Presentation on theme: "Solutions to Text book HW"— Presentation transcript:
1Solutions to Text book HW Chapter 5 – Part 3Solutions to Text book HWProblems 2, 7, 6
2Review: Components in Series Part 1Part 2Both parts needed for system to work.RS = R1 x R2 = (.90) x (.87) =.783
3Review: (Some) Components in Parallel RBU =.85R2 =.90
4Review: (Some) Components in Parallel System has 2 main components plus a BU Component.First component has a BU which is parallel to it.For system to work,Both of the main components must work, orBU must work if first main component fails and the second main component must work.
5Review: (Some) Components in Parallel A = Probability that 1st component or its BU works when neededB = Probability that 2nd component works or its BU works when needed= R2RS = A x B
6Review: (Some) Components in Parallel A = R1 + [(RBU) x (1 - R1)]= [(.85) x ( )]= .9895B = R2 = .90Rs = A x B= x .90= .8906
7Problem 2 A jet engine has ten components in series. The average reliability of each componentis What is the reliability of the engine?
8Solution to Problem 2 RS = reliability of the product or system R1 = reliability of the first componentR2 = reliability of the second componentand so onRS = (R1) (R2) (R3) (Rn)
9Solution to Problem 2 R1 = R2 = … =R10 = .998 RS = R1 x R2 x … x R10 = (.998) x (.998) x x (.998)= (.998)10=.9802
10Problem 7An LCD projector in an office has a main light bulb with a reliability of .90 and a backup bulb, the reliability of which is .80.R1 =.90RBU =.80
11Solution to Problem 7 RS = R1 + [(RBU) x (1 - R1)] 1 - R1 = Probability of needing BU component= Probability that 1st component fails
12Solution to Problem 7 RS = R1 + [(RBU) x (1 - R1)] RS = [(.80) x ( )]= [(.80) x (.10)]=RBU = .80R1 = .90
13Problem 6What would the reliability of the bank system above if each of the three components had a backup with a reliability of .80? How would the total reliability be different?
15Solution to Problem 6 – With BU First BU is in parallel to first component and so on.Convert to a system in series by finding the probability that each component or its backup works.Then find the reliability of the system.
16Solution to Problem 6 – With BU A = Probability that 1st component or its BU works when neededB = Probability that 2nd component or its BU works when neededC = Probability that 3rd component or its BU works when neededRS = A x B x C
17Solution to Problem 6 – With BU A = R1 + [(RBU) x (1 - R1)]= [(.80) x ( )]= .98
18Solution to Problem 6 – With BU B = R2 + [(RBU) x (1 - R2)]= [(.80) x ( )]= .978
19Solution to Problem 6 – With BU C = R3 + [(RBU) x (1 - R3)]= [(.80) x ( )]= .99
20Solution to Problem 6 – With BU .98.978.99RS = A x B x C= .98 x .978 x .99=.9489
21Solution to Problem 6– No BUs RS = R1 x R2 x R3= (.90) x (.89) x (.95)= .7610
22Solution to 6 - BU vs. No BU Reliability of system with backups =.9489 Reliability of system with backups is 25% greater than reliability of system with no backups:( )/ = .25