Manufacturing Systems Technology

Instructor: Shantanu Bhattacharya

Review of previous lecture

• Setting up the forms for recording the data.• Checklist necessary for X and R charts.• Starting the control charts and plotting both the charts.• Conclusions drawn from the chart.• Possible relationships of a process in control to upper and lower

control limit.• Possible relationship of a process in control to a single specification.• Milling of a slot in an aircraft terminal block.

Revision of theory of ProbabilityDefinition:•Probability is concerned with the likelihood of an event occurring. The scale of probability of an event varies from 0 to1.•If an event cannot occur on a trial, then the probability of its occurrence is 0. •If another event is certain to occur then its probability of occurrence is 1.

•As an example suppose that a trial is drawing a piece at random from some production line, and that the event in question is that the piece drawn is a defective or non conforming one.•Let us suppose that the probability of a defective is 0.05. This means that 5% of the time when we draw a random piece from the line, it is defective.•The complementary event is that the piece drawn is a good one. It’s probability is .95.P(good or defective) = 1Occurrence ratio•Suppose that we have a production process for which the probability of a piece containing at least one “ minor defect” is constantly 0.08. •We then say that the probability is 0.08 for a “minor defective”.•Now what happens to the observed proportion of minor defectives as we continue to sample? This observed proportion of minor defectives is what we know as the occurrence ratio.

Theory of probability•Let p’ = constant probability of a minor defective

Where the letter ‘p’ means the probability and the prime means population probability.

d = no. of defectives observedn = no. of pieces inspected or testedp = d/n = sample proportion of defectives.

Let us see how p = d/n behave as we sample more and more, that is, increase ‘n’?

Would we not expect that the observed occurrence ratio p= d/n would tend to approach p’ = 0.08.

We start with five samples of 10, then samples of 50.

The first 2 columns are for current sample of 10 or of 50.

The 3rd and the 4th column are for the total sample size and the cumulative total no. of defectives.

The 5th column is based on the 3rd and 4th columns and gives the current overall proportion defectives and the occurrence ratio (total defective/ total inspected).

Sample Totals Occurrence ratio

n D Σn Σd P = Σd/ Σn

10 0 10 0 .0000

10 1 20 1 .0050

10 1 30 2 .0667

10 1 40 3 .0750

10 0 50 3 .0600

50 5 100 8 .0800

50 4 150 12 .0800

50 6 200 18 .0900

50 4 250 22 .0880

50 8 300 30 .1000

50 1 350 31 .0886

50 3 400 34 .0850

50 5 450 39 .0867

50 3 500 42 .0840

50 5 550 47 .0855

50 5 600 52 .0867

50 5 650 57 .0877

50 4 700 61 .0871

50 3 750 64 .0853

50 6 800 70 .0875

50 4 850 74 .0871

50 7 900 81 .0900

50 4 950 85 .0895

50 4 1000 89 .0890

The proportion defective only tends to approach p’ =0.08.

Sometimes it gets closer, sometimes it backs away from p’.

Before the total sample size was 100, the occurrence ratio was below .08.

Between 100 and 150 it was 0.08 and thereafter above.

Principle:

We can say that the p= d/n is an estimate of the constant probability of population p’. How close the estimate will be depends on sample size n, the value of p’ and also upon chance.

Probability laws•Consider again the production line producing pieces with a constant probability .08 of the piece being a minor defective, and such that each piece is independent of the other produced.

•This means that the probability of the next piece being a defective is 0.08 and the piece being good is .92 irrespective of the preceding piece.

•Now let us suppose that we draw a sample of two pieces and inspect them.

•The outcomes are that the sample may contain 0, 1 or 2 defectives. Let us find the probabilities of this outcome or events.

•For the probability of the samples containing no defectives, we must have good pieces on both draws.

Probability Laws

Probability Laws•If 2 events A and B might occur on a trial or experiment, but the occurrence of either one prevents the occurrence of the other, then events A and B are called mutually exclusive.

•For two such events P(A) + P (B) = P( A or B, mutually exclusive events)

•We have seen this in the earlier example in the 1 good case P (1 Good) = .0736 +0.0736

•If one of the two events A and A is certain to occur on a trial, but both cannot simultaneously occur, then A and A are called complementary events.

