Question

help on these two questions please!

0.975 MITF a: the lifeti оро- 9.34 S 20 Consider the system shown in the accompa- ng figure. The reliability of each component is ovided in the figure. Assuming that the compo- onts operate independently, calculate the system cycles o 5 of the replaced reliability. Estimat 0.995 0.980 0.995 9.35 when The fai are 950 these u 9.36 and th respec withou the fai 0.950 пра- 0.999 ntial ates urs. End- 9.37 and th 9.31 Consider the system shown in the accompa- nying figure. The reliability of each component is provided in the figure. Assuming that the compo- nents operate independently, calculate the system reliability. respe and I Estim 0.990 at 5,0 expo vals 0.995 0.999 0.995 0.995 0.995 0.985 the an sion unts

Answer 1

9.30

Reliability in series is calculated by multiplying the reliability values with one another. On the other hand, parallel settings calls for multiplication of (1-reliability) and then subtraction from 1.

In this case we see three module.

0.995 In parallel

0.980 And 0.950 in parallel

The previous two in series and the series is in parallel with 0.999

We need to calculate the values for each of them.

0.995 in parallel has a reliability value of 1 – (1-0.995)*(1-0.995) = 0.999975

0.980 And 0.950 in parallel has a value of 1 – (1-0.980)*(1-0.950) = 0.999

These two in parallel creates a reliability of the upper region as 0.999975*0.999 = 0.998975. Now this has a parallel component with 0.999.

The overall system reliability is 1 – (1-0.998975)*(1-0.999) = 0.999999

The reliability of the system is 0.999999

9.31

Using the above logic, the reliability of first part is

1 – (1-0.995)*(1-0.995) = 0.999975

The reliability of the second part is

1 – (1-0.990)*(1-0.999)*(1-0.995) = 0.99999995

The reliability of the first major part (before the 0.995 on the right hand side) is

1 – (1-0.99999995)*(1-0.985) = 0.999999999

The reliability of the system is

0.999999999*0.995 = 0.9949

The reliability of this system is 0.9949