suitable g gfor this measuring application.

18

4.2 The Gage capability Cgage

The distribution of is usually well approximated by a normal distribution (Fig 4-2). Then, 6 is a good estimate of the

gage

gagedistribution (Fig 4 2). Then, 6 is a good estimate of the distribution range.

gage

While the process capability ratio Cp for a quality characteristic xis

pLSLUSLC

6

(4-15)total

p 6

the gage capability isg g p y

gageLSLUSLC

6

(4-16)19

gageg g 6

( )

A large value of Cgage indicates that the gage error is li ibl d t th d i t l d illnegligible compared to the design tolerance, and will

not significantly influence the total error.

Suppose, in example 4-1, the part has USL = 1077 and LSL = 1002 and σ = 1 039LSL 1002, and σgage 1.039

03.1210021077

C (4-17)03.12

039.16gageC ( )

h di ib i f i l lf h fHere, the distribution range of is only one twelfth of the design tolerance. The gage error is negligible.

gage

As a generally rule, Cgage 10 often implies adequate gage capability and the gage is suitable for the measuring

20

capability, and the gage is suitable for the measuring application.

Summary

ilityreproducibityrepeatabilproduction

gageproductiontotal

ypypp

222

222

gageproductiontotal

222

ilityreproducibityrepeatabilproduction

σproduction attributable to the variability of the manufacturing process producing the part.

σrepeatability attributable to the inherent error of the gage d t th tused to measure the part.

σ attributable to the variability of the inspector21

σreproducibility attributable to the variability of the inspector measuring the part.

After the components of errors have been separated, di ti b t k t d th tcorresponding actions can be taken to reduce the most

dominant component(s).

σproduction can be reduced by improving the process and using machines with higher precisionusing machines with higher precision.

σrepeatability can be reduced by using more precise gage.repeatability y g p g g

σreproducibility can be reduced by training the inspectors or p ysetting more uniform regulation for using the gage.

22

5 ACCEPTANCE SAMPLING PLAN

A li l i d i h i i dAcceptance sampling plan is concerned with inspection and decision making regarding the lots of products. Usually, this decision is either to accept or to reject the lotdecision is either to accept or to reject the lot.

Acceptance sampling plan is used primarily for incoming or receiving inspection. But, frequently a manufacturer will apply the acceptance sampling to its own product between various t f d tistages of production.

1

The number N of units in a lot is called lot size.

If every unit in a lot is inspected, it is the 100% inspection.

If only a sample of n units (n < N) are inspected, it is the sampling inspection For example suppose lot size N =sampling inspection. For example, suppose lot size N 100 and sample size n = 10. For every lot, 10 units will be randomly selected and, then, inspected.y , , p

The sampling inspection is used widely to replace the 100% inspection, because it can save the amount of inspections, and is especially useful in destructive testing.

2

The fraction nonconforming of a lot is equal to

P = D / N(5-1)

h D i th t t l b f f i it i th l twhere, D is the total number of nonconforming units in the lot. Meanwhile, the fraction nonconforming can be estimated by

= d / n (5-2)p

where, d is the total number of nonconforming units found in a sample of size n.p

3

Since the sampling inspection makes a conclusion about the quality of the whole lot based on the result of inspecting aquality of the whole lot based on the result of inspecting a sample of only n units, it may make two types of errors.

In an example N = 500 n = 5In an example, N = 500, n = 5.

(1) Suppose a lot contains 5 nonconforming units (D = 5), p = D/N = 5/500 = 0.01, the quality is quite good.

Now, 5 units are randomly picked for sampling inspection. , y p p g pAccidentally, they may be just the 5 nonconforming units in the lot (d = 5). The estimate fraction nonconforming is . %1005/5/ˆ ndpIt must mislead the inspector to reject the lot, he may believe that the lot quality is very bad and reject the lot.

4This case implies a type of error called the producer’s risk.

(2) Suppose another lot contains 250 nonconforming units (D = 250), p = D/N = 250/500 = 0.50, the quality is quite poor.

However if the 5 units randomly picked for the samplingHowever, if the 5 units randomly picked for the sampling inspection are, by chance, all conforming ones (d = 0), the estimate fraction nonconforming is The05/0/ˆ ndpestimate fraction nonconforming is . The inspector will also be misled and believes that the lot quality is extremely good and accept the lot.

05/0/ ndp

quality is extremely good and accept the lot.

This case implies another type of error called the p ypcustomer’s risk.

5

(1)Producer’s risk: a lot is rejected hen q alit is good The(1)Producer’s risk: a lot is rejected when quality is good. The probability of this risk is denoted by α.

(2)Customer’s risk: a lot is accepted when quality is bad. The probability of this risk is denoted by β.probability of this risk is denoted by β.

The performance of a sampling plan is usually measured by α and 6

p p g p y yβ. An effective acceptance sampling plan should minimize both α and β.

5.1 Single Sampling Plan

A i l li l k j d i i b dA single sampling plan makes an accept or reject decision based on the inspection of a single sample.

Two parameters determine a single sampling plan.

Sample size: n (number of units in a sample)Sample size: n (number of units in a sample)

Acceptance number:c

Let d be the number of nonconforming units found in a sample,

If d c accept the lot;If d c, accept the lot;

If d > c, reject the lot. (5-3)

The performance of a single sampling plan is totally determined by n and c. The lot size N is not an influential factor, as long as n

7<< N.

