SEQUENTIAL TEST PLANS

Sequential test plans can also be used for variables demonstration tests. The sequential test leads to a shorter average number of part hours of test exposure if the population MTBF is near _s , _d (i.e., close to the specified or design MTBF).

EXPONENTIAL DISTRIBUTION SEQUENTIAL TEST PLAN

This test plan can be used when:

1. The demonstration test is based upon time-to-failure data.

2. The underlying probability distribution is exponential.

The method to be used for the exponential distribution is to:

1. Identify _s , _d , ± , and ²

2. Calculate accept/reject decision points

   

Evaluate the following expression for the exponential distribution:

where t _i = time to failure of the ith unit tested and n = number tested .

If L > , the test is failed.

If L < , the test is passed.

If , the test should be continued . -a a

A graphical solution can also be used by plotting decision lines:

nb - h ₁ and nb + h ₂

where n = number tested; .

Let t _i equal time to failure for the ith item. Make conclusions based on the following:

If < nb “ h ₁ , the test has failed.

If ‰ nb + h ₂ , the test is passed.

If nb “ h ₁ ‰ < nb + h ₂ , continue the test.

Assume you are interested in testing a new product to see whether it meets a specified MTBF of 500 hours with a consumer's risk of 0.10. Further, specify a design MTBF of 1000 hours for a producer's risk of 0.05. Run tests to determine whether the product meets the criteria.

Determine D based on the known criteria:

Then calculate

Now solve for b

Using these results, we can determine at which points we can make a decision, by using the following:

If < nb - h ₁ , 693n - 2890, the test has failed.

If ‰ nb + h ₂ , 693n + 2251, the test is passed.

 

If nb - h ₁ ‰ < nb + h ₂ , 693n - 2890 ‰ < 693n + 2251, continue the test.

WEIBULL AND NORMAL DISTRIBUTIONS

Sequential test methods have also been developed for the Weibull distribution and for the normal distribution. If you are interested in pursuing the sequential tests for either of these distributions, see the selected bibliography

INTERFERENCE (TAIL) TESTING

Interference demonstration testing can sometimes be used when the stress and strength distributions are accurately known. If a random sample of the population is obtained, it can be tested at a point stress that corresponds to a specific percentile of the stress distribution. By knowing the stress and strength distributions, the required reliability, the desired confidence level, and the number of allowable failures, it is possible to determine the sample size required.

RELIABILITY VISION

Reliability is valued by the organization and is a primary consideration in all decision making. Reliability techniques and disciplines are integrated into system and component planning, design, development, manufacturing, supply, delivery, and service processes. The reliability process is tailored to fit individual business unit requirements and is based on common concepts that are focused on producing reliable products and systems, not just components .

RELIABILITY BLOCK DIAGRAMS

Reliability block diagrams are used to break down a system into smaller elements and to show their relationship from a reliability perspective. There are three types of reliability block diagrams: series, parallel, and complex (combination of series and parallel).

1. A typical series block diagram is shown in Figure 7.2 with each of the three components having R1, R2, and R3 reliability respectively.

   
   Figure 7.2: A series block diagram.

   The system reliability for the series is

   R _total = (R ₁ ) (R ₂ ) (R ₃ ) ... (Rn)

If the reliability for R1 = .80, R2 = .99, and R3 = .99, the system reliability is: R _total = (.80)(.99)(.99) = .78. Please notice that the total reliability is no more than the weakest component in the system. In this case, the total reliability is less than R1.

2. A parallel reliability block diagram shows a system that has built-in redundancy. A typical parallel system is shown in Figure 7.3. The system reliability is

   
   Figure 7.3: A parallel reliability block diagram.

   R _total = 1 - [1 - R ₁ (t) (1 - R ₂ (t) (1 - R ₃ )(t) ... (1 - R _n (t)]

 

If the reliability for R1 = .80, R2 = .90, and R3 = .99, the system reliability is: R _total = 1 - [(1 - .80)(1 - .90)(1 - .99)] = .9998 Please notice that the total reliability is more than that of the strongest component in the system. In this case, the total reliability is more than the R3.

3. Complex reliability block diagrams show systems that combine both series and parallel situations. A typical complex system is shown in Figure 7.4. The system reliability for this system is calculated in two steps:

   • Step 1. Calculate the parallel reliability.

   • Step 2. Calculate the series reliability ” which becomes the total reliability.

   
   Figure 7.4: A complex reliability block diagram.

 

If the reliability for R1 = .80, R2 = .90, R3 = .95, R4 = .98, and R5 = .99, what is the total reliability for the system?

