In planning future repairs, allocating parts, tools and skilled workforce, the expected number of future failures plays a crucial role. Let M ( t ) denote the expected number of failures in the interval ( 0 , t ) , then its determination depends on the types of repairs being used. The most important characteristic of any random failure is the time when it occurs; let F ( t ) be the Cumulative Distribution Function (CDF) of the time of first’ failure, f ( t ) the probability density function (pdf), and failure rate ρ ( t ) = f ( t ) / ( 1 − F ( t ) ) . The formal definition of the failure rate is as Equation (1).

ρ ( t ) = lim Δ t → 0 P ( t < X < t + Δ t | t < X ) Δ t (1)

where X is the time of the first failure. The numerator is the conditional probability that the object will break down during the next Δt time periods given that it is working at time t. The list of parameters is shown in Table 1.

In the case of minimal repairs, it is well known that the expected number of failures is the solution of the Volterra-type integral equation:

M ( t ) = M ( t ) F ( t ) + F ( t ) − ∫ 0 t M ( τ ) f ( τ ) d τ (2)

which can be shown by substitution to be:

M ( t ) = ∫ 0 t ρ ( τ ) d τ (3)

which is the cumulative failure rate. If the equipment is replaced after each failure, then:

M ( t ) = F ( t ) + ∫ 0 t M ( t − τ ) f ( τ ) d τ (4)

In the case of partial repairs, the form of the integral equation depends on the types of the repairs ‎ [26] . This paper introduces a general model which generalizes several earlier ones incorporating effects which were ignored in the earlier studies.

The earlier models assumed instantaneous repairs. Here we assume that each repair takes T time periods, when the object is idle, and the degradation rate of the object is less than that of the working one. Mathematically it is assumed that a repair results in an increase of the effective age by dT ( 0 < d < 1 ).

Partial repairs also make the state of equipment better by decreasing the expected number of future failures by a factor α ( 0 < α < 1 ). Parameter α depends on the actual steps of the repair process including which parts are replaced, renewed, maintained, etc, while the value of d is a function of conditions (like temperature, humidity, etc.) under which the repairment is performed. In this model CDF F ( t ) , pdf f ( t ) and parameters α and d are assumed to be known. Let t > 0 be a future time. For the value of M ( t ) we can generalize the integral Equations (2) and (4) based on the well-known idea of expectation by conditioning. Assume that the first failure occurs at time S 1 , then repair is finished at time S 1 + T , when the effective age of the object becomes S 1 + d T . Therefore, the effective age at time t becomes t − T + d T , since during T time periods the effective age increases with dT instead of T. The assumptions are illustrated in Figure 3. Expected number of failures and effective age of a single system. Assuming that the value of S 1 is fixed, then:

E ( X t | S 1 ) = { 0 if   t < S 1 1 if   S 1 ≤ t < S 1 + T 1 + α [ M ( t − T + d T ) − M ( S 1 + d T ) ] if   t ≥ S 1 + T (5)

where X t is the true number of failures in interval ( 0 , t ) . The first case shows that no other failure can occur before the first failure. In the second case the first failure already occurred, and repair is not finished before time S 1 + T . In the third case notice that in effective age interval ( S 1 + T , t − T + d T ) ), the expected number of failures would be the difference M ( t − T + d T ) − M ( S 1 + T ) , which is multiplied by α as the result of repair. And now the expectation of this function has to be determined with respect to S 1 :

M ( t ) = ∫ t ∞ 0 f ( s ) d s + ∫ t − T t 1 f ( s ) d s     + ∫ 0 t − T { 1 + α [ M ( t − T + d T ) − M ( s + d T ) ] } f ( s ) d s (6)

which can be simplified as

M ( t ) = F ( t ) + α M ( t − T + d T ) F ( t − T ) − α ∫ 0 t − T M ( s + d T ) f ( s ) d s (7)

if t ≥ T . Otherwise, if t < T , then only one failure may occur before time t because repair cannot be finished before or at time T. The number of failures in interval ( 0 , t ) is either 1 or 0 with probability F ( t ) or 1 − F ( t ) , respectively. Consequently M ( t ) = F ( t ) in this case. A numerical integration-based algorithm can be suggested to find the value of function M in a grid between 0 and any t > 0 .

The trapezoidal rule calculates the area beneath a curve by subdividing it into

multiple trapezoidal shapes and then summing the areas of these individual trapezoids. Therefore, this method is used as a numerical integration-based algorithm to find the expected number of failures.

