Redundancy and Simplification in Assembly Reliability: 22-Mec-B4 Integrated Manufacturing Systems, Undated paper, Q4

Question 4: Redundancy and Simplification in Assembly Reliability (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Reliability of the lightweight component, part (a)	$R$	0.70
Number of like units after adding two redundant ones	$n$	3, in active parallel
Reliability of each component of the assembly, part (b)	$r$	0.98
Components before and after simplification	—	6 in series, then 3 in series

Find. (a) The compound reliability of the three parallel units. (b) The change in assembly reliability produced purely by removing three components from a series chain, the individual reliabilities being unaltered.

Reliability block diagrams for the two parts. In (a) the three units are in active parallel, so the assembly survives unless all three fail; in (b) the components are in series, so the assembly survives only if every one of them survives.

Approach. Draw the reliability block diagram for each configuration, then apply the two multiplication rules that follow from independence — multiply reliabilities in series, multiply unreliabilities in parallel.

Check: assumptions carried into both parts. The components are taken as statistically independent, the two added units are identical to the original and are energised continuously (active, not standby, redundancy, since no switching arrangement is described), the switching or summing element that combines them is itself perfect, and the reliabilities quoted apply over the same mission time. Part (b) additionally assumes the three surviving components still perform the whole function, which is what “through simplification” means — the function is achieved with fewer parts, not curtailed.

Part (a) — write the parallel rule from the failure side. A parallel set fails only when every path fails, so it is the unreliabilities that multiply. With $q = 1 - R = 0.30$ for each unit, $$R_{\text{system}} = 1 - q^{n} = 1 - (1-R)^{3}.$$
Substitute and evaluate. $$R_{\text{system}} = 1 - (0.30)^{3} = 1 - 0.027 = \boxed{R = 0.973\ \text{for the three units together.}}$$ Written out the long way as the sum of the mutually exclusive survival cases, $3Rq^{2} + 3R^{2}q + R^{3} = 0.189 + 0.441 + 0.343 = 0.973$, which confirms the short form.
Read the engineering message out of the numbers. Reliability rises from 0.70 to 0.973, a gain of 0.273 or 39 % — but the sharper statement is on the failure side, where unreliability falls from 0.30 to 0.027, a factor of eleven. Note also the diminishing return: the first redundant unit buys $0.70 \rightarrow 0.91$, the second only $0.91 \rightarrow 0.973$. Because the question states that the weight penalty is tolerable, redundancy is the correct route here; where weight or cost is constrained, improving the single unit is usually cheaper than triplicating a poor one.
Part (b) — write the series rule. A series chain survives only if every component survives, so reliabilities multiply: $$R_{\text{series}} = \prod_{j=1}^{m} r_j = r^{m} \quad\text{for } m \text{ identical components.}$$
Evaluate before and after the simplification. $$R_6 = (0.98)^{6} = \boxed{0.8858},\qquad R_3 = (0.98)^{3} = \boxed{0.9412.}$$ A useful check: for identical components in series, halving the count takes the square root of the reliability, and $\sqrt{0.8858} = 0.9412$.
State the change, both ways. $$\Delta R = R_3 - R_6 = 0.9412 - 0.8858 = \boxed{+0.0553,\ \text{an improvement of } 6.25\ \%.}$$ Equivalently, assembly unreliability falls from 0.1142 to 0.0588 — it is very nearly halved, which is the way a reliability engineer would report it, because failures, not successes, are what get counted in service.

Taken together the two parts contrast the only two structural levers available once component reliabilities are fixed. Part (a) adds parts and buys reliability at the cost of weight, volume and unit cost; part (b) removes parts and buys reliability for nothing — indeed it usually reduces cost, assembly labour and inventory at the same time. That is why design for assembly, part-count reduction and integration of functions into single moulded or machined parts are treated as reliability measures and not merely as cost measures. Simplification is the first thing to try; redundancy is what remains when simplification has been exhausted and the required reliability is still not met.

Final results.

Quantity	Result
(a) Unreliability of one unit	0.30
(a) Compound reliability of three units in active parallel	0.973
(a) Gain over the single unit	+0.273 (unreliability cut about elevenfold)
(b) Reliability with six components in series	0.8858
(b) Reliability with three components in series	0.9412
(b) Change in assembly reliability	+0.0553, i.e. +6.25 per cent
(b) Assembly unreliability	0.1142 falling to 0.0588 — nearly halved

Topic. Series and active-parallel configurations in reliability block diagrams, and the two design levers of redundancy and part-count reduction.

Key relations. Series: $R_s=\prod_j r_j$, so $R_s=r^{m}$ for $m$ identical components. Active parallel: $R_p=1-\prod_j (1-r_j)=1-q^{n}$ for $n$ identical units. For $m$ identical components in series, $R$ scales as the $m$-th power, so halving the count takes the square root of the reliability.

Why this works. Both rules are statements about independent events. A series chain is a logical AND, so the probabilities of the individual survivals multiply, and because each factor is below one the chain is always less reliable than its worst member — reliability decays geometrically with part count, which is why a system of a thousand very good parts can still be poor. A parallel set is a logical OR, and the cleanest way to handle an OR of independent events is to complement it into an AND of the failures, which is exactly what the 1 minus q-to-the-n form does. The diminishing return on successive redundant units follows directly: each added unit multiplies the residual unreliability by q again, so the absolute gain shrinks geometrically even though the proportional gain in unreliability stays constant.

Common pitfall. Adding parallel reliabilities (0.70 + 0.70 + 0.70) instead of complementing unreliabilities — which produces the impossible answer 2.1 and is the commonest error on part (a). On part (b) the trap is reporting only the reliability change and not noticing that the unreliability nearly halves, and a second is assuming active redundancy where the question describes a standby arrangement, since a standby set gives a different answer that depends on the switch reliability $R_{sw}$ and on the failure-time distribution, not on $R$ alone.

Source. Ebeling, An Introduction to Reliability and Maintainability Engineering, 3rd ed., ch. 6 (Reliability of Systems — series, parallel and standby configurations); Buffa & Sarin, Modern Production / Operations Management, 8th ed., ch. 15 (product design and reliability).

Question 4 of 7: Redundancy and Simplification in Assembly Reliability

Question 4: Redundancy and Simplification in Assembly Reliability (20 marks)