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2 not aear to be so clear, at the cost of a longer and more exensive rocess (i.e., drawing the second samle). Consider the following examle. Assume that we have a device with mission time T (say, 2 hours), which requires (H ) a reliability of say,.9, with confidence.95. To test this hyothesis, we lace n (say 2) items on test for the T hours, and then count how many items survive. The number of survivals X is distributed as a Binomial (n 2, ), where is the reliability of the device; that is, the robability that such device survives beyond mission time T. We then exress such a Binomial robability model as: c x n-x P{X > c} 1- P{X c} 1- C n x (1- ) x 1 Now, assume that a reliability of.8 or less is unaccetable (H 1 ) and define the double samling Plan S (n 1 2, n 2 2, c 1 14, c 2 15, c 3 33), as described. We draw a first samle of size n 1 2 and count the number of survivals X. If X > 15, we don t reject H. If X < 14 we reject H (reliability is unaccetable). If X is 14 or 15, we draw a second samle of size n 2 2 and count the number of survivals (Y). Then, only if X + Y < 33 we reject H (and decide that the device reliability is unaccetable). Plan S (n 1 2, n 2 2, c 1 14, c 2 15, c 3 33) is constructed in the following way. For the first samle (n 1 ) we selected, from the Binomial (n 2,.9) tables, number c 2 15, because the robability of accetance (survivals X > 15) of a good batch (reliability >.9) is P.9 {X > 16}.957. We then selected number c 1 14 because the robability of rejecting a good batch P.9 {X < 13}.2. Then, from the Binomial (n 2,.8) table, the robability of acceting a bad batch (reliability.8 and X > 16) is 1- P.8 {X < 15} , and that of rejecting a bad batch is P.8 {X < 13}.87. If the number of survivals is X 14 or 15, results are considered inconclusive. The robability of an inconclusive result, when the batch is good, is.41, and when the batch is bad, is.284. If so, we take a second samle of n 2 2 and define number c We accet the batch if the total survivals are 33 or more; and reject it if the survivals are less than 33. We don t claim that lan S is otimal; but it rovides a good illustration of the construction aroach. The robability of accetance for such double samling lan S, for any, is given by the following equation. P{Acceting Batch} P{Acceting Initially} + P{Initially Inconclusive Then Accet at 2nd} P{First Successes > 16} + P{First Successes 14 or 15 and Combined Successes > 33} 2 Bin(x; n 2, ) + Bin(x 14; n 2, ) x 16 Table 1. Individual and Cum. Probs. for Binomial (n 2;.9 and.8) Cum. For Cum. For Surv. P.9 P.8 P 9 P x [Bin(x 19; n 2, ) + Bin(x 2; n 2, )] + Bin(x 15; n 2, ) x [Bin(x 18; n 2, )] + Bin(x 19; n 2, ) + Bin(x 2; n 2, ) Notice that, to obtain a result of 33 survivals or more, in the combined first and second samles, we have to obtain 14 or 15 successes (inconclusive results) in the first samle and then, enough successes in the second (say 18, 19, or 2) to add u to 33 or more. For the case where reliability.9, the double samling lan accetance is.982, instead of just.957 for a fixed samle test with n 2 and c 16: P{Acceting Batch} P{Acceting Initially} + P{Initially Inconclusive Then Accet at 2nd} x ( ) x ( ).982 The robability of rejection of our double samling lan S is obtained just by substituting accetance for rejection, in the equations. For a true reliability.9, the robability of incorrect rejection is.18, instead of just P{X < 16} , which is the corresonding robability for a fixed samle lan with n 2 and c 16: P{Rejecting Batch} P{Rejecting Initially} + P{Initially Inconclusive Then Reject at 2nd} P{First Successes < 13} + P{First Successes 14 or 15 and Combined Successes < 33} 13 Bin(X; n 2, ) + Bin(x 14; n 2, ) x x {1 - [Bin(x 19; n 2, ) + Bin(x 2; n 2, )]} + Bin(x 15; n 2, ) x {1 - [Bin(x 18; n 2, ) + Bin(x 19; n 2, ) + Bin(x 2; n 2, )]} x (1 - ( )) x (1 - ( )).18 2

