ch15 - Chapter 15 APPLICATIONS TO RELIABILITY AND LIFE...

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Unformatted text preview: Chapter 15 APPLICATIONS TO RELIABILITY AND LIFE TESTING 15.1 Since this is a series system , we need toA find R such that R8 = .95 so R = (95)”8 = .9936. 15.2 Since this is a parallel system , we need to find R such that 1— .96 = (1 — R)? SO R: 1 — (1 ~— .96)1/5 = .475. 15.3 The reliability of the B, C parallel segment is : BBC = 1 — (1 ~ .80)(1 — .90) = .980. .587 588 Chapter 15 RELIABILITY AND LIFE TESTING The reliability of the E. F parallel segment is : REF =1—(1—.90)(1—.85):.985. The reliability of the A. B, C combination segment is : Rm = RA BBC = (.95)(.980) = .931. The reliability of the D, E. F combination segment is : RDEF = RD REF = (.99)(.985) = .97515. Thus. the reliability of the system is: R 1— (1 — RABC)(1— RDEF) 1— (1 — .931)(1 — .97515) = .9983. 15.4 The diagram of the system is shown in Figure 15.1. B1 _____.J C1 C2 ~— BS Figure 15.1: A Diagram of the System for Exercise 5.4. (a) The reliability of the B component is : 589 R3 = 1—(1—.9995)(1—.999)(1—.20)=.9999996. ,The reliability of the C component is : RC =1—(1—.95)(1—.85)=.9925. Thus, the reliability of the system is: R = BA BB RC = (.9999)(.9999996)(.9925) = .9924. (b) In this case, the reliability of the B component is BB 2 1~(1-.9995)(1—.999)(1~.99) = 1— 5 x10‘9. Thus, the reliability of the system is: R 2 RA RB RC = (.9999)(1— 5 x 10'9)(.9925) = .9924. (c) The reliability of the B component ( before the increase in reliability of B3 ) is : R3 = 1 — (1 — .9995)(1 — .20) = .9996. Thus, the reliability of the system is: R = (.9999)(.9996)(.9925) = .9920. After the increase, the reliability of the component B3 is : RB : 1 a (1 —— .9995)(1 . .99) = 999995. 590 Chapter 15 RELIABILITY AND LIFE TESTING so the reliability of the system is R = (.9999)(.999995)(.9925) = .9924. The reliability of the system has increased by .0004 . (d) The reliability of the C component is now RC = 1 — (1 — .95)(1 — .85)(1 — .80) = .9985 so the reliability of the system is R = (.9999)(.9999996)(.9985) = .9984 which is an increase of .006 over the original system, 15.5 (a) We have that f(t) =Z(t)emp[——/O 2e) dt]. where _ B(1-t/a)for0<t<a. Z“) _ { 0 elsewhere. Thus elsewhere. m): { 5(1_t/a)€$pl—féfi(1—x/a)d:c] for 0 < t < a. so m) = { gm — t/a> exp [ we — t2/(2a) )1 3:63;: a. The distribution function is 591 _ ftf(17)d;r for0<t<a. F(t)_{ fgaflz) d1“ fort>a. OI‘ F(t)_ l—el'pl—Bfi—tQ/(Qafll for0<t<a. _ l—ezz‘pl—afl/Ql fort>a. (b) the probability of initial failure is F(a) = 1 —— 6‘03/2 as shown in the previous part. 15.6 (a) __ 91+féflf§1dtfora<t<fl Fm — {1—62 B fortzfi. _ 91+L156—1g—f'2-(t—‘92 fora<t<fl _ 1—92 fortzfi. (W Z“) _ f(t) _ (Maren/mm _ 1*F(t)—1—61—(1—61~62)(t—a)/(3—a) 1—91—62 (3-600-90-(Pam-91492) H (c) Figure 5.2 gives the plot of .2 W , 100<t<15,000 14,900(.95)+(.2)(t—100) — — 15.7 (a) Since the failure rate is constant, f (t) = tome—(.02): Chapter 15 RELIABILITY AND LIFE TESTING Failure Rate 1.30 9153 1.20 0 5000 10000 15000 Hours Figure 15.2: Failure Rate. Exercise 5.6 and F(t) 2 1 — e'iom. The unit of time is 1000 hours. The probability that the chip will last longer than 20,000 hours is 1 — F(20) = e—(‘02)'20 = .6703. (b) The 5000—h0ur reliability of four such chips in a series is R z [1 — F(5)]4 = e—4<-°2>‘5 = .6703. 