Sequential life testing with underlying normal and Weibull sampling distributions. D. I. De Souza Jr. Abstract. 1 Introduction

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1 Sequential life testing with underlying normal and Weibull sampling distributions D. I. De Souza Jr. Civil Engineering Department & Industrial Engineering Department, Fluminense Federal University& North Fluminense State Brazil Abstract The sequential life testing approach gathers sample information only until there is enough to allow a decision with a desirable degree of cofildence. The sample size is a random variable and is determined by the result of the analysis of the observed data, It happens that even with the use of a sequential life testing approach, sometimes the number of items necessary to reach a decision about accepting or rejecting a null hypothesis is quite large, as shown by De Souza [1]. In situations like that, the development of a truncation mechanism is essential to guarantee the major advantage of using sequential life testing; that is, small sample sizes. In this work, we will develop a sequential life testing approach in which the underlying sampling distributions are the normal and the Weibull models, We will use the two underlying models to analyze a life testing situation, comparing the results obtained from both. We will also develop a truncation mechanism for the Weibull and Normal models. We will provide rules to truncate a sequential life testing situation making one of the two possible decisions at the moment of truncation; that is, acceptor reject the null hypothesis HO, An example will develop the proposed truncated sequential life testing approach for the Weibull and Normal models. 1 Introduction The two-parameter Weibull distribution has a shape parameter ~, which specifies the shape of the distribution, and a scale parameter 0, which represents the characteristic life of the distribution, Both parameters are positive.

2 376 Risk Analysis III The normal distribution has been widely used as a failure model. It has two parameters: a shape parameter a and a scale parameter p. The Weibull density fimction is given by: The normal density fimction is given by The hypothesis testing situations will be given by Kapur and Lamberson [2], De Souza [1]: a) In the Weibull Case: 1. Forthescale parameter O: HO : 026.; Hl: 6 <60 The probability of accepting HO will be set at (1-et) if O = O.. Now, if 6 = 01 where the probability of accepting HOwill be set at a low level y. 2, Fortheshapeparameter~:Ho : ~>~o; H,: ~<~o The probability of accepting Ho will be set at (1-u) if 13= PO.Now, if ~ = ~, where ~ l<~o, then the probability of accepting HOwill also be set at a low level y. b) In the Normal Case: 1, Forthescale parameter~: HO : 1.L2VO; HI: I.L<VO The probability of accepting Ho will be set at ( 1-a) if v = Vo.Now, if v = I.L1, where pl < PO,then the probability of accepting Ho will be set at a low level y. 2. Fortheshape parametercr: HO : cr>oo; Hl: O<ISO The probability of accepting Ho will be set at (1-et) if cs= cro. Now, if G = al where 01<00, then the probability of accepting Ho will also be set at a low level y. When the decisions about these quantities O., Ql, PO, ~,, PO, pl, CSo,crl, ct and y are made, the sequential test is totally defined. 2 Methodology The development of a sequential test uses the likelihood ratio (LR) given by the following relationship proposed by Kapur and Lamberson [2] e De Souza [3] LR= Ll,~ Lo;n,/ The sequential probability ratio (SPR) will be given by SPR = Ll,l,n /Lo,o,n a) In the Weibull case, the SPR will be:

3 b) In the Normal case, we will have: Risk Analysis III 377 The continue region will become A< SPR < B, where A = y/(1-a) and also B = ( l-y)/cl, We will accept the null hypothesis HOif SPR 2 B and we will reject Ho if SPR < A. Now, if A <SPR< B, we will take one more observation, Then, we will have: a) In the Weibull case; By taking the natural logarithm of each term in the above inequality and rearranging thep we get where y$l=~ *-* b) In the Normal case; [1 t!l t!, I=161OO 1 0 $l)-pl)~ln(t,) i=l (4) Again, by taking the natural logarithm of each term in the above inequality and rearranging them, we get (5) (6) Using eqns (1) to (6), an example will now illustrate the proposed sequential life testing approach, comparing the results obtained from both the Weibull and Normal models.