Considering Figure 4. Trapezoidal Rule, 0 , h , 2 h , ⋯ , N h = t be the grid points and assume that the small value of h is selected so that d T = K h and T = L h with K and L being positive integers. The value of h can be selected so that dT is an integer multiple of h. If T is not an integer multiple of h, then select L so that L h ≤ T < ( L + 1 ) h . For l = 0 , 1 , 2 , ⋯ , L clearly M ( l h ) = F ( l h ) since l h ≤ T . If l > L , then using the trapezoidal rule for approximating the integral in Equation (7) we get the approximation:

M ( N h ) = F ( N h ) + α M ( ( N − L + K ) h ) F ( ( N − L ) h )   − α h { 1 2 M ( ( N − L + K ) h ) f ( ( N − L ) h ) + ∑ l = 1 N − L − 1 M ( ( l + K ) h ) f ( l h ) } (8)

Notice that in the right-hand side the value of function M appears at earlier grid points than Nh, since L > K . So given the initial values of M at poins l h , l = 0 , 1 , 2 , ⋯ , L , the further values of M(lh) can be obtained by using equation (8) in the order of l = L + 1 , L + 2 , ⋯ , N .

Any other integral approximation can be used similarly. The most popular formulas is discussed in [28] .

Assume first that there is no degradation during repair, when d = 0 . Then Equation (7) reduces to

M ( t ) = F ( t ) + α M ( t − T ) F ( t − T ) − α ∫ 0 t − T M ( s ) f ( s ) d s (9)

and if minimal repair is assumed, then α = 1 , so in this simpler case:

M ( t ) = F ( t ) + M ( t − T ) F ( t − T ) − ∫ 0 t − T M ( s ) f ( s ) d s (10)

which is the same as given in [26] . If this model is further simplified by

assuming instantaneous repairs with T = 0 , then the well-known model in Equation (2) is obtained.

Assume next failure replacement. A similar model to Equation (7) can be derived similarly. Using the earlier notation:

E ( X t | S 1 ) = { 0 if   t < S 1 1 if   S 1 ≤ t < S 1 + T 1 + E ( X t − S 1 − T ) if   t ≥ S 1 + T (11)

since after repair t − S 1 − T periods are needed to reach t. So

M ( t ) = ∫ t ∞ 0 f ( s ) d s + ∫ t − T t 1 f ( s ) d s + ∫ 0 t − T { 1 + M ( t − s − T ) } f ( s ) d s F ( t ) + ∫ 0 t − T M ( t − s − T ) f ( s ) d s (12)

for t ≥ T . If t < T ,, then M ( t ) = F ( t ) . Assume again instantaneous repairs with T = 0 , then the equation reduces to Equation (4) as it should.

Assume next that the partial repairs decrease the effective age of the object by a fixed factor α < 1 as is shown in Figure 5. S 1 be the time of the first failure, then at time S 1 the effective age is clearly S 1 , at time S 1 + T it is S 1 + d T and as the result of repair it decreases to α ( S 1 + d T ) . Without failure and repair it would be S 1 + T , so the effective age is decreased by S 1 + T − α ( S 1 + d T ) in comparison to the no failure case.

So, the effective age at time t has to be

t − ( S 1 + T − α ( S 1 + d T ) ) = t − ( 1 − α ) S 1 − T + α d T (13)

implying that

E ( X t | S 1 ) = { 0 if   t < S 1 1 if   S 1 ≤ t < S 1 + T 1 + [ M ( t − ( 1 − α ) S 1 − T + α d T ) − M ( α ( S 1 + d T ) ) ] if   t ≥ S 1 + T (14)

so

M ( t ) = F ( t ) + ∫ 0 t − T [ M ( t − ( 1 − α ) s − T + α d T ) − M ( α ( s + d T ) ) ] f ( s ) d s (15)

Assuming minimal repair with α = 1 , then Equation (15) becomes

M ( t ) = F ( t ) + ∫ 0 t − T [ M ( t − T + d T ) − M ( s + d T ) ] f ( s ) d s = F ( t ) + M ( t − T + d T ) F ( t − T ) − ∫ 0 t − T M ( s + d T ) f ( s ) d s (16)

which is identical to Equation (7) with α = 1 .

In the further specialized case when d = 0 we have a simpler Equation:

M ( t ) = F ( t ) + M ( t − T ) F ( t − T ) − ∫ 0 t − T M ( s ) f ( s ) d s (17)

which gives back Equation (2) in the case of instantaneous repairs with T = 0 . If T > 0 , then the integral terms have t − T as the upper bound for the interval. Therefore, a similar algorithm to Equation (8) can be developed, since in the right-hand sides only earlier values of function M are used. However, if T = 0 , then the right-hand side also has the value of M ( t ) so the same successive method cannot be used. Formulating the approximations of the integral equations for all M ( l h ) values for L + 1 ≤ l ≤ N , a system of linear algebraic equations is obtained for the unknown M ( l h ) which can be then easily determined as follows. The equation for l = L + 1 has only one unknown M ( L + 1 ) , so it can be solved for this unknown. After M ( L + 1 ) is determined the equation for M ( L + 2 ) gives the value of M ( L + 2 ) , and so on.

