Let's work through the two questions based on the diagram you've provided.

Question 1a: System Reliability for Identical Components

• The diagram in Question 1 represents a reliability block diagram where each block has a reliability of $R$.
• There are two parallel paths from the start to the end, each path containing two components in series.

To calculate the system reliability:

1. For each series path:

   • In a series system, the total reliability $R_{\text{series}}$ is the product of the individual reliabilities.
   • For a series of two components, $R_{\text{series}} = R \times R = R^2$.

2. For the overall system (parallel paths):

   • In a parallel system, the system reliability $R_{\text{system}}$ is given by: $R_{\text{system}} = 1 - \left( (1 - R_{\text{series}}) \times (1 - R_{\text{series}}) \right)$
   • Substituting $R_{\text{series}} = R^2$: $R_{\text{system}} = 1 - (1 - R^2)^2$

Question 1b: System Reliability when $R = 0.9$

• Now, substitute $R = 0.9$ into the formula: $R_{\text{series}} = 0.9^2 = 0.81$ $R_{\text{system}} = 1 - (1 - 0.81)^2 = 1 - (0.19)^2 = 1 - 0.0361 = 0.9639$ So, the system reliability when $R = 0.9$ is 0.9639 or approximately 96.39%.

Question 2: System Reliability for Given Component Reliabilities

• The diagram in Question 2 consists of a combination of series and parallel components. We need to follow similar steps as in Question 1, computing the reliability of each section.

1. Top path (parallel block):

   • There are two parallel components, with reliabilities $0.90$ and $0.95$.
   • Parallel reliability formula: $R_{\text{parallel}} = 1 - ( (1 - 0.90) \times (1 - 0.95) )$ $R_{\text{parallel}} = 1 - (0.10 \times 0.05) = 1 - 0.005 = 0.995$

2. Middle path (parallel):

   • The two parallel components have reliabilities $0.75$ and $0.85$.
   • Applying the parallel formula: $R_{\text{parallel}} = 1 - ( (1 - 0.75) \times (1 - 0.85) )$ $R_{\text{parallel}} = 1 - (0.25 \times 0.15) = 1 - 0.0375 = 0.9625$

3. Bottom path (series):

   • There is one series component with reliability $0.93$.

4. Combine the system (series connections):

   • Now that we have the parallel results from the top and middle paths, and the series component from the bottom, we multiply these together since they are in series: $R_{\text{system}} = R_{\text{top}} \times R_{\text{middle}} \times R_{\text{bottom}}$ $R_{\text{system}} = 0.995 \times 0.9625 \times 0.93$ Let's calculate this:

I'll compute it for you now.The system reliability for Question 2 is approximately 0.8906 or 89.06%.

Would you like further details or have any questions?

Here are five related questions to deepen understanding:

1. How does a parallel system improve overall reliability compared to a series system?
2. If a component in the system fails, how does the configuration impact overall failure?
3. What would happen to the system reliability in Question 1 if the number of parallel paths increases?
4. How would you calculate the system reliability if more than two components are in series within one path?
5. Can you apply these concepts to a practical real-world example, such as the reliability of a computer network?

Tip: In parallel systems, adding more components typically increases the overall reliability, since the system can still function if some components fail.