Final_Review_Solutions

# Final_Review_Solutions - ESI 6321 APPLIED PROBABILITY...

This preview shows pages 1–5. Sign up to view the full content.

Prof. Edwin Romeijn ESI 6321 Applied Probability Methods For Engineers Final Review Spring 2008 1/11 ESI 6321 APPLIED PROBABILITY METHODS FOR ENGINEERS Final review – Spring 2008 Exercise 1 What is the reliability of the following system (assume that the values given are component reliabilities)? Solution: r(t) = .98 × [1 - {(1 - (.95)(.90)) × (1 - .85) × (1 - (.95)(.90))} × .85 =.8304 Exercise 2 The PDF for the time-to-failure of an appliance is f ( t ) = 32 ( t + 4) 3 where t is in years. (a) Find the reliability function of the appliance: r(t) . (b) Find the failure rate: λ (t) . 0.90 0.95 0.85 0.90 0.95 0.98 0.85

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Prof. Edwin Romeijn ESI 6321 Applied Probability Methods For Engineers Final Review Spring 2008 2/11 Solution: (a) ( ) 1 ( ) r t F t = - where 3 0 2 0 2 32 ( ) ' ( ' 4) 16 ( ' 4) 16 1 ( 4) t t F t dt t t t = + = - + = - + therefore 2 2 16 ( ) 1 1 ( 4) 16 ( 4) r t t t = - - + = + (b) 3 2 ( ) ( ) ( ) 32 ( 4) 16 ( 4) 2 / ( 4) f t t r t t t year t λ = + = + = + Exercise 3 Suppose that the PDF for time-to-failure for a single unit is given as: f ( t ) = 1/ T , for 0 < t < T 0, otherwise (a) Find r(t) for two units in active parallel.
Prof. Edwin Romeijn ESI 6321 Applied Probability Methods For Engineers Final Review Spring 2008 3/11 (b) Find r(t) for two units in standby parallel. Solution: (a) For two units in active parallel unit, note that for units 1 and 2 1 1 , 0 ( ) 0, t t T r t T otherwise - < < = 2 1 , 0 ( ) 0, t t T r t T otherwise - < < = Using these above and the expression on Slide 34 of the notes, we have 1 2 1 2 2 2 2 2 ( ) ( ) ( ) ( ) ( ) 1 1 1 1 2 2 2 1 1 . r t r t r t r t r t t t t t T T T T t t t T T T t T = + -      = - + - - - -           = - - + - = - Therefore 2 2 1 , 0 ( ) . 0, t t T r t T otherwise - < < = (b) For two units in active standby unit, note that for units 1 and 2: 1 1 , 0 ( ) 0, t t T r t T otherwise - < < = 2 1 , 0 ( ) 0, t t T r t T otherwise - < < = .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Prof. Edwin Romeijn ESI 6321 Applied Probability Methods For Engineers Final Review Spring 2008 4/11 Using these quantities and the expression on Slide 35 of the notes, we have: 1 2 1 0 0 2 2 0 0 0 2 2 2 2 2 2 ( ) ( ) ( ) ( ) 1 1 1 1 1 1 2 1 2 t t t t t r t r t r t s f s ds t t s ds T T T t t s ds ds ds T T T T t t t t T T T T t T = + - -     = - + -         = - + - + = - + - +
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern