Maintenance action is characterized by the degree to which a system can be restored. The most common assumptions on maintenance efficiency are known as minimal repair, or “As Bad As Old” (ABAO), and perfect repair, or “As Good As New” (AGAN). In the ABAO case, each maintenance leaves the system in the state it was before maintenance. In the AGAN case, each maintenance is perfect, and leaves the system as if it were new. It is common for true experience to result between these behaviors: standard maintenance reduces failure intensity, but does not leave the system as good as new. This situation is known as imperfect maintenance. One of the most popular is the Generalized Renewal Process (GRP) :

 V[i] = V[i−1] + q[i]t[i]                                               

In this equation the t[i] is the length between (i-1) and i-th events,  V[i] is the virtual age of the item after the i-th maintenance action, q[i]  is the restoration factor of imperfect maintenance after i-th event (0

See article 

General Repair Models for Maintained Systems

A.Syamsundar, K. Muralidharan and V.N.A. Naikan 

Dear Alex Yiolkin , thanks a lot for your answer. I will carefully read this article. Regards, Sergey.

Answer for your question " How did you succeed in estimating correctly the Weibull parameters, IF the data come from GRP? " : for example, see paper on the RG

https://www.researchgate.net/publication/307639316_Is_it_Correct_to_Use_MLE_Method_for_GRP_Parameter_Estimation

Really we can correct estimate ONLY Weibull parameters, we could not estimate restoration parameters

Negative Results Is it Correct to Use MLE Method for GRP Parameter Estimation ?

Dear Fausto, you can extract data from article [5] (attached), table 1. Regards, Sergey

Dear Fausto , thanks a lot for your answer. I will carefully read your article. Regards, Sergey.

Dear Fausto, thanks for your answer. I should remember this research (I has performed it year ago !), so I will answer you ASAP. Regards, Sergey.

Dear Fausto, I don't understand, why " the CM data increase the System MTTF and therefore the estimate of "tetha". Input data are not only according Teta and Beta, they according fully imperfect maintenance (i.e. CM and PM, GRP with Teta, Beta, q_pm and q_cm). So, after MLE estimation we should (asymptotically, i.e. for large amount of events) get same values, as used for simulation. See, please, chapter 2.2 of my white paper "Is it Correct to Use MLE Method for GRP Parameter Estimation" (Analysis of Estimations, based on Other Statistics). 

Dear Fausto, Hi! First, we don't know q_pm and q_cm and have to estimate them. But even if we know values of q_pm and q_cm, expressions are more complicated. We could not multiple for q_pm or q_cm last time interval.   Please, see chapter 7 (7. Maximum likelihood estimation of GRP parameters) of the article Yanez M., Joglar F. and Modarres M. “Generalized renewal process for analysis of repairable systems with limited failure experience”, Reliab. Eng. Syst. Saf., vol.77, pp. 167–180, 2002.

No,  we could not multiple for q_pm or q_cm the last time interval.   Please, see chapter 7 of the attached article. Regards, Sergey

Sorry, I don't understand, what do you mean for " datum with q_pm included". If you mean, that this interval depends of q_pm or q_cm, answer will be YES. But situation more complicated.

1    1.1873 is PM; PM is AFTER 1.1873, so this interval does not depend on q_PM

2    0.3650 is CM; CM is AFTER 0.3650, so 0.3650 depends on previously q_PM

3    0.0913 is CM; Current CM is AFTER 0.0913, but 0.0913 depends on previously q_PM and q_CM.

4    0.0183 is PM; Current PM is AFTER 0.0813, but 0.0813 depends on previously q_PM, q_CM and q_CM

...

Please, use expressions from Chapter 7. 

Dear Fausto, sorry, I will out of office until 7 October. Regards, Sergey.  

Dear Fausto, Hi! Sorry, I have answered for your question 12 day ago (see above) and after this did not receive any additional questions. Below I repeat my answer (perhaps, you could not read it?).

Sorry, I don't understand, what do you mean for " datum with q_pm included". If you mean, that this interval depends of q_pm or q_cm, answer will be YES. But situation more complicated.

1    1.1873 is PM; PM is AFTER 1.1873, so this interval does not depend on q_PM

2    0.3650 is CM; CM is AFTER 0.3650, so 0.3650 depends on previously q_PM

3    0.0913 is CM; Current CM is AFTER 0.0913, but 0.0913 depends on previously q_PM and q_CM.

4    0.0183 is PM; Current PM is AFTER 0.0813, but 0.0813 depends on previously q_PM, q_CM and q_CM

...

Please, use expressions from Chapter 7 (from early attached article). Sergey.

It is fully OK, don't worry! Regards, Sergey.

