control charts cusum charts

Contents

Control charts CUSUM charts

22S39: Class Notes / October 14, 2000 back to start 1

Control charts


Control charts

• Aim to develop tools for checking whether or not a process is under (statistical)control, i.e., whether or not the population distribution remains the same overtime.


Control charts


• First, check whether or not the population mean is constant over time.


Control charts


• First, check whether or not the population mean is constant over time.

• This can be done by taking samples of size n over every time unit (hour, shift,day, etc.) For each sample, compute the sample means, standard deviationsand range.


• Example: consider the following samples on the overall length (actuallength-5.00mm) of a cigarette lighter body used in an automobile application:

obs1 obs2 obs3 obs4 xbar range s1 15 10 8 9 10.50 7 3.1102 14 14 10 6 11.00 8 3.8303 9 10 9 11 9.75 2 0.9574 8 6 9 13 9.00 7 2.9405 14 8 9 12 10.75 6 2.7506 9 10 7 13 9.75 6 2.5007 15 10 12 12 12.25 5 2.0608 14 16 11 10 12.75 6 2.7509 11 7 16 10 11.00 9 3.74010 11 14 11 12 12.00 3 1.410

xbar bar = 10.875range bar =5.9 and s bar = 2.604

11 13 8 9 5 8.75


12 10 15 8 10 10.75


12 10 15 8 10 10.75

• Assume the length, X, of a random lighter distributes as N(µ, σ2). Then, thesample means X̄ distribute as N(µ, σ2/n) or nearly so by the Central LimitTheorem if X is non-normally distributed.


12 10 15 8 10 10.75


• There is 0.9973 probability that the X̄’s lie within µ± 3σ/√n.


12 10 15 8 10 10.75



• Control charts: plot the X̄’s against the order the sample is collected andsuperimposed on the chart three horizontal lines at µ− 3σ/

√n, µ and

µ+ 3σ/√n.


12 10 15 8 10 10.75




√n, µ and

µ+ 3σ/√n.

• The lowest and highest lines form respectively the lower and upper controllimits. Observations that lie outside these two control limits are rareobservations that may indicate that the process is out of control.


12 10 15 8 10 10.75




√n, µ and

µ+ 3σ/√n.

• The lowest and highest lines form respectively the lower and upper controllimits. Observations that lie outside these two control limits are rareobservations that may indicate that the process is out of control.

• We then need to identify possible sources leading to these ”out-of-control”observations. For example, are they due to new suppliers or new workers?


• When µ and σ are unknown, they can be estimated from samples taken fromperiods when the process is known to be under control: µ is estimated by thegrand mean ¯̄X = (X̄1 + · · ·+ X̄m)/m.



• Two methods for estimating σ:




1. Define S̄ = (S1 + · · ·+ Sm)/m and use it to estimate σ. To allow for theerror due to estimating σ, the control limits become ¯̄X ±A3S̄ where A3

depends on the sample size and can be looked up from table C.1.





depends on the sample size and can be looked up from table C.1.2. For the lighter example, the control limits are

10.875± 1.63 ∗ 2.604 = (6.630, 15.12).






10.875± 1.63 ∗ 2.604 = (6.630, 15.12).3. Define the average range R̄ = (R1 + · · ·Rm)/m. Now, replace the control

limits by ¯̄X ±A2R̄. In other words, A3S̄ ≈ A2R̄ ≈ 3σ/√n.








4. For the lighter example, the control limits are10.875± 0.73 ∗ 5.9 = (6.568, 15.182).








4. For the lighter example, the control limits are10.875± 0.73 ∗ 5.9 = (6.568, 15.182).

5. The two methods yield very similar control limits. In practice, the lattermethod may be preferred for its simplicity.


2 4 6 8 10 12

05

1015

20

1:12

xbar

Upper control limit

Lower control limit

Figure 1: Control chart for the length of cigarette lighter bodies


• Similarly, we may want to monitor whether or not the population variance staysthe same over time.



• This can be done by plotting the R’s and see whether or not that are withinthe control limits D3R̄ to D4R̄. The constants D3 and D4 depend on n andcan be looked up from table C.1.



• This can be done by plotting the R’s and see whether or not that are withinthe control limits D3R̄ to D4R̄. The constants D3 and D4 depend on n andcan be looked up from table C.1.

• Lighter example: n = 4 so D3 = 0, D4 = 2.28 Hence the lower control limit is0 and the upper limit is 2.28 ∗ 5.9 = 13.45.


