Manufacturing Technology (ME461) Lecture24
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Transcript of Manufacturing Technology (ME461) Lecture24
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Manufacturing Technology
(ME461)
Instructor: Shantanu Bhattacharya
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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.
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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.
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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)
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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