SKEDSOFT

If a system consists of 3 components A, B, C in series, then the reliability of the system,

R_S = R_A . R_B . R_C

System Connected in Parallel

Here the function of A can be done by B or vice versa. If the system consists of components A and B in parallel with reliability RA and RB, then the reliability of the system,

R_S = [1 – (1 – R_A)(1 – R_B)]

If n components are connected in parallel,

R_S = [1 – (1 – R_A)(1 – R_B) ...... (1 – R_n)]

If R_A = R_B = ...... R, then

R_S = [1 – (1 – Rⁿ)]

Also (1 – R_S) = (1 – R^_n).

Prove that the failure rate of the system = the sum of the failure rates of the components of the system.

Or

P.T. for a series system, the failure rates are additive :

i.e., P.T. λ_S = λ_A λ_B ...... λ_n.

Proof.

Let R_S, R_A, R_B ...... be reliability of the system and its component parts A, B and C so on.

For independent components in series and exponentially distributed failure rate,

R_S = R_A . R_B . R_C ...(1)

If λS, λA, λB, λC ...... are the failure rate of the system and its component parts for a mean time ‘tm’.

From eqn. (1),

e^{− λ_Stm} = (e^{− λ_Atm} ) . (e^{− λ_Btm} ) . (e^{− λ_Ctm} ) ......

e^{− λ_Stm} = e^{− λtm (λ_A λ_B λ_C ......)}

e_λS = e^{(λ_A λ_B λ_C ......)}

Since the bases are save, by comparing L.H.S. and R.H.S., we equal the powers,

λ_S = λ_A λ_B λ_C ......

Hence proved.