Manufacturing Technology (ME461) Lecture24

8/12/2019 Manufacturing Technology (ME461) Lecture24

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Manufacturing Technology

(ME461)

Instructor: Shantanu Bhattacharya


2/15

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 toupper and lower control limit.

Possible relationship of a process in control to asingle specification.

Milling of a slot in an aircraft terminal block.


3/15

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. Its probability is .95.

P(good or defective) = 1

Occurrence 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.


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T eory o pro a i ityLet 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 tested

p = 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 3rdand the 4thcolumn are for the total sample size and the cumulative total no. of

defectives.

The 5thcolumn is based on the 3rdand 4thcolumns and gives the current overall proportion

defectives and the occurrence ratio (total defective/ total inspected).

S O


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Samp e Tota s Occurrence

ratio

n D n d P = d/ n

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, thevalue of p and also upon chance.


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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.


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Probability Laws


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Probability LawsIf 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 cannotsimultaneously 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.


9/15

Laws of ProbabilityTwo 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 theprobability 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)


10/15

Example of Dependence and Equal

LikelihoodAs 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 thesample. 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=1isdefective. 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 need

P(2 good) = P(good, good) = P(good on first draw). P(good on 2nddraw given good on 1st

draw) = 5/6. 4/5= 2/3


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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 samples

g1g2, 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 of this unordered. (45)


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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

formulae

P(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 thegenera formulae C(n,r) = n!/ r! (n-r)! In earlier case this would be = 10!/ 3!. 7! = 10.9.8/

3.2.1 = 120.


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Counted data: Defects and defectivesInspecting 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 good

d= 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.

i i l di ib i f d f i


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Binomial distribution for defectivesFirst 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


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Binomial distribution for DefectivesThe 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 dthterm of a Binomial distribution of n =4 and the p =0.9

and q= 0.1.

In general

P(d) = C(n,d) (p)n-d

(q)d

d P(d) P= d/n

0 .6561 .001 .2916 .25

2 .0486 .50

3 .0036 .75

4 .0001 1.00

Total 1.0000

Manufacturing Technology (ME461) Lecture24

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Transcript of Manufacturing Technology (ME461) Lecture24