Remote Telecommunications System Case Study for BlockSim 7

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BlockSim 7 Case Studies

Case Study RC5 - Remote Telecommunications System

Software Used:
BlockSim 7

Download Example Files (*.rbp)
Steps 1, 2 and 3 Example File
Step 4 Example File

A telecommunications system is to be constructed in an uninhabited stretch of jungle. The system consists of a transmitter and receiver with six relay stations to connect them.

The relays are situated so that the signal originating from one station can be picked up by the next two stations down the line. For example, a signal from the transmitter can be received by Relay 1 and Relay 2, a signal from Relay 1 can be received by Relay 2 and Relay 3 and so forth. Thus, this arrangement would require two consecutive relays to fail for the system to fail. (This is also known as a consecutive-k-out-of-n:F system.)

Step 1: Setup the Master RBD
Draw the RBD for this system.

Figure 1 [Enlarge]

Step 2 : Basic Reliability Analysis
Furthermore, consider the following:

The transmitter and receiver are made up of three subassemblies each, while the relay stations have two subassemblies each (all in series). Specifically:
- Subassembly SPS1 (solar power supply) is common to all.
- The transmitter has two additional subassemblies, TRC1 and TRC2.
- The receiver also has two additional subassemblies, RCR1 and RCR2.
- Relay stations have a subassembly RLYC1 in addition to SPS1.
Assume the failure distributions for each subassembly, as given in Table 1.

Table 1

table 1

Step 2 Tasks

2.1 Determine the reliability of the system after 1000 hrs.
2.2 Does the failure of any of the redundant relays (1 through 6) have the same effect on the system reliability? In other words, does it matter from a reliability perspective which specific relay fails?

Step 2 Solution

Subdiagram RBDs are used for each element (transmitter, relay and receiver) and the appropriate failure distributions are set at the subdiagram level. (See the actual subdiagram setup used...)

2.1 Once the model has been set up, the Reliability at 1,000 hrs is determined to be 97.675%. (See more details in computing this...)
2.2 Yes. Even though the relays are reliability-wise identical, their position within the diagram matters. The greatest impact on reliability is if 2 or 5 fails followed by 4 or 3 and then by 1 or 6. (See more details in computing this...)

Step 3: Introduce Basic Repairs/Maintenance
Further assume that the components given in Table 1 are line replaceable units and each repair involves the replacement of the existing component with a new component. (Note that if this were not the case, i.e. if we had assumed imperfect repair, then restoration factors could have been utilized).

Additionally, each component has a probabilistic repair (replacement) duration, as given in Table 2. For this step we will assume that repair begins immediately upon failure and that when the system is failed components do not age. We will not utilize (include) any other constraints.

Table 2

table 2

Step 3 Tasks

3.1 Determine the point availability of the system after 1 year of operation (8760 hrs).
3.2 Determine the average availability of the system after 1 year of operation.
3.3.Estimate the Mean Time to First Failure of the system.
3.4 Rank the components in terms of their Failure Criticality Index (RS FCI).
3.5 Estimate the expected number of spare parts required for each component.
3.6 What is the expected system downtime?

Step 3 Solution

To obtain the results of interest, discrete event simulation was utilized with 10,000 runs.

3.1 Point availability after 1 year of operation, A(t = 8760 hrs), is: 0.9986 (See more details...)
3.2 Average availability after 1 year of operation is 0.9993. (See more details...)
3.3.The Mean Time to First Failure of the system is 16397. (See more details...)
3.4 Table 3 shows that 99.9% of system failures were either due to the transmitter or receiver. From that, 54% were due to the transmitter and 20.5% of the transmitter failures were due to the SPS1 component. (See more details...)

Table 3: RS FCI by Subsystem and Component

Table 3: RS FCI by Subsystem and Component

3.5 The expected number of spare parts can be obtained by looking at the expected number of failures. Obviously, a spare part would be required for each failure. Table 4 shows these numbers. For SPS1, 0.9579 failures are expected. Another way to look at this is to say that there is a 96% chance that I will need a spare part for SPS1. Now the choice as to whether to keep spare parts around is based on additional economic and logistic information (e.g. How quickly can I get a part? How much does it cost me to keep it around? etc.) (See more details...)

Table 4: Expected Number Failures

Table 4: Expected Number Failures

3.6 The total system downtime is 6.16. (See more details...)

Step 4: Introduce Additional Constraints
To further analyze the system, let's introduce additional constraints/parameters to this analysis.

A subcontractor is available to repair the system. It takes on the average 36 hours (standard deviation of 6 ~ normal distribution) for the subcontractor to get a technician to the site and the subcontractor only has two technicians qualified to service this system. Also, the subcontractor keeps a single spare for SPC1 and RLYC1 on site. On-site spares (if in stock) are available immediately. Any other parts needed must be ordered and shipped.

Specifically, assume that:
A spare part that is ordered and shipped arrives based on a normal distribution with a mean of 72 hours and a standard deviation of 12.
One SPC1 and one RLYC1 are kept on site. When one is used, an additional spare is ordered.

Step 4 Tasks