•For any such pair of events

• We have seen this in the example of a single draw where p’ was given as 0.08.

Laws of Probability•Two events A and B are independent if the occurrence or non occurrence of A does not affect the probability of B occuring.

•Whenever a process produces defectives independently, or at random, so that the probability of a defective on the next piece does not depend upon what the preceding pieces were like, then we have the case of independent events. Such a process is said to be in control, that is stable, even though some non conformity is produced.

•Not all processes do behave in this manner.•For example: We consider the production of 3000 piston ring castings. The sample of 100 contain 25 defectives whereas the remaining 2900 only 4. This was because the defectives occur in bunches from a certain defect producing condition.•Particularly in this case it was found that the castings were made from stacks of molds, and if the iron is not hot enough when poured into a stack, many castings may be defective. Under such conditions, whether a piece is good or defective does have an influence on the probability of the next one being defective.

•If two events A and B are independent, then we have :P(Both A and B occurring) = P(A). P(B)

Example of Dependence and Equal Likelihood

•As a second example of probability, let us consider drawing without replacement from a lot of N=6 speedometers, of which 1 is defective

•Let N = no. of pieces in one lot and D= no. of defective pieces in a lot

•Now consider the very simple case in which we just draw a random sample of 1 from a lot of 6. Random means that each of the six meters is equally likely to be chosen for the sample. Probability of each is 1/6.

•There are only two kind of meters “good “ and “defective” with P(good) = 5/6 and P(defective) = 1/6.

•Now next consider drawing a sample of n= 2 from the lot having N =6, of which D=1 is defective. This may contain no. of defectives either d=0 or 1.

•This is a case of two consecutive drawings which are not independent. Take first the case of the sample yielding no defectives, that is, two good meters. We needP(2 good) = P(good, good) = P(good on first draw). P(good on 2nd draw given good on 1st draw) = 5/6. 4/5 = 2/3

Counting samples (Permutations and combinations)•In combinations we consider for example we have “n” objects which we can distinguish between.•Now how many distinct samples, each of one, can we draw from a lot of N=10? Obviously the answer is 10. So, we call this a combination of N objects taken 1 at a time, or in symbols :

C(N,1) = N.

•Next consider samples of two, from say four good pieces (g1, g2, g3 and g4).•Then the number of distinct unordered samples may be found from the number of distinct ordered samples. For example: for ordered samplesg1g2, g2g1, g1g3, g3g1, g1g4, g4g1, g2g3, g3g2, g2g4, g4g2, g3g4, g4g3

•The no. of unordered samples are only half as much , that is, six, because, for example, the one unordered pair g2g4 corresponds to two ordered pairs g2g4 and g4g2.

•Now let us consider lots of 10 distinct pieces. The number of possible ordered samples, each of two is 10.9, because there are 10 choices for the first piece and having made a choice their remain 9 choices for the second piece. So we have 90 ordered samples and exactly ½ of this unordered. (45)

Counting samples (Permutations and combinations)•Now let us go to a sample of 3 from a lot of 10. The no. of distinct ordered samples is 10.9.8; 10 choices for the first, 9 choices for the second and 8 choices for the third. But now six of this ordered samples correspond to just one unordered sample. For example:g1g4g6, g1g6g4, g4g1g6, g4g6g1, g6g1g4, g6g4g1 all correspond to g1g4g6 unordered sample.

•Hence the number of distinct or unordered samples or combinations is 10.9.8/6 = 120

•Ordered samples are also called permutations and they are calculated by the general formulaeP(n,r) = n! / (n-r)! In the earlier case P(10,3) = 10!/ 7! = 10.9.8 = 720

•We call the number of distinct unordered samples a combination and is given by the genera formulae C(n,r) = n!/ r! (n-r)! In earlier case this would be = 10!/ 3!. 7! = 10.9.8/ 3.2.1 = 120.

Counted data: Defects and defectives• Inspecting or testing ‘n’ pieces , we may search for defects or non conformances in the

‘n’ pieces and record the total number of such defects. This is measuring quality by a count of defects.