• Step 1. The parallel reliability for R3 and R4 is

  • R _total = l - [1 - R ₁ (t) (1 - R ₂ (t) = 1 - [1 - .95) (1 - .98)] = .999

• Step 2. The series reliability for R1, R2, (R3 & R4), and R5 is

  • R _total = (R ₁ ) (R ₂ ) (R ₃ & R ₄ ) (R ₅ ) = (.80)(.90)(.999)(.99) = .712

Please notice that the parallel reliability was actually converted into a single reliability and that is why it is used in the series as a single value.

 

WEIBULL DISTRIBUTION ” INSTRUCTIONS FOR PLOTTING AND ANALYZING FAILURE DATA ON A WEIBULL PROBABILITY CHART

This technique is useful for analyzing test data and graphically displaying it on Weibull probability paper. The technique provides a means to estimate the percent failed at specific life characteristics together with the shape of the failure distribution. The following procedure presents a manual method of conducting the analysis, but many computer programs can do the same calculations and also plot the Weibull curve. Weibull analysis is one of the simpler analytical methods, but it is also one of the most beneficial. The technique can be utilized for other than just analyzing failure data. It can be used for comparing two or more sets of data such as different designs, materials, or processes. Following are the steps for conducting a Weibull analysis.

1. Gather the failure data (it can be in miles, hours, cycles, number of parts produced on a machine, etc.), then list in ascending order. For example: We conduct an experiment and the following failures (sample size of 10 failures) are identified (actual hours to failure): 95, 110, 140, 165, 190, 205, 215, 265, 275, and 330.

2. Using the table of median ranks (Table 7.2), find the column corresponding to the number of failures in the sample tested. In our example we have a sample size of ten, so we use the "sample size 10" column. The "% Median Ranks" are then read directly from the table.

3. Match the hours (or some other failure characteristic that is measured) with the median ranks from the sample size selected. For example:

   Actual Hours to Failure	% Median Ranks
   95	6.7
   110	16.2
   140	25.9	Sample size of 10 failures
   165	35.5
   190	45.2
   205	54.8
   215	64.5
   265	74.1
   275	83.8
   330	93.3

4. In constructing the Weibull plot, label the "Life" on the horizontal log scale on the Weibull graph in the units in which the data were measured. Try to center the life data close to the center of the horizontal scale (Figure 7.5).

   
   Figure 7.5: The Weibull distribution for the example.

5. Plot each pair of "actual hours to failure" (on the horizontal logarithmic scale) and "% median rank" (on the vertical axis, which is a log-log scale) on the graph. The matching points are shown as dots (" s") on Figure 7.5. Draw a "line of best fit" ( generally a straight line) as close to the data pairs as possible. Half the data points should be on one side of the line, and the other half should be on the other side. No two people will generate the exact same line, but analysts should keep in mind that this is a visual estimate. (If the line is computer generated, it is actually calculated based on the "best fit" regression line.)

6. After the line of "best fit" is drawn, the life at a specific point can be found be going vertically to the "Weibull line" then going horizontally to the "Cumulative % Failed." In other words, this is the percent that is expected to fail at the life that was selected. In the example, 100 was selected as the life, then going up to the line and then across, we can see the expected % failed to be 10%. In this case, the life at 100 hours is also known as the B ₁₀ life (or 90% reliability) and is the value at which we would expect 10% of the parts to fail when tested under similar conditions. (Please note that there is nothing secret about the B ₁₀ life. Any B _x life can be identified. It just happens that the B ₁₀ is the conventional life that most engineers are accustomed to using.) In addition, we can plot the 5% and the 95% confidence using Tables 7.3 and 7.4 respectively.

   Table 7.3: Five Percent Rank Table

   	Sample Size (n)
   1	1	2	3	4	5	6	7	8	9	10
   1	5.000	2.532	1.695	1.274	1.021	0.851	0.730	0.639	0.568	0.512
   2		22.361	13.535	9.761	7.644	6.285	5.337	4.639	4.102	3.677
   3			36.840	24.860	18.925	15.316	12.876	11.111	9.775	8.726
   4				47.237	34.259	27.134	22.532	19.290	16.875	15.003
   5					54.928	41.820	34.126	28.924	25.137	22.244
   6						60.696	47.820	40.031	34.494	30.354
   7							65.184	52.932	45.036	39.338
   8								68.766	57.086	49.310
   9									71.687	60584
   10										74.113