3 The receding shows how the double samling scheme, when comared with a single samle test, not only increases the robability of acceting a good batch but also reduces the robability of rejecting a good one, even if the initial test results are inconclusive. This characteristic is their strongest advantage that, in many cases, far outweighs the extra cost and effort involved in imlementing such double samling schemes. The Sequential Probability Ratio Test (SPRT) Assume now that we want to test that the accetable mean life of a device (MTTF) is 2 hours or more (null hyothesis H ) versus that it is 1 hours or less (H 1 ). If the device lives are distributed Exonential (with mean µ) the distribution (CDF) and density (df) of the random variable (r.v.) life of a device, denoted X, under H i : µ i, for i, 1 are: -X -X µ 1 F (X) 1- e i µ ; f(x) e i µ ; i.1; µ 2; µ 1 1 i µ i Assume that we lace such devices, sequentially on test, one at a time, for a test duration of T 2 hours, after which we assess whether each device is still working (Pass) or not (Fail). The robabilities of such results, under the two above hyotheses H i, for i, 1 are: i P{Success Under Hi} Pµ {DeviceOutlives T 2} i -2 µ P {X > 2} e i µ i,1-2 P{Success Under H } e 2 e P{Success Under H } e1 e We can transform these relationshis into the resective equivalent hyotheses: H :.95 and H 1 : In such a setting, the test of each device for 2 hours is assumed to be an indeendent trial, with identical robability of success i. Therefore y, the cumulative number of successes, out of n trials, is distributed as a Binomial (n, i ), for i, 1. Define the Probability Ratio (PR) as that of the Binomial distributions, under H and H 1 : P{"y"Successes Under H1 } P{"y"Successes Under H} ; i Binomial (y; n, Binomial (y; n, y y K n-y n-y 1 (1-1) 1 (1-1) y n-y y n-y K (1- ) (1- ) 1) ) K is the number of SPRT-feasible ways that one can obtain y successes out of n trials. Now, define the two hyothesis test errors: α (roducer s risk, or robability of rejecting a device with accetable life) and β (consumer s risk, or robability of acceting a device with unaccetable life) and let α β.128. Then, we can find two values A and B such that, at any stage n, that is having tested n devices sequentially (one at a time) and having obtained y cumulative successes, the PR fulfills the robability: y (1- ) n-y P{PR > A} P 1 1 > A β y n-y (1- ) y (1- ) n-y P{PR < B} P 1 1 < B 1- α y n-y (1- ) Thus, we define S (B,A), the Sequential Probability Ratio Test (SPRT), by the receding equations, as one that comares PR with values A and B at every stage n, and decides: (i) to accet H if PR < B; (ii) accet H 1 if PR > A; or (iii) continue testing if B < PR < A. For examle, at any given stage n, say after lacing the tenth (n 1) device on test, we obtain y successes (say y 6). We thus obtain the Binomial SPRT test result: P{y 6 Successes Under H1 } P{y 6 Successes Under H} Binomial (6;1, Binomial (6;1, (1-.819) 3.24E (1-.91) E - 5 1) ) Then, we comare value 7.23 with adequate values for A and B and decide: (i) to sto testing and accet (H 1 ) that robability of success is.819, and hence the Exonential mean is 1 hours or less, if the PR value 7.23 is greater than A; (ii) to sto and accet (H ) that the robability of success is.95, and hence the Exonential mean is 2 hours or more, if the PR value 7.23 is smaller than B; or (iii) to take another samle and reeat the the rocess, if 7.23 is between the values of B and A. We can simlify the rocess and equations, by taking the Logarithms in the PR inequality below, which defines the region leading to the continuation of the test: y (1- ) n-y y n-y 1- B < A y n-y < (1- ) 1-3

