C 4 = 241,570(0.17) = 41,067. Thus, the number of particles that reach d by way OfC 1 , C 2 ,
and either C 3 or C 4 = 42,630 + 41,067 - 83,697.
As an alternative calculation, the number of particles that fail to penetrate C 3 =
284,200(0.85)-241,570.
Note that the proportion of particles that traverse the system by the indicated route is
83,697/1,000,000 = 0.08370 (to five decimal places), and this is the reliability OfD 3 in
Fig. 3Ic.
- Compute the number of particles that traverse the system
by way of either C 5 or C 6
Refer to Fig. 32. The number of particles that fail to penetrate a component is: for C 1 ,
1,000,000 - 580,000 = 420,000; for C 2 , 580,000 - 284,200 = 295,800; for C 4 , 241,570 -
41,067 - 200,503. The total is 420,000 + 295,800 + 200,503 = 916,303. These particles
return to a in Fig. 3 Ia and then proceed to c. Assume that they now enter C 5. The number
of particles that penetrate C 5 = 916,303(0.21) = 192,424, and the number that fail to pene-
trate C 5 = 916,303 - 192,424 - 723,879. The latter enter C 6 , and the number that penetrate
C 6 = 723,879(0.30) = 217,164. Thus, the number of particles that reach dby way of either
C 5 or C 6 - 192,424 + 217,164 = 409,588.
Note that the proportion of particles that traverse this route is 409,588/916,303 =
0.44700, and this is the reliability of D 2 in Fig. 3 Ic. - Compute the reliability of the system
From steps 1 and 2 (or from Fig. 32), the number of particles that traverse the system
from a to d = 83,697 + 409,588 = 493,285. The proportion of particles that traverse the
system = 493,285/1,000,000 = 0.49329, and this is the reliability of the system. This re-
sult is consistent with that in the preceding calculation procedure.
Alternatively, find the number of particles that traverse the system thus: A particle
FIGURE 32