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16 Quality Control Methods

Copyright © Cengage Learning. All rights reserved.

16.5 CUSUM Procedures

3

CUSUM ProceduresA defect of the traditional X chart is its inability to detect a relatively small change in a process mean.

This is largely a consequence of the fact that whether a process is judged out of control at a particular time depends only on the sample at that time, and not on the past history of the process.

Cumulative sum (CUSUM) control charts and procedures have been designed to remedy this defect.

4

CUSUM ProceduresThere are two equivalent versions of a CUSUM procedure for a process mean, one graphical and the other computational.

The computational version is used almost exclusively in practice, but the logic behind the procedure is most easily grasped by first considering the graphical form.

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The V-Mask

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The V-MaskLet 0 denote a target value or goal for the process mean, and define cumulative sums by

S1 = x1 – 0

S2 = (x1 – 0) + (x2 – 0) = (xi – 0)

. . . S1 = (x1 – 0) + . . . + (x1 – 0) = (x1 – 0)

(in the absence of a target value, x is used in place of 0).

7

The V-MaskThese cumulative sums are plotted over time. That is, at time l, we plot a point at height Sl.

At the current time point r, the plotted points are (1, S1), (2, S2), (3, S3),…, (r, Sr).

8

The V-MaskNow a V-shaped “mask” is superimposed on the plot, as shown in Figure 16.7.

Figure 16.7

(b)

CUSUM plots: (a) successive points (I, Sl) in a CUSUM plot; (b) a V-mask with 0 = (r, Sr);

(a)

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The V-Mask

Figure 16.7

(d)

CUSUM plots: (c) an in-control process; (d) an out-of-control process

(c)

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The V-MaskThe point 0, which lies a distance d behind the point at which the two arms of the mask intersect, is positioned at the current CUSUM point (r, Sr).

At time r, the process is judged out of control if any of the plotted points lies outside the V-mask—either above the upper arm or below the lower arm.

When the process is in control, the xi’ will vary around the target value 0, so successive Si’s should vary around 0. Suppose, however, that at a certain time, the process mean shifts to a value larger than the target.

11

The V-MaskFrom that point on, differences xi’ – 0 will tend to be positive, so that successive Sl’s will increase and plotted points will drift upward.

If a shift has occurred prior to the current time point r, there is a good chance that (r, Sr) will be substantially higher than some other points in the plot, in which case these other points will be below the lower arm of the mask.

Similarly, a shift to a value smaller than the target will subsequently result in points above the upper arm of the mask.

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Any particular V-mask is determined by specifying the “lead distance” d and “half-angle” , or, equivalently, by specifying d and the length h of the vertical line segment from 0 to the lower (or to the upper) arm of the mask.

One method for deciding which mask to use involves specifying the size of a shift in the process mean that is of particular concern to an investigator.

Then the parameters of the mask are chosen to give desired values of and , the false-alarm probability and the probability of not detecting the specified shift, respectively.

The V-Mask

13

The V-MaskAn alternative method involves selecting the mask that yields specified values of the ARL (average run length) both for an in-control process and for a process in which the mean has shifted by a designated amount.

After developing the computational form of the CUSUM procedure, we will illustrate the second method of construction.

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Example 8A wood products company manufactures charcoal briquettes for barbecues. It packages these briquettes in bags of various sizes, the largest of which is supposed to contain 40 lbs. Table 16.4 displays the weights of bags from 16 different samples, each of size n = 4.

Table 16.4Observations, x’s and Cumulative Sums for Example 8

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Example 8The first 10 of these were drawn from a normal distribution with = 0 = 40 and = .5. Starting with the eleventh sample, the mean has shifted upward to = 40.3.

Figure 16.8 displays an x chart with control limits0 3 x = 40 3 (.5/ ) = 40 .75

Figure 16.8

X control chart for the data of Example 8

cont’d

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Example 8No point on the chart lies outside the control limits. This chart suggests a stable process for which the mean has remained on target.