   	Sample Size (n)
   j	11	12	13	14	15	16	17	18	19	20
   1	0.465	0.426	0.394	0.366	0.341	0.320	0.301	0.285	0.270	0.256
   2	3.332	3.046	2.805	2.600	2.423	2.268	2.132	2.011	1.903	1.806
   3	7.882	7.187	6.605	6.110	5.685	5.315	4.990	4.702	4.446	4.217
   4	13.507	12.285	11.267	10.405	9.666	9.025	8.464	7.969	7.529	7.135
   5	19.958	18.102	16.566	15.272	14.166	13.211	12.377	11.643	10.991	10.408
   6	27.125	24.530	22.395	20.607	19.086	17.777	16.636	15.634	14.747	13.955
   7	34.981	31.524	28.705	26.358	24.373	22.669	21.191	19.895	18.750	17.731
   8	43.563	39.086	35.480	32.503	29.999	27.860	26.011	24.396	22.972	21.707
   9	52.991	47.267	42.738	39.041	35.956	33.337	31.083	29.120	27.395	25.865
   10	63.564	56.189	50.535	45.999	42.256	39.101	36.401	34.060	32.009	30.195
   11	76.160	66.132	58.990	53.434	48.925	45.165	41.970	39.215	36.811	34.693
   12		77.908	68.366	61.461	56.022	51.560	47.808	44.595	41.806	39358
   13			79.418	70.327	63.656	58.343	53.945	50.217	47.003	44.197
   14				80.736	72.060	65.617	60.436	56.112	52.420	49.218
   15					81.896	73.604	67.381	62.332	58.088	54.442
   16						82.925	74.988	68.974	64.057	59.897
   17							83.843	76.234	70.420	65.634
   18								84.668	77.363	71.738
   19									85.413	78.389
   20										86.089

   Table 7.4: Ninety-five Percent Rank Table

   	Sample Size (n)
   j	1	2	3	4	5	6	7	8	9	10
   1	95.000	77.639	63.160	52.713	45.072	39.304	34.816	31.234	28.313	25.887
   2		97.468	86.465	75.139	65.741	58.180	52.070	47.068	42.914	39.416
   3			98.305	90.239	81.075	72.866	65.874	59.969	54.964	50.690
   4				98.726	92.356	84.684	77.468	71.076	65.506	60.662
   5					98.979	93.715	87.124	80.710	74.863	69.646
   6						99.149	94.662	88.889	83.125	77.756
   7							99.270	95.361	90.225	84.997
   8								99.361	95.898	91.274
   9									99.432	96.323
   10										99.488

   	Sample Size (n)
   j	11	12	13	14	15	16	17	18	19	20
   1	23.840	22.092	20.582	19.264	18.104	17.075	16.157	15.332	14.587	13.911
   2	36.436	33.868	31.634	29.673	27.940	26.396	25.012	23.766	22.637	21.611
   3	47.009	43.811	41.010	38.539	36.344	34.383	32.619	31.026	29.580	28.262
   4	56.437	52.733	49.465	46566	43.978	41.657	39.564	37.668	35.943	34366
   5	65.019	60.914	57.262	54.000	51.075	48.440	46.055	43.888	41.912	40.103
   6	72.875	68.476	64.520	60.928	57.744	54.835	52.192	49.783	47.580	45.558
   7	80.042	75.470	71.295	67.497	64.043	60.899	58.029	55.404	52.997	50.782
   8	86.492	81.898	77.604	73.641	70.001	66.663	63.599	60.784	58.194	55.803
   9	92.118	87.715	83.434	79.393	75.627	72.140	68.917	65.940	63.188	60.641
   10	96.668	92.813	88.733	84.728	80.913	77.331	73.989	70.880	67.991	65.307
   11	99.535	96.954	93.395	89.595	85.834	82.223	78.809	75.604	72.605	69.805
   12		99.573	97.195	93.890	90.334	86.789	83.364	80.105	77.028	74.135
   13			99.606	97.400	94.315	90.975	87.623	84.366	81.250	78.293
   14				99.634	97.577	94.685	91.535	88.357	85.253	82.269
   15					99.659	97.732	95.010	92.030	89.009	86.045
   16						99.680	97.868	95.297	92.471	89.592
   17							99.699	97.989	95.553	92.865
   18								99.715	98.097	95.783
   19									99.730	98.193
   20										99.744

   The confidence lines are drawn for our example in Figure 7.5. The reader will notice that the confidence lines are not straight. That is because as we move in the fringes of the reliability we are less confident about the results.

