Maintenance Interval Optimization: Method

Maintenance Interval Optimization

What the tool answers

How often should an item be replaced, tested or inspected? The answer depends on how the item fails (its life distribution), on what a failure costs compared with planned work, and on the kind of task. The tool treats four tasks: replacement at a fixed age, replacement at fixed calendar intervals (one item or a group), a failure-finding test of a hidden function, and an on-condition inspection. All times are in one unit chosen on the form and all costs in one currency; nothing is converted.

Life distribution

The time to failure of a new item is a Weibull, R(t) = exp[−(t/η)^β], an exponential, R(t) = exp(−t/MTBF), or a lognormal, R(t) = 1 − Φ[(ln t − ln t₅₀)/σ]. F = 1 − R is the fraction failed, f the density and h = f/R the failure rate. The mean life limited to an age t, ∫₀^t R(x) dx, appears in every model and is closed for all three: η Γ(1 + 1/β) P(1/β, (t/η)^β) with the regularised incomplete gamma function P, MTBF [1 − exp(−t/MTBF)], and t R(t) + MTTF Φ[(ln t − ln t₅₀ − σ²)/σ].

A Weibull can also be fitted to pasted life data. Each line is a time with F (failure) or S (suspension: an item that had not failed at that age) and an optional count. The maximum-likelihood shape solves Σ t^β ln t / Σ t^β − 1/β = (1/r) Σ_failures ln t, with both sums of the first term over all units and r the number of failures, and η = (Σ t^β/r)^1/β. With few failures the shape is uncertain and biased high; the sensitivity tables show what that does to the interval.

Replacing an item before it fails can only pay when its failure rate rises with age. With a constant rate (exponential, β = 1) a survivor is as good as a new item, and with a falling rate (β below 1) it is better than a new one, so the tool reports that no optimum exists. Infant mortality is a matter for burn-in before service, which this tool does not size.

Age replacement

The item is replaced when it fails or when it reaches the age t. A cycle ends with a planned replacement (cost C_p) with probability R(t) and with a failure (cost C_f) with probability F(t); its mean length is ∫₀^t R. By the renewal-reward argument the long-run cost per unit time is the mean cost of a cycle over its mean length:

C(t) = [C_p R(t) + C_f F(t)] / ∫₀^t R(x) dx

Setting the derivative to zero gives h(t) ∫₀^t R(x) dx − F(t) = C_p/(C_f − C_p). The left side grows from zero when the failure rate rises, so there is one root for a Weibull with β above 1; the tool brackets the minimum on a logarithmic grid of ages and solves this equation by bisection. As t grows the cost tends to the run-to-failure value C_f/MTTF. For a lognormal life the failure rate eventually falls, and an optimum exists only if the cost dips below that value.

NAVAIR 00-25-403 (Appendix B 1.3.4) judges a hard-time task by the cost-benefit ratio CBR = {[C_BF N_S + C_AF (1 − N_S)] / [N_S t + (1 − N_S) MTTF_P]} / (C_AF/MTBF), where N_S is the fraction surviving to t and MTTF_P the mean age of those that fail earlier. Since N_S t + (1 − N_S) MTTF_P = ∫₀^t R, the ratio is C(t) divided by the run-to-failure cost; a value below 1 means the task is cost-effective.

Failures occur at the rate F(t)/∫R and planned replacements at R(t)/∫R per unit of operating time; multiplied by the use per year they give the yearly events, the items consumed (the demand for spares) and the yearly cost.

Minimum downtime. If a planned replacement takes T_p and a failure replacement T_f, the fraction of time down is D(t) = [T_p R + T_f F] / [∫R + T_p R + T_f F]. D is an increasing function of [T_p R + T_f F]/∫R, so the age of least downtime (highest availability) is the cost optimum with the downtimes in place of the costs.

Block and group replacement

Under block replacement the item is replaced at the fixed times T, 2T, … and at every failure in between. The expected number of failures in an interval is the renewal function M(T), which satisfies the renewal equation M(T) = F(T) + ∫₀^T M(T − x) dF(x). There is no closed form for a Weibull or lognormal life, so the equation is discretised on a uniform grid t_i = i h with the increments of M weighted by F at the midpoints (the Riemann-Stieltjes scheme of Xie 1989):

M_i [1 − F(h/2)] = F(t_i) + Σ_j=1ⁱ⁻¹ F((i − j + ½) h) (M_j − M_j−1) − F(h/2) M_i−1

The scheme uses only F, so it also works where the density is infinite at zero (β below 1). For large T, M(T) approaches T/MTTF + (σ² − MTTF²)/(2 MTTF²). For N identical items replaced together, with a set-up cost C_s for the visit,

C(T) = [C_s + N C_p + N C_f M(T)] / T

and the minimum is taken on the grid and refined by a parabola through the three lowest points. Block replacement wastes the remaining life of items replaced recently at a failure, so at equal costs age replacement is never dearer; block replacement needs no age records and one visit serves the whole group. The comparison table charges age replacement the set-up cost at every planned replacement, which is what makes group replacement attractive. Downtime per position is D(T) = [T_p + T_f M(T)]/(T + T_p).