4 The result roduces a linear equation that is a function of the number of successes y, out of a given number of trials (stage) n, and is bounded by the Logarithms of values A and B: < 1 1- ln(b) y ln + 1 (n - y) ln n ln 1 y ln 1 - ln 1 < ln(a) ln(b) < an + by < ln(a); with :a ln b ln - ln 1 1- The coefficients a and b of these equations are, obviously, functions of i, i, 1. It can be shown (References 4, 5, and 6) that the constants A and B are aroximated by: β 1- α In our examle:.95, 1.819, n 1, y 6 and α β.128. Hence, the SPRT coefficients a and b can be calculated and values A and B can be aroximated: a ln 1 ln ln(1.96).6446; b ln 1 - ln 1 ln - ln ; (1- β) A ; B α (1- β) (1-.128) A 6.813; ln(a) ln(6.813) α.128 β.128 B.147; ln(b) ln(.147) α For our examle, at the SPRT 1 th stage (or n 1 trials), with y 6 successes, we get: ln(b) <.6446n y < ln(a) But: x x > Since is above bound (PR > A) we sto testing and accet H 1 that (hence, that device MTTF < 1 hours). Had the equation.6446n y yielded a value below , we would have acceted H that.95 (and hence, that device MTTF > 2 hours). Had the equation results been between and (the continuation region), we would have roceeded with the samling. Figure 1 reresents these choices. The described rocedure, even when accurate and correct, is difficult to follow. To better track the SPRT test values, we need an equation comaring y, the number of successes, directly with stage boundaries that are a function of the number of trials n : ln(b) b a - b ln(b) < an + by < ln(a) ln(a) a n h1 + sn > y > - n h + sn b b For our current examle the values are: ln(b) ; ln(a) 1.919; b ; a h 1 + sn - n > y > - n h + sn r n h 1 + sn n > y > n h + sn a n By letting n, the number of stages, run from 1, 2,, we obtain (a n ; r n ) the SPRT decision boundaries (or accetance and rejection numbers) for our examle are shown in Figure 2. Z an + by Rejection Region MTTF < 1 hours Continuation Region, Test another samle Trials: n Accetance Region MTTF > 2 hours Figure 1. Reresentation of the SPRT Test for Horizontal Regions 4

5 Row Stage Reject Accet Success y Accet Device: MTTF > 2 hrs y n Continue Testing y n Reject Device: MTTF < 1 hrs Trials n Notice how it is imossible to reject the device (accet H 1 : 1.819, or MTTF < 1 hrs.) until having tested at least n 4 devices, all of which must have failed. It is also imossible to accet the device (H :.95, or MTTF > 2 hrs.) until having tested at least n 2 devices, without observing a single failure (these critical boundary values are bolded in Figure 2). At any stage, the robability of decision error is α β.128. In addition, notice how the boundary equation arameters (sloe and intercet) deend on SPRT errors α and β, and on robabilities, 1, which in turn deend on the device MTTFs: β ln ln(b) 1- Intercets : h α 1 ; h b 1- ln 1 - ln 1 1- Figure 2. Reresentation of SPRT ln(a) b Notice how, in the receding equations for the sloe (s) and intercets (h 1, h ), the smaller the errors (α, β), the larger the intercets (in absolute value). This means that, on the average, it will take the SPRT test longer (more stages) to arrive its final decision. This result is intuitive. Since we are demanding larger assurances (smaller errors) from the SPRT test rocedure, the SPRT will necessarily require more information (more stages) to be able to rovide a decision that fulfills such errors α, β. Some examles are shown in Table 2. Table 2. Intercets for the Lower/Uer Increasing Bounds of SPRT Lower Bound Intercet Uer Bound Intercet Beta/Alha (1- β) ln α 1 1- ln - ln ln 1 a 1- Sloe : s - - > b 1- ln 1 - ln 1 1- The sloe s, which is common to both boundary lines, deends on the two hyothesized robabilities and 1, which in turn deend on the hyothesized MTTF µ and µ 1 (but not on the aforementioned test errors α, β). The continuation region, defined by the sloe s and the intercets h 1 and h, characterize the seed at which decisions are taken at every ste. So, all other factors being equal, the further aart the robabilities, 1 (and hence the two MTTFs µ, µ 1 ), the more comfortably we can discriminate between them as shown in Table 3. 5