•In inspecting or testing ‘n’ pieces, we may consider whether each of ‘n’ pieces does contain any defects. Each piece having one or more defects is called a defective. This measure is known as the count of defectives.

•We shall consistently use the following symbols here and in later discussions:

n = number of units in a sample.d = number of defective units in the sample of n units.p = d/n= sample of fraction defective = proportion of defective units in the sample.q= 1-p = sample fraction goodd= np = no. of defective units in the sample.

Binomial distribution for defectives:

Suppose we had a process with population fraction defective of p’ = .10. For a sample size of 4 there could be

Either 0, 1, 2, 3 or 4 defectives can exist.

Binomial distribution for defectives• First find the probability of drawing a sample with all pieces good.

So for P(d=0) = P(4 good) = [P(good)]4 = (0.9)4 = .6561

Thus about 2/3 of the time, we draw a sample of n=4 pieces from the process, all four will be good ones.

•Now next we seek P (3 good, 1 defective) .

P (3 good, 1 defective) = P(g,g,g,d) + P(g,g,d,g) + P (g,d,g,g) + P (d,g,g,g) = (.9)(.9)(.9)(.1) + (.9) (.9)(.1)(.9) + (.9) (.1)(.9)(.9) + (.1) (.9)(.9)(.9) = 4(.9)^3 (.1) = .2916.

•Next consider samples with d=2; they have two defectives and two good ones. How many distinct orders are there for such samples? Six: ggdd. gdgd, gddg, dgdg, ddgg, dggd, the probability for each one of these sample outcomes is (0.9)2(0.1)2 . P(d=2) = 6 (0.9)2(0.1)2 = .0486

•Similarly for d=3 there are only 4 orders of sampling results P(d=3) = 4 (0.9)2(0.1)2 =.0036

•Finally for d =4, all four must be defective•P(d=4) = .0001

Binomial distribution for Defectives

• The sum of all the 4 outcomes are 1.

•We can also represent the various coefficients viz, 4,6,4 of the products of the powers of .9 and .1 as

C(4,1) =4, C(4,2) = 6, C(4,3)= 4 respectively.

•This reasoning enables us to write all the four probabilities in 1 formulae

P(d) = C(4,d) (0.9)4-d(0.1)d

•This is also a representative of the dth term of a Binomial distribution of n =4 and the p =0.9 and q= 0.1.•In generalP(d) = C(n,d) (p’)n-d(q’)d

d P(d) P= d/n

0 .6561 .00

1 .2916 .25

2 .0486 .50

3 .0036 .75

4 .0001 1.00

Total 1.0000

Mean and standard Deviation of a Binomial distribution

•Where many sets of trial are made of an event with constant probability of occurrence ‘p’, the expected average number of occurences in the long run is ‘np’.

•So ‘np’ is the expected value, or mathematical expectation, of x where x=0 if an item is acceptable or x=1 if it is rejectable.

•Mathematical expectation is an operator expressed as follows:

An experimental example of the meaning of average and standard deviation of Binomial distribution

No. of non –conforming items

Frequency

4 3

3 7

2 9

1 16

0 5

Total 40

•The table on the left may be used to clarify certain aspects of the mean and standard deviation of the Binomial distribution.•Column on the extreme left gives the number of non conforming items in each sample of 10.•The result of 40 samples may be arranged into a frequency distribution .•The average calculated by normal method results in 1.675 and the standard deviation by a similar method result in 1.127.•If we use the formulae for the mean and deviation of a binomial distribution the expected average

• The expected standard deviation

•So, the observed values seem fairly close to the expected value.

Control charts for attributes

There are several different types of control charts that may be used. These are the following:1. The p chart, the chart for fraction rejected as non conforming to specifications.2. The np chart, the control chart for number of non conforming items.3. The c chart, the control chart for non conformities.4. The u chart, the control chart corresponding to the non conformities per unit.

Control charts for fraction rejected

•The most versatile and widely used attributes control chart is the ‘p’ chart. This is the chart for the fraction rejected as non conforming to specifications. The principle advantage that these charts provide is that only 1 chart is sufficient to describe 10’s, 100’s or 1000’s of quality characteristics by classifying the item as accepted or rejected.