4 378 Risk Analysis III 3 Expected sample size of a sequential life testing According to Mood and Graybill [4], an approximate expression for the expected sample size E(n) of a sequential life tes}ing will be given by E(n) = L! E :, (7) E(w) where the expected value of the variate W. depends on the random w s and the random variate Q. Here, Y is given by w=ln f(t; el, pl ) f(t; eo, po ) The variate W*. takes on only values in which W. exceeds In A or falls short of in B. If one ignores the amounts by which W*. exceeds lna or falls short of in B, he or she may say that W*. takes essentially only two values, in A and in B. When the true distribution is f (t;e,~), the probability that W. takes the value In A is P(6,~), while the probability that it takes the value in B is 1 P(6,~). Then, according to Mood and Graybill [4], the expression for the expected value of the variate W*. will be given by Hence, eqn (9) becomes E(W~) z p(6, P)ln A + [1-P(o, P)] lnb (9) E(n) z P(f3,~)ln A + [1-P(O, ~)]lnb E(w) where A = y /(1-cL) and B = (l-y)/u. Eqn (1O) enables one to compare sequential tests with fixed sample size tests. The proofs of the existence of eqns (7) to (10) can be found in Mood and Graybill [4], pp a) In the Weibull case; For a Weibull sampling distribution, eqn (8) will be transformed into (8) (lo) The expected value of w, E(w), will then be given by (12)

5 Risk Analysis III 379 E[ln(t)]=ln(6)+~ X: x $ ln(ui )e-ui x (1,20r4) (14) {[ 1} To fmd the E[ln(t)] some numerical integration procedure (Simpson s 1/3 rule in this work) will have to be used. When the decisions about the quantities 6., 81, PO, PI, a, y and P(O, ~) are made, and after the E(n) is calculated, the sequential test is totally defined, b) In the Normal case; For a Normal sampling distribution, eqn (8) will be transformed into [[1[ 1! ~o (+2 (-0)2 Oryet z=ln exp 1 20; 202 o Z=wo)-ln(d+; (X-$2-(X-$ [ The expected value of z, E(z), will then be given by E(z)= ln(o, )- ln(~ ~) (13) Now, with E(x) = V; E(x2)=02+~2, the above eqn becomes: Using eqns (3) to (6), and (11) to (15), an example will develop the proposed truncated sequential life testing approach. 4 Example A low alloy high strength steel product will be life tested. Since this is a wellknown product, there is the knowledge that it could be represented by an underlying Normal sampling distribution having a scale parameter v of 3,160,000 cycles and a shape parameter o of 1,050,000, which we believe are the true values for these parameters. Based on our previous experience with this steel product, we also believe that it could be represented by a Weibull sampling distribution having a scale parameter 0 of 3,500,000 cycles and having a shape parameter ~ of 3.5. It was decided that cx= 0,05 and y = We will use these

6 38(I Risk Analysis III two underlying sampling models to life testing the low-alloy high strength steel product under analysis, comparing the results obtained from both models, Initially, using the Weibull sampling model, we elect the null hypothesis parameters to be (30= 3,500,000 cycles; PO= 3.5; with ct = 0.05 and y = 0,10 and choose the value of 3,000,000 cycles for the alternative scale parameter 61 and the value of 3,0 for the alternative shape parameter 13~, Then, using eqns (3) and (4) we will have: 3.0 nln x3500~~035)- ~n[(1~~~0)]=n [ 3,000, ,0 3,500, nln )+n[(l~~ 5)]= nx [ 3,000, x 3.5 t 3.0 t3.5 ~=~ + 0,5x$ ln(ti) i=l 3,000~ ,500~ i=l [ 1 Then, we get: n x < W <n x 7, (16) The procedure is defined by the following rules: 1. If W 2 n x , we will accept HO. 2. If W < n x ,8904, we will reject HO, 3, If n x < W <n x , we will take one more observation, Now using the Normal sampling model, we elect the null hypothesis parameters to be po= 3,500,000 cycles; cro= 1,050,000; with cx= 0.05 and y = 0.10 and choose the value of 2,500,000 cycles for the alternative scale parameter VI and the value of 950,000 for the alternative shape parameter o,. Then, using equations 5 and 6, we will have = nxo = nxo N.~~[(xi~~~~~~OO~ -[xi~~$~~oo~] (17) The procedure here is defined by the following rules: 1, IfN 2 n x we will accept HO. 2. IfN < n x ,8904, we will reject HO. 3, Ifn x 0, <N <n xo.1oo , we ll take one more item. After a sequential test graph has been developed for this life-testing situation, a random sample is taken. The failure times obtained in this life testing(cycles to failure) were the following: 3,282,070; 2,038,658; 3,842,361; 4,441,792; 1,840,065; 4,388,466; 3,467,202; 2,807,120; 3,749,865; 2,985,436; 1,693,218;

