Reliability Growth: Cumulative Versus Instantaneous MTBF
Three prototype units of a new servo drive are on a reliability growth test. At 4,800 accumulated unit-hours the team has logged eight failures, at cumulative test times of 120, 340, 780, 1,250, 1,900, 2,700, 3,600 and 4,500 hours. Five of those failures had occurred by the 2,400-hour mark, and at that point all corrective actions identified so far had been incorporated into the units. Fixes for three of the eight failures - the ones at 2,700, 3,600 and 4,500 hours - are still in work and have not yet been validated in the test. The program milestone requires a demonstrated MTBF of 1,000 hours at the 6,000-hour test point, where a formal demonstration test will begin. The chief engineer wants to stop growth testing now and advertise the 600-hour cumulative MTBF as the demonstrated value, arguing that the number is conservative. The customer's reliability engineer is asking for the current, not the historical, MTBF, and for an estimate of when the 1,000-hour target will actually be reached.
1. Estimate the Duane growth slope, the instantaneous MTBF now, and the test time needed to reach a 1,000-hour instantaneous MTBF. Is the program on track for its 6,000-hour milestone?
Read the worked answer
Estimate the growth from two points, then project. Cumulative MTBF is total time divided by cumulative failures.
- At 2,400 h: 2,400/5 = 480 h. At 4,800 h: 4,800/8 = 600 h.
- Duane slope b = ln(600/480) / ln(4,800/2,400) = 0.2231/0.6931 = 0.322.
- Cumulative model: theta_c(T) = 600 x (T/4,800)^0.322.
- Instantaneous MTBF = theta_c/(1 - b), so at 4,800 h, theta_i = 600/0.678 = 885 hours.
At the 6,000-hour milestone: theta_c = 600 x 1.25^0.322 = 662 h, and theta_i = 662/0.678 = 977 hours. That is about 2% short of 1,000 - close enough to be inside the noise of an eight-failure estimate, but it is not a demonstration.
When is 1,000 hours actually reached? Set theta_i = 1,000, so theta_c = 1,000 x 0.678 = 678 h. Then T = 4,800 x (678/600)^(1/0.322) = 4,800 x 1.462 = about 7,020 hours, roughly 2,200 hours beyond the milestone and 46% more test time than planned.
The program is growing, but not fast enough to hit the milestone on schedule. The options are to extend the test, add units in parallel to accumulate test time faster, or attack the highest-occurrence failure modes so the slope itself improves. At b = 0.32, waiting longer at the same rate will not close a 46% shortfall in a reasonable time.
2. The chief engineer wants to advertise the 600-hour cumulative MTBF as the demonstrated value, calling it conservative. Evaluate that position and explain what each metric predicts about fielded units.
Read the worked answer
The two metrics measure different things, and only one predicts the field.
Cumulative MTBF averages the entire history, including the early period when the design was immature. It lags the current design and is not a property any shipped unit possesses - no customer receives a product with the average reliability of a whole development program. Instantaneous MTBF, theta_c/(1 - b), estimates the failure rate of the current configuration, which is what fielded units will actually experience if the fixes are in the build.
The claim that 600 hours is conservative is backwards. It is not a conservative estimate of the current design's reliability; it is an estimate of a design that largely no longer exists. Advertising it creates two risks: the customer may reject a number that understates the product, or, worse, may later measure the fielded population against it and be surprised in either direction depending on which fixes actually reached production.
Legitimate use of cumulative MTBF: program tracking. It is stable and monotone, and it makes the trend visible to management and to the customer's reliability engineer, which is exactly what a growth curve is for.
Correct reporting: present both, clearly labelled - cumulative MTBF of 600 h as growth tracking, instantaneous MTBF of about 885 h as the current reliability estimate - and state the exact configuration each refers to. If the fixes for the last three failures are not in the current build, even the 885-hour figure describes a configuration that does not yet exist.
3. What corrections must be made before either number is quoted to the customer, given the three unvalidated fixes and the small number of failures?
Read the worked answer
Two corrections are needed before any number leaves the room.
Open fixes. The growth slope fitted from the raw failure times assumes every observed failure was followed by an effective corrective action incorporated into the test units. Three of the eight - at 2,700, 3,600 and 4,500 hours - have unvalidated fixes, so those failures do not represent improvement and the fitted slope is optimistic. The honest treatment is either to fit only the interval where fixes were complete, or to refit after the fixes are validated. The first option usually lowers the projected MTBF, which pushes the required test time beyond the 7,020 hours estimated earlier.
Statistical uncertainty. The estimate rests on eight failures. Treating the failure process as Poisson, a roughly 90% confidence interval on the current MTBF spans about 0.5x to 1.8x the point estimate - for 885 hours, roughly 460 to 1,600 hours. A point estimate of 977 hours at the milestone is not a demonstration of 1,000 hours; it is statistically indistinguishable from 700 or 1,300 hours.
What to do instead: declare the configuration, report the current estimate with its interval, and conduct a proper demonstration test at the milestone with a pre-agreed zero-failure or failure-terminated plan, so producer and consumer risk are fixed before the data are seen. Converting a growth trend into a contractual claim after the fact, on eight failures with three open fixes, is the kind of statement that a competent customer's reliability engineer will reject - and rightly.