Example 5 1 A single sampling plan (n = 89 c = 2) is usedExample 5-1 A single sampling plan (n = 89, c = 2) is used to check the incoming lots, each of them contains 1000 washers (N = 1000 N >> n)washers (N 1000, N >> n).

From each lot of 1000 units, a random sample of 89 units o eac ot o 000 u ts, a a do sa p e o 89 u tsare inspected and the number d of nonconforming units is counted. If d is less than or equal to 2, the lot is accepted; if q pd is greater than 2, the lot is rejected.

Suppose the values of d are equal to 2, 0, 3, 1, 5, respectively, for the samples from the first five lots, then l 1 2 d 4 d d l 3 d 5 j dlots 1, 2 and 4 are accepted, and lots 3 and 5 are rejected.

8

It is noted that during the use of a sampling plan the twoIt is noted that, during the use of a sampling plan, the two parameters n and c are fixed, but the number d is different from sample to sample. d is a random number and p pfollows a binomial distribution.

For given n and p (p is fraction nonconforming), the probability for d equal to an integer i is

iniini nnd

)1(!)1()( iniini

d ppini

nppi

iPid

)1(

)!(!!)1()(Pr (5-4)

9

Operating-Characteristic (OC) curve (Figure 5-1) p g ( ) ( g )

The OC curve is a very useful tool to show the overall performance characteristic of a sampling plan. It plots Paversus the fraction nonconforming p. Pa is the probability that a lot is accepted.

E l A t k d li 1000 l t f h tExample: A truck delivers 1000 lots of washers to Company A. The company carried out a single sampling inspection on each lot If 98 lots are rejected by theinspection on each lot. If 98 lots are rejected by the sampling plan,

902.01000

981000

aP

(5-5)

10

11Figure 5-1 OC curve of the single sampling plan (f13-2 p614)

Pa = Prob(a lot is accepted) = Prob(d c)

If p increases d is likely to increase

it is less likely (d c) Pa is likely to decrease.

So, if p increases, Pa decreases; if p decreases, Paincreasesincreases.

12

Under a single sampling plan (n, c), the value of Pa for a given p value is calculated as follows:

a

dbdbdbcdobP

P1P0PPr

ddd cPPPcdobdobdob

)()1()0(Pr1Pr0Pr

(5-6)

c

i

inic

id pp

ininiP

00

)1()!(!

!)(

The entire OC curve is drawn by connecting several points (Pa, p).

13

Example 5-2 For the single sampling plan where n = 89 and, c = 2.

When p = 0.01, using Eq (5-6),

)990()010(!89289 iiP

)990()010(!89)990()010(!89)990()010(!89

)99.0()01.0()!89(!

872881890

0

89

i

iia ii

P

9397.0

)99.0()01.0(!87!2

)99.0()01.0(!88!1

)99.0()01.0(!89!0

14

When p = 0.05,

)95.0()05.0()!89(!

!892

0

89

i

iia ii

P

)95.0()05.0(!87!2

!89)95.0()05.0(!88!1!89)95.0()05.0(

!89!0!89

)(872881890

0

i

1721.0

After more points of (Pa, p) are calculated and plotted, the OC curve can be drawncurve can be drawn.

Different sampling plans have different performance, and therefore, different OC curves, depending on the parameters n

15and c. Fi 5 2 OC f diff t li l

16Figure 5-2 OC curves for different sampling plans

Acceptable Quality Level AQL

AQL is a small value for p, representing a high quality level. If p = AQL, the product quality is considered good and the lots should be accepted.

For example, suppose AQL is set as 0.01. If p = AQL = 0.01, the p , pp Q p Q ,product quality is considered good and the lots should be accepted.

Producer’s risk α is the probability that a lot is rejected when quality is good. Here, “quality is good” means (p = AQL). So,q y g , q y g (p Q ) ,

α= prob(lot is rejected) | quality is good = prob(lot is rejected) | p = AQLp ( j ) | quality is good p ( j ) | p AQL

= 1 - prob(lot is accepted) | p = AQL = 1 - Pa | p = AQL (5-7)

17α is smaller when Pa | p = AQL is larger.

In Fig 5-1, AQL = 0.01,g , Q ,

, α = 1 – 0.9397 = 0.0603 = 6.03%9397.0AQLpaP

Discussion: At p = AQL = 0.01, the product quality is

AQLpa

considered good. The producer hopes that all of the lots are accepted. However, if 10000 lots are inspected by the

li l 6 03% f th (603 l t ) ill b j t dsampling plan, 6.03% of them (603 lots) will be rejected, on average.

10000 = 10000 0.0603 = 603

A large α means that the probability that lots of good quality are rejected is high. It is a risk (or loss) suffered by the

18

j g ( ) yproducer.

Rejectable Quality Level RQL

RQL is a large value of p, representing a low quality level. If p = RQL, the product quality is considered bad and the lots should be rejected.

For example, suppose RQL is set as 0.05. If p = RQL = 0.05, theFor example, suppose RQL is set as 0.05. If p RQL 0.05, the product quality is considered bad and the lots should be rejected.

