chapter 36 exponentially weighted moving average (ewma...

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Chapter 36

Exponentially Weighted Moving Average

(EWMA) and

Engineering Process Control (EPC)

Introduction

• Under the Shewhart model for control charting, it is assumed

that the mean is constant. Also, errors are NID(0, 𝜎2). In

many applications this assumption is not true.

• Exponentially weighted moving average (EWMA) techniques

offer an alternative based on exponential smoothing

(sometimes called geometric smoothing).

• The computation of EWMA as a filter is done by taking the

weighted average of past observations with progressively

smaller weights over time.

• EWMA has flexibility of computation through the selection of

a weight factor and can use this factor to achieve balance

between older data and more recent observations.

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Introduction

• EWMA techniques can be combined with engineering

process control (EPC) to indicate when a process should be

adjusted.

• Application examples for EWMA with EPC include the

monitoring of pans produced by a tool that wears and needs

periodic sharpening, adjustment, or replacement.

• Much of the discussion in this chapter is a summary of the

discussion of Hunter (1995, 1996).

36.1 S4/IEE Application Examples:

Three-way Control Chart

• Manufacturing 30,000-foot-level metric: An S4/IEE project was to

decrease the within- and between-part thickness

capability/performance of manufactured titanium metal sheets. A

KPIV to this process is the acid concentration of a chemical-

etching step (pickling), which removes metal from the sheet. As

part of the control phase of the project, an EWMA with EPC

procedure was established that would identify when additional acid

should be added to the etching tank.

• Manufacturing 30,000-foot-level metric: An S4/IEE project was to

decrease the within- and between-part variability of the diameter of

a metal part, which was ground to dimension. As part of the

control phase of the project, an EWMA with EPC procedure was

established that would identify when adjustments should be made

to the grinding wheel because of wear.

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36.2 Description

• Consider a sequence of observations 𝑌1, 𝑌2, 𝑌3, . . . , 𝑌𝑡. We

could examine these data using any of the following

procedures, with the differences noted:

• Shewhart--no weighting of previous data

• CUSUM--equal weights for previous data

• Moving average—weight, for example, the five most

recent responses equally as an average

• EWMA—weight the most recent reading the highest and

decrease weights exponentially for previous readings

36.2 Description

• A Shewhart, CUSUM, or moving average control chart for

these variables data would all be based on the model

𝑌𝑡 = 𝜂 + 𝑚𝑡

where the expected value of the observations 𝐸(𝑌𝑡) is a

constant 𝜂 and 𝑚𝑡, is NID(0, 𝜎𝑚2).

• For the Shewhart model the mean and variance are both

constant, with independent errors.

• Also with the Shewhart model, the forecast for the next

observation or average of observations is the centerline of

the chart (𝜂0).

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36.2 Description

• An EWMA is retrospective when plotted under Shewhart

model conditions. It smoothes the time trace, thereby

reducing the role of noise which can offer insight into what

the level of the process might have been, which can be

helpful when identifying special causes.

• Mathematically, for 0 < 𝜆 < 1 this can be expressed as

𝐸𝑊𝑀𝐴 = 𝑌 𝑡 = 𝜆𝑌𝑡 + 𝜃𝑌 𝑡−1 𝑤ℎ𝑒𝑟𝑒 𝜃 = 1 − 𝜆

• At time 𝑡 the smoothed value of the response equals the

multiple of lambda times today's observation plus theta

times yesterday’s smoothed value.

36.2 Description

• A more typical plotting expression for this relationship is

𝐸𝑊𝑀𝐴 = 𝑌 𝑡+1 = 𝑌 𝑡 + 𝜆𝑒𝑡 𝑤ℎ𝑒𝑟𝑒 𝑒𝑡 = 𝑌𝑡 − 𝑌 𝑡

• The predicted value for tomorrow equals the predicted value

of today plus a “depth of memory parameter“ (lambda) times

the difference between the observation and the current day's

prediction.

• For plotting convenience, EWMA is often put one unit ahead

of 𝑌𝑡.