Control limits for the p-chart

Problems introduced by variable subgroup size•One practical difference between the X chart and p chart is the variability in subgroup size in the p chart.•P charts typically use data taken for other purposes than the control chart; where subgroups consist of daily or weekly production and tend to vary a lot.•Whenever subgroup size is expected to vary, a decision must be made as to the way in which control limits are to be shown on the p chart. There are three solutions to all these problems.

1. Compute new control limits for every subgroup, and show these fluctuating limits on the control chart. This is illustrated in the first 2 months of the 4 months.

2. Estimate the average subgroup size for the immediate future. Compute one set of limits for this average, and draw them on the control chart. Whenever the actual subgroup size is substantially different from this estimated average, separate limits may be computed for individual subgroups.

3. Draw several sets of control limits on the chart corresponding to different subgroup sizes. A good plan is to use three sets of limits, one for expected average, one close to the expected minimum and one close to the expected maximum.

•This example shows a 4 month record of daily 100% inspection of a single critical quality characteristics of a part of an electrical device.•When after a change in design, the production of this part was started early in June, the daily fraction rejected was computed and plotted on a chart.•At the end of the month the average fraction rejected P was computed.•Trial control limits were computed for each point. A standard value of fraction rejected p0 was then established to apply to future production.•During July new control limits were computed and plotted daily on the basis of the no. of parts ‘n’ inspected during the day.

Table 1

•A single set of control limits was established for August, based on the estimated average daily production.•At the end of August, a revised p0 was computed to apply to September, and the control chart was continued during September with this revised value.

Calculation of the trial control limits:•Table 1 shows the number inspected and no. rejected on each day. The fraction rejected on each day is the no. of parts rejected each day divided by the number inspected that day.•For example: on June 6th, Pi = 31/3350 = 0.0093.•At the end of the month the average P is computed

•The standard deviation is calculated on the basis of this observed value P

Table 2

• Determination of P0: If all the points fall within the trial control limits, the standard value P0 may be assumed to be equal to P.•Here point fell outside the control limit. Thus a revised value of P is calculated.•With the days June 7, 12, 13 and 22 eliminated, the remainder no. of rejects is 290 and no. inspected is 46399. The revised P0 = 0.0063.•After considering this and the previous record on similar parts of slightly different design, it was decided to assume p0 = 0.0065.•Table 2 is calculated on the basis of p0.

Table 3

Establishment of control limit based on expected average subgroup size:

•Although the correct position of 3-sigma control limits on a p chart depends on subgroup size, the calculation of new limits for each new subgroup consumes some time and effort.•Where the variation in subgroup size is not too great , the calculation of each new subgroup consumes time and effort.•Where the variation of subgroup size is not more than 25%, it may be good enough for practical purposes to establish a single set of control limits based on the expected average subgroup size.•For the month of august (Table 3) the estimated average daily production was 2600.

•Table ‘a’ is based on 33 years of records for 10 rainfall stations widely scattered throughout the mid-western United States.•It gives a number of 10-min periods in a year having half an inch or more of rainfall, c is the no. of such cloudbursts in a station year ; the frequency is the no. of station years having respectively 0,1,2,3,4, and 5 such excessive rainstorms.•The average no. of storms per station year is 1.2.•Table ‘b’ represents the number of cavaliers killed by fatal kick of horses in each of the 14 cavalry corps in each of the 20 successive years. Here ‘c’ is the no. of Prussians killed in this way in a corps year. •The frequency is the no. of corps year having exactly c cavalry men killed

•Table ‘c’ is based on the n. of articles turned in per day to the lost and found bureau of a large office building. The frequency is the number of days with exactly ‘c’ lost articles turned in.•Table ‘d’ shows the vacancies in the United States Supreme Court, either by death or resignation of members, from 1837 to 1932. The frequency is the no. of years in which there were exactly ‘c’ vacancies.•Table ‘e’ shows the no. of telephone calls per 5-min interval in a group of six coin-box telephones in a large railway terminal. The data were taken for the period from noon to 2 P.M. for seven days. The frequency no. is the number of 5-min. intervals in which exactly ‘c’ calls were made.