Customer’s risk β is the probability that a lot is accepted whenCustomer’s risk β is the probability that a lot is accepted when quality is bad. Here, “quality is bad” means (p = RQL). So,

β= prob(lot is accepted) | quality is bad

= prob(lot is accepted) | RQL = P | RQL (5-8) prob(lot is accepted) | p = RQL Pa | p = RQL (5 8)

β is smaller when Pa | p = RQL is smaller.

19

In Fig 5-1, RQL = 0.05,

, β = 0.1721 = 17.21%1721.0RQLpaP

Discussion: At p = RQL = 0.05, the product quality is id d b d Th t h th t ll f th l tconsidered bad. The customer hopes that all of the lots are

rejected. However, if 10000 lots are inspected by the sampling plan 17 21% of them (1721 lots) will be accepted on averageplan, 17.21% of them (1721 lots) will be accepted, on average.

10000 β = 10000 0 1721 = 172110000 β 10000 0.1721 1721

A large β means that the probability that lots of bad quality are

g β p y q yaccepted is high. It is a risk (or loss) suffered by the customer.

20

F ff i li l i OC h ldFor an effective sampling plan, its OC curve should display (see Fig (5-1)):

(1) A high value. It means that lots of good quality are very likely to be accepted or the producer’s

AQLpaP

quality are very likely to be accepted, or the producer s risk α is low.

(2) Also, a low value. It means that lots of poor quality are very unlikely to be accepted, or the customer’s

RQLpaP

q y y y prisk β is low.

The values of α and β of a single sampling plan are determined by the parameters n and c.

21

(a) When n is fixed, increasing c implies that it is more likely that d ≤ c (or that, Pa = Prob(d ≤ c) becomes higher). It will reduce α, but increase β (Fig 5-3 (a)).

(b) Conversely, decreasing c will reduce β, but increase α (Fig 5-3 (b) Co ve se y, dec eas g c w educe β, but c ease α ( g 5 3(b)).

(c) When n is allowed to be increased both α and β can be reduced(c) When n is allowed to be increased, both α and β can be reduced simultaneously. But the cost of the sampling inspection will become higher (Fig 5-3 (c))become higher. (Fig 5 3 (c)).

22Figure 5-3 OC curves for different sampling plans

How is the lot formed ?

I i fl h ff i f h liIt can influence the effectiveness of the acceptance sampling plan.

(1) All lots inspected by a same sampling plan should be homogeneous (the p values for all lots are almost identical). The units should be produced by the same machines and the same operators, from the common raw materials, and approximately at th tithe same time.

An acceptance sampling plan is designed based on a specified (or predetermined) p value and works most effectively at this p level (or it produces least α and β). If lots are nonhomogeneous with different p values, the acceptance sampling plan may not function as effectively as it could.

23Furthermore, nonhomogeneous lots make it more difficult to identify and eliminate the source of the problems.

(2) Larger lots are preferred over smaller ones. It can be shown that larger lot is usually more effective in sense of αg yand β.

For example in a production line 3000 units are producedFor example, in a production line, 3000 units are produced daily. The available resource allows 300 (10%) units to be inspected. The following two schemes may be used. Bothinspected. The following two schemes may be used. Both schemes inspect 300 units per day. However, scheme 1 uses a larger lot size than scheme 2. Therefore, the former gis more effective (producing smaller α and β) than the latter.

24

(3) Lots should be comfortable to the material-handling systems.For example if each box contains 500 units it is moreFor example, if each box contains 500 units, it is more

comfortable to have a lot size of 500 (i.e., each box is a lot) than a lot size of 400.a lot size of 400.

(4) The units selected for inspection from the lot should be chosen

25at random, and should be representative of all the units in the lot.

Most of the acceptance sampling plans do not provide any directimprovement of the product quality The quality of the acceptedimprovement of the product quality. The quality of the accepted lots are no better than the quality of the rejected ones.

For example, N = 1000, n = 100 and c = 9.

Lot A: contains 120 nonconforming units, pA = 0.12

Randomly inspect 100 units, and d = 9 (accepted)

Lot B: contains 120 nonconforming units, pB = 0.12

R d l i t 100 it d d 11 ( j t d)Randomly inspect 100 units, and d = 11 (rejected)

Lot A is accepted and lot B is rejected It happens just by chance26

Lot A is accepted and lot B is rejected. It happens just by chance. In fact, lot B has the same quality level as lot A, as pA = pB.

The most effective way is to use an acceptance sampling plan as an audit tool.as an audit tool.

An effective sampling plan accepts most of the lots if the p g p pproduct quality is good, and rejects almost all lots if the product quality is poor (see Fig 5-1). By this way, “good quality” is rewarded, and “poor quality” is penalized. The supplier is under high pressure to improve and ensure the quality of his products.

27

Example: Our company uses a single sampling plan (n = 89, c = 2) to inspect the parts supplied by vendors A and B One day each ofto inspect the parts supplied by vendors A and B. One day, each of the two vendors delivers 10000 lots of parts to our company.

Th d f d A h li f 0 01 h PThe products of vendor A has a quality of p = 0.01, then Pa = 0.9397 (Fig 5-1). Consequently, majority of the lots (9397) have been accepted only 603 lots are rejected Vendor A is quitebeen accepted, only 603 lots are rejected. Vendor A is quite satisfied, because he has been rewarded for the high quality of his productsproducts.