• Under certain conditions, as described later, EWMA can be

used as a forecast.

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36.2 Description

• The three sigma limits for an EWMA control chart are

±3𝜎𝐸𝑊𝑀𝐴 = 𝜆/(2 − 𝜆)[±3𝜎𝑆ℎ𝑒𝑤ℎ𝑎𝑟𝑡]

• When there are independent events, an EWMA chart with 𝜆

= 0.4 yields results almost identical to the combination of

Western Electric rules, where the control limits are exactly

half of those from a Shewhart chart (Hunter 1989b).

• The underlying assumptions for a Shewhart model are often

not true in reality.

• An EWMA can be used to model processes that have linear

or low-order time trends, cyclic behavior, and a response

that is a function of an external factor, nonconstant variance,

and autocorrelated patterns.

36.3 Example 36.1:

EWMA with Eng. Process Control

• The data (Wheeler l995b; Hunter 1995) are

the bearing diameters of 50 camshafts

collected over time.

• A traditional 𝑋𝑚𝑅 chart indicates that there

are many out-of-control conditions.

• However, this example illustrates how

underlying assumptions for application of

the 𝑋𝑚𝑅 chart to this data set are probably

violated.

• EWMA and EPC alternatives are then

applied.

Dia.

1 50

2 51

3 50.5

4 49

5 50

6 43

7 42

8 45

9 47

10 49

11 46

12 50

13 52

14 52.5

15 51

16 52

17 50

Dia.

18 49

19 54

20 51

21 52

22 46

23 42

24 43

25 45

26 46

27 42

28 44

29 43

30 46

31 42

32 43

33 42

34 45

Dia.

35 49

36 50

37 51

38 52

39 54

40 51

41 49

42 50

43 49.5

44 51

45 50

46 52

47 50

48 48

49 49.5

50 49

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36.3 Example 36.1:


36.3 Example 36.1:


• First we will check for non-independence. To do this we will

use time series analysis techniques. If data meander, each

observation tends to be close to the previous observation and

there is no correlation between successive observations. That

is, there is no autocorrelation (i.e., correlation with itself).

• If observations are independent of time, their autocorrelation

should equal zero. A test for autocorrelation involves

regressing the current value on previous values of the time

series to determine if there is correlation.

• The term lag quantifies how far back comparisons are made.

• Independence of data across time can be checked by

estimation of the lag autocorrelation coefficients 𝑝𝑘, where k =

1, 2, . . . .

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36.3 Example 36.1:


Dia. Lag1 Lag2 Lag3 Lag4 Lag5 Lag6 Lag7 Lag8 Lag9 Lag10

1 50

2 51 50

3 50.5 51 50

4 49 50.5 51 50

5 50 49 50.5 51 50

6 43 50 49 50.5 51 50

7 42 43 50 49 50.5 51 50

8 45 42 43 50 49 50.5 51 50

9 47 45 42 43 50 49 50.5 51 50

10 49 47 45 42 43 50 49 50.5 51 50

11 46 49 47 45 42 43 50 49 50.5 51 50

12 50 46 49 47 45 42 43 50 49 50.5 51

13 52 50 46 49 47 45 42 43 50 49 50.5

14 52.5 52 50 46 49 47 45 42 43 50 49

15 51 52.5 52 50 46 49 47 45 42 43 50

16 52 51 52.5 52 50 46 49 47 45 42 43

17 50 52 51 52.5 52 50 46 49 47 45 42

18 49 50 52 51 52.5 52 50 46 49 47 45

19 54 49 50 52 51 52.5 52 50 46 49 47

20 51 54 49 50 52 51 52.5 52 50 46 49

21 52 51 54 49 50 52 51 52.5 52 50 46

22 46 52 51 54 49 50 52 51 52.5 52 50

23 42 46 52 51 54 49 50 52 51 52.5 52

24 43 42 46 52 51 54 49 50 52 51 52.5

25 45 43 42 46 52 51 54 49 50 52 51

26 46 45 43 42 46 52 51 54 49 50 52

27 42 46 45 43 42 46 52 51 54 49 50

28 44 42 46 45 43 42 46 52 51 54 49

29 43 44 42 46 45 43 42 46 52 51 54

30 46 43 44 42 46 45 43 42 46 52 51

31 42 46 43 44 42 46 45 43 42 46 52

32 43 42 46 43 44 42 46 45 43 42 46

33 42 43 42 46 43 44 42 46 45 43 42

34 45 42 43 42 46 43 44 42 46 45 43

35 49 45 42 43 42 46 43 44 42 46 45

36 50 49 45 42 43 42 46 43 44 42 46

37 51 50 49 45 42 43 42 46 43 44 42

38 52 51 50 49 45 42 43 42 46 43 44

39 54 52 51 50 49 45 42 43 42 46 43

40 51 54 52 51 50 49 45 42 43 42 46

41 49 51 54 52 51 50 49 45 42 43 42

42 50 49 51 54 52 51 50 49 45 42 43

43 49.5 50 49 51 54 52 51 50 49 45 42

44 51 49.5 50 49 51 54 52 51 50 49 45

45 50 51 49.5 50 49 51 54 52 51 50 49

46 52 50 51 49.5 50 49 51 54 52 51 50

47 50 52 50 51 49.5 50 49 51 54 52 51

48 48 50 52 50 51 49.5 50 49 51 54 52

49 49.5 48 50 52 50 51 49.5 50 49 51 54

50 49 49.5 48 50 52 50 51 49.5 50 49 51

Correl 0.7431 0.54573 0.3426 0.239356 0.131113 -0.02377 -0.09443 -0.22902 -0.22966 -0.34415

36.3 Example 36.1:

EWMA with Eng. Process Control Dia. Lag1 Lag2 Lag3 Lag4 Lag5 Lag6 Lag7 Lag8 Lag9 Lag10

1 50

2 51 50

3 50.5 51 50

4 49 50.5 51 50

5 50 49 50.5 51 50

6 43 50 49 50.5 51 50

7 42 43 50 49 50.5 51 50

8 45 42 43 50 49 50.5 51 50

9 47 45 42 43 50 49 50.5 51 50

10 49 47 45 42 43 50 49 50.5 51 50

11 46 49 47 45 42 43 50 49 50.5 51 50

12 50 46 49 47 45 42 43 50 49 50.5 51

13 52 50 46 49 47 45 42 43 50 49 50.5

14 52.5 52 50 46 49 47 45 42 43 50 49

15 51 52.5 52 50 46 49 47 45 42 43 50 : : : : : : : : : : : : : : : : : : : : : : : :

45 50 51 49.5 50 49 51 54 52 51 50 49

46 52 50 51 49.5 50 49 51 54 52 51 50

47 50 52 50 51 49.5 50 49 51 54 52 51

48 48 50 52 50 51 49.5 50 49 51 54 52

49 49.5 48 50 52 50 51 49.5 50 49 51 54

50 49 49.5 48 50 52 50 51 49.5 50 49 51

Correl 0.7431 0.5457 0.3426 0.2393 0.1311 -0.0237 -0.0944 -0.2290 -0.2296 -0.3441

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36.3 Example 36.1:


• The estimate of the autocorrelation coefficients are:

𝑟1 = 0.74 𝑟6 = −0.02

𝑟2 = 0.55 𝑟7 = −0.09

𝑟3 = 0.34 𝑟8 = −0.23

𝑟4 = 0.24 𝑟9 = −0.23

𝑟5 = 0.13 𝑟10 = −0.34

• For the hypothesis that all 𝑝𝑘 = 0 the approximate standard

error of 𝑟𝑘 is 1/ 𝑛, which leads to an approximate 95%

confidence interval for 𝜌1 of 𝑟1 ± 2/ 𝑛, which results in

0.74 ± 0.28. Because zero is not contained within this interval,

we reject the null hypothesis. The implication of correlation is

that the moving-range statistic does not provide a good

estimate of standard deviation to calculate the control limits.