Figure 16.9 shows CUSUM plots with a particular V-mask superimposed.

cont’d

17

Example 8The plot in Figure 16.9(a) is for current time r = 12. All points in this plot lie inside the arms of the mask. However, the plot for r = 13 displayed in Figure 16.9(b) gives an out-of-control signal.

(a) (b)

Figure 16.9

CUSUM plots and V-masks for data of Example 8: (a) V-mask at time r = 12, process in control; (b) V-mask at time r = 13, out-of-control signal

cont’d

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Example 8The point falling below the lower arm of the mask suggests an increase in the value of the process mean.

The mask at r = 16 is even more emphatic in its out-of-control message. This is in marked contrast to the X chart.

cont’d

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A Computational Version

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A Computational VersionThe following computational form of the CUSUM procedure is equivalent to the previous graphical description.

Let d0 = e0 = 0, and calculate d1, d2, d3, . . . , and e1, e2, e3, . . . , recursively, using the relationships

dl = max[0, dl – 1 + (xl – (0 + k))]

el = max[0, el – 1 – (xl – (0 – k))] (l = 1, 2, 3, . . .)

Here the symbol k denotes the slope of the lower arm of the V-mask, and its value is customarily taken as /2 (where is the size of a shift in on which attention is focused).

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A Computational VersionIf at current time r either dr > h or er > h, the process is judged to be out of control.

The first inequality suggests that the process mean has shifted to a value greater than the target, whereas er > h indicates a shift to a smaller value.

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Example 9Reconsider the charcoal briquette data displayed in Table 16.4 of Example 8.

Table 16.4Observations, X’s and Cumulative Sums for Example 8

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Example 9The target value is 0 = 40, and the size of a shift to be quickly detected is = .3. Thus

k = = .15 0 + k = 40.15 0 – k = 39.85

so

dl = max[0, dl – 1 + (xl – 40.15)]

el = max[0, el – 1 – (xl – 39.85)]

cont’d

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Example 9Calculations of the first few dl’s proceeds as follows:

d0 = 0

d1 = max[0, d0 + (x1 – 40.15)]

= max[0, 0 + (40.20 – 40.15)]

= .05

cont’d

25

Example 9 d2 = max[0, d1 + (x2 – 40.15)]

= max[0, .05 + (39.72 – 40.15)]

= 0

d3 = max[0, d2 + (x3 – 40.15)]

= max[0, 0 + (40.42 – 40.15)]

= .27

cont’d

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Example 9The remaining calculations are summarized in Table 16.5.

Table 16.5

CUSUM Calculations for Example 9

cont’d

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Example 9The value h = .95 gives a CUSUM procedure with desirable properties—false alarms (incorrect out-of-control signals) rarely occur, yet a shift of = .3 will usually be detected rather quickly.

With this value of h, the first out-of-control signal comes after the 13th sample is available. Since d13 = 1.17 > .95, it appears that the mean has shifted to a value larger than the target.

cont’d

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Example 9This is the same message as the one given by the V-mask in Figure 16.9(b).

(b)

Figure 16.9

CUSUM plots and V-masks for data of Example 8: (b) V-mask at time r = 13, out-of-control signal

cont’d

29

A Computational VersionTo demonstrate equivalence, again let r denote the current time point, so that x1, x2 . . . , xr are available. Figure 16.10 displays a V-mask with the point labeled 0 at (r, Sr).

Figure 16.10

A V-mask with slope of lower arm = k

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A Computational VersionThe slope of the lower arm, which we denote by k, is h/d. Thus the points on the lower arm above r, r – 1, r – 2,. . . are at heights Sr – h, Sr – h – k , Sr – h, Sr – h – 2k and so on.

The process is in control if all points are on or between the arms of the mask. We wish to describe this condition algebraically.

To do so, let

Tl = [ xi – (0 + k)] l = 1, 2, 3, . . . , r

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A Computational VersionThe conditions under which all points are on or above the lower arm are

Sr – h Sr (trivially satisfied) i.e., Sr Sr + h

Sr – h – k Sr – 1 i.e., Sr Sr – 1 + h + k

Sr – h – 2k Sr – 2 i.e., Sr Sr – 2 + h + 2k

. . . . . .