593 15.8 The probability of failure during the first 365 days of operation is: 1 _ 6400050665) = .167- (b) The probability that two independent components will survive the first 365 days of operation is: 6—(.0005>(365) .€A<-0005><365> = .694. 15.9 Using the formula for the mean time between failures 1 ‘1— 1 (1+1++ +1) 2 ‘n‘gxio—4 2 n‘ Since the failure rate is the inverse of the mean time between failures7 in this case we require that up 2 1/(4 X 10—4). Thus we must find n such that <(1+1' +1) - 2T n- #‘Alko A trial and error search yields n = 5 and the failure rate is 9.0 x 10‘4 = .4 1—4. 2.2833 39 X 0 15.10 The mean time between failures for a series system is: .MTBF = .11628 1.9 + 3.1 + 1.4 + 2.2 or 1,162.8 hours. 594 Chapter 15 RELIABILITY AND LIFE TESTING 15.11 (a) The probability that a part will last 500 hours is: 1 — H500) = 6500/1000 2 .6065. (b) The probability that at least one will fail is one minus the probability that none will fail or 1 _ 8~3<1000)/1000 : .9509 (C) The probability that a particular part fails is: 1 — 6600/1000 2 .4512. The probability that exactly two parts fail is given by the binomial distribu— tion with p = .4512 or < ‘21 > (.4512)2 (1 — .4512)? = .3679. 15.12 The VVeibull density is f(t) = afitfi‘le‘a‘fi for Oz,fi,t > 0. Thus, F(t) = 1 — as”. The probability that the component will operate at least 5,000 hours is 1— F(5000) = e-<-°°5><50°°>'8° = .0106 for a, 5,: > 0. 15.13 (a) The probability of failure on any trial is p and the probability of no failure is 1 ~ 17 provided failure has not occurred previously. thus, failure on the ar’th trial means no failure for a: — 1 trials and then a failure. Since the trials are (C) 15.14 (a) independent f(1:)=(1—plx_1p for $21.23,... I ’ 1—1 Ft?) = ZO-W‘UmpZU—pll 1:1 1:0 1—11— I = p (p p) = —(1—p)x for rc=1,2,3.-~- The probability that the switch survives 2,000 cycles is 1 — F(2000) = (1— 6 x 10—62000 = .301. TT 2 ntr 2 50-760 2 38, 000. Since x2975 with 2 - 7" = 16 degrees of freedom is 6.908 and X2025 with 16 degrees of freedom is 28.845, the 95 percent confidence interval for u is 2 - 38000 < < 2 - 38000 28.845 “ 6.908 or 2,635 < u <11,002 The null hypothesis is ,u = 10,000 and the alternative is u < 10,000. Since X295 with 2- r = 16 degrees of freedom is 7.962 , we reject the null hypothesis at the .05 level if 1 1 Tr < 511096295 = §(10,000)(7.962) = 39,810. Since Tr 2 38,000 , we reject the null hypothesis at the 0.05 level of signifi— cance. 596 15.15 Chapter 15 RELIABILITY AND LIFE TESTING ll Zt, + (n — r)t,. 1:1 250 + 380 ’,' 610 + 980 +1250 + 30- 1250 = 40, 970. ll Since $995 with 10 degrees of freedom is 2.156 and Xfi305 with 10 degrees of freedom is 25.118, the 99 percent confidence interval is 2 - 40. 970 25.118 2 ~ 40,970 < p < 2.156 CI 3. 2531 < ,u < 33,0055 The null hypothesis is 11 = 5. 