The products of vendor B has a quality of p = 0.05, then Pa = 0 1721 A l f h l (8279) h b j d0.1721. As a result, most of the lots (8279) have been rejected, only 1721 lots are accepted. When the large amount of the rejected lots are sent back vendor B is badly disappointed He has to trylots are sent back, vendor B is badly disappointed. He has to try every way to improve the quality of his products, otherwise he cannot stay in business

28

cannot stay in business.

It is the effect of the acceptance sampling plan.

5 2 Double Sampling Plan5.2 Double Sampling Plan

In a double-sampling plan, a second sample is sometimesp g p , prequired before the lot can be sentenced (accepted or rejected).

A double-sampling plan is determined by four parameters.

nl = sample size of the first samplecl = acceptance number of the first sample

l i f th d ln2 = sample size of the second sample c2 = acceptance number for both samples (c2 > c1)

Let d1 and d2 be the numbers of the nonconforming units found in the first and second samples respectively

29

units found in the first and second samples, respectively.

The first sample of size n1 is drawn:

Case 1:If d1 c1 the lot is accepted immediately by the first sample, and the second sample is not needed.

C 2 If d > th l t i j t d i di t l b th fi tCase 2:If d1 > c2, the lot is rejected immediately by the first sample, and the second sample is not needed.

Case 3:If c1 < d1 c2, a second sample of size n2 is drawn.

After the second sample is inspected:

(i) If d1 + d2 c2, the lot is finally accepted.

30(ii) If d1 + d2 > c2, the lot is finally rejected.

F 1 d 2 l l i d d f 3For cases 1 and 2, only one sample is needed; for case 3, two samples are used.

While the first sample of size n1 must be drawn unconditionally the second sample may or may not be drawnunconditionally, the second sample may, or may not, be drawn.

It is also noted that, in the second sample, c2 is used to checkIt is also noted that, in the second sample, c2 is used to check (d1 + d2), rather than d2 only.

31

Example 5-3 nl = 50, cl = 1, n2 = 100, c2 = 3 (Note, cl < c2)p l , l , 2 , 2 ( , l 2)

Fi 5 4 D bl li l32

Figure 5-4 Double sampling plan

(1) If d1 1 the lot is accepted by the first sample The(1) If d1 1, the lot is accepted by the first sample. The second sample is not needed.

(2) If d1 > 3, the lot is rejected by the first sample. The second sample is not needed.

(3) If 1 < d1 3, the second sample of size n2 is drawn.

(i) If d1 + d2 3, the lot is finally accepted.

(ii) If d1 + d2 > 3, the lot is finally rejected.

It is noted that, after the second sample has been taken, there must be a definite sentence (accept or reject) for every lot

33

must be a definite sentence (accept or reject) for every lot.

For a given double sampling plan (with parameters n1, n2, c1 and c2), we may identify an equivalent single sampling plan (withc2), we may identify an equivalent single sampling plan (with parameters n and c) which has an OC curve very close to the OC curve of the double sampling plan. It means that the overall p g pperformance of the two equivalent sampling plans are very much the same.

It can be shown that, for a pair of equivalent double and single sampling plans,sampling plans,

n1 < n < n1 + n2 (5-9)

The potential advantage of using a double sampling plan is that it may reduce the average amount of inspections, compared to the equivalent single sampling plan. It means, while the double sampling plan performs as effectively as the equivalent single

li l th f i l i t d t th th34

sampling plan, the former is less expensive to conduct than the latter.

Figure 5 5 Equivalent double and single sampling plans35

Figure 5-5 Equivalent double and single sampling plans

For a single sampling plan, the number of inspected units in each sample is consistently equal to the sample size n.sample is consistently equal to the sample size n.

For a double sampling plan, if a lot is accepted or rejected on the first sample the number of the inspected units is equal to thefirst sample, the number of the inspected units is equal to the sample size n1 of the first sample. Since n1 < n, using the double sampling plan for this lot will save the amount of inspectionsampling plan for this lot will save the amount of inspection.

On the other hand, if a decision cannot be made on the first sample ( d l i d d) th b f th i t d it i(second sample is needed), the number of the inspected units is equal to (n1+n2). Since (n1+n2) > n, using the equivalent single sampling plan for this lot will save the amount of inspectionsampling plan for this lot will save the amount of inspection.

It means that, by using a double sampling plan, the number of the inspected units required by different lots will be different. Consequently, we may have to consider the average number of i t d it l t

36inspected units per lot.

A Story: Comparing the average incomes AI of anA Story: Comparing the average incomes, AI, of an engineer and an insurance agent.

For the engineer, monthly income is a constant, equal to $5000.

AI 5000 (5-10)

37

For the agent, monthly income is not a constant. Suppose, there are only two possible events:only two possible events:

Event I (E1): the agent earns $10000 a month. The probability of Event I is 0 3Event I is 0.3.

Event II (E2): the agent earns $2500 a month. The probability of E t II i (1 0 3)Event II is (1 - 0.3).

Therefore, the AI for the agent is

AI = (income for E1) (prob of E1) + (income for E2) (prob of E2)

= 10000 0 3 + 2500 (1 0 3) = 4750 (5 11)

= 10000 0.3 + 2500 (1 - 0.3) = 4750 (5-11)

By comparing the AI values, it can be concluded that the engineer k h h i

makes more money, on average, than the insurance agent.

