Chapter 8: Cusum

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Chapter 8 Cusum EWMA Charts
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Cusum charts

• Shewhart charts are not always sensitive to
  shifts in parameter values
• The Cusum technique is sensitive
• We will look at Cusum charts for changes in the
  mean

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Cusum charts

• Deviation from a reference value, k, is
  maintained
• Xi k, where k is a constant
• C1 X1 k
• C2 (X2 k) (X1 k) (X2 k) C1
• C3 (X3 k) C2
• Cm (Xm k) Cm-1
• The Ci values are plotted to form a rudimentary
  Cusum chart

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Cusum charts

• Consider a desired level of a process at its mean
  m0
• If the mean output of a process, Xbar stays
  around m0, the cusum will be roughly horizontal
• With approximately the same number of values
  above and below m0

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Example

• Given 20 values from a N(0, 1) followed by 20
  values from a N(1, 1)
• These observations are sample means
• Assume the reference value is zero
• First, these values are plotted on a Shewhart
  chart and an R chart
• Then, they are plotted on a rudimentary Cusum
  chart

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Note on Shewhart chart

• Although there are two observations that seem
  somewhat high, the shift is not detected

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Rudimentary Cusum chart
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Notes on rudimentary Cusum chart

• There is no slope on the first 20 samples
• But, after the first 20 samples, the slope is
  definitely steep
• If the alarm value (h) is 5, a call to action
  would have been signaled on the 24th observation
• What is the appropriate value of h?

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One-sided Cusum

• We have been talking about two-sided Cusum charts
• Initially, Cusum charts were one-sided
• A variation of Cm S(Xbari k) is plotted where
  k is the reference value
• Suppose that there is a quality level, m0, that
  is considered acceptable and another level, m1,
  that is considered rejectable

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One-sided Cusum

• Reference value, k
• k (m0 m1)/2
• If Cm falls below zero, reset to zero
• This is the variation
• Cm gt h is a signal that the process mean has
  shifted to some value greater than k

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One-sided Cusum

• The proper value of h
• Base it on the ARL
• The ARL should be large if the process mean is
  stable at m0
• The ARL should be small if the process mean has
  shifted to m1
• ARL at m0, L0
• ARL at m1, L1

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ARLs for several Cusum schemes
   
A h SQRT(n)/s B ABS(k m0)SQRT(n)/s
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Example

• Assume m0 10 and m1 10.4
• Given s .6
• Find a Cusum scheme that comes close to L0 500
  and L1 3
• From the Table
• B 1.04
• A 2.26

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Example, cont.

• K (10 10.4)/2 10.2
• B .2 SQRT(n)/.6
• n 9.7 10
• A h SQRT(10/.6 2.26
• h .43
• Summary Take samples of n 10, and when Qm gt
  .43 that is the signal that the process is out of
  control. When Qm lt 0, reset it to zero.

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Using a nomogram

• Procedure
• Connect the desired L0 and L1
• Results in a point on the B scale
• Determine n from
• n Bs/ABS(k m0)2
• Usually round n up unless it is slightly above an
  integer
• Recompute B using the rounded n

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Using a nomogram

• Procedure
• Connect the new value of B to the desired value
  on the L0 scale
• Note the value on the L1 scale
• Determine h from the value of A
• The Cusum scheme is now specified

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Using a nomogram

• Procedure
• Alternate Cusum scheme is obtained by connecting
  a point on the B scale to the desired value on
  the L1 scale and noting the value on the L0 scale
• The final value on the A scale is read resulting
  in another Cusum scheme

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Using a nomogram

• Procedure
• Two additional Cusum schemes may be obtained by
  rounding n in the other direction
• There will be four Cusum schemes
• Choose on the basis of how close the schemes come
  to the desired ARL values
• (If n happens to be an integer, there will only
  be one Cusum scheme)

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Example

• One-sided Cusum scheme with ARL0 400 when the
  mean is 80 (acceptable quality) and ARL1 5
  when the mean is 100 (rejectable quality)
• Process output is normally distributed
• Standard deviation is 20

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Example, cont.

• k (100 80)/2 90
• Connect L1 5 and L0 400
• Read B .722
• 10 SQRT(n)/20 .722
• n 2.08
• Round n to 2
• 10 SQRT(2)/20 .707

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Example, cont.

• Connect B .707 and L0 400 reading A
  3.16
• h SQRT(2)/20 3.16
• h 44.69
• Summary Compute S(Xbari 90). If this value
  becomes negative, start anew
• If the summation exceeds 44.69, the process is
  out of control

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Example, cont.

• The line connecting L0 400 and A 3.16 also
  intersects L1 5.2
• The scheme k 90, h 44.69, n 2 yields the
  desired ARL at m0 80, but a slightly worse ARL
  at m1 100

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Example, cont.

• Alternate
• Connecting B .707 and L1 5, we could have
  found a scheme that holds the ARL at m1 100,
  but has L0 300 at m0 80

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Example, cont.

• More alternates
• Since n 2.08 and was rounded down to 2, a
  conservative approach would be to round n up to 3
• 10 SQRT(3)/20 .866
• Connect B .866 to L0 400
• h SQRT(n)/s 2.6
• h 2.6 (20)/SQRT(3) 30.02

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Example, cont.

• More alternates
• The line intersects L1 3.8 which is better than
  the called for ARL at m0 80
• Connecting the points B .866 and L1 5 yields
  an extremely large ARL at m1 100
• Which scheme is the best?
• Probably the very first scheme

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First example

• Consider the 40 values, 20 from N(0, 1) followed
  by 20 from N(1,1)
• We are concerned about increases from m0 0
  to m0 1 with L0 500
• Here n 1 and s 1
• B .5 SQRT(1)/1 .5
• A h 4.42 and L1 9.5 (compared to 44 on a
  Shewhart chart)

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First example, 2-sided

• Suppose that we are concerned with decreases to
  m2 -1 as well as increases to m1 1
• We just determined that h 4.42
• The ARLs will be
• 1/L0 1/500 1/500 giving L0 250
• 1/L1 1/9.5 1/9.5 giving L1 4.75

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8-1.2 The Tabular or Algorithmic Cusum
for Monitoring the Process Mean (two
sided)

• Accumulate derivations from the target ?0 above
  the target with one statistic, C
• Accumulate derivations from the target ?0 below
  the target with another statistic, C
• C and C- are one-sided upper and lower cusums,
  respectively.

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8-1.2 The Tabular or Algorithmic Cusum
for Monitoring the Process Mean

• The statistics are computed as follows
• The Tabular Cusum
• starting values are
• K is the reference value (or allowance or slack
  value)
• If either statistic exceeds a decision interval
  h, the process is considered to be out of
  control. Often taken as h 5?

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8-1.2 The Tabular or Algorithmic Cusum
for Monitoring the Process Mean

• Example 8-1
• ?0 10, n 1, ? 1
• Interested in detecting a shift of 1.0?
  1.0(1.0) 1.0
• Out-of-control value of the process mean ?1 10
  1 11
• k ½ and h 5? 5 The equations for the
  statistics are then

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8-1.2 The Tabular or Algorithmic Cusum
for Monitoring the Process Mean

• Example 8-1
• If an adjustment has to be made to the process,
  may be helpful to estimate the process mean
  following the shift.
• The estimate can be computed from
• N, N- are counters, indicating the number of
  consecutive periods that the cusums C or C- have
  been nonzero.

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Example 8-1

• Pgs. 411-414
• Note on page 414, the new mean is estimated as m0
  k C29/N
• 10 .5 5.28/7 11.25

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8-1.2 The Tabular or Algorithmic Cusum
for Monitoring the Process Mean

• Example 8-1
• The cusum control chart indicates the process is
  out of control.
• The next step is to search for an assignable
  cause, take corrective action required, and
  reinitialize the cusum at zero.
• If an adjustment has to be made to the process,
  may be helpful to estimate the process mean
  following the shift.

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8-2. The Exponentially Weighted Moving
Average Control Chart

• The Exponentially Weighted Moving Average Control
  Chart Monitoring the Process Mean
• The exponentially weighted moving average (EWMA)
  is defined as
• where 0 lt ? ? 1 is a constant.
• z0 ?0 (sometimes z0 )

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8-2.1 The Exponentially Weighted Moving
Average Control Chart Monitoring the Process Mean

• The control limits for the EWMA control chart are
• where L is the width of the control limits.

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8-2.1 The Exponentially Weighted Moving
Average Control Chart Monitoring the Process Mean

• As i gets larger, the term 1- (1 - ?)2i
  approaches zero.
• This indicates that after the EWMA control chart
  has been running for several time periods, the
  control limits will approach steady-state values
  given by

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8-2.2 Design of an EWMA Control Chart

• The design parameters of the chart are L and ?.
• The parameters can be chosen to give desired ARL
  performance.
• In general, 0.05 ? ? ? 0.25 works well in
  practice.
• L 3 works reasonably well (especially with the
  larger value of ?.
• L between 2.6 and 2.8 is useful when ? ? 0.1
• Similar to the cusum, the EWMA performs well
  against small shifts but does not react to large
  shifts as quickly as the Shewhart chart.
• EWMA is often superior to the cusum for larger
  shifts particularly if ? gt 0.1

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Example 8-2

• Pgs. 428-431

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First example in the notes for this chapter

• N(0,1) to N(1,1), 2-sided, l .2, L 3
• See next slide

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Out-of-control on 25th sample
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8-2.4 Robustness of the EWMA to
Non-normality

• As discussed in Chapter 5, the individuals
  control chart is sensitive to non-normality.
• A properly designed EWMA is less sensitive to the
  normality assumption.

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Assignment

• Suggestion 8-1, 8-7, 8-15, 8-19

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End