7. The graph can be used for estimating the cumulative % failure at a specified life, or it can be used for determining the estimated life at a cumulative % failure. In the example, we would expect 63.2% of the test units to fail at 222 hours. This value at 63.2% is also known as the characteristic life or the mean time between failures (MTBF) for the example distribution. Or looking at the chart another way, we would like to estimate the failure hours at a specified % failure. For example at 95% cumulative % failed, the hours to failure are 325 hours. Once the Weibull plot is determined, an analyst can go either way.

8. The Weibull graph can also be used to estimate the reliability at a given life, using the equation of R(t) = 1 - F(t). A designer who wishes to estimate the reliability of life at 200 hours would go vertically to the Weibull line, then go horizontally to 52%, which is the percent expected to fail. The estimated reliability at 200 hours would be 1 - 0.52 = 0.48 or 48%. At 80 hours it would be 1 - 0.056 = 0.944 or 94.4%. The slope is obtained by drawing a line parallel to the Weibull line on the Weibull slope scale that is in the upper left corner of the chart.

9. If a computer program is used, the calculation for the line of best fit is determined by the computer. Some programs draw the graph and show the paired points, the line of best fit (using the least squares method or the maximum likelihood method), the reliability at a specified hour (or other designated parameter), and the slope of the line.

10. One of the interesting observations regarding the Weibull graph is the interpretations that can be made about the distribution by the portrayal of the slope. When the slope is:

    • Less than 1, this indicates a decreasing failure rate, early life, or infant mortality

    • Approximately 1, the distribution indicates a nearly constant failure rate (useful life or a multitude of random failures)

    • Exactly 1, the distribution has an exponential pattern

    • Greater than 1, the start of wear out

    • Approximately 3.55, a normal distribution pattern,

11. Weibull plots can be made if test data also include test samples that have not failed. Parts that have not failed (for whatever reason during the testing) can be included in the calculations together with the failed parts or assemblies. The non-failed data are referred to as suspended items. The method of determining the Weibull plot is shown in the next set of instructions.

INSTRUCTIONS FOR PLOTTING FAILURE AND SUSPENDED ITEMS DATA ON A WEIBULL PROBABILITY CHART

1. Gather the failure and suspended items data, then including the suspended items, list in ascending order.

   Item Number	Hours to Failure or Suspension	Failure or Suspension Code [a]
   1	95	F1
   2	110	F2
   3	140	F3
   4	165	F4	Sample Size 13 10 failures 3 suspensions
   5	185	S1
   6	190	F5
   7	205	F6
   8	210	S2
   9	215	F7
   10	265	F8
   11	275	F9
   12	330	F10
   13	350	S3
   [a] Code items as failed (F) or suspended (S).

2. Calculate the mean order number of each failed unit. The mean order numbers before the first suspended item are the respective item numbers in the order of occurrence, i.e., 1, 2, 3, and 4. The mean order numbers after the suspended items are calculated by the following equations.

   Mean order number = (previous mean order number) + (new number)

   where, new increment =

   

   and N = total sample size.

   For example, to compensate for S1 (first suspended item), new increment = [(13 + 1) -4]/(1 + 8) = 1.111 and the mean order number of F5 (fifth failed item) = 4 + 1.111 = 5.111.

   Note  

   Only one new increment is found each time a suspended item is encountered . Mean order number of F6 = 5.111 + 1.111 = 6.222.

   New increment for mean order number of F7 = [(13 + 1) - 6.222] (1 + 5) = 1.296.

   Then, the mean order number of F7 (seventh failed item) is 6.222 + 1.296 = 7.518 (and so on for F8, F9, and F10).

   This new increment also applies to mean order numbers:

   Item Number	Hours to Failure or Suspension	Failure or Suspension Code	Mean Order Number
   1	95	F1	1
   2	110	F2	2
   3	140	F3	3
   4	165	F4	4
   5	185	S1	”
   6	190	F5	5.111
   7	205	F6	6.222
   8	210	S2	”
   9	215	F7	7.518
   10	265	F8	8.815
   11	275	F9	10.111
   12	330	F10	11.407
   13	350	S3	”

3. A rough check on the calculations can be made by adding the last increment to the final mean order number. If the value is close to the total sample size, the numbers are correct. In our example, 11.407 + [11.407 - 10.111] = 11.407 + 1.296 = 12.702, which is a close approximation to the sample size of 13.

4. Using the table of median ranks for a sample size of 13 we can determine the median rank for the first four failures, or we can use the approximate median rank formula.