000 and the alternative is 11 > 5, 000. Since X205 with 2 - 7“ = 10 degrees of freedom is 18.307, we reject the null hypothesis at the .05 level if 1 1 T, > iuoxios = §(5,000)(18.307) = 45, 767.5 Since T, = 40,970 , we cannot reject the null hypothesis at the .05 level of significance. We cannot be sure the manufacturer’s claim is true. 15.16 We calculate the successive values of total time on test T1 :11 + (n ~1)t1 = 250 + 34(250) = 3, 750 T2 :11 HQ + (n — 2% = 250 + 380 + 33(380)=13,170 ' Continuing, we use Ti=tl+t2+"'+ti+(n—7)ti to find T3 = 20,760, T4 = 32,600 and, T5 = 40,970. The points 597 are graphed as a total time on test plot in Figure 15.3. 1.0 1 Tl / Tr 0.6 _ 0.2 0.2 0.4 0.6 0.8 1.0 i/r Figure 15.3: Total Time on Test Plot. Exercise 5.16 15.17 (a) TT = Ztfl-(n—rfir 1—1 211+ 350 + 384 + 510 + 539 + 620 + 715 = 3, 329., Since X2975 with 14 degrees of freedom is 5.629 and X2025 with 14 degrees of freedom is 26.1197 the 95 percent confidence interval is 2-3329< <2-3329 26.119 “ 5.629 01‘ 254.9 < 11 < 1,182.8 598 Chapter 15 RELIABILITY AND LIFE TESTING (b) The null hypothesis is 11 = 500 and the alternative is h 51$ 500. Since Nib with 14 degrees of freedom is 23.658 and xi.) with 14 degrees of freedom is 6.571. we reject the null hypothesis at the .10 level if 1 1 T. < 5110/1115 = 3(501))(6571): 1, 642.75 or if 1 1 T. > éuoxts = 3(500)(23.685) = 5,921.25 Since TT 2 3,329 , we cannot reject the null hypothesis at the .10 level of significance. 15.18. (a) From Exercise 15.15 we have that Tr 2 40,970 and X205 2 18307 for 10 degrees of freedom. Thus, the one—sided 95 percent tolerance limit is _‘) . . f 2 1 40,970 ln(.8) 2998.8 18.307 (b) From Exercise 15.17 we have that Tr : 3,329. Since X201 2 29.141 for 14 degrees of freedom, the one—sided 99 percent tolerance limit is —2 - 332911119) * = —————— = 24. 7. t 29.141 0 15.19 Since 2Tr/,u is a X2 random variable with 27" degrees of freedom , 2 T 2TT T and u u Pl Xi—a/2 < < Xi/2 l = 04, where xfmfl and xi” are the chi-square quantiles for 27" degrees of freedom. Multiplying the first inequality by ,LL/Xf_a/2 and the second by 11/ X: /2 gives: 2T, and 2T, X1152 Xi/g P[/r< <u]=a. 599 Thus the (1 — (1)100 percent confidence interval is: 2T. 2T, ,2 < H‘ < 2 \0/2 \l—a/2 15.20 We solve the equation for n = 100 , 7‘ = 10 and the first 10 failure times 7.07 14.1. 18.9, 31.6, 52.8, 80.0, 164.5, 355.4, 451.0, 795.1. Using a computer, we obtain the solution B = 43449 so that 1 1 l r t5 + (n — mt? _ 11—0 2.21 #3449 + 90(795.1)-43449 7‘ i=1 1 = .005818. 61 z The failure rate Z(t) : afit5"1 , at t = 1000 , is estimated by zit) = Milt/3’1 = (.005818)(.43449)(1000)‘43449‘1 = .0508 x 10—3 hours. If the exponential model had been used, the estimate of the mean would have T. _ 73,5294 — = 7, 352.94. 7" 10 ,1: Thus the estimate of the constant failure rate is 1/11 = .0136 X 10"3 hours. 15.21 There are 7‘ = 3 failures and total time on test is T3 = 2076 + 3667 + 9102 + 197(9102) = 1, 807, 939 Since X205 = 12.592 for 2r : 6 degrees of freedom, the 95 percent lower confidence 600 Chapter 15 RELIABILITY AND LIFE TESTING boundis 2T3 _ 2(1.807,939l X205 _ 12.592 = 287,1568 15.22 We solve the equation Zlet';3lnti+(n—r)t3ln1r 1 1* 7" , lnt- = 0 i=1 ti? + (n ‘ ")tij 5 i=1 1 for n = 60 , 7“ = 9 and the first 9 failure times 3.6, 6.9, 9.5, 15.77 27.3, 41.2, 81.7. 