Quantifying What a Monitoring Scheme Can Be Expected to Do: the ARL ConceptIE 361 Module 15The Average Run Length ConceptReading: Section 3.5 of Statistical Quality Assurance Methods forEngineersProf. Steve Vardeman and Prof. Max MorrisIow a State UniversityVardeman and Morris (I owa State Univ ersity) IE 361 Module 15 1 / 15Quantifying What a Monitoring Scheme Can Be Expectedto Do: the ARL ConceptThe general question addressed here is "How does one quantify the likelyperformance of a process monitoring scheme?"Once one begins to contemplate alternative schemes for issuingout-of-control signals based on process-monitoring data, the need quicklyarises to quantify what a given scheme might be expected to do. Forexample, what are the pros and cons of adding the Western Electric set ofalarm rules to a control charting scheme? The most e¤ective meansknown for making this kind of prediction is the "Average Run Length"(ARL) notion.Vardeman and Morris (I owa State Univ ersity) IE 361 Module 15 2 / 15The ARLIt is useful to adopt the notationT = the perio d at wh ich a process-monitoring scheme …rst signalsT is called the (random) run length for the scheme. The probabilitydistribution of T is called the run length distribution, and the mean oraverage value of this distribution is called the Average Run Length(ARL) for the process-monitoring scheme. That is,ARL = µTWhen one is setting up a process monitoring scheme, it is desirable that itproduce a large ARL when the process is stable at standard values forprocess parameters and small ARLs under other conditions.Vardeman and Morris (I owa State Univ ersity) IE 361 Module 15 3 / 15The ARL (the Simplest Case)Evaluating ARLs is not usually elementary. But there is one circumstancewhere an explicit formula for ARLs is possible and we can illustrate themeaning and usefulness of the ARL concept in elementary terms. That isthe situation wherethe process-monitoring scheme employs only the single alarm rule"signal the …rst time that a point Q plots outside control limits," andit is sensible to think of the process as physically stable (thoughperhaps not at standard values for process parameters).Under the second condition, the values Q1, Q2, Q3, . . . can be modeled asrandom draws from a …xed distribution, and the notationq = P[Q1plots outside control limits]will prove useful. In this simplest of cases, it follows thatARL =1qVardeman and Morris (I owa State Univ ersity) IE 361 Module 15 4 / 15The ARLExample 15-1 (Some ARLs for Shewhart x-bar Charts)Consider …nding ARLs for a standards given Shewhart x chart based onsamples of size n = 5. If standard values for the pro cess mean andstandard deviation are respectively µ and σ, the relevant control limits areLCLx= µ 3σp5and UCLx= µ + 3σp5Thusq = Px < µ  3σp5or x > µ + 3σp5First suppose "all is well" and the process is stable at standard values ofthe process parameters. Then µx= µ and σx= σ/p5 and if the processoutput is normal, so also is the random variable x. ThusZ =x  µσ/p5is standard normal.Vardeman and Morris (I owa State Univ ersity) IE 361 Module 15 5 / 15The ARLExample 15-1 (x-bar All OK ARL)Thusq = 1  Pµ  3σp5< x < µ + 3σp5= 1  P3 <x  µσ/p5< 3and using a normal table (with an additional signi…cant digit beyond whatis typical) it is possible to establish thatq = 1  P[3 < Z < 3] = .0027to 4 digits. Therefore, it follows thatARL =1.0027= 370The interpretation is that when all is OK (i.e., the process is stable andparameters are at their standard values), the x chart will issue (falsealarm) signals on average only once every 370 plotted points.Vardeman and Morris (I owa State Univ ersity) IE 361 Module 15 6 / 15The ARLExample 15-1 (One "Not OK" x-bar Case)In contrast, consider the possibility that (while the process standarddeviation is at its standard value) the process mean is one processstandard deviation above its standard value. In these circumstances onestill has σx= σ/p5, but now µx= µ + σ (µ and σ are still the standardvalues of respectively the process mean and standard deviation). Then,q = 1 Pµ  3σp5< x < µ + 3σp5= 1 P"µ  3σp5 (µ + σ)σ/p5<x  (µ + σ)σ/p5<µ + 3σp5 (µ + σ)σ/p5#= 1 P[5.24 < Z < .76]= .2236Vardeman and Morris (I owa State Univ ersity) IE 361 Module 15 7 / 15The ARLExample 15-1 (One "Not OK" x-bar Case)The following …gure illustrates the calculation being done here and showsthe roughly 22% chance that under these circumstances the sample meanwill plot outside x chart control limits.Figure: q Where µx= µ + σ (Distribution of x Pictured)So for this case, ARL = 1/.2236 = 4.5. That is, if the process mean is o¤target by as much as one process standard deviation, then it will take onaverage only 4.5 samples of size n = 5 to detect this misadjustment.Vardeman and Morris (I owa State Univ ersity) IE 361 Module 15 8 / 15The ARL (Generalities)Example 15-1 should agree completely with intuition about "how thingsshould be." It says that when a process is on target, one can expect longperiods between signals from an x chart. On the other hand, should theprocess mean shift o¤ target by a substantial amount, there will typicallybe quick detection of that change.It is possible to produce elementary formulas for ARLs only for very simplecases. Where the rules used to convert observed values Q1, Q2, Q3, . . .into out-of-control signals or the probability model for these variables aremore complicated than the combination of the "one point outside controllimits" rule and the stable process model, elementary computations arerare. But it is not necessary to understand the more advanced detailsneeded to produce ARLs in order to appreciate and interpret what an ARLsays about a monitoring scheme.Vardeman and Morris (I owa State Univ ersity) IE 361 Module 15 9 / 15The ARL (Generalities)A result that should be mentioned here concerns ARLs for control chartswhen some of the "special checks" for patterns discussed earlier are used.Champ and Woodall in a 1987 Technometrics paper gave ARLs formonitoring schemes that use various combinations of the four WesternElectric alarm rules. As an example of conclusions that were reached,consider ARLs for an x chart. We have seen that the "all OK" ARL for ascheme using only the "one point outside 3σxcontrol limits" rule is about370. When all four Western Electric rules are employed simultaneously,Champ and Woodall found