7 Risk Analysis III 381 2,432,809; 3,008,410; 2,246,590; 4,018,243, Tables (1) and (2) show, respectively, the results of this test for the Weibull and Normal models case. Table 1. Sequential test results for the Weibull model case, P1 = 3.0; 61= 3,000,000p,= 3,5; e,= 3,500,000.<. Unit Lower Upper Value Number Limit L;tit of w ,094 8, < , , < Unit Lower Upper Value Number.. Limit L;~t Ofw 9 67, , Table 2. Sequential test results for the Normal model case. Bl= 950,000; 01= 2,500,000 ~,= 1,050,000; 6.= 3,500,000 Unit Lower Upper Value Unit Low& Upper Value..... Number Limit. Limit of.. w. -.,.,.. Number,,,,,, Limit. Limit.... of w ,790 2, =67 2-2,690 2, , , , , , ,190 2, < In this case, even after the observation of 15 times to failure, it was not possible to make the decision to accept or reject the null hypothesis Ho, Now using equations (11) to (15), we can calculate the expected sample size E(n) of this sequential life testing under analysis. In the Weibull case, with; ~1 =3.0; 6 =00= 3,500,000 cycles; el = 3,000,000 cycles; PO= 3.5; a = 0.05; y = 0.10; and electing P(8, ~) to be 0,01, we will have: E(W) = -p o ) E[ln(t)] ~ (t l)+*e(t O 1= E(W~)=P(6, ~)ln A + [1-P(6, P)] lnb =-O.OIX 2, x = E(n) = p(o,13)lna + [1-p(6,13)]ln B _ = 8.3 s 9 items, E(w) So, we could make a decision about accepting or rejecting the null hypothesis Ho afier the analysis of observation number 9.

8 382 Risk Analysis III In the Normal case, with o = cro=1,050,000; crl =950,000; p = po= 3,500,000 cycles; pl = 2,500,000 cycles; ci = 0.05; y = 0.10; and electing P(6, ~) to be 0.01, we will have: E(z)= ) E(z) = O.1OOO8+5.O25O9X10-25 x 1,32302x 1024 = 0.765; Then: E(n) = P(6, ~)lna + [1- P(O,~)]ln B = =3.71 >4 items. E(w) So, we could make a decision about accepting or rejecting the null hypothesis Ho after the analysis of observation number 4. Both models were able to effectively represent the low-alloy high strength steel product being life tested, In fact, when the value of the Weibull shape parameter ~ approachs 3,6, as in our example were (30 value was 3.5, the resulting distribution tends toward the Normal model, This situation was also described by Dodson (Dodson 1994), 5 A procedure for early truncation According to Kapur (Kapur & Lamberson 1977), when the truncation point is reached, a line partitioning the sequential graph can be drawn as shown in Figures 1 and 2 below. 120 r Q &u&l 20 K,,...,,,,,,, [ NUMBER OF ITEMS TESTED 1! TRUNCATION POINT I Figure 1. A truncation procedure for the sequential testing - Weibull model case.

