Systems are made of component parts, and the reliabil­ity of each component part is either known or can be predicted. The reliability of the overall system is what is desired. The reliability of the system can be predicted as a function of the reliabilities of its various component parts by applying the logic of success—failure events in the system.

The methods discussed in this section are “decision tree” logic, Boolean algebra, conditional probability, mini­mal cuts, binomial theorem, and availability analysis All the methods of probabilistic manipulation described here are suitable for hand calculation and some can be used as the basis for computer calculation.

(a) Decision Tree Logic. The “decision tree” is a systematic way of accounting for all the system paths to success and of giving each its proper probabilistic weight. The success diagram of Fig 116 represents a physical

Fig. 11.6—Reliability block diagram for a series—parallel system.

system in which two amplifiers are dependent on one power supply. Success is assured if the power supply and at least one amplifier are operating

The decision tree used to calculate system success is shown in Fig 11.7. Each branch in the diagram, reading

from left to right, represents a decision, go or no-go, on the success or failure of that particular component. By conven­tion, good outcomes branch upward and bad outcomes branch downward. The components are considered in order A good outcome on A and a good outcome on В ensures success, so this branch terminates on the vertical bar labeled “success ” Likewise, a good outcome on A followed by a good outcome on C ensures success. A good outcome on A followed by bad outcomes on both В and C ensures failure. Finally, a bad outcome on A ensures failure regardless of the state of В and C since the amplifiers cannot operate without power

Fig. 11.7 —Decision tree for predicting system success for the series—parallel system of Fig. 11.6.

A probability must be assigned to each branch based on the probability of success and failure of the components in that branch. Assume that the probabilities of success and failure for each component are as follows

Component A Probability of success = 0 999, prob­ability of failure = 0.001.

Component В Probability of success = 0.99, prob­ability of failure = 0.01

Component C Probability of success = 0.99, prob­ability of failure = 0 01

The probability of success may be computed by tracing each success path back to the origin, taking the product of the probabilities along each path and summing the result For success path 1, find the product of (0 99) X (0.999) = 0 98901 For success path 2, find the product of (0 99) X (0.01) X (0 999) = 0.0098901 The total probability of success is the sum of the two products, or 0.9989001.

These rules are essential

1 The sum of the probabilities at each branch must be unity.

2. The components of the system must be considered in order until success or failure is ensured without regard to the state of any of the remaining components

3. The various component-failure events must be statis­tically independent

The least redundant component should always start the first branch and the most redundant components should be considered last to reduce the number of branches on the tree.

The decision tree can handle events with more than two states. It can also accommodate certain physical interdepen­dencies. Consider the success diagram of Fig. 11.8. This figure shows an amplifier driving an output transistor whose

AMPLIFIER TRANSISTOR Fig. 11.8—System with multiple component failure states

output is connected in parallel with the output transistor from another amplifier One amplifier can fail without disturbing the other The output transistor can be good, fail open, or fail short. If it fails short, it voids the output of the parallel output transistor and failure is certain.

The decision tree for Fig. 11.8 is shown in Fig. 11.9. Note that three branches are shown for each output transistor decision good, open, short. Note also that any short causes failure whereas an open only leads to failure in

Fig. 11.9—Decision tree for the system with multiple com­ponent failure states shown in Fig, 11.8.

combination with other failures. As in the previous ex­ample, a probability is assigned to each branch, summing to unity. The probability of success is the sum of the products of probabilities along each success path.

As numerical check, it is advisable to calculate the probability of system failure m the same manner and check to see that the sum of the success and failure probabilities is unity. Such a check does not guarantee against errors in the way the decision nee is branched to represent the problem. Consequently it is essential to exercise care in constructing the tree to be sure that it represents the physical system.

(b) Boolean Algebra. The techniques of using Boolean algebra as an adjunct to probabilistic calculations are documented in many sources.6 8 For those not familiar with Boolean algebra, a simple, but often overlooked, technique will be described.

Consider a two-out-of-three, or majority, logic configu­ration as represented in the success diagram of Fig. 11.10.

Fig. 11.10—Majority logic success model

If any two of the components are good, the system is good. In Boolean algebra, the event “success” is described symbolically as

S = AB + AC + BC (11.3)

where AB means A and В and + means or In words, Eq. 11.3 says, “The system is successful if A and В are good or if A and C are good or if В and C are good.” The negation of A is denoted by the symbol A, which means “not A” or, in this case, “A fails.”

