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1.0 Cumulative Sum Control Chart .................................................. 3

1.1 Tabular or Algorithmic CUSUM for Monitoring the Mean of

the Process ...................................................................................... 4

1.2 V-Mask for Monitoring the Mean of the Process ...................... 6

1.3 Standardized Cumulative Sum Control Chart ........................ 10

1.4 ARL of Cumulative Sum Control Chart .................................. 11

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Figure 1: Showing the deviation of xi value from targeted

iC value,

iC value,

decision interval, and counter values ..................................................... 5

Figure 2: CUSUM control chart plotted with data shown in Fig. 1 ...................... 6

Figure 3: Illustration of V-mask parameters .......................................................... 7

Figure 4: An illustration of V-mask showing out of control point ........................ 8

Figure 5: Calculated CUSM values for the sample batches .................................. 9

Figure 6: The V-mask results placed (a) at batch number 4, (b) at batch number

8, (c) at batch number 10, and (d) at batch number 12 ........................ 10

Figure 7: ARL for CUSUM with k = 0.5 and h = 4 and h = 5 ............................. 12

Figure 8: ARL of CUSUM chart for a given standardized k = 0.5 and h = 4 and

5, and ARL of Shewart X control chart .............................................. 13

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1.0 Cumulative Sum Control Chart

A cumulative sum CUSUM control chart is a time-weighted control chart that

displays the cumulative sum of the deviation of each sample value from the target

value. Owing to the factor that it is cumulative, a minor drifting in the process mean

will lead to steadily increasing or decreasing cumulative deviation value. It was

developed by E. S. Page of the University of Cambridge. Cumulative sum CUSUM

control chart has been shown to be more efficient in detecting small shift in the mean

of a process. In analyzing the average run length ARL, CUSUM control chart shows

a better result than Shewhart control chart when it is desired to detect shift in the

mean by less than two standard deviations. However, CUSUM control chart is

relatively slow to respond to large shift and hard to detect and analyze special trend

patterns.

Let’s collect n sample batches, each of sample size m, and calculate the mean

of each sample batch

m

j

iji

1

xx . The cumulative sum CUSUM control chart is

formed by plotting one of the following quantities.

n

i

in

1

0ˆxC (1)

where 10 CˆxC nin .

or standardized cumulative sum

n

i

in

i1

0

x

' ˆx1

C (2)

where '

1

x

0' Cˆx

C -ni

n

i

. Cn is called cumulative sum up to and including the ith

sample. 0̂ is the estimate of the in-control mean and ix

is the known or estimated

of standard deviation of the sample mean ix . The choice of which of these two

quantities is plotted is usually determined by the statistical software package. In

either case, as long as the process remains in control centered at 0̂ , the CUSUM

plot will show variation in a random pattern centered about zero. If the mean of the

process shifts upward, the plotted CUSUM points will eventually drift upward and

vice versa if the mean of the process decreases.

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There are two basic ways to present CUSUM control chart, which are tabular

or algorithmic CUMSUM and V-mask. Let’s discuss the way to establish them and

use them to analyze the process.

1.1 Tabular or Algorithmic CUSUM for Monitoring the

Mean of the Process

Let’s use cumulative sum up Cn to establish CUSUM control chart. For in-control

process, the xi observation, normal distribution with mean µ1 and standard deviation

. The Tabular CUSUM limits are defined as

10 C)(x,0MinC iii k (3)

and

10 C)(x,0MaxC iii k (4)

where µ0 is the targeted mean. The starting values of

iC and

iC are equal to 0. The

statistic

iC and

iC are called one sided lower and one sided upper CUSUM

respectively.

k is the reference value chosen from half of the between the mean µ1 of the

subsequent sample and targeted mean µ0. i.e. 2

01 k . If 01 then

2

k

. If either

iC or

iC value exceeds the decision interval H, which usually defined as

equal H = 5 then the process is considered to be out of control.

Plotting the values of

iC and

iC versus the sample number would show the

CUSUM status chart. The CUSUM is particularly useful to identify the assignable

cause. The counters are N+ and N- for one sided lower CUSUM and one sided upper

CUMSUM respectively. They are used to register the count number when

iC value

starts to descend from zero to less than zero and

iC value begins to ascend from

zero to more than zero. The count number increases as long as it is either in

descending or ascending trend. It resets to zero when either

iC or

iC value crossing

x-axis line. The count enables one to find the interval when the process is in control

and when it is not in control by identifying the process shift.