2 4 6 8 10 12

05

1015

1:12

rang

e

UCL

Figure 2: Control chart for the range of the length of cigarette lighter bodies


• When a process is under control, it means that its associated distributiondoesn’t change over time.



• The next step is to make sure that the distribution conforms to thespecifications. For example, if we want 99.73% of the lighters to be withinµ± δ, with δ equal to, say, 5. This translates to requiring that 3σ ≤ δ.




• Recall that A2R̄ ≈ 3σ/√n so that A2R̄

√n ≈ 3σ. Hence a process that is

under control will also meet the specification if ¯̄x±A2R̄√n is inside the

interval µ± δ!







interval µ± δ!

• Lighter example, A2R̄√n = 0.73 ∗ 5.9 ∗

√4 = 8.614 > 5. Hence, even though

the manufacturing process is under control, not all its products are meeting thespecification.







interval µ± δ!

• Lighter example, A2R̄√n = 0.73 ∗ 5.9 ∗

√4 = 8.614 > 5. Hence, even though

the manufacturing process is under control, not all its products are meeting thespecification.

• What is the defective rate?


• A weakness of control chart is that it is insensitive to small shift in the mean.



• To illustrate this point, consider taking samples of size n = 4. Suppose µ isshifted to µ+ σ/

√n. Hence, X̄ ∼ N(µ+ σ/

√n, σ/

√n).





√n, σ/

√n).

• Then P (µ− 3σ/√n < X̄ < µ+ 3σ/

√n) = µ−3σ/

√n−(µ+σ/

√n)

σ/√n

<X̄−(µ+σ/

√n)

σ/√n

< µ−3σ/√n−(µ+σ/

√n)

σ/√n





√n, σ/

√n).

• Then P (µ− 3σ/√n < X̄ < µ+ 3σ/

√n) = µ−3σ/

√n−(µ+σ/

√n)

σ/√n

<X̄−(µ+σ/

√n)

σ/√n

< µ−3σ/√n−(µ+σ/

√n)

σ/√n

= P (−4 < Z < 2) = 0.9772.





√n, σ/

√n).

• Then P (µ− 3σ/√n < X̄ < µ+ 3σ/

√n) = µ−3σ/

√n−(µ+σ/

√n)

σ/√n

<X̄−(µ+σ/

√n)

σ/√n

< µ−3σ/√n−(µ+σ/

√n)

σ/√n

= P (−4 < Z < 2) = 0.9772.

• In other words, the probability that the sample mean falls outside the controllimits is 1− 0.9772 = 0.0228.


• Here is a simulated example with n = 4 and σ = 2.

-1.15 -0.15 -0.31 2.77 2.17 0.20 0.39 1.27 -1.01 2.00 0.94 2.051.66 -0.77 2.42 0.52 2.45 -1.25 2.76 1.35 -0.34 0.11 0.78 1.222.09 3.47 1.68 1.53 0.19 3.11 0.41 1.41 -0.63 0.41 -0.55 -0.470.89 -0.84 0.79 0.67 1.67 -0.32 0.56 2.15 1.19 2.14 1.29 0.910.95 -0.64 0.86 1.27 1.27 -0.04 1.16 0.94 0.95 1.22 -0.85 2.52

This takes 26 runs of experiments to detect the change.a second simulation

0.38 0.31 -0.90 0.85 1.74 -0.26 3.61 0.89 0.94 -0.62 1.06 1.92-0.20 1.14 0.53 0.46 0.75 1.91 2.46 0.90 1.70 1.43 1.48 2.381.24 1.50 1.57 1.26 2.25 0.10 -1.23 0.26 1.16 -0.24 0.82 0.862.67 0.94 2.35 0.71 1.39 1.54 0.03 0.48 -0.32 0.81 0.23 0.760.39 -1.03 -0.26 1.84 0.77 2.59 1.51 2.35 2.30 -0.48 0.11 2.42

This takes 7 runs of experiments to detect the change.


• How many runs of experiments will it take to have a sample mean outside thecontrol limits?



• This is like flipping a biased coin with P (H) = p = 0.0228 and finding themean number of flips needed to get the first H.




• Let Y =number of flips to get the first H.

y 1 2 · · · y · · ·f(y) p qp · · · qy−1p · · ·





y 1 2 · · · y · · ·f(y) p qp · · · qy−1p · · ·

• It can be verified that E(Y ) = 1/p.





y 1 2 · · · y · · ·f(y) p qp · · · qy−1p · · ·

• It can be verified that E(Y ) = 1/p.

