system, and one that combines series and parallel arrangements is a composite system. A
composite system may be regarded as composed of subsystems, with a subsystem being a
set of components arranged solely in series or parallel formation. A subsystem can be re-
placed with its resultant, which is a single component that has a reliability equal to that of
the subsystem. Therefore, the conventional method of analyzing a composite system con-
sists of resolving the system into sub-systems and replacing the subsystems with their re-
sultants, continuing the process until the given system has been transformed to an equiva-
lent simple system.
Let D 1 and D 2 denote the resultants of C 3 and C 4 and of C 5 and C 6 , respectively. Apply
Eq. 18 to obtain these reliabilities: for D 1 , 1 - (0.85)(0.83) = 0.29450; for Z) 2 , 1 -
(0.79)(0.70) = 0.44700. Replace these components with their resultants, producing the
equivalent composite system shown in Fig. 3IZ?.
- Perform the second cycle
Let D 3 denote the resultant of C 1 , C 2 and D 1. Apply Eq. 17: reliability of D 3 =
(0.58)(0.49)(0.29450) = 0.08370. Replace these components with D 3 , producing the
equivalent simple system shown in Fig. 3 Ic. - Compute the reliability of the system
Apply Eq. 18. The reliability of the system in Fig. 31c is Rs(t) = 1 - (0.91630)(0.55300) =
0.49329. This is also the reliability of the original system.
ANALYSIS OF COMPOSITE SYSTEM
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 1 , C 2 , and either C 3 or C 4
The alternative method of analyzing a composite system is based on this conception: Dur-
ing a given interval, a certain number of particles enter the system at one terminal and
seek to move through the system to the other terminal. Each component offers resistance
to the movement of these particles, and the proportion of particles that penetrate a compo-
nent is equal to the reliability of that component. If a particle is obstructed at any point
along its path, it returns to an earlier point and then proceeds along an alternative path if
one is available. However, a particle can enter a component in a given direction only
once. The reliability of the system is equal to the proportion of particles that traverse the
system.
Consider that during a given interval 1,000,000 particles arrive at point a in Fig. 3 Ia,
seeking a path to d. They can reach d by passing through C 1 , C 2 , and either C 3 or C 4 , or by
passing through C 5 or C 6. Assume that the particles attempt passage by the first route, and
refer to the schematic drawing in Fig. 32. The number of particles that penetrate C 1 =
1,000,000(0.58) = 580,000, and the number that then penetrate C 2 = 580,000(0.49) =
284,200. Assume that these particles now enter C 3. The number of particles that penetrate
C 3 = 284,200(0.15) = 42,630. The number that fail to penetrate C 3 = 284,200 - 42,630 =
241,570; these particles return to b in Fig. 3 Ia and enter C 4. The number that penetrate