Step-by-Step Examples

To better understand each of the previously mentioned options, three step-by-step examples are presented in this section. Each simple example is intended to illustrate the usage and assumptions/rules used in each option. Furthermore, in addition to the examples, a detailed list of rules (and/or assumptions) made are provided in the next section.

Example Using the Continue Simulation Option

To illustrate the Continue Simulation option, consider the deterministic scenario represented in Figures 11.5 and 11.6.

Figure 11.5 RBDs associated with each phase for the example illustrating the Continue Simulation option.

Figure 11.6 The phase diagram associated with each phase for the example illustrating the Continue Simulation option.

Table 1. The properties/parameters table associated with the example illustrating the Continue Simulation option. Note that all times are in hours.

In this example, a system has two components: Block A and Block B. The system undertakes a mission that can be divided into four phases. The first phase is an operational phase of duration 1000 hrs with both the components in series. In this phase Block A fails every 550 hrs while Block B fails every 750 hrs. Corrective maintenance actions on each of the components in this phase require 50 hrs to be completed. The second phase of the mission is also an operational phase in which the two components function in a parallel configuration. In this phase Block A fails every 450 hrs while Block B fails every 800 hrs. No maintenance actions can be performed on the components in this phase. The duration of the second phase is 700 hrs. The third phase for the system is a maintenance phase where corrective maintenance is performed on the failed components and preventive maintenance is performed on the non-failed components. The duration of the corrective maintenance is 100 hrs while the duration for the preventive maintenance is 20 hrs. The fourth phase of the mission is an operational phase which is identical to the second phase, but has a duration of 200 hrs.

All maintenance actions during the entire mission of the system have a Type II restoration factor of 1. All operational phases have the Continue Simulation option selected for the On System Failure property.

The system behavior from 0 to 2000 hrs is shown in Figure 11.7 and described next.

Figure 11.7 System behavior for Continue Simulation option example.

  1. Phase 1 begins at 0 hrs. The duration of phase 1 is 1000 hrs.

  2. At 550 hrs Block A fails in phase 1. Block A is in series with Block B in this phase. Hence the failure of Block A results in a system failure at 550 hrs. Since the Continue Simulation option is selected for this phase, repairs begin on Block A in phase 1 itself. The duration of corrective maintenance for Block A in phase 1 is 50 hrs. Block A is repaired from 550 hrs to 600 hrs.

  3. At 800 hrs Block B fails in phase 1. This again leads to a system failure at 800 hrs as Block B is in series with Block A in phase 1. Under the Continue Simulation option, repairs begin on Block B in phase 1. Block B is repaired from 800 hrs to 850 hrs.

  4. At 1000 hrs phase 1 ends and phase 2 begins. The duration of phase 2 is 700 hrs.

  5. At 1450 hrs Block A fails in phase 2, having accumulated an age of 450 hrs. Block A is in a parallel configuration with Block B in this phase. Thus the failure of Block A does not lead to a system failure. Since Block A cannot be repaired in phase 2, it remains failed until the end of the phase.

  6. At 1700 hrs phase 2 ends and phase 3 begins. Phase 3 is a maintenance phase. Since Block A enters phase 3 in a failed state, it is subjected to a corrective maintenance of 100 hrs duration from 1700 hrs to 1800 hrs. Block B does not enter phase 3 in a failed state and hence undergoes preventive maintenance of 20 hrs duration from 1700 hrs to 1720 hrs.

  7. At 1800 hrs all maintenance actions in phase 3 are completed and phase 3 ends. Phase 4 begins at this time. The duration of phase 4 is 200 hrs.

  8. At 2000 hrs phase 4 ends. This also marks the end of the mission.

Age Transfer Across Phases Using Cumulative Damage

The age transfer principle is crucial in understanding the concept of phases. Components accumulate age (wear-out) as they go from phase to phase, and their accumulated age is directly related to their probability of failure during the next phase. (Obviously, if the simplifying assumption of an exponential distribution (i.e. no wear-out) is used, then accumulated age does not affect this probability of failure. However, this simplifying assumption is rarely justified.) In other words, any damage the component may have accumulated in the previous phase is carried to the next phase (unless the component was restored). First it is important to note that a component (a block) may have a different failure distribution in different phases, and may have accumulated a different level of damage in each phase. Thus, to correctly transfer the age of such a block across phases, the principle of cumulative damage is used. The principle and its implementation in phase diagrams is best illustrated by example. Assume that you have a single component A subjected to two phases of 500 hrs and 200 hrs duration respectively. Furthermore, assume that  follows a Weibull distribution in phase 1 with β = 1.5 and η = 650, while in phase 2 it follows a Weibull distribution with β = 1.5 and η = 700.