   Median rank = [J - .3]/[N + .4]

   where J = mean order number and N = total sample size.

   For example, the median rank of F5 is:

   = 0.359

   and, the remainder of the failures:

   = 0.442

   and so on.

   Item Number	Hours to Failure or Suspension	Failure or Suspension Code	Mean Order Number	% Median Rank
   1	95	F1	1	5.2
   2	110	F2	2	12.6
   3	140	F3	3	20.0
   4	165	F4	4	27.5
   5	185	S1	”	”
   6	190	F5	5.111	35.9
   7	205	F6	6.222	44.2
   8	210	S2	”	”
   9	215	F7	7.518	53.9
   10	265	F8	8.815	63.5
   11	275	F9	10.111	73.2
   12	330	F10	11.407	82.9
   13	350	S3	”	”

5. Label the "Life" on the horizontal log scale on the Weibull graph in the units in which the data were measured. Try to center the life data close to the center of the horizontal scale.

6. Plot each pair of "actual hours to failure" (on the horizontal scale) and "% median rank" (on the vertical scale) on the graph. Draw a "line of best fit" (generally a straight line) as close to the data pair as possible. Half the data points should be on one side of the line, and the other half should be on the other side.

7. Once the line is drawn, the life at a specific point can be found by going vertically to the "Weibull line" then going horizontally to the "Cumulative % failed." In other words, this is the percent that is expected to fail at the life that was selected. In the example, 200 hours was selected as the life, then going up to the line and then across, we can see the expected % failed to be 40%.

8. Other reliability parameters that can be read from the Weibull plot are:

   MTBF	=	240 hours
   B 10	=	105 hours
   B	=	2.5

   Reliability at 100 hours is 1 - 0.09 = 0.91 reading from the graph, or using the Weibull equation

   

9. Comparing the two examples shows that the analysis with suspended items results in a slightly higher reliability characteristics. This is using the same failure data plus the three suspended items.

ADDITIONAL NOTES ON THE USE OF THE WEIBULL

1. Weibull plotting is an invaluable tool for analyzing life data; however, some precautions should be taken. Goodness-of-fit is one concern. This can be tested with various tests such as the Kolmogorov-Smirnov or Chisquare. The use of an adequate sample size is another concern. Generally a sample size should be greater than ten, but if the failure rate is in a tight pattern (with relatively low variability), this generality may be relaxed . Be suspicious of a curved line that best fits the data. This may indicate a mixed sample of failures or inappropriate sampling.

2. If the Weibull plot is made and a curvilinear relation develops for the connecting points, it usually indicates that two or more distributions are making up the data. This may be due to infant mortality failures being mixed with the data, failures due to components from two different machines or assembly operations, or some other underlying cause. If a curved relationship is indicated, the analyst should revisit the data and try to determine if the data are made up of two or more distributions and then manage each distribution separately.

3. There is another parameter in the Weibull analysis that was not discussed. Beside the shape or slope (b) of the Weibull line and the scale or characteristic life (the mean life or MTBF at the 63.2% cumulative percentage), there is the "location parameter." In most cases it is usually zero and should be of little concern. In effect, it states that the distribution of failure times starts at zero time, which is more often the case because it is difficult to imagine otherwise . The characteristic life splits the distribution in two areas of 0.632 before and 0.368 ( R ( ) = = e ^-1 = .368) after.

4. One of the advantages of using the Weibull is that it is very flexible in its interpretations. A wealth of information can be derived from it. If the Weibull slope is equal to one, the distribution is the same as the exponential, or a constant failure rate. If the slope is in the vicinity of 3.5, it is a "near normal distribution." If the slope is greater than one, the plot starts to represent a wear out distribution, or an increasing hazard rate. A slope less than one generally indicates a decreasing hazard rate, or an infant mortality distribution.

5. Analysts should be careful about extrapolating beyond the data when making predictions . Remember that the failure points fall within certain bounds and that the analyst should have a valid reason when venturing beyond these bounds. When making projections over and above these confines, sound engineering judgment, statistical theory, and experience should all be taken into consideration.

6. The three-parameter Weibull is a distribution with non-zero minimum life. This means that the population of products goes for an initial period of time without failure. The reliability function for the three-parameter Weibull is given by

   R(t) = , t ‰

   where t = time to failure (t ‰ ); = minimum life parameter ( ‰ 0); ² = Weibull slope ( ² > 0); and = characteristic life ( ‰ ).

   For a given reliability

   t = +( - ) —

   and the B ₁₀ life is

   B ₁₀ = +( - ) —