38

Average Sample Number, ASN --- the average number of inspected units per lot.

For the equivalent single sampling plan, sample size n is a constant For every lot n units are inspectedconstant. For every lot, n units are inspected.

ASN n (5-12)ASN n (5-12)

39

For the double sampling plan, sometimes n1 units are inspected; and other times (n1+n2) units are inspected.

Event I (E1): a decision about a lot is made on the firstEvent I (E1): a decision about a lot is made on the first sample and the number of inspected units is n1. Let the probability of Event I be PI. p y I

Event II (E2): a decision about a lot cannot be made on the first sample (the second sample is needed) and the numberfirst sample (the second sample is needed) and the number of inspected units is (n1+n2). The probability of Event II is (1-PI). Referring to Eq (5-11). (5-13)(1 PI). Referring to Eq (5 11).

ASN = (number of inspected units for E1) (prob of E1) + (number of inspected units of E2) (prob of E2)

(5 13)

(number of inspected units of E2) (prob of E2)

)1)((211 II

PnnPn

40)1(

21 IPnn

How to determine PI ?

PI = Pr{lot is sentenced on the 1st sample}

= Pr{lot is accepted on the 1st sample} + Pr{lot is rejected on the 1st sample}{ j p } 2111 PrPr cdcd (5-14)

inic

iddd pp

inincPPPcd

11

)1()!(!

!)()1()0(Pr0 1

1111

i ini )!(!0 1

ddd cPPPcdcd )()1()0(1Pr1Pr 22121 ini

c

i

ddd

ppini

ncPPPcdcd

12

)1()!(!

!1

)()1()0(1Pr1Pr

0 1

1

22121

(5-15)

41

i ini )!(!0 1

Example 5 3Example 5-3

A single sampling plan is used to inspect a can drink TheA single sampling plan is used to inspect a can drink. The estimated defect rate (fraction nonconforming) is p = 0.02. The sample size n = 85, and the acceptance number c = 2. The sample size n 85, and the acceptance number c 2. From Eq (5-12),

ASNsingle = n = 85

An equivalent double sample plan is also designed with

50 1 100 3nl = 50, cl = 1, n2 = 100, c2 = 3

42

For the double sampling plan (see Eqs (5-13), (5-14) and (5-15)):

7358.0)02.01(02.0)!50(!

!50Pr 501

011

ii

i iicd

0177.0)02.01(02.0)!50(!

!501Pr 503

021

ii

i iicd

7535.00177.07358.0

PrPr 2111

cdcdPI

7535.00177.07358.0

)1(21 IPnnASN65.74)7535.01(10050

Compare to (ASNsingle = 85), the double-sampling plan will save the number of inspections (by 10.35 units per lot) in thi l

43this example.

The comparison in above example is conducted based on a p pgiven value of p (p = 0.02). In fact, the ASN value of the double sampling plan is a function in terms of p. Figure 5-6 compares the ASN value of a double sampling plan with the ASN values of the equivalent single sampling plan over different values of p.

(1) When p is very small, the double sampling plan has ll ASN th th i l t i l li l Bsmaller ASN than the equivalent single sampling plan. Because,

most of the lots are accepted by the first sample and, therefore, only have to inspect n unitsonly have to inspect n1 units.

(2) When p is very large the double sampling plan also has(2) When p is very large, the double sampling plan also has smaller ASN. Because, most of the lots are rejected by the first sample and, again, only have to inspect n1 units.

44

p , g , y p 1

Figure 5-6 ASN curve for single and double sampling plans45

g g p g p

(3) When p is in the middle range, the double sampling plan has larger ASN than the equivalent single sampling plan. Because, many lots are neither accepted nor rejected by the first sample, and the second sample has to be drawn. All of these lots have to inspect (n1 + n2) units.

Above discussion means that using the double sampling plan may not always save the amount of inspection It is a bettermay not always save the amount of inspection. It is a better choice only when p is either very small or very large.

But, a curtailment technique discussed below may ensure that the double sampling plan requires less amount of inspection p g p q pthan the equivalent single sampling plan.

46

Curtailment

As soon as the total number of observed nonconforming units exceeds the second acceptance number c2, the inspection of the double sampling plan is terminated and the lot is rejected. Because further inspection will not change sentence of rejection.

Example 5-4, for a double sampling plan

nl = 50, cl = 1, n2 = 100, c2 = 3

During the inspection of a lot, in the first sample, units 4, 25 and 43 are nonconforming, d1 = 3. Since c1 < d1 c2, the second sample43 are nonconforming, d1 3. Since c1 d1 c2, the second sample has to be drawn.

47In the second sample, units 2, 30, 45, 75 and 89 are nonconforming, d2 = 5.

Figure 5-7 Curtailment

For an ordinary double sampling plan

Figure 5 7 Curtailment

For an ordinary double sampling plan

Since d1 + d2 = 3 + 5 = 8 > c2, the lot is rejected. The48

Since d1 + d2 3 + 5 8 c2, the lot is rejected. The total number of inspections is equal to (n1 + n2 = 150).

If the curtailed double sampling plan is used, the first sampling proceeds as before.proceeds as before.

