A transportation system for carrying ore from stock pile to railway siding through an ore bin is shown. The time between failure of each conveyor belt follows an exponential distri... A transportation system for carrying ore from stock pile to railway siding through an ore bin is shown. The time between failure of each conveyor belt follows an exponential distribution with mean time between failure of 700 hours. The system is considered to be a ‘success’ if ore transports from stock pile to siding by any combination of belts. The reliability of the system for 350 hours of continuous successful operation is ________ (round off up to 2 decimals). Read more

Steps to Solve

1. Understanding Exponential Distribution The time between failures (T) for each conveyor belt follows an exponential distribution with a mean ($\mu$) of 700 hours. The reliability function for an exponential distribution is given by:
   $$ R(t) = e^{-\frac{t}{\mu}} $$
   Where $t$ is the time of interest (350 hours).

2. Calculate Reliability for Each Belt Using the reliability function, we can compute the reliability for a single belt operating for 350 hours:
   $$ R(350) = e^{-\frac{350}{700}} $$
   This simplifies to:
   $$ R(350) = e^{-0.5} $$

3. Compute Reliability Value Now, calculate the actual value:
   Using a calculator, we find:
   $$ R(350) \approx 0.6065 $$

4. Reliability of the System with Multiple Belts In this case, we have four belts. For the system to function, at least one belt needs to be operational. Therefore, the system reliability is given by the formula:
   $$ R_{system}(t) = 1 - (1 - R_b)^n $$
   Where $n$ is the number of belts and $R_b$ is the reliability of a single belt.

5. Substituting Values Here, $n = 4$ and $R_b \approx 0.6065$:
   $$ R_{system}(350) = 1 - (1 - 0.6065)^4 $$

6. Calculate System Reliability Now, compute the system reliability:
   $$ R_{system}(350) = 1 - (0.3935)^4 $$
   Calculate $(0.3935)^4 \approx 0.0254$, so:
   $$ R_{system}(350) \approx 1 - 0.0254 = 0.9746 $$

7. Round to Two Decimal Places Finally, round the result to two decimal places: $$ R_{system}(350) \approx 0.97 $$