178.3, 227.1 . Using a computer, we obtain the solution 6 = .5062 so that 1 1 &=—————1————————-:———_——,—=.0105. 1 1:1 t? + (n — 70215 521:1 11°06? + 5102703062 The estimate of the mean life is 11 = «~1/fir(1+i)=<.om5)-1/~50621“(1+ —1 )=15,863-99 fl .5062 If the exponential model had been used, the estimate of the mean would have Tr _ 12,1734 ,1 = — =1,352.6. T which is far too low. 15.23 The probability the diaphragm valve will perform at least 150 hours is 1 minus the probability that it fails before 150 hours. Since 1— F(150) = 1 — (1 — e"°‘(150)5), d = .0105 and B = .5062, we estimate this probability by e—a(150)5 : e—(.0105)(150)~5062 = .8753 601 15.24 The VVeibull density is f(zf) = aflta—le—a‘a Thus, t t F(t) :/ f(;r) dr 2 / Q3Ifi~1€_urd dx. 0 . 0 Using the change of variables u = orj, we have 3 at J F(t)=/ e‘“du=1—e_at. 0 Since RU) = 1 — F(t), we have 15.25 Suppose T has the Weibull distribution with parameters a and B. In the text it was shown that 1 1 Since Var(T) = E(T2) — [E(T)]2, we must find E(T2). The Weibull density is Ody—16"!” SO E(T2) = foo tgafltfi_le_atfi dt. 0 Let u = atfi. Then 1 00 1 2 2 _ 25—11 _ E(T)—a2/B/(; u/e du—OQ/EFU-i-B) 602 Chapter 15 RELIABILITY AND LIFE TESTING so that , ‘ 1 ‘ 2. 1 #1 2 l/a7(T) = a2/3F(1T73) 1 — ——1 ir<1+3>~<r<1+3>>21 — a?” 5’ 15.26 Since this is a parallel system , we need to find R such that 1— .90 = (1— R)7. SO R = 1 — (1 — .90)”7 = .28. 15.27 The probability of failure during the first 250 hours of operation is: 1 — e_('0045)’250 = .6753. (b) The probability that two independent components will survive the first 100 hours of operation is: e-(.oo45)-100 . e—(.0045)-100 = .4066. 15.28 Using the formula for the mean time between failures «1(1+1++ +1)— 1 (1+1++ +1) “Fly 2 n—2x10‘4 2 n' Since the failure rate is the inverse of the mean time between failures, in this case we require that up 2 1/(5 X 10'5). Thus we must find n such that 1 4g(1+§+~--+ ). 1 n A trial and error search yields 71. = 31 and the failure rate is 2.0 x 10‘4 = 4.966 10—5. 1.027 X 15.29 The mean time between failures for a series system is: Z W =. 33 tl h 1.8+2i4+2_0+1.3+3.0+1‘5 08 1ousand ours or 83.3 hours. 1530 (a) T. = :1.- + (n — mt. = 16.5 + 19.2 + 20.8 + 37.3 + (11)37.3 = 504.1. i=1 603 Since X295 With 8 degrees of freedom is 2.733 and X205 with 8 degrees of freedom is 15507, the 90 percent confidence interval is 2504.1< <2504.1 15.507 ” 2.733 01‘ 65.0 < ,u < 368.9 (b) A failure rate of .004 failures per hour corresponds to a mean life of 1/ .004 = 250 hours . Thus, we test the null hypothesis ,u = 250 against the alternative 11 > 250. Since X201 with 2 - r = 8 degrees of freedom is 20.090, we reject the null hypothesis at the .01 level if 1 T. > $110th = 5(250)(20.090) = 2,511.25 604 Chapter 1.5 RELIABILITY AND LIFE TESTING Since T, = 504.1 , we cannot reject the null hypothesis at the .01 level of significance. 15.31 There are r = 4 failures and total time on test is T; = 3582 + 8482 + 8921+16303 ~1— "296(16303) = 4, 862. 976 Since flog) = 15.507 for 27" = 8 degrees of freedom. the 95 percent lower confidence boundis 2T 2 4.862. 