Importance of Poisson’s Distribution•At first impression, these six Poisson illustrations may seem to have very little similarity to one another.•If we look at all these more critically, they all have definite characteristics in common. In each a count was made of the no. of occurrences of an event that had many opportunities to occur but was extremely unlikely to occur at any given opportunity.•For example there were many 10-min periods in a year; it was unlikely that any particular one would bring a cloudburst of half an inch or more of rain.•Similarly, there were many contacts between a cavalryman and a horse during a year of history of a cavalry corps; it was unlikely that the horse would make of a fatal kick at any particular contact.•Similarly, there were many people passing through the office building throughout the day; it was unlikely that any one person would find a lost article and turn it in the lost an found section .•So, in all these cases, there is a concept of the existence of a large ‘n’ and a small ‘p’, even though it may be impossible to assign a definite value to either ‘n’ or ‘p’. Even in cases where it is not impossible to assign a definite value to either ‘n ‘ or ‘p’, it may simply be of no advantage to determine ‘n’ and ‘p’ and to calculate Binomial based on them. •So, Poisson’s is definitely an advantage in such cases.

The Poisson law as a probability distribution

•Certain types of frequency distributions occur in nature, both in quality control work and elsewhere, that are closely fitted by a formulae known as Poisson’s law.•If ‘c’ is the count of the occurrences of some event of interest, and μc is the parametric value of the rate of occurrence, then the Poisson’s probability density function may be stated as :

The Poisson’s Distribution as an approximation to Binomial

•Calculation involving the use of binomials are often burdensome; this is particularly true if many terms are involved and if ‘n’ is large.•Fortunately, a simple approximation may be obtained to any term of the binomial. •This approximation is often called Poisson’s exponential binomial limit. The larger the value of ‘n’ and the smaller the value of ‘p’, the closer the Poisson approximation.•The Poisson formula may be derived from the binomial theorem in the following manner. In the Binomial formula, let c=np, then

Use of tables and computer programs for solution of Poisson Problems

Standard tables provide summation terms of the Poisson’s c.d.f. upto three decimal places.

Individual Poisson’s terms must be obtained by subtracting terms of the adjacent summation terms. Thus

Average and standard deviation of the Poisson

•The average and standard deviation of the Poisson distribution are respectively.

• The table on the right provides the numbers of errors of alignment observed at final inspection of a certain model of airplane. The alignment errors observed on each airplane constitute one subgroup for this chart.

• The total no. of alignment errors in the first 25 planes was 200. The average C is 200/25 =8.0. The trial control limits computed from this average are as follows:

UCL = C + 3[C] * ½ = 16.5

LCL = C - 3[C] * ½ = negative, therefore no LCL.• Whenever, calculations give a negative value of

the lower control limit of a control chart for attributes, no lower control limit is used. In effect, the chart will exhibit only an upper control limit.

• As none of the first 25 points on this chart is outside the trial control limits based on these points, the standard number of defects C0 may be taken as equal to C and the control chart continued for the following period with a central line of 8.0 and an UCL of 16.5.

• One point (airplane no. 236) out of the next 25 is above the upper control limit.

• The average during this period was 236/25 =9.44 (even omitting the out-of-control value, the average is 9.08).

• Of the final 16 points corresponding to airplanes 235 to 250, 12 are above the standard c0 , 3 are exactly at the standard, and only 1 is below.

• It seems that there has been a slight but definite deterioration in quality during this period.

Plot of the C-chart

Normal curve as an estimate to Binomial distribution

•One of the several common derivations of the normal curve is as a limit of the binomial distribution as ‘n’ is increased indefinitely.•The greater the value of ‘n’ the better the estimate of the area that can be made from a normal curve area table.•One important point to be mentioned is that the normal curve is a symmetrical distribution.•The binomial is symmetric only in special cases where p =1/2.•So, for a given value of ‘n’ the normal curve is able to predict accurately the binomial distribution of p value close to ½.