However in the second sampling, after unit 2 has been inspected, the total number of nonconforming units is (d + 1 = 3 + 1 = 4)the total number of nonconforming units is (d1 + 1 = 3 + 1 = 4). It is already larger than c2 and the lot must be rejected even if all of the remaining units are conforming Thus the secondof the remaining units are conforming. Thus, the second sampling is terminated immediately and the lot is sentenced to be rejected.rejected.

The total number of inspections is only equal to (n1 + 2 = 52). It is much smaller than 150 units required by the double samplingis much smaller than 150 units required by the double sampling plan without curtailment.

The curtailed double sampling plan always reaches the same conclusion (accept or reject) as the double sampling plan without

t il t B t th f i bl t l ASN i ifi tl49

curtailment. But the former is able to lower ASN significantly (see Figure 5-6).

Double sampling plan has two potential disadvantagesDouble sampling plan has two potential disadvantages.

(1)Unless the fraction nonconforming p is very small or(1)Unless the fraction nonconforming p is very small or very large (in which, most of the lots are accepted or rejected in the first sample), the double sampling plan may ejected t e st sa p e), t e doub e sa p g p a ayrequire even more inspections on average than the equivalent single sampling plan.q g p g p

However, the use of curtailment may significantly lower ASN in the double sampling plan.

(2)I i d i i i l l h d bl(2)It is administratively more complex to use the double sampling plan, compared to the single sampling plan.

50

5.3 Rectifying Inspection

Rectifying inspection will improve the final quality of theRectifying inspection will improve the final quality of the outgoing product. It is usually used for in-plant inspection, in which the manufacturer wishes to ensure a required quality levelwhich the manufacturer wishes to ensure a required quality level of the products before they are sent to the next stage.

I tif i i ti f h l t i l li i tiIn rectifying inspection, for each lot, a single sampling inspection is conducted first.

Event I:If the lot is accepted by the single sampling inspection, it is sent to the output point without any further inspection.

Event II:If the lot is rejected by the single sampling inspection, a 100% inspection is applied to the lot.p pp

All discovered nonconforming units (either in the single sampling inspection or in the 100% inspection) will be replaced by

51

inspection or in the 100% inspection) will be replaced by conforming ones.

Incoming fraction nonconforming: p0

Outgoing fraction nonconformingOutgoing fraction nonconforming (Average Outgoing Quality): AOQ

AOQ < p0 (5-16)

Figure 5-8 Rectifying inspection (f13-10 p621 modify)

52

Figure 5 8 Rectifying inspection (f13-10 p621 modify)

Average outgoing quality (AOQ)

Lot size N, incoming fraction nonconforming p0

For the single sampling inspection:

sample size n, acceptance number c.

Let Pa be the probability of acceptance in the single sampling inspection.p g p

a cdP Pr

c

iic

ddd

ncPPP

!)()1()0(

(5-17)

53

i

ini

id pp

ininiP

000

0

)1()!(!

!)(

Event I (E1): the lot is accepted by the single sampling inspection.inspection.

The n inspected units are all conforming, because all discovered nonconforming units have been replacednonconforming units have been replaced.

But, the (N–n) non-inspected units may contain nonconforming units. Since the rate of nonconforming is p0, the number of nonconforming units contained in this portion is p0(N - n).

Therefore, the average number of nonconforming units in an accepted lot is

= 0 + p0(N - n) = p0(N - n) (5-18)Id

The probability of Event I is Pa.

54

Example: if N = 10 000 n = 89 c = 2 p = 0 01Example: if N = 10,000, n = 89, c = 2, p0 = 0.01

For an accepted lot 89 units have been inspected in theFor an accepted lot, 89 units have been inspected in the single sampling plan. Any nonconforming units have been replaced by the conforming ones. So, after the single ep aced by t e co o g o es. So, a te t e s g esampling plan, there is no nonconforming units in these 89 units.

9911 (= 10000 – 89) units have not been inspected in the single sampling plan. The average number of nonconforming units in this portion is 99.11 (= 0.01 × 9911).

Consequently, in this accepted lot, the average number of nonconforming units is also equal to 99 11

55

nonconforming units is also equal to 99.11.

Event II (E2): the lot is rejected by the single sampling inspection.

The whole lot contains no nonconforming units, because every unit has been inspected either in the single sampling plan or in th 100% i ti d ll di d f i itthe 100% inspection, and all discovered nonconforming units have been replaced by the conforming ones. The average number of nonconforming units in a rejected lot is thereforenumber of nonconforming units in a rejected lot is, therefore,

= 0 (5-19)IId 0 (5 19)

The probability of Event II is (1 - Pa).

IId

p y ( a)

56

Now, combining both Events I and II, the overall average number of nonconforming units in a lot is

( f E1) ( b f E1) + ( f E2) ( b f E2)d d d = ( for E1) (prob of E1) + ( of E2) (prob of E2)d Id IId

)1(0)(0 PPnNp aa

(5-20))(0 nNpPa

So, the outgoing fraction nonconforming isnNpPd )(

(5-21)NnNpP

NdAOQ a )(0

If N is much larger than n, N – n ≈ N,

57(5-22)0

0 pPN

NpPAOQ aa

E l 5 5Example 5-5

A rectifying inspection plan is used to check the lots ofA rectifying inspection plan is used to check the lots of computer components. The incoming fraction nonconforming is: p = 0 01is: p0 0.01

N = 10,000, n = 89, c = 2.N 10,000, n 89, c 2.