976 24 = = 627.1975 X105 1050/ 15.32 We solve the equation Ltglti —t311, 1 1T _ _ _ _ 2 Int, 2 [:1 t, ‘i‘ (TL — T)tr fl T i=1 for n = 100 , r = 10 and the first 10 failure times 28, 46, 50, 63, 81, 101, 116, 137, 159, 175. Using a computer, we obtain the solution = 1.3665 so that 1 1 A = w— 2 W = .0000909. a l 3:, t? + (n — fit? fi 212, 121-3793 + 90(175)1-3665 The estimate of the mean life is A 1 1 ‘ = “WM 7 = .0000909 *1/1-3665r1 __.._ =829_8 M ( + 13) ( ) ( +1.3665) If the exponential model had been used, the estimate of the mean would have T, ‘ 12,331 r " 10 [1: = 1,233.1. 15.33 The probability the circuit will perform at least 100 hours is 1 minus the probability 605 that it fails before 100 hours. Since 1— H100) = 1 — (1 — ammo”) 6x = .0000909 and = 1.3665, we estimate this probability by —d(100)5 1.3665 _ 6 .0000909)(100) 93.20. :6_( :_. 15.34 (a) Calculate the successive values of total time on test T1=t1 + (n —1)t1= 28 + 74(28) = 2,100 T2 =t1+t2 + (n — 2)::2 = 28 +46 + 73(46) 2 3,432 Continuing, use n=t1+t2+"'+ti+(n_i)ti to find T3 = 3,724, T4 9,801, T9 = 11,275, T10 4,660, T5 = 5,938, T6 2 7,338, T7 = 8,373, T8 = 12,331. The points H are graphed in Figure 15.4 as a total time on test plot. A (b) Using the approximation F(t,~) = i/(n+1), with n = 75 , calculate the 7" = 10 pairs (23,, yi) where ac,- = In t,- and 606 Chapter 15 RELIABILITY AND LIFE TESTING 01 -~ 0 ‘— 0 CO 0" ° L o |-—<o' \O“ ' *— o Vt- O o o o “1 CL. "'1 1 T'— V.— 0.2 0.4 0.6 0.8 1.0 i/r Figure 15.4: Total Time on Test Plot. Exercise 5.34 F021 ti 51% EH .0132 28 3.33 —4.32 .0263 46 3.83 —3.62 .0395 50 3.91 —3.21 .0526 63 4.14 —2.92 .0658 81 4.39 —2.69 .0789 101 4.62 -2.50 .0921 116 4.75 —2.34 .1053 137 4.92 -2.20 .1184 159 5.07 -2.07 .1316 175 5.16 —1.96 In Figure 15.57 these points are graphed as a Weibull plot. 607 O w‘i .° » .1 O U) 9 O 8 o._ - w C? g o .D .6 A g o O. “is :5 1 T— l 35 40 4.5 50 Wt) Figure 15.5: Weibull Plot. Exercise 5.34 15.35 (a) we are given that X has = 1 ~ 6"le and flat) = file—'0” and Y has distribution C(y) = 1 — e"005y and g(y) = 0056—0059. Consequently, ()0 PW > X] =/ F(y)9(y) dy —00 R [)0 [1 ~ 6—'01y].005€_’005y dy "OO /°° .0056'005ydy — foo (.005)e-(‘°°5+-01>y dy ~—oo ‘00 II 608 Chapter 15 RELIABILITY AND LIFE TESTING _ .005 _ .015 "' — .6667. (b) We are given that has F(;r) = 1 — 6—0051" and fix) = 0058—0053: and Y has distribution C(y) = 1 — 6"0053’ and g(y) = 005500574. Consequently, R z P[Y>Xi=/_OOF(y)g(y)dy : / [1 _ 6—.005y](.005)e-.005y dy —oo = / (005)6—0051/ .0056—(i005+.005)y -00 _00 .005 .010 5 (G) Since lnX and lnY are independent and each has a normal distribution, lnY — lnX has a normal distribution with mean My — pm 2 80 — 60 = 20 and variance 0: +0: 2 52 +52 = 50. To avoid numerical integration7 we note that R = P[Y>X]=P[1nY>1nX]=P[1nY—1nX>O] —20 =1—F——— =1—F—2.83 (m) ( ) where 1 — F(——2.83) = F(2.83) = .9977 is obtained from the standard normal distribution. ...
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ch15 - Chapter 15 APPLICATIONS TO RELIABILITY AND LIFE...

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