Average Product of Sawtooth Functions

Beginning from a non-failed state at time t = 0, a component with a constant failure rate λ has probability P(t) = 1 – e^−λt of being failed at time t. If λt is much less than 1 (as is often the case for highly reliable components), this is closely approximated by P(t) = λt. Hereafter we focus on this case. If we inspect and repair the component every T hours, then the probability of being failed at any time t is a sawtooth function that repeats every T hours as depicted below.

The average value of this sawtooth function is just the average over any single cycle, so we have

Since the maximum value is P_max = λT, this shows that for a single component we have P_ave/P_max = 1/2.

For N components, with constant failure rates λ₁, λ₂, …, λ_N respectively, the joint probability of all N being failed at time t (assuming no repairs) is the product

If all the components are inspected and repaired (synchronously) every T hours, then the joint probability has a periodic shape with maximum value P_max = P_prod(T). This is illustrated for the case N=3 in the figure below.

The average of the product is (again) the average over any single cycle, so we have

For a less trivial case, suppose the individual components are inspected and repaired at different intervals. Consider the case of two components, with failure rates λ₁ and λ₂, and suppose they are inspected/repaired at intervals of T and n₁T hours, respectively (beginning from a common initial time) for some integer n₁. In this case the maximum probability is P_max =  P₁(T)P₂(n₁T). Of course, if we make P₁ periodic with interval T, this is formally equivalent to P_max = P_prod(n₁T), but we typically calculate using the continuous basic functions. The resulting joint probability given by the product of these two sawtooth functions is shown below for the case n₁ = 2.

The average of the function equals the average over the full cycle from t = 0 to n₁T, and we can evaluate this by noting that the product over the first segment of length T is P₁(t)P₂(t) as t ranges from 0 to T, and over the second segment it is P₁(t)P₂(t+T) as t ranges from 0 to T, and over the third segment it is P₁(t)P₂(t+2T) as t ranges from 0 to T, and so on, up to the (n₁)th segment with P₁(t)P₂(t+(N−1)T) as t ranges from 0 to T. Thus the average is

Dividing by P_max = (λ₁T)(λ₂ n₁T) gives the result for N=2 components

With n₁ = 1 we have equal inspection intervals and this equation gives the ratio 1/3 as we saw previously for N=2 components. As the value of n₁ increases, the ratio asymptotically approaches 1/4.

Now consider the case of N=3 components with inspection/repair intervals of T, n₁T, and n₁n₂T respectively. We could proceed in the simple ad hoc manner as above, but to generalize to higher cases it’s useful to establish a more systematic approach. In this case we need to integrate n₁n₂ terms that can be written in the form

where k₁ cycles from 0 to n₁−1 and k₂ consists of n₂ blocks of n₁ numbers, and the numbers in the first block are all 0, and in the second block are all n1, and in the third are all 2n₁, and so on, up to (n₂−1)n₁. For example, with n1 = 3 and n₂ = 4 we have the k₁ values 0,1,2,0,1,2,0,1,2,0,1,2 and the k₂ values 0,0,0,3,3,3,6,6,6,9,9,9 respectively

We want to integrate the sum of the n₁n₂ terms of the above form, and to determine the coefficients of t³, etc., we need to determine the sum of (1), and the sum of the values of k₁, and the sum of the values of k₂, and the sum of the values of k₁², and the sum of the values of k₁k₂. In general we have

Inserting these expressions we can evaluate the average probability as

Recalling that in this case the maximum value of the product function is

we have the result for N=3 components

As expected, with n₁ = n₂ = 1 this ratio is 1/4, in agreement with the formula for equal intervals with N=3. On the other hand, if either n₁ or n₂ is 1 and the other is increased, the ratio asymptotically approaches 1/6, and if both n₁ and n₂ are large the ratio asymptotically approaches 1/8. This ratio happens to be symmetrical in n₁ and n₂, but P_max is not, so neither is P_ave.

Now consider the case of N=4 components with inspection/repair intervals of T, n₁T, n₁n₂T, and n₁n₂n₃T respectively. In this case we need to integrate n₁n₂n₃ terms that can be written in the form

where

The values of k₁ cycle from 0 to n₁−1 and k₂ consists of n₂ blocks of n₁ numbers, and the numbers in the first block are all 0, and in the second block are all n1, and in the third are all 2n₁, and so on, up to (n₂−1)n₁. The values of k3 consist of n3 blocks of n1n2 numbers, and the numbers in the first block are 0, and in the second are n₁n₂, and in the third are 2n₁n₂, and so on, up to (n₃−1)n₁n₂.

We want to integrate the sum of the n₁n₂n₃ terms of the above form, and to determine the coefficients of t³, etc., we need to determine the sum of (1), and the sum of the products of the k_j values. In general we have

with the stipulation that 0⁰ = 1. Thus if α_j = 0 the corresponding summation is simply n_j. For example, the coefficient of t₄ is n₁n₂n₃. We can express each summation in closed form using the sum of powers identities

Making use of these expressions we can evaluate the average probability as

Recalling that, in this case, the maximum value of the product function is

we have the result for N=4 components

As expected, with n₁ = n₂ = n₃ = 1 this ratio is 1/5, in agreement with the formula for equal intervals with N=4. On the other hand, for large values of n₁, n₂, and n3 the ratio asymptotically approaches 1/16.

The ratio can also be written in the equivalent form

The first term on the right side in parentheses is half the ratio for N=3, and the value of n₃ appears only in the denominator of the second term on the right, so that terms approaches zero for sufficiently large n₃. This is as expected, since the 4^th component would just be uniformly ramping up from zero to its maximum value as the other three components proceed as in the N=3 case, resulting is a simple factor of 1/2. Likewise we can write the lower order solutions as

Thus each of the ratios splits naturally into a sum of half the lower order ratio and a quantity inversely proportional to the highest-order time factor.

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