• Hence, if p = 0.0228, it takes on the average 1/0.0228 = 43.9 runs before thefirst sample mean goes beyond the control limits, signifying that there may be achange in the population mean.


• Above, 44 is the so-called average run length for detecting a shift of σ/√n in

the mean where n is the sample size in each run of experiment.


• Above, 44 is the so-called average run length for detecting a shift of σ/√n in

the mean where n is the sample size in each run of experiment.

• It is desirable to devise a procedure which will take a long average run length togive a false alarm, but a short mean run length to detect a mean shift.


cusum chart


cusum chart

• The idea of the cumsum (cumulative sum) chart is that by adding up thesample means, a small shift in the mean will result in a trend that is easier todetect.


cusum chart


• To illustrate, we simulate 10 N(0, 0.12) values followed by another 10N(0.1, 0.12) : 0.03, 0.11, 0.08, 0.10, 0.00, 0.09, 0.03,−0.05,−0.09,−0.10, 0.14−0.02, 0.07,−0.08, 0.14, 0.01, 0.17, 0.10, 0.09, 0.03


cusum chart


• To illustrate, we simulate 10 N(0, 0.12) values followed by another 10N(0.1, 0.12) : 0.03, 0.11, 0.08, 0.10, 0.00, 0.09, 0.03,−0.05,−0.09,−0.10, 0.14−0.02, 0.07,−0.08, 0.14, 0.01, 0.17, 0.10, 0.09, 0.03

• We now sum up the data cumulatively, i.e,.0.03, 0.03 + 0.11, 0.03 + 0.11 + 0.08, and so on to get0.03, 0.14, 0.22, 0.32, 0.32, 0.41, 0.44, 0.39, 0.30, 0.20, 0.34, 0.32, 0.39, 0.31, 0.450.46, 0.63, 0.73, 0.82, 0.85.


• Note that around the tenth sample, the cumulative sum takes off to infinity,suggesting a small positive mean shift.



• This can be verified as follows: let the population mean be 0 fromt = 1, 2, · · · , 10 and then it shifts to δ > 0. DefineS0 = 0, S1 = X̄1, S2 = X̄1 + X̄2, and so on. ThenE(S1) = 0, E(S2) = E(X̄1 + X̄2) = 0 + 0 = 0 and so on until t = 11,E(S11) = E(S̄10 + X̄11) = 0 + EX̄11 = δ andE(S12) = E(S11 + X̄12) = δ + δ = 2δ, and E(St) = (t− 10) ∗ δ, t ≥ 10.




• This shows that if the population mean has shifted from zero to a larger value,the cumumlative sums will follow an upward trend pattern soon after the timeof the shift.





• Suppose that we want to detect a positive shift in the mean.





• Suppose that we want to detect a positive shift in the mean.

• Let µ0 be the Acceptable Mean Level (AQL) and if the mean is shifted theRejectable Mean Level (RQL), µ1 > µ0.


• Let k = (µ0 + µ1)/2. Define the cumulative (partial) sums of the (k-deleted)sample means S1 = (X̄1 − k), S2 = (X̄1 − k) + (X̄2 − k), and so on.



• If the true population mean is µ0, then E(S1) = µ0 − k =−(µ1 − µ0)/2, E(S2) = −2(µ1 − µ0)/2, E(S3) = −3(µ1 − µ0)/2, and so on.




• In other words, the means of the cumulative sums are increasingly negative, ifthe true population mean stay at µ0.





• So if the process is under control with the population mean staying at µ0, thepartial sums will tend to −∞.






• On the other hand, if the process shifts to µ1 at say, t = 11, thenE(S11) = E(S10 + X̄11 − k)






• On the other hand, if the process shifts to µ1 at say, t = 11, thenE(S11) = E(S10 + X̄11 − k) = E(S10) + E(X̄11 − k)






• On the other hand, if the process shifts to µ1 at say, t = 11, thenE(S11) = E(S10 + X̄11 − k) = E(S10) + E(X̄11 − k) = E(S10) + µ1 − k =−10(µ1− µ0)/2 + (µ1− µ0)/2 = −9(µ1− µ0)/2 which is larger than E(S10).






• On the other hand, if the process shifts to µ1 at say, t = 11, thenE(S11) = E(S10 + X̄11 − k) = E(S10) + E(X̄11 − k) = E(S10) + µ1 − k =−10(µ1− µ0)/2 + (µ1− µ0)/2 = −9(µ1− µ0)/2 which is larger than E(S10).

