Any reliabillity engineering geek?
in [Matlab]
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From: Similione Kasan on 19 May 2010 02:26 Hi, if suppose I have 20 identical samples and the required test time for them without failure is 100 hours. Now if out of these 20, I take 10 samples and test them for a time of 50 hours instead of 100 and the next 10 for a time of 200 hours. Then in the end what would be my total reliability? Assume confidence level of 80%
From: us on 19 May 2010 02:36 "Similione Kasan" <mech.engr(a)hotmail.com> wrote in message <ht009t$se6$1(a)fred.mathworks.com>... > Hi, > > if suppose I have 20 identical samples and the required test time for them without failure is 100 hours. > Now if out of these 20, I take 10 samples and test them for a time of 50 hours instead of 100 and the next 10 for a time of 200 hours. Then in the end what would be my total reliability? Assume confidence level of 80% ok, nice, now where's your question re ML(?)... us
From: Similione Kasan on 19 May 2010 02:44 "us " <us(a)neurol.unizh.ch> wrote in message <ht00sk$61d$1(a)fred.mathworks.com>... > "Similione Kasan" <mech.engr(a)hotmail.com> wrote in message <ht009t$se6$1(a)fred.mathworks.com>... > > Hi, > > > > if suppose I have 20 identical samples and the required test time for them without failure is 100 hours. > > Now if out of these 20, I take 10 samples and test them for a time of 50 hours instead of 100 and the next 10 for a time of 200 hours. Then in the end what would be my total reliability? Assume confidence level of 80% > > ok, nice, now where's your question re ML(?)... > > us ----- Thanks for the response... I could not get you .... ok let's go this way.... Out of these 20 identical samples, I did testing for the first 10 for about 50 hours (which means lifetime ratio is different now). As the required testing time was 100 but I did for 50 hours and also my sample size is reduced to 10 so my reliability will be low now. And for the next 10 samples, I tested for 200 hours so my reliability will increase for these 10 samples. BUT how can I find the final reliability of these complete 20 samples after testing the first 10 for 50 hours and the next 10 for 200 hours? I am now only struck at this last point.... Got it or still need explanation?
From: us on 19 May 2010 03:00 "Similione Kasan" <mech.engr(a)hotmail.com> wrote in message <ht01bo$5u6$1(a)fred.mathworks.com>... > "us " <us(a)neurol.unizh.ch> wrote in message <ht00sk$61d$1(a)fred.mathworks.com>... > > "Similione Kasan" <mech.engr(a)hotmail.com> wrote in message <ht009t$se6$1(a)fred.mathworks.com>... > > > Hi, > > > > > > if suppose I have 20 identical samples and the required test time for them without failure is 100 hours. > > > Now if out of these 20, I take 10 samples and test them for a time of 50 hours instead of 100 and the next 10 for a time of 200 hours. Then in the end what would be my total reliability? Assume confidence level of 80% > > > > ok, nice, now where's your question re ML(?)... > > > > us > > > ----- > Thanks for the response... > > I could not get you .... ok let's go this way.... > > Out of these 20 identical samples, I did testing for the first 10 for about 50 hours (which means lifetime ratio is different now). As the required testing time was 100 but I did for 50 hours and also my sample size is reduced to 10 so my reliability will be low now. And for the next 10 samples, I tested for 200 hours so my reliability will increase for these 10 samples. > BUT how can I find the final reliability of these complete 20 samples after testing the first 10 for 50 hours and the next 10 for 200 hours? > > I am now only struck at this last point.... Got it or still need explanation? well... you obviously don't get it: this NG (CSSM) deals with matters around and about the computer language and programming environment MATLAB... http://www.mathworks.com/products/matlab/ you're better off in a stats NG... us
From: Samoline1 Linke on 19 May 2010 06:35
"us " <us(a)neurol.unizh.ch> wrote in message <ht02a6$5t7$1(a)fred.mathworks.com>... > "Similione Kasan" <mech.engr(a)hotmail.com> wrote in message <ht01bo$5u6$1(a)fred.mathworks.com>... > > "us " <us(a)neurol.unizh.ch> wrote in message <ht00sk$61d$1(a)fred.mathworks.com>... > > > "Similione Kasan" <mech.engr(a)hotmail.com> wrote in message <ht009t$se6$1(a)fred.mathworks.com>... > > > > Hi, > > > > > > > > if suppose I have 20 identical samples and the required test time for them without failure is 100 hours. > > > > Now if out of these 20, I take 10 samples and test them for a time of 50 hours instead of 100 and the next 10 for a time of 200 hours. Then in the end what would be my total reliability? Assume confidence level of 80% > > > > > > ok, nice, now where's your question re ML(?)... > > > > > > us > > > > > > ----- > > Thanks for the response... > > > > I could not get you .... ok let's go this way.... > > > > Out of these 20 identical samples, I did testing for the first 10 for about 50 hours (which means lifetime ratio is different now). As the required testing time was 100 but I did for 50 hours and also my sample size is reduced to 10 so my reliability will be low now. And for the next 10 samples, I tested for 200 hours so my reliability will increase for these 10 samples. > > BUT how can I find the final reliability of these complete 20 samples after testing the first 10 for 50 hours and the next 10 for 200 hours? > > > > I am now only struck at this last point.... Got it or still need explanation? > > well... you obviously don't get it: this NG (CSSM) deals with matters around and about the computer language and programming environment MATLAB... > > http://www.mathworks.com/products/matlab/ > > you're better off in a stats NG... > > us ---------------- Yes but still reliability engineering related issues can be done in matlab e.g. Reliability R(t) = (1 - P)^(1/(Lv^b).n)....... eq(1) where R(t1) = reliability required at 50 hours P = confidence level = 80% n1 = 10 Lv = 50 /100 Putting these values in eq (1) one can get the value of R(t1) i.e. for 10 samples tested for 50 hours Similarly R(t2) can be determined based on above formula with n2 = 10 and Lv = 200/100 ... But I dont know how will I measure FINAL TOTAL RELIABILITY...shall I add R(t1) and R(t2) or multiply those.... |