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MIL-HDBK-338B §7.5.5.3: cold standby redundancy to reach R(100 h) = 0.98
| Tool | System Reliability |
| Source | U.S. Department of Defense (1998). Electronic Reliability Design Handbook. MIL-HDBK-338B, 1 October 1998. Section 7.5.5.3 and Figure 7.5-9. |
| Level | Undergraduate |
Problem
Identical units with an exponential life of θ = 100 hours are placed in cold standby with perfect switching. How many units are needed for a mission reliability of 0.98 at 100 hours, and what is the MTBF of the standby group?
Inputs to Enter
- Components: G mtbf 100; H mtbf 100; I mtbf 100; J mtbf 100
- Structure: standby(G, H, I, J); mission time 100 hours
Answers
- n = 4: R(100) = e^−1 (1 + 1 + 1/2 + 1/6) = 0.9810, which meets 0.98; n = 3 falls short at 0.9197; n = 2 gives 2e^−1 = 0.7358
- MTBF of four units in standby = n θ = 400 hours (Figure 7.5-9 ratios 1.50, 1.83, 2.08, 2.28 for 1-of-n active parallel are also reproduced)
- Standby beats the same two units in active parallel by e^−2 at t = θ
Notes
Replace the default components and structure; the order of the units in standby() does not change the result (the handbook's remark on which element is primary).