In very special cases, like for equation (4), Laplace transforms can be used. Let F * , f * and M * denote the Laplace transforms of F, f and M, respectively, then from Equation (4),

M * ( s ) = F * ( s ) + M * ( s ) f * ( s ) (18)

so M * ( s ) is now given and M ( t ) is determined by using inverse Laplace transforms.

The CDF of first failures F ( t ) can be obtained from the first failures data. After each repair the failure rate as well as the CDF changes. We can however transform the later failure times into effective ages of the object if F ( t ) would not change. A successive algorithm can be offered to find the ( F ( t ) − ) effective ages at each later failure.

Let t 1 denote the time of first failure, and before this event F ( t ) is the CDF. Before repair the effective age is t ¯ 1 = t 1 , which becomes t ¯ 1 = t 1 + d T at calendar time t 1 + T .

Consider now the k^th failure of an object, which occurs at time t k and F k ( t ) is the CDF before this failure occurs. Repair is finished at time t k + T . If t ¯ k is the ( F ( t ) − ) effective age at the time of failure, then at the time t k + T it becomes t ¯ k + d T . If ρ k ( t ) is the failure rate before breakdown, then after repairing it becomes α ρ k ( t ) , since

F k ( t ) = 1 − e − ∫ 0 t ρ k ( τ ) d τ (19)

after repair the CDF becomes

F k + 1 ( t ) = 1 − e − a ∫ 0 t ρ k ( τ ) d τ (20)

so, the corresponding reliability functions are as follows:

R k ( t ) = e − ∫ 0 t ρ ( τ ) d τ ,     R k + 1 ( t ) = e − α ∫ 0 t ρ ( τ ) d τ = [ R k ( t ) ] α (21)

So

F k + 1 ( t ) = 1 − R k + 1 ( t ) = 1 − [ 1 − F k ( t ) ] α (22)

Under F k + 1 ( t ) object is working from effective age t ¯ k + d T at calendar age t k + T to calendar age t k + 1 , so the ( F k + 1 ( t ) − ) effective age of the object is t ¯ + d T + t k + 1 − t k − T , and if the ( F ( t ) − ) effective age is t ¯ k + 1 , then

F ¯ k ( t ¯ k + d T + t k + 1 − t k − T ) = F ( t ¯ k + 1 ) (23)

which always has a unique solution in F ( t ) is strictly increasing. The solution t ¯ k + 1 gives the ( F ( t ) − ) effective age at time t k + 1 .

The repeated application of this process gives the ( F ( t ) − ) effective ages t ¯ 1 , t ¯ 2 , ⋯ of the subsequent failure times. If several identical objects are observed then for each of them the procedure can be repeated, so a set of failure ( F ( t ) − ) effective ages become available. To test if the assessed parameter values α and d are acceptable we can use a hypothesis test to see if the obtained data set is homogeneous, or data of the first failures and that of the later failures belong to the same statistical family, that is, have the same distribution.

Assume Weibull distribution for time to failure with parameters λ = k = 2 , so

F ( t ) = 1 − e − ( t λ ) k = 1 − e − t 2 4 ,     f ( t ) = k λ ( t λ ) k − 1 e − ( t λ ) k = t 2 e − t 2 4 (24)

The parameter values α = 0.9 , d = 0.1 and T = 1 are chosen. The grid step size was selected as h = 0.01 . We considered interval (0, 10). Therefore K = 10 , L = 100 and t = N h with N = 1000 . Equation (8) was used to find the values of M ( T ) for t = l h for l = 0 , 1 , 2 , ⋯ , 1000 . Table 2 shows the computer results for integer values of t and Figure 6. The expected number of failures in the time interval (0,10) shows the expected number of failures within the time interval.

M ( t = 0.01 ) = F ( 0.01 )

M ( t = 0.02 ) = F ( 0.02 )

⋮

M ( t = 1 ) = F ( 1 )

M ( t = 1.1 ) = F ( 1.1 ) + α M ( 0.11 ) F ( 0.01 ) − α h ( 1 / 2 M ( 0.11 ) f ( 0.01 ) )

M ( t = 1.2 ) = F ( 1.2 ) + α M ( 0.12 ) F ( 0.02 )   − α h ( 1 / 2 M ( 0.12 ) f ( 0.02 ) + M ( 0.11 ) f ( 0.01 ) )

M ( t = 1.3 ) = F ( 1.3 ) + α M ( 0.13 ) F ( 0.03 ) − α h ( 1 / 2 M ( 0.13 ) f ( 0.03 )   + M ( 0.11 ) f ( 0.01 ) + M ( 0.12 ) f ( 0.02 ) )

