## Components in Series

If two components (that comprise a system) are placed in a ''series'' reliability configuration, as in Figure 11.4(a), it means that both must be operative for the system to be working properly. The reliability of the system is therefore

This is the basis for the simple addition of failure rates for components when considering the reliability of a system.

Because the MTBFs and failure rates are reciprocals of one another,

Component A

Component A

Figure 11.4. Two reliability configurations.

Figure 11.4. Two reliability configurations.

MTBF MTBF MTBF,

as described in Chapter 8, Section 8.7.2. Extension to many components in a series reliability configuration is immediate.

Example. The probability that the system described in Figure 8.3 will survive without failure for 500 hours can be computed by adding the given failure rates of 0.0004, 0.0005, 0.0006, and 0.0005, yielding a system failure rate of 0.002. The reliability of the system is therefore exp (-.002/), where, in this case, t = 500 hours. This then reduces simply to R = exp (-1) = 0.368. We note that this is the result for any simple exponential system in terms of failure-free operation to its MTBF.

### 11.11.2 Components in Parallel

A parallel reliability configuation, as in Figure 11.4(b), means that at least one of the components must be operative in order for the system to be working. For two components in parallel, the system reliability therefore can be expressed as

and the failure rates are not additive for such a system. The parallel configuration introduces redundancy, and thus improves the reliability of the system, with the penalty being the addition of the redundant component. This is necessary when it is extremely important to keep a system on the air, such as with a manned spacecraft or an air traffic control system.

Example. If we take the system in the previous example and place it in a redundant configuration, the reliability then becomes

Rs = 1 - (1 - 0.368)(1 - 0.368) = 1 - (0.632)2 = 0.6

Thus, the reliability has improved from 0.368 to 0.6 by adding simple redundancy, for which one pays the price of duplicating this piece of equipment.

Example. What is the probability of succesful operation for 100 hours for a system with two subsystems with MTBFs of 200 and 300 hours when (a) the two subsystems are in ''series,'' and (b) the two subsystems are in ''parallel''? For part (a), R1 = exp (-100/200) = 0.6065 and R2 = exp (-100/300) = 0.7168. The product of these represents the series case, which yields the result 0.434. In part (b), we have an overall reliability of 1 - (1 - 0.6065)(1 - 0.7168) = 1 - (0.3935)(0.2852) = 0.8885.

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