)1(!Pr 00 c

ini ppncdP

)01.01(01.0)!89(!

!89

)1()!(!

Pr

289

000

ii

ia

ii

ppini

cdP

9397.0

)()!89(!0

i ii

58

Using Eq (5-21)

0093.010000

)8910000(01.09397.0)(0

N

nNpPAOQ a

Since N >> n, using Eq (5-22),

0094.001.09397.00 pPAOQ a

Comparing AOQ with p0, it can be seen that the out-going fraction nonconforming is smaller than the incoming onenonconforming is smaller than the incoming one.

The reason is that some nonconforming units have been gdiscovered during either the single sampling inspection or the 100% inspection, and have been replaced by the conforming ones.

59

AOQ is a function in terms of p (see Fig 5 9) sinceAOQ is a function in terms of p0, (see Fig 5-9), since

NnNpPAOQ a )(0

c

inia pp

ininP 00 )1(

)!(!!

(5-23)NQ

i ini0 )!(! (5-23)

(1) When p0 is very small, AOQ is also small. That means, h th i i lit i d th t iwhen the incoming quality is very good, the average outgoing

quality is very good too.

(2) When p0 is very large, AOQ is again very small. Because, when the incoming quality is very bad most of the lots are rejectedwhen the incoming quality is very bad, most of the lots are rejected by the single sampling plan and subject to 100% inspection. Then, during the 100% inspection, all nonconforming units are replaced g p , g pby the conforming ones. But, when p0 is large, the rectifying inspection is very costly, because it requires a huge amount of

60inspections and replacements.

61Figure 5-9 AOQ curve

(3) I b b h AOQ i(3) In between above two extremes, the AOQ curve rises, passes through a maximum, and then descends.

The maximum AOQ value is called the Average Outgoing Quality Limit (AOQL) which representsOutgoing Quality Limit (AOQL), which represents the worst possible average quality (or the highest possible value of AOQ) that would result from thepossible value of AOQ) that would result from the rectifying inspection. The outgoing lots are warranted to have an AOQ value no higher than AOQL, regardless Q g gof the incoming quality p0.

It is a very useful feature of the rectifying inspection.

62

For example, in a production line, it is required that the incoming fraction nonconforming to stage 2 be smaller than 0.016.g g

A rectifying inspection plan in designed: n = 89, c = 2, N = 10000. Correspondingly, AOQL = 0.0155 (Fig 5-9).

Figure 5-10 Application of AOQL63

g pp Q

This rectifying inspection is inserted between stage 1 and y g p gstage 2. Then, the outgoing quality of stage 1 is the incoming fraction nonconforming p0 for the rectifying inspection.0

Now, regardless of the value of p0, the AOQ of the rectifying inspection (or the incoming p0 of stage 2) is always smaller than 0.0155. As a result, the requirement that the incoming f ti f i t t 2 b ll th 0 016 ifraction nonconforming to stage 2 be smaller than 0.016 is satisfied.

64

5.4 Sampling Plan Standard

MIL STD 105E is a widely used acceptance sampling system (standard). It is an overall strategy specifying thesystem (standard). It is an overall strategy specifying the way in which several sampling plans are to be used together. Particularly, the standard will aid the QA g y, Qengineers to determine the parameters (e.g., n, c) of the sampling plan that will satisfy the inspection requirements.

The primary focal point of MIL STD 105E is the acceptable quality level (AQL). Or in other words, the producer’s risk α is warranted at a specified low level. But the customer’s i k β i t t d Th l f β d d thrisk β is not guaranteed. The value of β depends on the

sample size n. If the sample size is larger (i.e., more effort will be spent on the sampling inspection) the β value will

65

will be spent on the sampling inspection), the β value will be smaller. 66Figure 5-11 Inspection levels and customer’s risk β

The sample size n used in MIL STD 105E is mainly decided by the two factors: the lot size N and the inspection level.p

(1) Lot size N. The higher the N, the larger the n should be.

(2) Inspection level. It is an indication of the effort spent on the sampling inspection. The higher inspection level requires larger

l i d ill lt i ll t ’ i k βsample size, and will result in smaller customer’s risk β.

General inspection level

Level I: requires a sample size about one-half of that for inspection Level II. It is used when higher β can be tolerated.

Level II is designated as normal. g

Level III requires a sample size about twice as that for inspection Level II It is used when β must be low

67

Level II. It is used when β must be low.

S i l i i l lSpecial inspection level

S 1 S 2 S 3 and S 4S-1, S-2, S-3, and S-4.

They use very small n (even smaller than in level I) andThey use very small n (even smaller than in level I), and should be employed only when the small sample sizes are necessary and high β can be tolerated.necessary and high β can be tolerated.

68

Table 5-1 Sample size code letters

69

Table 5-1 associates a Sample Size Code (SSC) letter with the lot size N and inspection level. The higher the alphabetical order of the SSC letter, the large the sample size will be.

The alphabetical order of the SSC letter increases from top to b tt ( l ith th i f th l t i ) d f l ftbottom (along with the increase of the lot size) and from left to right (along with the increase of inspection level)

Suppose, if the lot size N = 2000 and a level II inspection is used the SSC letter is K The actual value of the sample sizeused, the SSC letter is K. The actual value of the sample size can be found based on the SSC letter from Tables 5-3 to 5-5.