36.3 Example 36.1:


• George Box (Hunter 1995; Box and Luceno 1997) suggests

using a variogram to check adequacy of the assumptions of

constant mean, independence, and constant variance. It

checks the assumption that data derive from a stationary

process. A variogram does this by taking pairs of observations

1, 2, or m apart to produce alternative time series. When the

assumptions are valid, there should be no difference in the

expectation of statistics obtained from these differences.

• For the standardized variogram

𝐺𝑚 =𝑉𝑎𝑟(𝑌𝑡+𝑚 − 𝑌𝑡)

𝑉𝑎𝑟(𝑌𝑡+1 − 𝑌𝑡)

the ratio 𝐺𝑚 equals 1 for all values of 𝑚, if the data have a

constant mean, independence, and constant variance.

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36.3 Example 36.1:


• For processes that have an ultimate constant variance (i.e.,

stationary process), 𝐺𝑚 will increase at first but soon become

constant.

• When the level and variance of a process grows without limit

(i.e., nonstationary processes), 𝐺𝑚 will continually increase.

• For processes that increase as a straight line, an EWMA gives

a unique model.

• A simple method to compute 𝐺𝑚 is to use the moving-range

computations for standard deviations.

36.3 Example 36.1:


Dia. m=1 m=2 m=3 m=4 m=5 m=6 m=7 m=8 m=9 m=10

1 50

2 51 1

3 50.5 0.5 0.5

4 49 1.5 2 1

5 50 1 0.5 1 0

6 43 7 6 7.5 8 7

7 42 1 8 7 8.5 9 8

8 45 3 2 5 4 5.5 6 5

9 47 2 5 4 3 2 3.5 4 3

10 49 2 4 7 6 1 0 1.5 2 1

11 46 3 1 1 4 3 4 3 4.5 5 4

12 50 4 1 3 5 8 7 0 1 0.5 1

13 52 2 6 3 5 7 10 9 2 3 1.5

14 52.5 0.5 2.5 6.5 3.5 5.5 7.5 10.5 9.5 2.5 3.5

15 51 1.5 1 1 5 2 4 6 9 8 1

16 52 1 0.5 0 2 6 3 5 7 10 9

17 50 2 1 2.5 2 0 4 1 3 5 8

18 49 1 3 2 3.5 3 1 3 0 2 4

19 54 5 4 2 3 1.5 2 4 8 5 7

20 51 3 2 1 1 0 1.5 1 1 5 2

21 52 1 2 3 2 0 1 0.5 0 2 6

22 46 6 5 8 3 4 6 5 6.5 6 4

23 42 4 10 9 12 7 8 10 9 10.5 10

24 43 1 3 9 8 11 6 7 9 8 9.5

25 45 2 3 1 7 6 9 4 5 7 6

26 46 1 3 4 0 6 5 8 3 4 6

27 42 4 3 1 0 4 10 9 12 7 8

28 44 2 2 1 1 2 2 8 7 10 5

29 43 1 1 3 2 0 1 3 9 8 11

30 46 3 2 4 0 1 3 4 0 6 5

31 42 4 1 2 0 4 3 1 0 4 10

32 43 1 3 0 1 1 3 2 0 1 3

33 42 1 0 4 1 2 0 4 3 1 0

34 45 3 2 3 1 2 1 3 1 0 2

35 49 4 7 6 7 3 6 5 7 3 4

36 50 1 5 8 7 8 4 7 6 8 4

37 51 1 2 6 9 8 9 5 8 7 9

38 52 1 2 3 7 10 9 10 6 9 8

39 54 2 3 4 5 9 12 11 12 8 11

40 51 3 1 0 1 2 6 9 8 9 5

41 49 2 5 3 2 1 0 4 7 6 7

42 50 1 1 4 2 1 0 1 5 8 7

43 49.5 0.5 0.5 1.5 4.5 2.5 1.5 0.5 0.5 4.5 7.5

44 51 1.5 1 2 0 3 1 0 1 2 6

45 50 1 0.5 0 1 1 4 2 1 0 1

46 52 2 1 2.5 2 3 1 2 0 1 2

47 50 2 0 1 0.5 0 1 1 4 2 1

48 48 2 4 2 3 1.5 2 1 3 6 4

49 49.5 1.5 0.5 2.5 0.5 1.5 0 0.5 0.5 1.5 4.5

50 49 0.5 1 1 3 1 2 0.5 1 0 2

Mean 2.0816 2.59375 3.2553 3.391304 3.688889 4.045455 4.209302 4.392857 4.792683 5.2375

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36.3 Example 36.1:

EWMA with Eng. Process Control Dia. m=1 m=2 m=3 m=4 m=5 m=6 m=7 m=8 m=9 m=10

1 50

2 51 1

3 50.5 0.5 0.5

4 49 1.5 2 1

5 50 1 0.5 1 0

6 43 7 6 7.5 8 7

7 42 1 8 7 8.5 9 8

8 45 3 2 5 4 5.5 6 5

9 47 2 5 4 3 2 3.5 4 3

10 49 2 4 7 6 1 0 1.5 2 1

11 46 3 1 1 4 3 4 3 4.5 5 4

12 50 4 1 3 5 8 7 0 1 0.5 1

13 52 2 6 3 5 7 10 9 2 3 1.5 : : : : : : : : : : : : : : : : : : : : : : : :

45 50 1 0.5 0 1 1 4 2 1 0 1

46 52 2 1 2.5 2 3 1 2 0 1 2

47 50 2 0 1 0.5 0 1 1 4 2 1

48 48 2 4 2 3 1.5 2 1 3 6 4

49 49.5 1.5 0.5 2.5 0.5 1.5 0 0.5 0.5 1.5 4.5

50 49 0.5 1 1 3 1 2 0.5 1 0 2

Mean 2.0816 2.5937 3.2553 3.3913 3.6888 4.0454 4.2093 4.3928 4.7926 5.2375

36.3 Example 36.1:


m 𝑀𝑅 𝑑2 𝜎𝑌𝑚 = 𝑀𝑅/𝑑2 𝜎𝑌𝑚

2 𝐺𝑚 =

𝜎𝑌𝑚2

𝜎𝑌12

1 2.08163 1.128 1.8454 3.40557 1.0000

2 2.59375 1.128 2.2994 5.28735 1.5526

3 3.25532 1.128 2.8859 8.32854 2.4456

4 3.39130 1.128 3.0065 9.03889 2.6541

5 3.68889 1.128 3.2703 10.69481 3.1404

6 4.04545 1.128 3.5864 12.86224 3.7768

7 4.20930 1.128 3.7317 13.92522 4.0890

8 4.39286 1.128 3.8944 15.16617 4.4533

9 4.79268 1.128 4.2488 18.05258 5.3009

10 5.23750 1.128 4.6432 21.55906 6.3305

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36.3 Example 36.1:


0.0000

1.0000

2.0000

3.0000

4.0000

5.0000

6.0000

7.0000

1 2 3 4 5 6 7 8 9 10

G(m)

36.3 Example 36.1:


• The plot of 𝐺𝑚 versus the interval 𝑚 is an increasing straight

line, which suggests that an EWMA model is reasonable.

• An estimate for 𝜆 can be obtained from the slope of the line.

Because the line must pass through 𝐺𝑚 = 1 and 𝑚 = 1, the

slope can be obtained from the equation

𝑏 = 𝑥𝑦

𝑥2= 𝑚− 1 [𝐺𝑚 − 1]

[𝑚 − 1]2

=0 0 + 1 .5526 + ⋯+ 9(5.3305)

02 + 12 +⋯+ 92=155.9408

285= 0.54716

• Use the following relationship to solve for 𝜆, 𝜆 = 0.76

𝑏 =𝜆2

1 + (1 − 𝜆)2

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36.3 Example 36.1:


• The limits of the EWMA control chart are determined by the

equation

±3𝜎𝐸𝑊𝑀𝐴 = 𝜆/(2 − 𝜆) ±3𝜎𝑆ℎ𝑒𝑤ℎ𝑎𝑟𝑡

=0.76

2 − 0.7653.74− 48.20 = 0.783 × 5.54 = 4.33

• The resulting control limits are then 52.53 and 43.87

(48.204.33).

