1. Structure Function
Write the structure function for the block diagram below. Identify minimum path vectors (X) (X) = x1x2 + x3 - x1x2x3 d = abc + aBc + Abc D = abC + aBC + AbC + ABc + ABC = (a+A)bc + aBc = bC + BC + ABc = bc + aBc = (b + aB)c = C + ABc 1 2 3 rd
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2. Exponential Function (CFR)
Write the density function: f(t) = e-t, t  0
Write E(T)? 1/ = MTTF
Write the failure rate. 
Write the reliability function. R(t) = e-t = 1 – F(t)
Write the hazard function. h(t) = f(t)/R(t) =  (CFR).
What property does #5 invoke? Memoryless
Lack of memory” means the probability of
failure in a specific time interval is the same regardless of the starting point of that time interval. rd
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3. Example
a) Find the probability of an item surviving until t = 100 units if the item is exponentially distributed with  = 1/80. R(100) = e-100/80 = b) Given that the item survived 200 units, what is the probability of survival until t = 300 units? What is the value of the hazard function at 200 units? at 300 units? P(T > 300|T > 200) = P(100) = h(t) =  = 1/80 = For f(t) = e-t, t  0; rd
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4. Systems Reliability
Given 5 components connected in series with respective probabilities [ ], specify which component to improve to enhance the system reliability. T or F System reliability for series connected components is always greater than its least reliable component. T or F System reliability RS = P((X) = 1) = E[(X)] Give an example of a standby system. Spare tire rd
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5. Path Vector Technique
Find (X) for the bock diagram using path vector techniques (X) = 1 – (1 – X1X2)(1 – X1X3)(1 – X2X3) The 4 path vectors are (1 1 0) (1 0 1) (0 1 1) (1 1 1) R(p)=E[(X)] = [p1p2(1-p3) + p1(1-p2)p3 + (1-p1)p2p3) + p1p2p3 = p1p2 + p1p3 + p2p3 – 2p1p2p3 R = 1 - ( * 0.95)3 = 1 3
(poly^n #(R ) 3)
#(R ) 
3p2 - 2p3
R = 1 – 3r r4 - r6 rd
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6. Simulation
Two independent constant failure rates  = 1 and  = ½ per month are connected in parallel. Find 10th percentile, mean lifetime and probability of failure within a month. (percentile 10 (mapcar #' max (sim-exponential ) (sim- exponential 1/ )))  month vs theoretical (mu (mapcar #' max (sim-exponential ) (sim-exponential 1/ )))  months is mean lifetime. 1/1 + 1/2 - 1/(1 + 2) = 7/3 (count-if #'(lambda (x) (< x 1)) (mapcar #' max (sim-exponential ) (sim-exponential 1/ )))  2509 => 0.25 theoretical
1/1 + 1/ 2 - 1/(1 + 2)
1 – R(t) = F(t) = 1 – (e-1 + e-1/2 - e-1 * e-1/2) = 0.1 and solve for t.
Let y = e-t/2 => 1 – y2 - y + y = 0 or (cubic )  ( )
e-t/2 = => t = 1
R(t) = e-1 + e-1/2 - e-1 * e-1/2 =
1/2 rd
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7. Continued
1/1 + 1/ 2 - 1/(1 + 2) /3 = 7/3 = – R(t) = F(t) = 1 – (e-1 + e-1/2 - e-1 * e-1/2) = 0.1 and solve for t for 10th percentile. Let y = e-t/2 => 1 – y2 - y + y = 0 or (cubic )  ( ) e-t/2 = => t = rd
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8. Survivor and Hazard Functions
Given the density function f(t) = 2e-2t, find the survivor and hazard lifetime distributions. rd
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9. Decomposition
Find system reliability of network. Assume Ri = 0.95 A C decompose with E working and E E not working B D RS = [2(0.95) – 0.952] [2 *0.952 – 0.954] 0.05 = = rd
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10. BRIDGE 1 4 3 2 5 1 4, 2 5, 1 3 5, 2 3 4 with 's = 1 with RS =0.99981
for t = 1 hours. \ 5 rd
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11. Systems Engineering and Analysis
Chapter 13 Design for Maintainability rd
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12. AFTER STUDYING THIS CHAPTER, YOU SHOULD BE ABLE TO:
Define and Explain maintainability;
Use Measures of maintainability – elapsed times, frequencies, labor hours, costs;
Apply availability and effectiveness factors
Apply maintainability in the life-cycle – allocation, component selection, design participation, design review
Apply maintainability analysis methods – trade-offs, level of repair analysis (LORA), maintenance task analysis (MTA) and total productive maintenance (TPM)
L E A R N I N G O B J E C T I V E S rd
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13. Maintainability
The relative ease and economy of time and resources with which an item can be retained in, or restored to, a specified condition when maintenance is performed by personnel having specified skill levels, using prescribed procedures and resources, at each prescribed level of maintenance and repair. A function of design Effective, Safe, ASAP, Least cost Minimum support (people, material, test equipment, facilities) Maintainability – design dependent parameter, kept in good condition, elapsed times, costs, frequencies rd