70

A product can be inspected by one of the three types of sampling plans.

Normal sampling plan to be used as long as the supplier is producing the product at AQL or better.

Example: n = 125, c = 2

(i) d c, accept (ii) d > c, reject

71

Tightened sampling plan to be used when there is an g p g pindication that the product quality becomes worse.

Example: n = 125, c = 1

(i) d c, accept (ii) d > c, reject

Th ti ht d l b t ll d t

The tightened plan uses same n but smaller c compared to the normal plan. Therefore, the condition of acceptance is more stringentmore stringent.

For example if d = 2 in a sample the corresponding lot willFor example, if d 2 in a sample, the corresponding lot will be accepted if a normal sampling plan is in use; but the same lot will be rejected if a tightened sampling plan is used.

72

j g p g p

Reduced sampling plan to be used when there is anReduced sampling plan to be used when there is an indication that the product quality has been significantlyimproved. Therefore, the effort of inspection can be relaxed p , p(n is reduced).

Example: n = 50, c = 1

r = 3 (rejection number)

(i) d t(i) d c, accept

(ii) d r reject

(ii) d r, reject

73

(iii) c < d < r, (5-24)

Accept the current lot, but switch to the normal samplingAccept the current lot, but switch to the normal sampling plan

74

Switch Rules

Th li i i ill diff fThe sampling inspection system will use different types of sampling plans at different time according to the current quality level of the productslevel of the products.

The sampling inspection usually starts with the normal sampling plan.

(1)Switch from normal inspection to tightened inspection( ) p g p

When two out of five consecutive lots have been rejected on the normal inspection It is an indication thatrejected on the normal inspection. It is an indication that product quality becomes worse.

(2)S i h f i h d i i l i i(2)Switch from tightened inspection to normal inspection

When five consecutive lots are accepted on the tightened 75

p ginspection.

(3)Discontinuance of inspection.

If ten consecutive lots remain on tightened inspection (i.e., cannot be switched back to normal inspection), the entire inspection should be stopped, and actions should be taken to improve the quality of submitted lots.

Figure 5-12 Switch between normal and tightened plans76

g g p

(4)Switch normal inspection to reduced inspection, when all of the following four conditions are satisfied (an indication that product quality has been significantly improved).

(i)The preceding 10 lots have been on normal inspection, and all ofthem have been accepted.

(ii)The total number of nonconforming units in the preceding 10samples is less than or equal to a specified limit Dsamples is less than or equal to a specified limit D.

(iii)Production is at a steady condition; that is no problems such(iii)Production is at a steady condition; that is, no problems such as machine breakdowns or material shortages.

(iv)Reduced inspection is considered desirable by the authorityresponsible for the sampling inspection.

77

Example for condition (ii): for a normal sampling planExample for condition (ii): for a normal sampling plan,

n = 89, c = 2,

In addition, D = 13 is specified as a limit for the sum of the nonconforming units found in the preceding 10 samples.

It is found from Table 5-2 that, all of the 10 lots are accepted, but, the total number of nonconforming units in the 10

l f th 10 l t i l t 15 l th D Th tsamples from these 10 lots is equal to 15, larger than D. That means, condition (ii) has not been satisfied.

Consequently, the sampling inspection is not allowed to be switched to the reduced sampling plan even if all other three

78

switched to the reduced sampling plan, even if all other three conditions are satisfied.

Table 5-2 Number of nonconforming units in the samples of 10 lots

79

(5) Switch from reduced inspection to normal inspection, when any of the following four conditions is met.

(i)A l t i j t d(i)A lot is rejected.

(ii)Production is irregular or delayed(ii)Production is irregular or delayed.

(iii)When c < d < r(iii)When c < d < r,

Referring to Eq (5-24), the current lot is accepted, g q ( ), p ,but normal inspection is reinstituted starting with the next lot.

(iv)Other conditions warrant that normal inspection be

80instituted.

Figure 5-13 Switch rules for three types of inspections81

g yp p

Example 5-6 A step-by-step procedure of designing and using a MIL STD l05E acceptance sampling system for an electronic product.

(1)Specify Acceptable Quality Level AQL.[AQL = 0.65%]

(2)Ch th i ti l l [l l II](2)Choose the inspection level. [level II]

(3)Decide the lot size N [N = 2000](3)Decide the lot size N. [N = 2000]

(4)Find the appropriate SSC letter(4)Find the appropriate SSC letter from Table 5-1. [K]

(5)Decide the appropriate type (single or double) of sampling plan. [single]

82(6)Check the appropriate tables [Tables 5-3, 5-4, 5-5]

(7)Determine the parameters of the sampling inspection plans.[[

Normal: n = 125, c = 2.

Tightened: n= 125, c = 1.

Reduced: n = 50, c = 1, r = 3.]]

Note: on Tables 5-3, 5-4 and 5-5, AQL = 0.65% is displayed as , , Q p y0.65

(8)Carry out the sampling inspections by following the switch rules in Figure 5-13.

83

Table 5 3 Table for normal inspectionTable 5-3 Table for normal inspection

84

T bl 5 4 T bl f i h d i iTable 5-4 Table for tightened inspection

85

Table 5 5 Table for reduced inspectionTable 5-5 Table for reduced inspection

86