Cusum Control Charts

17.6 Cusum Control Charts

The disadvantage of the Shewhart-type control charts, developed and illustrated in the preceding sections, lies in their inability to detect small changes in the mean. A quality control mechanism that has received considerable attention in the statistics literature and usage in industry is the cumulative sum (cusum) chart. The method for the cusum chart is simple and its appeal is intuitive. It should become obvious to the reader why it is more responsive to small changes in the mean. Consider a control chart for the mean with a reference level established at value

W . Consider particular observations X 1 ,X 2 ,...,X r . The first r cusums are

S 1 =X 1 −W S 2 =S 1 + (X 2 −W) S 3 =S 2 + (X 3 −W)

. .. S r =S r−1 + (X r − W ).

It becomes clear that the cusum is merely the accumulation of differences from the reference level. That is,

The cusum chart is, then, a plot of S k against time. Suppose that we consider the reference level W to be an acceptable value of the mean μ. Clearly, if there is no shift in μ, the cusum chart should be approximately horizontal, with some minor fluctuations balanced around zero. Now, if there is only a moderate change in the mean, a relatively large change in the slope of the cusum chart should result, since each new observation has a chance of contributing

a shift and the measure being plotted is accumulating these shifts. Of course, the signal that the mean has shifted lies in the nature of the slope of the cusum chart. The purpose of the chart is to detect changes that are moving away from the reference level. A nonzero slope (in either direction) represents a change away from the reference level. A positive slope indicates an increase in the mean above the reference level, while a negative slope signals a decrease.

Cusum charts are often devised with a defined acceptable quality level (AQL) and rejectable quality level (RQL) preestablished by the user. Both represent values of the mean. These may be viewed as playing roles somewhat similar to those of the null and alternative mean of hypothesis testing. Consider a situation where the analyst hopes to detect an increase in the value of the process mean. We shall

use the notation μ 0 for AQL and μ 1 for RQL and let μ 1 >μ 0 . The reference level is now set at

W=

The values of S r (r = 1, 2, . . . .) will have a negative slope if the process mean is at

μ 0 and a positive slope if the process mean is at μ 1 .

706 Chapter 17 Statistical Quality Control

Decision Rule for Cusum Charts

As indicated earlier, the slope of the cusum chart provides the signal for action by the quality control analyst. The decision rule calls for action if, at the rth sampling period,

d r > h,

where h is a prespecified value called the length of the decision interval and

d r =S r − min S i .

1≤i≤r−1

In other words, action is taken if the data reveal that the current cusum value exceeds by a specified amount the previous smallest cusum value.

A modification in the mechanics described above makes employing the method easier. We have described a procedure that plots the cusums and computes differ- ences. A simple modification involves plotting the differences directly and allows for checking against the decision interval. The general expression for d r is quite simple. For the cusum procedure where we are detecting increases in the mean,

d r = max[0, d r−1 + (X r − W )].

The choice of the value of h is, of course, very important. We do not choose in this book to provide the many details in the literature dealing with this choice. The reader is referred to Ewan and Kemp, 1960, and Montgomery, 2000b (see the Bibliography) for a thorough discussion. One important consideration is the expected run length. Ideally, the expected run length is quite large under μ = μ 0

and quite small when μ = μ 1 .

Review Exercises

17.1 Consider X 1 ,X 2 ,...,X n independent Poisson

Sample

9 2.3951 0.0068 the properties of moment-generating functions to show n

random variables with parameters μ 1 ,μ 2 ,...,μ n . Use

10 2.4215 0.0048 that the random variable

X is a Poisson random

13 2.4009 0.0077 variable with mean

μ i and variance

17.2 Consider the following data taken on subgroups 16 2.4107 0.0138 of size 5. The data contain 20 averages and ranges on

17 2.4109 0.0037 the diameter (in millimeters) of an important compo-

18 2.3944 0.0052 nent part of an engine. Display ¯

19 2.3951 0.0038 the process appear to be in control?

20 2.4015 0.0017 Sample

X- and R-charts. Does

17.3 Suppose for Review Exercise 17.2 that the buyer

has set specifications for the part. The specifications

require that the diameter fall in the range covered by

2.40000 ± 0.0100 mm. What proportion of units pro-

duced by this process will not conform to specifica-

17.4 For the situation of Review Exercise 17.2, give

numerical estimates of the mean and standard devia-

Review Exercises 707 tion of the diameter for the part being manufactured process producing a certain type of item that is consid-

in the process. ered either defective or not defective. Twenty samples are taken.

17.5 Consider the data of Table 17.1. Suppose that (a) Construct a control chart for control of proportion additional samples of size 5 are taken and tensile

defective.

strength recorded. The sampling produces the follow- ing results (in pounds per square inch).

(b) Does the process appear to be in control? Explain.

Number of Sample

Number of

Sample Items

14 9 4 19 3 (a) Plot the data, using the ¯

10 3 20 1 preliminary data of Table 17.1. (b) Does the process appear to be in control? If not,

X- and R-charts for the

17.9 For the situation of Review Exercise 17.8, sup- explain why.

pose that additional data are collected as follows:

Number of Defective Items 17.6 Consider an in-control process with mean μ = 25

Sample

1 3 and σ = 1.0. Suppose that subgroups of size 5 are √

2 4 used with control limits μ ± 3σ/ n, and centerline at

3 2 μ. Suppose that a shift occurs in the mean, and the

4 2 new mean is μ = 26.5.

5 3 (a) What is the average number of samples required

6 1 (following the shift) to detect the out-of-control sit-

7 3 uation?

8 5 (b) What is the standard deviation of the number of

9 7 runs required?

10 7 Does the process appear to be in control? Explain.

17.7 Consider the situation of Example 17.2. The fol- lowing data are taken on additional samples of size 5.

17.10 A quality control effort is being undertaken for Plot the ¯

a process where large steel plates are manufactured and were produced with the data in the preliminary sam- surface defects are of concern. The goal is to set up ple. Does the process appear to be in control? Explain

X- and S-values on the ¯

X- and S-charts that

a quality control chart for the number of defects per why or why not.

plate. The data are given below. Set up the appropri- ate control chart, using this sample information. Does

Sample

the process appear to be in control?

Number of

Number of

Sample Defects

17.8 Samples of size 50 are taken every hour from a 10 2 20 3

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Chapter 18

Bayesian Statistics