Equation 11.3 can be transformed into a form that is more useful m reliability calculations The following rela­tions from Boolean algebra are used in the transformation

X + Y = X + XY (114)

XY = X + Y = X + XY (115)

XX = X (116)

XX = 0 (11.7)

X = area inside circle (X) = areas b and c Y = area inside circle (Y) = areas c and d X = area outside circle (X) = areas a and d Y = area outside circle (Y) = areas a and b XY = overlap of circles (X) and (Y| = area c X + Y = area covered by circles (X) and (Y) — areas b, c and d XX = overlap of X and X = 0 (since there is no overlap) X + X = area covered by circle (X) and circle (X) = area covered by (X) XX = overlap of X and X = X (since overlap is same as X) X + X = area covered by X and X = total area (a, b, c, and d) = 1 by definition XY = area of overlap of X and Y = area b XY = area of overlap of X and Y = area d X + XY = area covered by X and XY = areas b, c, and d

Fig 11.11—Venn diagram illustrating basic Boolean equa­tions The diagram is a nonrigorous way to visualize the basic Boolean equations. From the above it can be seen that, e. g., XX = 0, YY = 0, X + X = X, Y + Y = Y, X + X = 1,

Y + Y = l, XX = X, YY = Y, and X + XY = X + Y Multi­plying corresponds to “overlapping” or “intersecting”, addi­tion corresponds to “covering” or “union” of the added elements. Multiplication can also be read as “and,” і e, XY = X and Y, addition can be read as “or”, i. e, X + Y = X

Boolean algebra and shows how they may be derived by a graphical representation (the Venn diagram)

Returning to Eq. 11.3, the expression is first broken into two terms

S = AB + [AC + BC]

and then, using Eq. 11.4, this is rewritten as S = AB + AB [AC + BC]

Similarly, the two terms inside the brackets are written according to the form of Eq. 11.4,

S = AB + AB [AC + (AC)(BC)] (118)

From Eq. 115 the terms AB and AC can be written as A + AB and A + AC, respectively. Substituting these into Eq. 11.8 yields

S = AB + (A + AB)[AC + (A + AC) BC]

= AB + AAC + AABC + AACBC + ABAC + ABABC + ABACBC

= AB + ABC + ABC (11.9)

In the final step the relations (from Eqs 11 6 and 11.7) AA = 0, AA = A, CC = 0, and AA = A have been used

In Fig 11.12, the result (Eq. 11.9) is shown graphically. The terms AB, ABC, and ABC are seen to be mutually exclusive, і e, the areas wherein two events intersect are all shown on the Venn diagram, but the areas representing the three terms of Eq. 11.9 do not overlap. Therefore the probability of success is simply the sum of the joint probabilities

P(S) = P(AB) + P(ABC) + P(ABC)

= P(A) P(B) + P(A) P(B) P(C)

+ P(A) P(B) P(C) (1110)

As a numerical example, assume P(A) = P(B) = P(C) = 0.99, then

P(S) = (0.991(0.99) + (0.01)(0 99)(0.99)

+ (0.99)(0.01)(0 99)

= 0.999702

In summary, the foregoing method is as follows

1. Write the Boolean expression that is the union of all possible success paths.

2. Separate the first term and intersect the negation of that term with the rest of the terms Continue down to the last term.

3. Express each negated success path as the union of mutually exclusive events

AB. N = A + AB + . . . ABN

4. Starting with the innermost enclosures, clear the expression using the Boolean relations AA = 0, A + A = A, and AA = A.

5. The probability of success is equal to the sum of the probabilities of the mutually exclusive events

Fig. 11.12—Venn diagram for majority logic case Note that the regions AB, ABC, and ABC are mutually ex­clusive, і e, they do not overlap each other

(c) Conditional Probability Frequently the complex lty of a probabilistic calculation can be reduced by the use of conditional probability It is especially useful if a given component is repeated throughout many branches of the system success model or if the component occupies a key position m the model that makes it difficult to evaluate Conditional probability may be expressed as

P(S) = P(S|A) P(A) + P(S|A) P(A) (11 11)

where P(S) = the probability of system success

P(S|A) = the probability of system success, given that component A is good

P(A) = the probability that component A is good P(S I A) = the probability of system success, given that component A is bad

P(A) = the probability that component A is bad

The method is illustrated by the solution to the reliability model resembling the bridge circuit of Fig 11 13(a) The probability of success can be expressed in the conditional sense as

P(S) = P(S|E) P(E) + P(S|E) P(I) (11 12)

The P(S|E) may be obtained from the easily computed diagram of Fig 11 13(b), where component E is replaced

(a)

(b) P(S|E)

(c) P(S|E)

Fig 11 13—Bridge type reliability model

by a solid line, indicating that E is perfect The P(S|E) may be obtained from Fig 11 13(c), where the path normally provided by component E is missing, indicating that E has failed

This method is described fully in Ref 9

(d) Minimal Cuts A cut is a collection of equipments belonging to a model such that if all these equipments fail, then successful completion of the mission phase repre sented by that model is precluded 10 A minimal cut is a unique set of failed equipment such that the deletion of any one piece of equipment from the cut restores the system to success

