We have seen that X and R charts can be used when quality can be measured in numbers, the p-chart can be tried to attain control when the quality is classified by defectives and non-defectives. But there are situations in which quality is given as number of defects per unit, e.g., the number of surface defects in a painted area or photographic film, the number of seeds in a glass bottle etc. The argument could be extended even to those manufactured parts which have large number of dimensions to be measured and part can’t be just rejected owing to only one dimension being out of order; under such condition the out of order dimensions can be expressed as number of defects over the number of dimensions measured.
In industry, no product can be absolutely perfect and flaws are bound to occur, though rarely. Our aim is to control and keep even these rare occurrences at a minimum possible. The c-chart techniques help us to keep these defects per unit at the lowest level. In the language of probability, it could be said that the opportunities for the occurrence of a defect are infinite: but the probability of occurrence at some chosen spot or time is very small (infinitesimal) and as a consequence a small (finite) number of defects actually occur. The methods to study such situations have been evolved by statistics, according to which the fluctuation from a standard (or a norm) are described what is known as a Poisson distribution if the following conditions are satisfied.
(a)There should be infinite opportunities for occurrence, with a low (but steady) probability of occurrence in an individual case, and a few occurrences do actually take place.
(b)The samples are chosen by random selection and are independent of one another.
( c )The choice of unit area (area of opportunities) should be the same from sample to sample.
In most of the cases, the above conditions are satisfied and where the’ ‘area of opportunity” varies, the use of c/n -chart can also be made.
According to the Poisson Law of Distribution, (which states that if m is the “standard” defects per unit, the sampling fluctuations can take a value x in the long run with the proportional frequency equal to f(x) =e−m mx ∕ x¡ (for x=1,2, …….α) the defects per unit have to be integers in actual counting of defects in an experiment, and that a large value of x can occur but has a low probability.
The various properties of the Poisson (frequency) distribution are summed up below:
(i) The mean of the distribution is equal to the “standard” defects per unit(m).
(ii) The standard deviation of the distribution = √mean = √m.
(iii) For large values of m, the Poisson distribution approaches the normal curve pattern. (For normal work if m ≥ 3, normal approximation for the Poisson distribution could be used but it is more satisfactory when m ≥ 6.)
(iv) Poisson distribution is completely specified by one parameter, i.e., m.
When the 3 conditions (a, b and c) given above are satisfied, the defects per unit (c) fluctuate in the Poisson Law of Pattern. Now if m is large Poisson Law of Pattern approaches normal curve pattern of variation and ± 3√m would contain most of the possible fluctuations unless an assignable cause affects our standard. If the sample size (area of opportunity) is kept constant, the values of m and 3√m can be used in the construction of c-chart. In actual practice the value of m is not known, which can be estimated by c, where
C= Number of defects in all units inspected ∕Total number of units inspected
So in c-chart. Central line = c
U.C.L. =c + 3√c
L.CL. = c-3√c
For c-chart the sample size may be set at one single unit of product e.g., 50 meters of rope, a dozen steel containers, a pair of chappals etc. In other words, the sample size should be a fixed number of units of product It may be noted that c-chart controls variation in number of defects per sample.
Where sub-group consists of n units, U-chart is used where
U=c ∕n and U Σc ∕ Σn
Central line =U
and U.C.L =U +3 √U ∕ √ n
and L.C.L =U−3 √U ∕ √ n
Example 19.20. The following table shows the number of defects observed on 25 similar big castings:
|Casting No.||No. of defects observed||Casting No.||No. of defects observed|
Find c and compute the trial control limits. What value of c’ can be suggested for subsequent period ?
Solution. Total No. of defects observed on 25 castings (added all the defects) = 328
.’. c= 328 ∕ 25 =13.12
Now 3√c=3√13.12 = 3×3.62= 10.86
.’. U.C.L. = c + 3√c = 13.12+ 10.58 = 23.98
and L.C.L. = 13.12-10.89 = 2.26
Observing the data it is seen that only casting No. 24 lies outside the U.C.L.
