Quality Control

Sum 12

For the subsystem with the components A and B , the reliability is

R1 = 1-( 1- Ra) (1- Rb) = 1- (1-0.96)(1-0.92) =0.9968

The reliability of the system with components E F and G is

R2 = 1- ( 1- ReRf) ( 1- Ra) = 1- ( 1-0.95* 0.88)(1-0.9)=0.983

The system reliability is given by

Rs = R1 *Rc *Rd*R2 =0.9968*0.94 *0.89*0.983 =0.8202

The component C and D are modified.

R3 = 1- (1-Rc) (1-Rd) = 1- (1-0.94)(1-0.89) = 0.9934

The system reliability Rs1 = R1 * R2 * R3 =0.9968* 0.983*0.9934 =0.9740

hence

The system reliability is improved

Sum 13

For the subsystem with the components A and B the reliability is

R1 =1-(1-e-0.005*1000) (1-e-0.005*1000) =0.8452

For the system with reliability

R2 =1- (1-e-λete-λft)(1-e-λgt) =0.0033

The reliability of the system is Rs = R1 R2* e-0.003*1000 * e-0.008*1000 =0.0009

The MTTF1 for A/B system = 1/0.0005 (1 +0.5) = 3000 hrs

The failure rate of e/ f system is 0.0064

The MTTF2 for E F G subsystem =1/0.0064(1=0.5) =234.375 hours

The system failure rate =1/3000 +1/234.375+0.003+0.0008 =0.0057

Thus the mean time to failure is 1/ system failure rate =175.44 hrs

Sum 14

The failure rate λ for each comp is 0.008/hr

The system reliability Rs =

=0.781

The mean time to failure 4+1/0.008 =6.25 hrs

Considering if the all the components are parallel

Thus the failure is 1- e-0.008*400 0.9592

The system reliability Rs = (1-0.9592)5 =0.188

The MTTF for the system is 1/λ(1+1/2+1/3+1/4+1/5) = 285.4 hrs

SUM 16

Sum23

We have ϴ0 = 1500 alpha = 0.05 , ϴ1= 600 beta = 0.10 .

ϴ1/ ϴ0 =0.4

If the sample size is 5 times the rejection number is 15

Thus T=0.136 / ϴ0 =204 hrs

So the random sample of 75 is selected and tested if it 15th failure occurs before the T then it is rejected if it occurs after or does not occur then it is accepted

Sum 25

Theta 0 =1400 , alpha =0.05 r=7 n=35

For alpha =0.05, r =7 and n=35

Calculating the test termination time = 0.103* 1400...