• Indeed, t = 11 is the turning point of the cumulative sums, as the summandsX̄t − k, t ≥ 11 have positive mean (µ1 − µ0)/2.


• In summary, the cumulative sums (CUSUM) tend to decrease monotonically ifthe process is under control, i.e., the population mean stays at the desirablevalue µ0, but they tend to increase monotonically after the population meanshifts to the RQL µ1.



• Thus, a shift of the mean to the RQL may be indicated if the CUSUMs reversethe downward trend.



• Thus, a shift of the mean to the RQL may be indicated if the CUSUMs reversethe downward trend.

• This reversal can be measured by dt = St −min(Si, i = 0, 1, 2, · · · , t), whichmeasures the height of the current cumumlative sum relative to the all time lowamong the current and all past CUSUMS.


• Below is an illustration with the AQL being 0 and the RQL equal to 1 so thatk = 0.5, with n = 4:

sample means-0.840 0.870 1.600 -1.100 -0.240 0.850 -0.240 -0.820 0.097 -2.9001.800 2.200 0.083 0.830 0.460 1.800 1.600 1.300 0.078 -0.940

sample means -0.5-1.300 0.370 1.100 -1.600 -0.740 0.350 -0.740 -1.300 -0.400 -3.4001.300 1.700 -0.420 0.330 -0.042 1.300 1.100 0.830 -0.420 -1.400

cumumlative sums of the (sample means -0.5)-1.340 -0.970 0.130 -1.470 -2.210 -1.860 -2.600 -3.920 -4.323 -7.723-6.423 -4.723 -5.140 -4.810 -4.850 -3.550 -2.450 -1.650 -2.072 -3.512the minimum of the partial sums-1.340 -1.340 -1.340 -1.470 -2.210 -2.210 -2.600 -3.920 -4.323 -7.723-7.723 -7.723 -7.723 -7.723 -7.723 -7.723 -7.723 -7.723 -7.723 -7.723the d’s0.000 0.370 1.470 0.000 0.000 0.350 0.000 0.000 0.000 0.000 1.300 3.0002.583 2.913 2.873 4.173 5.273 6.073 5.651 4.211


• There is a simpler formula for computing the d’s, namely,d1 = 0, dt+1 = min(dt + (xt+1 − k), 0), for t ≥ 1.


• There is a simpler formula for computing the d’s, namely,d1 = 0, dt+1 = min(dt + (xt+1 − k), 0), for t ≥ 1.

• The process is said to be out of control and the mean shifted to the RQL orhigher if the d’s exceed some threshold, denoted as h.


5 10 15 20

−3

−2

−1

01

2

t

sam

ple

mea

n

5 10 15 20

−8

−6

−4

−2

0

t

cum

sum

5 10 15 20

01

23

45

6

t

d

Figure 3: CUSUM chart for the above simulated data where h = 5.


• The threshold h and the sample size n should be chosen so that it takes a largeaverage run length to have a false alarm but a short average run length todetect a mean shift to the RQL.



• The nomograms on p.204 and 205 are useful for determing h and n.




• For example, if n = 4, σ = 2, h = 5, µ0 = 0 and µ1 = 1, then

k = 0.5, |k−µ0|√n

σ = 0.5, h√n

σ = 5.0 From Fig. 5.2-2(a), the average run lengthfor a false alarm is about 950.





k = 0.5, |k−µ0|√n

σ = 0.5, h√n


• Note that |µ1−k|√n

σ = |k−µ0|√n

σ = 0.5. Fig. 5.2-2(b) shows that it takes on theaverage of 10.5 runs to detect a shift from µ0 = 0 to µ1 = 1.





k = 0.5, |k−µ0|√n

σ = 0.5, h√n



σ = |k−µ0|√n


• Suppose that n = 4. How should we set h so that it takes on the average of500 runs for a false alarm?





k = 0.5, |k−µ0|√n

σ = 0.5, h√n



σ = |k−µ0|√n


• Suppose that n = 4. How should we set h so that it takes on the average of500 runs for a false alarm?

• From Fig. 5.2-2(a), h√n

σ = 4.4. Hence, h = 4.4σ/√n = 4.4.


• Now, if h = 4.4, from Fig. 5.2-2(b), it takes on the average 9 runs to detectthe shift from the AQL to the RQL.


control charts cusum charts

Documents