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A Computational VersionNow subtract rk from both sides of each inequality to obtain

Sr – rk Sr – rk + h i.e., Tr Tr + h

Sr – rk Sr – 1 – (r – 1)k + h i.e., Tr Tr – 1 + h

Sr – rk Sr – 2 – (r – 2)k + h i.e., Tr Tr – 2 + h . . . . . .

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A Computational VersionThus all plotted points lie on or above the lower arm if and only if (iff) Tr – Tr h, Tr – Tr – 1 h, Tr – Tr – 2 h, and so on. This is equivalent to

Tr – min(T1, T2, . . . , Tr) h

In a similar manner, if we let

Vr = [xi – (0 – k)] = Sr + rk

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A Computational Versionit can be shown that all points lie on or below the upper arm iff

max(V1, . . , Vr) – Vr h

If we now let

dr = Tr – min(T1, . . . , Tr)

er = max(V1, . . . , Vr) – Vr

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A Computational Versionit is easily seen that d1, d2, . . . , and e1, e2, . . .can be calculated recursively as illustrated previously. For example, the expression for dr follows from consideration of two cases:

1. min(T1 , . . . , Tr) = Tr , whence dr = 0

2. min(T1 , . . . , Tr) = min(T1 , . . . , Tr – 1), so that

dr = Tr – min(T1, . . . , Tr – 1)

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A Computational Version = xr – (0 + k) + Tr – 1 – min(T1, . . ., Tr – 1)

= xr – (0 + k) + dr – 1

Since dr cannot be negative, it is the larger of these two quantities.

37

Designing a CUSUM Procedure

38

Designing a CUSUM ProcedureLet denote the size of a shift in that is to be quickly detected using a CUSUM procedure. It is common practice to let k = /2.

Now suppose a quality control practitioner specifies desired values of two average run lengths:1. ARL when the process is in control ( = 0)

2. ARL when the process is out of control because the mean has shifted by ( = 0 + or = 0 – )

39

Designing a CUSUM ProcedureA chart developed by Kenneth Kemp (“The Use of Cumulative Sums for Sampling Inspection Schemes,” Applied Statistics, 1962: 23), called a nomogram, can then be used to determine values of h and n that achieve the specified ARLs.

This chart is shown as Figure 16.11.

Figure 16.11

The Kemp nomogram

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Designing a CUSUM ProcedureThe method for using the chart is described in the accompanying box.

Either the value of must be known or an estimate is used in its place.

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Designing a CUSUM ProcedureUsing the Kemp Nomogram

1. Locate the desired ARLs on the in-control and out-of-control scales. Connect these two points with a line.

2. Note where the line crosses the k scale, and solve for n using the equation

Then round n up to the nearest integer.

42

Designing a CUSUM Procedure3. Connect the point on the k scale with the point on the in-control ARL scale using a second line, and note where

this line crosses the h scale. Then h = ( / ) h.

The value h = .95 was used in Example 9. In that situation, it follows that the in-control ARL is 500 and the out-of-control ARL (for = .3) is 7.

43

Example 10The target value for the diameter of the interior core of a hydraulic pump is 2.250 in.

If the standard deviation of the core diameter is = .004, what CUSUM procedure will yield an in-control ARL of 500 and an ARL of 5 when the mean core diameter shifts by the amount of .003 in.?

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Example 10

Figure 16.11

The Kemp nomogram*

cont’d

Connecting the point 500 on the in-control ARL scale to the point 5 on the out of-control ARL scale and extending the line to the k scale on the far left in Figure 16.11 gives k = .74.

Thus

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Example 10So

The CUSUM procedure should therefore be based on the sample size n = 4.

cont’d

46

Example 10Now connecting .74 on the k scale to 500 on the in-control ARL scale gives h = 3.2, from which

h = (/ ) (3.2)

= (.004 / )(3.2)

= .0064

An out-of-control signal results as soon as either dr > .0064 or er > 0064.

cont’d