9 Risk Analysis III 383,,.,...- t -v p / ~. o Poll. MT! E.Z 4 [~ w 1 J I I 1 i 1 { 4 7 %78S M $6 TRUNCATION ( NUMBE:R OF ITEMS TESTED 1 I EEl I Figure 2, A truncation procedure for the sequential testing Normal model case. This line is drawn through the origin of the graph parallel to the accept and reject lines. The decision to accept or reject Ho simply depends on which side of the line the final outcome lies. Obviously this procedure changes the levels of cx and y of the original test; however, the change is slight if the truncation point is not too small. Figures 1 and 2 above show, respectively, the sequential test graphs developed for this example, As we can see in Figures 1 and 2, the null hypothesis Ho should be accepted since the final observation in the Weibull model case (observation number 9) and in the normal model case (observation number 4) lie on the side of the line related to the acceptance of Ho. 6 Conclusion The sequential life testing approach developed in this paper shows that the Normal and Weibull models could both effectively represent the low alloy high strength steel product being life-tested in the above example. As reported by Dodson [5], when the value of the Weibull shape parameter ~ approachs 3.6, the resulting distribution tends toward the Normal model. In our example, the Do value was 3,5. In fact, the Weibull model is rich in shape which suggests that it fits the pattern of variability of so many industrial products and processes, so well. Because of this versatility in shape, this distribution has been widely used as a failure model, particularly for metallurgical and mechanical components. If we now consider that when the shape parameter 13of the Weibull model is equal to 1, the resulting distribution is the exponential; then we can see that the

10 384 Risk Analysis III Weibull model is the most widely used life-testing model in industry. We should also remember that, for the Weibull model, the coefficient of variation depends only on the shape parameter ~, as shown by De Souza and Lamberson [6]. This versatility presented by the Weibull model makes it the adequate one to be used in ahnost all life testing experiments, We should remember, however, that each product, each chemical composition, or even each heat treatment, will produce a Weibull curve with different shape and scale parameters. Therefore one should be well knowledgeable about the product to be sequentially life tested when using a Weibull underlying model. The major advantage of a sequential life testing approach in relation to the fixed size approach is to keep the samples size small, with a resulting savings in cost, It happens that even with the use of a sequential life testing approach, sometimes the number of items necessary to reach a decision about accepting or rejecting a null hypothesis is quite large, De Souza [1]. Thus, the test must be truncated after a fixed time or number of observations. The truncation mechanism for a sequential life testing approach developed in this work provides rules for working with the null hypothesis Ho in situations where the underlying sampling distributions are the Weibull and the normal models. This fact shows the advantage of such a truncation mechanism to be used in a sequential life test approach. 7 References [1] De Souza, Daniel I., Further thoughts on a sequential life testing approach using a Weibull model. Foresight and Precaution, ESREL 2000 Congress, eds. Cottam, Harvey, Pape & Tait, Balkema: Edinburgh, Scotland, 2: pp , [2] Kapur, K. & Lamberson, L. R,, Reliabili@ in Engineering Design, John Willey & Sons, Inc., New York:, [3] De Souza, Daniel I., Sequential life testing with an underlying Weibull sampling distribution., Proceedings of the ESREL 99 International Conference on Safety and Reliability, eds. Schueller & Kafka, Balkema: Garching, Germany, 2: pp , [4] Mood, A, M. & Graybill, F. A., Introduction to the Theo~ of Statistics. Second Edition, McGraw-Hill, New York, 1963, [5] Dodson, Brian, WeibullAnalysis, ASQ - Quality Press, Minneapolis:,1994. [6] De Souza, Daniel L & Lamberson, L. R., Bayesian Weibull Reliability Estimation, IIE Transactions, 27: pp , 1995.

Truncated Life Test Sampling Plan Under Odd-Weibull Distribution

International Journal of Mathematics Trends and Technology ( IJMTT ) Volume 9 Number 2 - July Truncated Life Test Sampling Plan Under Odd-Weibull Distribution G.Laxshmimageshpraba 1, Dr.S.Muthulakshmi