⋮

M ( t = 10 ) = F ( 10 ) + α M ( 9.11 ) F ( 9 ) − α h ( 1 / 2 M ( 9.11 ) f ( 9 )   + M ( 0.11 ) f ( 0.01 ) + ⋯ + M ( 9.1 ) f ( 8.99 ) )

The expected time to failure is λ Γ ( 1 + 1 k ) = 2 Γ ( 1 + 1 2 ) = Γ ( 1 2 ) = π which is approximated by 1.772. So, if after each repair brand new object would start working, then we might think that:

10 / 1.772 ≈ 5.643

is the expected number of failures. Since in time the object’s degradation speeds up, it is clear that for larger values of t, the value of M(t) becomes larger than this ratio.

BNSF railway is a significant North American network of freight railroads. The data set from BNFS 2007 to 2013 is used in this study. In the data set, a track can fail geometrically in one of three ways as follows [16] :

• The first is cross level failure mode, which assesses the variation in top surface elevation between two rails at any particular location along the railroad track. Since the rails can move up or down when under load, the cross-level measurement is typically done while they are in motion (Figure 7).

• The second is the surface failure mode, which assesses any irregularities in a rail’s top surface. When there is a hump or a dip, the surface measurement can be positive or negative (Figure 8).

• DIP, which measures a decrease or increase in the track’s centerline, is the third failure mode (Figure 9).

To model the time to failure for each failure mode, we use the Weibull distribution. The Weibull distribution as one of the most flexible and powerful distributions can be used to model both increasing and decreasing failure. The Weibull distribution has been used in [29] to model rail geometry defect time to failure. Equations (6) and (7) show the Weibull CDF and PDF for the failure at time t, respectively. The k and λ show shape and scale parameter of Weibull distribution parameters. For each failure mode the Weibull parameters (shape and scale) has been estimated and their standard deviation has been calculated based on the BNSF data and the result is illustrated in Table 3.

parameter values α = 0.8 , d = 0.1 and T = 1 are chosen. The grid step size is selected as h = 0.01 . We consider an interval of two years or (0, 730) days to study the number of failures within the selected interval for each of failure mode. Therefore K = 1 , L = 10 and t = N h with N = 7300 . The proposed algorithm is used to find the values of M ( T ) for t = l h for l = 0 , 1 , 2 , ⋯ , 7300 . Figure 10 shows the computer results for M ( t ) for different type of failures.

To evaluate our model, we conduct a Monte Carlo (MC) simulation study to examine different failure modes. We take inspiration from [14] particularly sections 5.1 and 5.3.

Step 1: Generating Random Values

In this step, we assume a Weibull distribution for inter-arrival failures. To represent F ( t i ) , we generate random values from a uniform distribution ranging between 0 and 1.

Step 2: Accumulating Time Intervals

As part of the simulation process, we repeatedly generate random t i values. Each generated t i is cumulatively added to the sum of previously generated t i values.

Step 3: Comparing with Time Interval

We compare the cumulative sum of these generated t i values with a predefined time interval. In our study, this time interval is set at 2 years (730 days).

Step 4: Determining Simulation Outcome

If the cumulative sum of all generated t s equal or exceed 730 days, the simulation algorithm terminates. At this point, we calculate the average number of failures generated at each step. This average represents the expected number of failures based on 10,000 simulation runs.

Step 5: Continuing Simulation

Conversely, if the cumulative sum of generated t s is still less than the study time interval (730 days), we continue generating additional t i values.

Step 6: Result Presentation

Finally, the result of this MC simulation is used to determine the number of failures occurring within a two-year timeframe, which is presented in Table 4.

The findings indicate that, within a two-year timeframe, the anticipated occurrences of cross level failures, surface failures, and DPI failures are 2.4, 3.8, and 5.8, respectively.

In this part the sensitivity of parameters regarding expected number of failures is discussed and cross level mode is selected as the failure analysis. Three parameters are considered to analyze the results. First, the shape parameter of Weibull distribution (i.e., k) is selected. As it is shown in Figure 11, by increasing value of k in range (0.5, 4.5) and keeping other parameters constant, the expected number of failures is increased dramatically. The next two parameters have impacts on the effective age of the system. Figure 12 shows by decreasing the value of α, the expected number of failures is dropped since the value of α = 0 shows the quality of repair is as good as new, while α = 1 shows the quality of repair is bad as old. Also, the effect of degradation when the system is idle (d) is studied. By increasing the value of d, the expected number of failures increased but not significantly. Therefore, we can conclude the effect of d on M(t) is negligible

in this case. When the degradation rate increases, the effective age of the system increases and consequently, the system gets old more than the other modes and the chance of facing the failure increases.