Eliminating it, the suggested value of c’ = 302 ∕ 24 = 12.58.
Example 19.21. Six hundred bicycle mudguards were observed for point defects-air bubble, scratch, foreign particle, etc. In all 1200 defects were found in 600 mudguards. What type of control chart be used here ? If sample size 10, what will be the limits.
Solution. For this application, since defects are observed in a samples of 10, U-chart will be used.
U= 200 ∕ 600 = 2
U.C.L. =U+ 3 σu =U +3 √U ∕ √ n= 2 +3 √2 ∕ √10=2 +3 ∕10 √20
=2 + 0.3 × 4.47 = 2 + 1.441 = 3.441
and L.C.L. = U −3 √U ∕ √ n = 2- 1.441 = 0.559.
Example 19.22. (a) What is the difference between a defect and defective? (b) The following tabulation gives defects observed in an aircraft sub-assembly operation and shows the number of units produced on each day during a week. Prepare an U-chart for this. Assume U’ = 4.2.
Solution, (a) A defective is an article that in some way fails to conform to one or more given specifications; whereas each instance of article’s lack of conformity to specifications is a defect. Obviously a defective can contain one or more defects, (b) The value of U .ie., control line on the control chart = 4.2. Control chart limits are computed as follows:
|Day||No. of defect||Unit producedn||Defects per unitu=c-|
|Defects per unit3a=3Vtf||U.CL. = i/’ + 3a||L.CJL. = U’-3a|
Example 19.23. How can you adapt c-chart for quality rating ? (Demerit Control Chart).
Solution. For some products, all types of defects are not of equal importance ; and to plot the sum of all defects, regardless of type, on one control chart is misleading and uneconomical. In such cases where some defects are more serious than others, it is advantageous to weigh defects according to some scale that measures their seriousness.
In such cases, the defects are classified into certain groups. One such classification is as follows :
Class A Defects- Critical (very serious).
Will render product totally unfit for service; useless or unsaleable.
Liable to cause personal injury or damage.
Class B Defects- Major (serious).
Will cause considerable consumer dissatisfaction, will shorten the life of the product, lessen its efficiency, or require exclusive servicing.
Class C Defects- Moderately serious.
Will have the same effects as those of class B, but to a lesser degree. It may include major defects ol appearance, finish or workmanship.
Class D Defects- Minor (Not serious).
These will not cause operating failure of unit in service. These include defects of appearance, finish or workmanship.
Each class of defect is given some weightage which will ordinarily be somewhat arbitrary, but is supposed to reflect the economic consequences of failure to correct each type of defect. The standard deviation is given by √Σ1rXi2Ci , where Xi is the weight assigned and c, is the expected number of defects per sample and r is the number of classes of defects. With some product, it may be required to have a single index of its product quality. Such a quality index may be obtained from use of demerit-per-unit data as follows :
Quality Index= Observed demerits per unit ∕ Expected demerits per unit
Example 19.24. Four types of defects are classified in a telephone assembly industry. The weightage to each defect is shown below. A sample of 25 units is inspected and number of defects of each class observed is tabulated below. Determine the value of U.C.L. and L.CL. for demerit control chart.
Class of defect Weightage x, No. of defects £ a
1 0.75 5
2 0.60 15
3 0.20 75
4 0.05 50
Solution. The control limits are calculated as under:
Std. deviation = √Σi4 ×Xi2 Ci= √0.455=0.675
Central tine should be placed a t Σ Xi Ci = 1.21
.’. U.CL. = 1.21 + 3 × 0.675 = 1.21 + 2.025 = 3.235
and L.CL. = 1.21 – 3 × 0.675 = 0. Ans.
|Class of defect||WeightageX,||No. of defects Defects for lc( Ci||Demerits per sample Xi Ci||Variance Xi2 C,|