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| ====Example 9====
| | #REDIRECT [[ReliaSoft Examples]] |
| A group of 20 units are put on a life test with the following results.
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| <center><math>\begin{matrix}
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| Number & State & State \\
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| in State & (F or S) & End Time \\
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| 3 & F & 9 \\
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| 1 & S & 9 \\
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| 1 & F & 11 \\
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| 1 & S & 12 \\
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| 1 & F & 13 \\
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| 1 & S & 13 \\
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| 1 & S & 15 \\
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| 1 & F & 17 \\
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| 1 & F & 21 \\
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| 1 & S & 22 \\
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| 1 & S & 24 \\
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| 1 & S & 26 \\
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| 1 & F & 28 \\
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| 1 & F & 30 \\
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| 1 & S & 32 \\
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| 2 & S & 35 \\
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| 1 & S & 39 \\
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| 1 & S & 41 \\
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| \end{matrix}</math></center>
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| Use the Kaplan-Meier estimator to determine the reliability estimates for each failure time.
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| =====Solution to Example 9=====
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| Using the data and Eqn. (kapmeier), the following table can be constructed:
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| <center><math>\begin{matrix}
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| State & Number of & Number of & Available & {} & {} \\
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| End Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & \tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{i}}} & \mathop{}_{}^{}\tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{i}}} \\
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| 9 & 3 & 1 & 20 & 0.850 & 0.850 \\
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| 11 & 1 & 0 & 16 & 0.938 & 0.797 \\
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| 12 & 0 & 1 & 15 & 1.000 & 0.797 \\
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| 13 & 1 & 1 & 14 & 0.929 & 0.740 \\
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| 15 & 0 & 1 & 12 & 1.000 & 0.740 \\
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| 17 & 1 & 0 & 11 & 0.909 & 0.673 \\
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| 21 & 1 & 0 & 10 & 0.900 & 0.605 \\
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| 22 & 0 & 1 & 9 & 1.000 & 0.605 \\
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| 24 & 0 & 1 & 8 & 1.000 & 0.605 \\
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| 26 & 0 & 1 & 7 & 1.000 & 0.605 \\
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| 28 & 1 & 0 & 6 & 0.833 & 0.505 \\
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| 30 & 1 & 0 & 5 & 0.800 & 0.404 \\
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| 32 & 0 & 1 & 4 & 1.000 & 0.404 \\
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| 35 & 0 & 1 & 3 & 1.000 & 0.404 \\
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| 39 & 0 & 1 & 2 & 1.000 & 0.404 \\
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| 41 & 0 & 1 & 1 & 1.000 & 0.404 \\
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| \end{matrix}</math></center>
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| As can be determined from the preceding table, the reliability estimates for the failure times are:
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| <center><math>\begin{matrix}
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| Failure Time & Reliability Est. \\
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| 9 & 85.0% \\
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| 11 & 79.7% \\
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| 13 & 74.0% \\
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| 17 & 67.3% \\
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| 21 & 60.5% \\
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| 28 & 50.5% \\
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| 30 & 40.4% \\
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| \end{matrix}</math></center>
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| ====Example 10====
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| A group of 55 units are put on a life test during which the units are evaluated every 50 hours, with the following results:
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| <center><math>\begin{matrix}
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| Start & End & Number of & Number of \\
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| Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} \\
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| 0 & 50 & 2 & 4 \\
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| 50 & 100 & 0 & 5 \\
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| 100 & 150 & 2 & 2 \\
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| 150 & 200 & 3 & 5 \\
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| 200 & 250 & 2 & 1 \\
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| 250 & 300 & 1 & 2 \\
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| 300 & 350 & 2 & 1 \\
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| 350 & 400 & 3 & 3 \\
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| 400 & 450 & 3 & 4 \\
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| 450 & 500 & 1 & 2 \\
