- Compute the probability of a type 1 failure
To simplify the notation, let R = reliability of a component, with a subscript identical with
that of the component, and Rs = reliability of the system.
As previously stated, if two events are independent of each other, the probability that
both will occur is the product of their respective probabilities. Consider a type 1 failure.
The probability that C 2 fails is 1 - R 2. The probability that both C 1 , and C 4 operate is R 1 ,
R 4 thus, the probability that either or both components fail is 1 - RI, R 4. Similarly, the
probability that C 3 or C 5 fails or both fail is 1 - -R 3 -K 5. Thus, the probability of a type 1
failure is P(type 1) = (1 - ,R 2 )(I - ,M 4 )(I - R 3 R 5 ), or P(type 1) = (0.47)[1 - (0.62)
(0.39)][1 - (0.41)(0.55)] = 0.27600. - Compute the probability of a type 2 failure
Multiply the probabilities of the three specified events, giving P(type 2) = ^ 2 (I - R 4 )
x (1 -^ 3 ) or />(type 2) = (0.53)(0.61)(0.45) = 0.14549. - Compute the reliability of the system
The probability that either of two mutually exclusive events will occur is the sum of their
respective probabilities. From steps 2 and 3, the, probability that the system will fail is
0.27600 + 0.14549 = 0.42149. Then Rs= 1 - 0.42149 = 0.57851.
ANALYSIS OF SYSTEM WITH SAFEGUARD
BYALTERNATIVE METHOD
With reference to the preceding calculation procedure, find the reliability of the system by
the moving-particle method.
Calculation Procedure:
- Compute the number of particles that traverse the system
by way of C 2
Consider that during a given interval 1,000,000 particles arrive at point a in Fig. 33, seek-
ing a path to c. They can reach c by any of three routes: C 2 and either C 4 or C 5 ; C 1 and C 4 ;
C 3 and C 5. Assume that the particles attempt passage by the first route, and refer to Fig. - The number of particles that penetrate C 2 = 1,000,000(0.53) = 530,000. Assume that
these particles now enter C 4. The number of particles that penetrate C 4 = 530,000(0.39) =
206,700, and the number that fail to penetrate C 4 = 530,000 - 206,700 = 323,300. The lat-
ter return to b in Fig. 33 and then enter C 5. The number that penetrate C 5 = 323,300(0.55)
= 177,815. Thus, the number of particles that reach c by way of C 2 and either C 4 or C 5 =
206,700 +177,815 = 384,515. - Compute the number of particles that traverse the system
by way of C 1 and C 4
Refer to Fig. 34. The number of particles that fail to penetrate C 2 = 1,000,000 - 530,000 =
470,000. These particles return to a in Fig. 33; assume that they then enter C 1. The num-
ber of particles that penetrate C 1 = 470,000(0.62) = 291,400, and the number that then
penetrate C 4 = 291,400(0.39) = 113,646. Thus, 113,646 particles reach c by way of C 1
and C 4. - Compute the number of particles that traverse the system
by way of C 3 and C 5
From step 2, the number of particles that fail to penetrate C 1 = 470,000 - 291,400 =