# Non Homogeneous Poisson Process, failure rates and analysis

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TQ2

1. Suppose the failure time history of a component is given (where downward -facing arrows indicate failure times):

a. Assume the failure process follows a NHPP with λ(t) = exp(α+βt). Estimate the model parameters α and β using the MLE.

b. What is the expected number of failures from the current time until 72 months?

c. Check if the failure process has a time trend [α = 0.1]. If it does not, find the ROCOF.

2. Suppose the following data were observed in a repairable system:

a. Assume the failure process follows a renewal process and the interarrival time have

Estimate the parameter α based on the failure data. (Hint: find the MLE)

b. Now, assume that the failure process follows a renewal process with an exponentially distributed interarrival time Estimate the value of α.

3. The failure time distribution of a product can be described by the following CDF:

Where b is the parameter to be estimated. We have observed the following failure times (note + means that the time corresponds to a censored unit - i.e., the component has not failed at the point in time): 10, 12, 15, 18+, 23, 28, 35, 44, 45+, 51, 56+. Construct the likelihood function and the MLF of b.

4. Seven Pumps have failure times (in months) of 15.1, 10.7, 8.8, 11.3, 12.6, 14.4, and 8.7. Assume the pump failure follows an exponential distribution.

a. Find the MLF of λ.

b. Estimate the reliability of a pump at t = 12 months.

c. Calculate the 95% two-sided interval of λ

5. The following data were collected by Frank Proschan in 1983. Operating hours to first failure of an engine cooling part in 13 aircrafts were:

Do these data support an increasing, decreasing, or constant failure rate assumption?

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#### Solution Preview

All questions answered but I haven't had a chance to check my work, thanks

Suppose the failure time history of a component is given (where downward -facing arrows indicate failure times):

Assume the failure process follows a NHPP with λ(t) = exp(α+βt). Estimate the model parameters α and β using the MLE.

What is the expected number of failures from the current time until 72 months?

Check if the failure process has a time trend [α = 0.1]. If it does not, find the ROCOF.

(a) A NHPP is defined by a random process which has varying pdfs defined by intensity functions which are a function of time for different time periods. The pdf defining time to first failure (observed over a time period T) for a NHPP is

f_1 (t)=λ(t).e^(-∫_0^T▒〖λ(t^' ).dt^' 〗)

The likelihood function is defined by the product of n elemental pdfs (f_1 (t_i))for failures of first component, thus

L=∏_(i=1)^n▒〖f_1 (t_i ) 〗

Making substitutions λ(t)=e^(α+βt)=e^α e^βt

∫_0^T▒〖λ(t^' ).dt^' 〗=e^α ∫_0^T▒〖e^(βt^' ).dt^' 〗=e^α/β (e^βT-1)

∏_(i=1)^n▒〖e^α e^(βt_i ).e^(-e^α/β(e^βT-1)) 〗

L=∏_(i=1)^n▒〖e^α e^(βt_i ).e^(e^α/β(1-e^βT)) 〗

L=exp{nα+β∑_(i=1)^n▒〖t_i+exp(α)/β[1-exp(βT]〗}

Taking natural logarithms to simplify

Λ=Ln(L)=nα+β∑_(i=1)^n▒〖t_i+exp(α)/β[1-exp(βT)]〗

To find MLE for parameter α we find dΛ/dα and set this to zero

dΛ/dα=n+((1-e^βT))/β e^α=0

Thus

e^α=nβ/(e^βT-1) [1]

To find MLE for parameter β we find dΛ/dβ and set this to zero

dΛ/dβ=∑_(i=1)^n▒〖t_i-e^α/β^2 (1-e^βT ) 〗-(Te^α)/β e^βT=0

Thus

∑_(i=1)^n▒t_i =e^α/β^2 (1-e^βT )+ (Te^α)/β e^βT

Substituting the previous result e^α=nβ/(e^βT-1) in the above

∑_(i=1)^n▒t_i =nβ/(β^2 (e^βT-1) ) (1-e^βT )+nTβ/β(e^βT-1) e^βT

∑_(i=1)^n▒t_i =(nTe^βT)/(e^βT-1)-n/β

Dividing the first term (in the RHS of the above expression) both top and bottom by e^βT which is the equivalent of multiplying by 1 we obtain

∑_(i=1)^n▒t_i =(nTe^βT/e^βT)/((e^βT-1)/e^βT )-n/β

∑_(i=1)^n▒t_i =nT/(1-e^(-βT) )-n/β [2]

For a full mathematical treatment see article here

Equation [2] in the above cannot be solved analytically so we choose to solve for MLE of β value numerically. I complied an excel spread sheet and input test values for β that yielded the sum of t_i values

Examining the data we see we have n=7 in the data set,

∑_(i=1)^n▒t_i =5.2+10.8+18.3+28.8+33.3+44.1+49.2=189.7

Therefore I found

MLE β=0.0004115

From this value of β in equation [1] we therefore find the MLE for parameter α

MLE α=(7×0.0004115)/(e^(0.0004115×54)-1)=0.128195

Thus the process is modelled by intensity factor

λ(t)=e^(0.128195+0.0004115t)

(b) For the NHPP process we estimated in (a) that the intensity function is

λ(t)=e^(0.128195+0.0004115t)

The number of failures between 54 and 72 hours is given by the integral

N_72 (t≥54)=∫_54^72▒〖λ(t).dt=∫_54^72▒e^(0.128195+0.0004115t) .dt〗

N_72 (t≥54)=e^0.128195 (_54^72)|e^0.0004115t/0.0004115| =21

(c) The hazard function is given by

h(t)=(f(t))/(R(t))

For NHPP the pdf is given by f(t)=λ(t).e^(-∫_0^T▒〖λ(t).dt〗) and the reliability by

R(t)=e^(-∫_0^T▒〖λ(t).dt〗)

Hence for such a process the hazard function (equal to the intensity function) results

h(t)=(λ(t).e^(-∫_0^T▒〖λ(t).dt〗))/e^(-∫_0^T▒〖λ(t).dt〗) =λ(t)

For the particular case (rounding to one decimal place) α=0.1,β=0.0

h(t)=e^0.1 e^0.0t=e^0.1=1.1052

As β≃0 process has a ...

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