6 Table 3. Discrimination Between Two MTTFs: Sloe of the Increasing SPRT Boundaries (sloes in italics) Finally, the receding can be similarly alied to an SPC/Quality Control roblem. For examle, assume that we are interested in assessing a batch of incoming material and that the accetable level of quality (AQL) is defined by some maximum ercent of defectives, say 1% ( <.1). Also, assume another ercent defective defines an unaccetable quality (LTPD), beyond which we will not receive the batch (say 2% defectives or more). Then, define values , and , for hyotheses H and H 1. Now assume that for rocedural ease, cost or other ractical reason, it is decided to test the lot by taking each item sequentially, one at a time, instead of taking a single samle of fixed and redetermined size n, all at one time. Then, for the accetance samling roblem described, all the reviously described SPRT derivations and results are alicable, with the ertinent modifications. The Average Samle Number (ASN) The main advantage of multile stage samling lans is the reduced long run or average samle size, required to arrive to a good decision. For now, the random variable samle size is a robabilistic outcome (varies with every case). Its Exected Value, known as ASN or Average Samle Number, deends on the value of the real arameter under test, be it the ercent defective, reliability, or any other arameter of interest. The ASN is obtained following the definition of Exected Value. For double samling: ASN E{SN} SN x P{SN} SN n1 x P(n1) + (n1 + n2) x P(n1 + n2) In the double samling scheme, SN (samle number) can be only n 1 or n 1 + n 2. P (n 1 ) is the robability of drawing a first samle only, which occurs when arriving at a decision at the first samle (with robability 1 - P{c 1 < Y<c 2 }). The robability P{n 1 + n 2 } of having to draw a second samle, totaling a size of n 1 + n 2, occurs when we had an inconclusive outcome from the first samle (i.e., with robability: P {c 1 < Y<c 2 }). We illustrate this case using our double samling examle S (n 1 2, n 2 2, c 1 14, c 2 15, c 3 33), described earlier. Let the true reliability, be.9, and let Y be the number of survivals obtained in the first samle of size n 1 2. Then, the robability of taking no decision on the first samle, when.9, is P (c 1 < Y<c 2 ) P (14 < Y<15) This yields ASN 2.81, barely larger than the exact n 2 elements that would be required by a single samle lan, of fixed size n 1 2: ASN n1 x P(n1) + n2 x P(n2) 2 x [1- ( )] + 4 x ( ) Table 4 resents the ASN values for the double samling lan S, described in the first section, calculated at selected values of the (reliability) arameter. Table 4. Comarison of ASN for Double Samling, Given Reliability ASN In Figure 3, we show grahically the relationshi between the reliability arameter and the corresonding double samle ASN ASN Reliability Figure 3. ASN for Double Samling 6

8 racticing engineers, and written a series of articles on statistics and data analysis for the AMPTIAC Newsletter and RAC Journal. Other START Sheets Available Many Selected Toics in Assurance Related Technologies (START) sheets have been ublished on subjects of interest in reliability, maintainability, quality, and suortability. START sheets are available on-line in their entirety at <htt://rac. alionscience.com/rac/js/start/startsheet.js>. For further information on RAC START Sheets contact the: Reliability Analysis Center 21 Mill Street Rome, NY Toll Free: (888) RAC-USER Fax: (315) or visit our web site at: <htt://rac.alionscience.com> About the Reliability Analysis Center The Reliability Analysis Center is a world-wide focal oint for efforts to imrove the reliability, maintainability, suortability and quality of manufactured comonents and systems. To this end, RAC collects, analyzes, archives in comuterized databases, and ublishes data concerning the quality and reliability of equiments and systems, as well as the microcircuit, discrete semiconductor, electronics, and electromechanical and mechanical comonents that comrise them. RAC also evaluates and ublishes information on engineering techniques and methods. Information is distributed through data comilations, alication guides, data roducts and rograms on comuter media, ublic and rivate training courses, and consulting services. Alion, and its redecessor comany IIT Research Institute, have oerated the RAC continuously since its creation in

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