Dear Fausto, Hi! In our real life we usually don't see monotonically reduction of the intervals between new data, because noise exist. Data from Table 1 are obtained by means of Random Generator, so we also don't see monotonically reduction of the intervals between new data. But trend for reduction exist! I have checked it based on mean calculation for data with numbers [1...10], [11...20], [21...30],...[91...100]. Results are:

mean(D_PM_0_CM_1(1:10)) = 0.2335;

mean(D_PM_0_CM_1(11:20)) = 0.1628;

mean(D_PM_0_CM_1(21:30)) = 0.0979;

mean(D_PM_0_CM_1(31:40)) = 0.0812;

mean(D_PM_0_CM_1(41:50)) = 0.0366;

mean(D_PM_0_CM_1(51:60)) = 0.0395;

mean(D_PM_0_CM_1(61:70)) = 0.0423;

mean(D_PM_0_CM_1(71:80)) = 0.0408;

mean(D_PM_0_CM_1(81:90)) = 0.0746;

mean(D_PM_0_CM_1(91:100)) = 0.0456;

Dear Fausto, you are fully right. Sorry, it was error of data insertion. Thanks a lot for your correction! But in any case my explanation is right . These data are according Random Generator. In this time-series trend of reduction of "time interval between events" is low, in next time series trend may be large or, opposite, low... Sometimes may be 10 times (step-by-step) coin will be on same side - and it isn't error! I have checked results of Arkanzas by means of generation of own large Data Sets (1000 events). Output results are same.

Dear Fausto, please, see attached Matlab file. Regards, Sergey.

OK, I will do this transformation ASAP and will send to you EXCEL file. regards, Sergey.

See, please, attached file. Regards, Sergey.

Dear Fausto, thanks a lot for your answer. I will carefully analyzed it. Regards, Sergey.

Dear Fausto, Hi! First, thanks a lot for your interest for my work. Second, let's discuss more detail your conclusion. Why do you suppose, that "...From the equality Random Number (probability!) = R(t|s), [s=age of the system, and t>s] did you (and Arkansas people) find t AND then reported the length of the "interval t-s"? IF this is the way you did you should find that the "distribution of the reliabilities R(t|s)" is uniform..."

OK, THANKS. For GRP Data set generation I have used following expressions:

 dt[i]=( V[i]^b - log(1-rand[i])/a )^(1/b) - V[i]; V[i+1] = V[i] + Q*dt[i]; Where a and b are Weibull parameters, Q is Restoration Factor, rand[i] are uniform distributed random values, V[i] is virtual aging. I will calculate rand[i] from my generated statistics (1000 points) according this expressions and check, are they really uniform (as were for generation -:). Regards, Sergey.

I will carefully check this ASAP, but in our case these expressions are same - because a = 1.

For GRP :

R = 1 - exp( -a((V[i] + dt[i])^b – V[i]^b) )

So, dt[i]=(V[i]^b-log(1-R)/a)^(1/b)-V[i]

It is exactly same, but we use different abbreviations:

1. For me R is conditional Probability of failure during dt[i]

For you R is conditional Probability of ABSENT failure during dt[i] (i.e. (t - s) )

So, R_my = 1 - R_your

2. V[i] =s, V[i] +dt[i] = t, dt[i] = t - s.

Regards, Sergey.

Dear Sergey and Fausto, thank you very much for your discussion. It was very helpful for me to read it.

Dear Fausto, this week I too busy, but next week I ASAP will calculate values rand[i] from array dt[i] (which I have sent to you ) according expressions : dt[i]=( V[i]^b - log(1-rand[i])/a )^(1/b) - V[i]; V[i+1] = V[i] + Q*dt[i];

In my opinion, rand[i] will be uniform distributed. I will check this. Regards, Sergey.

Dear Fausto, Hi! I have performed calculations and have got conditional probabilities. See, please, attached file. Sheet 1 – input data, i.e. dt (I have sent it week ago). Sheet 2 – calculated conditional probabilities (you can compare them with your results). Sheet 3 – sorted calculated conditional probabilities. See, please, also figure of sorted calculated conditional probabilities (attached).

You are fully right – these data are not according uniform distribution. It is seen very well on figure – it is not line type y = x. Reason is following. For one type of maintenance (e.g., all events are CM) probabilities will be exactly uniform distributed, but I have generated events of two types (CM and PM) – see sheet 1 of attached file. So, really dt is min (generated time for PM, generated time for CM). And so calculated conditional probabilities are distributed according CDF for min (x, y).

Regards, Sergey.

Dear Fausto, my answers are below. Regards, Sergey.

1. Yes

2. Yes

3. It isn't my idea, I only have used approach from article "Parameter Estimation for a Repairable System under Imperfect Maintenance" (you have received it from me early). Generally speaking, we can use a lot of expressions to generate events for PM - fixed time from last event, fixed time from last PM, some random time,...

4. Yes

Dear Fausto, hi! In my opinion, our discussion was very useful both for me and for you! Thanks a lot for participation. Regards, Sergey.