06_Reliability+of+Systems_Problems.doc

# Low r r 3 k out of n redundancy total engines 4

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low R R 3. k-out of-n Redundancy Total engines = 4 Required for service = 3 The probability of exactly x components operating out of n components, x n x R R x n x P ) 1 ( ) ( , where, 97 . 0 R The probability of k or more successes out of n components, n k x s x P R ) ( 3 0.80 0.80 0.90 0.90 0.85 0.85 0.80 0.80 0.90 0.90 0.85 0.85

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9948 . 0 ) 03 . 0 ( 97 . 0 ! 0 !* 4 ! 4 ) 03 . 0 ( 97 ! 1 !* 3 ! 4 ) 97 . 0 1 ( 97 . 0 4 4 ) 97 . 0 1 ( 97 . 0 3 4 ) 1 ( 4 0 4 1 3 3 4 4 3 4 3 4 3 x x n x s R R x R 4. Assume the reliability of wheel cylinder subsystem is WC R and brake assembly is BP R . Using the method of decomposition, Case1: BP3 Fails & BP4 Operational This probability = BP BP I R R P ) 1 ( . Reliability for the fluid breaking system is, ]} 1 [ ] 1 [ 1 { 2 WC BP WC M f R R R R R Reliability for the mechanical (cable) subsystem = C R , since BP4 is operational. These two subsystems are in parallel, therefore the system reliability in case I, ] 1 ][ 1 [ 1 C f I R R R Case II: BP4 Fails and BP3 Operational Because of each wheel cylinder and brake pad have the same reliability, Case II is identical to Case I Case III: BP3 and BP4 Failed This probability is 2 ] 1 [ BP III R P . The cable system will not function, and therefore, } ] 1 [ 1 { 2 BP WC M III *R R R R Case IV: Both BP3 and BP4 are operational This probability is 2 BP IV R P . The reliability of the fluid system is, } ) 1 ]( ) 1 ( 1 {[ 2 2 WC BP WC M f R *R R R R Since the cable system will function as long as C functions, ] 1 ][ 1 [ 1 C f IV R R R Now, the overall system reliability can be found from, IV IV III III II II I I S R P R P R P R P R 4
Also, observe that, 1 IV III II I P P P P 5. Common mode failure System reliability, 2 1 1 * )] 1 )( 1 ( 1 [ ) ( R R R t R s t t t e e e 2 1 1 )]. 1 )( 1 ( 1 [ 9405 . 0

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• Three '17
• Bouanar Jack

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