16-9 CONTROL CHART PERFORMANCE 631That is, the control chart will require two samples to detect the process shift, on the aver-
age, so two hours will elapse between the shift and its detection (again on the average).
Suppose this approach is unacceptable, because production of piston rings with a mean di-
ameter of 74.0135 millimeters results in excessive scrap costs and delays final engine as-
sembly. How can we reduce the time needed to detect the out-of-control condition? One
method is to sample more frequently. For example, if we sample every half hour, only one
hour will elapse (on the average) between the shift and its detection. The second possibil-
ity is to increase the sample size. For example, if we use n10, the control limits in
Fig. 16-3 narrow to 73.9905 and 74.0095. The probability of falling between the control
limits when the process mean is 74.0135 millimeters is approximately 0.1, so p0.9, and
the out-of-control ARL isThus, the larger sample size would allow the shift to be detected about twice as quickly as the
old one. If it became important to detect the shift in the first hour after it occurred, two con-
trol chart designs would work:ARL1
p1
0.91.11XTable 16-6 Average Run Length (ARL) for an Chart with 3-Sigma
Control Limits
Magnitude of ARL ARL
Process Shift n 1 n 4
0 370.4 370.4
0.5 155.2 43.9
1.0 43.9 6.3
1.5 15.0 2.0
2.0 6.3 1.2
3.0 2.0 1.0XDesign 1 Design 2
Sample size: n 5 Sample size: n 10
Sampling frequency: every half hour Sampling frequency: every hourTable 16-6 provides average run lengths for an chart with 3-sigma control limits. The aver-
age run lengths are calculated for shifts in the process mean from 0 to 3.0and for sample
sizes of n1 and n4 by using 1p, where pis the probability that a point plots outside of
the control limits. Figure 16-18 illustrates a shift in the process mean of 2.Xμμσ+ 2Figure 16-18
Process mean shift
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