Example 1

Twenty measurements are taken from a process with targeted value µ0 equals to 325

and standard deviation equals to 1.072. Taking the value of k to be 0.536 and

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decision interval H to be 50, calculate the one sided lower and upper CUSUM

values, plot the graph and determine if there is any out of control for the process.

Solution

The deviation of xi value from targeted value and the one sided lower and upper

CUSUM values are calculated using equation (3) and (4) respectively together with

N- and N+ counter values and decision intervals are shown in Fig. 1.

No xi xi -µ0 CUSUM

iC

iC -H H N- N+

1 324.9 -0.10 -0.10 0.00 0.00 -5.36 5.36

2 324.6 -0.40 -0.50 0.00 0.00 -5.36 5.36

3 324.7 -0.30 -0.80 0.00 0.00 -5.36 5.36

4 324.3 -0.70 -1.50 -0.20 0.00 -5.36 5.36 1

5 325.3 0.30 -1.20 0.00 0.00 -5.36 5.36

6 325.2 0.20 -1.00 0.00 0.00 -5.36 5.36

7 324.1 -0.90 -1.90 -0.40 0.00 -5.36 5.36 1

8 325.2 0.20 -1.70 0.00 0.00 -5.36 5.36

9 325.2 0.20 -1.50 0.00 0.00 -5.36 5.36

10 324.6 -0.40 -1.90 0.00 0.00 -5.36 5.36

11 324.6 -0.40 -2.30 0.00 0.00 -5.36 5.36

12 325.1 0.10 -2.20 0.00 0.00 -5.36 5.36

13 328.3 3.30 1.10 0.00 0.60 -5.36 5.36 1

14 324.2 -0.80 0.30 -0.30 0.00 -5.36 5.36 1

15 327.8 2.80 3.10 0.00 2.60 -5.36 5.36 1

16 325.5 0.50 3.60 0.00 3.10 -5.36 5.36 2

17 324.6 -0.40 3.20 0.00 2.70 -5.36 5.36 3

18 325.7 0.70 3.90 0.00 3.40 -5.36 5.36 4

19 325.8 0.80 4.70 0.00 4.20 -5.36 5.36 5

20 325.3 0.30 5.00 0.00 4.50 -5.36 5.36 6

Figure 1: Showing the deviation of xi value from targeted

iC value,

iC value, decision interval, and

counter values

From the data shown in Fig. 1, the CUSUM control chart is plotted and shown in

Fig. 2.

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Figure 2: CUSUM control chart plotted with data shown in Fig. 1

From the results shown in Fig. 2, the process is in control because none of the

iC

and

iC value is lower than –H interval value or higher than H interval value.

The estimation of the mean of the process shift is defined as

N

Cˆ

0ik ; if

iC < -H (5)

and

N

Cˆ

0ik ; if

iC > H (6)

1.2 V-Mask for Monitoring the Mean of the Process

A visual procedure proposed by Alfred Barnard in 1959 known as V-mask is at time

used to determine if a process is out of control. Usually the tabulated form of V-

mark is preferred.

A V-mask is a V-shape overlay superimposed on the graph of the cumulative

sum. The origin point of the V-mask is placed on top of the latest cumulative sum

point so that past points can be examined to see if any point falls below or above the

V-shape overlay. As long as all the past points lie inside the V-shape overlay, the

process is said to be in control. Otherwise, even with one point lies outside the V-

shape overlay, the process is suspected of being out of control.

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The parameters of the V-mask as illustrated in Fig. 3 are defined as follows:

O is the origin.

d is the lead distance is equal to

1ln

22

d .

A2tan 1 , where

x

; is the desired shift to be detected; x is the

standard deviation of X ; A is the scale factor – the narrowness of the V-mask

and it is usually equal to A = x2 .

β is the type II error and usually equal to 0.01.

is the type I error and is equal to 0.0026.

h is the multiplier for decision interval. i.e. H = xh and it is equal to h = dk.

k is the multiplier for the slope of V-mask K = xk .