Minimal repair. If a failed item is repaired to the condition it had just before the failure (cost C_m) and replaced only at T, the failures in an interval follow the cumulative failure rate H(T) = −ln R(T), and C(T) = [C_p + C_m H(T)]/T. For a Weibull life the minimum is closed: T = η [C_p/(C_m (β − 1))]^1/β.

Failure-finding interval

A hidden function (a standby pump, a relief valve, a trip) fails without any sign and is found failed only by a demand or a test. Tested every T and restored when found failed, it is down from the failure to the next test, and its mean availability is A(T) = (1/T) ∫₀^T R(x) dx. For a constant failure rate A(T) = (MTBF/T)[1 − exp(−T/MTBF)] ≈ 1 − T/(2 MTBF), because a failure falls on average in the middle of the interval. This gives the widely used rule

FFI = 2 (1 − A) MTBF

The tool gives this linear form and the exact solution of A(T) = A, which differ noticeably only for low availabilities. With a wear-out distribution the exact interval is much longer, but only if the item is renewed at each test; a test that merely confirms function leaves the item at its age, and the exponential interval should be used.

Least cost. If demands on the function arrive on average every M_d, a demand meets a failed function (a multiple failure) at the rate [1 − A(T)]/M_d. With a test cost C_i and a multiple-failure cost C_mf, the cost per unit time is c(T) = C_i/T + (C_mf/M_d)[1 − A(T)], least where ∫₀^T x f(x) dx = C_i M_d/C_mf; for short intervals T = √(2 C_i MTBF M_d/C_mf). If C_i M_d/C_mf is not below the mean life, testing never pays.

Acceptable probability of a multiple failure. NAVAIR 00-25-403 (Appendix B 1.4) writes P_mf = P_hidden × P_additional with each P = 1 − exp(−t/MTBF), sets the MTBF required of the multiple failure from an acceptable probability over a known period, MTBF_mf = −L/ln(1 − P_acc), and iterates the equations for t. The tool solves [1 − exp(−t/MTBF)] [1 − exp(−t/M_d)] = 1 − exp(−t/MTBF_mf), which for short intervals is t = MTBF M_d/MTBF_mf. The manual does not write this last step out; it is this tool's reading. The mean-unavailability argument above gives twice that interval, 2 MTBF M_d/MTBF_mf; both are printed.

On-condition inspection

Many failures announce themselves: a crack, a vibration, a temperature rise. The P-F interval is the time from the point where the potential failure can be detected to the functional failure. NAVAIR 00-25-403 (Appendix B 1.2) puts n inspections in that interval, I = PF/n. If one inspection finds an existing potential failure with probability θ, all n miss it with probability (1 − θ)ⁿ; for an acceptable probability P_acc, n = ln(P_acc)/ln(1 − θ). For failures without safety consequences the manual gives the cost-optimal number

n = ln{ −(MTBF/PF) C_i / [(C_npm − C_pf) ln(1 − θ)] } / ln(1 − θ)

with C_i the cost of one inspection, C_pf the cost of correcting a potential failure and C_npm the cost of a functional failure without the task. It minimises C_i n/PF + [(1 − (1 − θ)ⁿ) C_pf + (1 − θ)ⁿ C_npm]/MTBF. By the manual's rules, n ≤ 0 means the task is not cost-effective, and for n between 0 and 1 the P-F interval itself is the task interval. The MTBF is the mean of the entered life distribution.

Sensitivity

The optimum of a replacement model is flat: an interval 10 % off costs under 1 % more in the default case. What moves it is the cost ratio C_f/C_p and the shape of the distribution. The first table repeats the optimisation for ratios from 2 to 100; the second varies the Weibull shape (or the lognormal σ) by ±15 % and ±30 % and shows what keeping the entered optimum would cost if the true value were different.

Limits

The models are long-run averages for one failure mode with perfect replacement; they ignore discounting, finite horizons, opportunistic maintenance, imperfect repair and inspection downtime. The cost models do not add the downtime to the cycle length. The default case is illustrative, not a published example.