BlockSim Example 2 - Optimized Reliability Allocation

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BlockSim Example 2 – Optimized Reliability Allocation

Download Example File for Version 10 (*.rsgz10) or Version 9 (*.rsr9)

Example

This example shows you how to use allocation analysis to improve the reliability of a system by improving the reliability of its individual components.

Figure 1: System RBD

Subsystem A can be broken down into two assemblies, A and B, as shown next.

Figure 2: RBD of Subsystem A

Subsystem B can also be broken down into two assemblies, C and D.

Figure 3: RBD of Subsystem B

Assembly A is made up of components 0 and 1.

Figure 4: RBD of Assembly A

Assembly B is made up of components 3, 4 and 5

Figure 5: RBD of Assembly B

Assembly C is made up of components 6 and 7.

Figure 6: RBD of Assembly C

Assembly D is made up of components 8 and 9.

Figure 7: RBD of Assembly D

Furthermore, assume that each component has the following life characteristics:
Component Life (Failure) Distribution
Comp. 1 Weibull Beta = 2 Eta = 5,000 hours
Comp. 2 Weibull Beta = 2 Eta = 4,500 hours
Comp. 3 Weibull Beta = 1.5 Eta = 8,965 hours
Comp. 4 Weibull Beta = 1 Eta = 8,000 hours
Comp. 5 Lognormal Log Mean = 10 hours Log Std = 1.4 hours
Comp. 6 Weibull Beta = 3 Eta = 10,000 hours
Comp. 7 Weibull Beta = 3 Eta = 6,000 hours
Comp. 8 Exponential Mean = 10,000 hours –
Comp. 9 Weibull Beta = 2.2 Eta = 7,000 hours

Analysis

Step 1: Estimate the system reliability of the configuration at 500 hours. The following picture shows that the estimated reliability is 92.56565%, as calculated in BlockSim.

Figure 8: System Reliability Calculated Using BlockSim

where:

  • Ci(Ri) is the penalty (or cost) function as a function of component reliability.
  • f is the feasibility (or cost index) of improving a component’s reliability relative to the other components in the system.
  • Rmin,i is the current reliability at the time at which the optimization is to be performed.
  • Rmax,i is the maximum achievable reliability at the time at which the optimization is to be performed.

We’ll set Subsystem A to f = 0.9 and Subsystem B to f = 0.7. The Rmax for both subsystems is 99.999% at 500 hours.

The optimum allocation scheme for each subsystem is the one that satisfies:

Step 2: Let’s assume that the target reliability for this system is 95% at 500 hours. Determine an optimum reliability allocation strategy, assuming that the cost/difficulty of increasing an item’s reliability is defined by the following function:
while minimizing

Solving this yields an optimum allocation for each subsystem. The solution can also be obtained via the allocation analysis tool in BlockSim.

BlockSim allows you to use a predefined feasibility value in the calculation. You select the value from a set shown on a scale from Easy (1) to Hard (9). The relationship between the scale values (SV) in BlockSim to the feasibility function in Eqn. (1) is described as:

Therefore, in the allocation analysis tool in BlockSim, set the value in the Feasibility column to 1 for Subsystem A and 3 for Subsystem B. Set the value in the Maximum Achievable Reliability column of both subsystems to 0.99999. The following picture shows the result of the analysis. It shows that in order to meet the target system reliability of 95% at 500 hours, the reliability of Subsystem A should be at least 98.6140% and the reliability of Subsystem B at least 96.3352%, at 500 hrs.

Figure 9: Allocation Analysis Tool in BlockSim.

The subdiagram blocks are displayed as hyperlinks. Click the name of the subdiagram block to add another tab to the allocation analysis folio and perform an analysis for the subassembly. The following table summarizes the results.
Feasibility (SV) Target R, 500 hrs
System 0.95
Subsystem A 1 0.986140
Assembly A 1 0.987318
Comp. 1 1 0.997046
Comp. 2 9 0.990243
Assembly B 9 0.979923
Comp. 3 1 0.986916
Comp. 4 1 0.939413
Comp. 5 1 0.996573
Subsystem B 3 0.963352
Assembly C 9 0.999297
Comp. 6 1 0.999875
Comp. 7 1 0.999422
Assembly D 1 0.964030
Comp. 8 1 0.966936
Comp. 9 1 0.996995