Figure 3: Illustration of V-mask parameters

From the figure, it is clear that the behavior of V-mask is decided by the distance k,

which is the lower slope of the lower arm and the rise distance h, which are the

design parameters of the V-mask. Note that we can also specify d and the vertex

angle as it is more common in the literature specified = 2θ = 2tan-1(1/2) = 530 as

the vertex angle.

Example 2

The desired shift is 1.0. The standard deviation of the mean is 1.0 and k = 0.5,

calculate the lead distance d and multiplier h of V-mask.

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Solution

The x

is 1.0/1.0 = 1.0.

The lead distance is

1ln

22

d =

0026.0

01.01ln

1

2= 11.89.

h = dk = 11.89x0.5 = 5.945.

The angle is

A2tan 1 =

89.11

94.5tan 1 = 26.50.

An illustration of V-mask with CUSUM data is shown in Fig. 4.

Figure 4: An illustration of V-mask showing out of control point

In Fig. 4 where the origin of the V-mask is placed at sample point 28, it shows an

out of control at sample point 9 and 23 because these points lie below the V-mask.

By sliding the V-mask backwards so that the origin point can cover other cumulative

sum data point, one can identify first sample point showing out of control situation.

This is useful for diagnosing the cause the out of control.

Example 3

The following 10 data points are the average values of a process parameter: 101.1,

99.8, 101.2, 103.1, 98.8, 101.7, 99.2, 102.4, 101.9, 99.1, 99.4, 98.7, 98.6, 98.3, and

97.8. Each the average is taken five samples. The targeted value of the process

parameter is 100. Construct and apply of a V-mask to analyze the data with desired

shift of one standard deviation.

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Solution

Based on the data standard deviation of the process is 1.67. Thus, the standard

deviation of the sample mean is 5

167.1= 0.522. The lead distance is

1ln

22

d =

0026.0

01.01ln

1

2= 11.89. The h value is h = dk = 11.89x0.522x0.5 = 3.103. The angle

of vortex is equation 2tan-1(3.103/11.89) = 29.20.

Figure 5 shows the raw data and CUSUM values for the sample batches.

No ix 0x i CUSUM

1 101.1 1.1 1.1

2 98.5 -1.5 -0.4

3 99.1 -0.9 -1.3

4 99.3 -0.7 -2.0

5 100.9 0.9 -1.1

6 101.7 1.7 0.6

7 101.6 1.6 2.2

8 99.7 -0.3 1.9

9 101.9 1.9 3.8

10 101.3 1.3 5.1

11 102.3 2.3 7.4

12 102.8 2.8 10.2

13 99.3 -0.7 9.5

14 99.2 -0.8 8.7

15 102.1 2.1 10.8 Figure 5: Calculated CUSM values for the sample batches

The V-mask results for placed at batch number 4, 8, 10, and 12 are respectively

shown in Fig. 6.

The results indicate that process is in control for desired shift of one standard

deviation.

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(a) (b)

(c) (d) Figure 6: The V-mask results placed (a) at batch number 4, (b) at batch number 8, (c) at batch number

10, and (d) at batch number 12

1.3 Standardized Cumulative Sum Control Chart

A number of CUSUM user prefers using standardized cumulative sum control chart,

which is defined as

0x

z ii . The standardized two-sided cumulative sum up are

defined as

1Cz,0MinC iii k (7)

and

1Cz,0MaxC iii k (8)

The advantages of the standardized CUSUM control chart are; it can have same

values of k and h and allow one to use CUSUM to control variability, which is

349.0

822.0z

i

i . Thus, the standard deviation CUSUM is defined as

1s,0Mins iii k (9)

and

1s,0Maxs iii k (10)

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1.4 ARL of Cumulative Sum Control Chart

One would like to see a high ARL lower side value when the process is on target

and a low ARL upper side value when the process mean shift to an unsatisfactory

level.

Let’s look at one sided ARL approximated calculation proposed by David

Siegmund in 1985. It is defined as

2

b2

2

1b2eARL

(11)

where k * for one sided lower CUSUM

iC and k * for one-sided upper

CUSUM

iC . k is the usual defined as 2

01 k . b is equal to b = h + 1.166 and *

is equal to

01* . For = 0, one can use ARL = b2. The quantity * represents

the shift in the mean in the units of , where ARL is to be calculated.

