Francisco Anes
December 8th, 2000, 06:05 PM
[Originally Posted: 10/31/00--Transferred by ReliaSoft Moderator]
I have four fuel feed pumps. I observed all at the same time and I stoped observation also at the same time. At starting time all four had accumuled operation time(do not know by observer). For all pumps is considered a renewal process. So from maintenance files I got the following failure and right censored data.
(a)Pump 1: 1709,16401(RC) (b)Pump 2: 18330(RC) (c)Pump 3: 4835, 1699, 2271, 1085(RC) (d)Pump 4: 911, 5990, 1085(RC)
As renewall processes I put all data together and I ajusted a Weibull distribution, according to the statements below:
(1)I treated data as six failures and four right censored data points. I got by MLE (90% confidence) to = 0 ; B = 0.825(0.5513 - 1.235); n = 9147(4683 - 17290).
(2)I treated times to first failures as censored left (1709, 4835 and 911)as well as failure data (three data points) and censored right (four data points). I got the following results: t0 = 0; b = 0.906(0.5193 - 1.581);n = 19480(7642 - 49660).
So I would get ansewers for the following questions:
(a)On (1) I treated the suposed left censored data as failure data. Do you agree? Why? (b)On (2)do you agree to treat left and right censored data as the same way?Why? (c)What method do you select (1) or(2)? Why? (d)What is really the best method to treat this kind of data?
I would like to thank all of you.
Francisco Anes
I have four fuel feed pumps. I observed all at the same time and I stoped observation also at the same time. At starting time all four had accumuled operation time(do not know by observer). For all pumps is considered a renewal process. So from maintenance files I got the following failure and right censored data.
(a)Pump 1: 1709,16401(RC) (b)Pump 2: 18330(RC) (c)Pump 3: 4835, 1699, 2271, 1085(RC) (d)Pump 4: 911, 5990, 1085(RC)
As renewall processes I put all data together and I ajusted a Weibull distribution, according to the statements below:
(1)I treated data as six failures and four right censored data points. I got by MLE (90% confidence) to = 0 ; B = 0.825(0.5513 - 1.235); n = 9147(4683 - 17290).
(2)I treated times to first failures as censored left (1709, 4835 and 911)as well as failure data (three data points) and censored right (four data points). I got the following results: t0 = 0; b = 0.906(0.5193 - 1.581);n = 19480(7642 - 49660).
So I would get ansewers for the following questions:
(a)On (1) I treated the suposed left censored data as failure data. Do you agree? Why? (b)On (2)do you agree to treat left and right censored data as the same way?Why? (c)What method do you select (1) or(2)? Why? (d)What is really the best method to treat this kind of data?
I would like to thank all of you.
Francisco Anes