Consider the reliability block diagram of Fig 11 14 The hand-calculation method is as follows Start by

Fig 11 14—Reliability block diagram to illustrate minimal cuts method

considering component A as failed and write all the minimal cuts that must include A By inspection, they are AD, AEG, and ABCG Then restore A to operation, and write all the minimal cuts that do not include A In this it is obvious that, for the system to fail, all minimal cuts must either include A or F, thus the remaining paths are FD, and FEG Adding together all the minimal cuts gives an approximate expression for the probability of failure

P(F) ^ AD + AEG + ABCG + FD + FEG (1113)

The approximation is very good provided the probability of failure of each individual component is *^1 For example, if all components have a probability of failure of 0 01, the system probability of failure is 0 000202 by the minimal — cuts method and 0 000201 by an exact method, an error of only 0 05%

In highly redundant systems, not all the minimal cuts need to be written if their total contribution to failure is small compared to the dominant paths For example, if

ABCG is known to be very small compared to AD or FD, ignore it

(e) Binomial Expansion. The bionomial expansion is useful in solving models using redundant components Let p be the probability of success and q the probability of failure of an individual component By definition,

P + q= 1

Likewise,

(p + q)n = 1

where n represents the level of redundancy For example, if n = 5,

(p + q)5 = p5 + 5p4q + 10p3q2 + 10p2q3 + 5pq4 + q5 = 1

The terms represent the various ways success and failure can be achieved The first term, p5, is the probability that all components succeed, there is only one way all of them can succeed The second term, 5p4q, is the probability that four components succeed and one fails, there are five combinations, including exactly four components succeed­ing and one failing There are ten combinations of three successes and two failures, or 10p3q2, etc The terms thus account for all the possible combinations of success and failure of the five components

As an example, consider the system with five relief valves If three or more must function, then the probability of success is

P(S) = p5 + 5p4q+ 10p3 q2

that is, either all five can function, or four function and one fail, or three function and two fail If the probability of a single valve functioning is p = 0 95, then the probability of a single valve failing is q = 0 05 and the probability of system success is

P(S) = (0 95)s + 5(0 95)4 (0 05) + 10(0 95)3 (0 05)2 = 0 9988

In general, the binomial expansion is n

(p + q)n=2(k)pkqnk (1115)

k-0

where

(f) Availability Analysis Senes Subsystems If there are n subsystems in series with failure rates Xi, X2 , X3 , ,

Xn and mean repair times of 0,, 02, $з> > $n and if

repair is instituted on each subsystem as soon as failure occurs, then the series configuration

reduces to

where

Xp — Xj + X2 + + Xn

and

+ X202 "t Xn0n

0T =—————— T—————-

Д-р

The system parameters are

Mean time between failure for the system = (1/Xp) Mean down time for system = 0p

Parallel Subsystems If there are two subsystems in parallel with failure rates X, and X2 and mean repair times 0, and 02, if either one or both subsystems in operation constitute an operating system, and if repair is instituted on each subsystem as soon as failure occurs, then the parallel configuration

reduces to

If there are n subsystems m parallel (partially redundant to each other), each having the same failure rate X and mean repair time в, and r out of n of these subsystems constitute an operating system, and if repair is instituted on each subsystem as soon as failure occurs, then this system

reduces to where

/n — 1 (n-1)’

ir — 1/ (r — l)'(n — r)1

в

n — г + 1

Example 1 Two out-of-three subsystems must operate, ie.,n= 3, r = 2.

Note that the unavailability (= 1 — availability) of the three-out-of-four system is twice that of the two-out-of — three system.

The foregoing relations were derived from Ref 9. In the derivations it is assumed that failure rates and repair times are exponentially distributed and that all в products are <1

By these techniques any simple series-parallel avail­ability model can be reduced to a single block with an equivalent failure rate and repair time. This technique is not applicable to a system with repeated components or components that bridge between series-parallel strings.

In many cases repair does not commence when failure occurs but rather when failure is discovered. This is particularly true of most of the failures of standby systems and of non-fail-safe failures on power-plant protection systems In these cases the mean repair time is equal to one-half the time interval between tests plus the actual repair time.

In the event that repair starts immediately on detection of the failure at a periodic test, the following substitution in the preceding formulas will yield useful results with little error

в* =-+в
2

where в* is a new equivalent repair time including the time elapse between failure and discovery. Frequently the actual repair time is short compared to the time between tests, so the above can be approximated by 0* — г/2

XT = 6X20

в в 1 + 1 2

Availability = 1 — Хт$т = 1 — 3(X0)2

Example 2 Three-out-of-four subsystems must oper­ate, i. e., n = 4, r = 3

(::;b

XT=^ = 12X

в в

1+1 2

Availability = 1 — Х-р^т = 1 — 6(Хб)2