The ARL for two sided CUSUM is defined as

1

ARL

1

ARL

1ARL

(12)

Example 4

If k = 1/2 , h = 5 , and * = 0 calculate the average run length.

Solution

Since * = 0 then 2/1 and b value is equal to 6.166.

The one side upper ARL is 2

b2

02

1b2eARL

=

2166.6)5.0(2

2/12

1166.6x5.0x2e

= 938.2.

Since * = 0, it means that the mean is on targeted value. Thus, the one sided lower

ARL is equal to one sided upper ARL. Therefore, the overall ARL is equal to 1

00 ARL

1

ARL

1ARL

=

1

2.938

1

2.938

1

= 469.1.

Example 5

If the mean is shift by 2, calculate the average run length for h = 4.

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Solution

If the mean is shift by 2 then * = 2, 5.2 for one side lower CUSUM, and

5.1 for one sided upper CUSUM. The b value is b = 4 + 1.166 = 5.166.

The one sided lower ARL is 2

b2

22

1b2eARL

2

166.5x5.2x2

)5.2(2

1166.5x5.2x2e

= 1.321x1010.

The one sided upper ARL is 2

b2

22

1b2eARL

2

166.5x5.1x2

)5.1(2

1166.5x5.1x2e

= 3.22.

Thus, the overall ARL is

1

00 ARL

1

ARL

1ARL

=

1

610x195.1

1

22.3

1

= 3.22.

Since equation (11) and (12) provided an approximation of ARL, the more

accurately results using computation is shown in Fig. 7.

Shift in Mean

multiple of h = 4 h = 5

0.00 168 465

0.25 74.2 139

0.50 26.6 38

0.75 13.3 17

1.00 8.38 10.4

1.50 4.75 5.75

2.00 3.34 4.01

2.50 2.62 3.11

3.00 2.19 2.57

4.00 1.71 2.01 Figure 7: ARL for CUSUM with k = 0.5 and h = 4 and h = 5

One may like calculate the average run length using standardized parameter ks and

hs are equal to n/

kk

s and

n/

hh

s respectively. Some literatures proposed

solving the integral equation that forms the basis of the exact solution using an

approximation of Systems of Linear Algebraic Equation SLAE. The computerized

solution for calculating ARL for a given the standardized h and k for some examples

are shown in Fig. 8.

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Standardized CUSUM

Shewart

X n/

hh

s

4 5

336 930 371.00

74.2 140 281.14

26.6 30 155.22

13.3 17 81.22

8.38 10.4 44.00

4.75 5.75 14.97

3.34 4.01 6.30

2.62 3.11 3.24

2.19 2.57 2.00

1.71 2.01 1.19 Figure 8: ARL of CUSUM chart for a given standardized k = 0.5 and h = 4 and 5, and ARL of Shewart

X control chart

If one compares the ARL of CUSUM control chart with ARL Shewart control chart,

one can draw the conclusion that Shewart chart is superior for detecting large shift

whereas CUSUM is fast to detect small shift. The results in Fig. 2.37 show that for

shift of one standard deviation ARLs are 8.38 and 10.4 respectively for h = 4 and 5,

whereas for a shift of one standard deviation for Shewart control chart is 44.0.

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A

Average run length ............................................... 1, 9

B

Barnard, Alfred ......................................................... 4

C

Cumulative sum ......................................... 1, 2, 3, 6, 7

Cumulative sum control chart............................1, 2, 3

Cumulative sum up ............................................... 1, 2

CUSUM control chart....See Cumulative sum control

chart

O

One sided lower CUSUM ..................................... 2, 9

One sided upper CUSUM ..................................... 2, 9

P

Page, E. S. ................................................................. 1

S

Shewhart control chart .............................................. 1

Siegmund, David ...................................................... 9

SLAE ........... See System of linear algebraic equation

Standard deviation CUSUM ..................................... 8

Standardized cumulative sum ................................... 1

Standardized cumulative sum control chart .............. 8

Systems of linear algebraic equation....................... 10

T

Tabular CUSUM limit .............................................. 2

V

V-mask .................................................................. 4, 6