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| 500 & 550 & 2 & 1 \\
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| 550 & 600 & 1 & 0 \\
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| 600 & 650 & 2 & 1 \\
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| \end{matrix}</math></center>
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| =====Solution to Example 10=====
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| The reliability estimates for the simple actuarial method can be obtained by expanding the data table to include terms used in calculation of the reliability estimates for Eqn. (simpact):
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| <center><math>\begin{matrix}
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| Start & End & Number of & Number of & Available & {} & {} \\
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| Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & 1-\tfrac{{{r}_{j}}}{{{n}_{j}}} & \mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{{{n}_{j}}} \\
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| 0 & 50 & 2 & 4 & 55 & 0.964 & 0.964 \\
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| 50 & 100 & 0 & 5 & 49 & 1.000 & 0.964 \\
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| 100 & 150 & 2 & 2 & 44 & 0.955 & 0.920 \\
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| 150 & 200 & 3 & 5 & 40 & 0.925 & 0.851 \\
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| 200 & 250 & 2 & 1 & 32 & 0.938 & 0.798 \\
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| 250 & 300 & 1 & 2 & 29 & 0.966 & 0.770 \\
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| 300 & 350 & 2 & 1 & 26 & 0.923 & 0.711 \\
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| 350 & 400 & 3 & 3 & 23 & 0.870 & 0.618 \\
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| 400 & 450 & 3 & 4 & 17 & 0.824 & 0.509 \\
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| 450 & 500 & 1 & 2 & 10 & 0.900 & 0.458 \\
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| 500 & 550 & 2 & 1 & 7 & 0.714 & 0.327 \\
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| 550 & 600 & 1 & 0 & 4 & 0.750 & 0.245 \\
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| 600 & 650 & 2 & 1 & 3 & 0.333 & 0.082 \\
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| \end{matrix}</math></center>
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| As can be determined from the preceding table, the reliability estimates for the failure times are:
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| <center><math>\begin{matrix}
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| Failure Period & Reliability \\
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| End Time & Estimate \\
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| 50 & 96.4% \\
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| 150 & 92.0% \\
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| 200 & 85.1% \\
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| 250 & 79.8% \\
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| 300 & 77.0% \\
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| 350 & 71.1% \\
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| 400 & 61.8% \\
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| 450 & 50.9% \\
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| 500 & 45.8% \\
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| 550 & 32.7% \\
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| 600 & 24.5% \\
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| 650 & 8.2% \\
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| \end{matrix}</math></center>====Example 11====
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| Find reliability estimates for the data in Example 10 using the standard actuarial method.
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| =====Solution to Example 11=====
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| The solution to this example is similar to that of Example 10, with the exception of the inclusion of the <math>n_{i}^{\prime }</math> term, which is used in Eqn. (standact). Applying this equation to the data, we can generate the following table:
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| <center><math>\begin{matrix}
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| Start & End & Number of & Number of & Adjusted & {} & {} \\
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| Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, n_{i}^{\prime } & 1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} & \mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} \\
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| 0 & 50 & 2 & 4 & 53 & 0.962 & 0.962 \\
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| 50 & 100 & 0 & 5 & 46.5 & 1.000 & 0.962 \\
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| 100 & 150 & 2 & 2 & 43 & 0.953 & 0.918 \\
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| 150 & 200 & 3 & 5 & 37.5 & 0.920 & 0.844 \\
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| 200 & 250 & 2 & 1 & 31.5 & 0.937 & 0.791 \\
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| 250 & 300 & 1 & 2 & 28 & 0.964 & 0.762 \\
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| 300 & 350 & 2 & 1 & 25.5 & 0.922 & 0.702 \\
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| 350 & 400 & 3 & 3 & 21.5 & 0.860 & 0.604 \\
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| 400 & 450 & 3 & 4 & 15 & 0.800 & 0.484 \\
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| 450 & 500 & 1 & 2 & 9 & 0.889 & 0.430 \\
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| 500 & 550 & 2 & 1 & 6.5 & 0.692 & 0.298 \\
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| 550 & 600 & 1 & 0 & 4 & 0.750 & 0.223 \\
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| 600 & 650 & 2 & 1 & 2.5 & 0.200 & 0.045 \\
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| \end{matrix}</math></center>
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| As can be determined from the preceding table, the reliability estimates for the failure times are:
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