36.3 Example 36.1:


Minitab:

Stat

Control Charts

Time-Weighted

Charts

EWMA

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36.3 Example 36.1:


• The limits of the EWMA control chart are determined by the

equation

𝑌 𝑡+1 = 𝑌 𝑡 + 0.76 𝑌𝑡 − 𝑌 𝑡 = 0.76𝑌𝑡 + 0.24𝑌 𝑡

• Let us now consider employing the fitted EWMA model. If we

let the target value for the camshaft diameters be 𝜏 = 50 and let

the first prediction be 𝑌 1 = 50, the fitted EWMA gives the

prediction values and errors in the following table

36.3 Example 36.1:


𝑌𝑡 𝑌 𝑡+1 𝑒𝑡 𝑌𝑡 − 𝜏 𝑌𝑡 −𝑌 𝑒𝑡 2

1 50.0 50.000 0.000 0 3.152159 0.000

2 51.0 50.000 1.000 1 7.703023 1.000

3 50.5 50.760 -0.260 0.25 5.177591 0.068

4 49.0 50.562 -1.562 1 0.601295 2.441

5 50.0 49.375 0.625 0 3.152159 0.391

6 43.0 49.850 -6.850 49 27.29611 46.922

7 42.0 44.644 -2.644 64 38.74525 6.991

8 45.0 42.635 2.365 25 10.39784 5.595

9 47.0 44.432 2.568 9 1.499567 6.593

10 49.0 46.384 2.616 1 0.601295 6.845

11 46.0 48.372 -2.372 16 4.948703 5.627

12 50.0 46.569 3.431 0 3.152159 11.770

13 52.0 49.177 2.823 4 14.25389 7.971

14 52.5 51.322 1.178 6.25 18.27932 1.387

15 51.0 52.217 -1.217 1 7.703023 1.482

16 52.0 51.292 0.708 4 14.25389 0.501

17 50.0 51.830 -1.830 0 3.152159 3.349

18 49.0 50.439 -1.439 1 0.601295 2.071

19 54.0 49.345 4.655 16 33.35562 21.665

20 51.0 52.883 -1.883 1 7.703023 3.545

21 52.0 51.452 0.548 4 14.25389 0.300

22 46.0 51.868 -5.868 16 4.948703 34.439

23 42.0 47.408 -5.408 64 38.74525 29.251

24 43.0 43.298 -0.298 49 27.29611 0.089

25 45.0 43.072 1.928 25 10.39784 3.719

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36.3 Example 36.1:


𝑌𝑡 𝑌 𝑡+1 𝑒𝑡 𝑌𝑡 − 𝜏 𝑌𝑡 − 𝑌 𝑒𝑡 2

26 46.0 44.537 1.463 16 4.948703 2.140

27 42.0 45.649 -3.649 64 38.74525 13.315

28 44.0 42.876 1.124 36 17.84697 1.264

29 43.0 43.730 -0.730 49 27.29611 0.533

30 46.0 43.175 2.825 16 4.948703 7.979

31 42.0 45.322 -3.322 64 38.74525 11.036

32 43.0 42.797 0.203 49 27.29611 0.041

33 42.0 42.951 -0.951 64 38.74525 0.905

34 45.0 42.228 2.772 25 10.39784 7.682

35 49.0 44.335 4.665 1 0.601295 21.764

36 50.0 47.880 2.120 0 3.152159 4.493

37 51.0 49.491 1.509 1 7.703023 2.276

38 52.0 50.638 1.362 4 14.25389 1.855

39 54.0 51.673 2.327 16 33.35562 5.414

40 51.0 53.442 -2.442 1 7.703023 5.961

41 49.0 51.586 -2.586 1 0.601295 6.687

42 50.0 49.621 0.379 0 3.152159 0.144

43 49.5 49.909 -0.409 0.25 1.626727 0.167

44 51.0 49.598 1.402 1 7.703023 1.965

45 50.0 50.664 -0.664 0 3.152159 0.440

46 52.0 50.159 1.841 4 14.25389 3.388

47 50.0 51.558 -1.558 0 3.152159 2.428

48 48.0 50.374 -2.374 4 0.050431 5.636

49 49.5 48.570 0.930 0.25 1.626727 0.865

50 49.0 49.277 -0.277 1 0.601295 0.077

Ave/Sum 48.225 775 613.0302 312.4701

36.3 Example 36.1:


𝑌 𝑡+1 = 48.225

(𝑌𝑡 − 𝜏)2= 775.00 → 𝑠𝜏 =775

50 − 1= 3.98

(𝑌𝑡 − 𝑌 )2= 613.03 → 𝑠𝑌 =613

50 − 1= 3.54

(𝑌𝑡 − 𝑌 𝑡)2= 312.47 → 𝑠𝜏 =

312.47

50 − 1= 2.53

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36.3 Example 36.1:


• From this it seems possible that use of the EWMA as a forecast

to control the process could result in a very large reduction in

variability (i.e., sum of squares from 775.00 to 312.47).

• A comparison of the autocorrelation coefficients of the original

observations with the residuals 𝑒𝑡, after fitting the EWMA model

is as follows:

• Original Observations Y, EWMA Residuals e,

36.3 Example 36.1:


• The residuals suggest independence and support our use of

EWMA as a reasonable model providing useful forecast

information of process performance.

• Consider now what can be done to take active control. We must

be willing to accept a forecast for where a process will be in the

next instant of time.

• When a forecast falls too distant from a target 𝜏, an operator

can then change some influential external factor 𝑋𝑡 to force the

forecast to equal target 𝜏. This differs from the Shewhart model

discussed above in that the statistical approach is now not

hypothesis testing but instead estimation.

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36.3 Example 36.1:


• The application of process controls from an external factor can

be conducted periodically when the response reaches a certain

level relative to the specification. However, for this example we

will consider that adjustments are made after each reading and

that full consequences of taking corrective action can be

accomplished within the next time interval.

• Table 36.5 summarizes the calculations, which can be

explained as follows. Let us consider that 𝑋𝑡 is the current

setting of a control factor, where 𝑌 𝑡+1 is the forecast.

• Also, we can exactly compensate for a discrepancy of

𝑧𝑡 = 𝑌 𝑡+1 − 𝜏 by making the change 𝑥𝑡 = 𝑋𝑡+1 − 𝑋𝑡. When

bringing a process back to its target, we set 𝑔𝑥𝑡 = −𝑧𝑡, where 𝑔

is the adjuster gain.

36.3 Example 36.1:


• The controlling factor is initially set to zero and the first forecast

is 50, the target. The table contains the original observations

and new observations, which differ from original observations

by the amount shown. The difference between the new

observation and the target of 50 is shown as 𝑒(𝑡), while

0.76 𝑒(𝑡) represents the new amount of adjustment needed.

EWMA is determined from the equation 𝑌 𝑡+1 = 0.76𝑌𝑡 + 0.24𝑌 𝑡

• We can see that the 𝑋𝑚𝑅 chart of the new observations shown

in Figure 36.4 is now in control/predictable. The estimated

implications of control to the process are as follows:

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36.3 Example 36.1:


𝑌𝑡 𝑌 𝑡+1 𝑒𝑡 𝜆𝑒𝑡 Adj. New

Obs. EWMA

1 50.0 50.000 0.000 0.000 0.000 50.000 50.000

2 51.0 50.000 1.000 0.760 0.000 51.000 50.000

3 50.5 50.760 -0.260 -0.198 -0.760 49.740 50.760

4 49.0 50.562 -1.562 -1.187 -0.562 48.438 49.985

5 50.0 49.375 0.625 0.475 0.625 50.625 48.809

6 43.0 49.850 -6.850 -5.206 0.150 43.150 50.189

7 42.0 44.644 -2.644 -2.009 5.356 47.356 44.839

8 45.0 42.635 2.365 1.798 7.365 52.365 46.752

9 47.0 44.432 2.568 1.951 5.568 52.568 51.018

10 49.0 46.384 2.616 1.988 3.616 52.616 52.196

11 46.0 48.372 -2.372 -1.803 1.628 47.628 52.515

12 50.0 46.569 3.431 2.607 3.431 53.431 48.801

13 52.0 49.177 2.823 2.146 0.823 52.823 52.320

14 52.5 51.322 1.178 0.895 -1.322 51.178 52.702

15 51.0 52.217 -1.217 -0.925 -2.217 48.783 51.544

16 52.0 51.292 0.708 0.538 -1.292 50.708 49.445

17 50.0 51.830 -1.830 -1.391 -1.830 48.170 50.405

18 49.0 50.439 -1.439 -1.094 -0.439 48.561 48.706

19 54.0 49.345 4.655 3.537 0.655 54.655 48.596

20 51.0 52.883 -1.883 -1.431 -2.883 48.117 53.200

21 52.0 51.452 0.548 0.417 -1.452 50.548 49.337

22 46.0 51.868 -5.868 -4.460 -1.868 44.132 50.257

23 42.0 47.408 -5.408 -4.110 2.592 44.592 45.602

24 43.0 43.298 -0.298 -0.226 6.702 49.702 44.834

25 45.0 43.072 1.928 1.466 6.928 51.928 48.534

𝑌𝑡 𝑌 𝑡+1 𝑒𝑡 𝜆𝑒𝑡 Adj. New Obs. EWMA

26 46.0 44.537 1.463 1.112 5.463 51.463 51.114

27 42.0 45.649 -3.649 -2.773 4.351 46.351 51.379

28 44.0 42.876 1.124 0.854 7.124 51.124 47.558

29 43.0 43.730 -0.730 -0.555 6.270 49.270 50.268

30 46.0 43.175 2.825 2.147 6.825 52.825 49.509

31 42.0 45.322 -3.322 -2.525 4.678 46.678 52.029

32 43.0 42.797 0.203 0.154 7.203 50.203 47.962

33 42.0 42.951 -0.951 -0.723 7.049 49.049 49.665

34 45.0 42.228 2.772 2.106 7.772 52.772 49.197

35 49.0 44.335 4.665 3.546 5.665 54.665 51.914

36 50.0 47.880 2.120 1.611 2.120 52.120 54.005

37 51.0 49.491 1.509 1.147 0.509 51.509 52.572

38 52.0 50.638 1.362 1.035 -0.638 51.362 51.764

39 54.0 51.673 2.327 1.768 -1.673 52.327 51.459

40 51.0 53.442 -2.442 -1.856 -3.442 47.558 52.118

41 49.0 51.586 -2.586 -1.965 -1.586 47.414 48.653

42 50.0 49.621 0.379 0.288 0.379 50.379 47.711

43 49.5 49.909 -0.409 -0.311 0.091 49.591 49.739

44 51.0 49.598 1.402 1.065 0.402 51.402 49.627

45 50.0 50.664 -0.664 -0.504 -0.664 49.336 50.976

46 52.0 50.159 1.841 1.399 -0.159 51.841 49.730

47 50.0 51.558 -1.558 -1.184 -1.558 48.442 51.334

48 48.0 50.374 -2.374 -1.804 -0.374 47.626 49.136

49 49.5 48.570 0.930 0.707 1.430 50.930 47.988

50 49.0 49.277 -0.277 -0.210 0.723 49.723 50.224

Avg 48.20 49.975

36.3 Example 36.1:


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36.3 Example 36.1:


• The estimated implications of control to the process are as

follows:

• It should be noted that for this situation a chart could be created

from the EWMA relationship that describes how an operator in

manufacturing should adjust a machine depending on its

current output.

Mean Std. Deviation

No Control 48.20 3.54

Every observation control 49.98 2.53

chapter 36 exponentially weighted moving average (ewma...

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