Figure 11.8 Illustrating age transfer across phases.

An example (one random scenario) of this system's behavior is shown in Figure 11.8 and explained as follows:

  1. At 0 hrs phase 1 begins. A uniform random number UR[0,1] = 0.2534 is generated and the age of Block A (T) in phase 1 is calculated using the reliability equation of the Weibull distribution W(1.5, 650) as follows:

The duration of phase 1 is 500 hrs. Thus Block A will not fail in phase 1 because the phase ends before the block reaches its time to failure.

  1. At 500 hrs phase 1 ends. The age of 500 hrs accumulated by Block A in phase 1 is represented by T1. The reliability of Block A at the end of phase 1 is:

  1. At 500 hrs phase 2 begins. At the start of phase 2, Block A accumulated a certain amount of damage in phase 1 on the account of 500 hrs of operation in phase 1. Now, given the fact that the distribution in phase 2 is different from phase 1 (i.e. the stress experienced in phase 2 is different), you need to compute an equivalent time of operation in phase 2 that would have caused the same (equivalent) amount of damage as if the block only operated in this phase 2. This is done by observing that the reliability of Block A at the beginning of phase 2 is the same as its reliability at the end of phase 1, since these two events occur at the same point in time. Thus the equivalent starting age of Block A (T2) in phase 2 can be calculated using the reliability value at the end of phase 1 and the reliability equation of the Weibull distribution W(1.5, 700) as shown next:

Thus Block A begins phase 2 with an equivalent age of 538.4924 hrs, instead of 500 hrs. In other words, 500 hrs in phase 1 causes the same damage as 538.4924 hrs in phase 2.

  1. Now for phase 2 we can draw another uniform random number UR[0,1] = 0.8828 to obtain another reliability value for this phase. Using conditional probability, we then compute a time to failure for this phase, conditioned on the accumulated age of the block. Thus, the time to failure of Block A in phase 2 is:

  1. Thus, at the time of 564.4230 hrs Block A fails during phase 2, after accumulating 64.4230 hrs in this phase and 500 hrs in phase 1.

Example Using the Start New Simulation Option

To illustrate the Start New Simulation option, consider the deterministic scenario represented in Figures 11.9 and 11.10.

Figure 11.9 RBDs associated with each phase for the example illustrating the Start New Simulation option.

Figure 11.10 The phase diagram for the example illustrating the Start New Simulation option.

Table 2. The properties/parameters table associated with the example illustrating the Start New Simulation option. Note that all times are in hours.

In this example, a system has two components: Block A and Block B. The system undertakes a mission that can be divided into three phases. The first phase is an operational phase of duration 1000 hrs with both the components in a parallel configuration. In this phase Block A fails every 550 hrs while Block B fails every 750 hrs. Corrective maintenance actions on each of the components in this phase require 50 hrs to be completed. The second phase of the mission is an operational phase of duration 700 hrs in which the two components are in series. In this phase Block A fails every 450 hrs while Block B fails every 800 hrs. The duration of the corrective maintenance tasks in this phase is 100 hrs. The third phase for the system is also an operational phase and is identical to phase 1.

All maintenance actions during the entire mission of the system have a Type II restoration factor of 1. All operational phases have the Start New Simulation option selected for the On System Failure property.

The system behavior from 0 to 2500 hrs is shown in Figure 11.11 and described next.

Figure 11.11 System behavior for Start New Simulation option example.

  1. At 0 hrs phase 1 begins. The duration of phase 1 is 1000 hrs.

  2. At 550 hrs Block A fails in phase 1. Block A is in a parallel configuration with Block B in this phase. Hence the failure of Block A does not result in a system failure. Repairs of duration 50 hrs begin on Block A. Block A is repaired from 550 hrs to 600 hrs.

  3. At 750 hrs Block B fails in phase 1. Block B is in a parallel configuration with Block A. Hence the failure of Block B does not result in a system failure. Repairs of duration 50 hrs begin on Block B. Block B is repaired from 750 hrs to 800 hrs.

