727. The device consists of two nodes connected in series. The first node contains two elements: $A, B$, which are connected in parallel. The second node contains one element $C$. The time of failure-free operation of the elements is distributed according to the exponential law with parameters equal to 0.04, 0.05, and 0.10, respectively. Find by the Monte Carlo method: a) the estimate $P$ * of the probability of failure-free operation of the device for a duration of 10 hours; b) the mean time of failure-free operation of the device. Conduct 50 trials.
Solution. a) Let's simulate the time (in hours) of failure-free operation of the elements using the formulas:

$$
\begin{aligned}
& t_{A}=-(1 / 0.04) \ln r_{1}=25\left(-\ln r_{1}\right) \\
& t_{B}=-(1 / 0.05) \ln r_{2}=20\left(-\ln r_{2}\right) \\
& t_{C}=-(1 / 0.10) \ln r_{3}=10\left(-\ln r_{3}\right)
\end{aligned}
$$

where \( r_{1}, r_{2}, r_{3} \) are random numbers. For example, for the first test, let's take three random numbers from the table in Appendix 9: \( 0.10, 0.09, 0.73 \) and use them to simulate the failure-free operation time of the elements:

$$
\begin{aligned}
& t_{A}=25(-\ln 0.10)=25 \cdot 2.30=57.50 \\
& t_{B}=20(-\ln 0.09)=20 \cdot 2.41=48.20 \\
& t_{C}=10(-\ln 0.73)=10 \cdot 0.32=3.2 .
\end{aligned}
$$

Elements \( A \) and \( B \) of the first node are connected in parallel, so for the node to work, at least one element must be operational. Therefore, in the first test, the first node will work for \( \max (57.50 ; 48.20)=57.50 \) hours.

The first and second nodes are connected in series, so the device works if both nodes are operational simultaneously. Therefore, in the first test, the device will work for \( \min (57.50; 3.2)=3.2 \) hours.

| Homep  Honute  ens | ![](https://cdn.mathpix.com/cropped/2024_05_22_1d1b16277022f98e68adg-311.jpg?height=93&width=324&top_left_y=850&top_left_x=171) |  |  | Failure-free operation time |  |  |  |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | \( \bar{A} \) | \( \bar{B} \) | \( \bar{c} \) | \( \bar{A} \) | \( \bar{B} \) | \( \overline{\boldsymbol{c}} \) | First node | Second node | Device |
| \( \mathbf{1} \)  \( \mathbf{2} \)  \( \mathbf{3} \)  \( \mathbf{4} \) | 0.10  0.25  0.52  0.86 | 0.09  0.33  0.01  0.34 | 0.73  0.76  0.35  0.67 | 57.50  34.75  16.25  3.75 | \( \left\|\begin{array}{l}48.20 \\ 22.20 \\ 92.00 \\ 21.60\end{array}\right\| \) | 3.2  2.7  10.5  4.0 | 57.50  34.75  92.00  21.60 | 3.2  2.7  10.5  4.0 | 3.2  2.7  10.5  4.0 |

Table 58

In Table 58, the results of only four tests are presented. If 50 tests are conducted, it will be found that in 18 tests, the device worked for 10 hours (or more).

The estimated reliability of the device (the probability of its failure-free operation for a duration of 10 hours) is \( P^{*}=18 / 50=0.36 \).

For comparison, let's provide the analytical solution. The probabilities of failure-free operation of the elements:

$$
\begin{aligned}
& R_{A}(10)=\mathrm{e}^{-0.04 \cdot 10}=\mathrm{e}^{-0.4}=0.67 \\
& R_{B}(10)=\mathrm{e}^{-0.05 \cdot 10}=\mathrm{e}^{-0.5}=0.61 \\
& R_{C}(10)=\mathrm{e}^{-0.10 \cdot 10}=\mathrm{e}^{-1}=0.37
\end{aligned}
$$

The probability of failure-free operation of the first node for a duration of 10 hours

$$
P_{1}=1-(1-0.67)(1-0.61)=0.87
$$

The probability of failure-free operation of the device for a duration of 10 hours

$$
P=P_{1} \cdot R_{c}(10)=0.87 \cdot 0.37=0.32
$$

The absolute error \( \left|P-P^{*}\right|=|0.32-0.36|=0.04 \).

b) Let's find the mean time of failure-free operation of the device, considering that in 50 tests it worked without failure for a total of 450 hours: \( \overline{i^{*}}=450 / 50=9 \).

For comparison, let's provide the analytical solution. The mean time of operation of the elements: \( \bar{t}_{A}=1 / 0.04=25, \bar{t}_{B}=1 / 0.05=20, \bar{t}_{C}=1 / 0.10=10 \).

The mean time of operation of the nodes: \( \bar{t}_{1}=\max (25,20)=25, \bar{t}_{2}=10 \).

The mean time of operation of the device: \( \bar{t}=\min (25,10)=10 \).

The absolute error \( \left|\bar{i}-\bar{t}^{*}\right|=10-9=1 \).