Need help on figuring out system reliability Consider the sy

Consider the system represented in the block diagram below. Suppose r_1 = 0.98, r_2 = r_3 = 0.91, r_4 = r_3 = 0.94, r_6 = 0.99, and r_7 = r_8 = r_9 = 0.78. Compute the system reliability.

Solution

In this diagram, the 2 and 3 are parallel

R₁=1-(1-r₂)(1-r₃)= 1-0.09*0.09 = 0.9919

Similarly, 4 and 5 are parallel

R₂=1-(1-r₄)(1-r₅)= 1-0.06*0.06 = 0.9964

7, 8 and 9 are parallel

R₃=1-(1-r₇)(1-r₈)(1-r₉)=1-0.22*0.22*0.22 = 0.9894

r₁ and R₁ is a series

R₁₁= r₁*R₁ = 0.9919*0.98 = 0.9720

R₂, r₆ and R₃ is a series

R₁₂= R₂*r₆*R₃ = 0.9964*0.99*0.9894 = 0.9760

Then the system reliability is

R_s= 1-(1-R₁₁) (1-R₁₂) because these are parallel

    =1-0.028*0.024=0.9993