  4. At 1000 hrs phase 1 ends and phase 2 begins. The duration of phase 2 is 700 hrs.

  5. At 1450 hrs Block A fails in phase 2. Block A is in a series configuration with Block B in this phase. Hence the failure of Block A results in a system failure at 1450 hrs. Since the Start New Simulation option is selected for this phase for the On System Failure property, the simulation ends at 1450 hrs and phase 3 is aborted.

Example Using the Go to Maintenance Phase Option

To illustrate the Go to Maintenance Phase option, consider a simple system with two components: Block A and Block B. The two components make up a phase diagram for the system represented in Figures 11.12 and 11.13.

Figure 11.12 RBDs associated with each phase for the example illustrating the Go to Maintenance Phase option.

Figure 11.12 The phase diagram for the example illustrating the Go to Maintenance Phase option.

Table 3. The properties/parameters table associated with the example illustrating the Go to Maintenance Phase option. Note that all times are in hours.

In this example, the first phase is an operational phase in which the two components are in series and follow failure distributions of W(1.5, 650) and W(2, 500) respectively. The duration of the first phase is 500 hrs. The second phase is an operational phase in which the components are in parallel. The failure distributions on the components in this phase are W(1.5, 700) and W(2, 1300) respectively. The duration of this phase is 700 hrs. In both the operational phases no repairs can be performed on the two components. A system failure in any of the two phases takes the system to the third phase, which is the maintenance phase. In this maintenance phase, corrective maintenance of 100 hrs duration is performed on the failed components, and preventive maintenance of 20 hrs duration is performed on the non-failed components. All maintenance actions during the maintenance phase have a Type II restoration factor of 1.

The system behavior for a simulation of 1400 hrs duration is shown in Figure 11.13 and explained next. Since this system is not deterministic, each simulation will give a different result according to the random failure times generated during the simulation.

Figure 11.13 System behavior for the Go to Maintenance Phase option example.

  1. Phase 1 begins at 0 hrs. The duration of this phase is 500 hrs.

  2. At 500 hrs Phase 1 ends. Since no failures occur in phase 1, it is executed to its total duration of 500 hrs. Phase 2 begins at 500 hrs. The duration of phase 2 is 700 hrs.

  3. At 564.423 hrs Block B fails in phase 2. The time to failure of Block B in phase 2 is 64.423 hrs. This time is calculated using the cumulative damage principle (see 11.6.1.1) as Block B follows a different distribution in phase 2. Since Block B is in a parallel configuration with Block A in this phase, the failure of Block B does not lead to a system failure, as Block A is still functioning. Since no repairs can be performed in phase 2, Block B remains in a failed state.

  4. At 739.409 hrs Block A fails in phase 2. The time to failure for Block A in this phase is calculated using the cumulative damage principle, since the block follows a different distribution here (see 11.6.1.1). Since Block A is in a parallel configuration with Block B in this phase, and Block B has already failed, the failure of Block A leads to a system failure. Under the Go to Maintenance Phase option selected for phase 2, phase 2 ends on this system failure before completing its duration of 700 hrs. Phase 3, the maintenance phase, begins at 739.409 hrs. Both the failed components, Block A and Block B are repaired in this phase.

  5. At 839.409 hrs all maintenance actions in the maintenance phase are completed and the blocks are restored to as-good-as-new condition. Phase 3 ends at 839.409 hrs. This also marks the completion of the first cycle of the phase diagram. At 839.409 hrs the second cycle begins with phase 1 being executed for the second time.

  6. At 1278.194 hrs Block B fails in phase 1. The accumulated age of Block B at this time is 438.785. This is the age accumulated since the beginning of the second cycle. No age is transferred from the first cycle because of the perfect repair in the last phase of the first cycle. Block B is in series with Block A in this phase. Thus, the failure of Block B leads to a system failure at 1279.194 hrs. Under the Go to Maintenance Phase option selected for phase 1, phase 1 ends on this system failure before completing its duration of 500 hrs. Phase 3, the maintenance phase, begins at 1278.194 hrs. In this phase Block B is repaired for a duration of 100 hrs while Block A, which has not failed, is maintained for 20 hrs duration.

  7. At 1378.194 hrs all maintenance actions in the maintenance phase are completed and phase 3 ends. This completes the second cycle. Phase 1 begins in the third cycle at 1378.194 hrs.

  8. At 1400 hrs the simulation ends.

See Also:
Reliability Phase Diagrams (RPDs)

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