22-Mec-B4 Integrated Manufacturing Systems · May 2017

Question 4 of 6: Redundancy and Component Reduction in Assembly Reliability

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2017 — 16-Mec-B4 Integrated Manufacturing Systems. Three hours, open book, any non-communicating calculator permitted. Six questions are printed and any five constitute a complete paper; all questions are of equal value, so each is treated below as a 20-mark question. Every question is solved here, because the set is a study resource rather than a sitting.

Reference texts. M. P. Groover, Automation, Production Systems, and Computer-Integrated Manufacturing, 5th ed. (process planning, CAPP, group technology, discrete control and programmable logic controllers); R. B. Chase and F. R. Jacobs, Operations and Supply Chain Management, 16th ed. (forecasting); S. Nahmias and T. L. Olsen, Production and Operations Analysis, 7th ed. (inventory models and lot sizing); C. E. Ebeling, An Introduction to Reliability and Maintainability Engineering, 3rd ed. (series and parallel reliability); S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology, 8th ed. (machinability data and cutting conditions).

Question 4: Redundancy and Component Reduction 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. Two independent reliability problems on the same electrical assembly, with the data below.

Given data
Part	Item	Value
(a)	Reliability of the lightweight component, $R_c$	0.70
(a)	Redundant units added	2, giving 3 units in total
(a)	Redundancy type	active (hot) parallel, weight penalty negligible
(b)	Reliability of each component, $R_u$	0.98
(b)	Components before simplification	6 in series
(b)	Components after simplification	3 in series

Find. (a) the compound reliability of the three parallel units, and (b) the change in assembly reliability when the series chain is cut from six components to three.

Figure 4.1 — Reliability block diagrams. In (a) the three units are in parallel, so the assembly survives if any one survives; in (b) the components are in series, so the assembly survives only if every one survives.

Approach. Draw the reliability block diagram for each case, apply the product rule for series elements and the product-of-unreliabilities rule for active parallel elements, and express the part (b) answer both as a change in reliability and as a change in the probability of failure, since the latter is what a design review actually cares about.

Part (a) — set up the parallel model. Three identical units are in active parallel and the assembly functions if at least one of them functions. It is easier to work with the complement: the assembly fails only if all three fail simultaneously. With independent units, $$Q_{\text{sys}} = \prod_{i=1}^{3}\left(1-R_i\right) = \left(1-R_c\right)^{3}.$$
Evaluate the unreliability. Substituting $R_c = 0.70$, each unit has an unreliability of $1 - 0.70 = 0.30$, so $$Q_{\text{sys}} = (0.30)^{3} = 0.027 .$$ It is worth noting the intermediate result for two units, $1-(0.30)^{2}=0.91$, because it shows where the diminishing return sets in: the first redundant unit buys 21 points of reliability and the second buys only 6.3 more.
Convert back to reliability. The compound reliability of the three units is the complement of that failure probability: $$\boxed{\;R_{\text{sys}} = 1-\left(1-R_c\right)^{3} = 1-0.027 = 0.973\;}$$ The redundancy has raised the reliability of that function from 0.70 to 0.973, a gain of 0.273, and it has cut the probability of losing the function from 30 % to 2.7 %, a factor of eleven.
Part (b) — set up the series model. A series chain functions only if every element functions, so for $n$ identical independent components $$R_{\text{assy}} = \prod_{i=1}^{n}R_i = R_u^{\,n}.$$ This is the reason component count is itself a reliability parameter: each additional part multiplies the assembly reliability by a number less than one.
Evaluate the six-component assembly. With $R_u = 0.98$ and $n = 6$, $$R_{6} = (0.98)^{6} = 0.885842 .$$ The six-part assembly therefore fails about 11.4 times in every hundred, even though every individual component is 98 % reliable — the standard illustration of how quickly series reliability erodes.
Evaluate the three-component assembly. With the same component reliability and $n = 3$, $$R_{3} = (0.98)^{3} = 0.941192 .$$
Report the change. Subtracting the two results gives the improvement obtained purely by simplifying the design: $$\boxed{\;\Delta R = R_{3}-R_{6} = 0.941192-0.885842 = 0.055350\;}$$ that is, an increase of 5.54 percentage points, or a relative improvement of 6.25 % on the original assembly reliability.
Restate the result as a failure probability. The more useful engineering statement is what happened to the failure rate. The unreliability falls from $1-R_6 = 0.114158$ to $1-R_3 = 0.058808$, so $$\frac{1-R_{6}}{1-R_{3}} = \frac{0.114158}{0.058808} = 1.94 .$$ Halving the part count has very nearly halved the assembly failure probability, which is the expected result: for small unreliabilities $1-R_u^{\,n}\approx n\left(1-R_u\right)$, so failure probability is approximately proportional to component count.

Final results — Question 4
Quantity	Result
(a) Unreliability of one unit	0.30
(a) Two units in parallel	0.910
(a) Probability all three units fail	0.027
(a) Compound reliability of the three units	0.973
(a) Gain over the single unit	+0.273
(b) Six components in series	0.885842
(b) Three components in series	0.941192
(b) Change in assembly reliability	+0.055350 (+5.54 percentage points)
(b) Relative improvement	+6.25 %
(b) Failure probability, before and after	0.114158 → 0.058808 (factor 1.94)

Check: the redundancy in part (a) is taken as active and the failures as independent. The question says the units can be provided with no weight penalty, which is the classic description of hot standby, and gives no switch reliability, so all three units are assumed energised and the result is the pure parallel value 0.973. If the arrangement were standby redundancy through a changeover switch of reliability R_s, the result would be R = R_c + R_sR_c(1 − R_c) + R_s²R_c(1 − R_c)², which reduces to 0.973 only for a perfect switch. Independence also assumes the three units do not share a common-cause failure such as one power supply or one connector; if they do, the achievable reliability is bounded by that shared element and the redundancy is worth much less than the arithmetic suggests.

Topic. System reliability from component reliability: series chains, active parallel redundancy, and the effect of part count.

Key relations. Series: R_s = product of R_i = R_uⁿ for identical components. Active parallel: R_p = 1 − product of (1 − R_i) = 1 − (1 − R)ⁿ. Small-unreliability approximation: 1 − R_uⁿ is approximately n(1 − R_u). Standby redundancy with an imperfect switch adds a factor R_s to each successive term.

Why this works. Both rules are statements about independent events. A series system needs every component to survive, and the probability of a conjunction of independent events is the product of their probabilities. A parallel system fails only when every path fails, so the same product rule is applied to the unreliabilities and the answer complemented. The two parts of this question are the two levers a designer has: add paths, which attacks unreliability multiplicatively and gives large gains on a weak component, or remove parts, which attacks it additively and is usually cheaper, lighter and also reduces assembly labour and inspection.

Common pitfall. Adding reliabilities in the parallel case, which can produce a "reliability" above 1.0 and is always wrong; the complement of the product of unreliabilities is the only correct route. In part (b) the frequent error is to report the ratio 0.941/0.886 as though it were the change in reliability; the question asks for the change, so state the difference of 0.0554 and offer the ratio as supporting information rather than as the answer.

Source. Ebeling, An Introduction to Reliability and Maintainability Engineering, 3rd ed., ch. 6 (Reliability of Systems), sections on series, parallel and standby configurations; Chase and Jacobs, Operations and Supply Chain Management, 16th ed., appendix on reliability in product design.

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