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14. Maintainability & Maintenance
Maintainability ~ ability of system to be maintained must be inherent or built into the design Maintenance ~ series of actions taken to restore or retain a system in specified state. Fix it. Is the result of the design. rd
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15. Maintainability
Design for maintainability requires a product that is serviceable (must be easily repaired) and supportable (must be cost-effectively kept in or restored to a usable condition) - better yet if the design includes reliability (absence of failures) Design for maintainability defines how long equipment will be down and unavailable. H(t) = 1 – R(t) = 1 – e-t = F(t) = P(T < t) for exponentially distributed repair times. rd
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16. Constant Failure Rate
Suppose a component can be repaired at a constant rate of 12 per 8-hour day. Find the probability that the time to fix one exceeds 2 hours. H(t) = 1 – R(t) = 1 – e-t/MTTF => P(T > 2) = 1 – H(t) = e-2/(2/3) = e-3 = rd
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17. Maintainability and Reliability
Inherent Availability = Ai = MTBF/(MTBF + MTTR)
Operational Availability = Ao = MTBM/(MTBM + MDT)
Achieved Availability = AA = MTBM/(MTBM + M-Bar) rd
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18. Measures of Maintainability
Corrective maintenance – restore system Preventive maintenance – scheduled maintenance to retain system in working order Adaptive maintenance – software Perfective maintenance – enhance software Downtime Mean Corrective Maintenance Time (Mct) or MTTR Failure detection, fault isolation, disassembly to gain access, repair. Active or maintenance times. Mean Active Corrective Maintenance Time (Mct) Mean Active Preventive Maintenance Time (Mpt) Availability = MTBF / (MTBF + Mct) rd
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19. Corrective Maintenance Times
(setf Table13.1 '( )) (mu-std-err table13.1)  (61.9, ) (histogram table13.1 7)  5 in (59/2 79/2) (* * * * *) 7 in (79/2 99/2) (* * * * * * *) 10 in (99/2 119/2) (* * * * * * * * * *) 12 in (119/2 139/2) (* * * * * * * * * * * *) 9 in (139/2 159/2) (* * * * * * * * *) z = (40 – 61.9)/16.3) = in (159/2 179/2) (* * * *) z = (50 – 61.9)/16.3) = in (179/2 199/2) (* * *) (del-phi )  (del-normal 61.9 (square ) 40 50)  or 50*  7 (del-normal 61.9 (square ) 46 78)  or 50*  34
X ~ N(62, 162) assumed; X-bar ~ N(62, 162/n) rd
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20. Corrective Maintenance Timesrd
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21. Stem and Leaf of Maintenance Times
Cum Stem Leaf N = 50
(phi -1.34)  0.09
(inv-phi 0.09)  -1.34
(mu-std-err table13.1)  Mct = 61.9; σ = 16.3)
Z40 = (40 – 61.9)/ = -1.34 rd
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22. Standard Z-Scores for Mcti
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23. Median
Discrete population: If sample size is odd, the middle value of the sorted population is the median. If sample size is even, the mean of the two middle values is the median. Example: (sort '( ) #' <)  ( ) => 9.5 is the median Continuous distribution: Integrate up to the median for 50% probability Example: Find the median for f(x) = 2x on [0, 1]. rd
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24. Availability
Inherent – excludes preventive, scheduled maintenance, logistics delay, admin delay. Ai is the largest availability value.
Inherent  Ai = MTBF
(MTBF +
Achieved  Aa = MTBM includes Mpreventive (MTBM )
Operational  Ao = MTBM (MTBM + MDT)
MDT = mean maintenance downtime = MTTR.
= Mean active maintenance time.
= mean active corrective maintenance time
As reliability decreases (i.e., MTTF becomes smaller), better maintainability (i.e., shorter MTTR) is needed to achieve the same availability. rd
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25. Operational Availability
Availability is given by If the failure rate is 0.01 failure per hour and the mean- time-to-repair (MTTR or MDT) distribution is continuous uniform on [2, 8] hours, find the system availability. Given: MTBF = 1/0.01 = 100; E(MDT) = (2 + 8)/2 = 5 A = 100/( ) = rd
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26. Incorrect Specifications
The system shall have an operational availability of with a MTBF of 500 hours and a MDT of 6 hours. A = 500/( ) = < (specified) conflicting specs rd
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27. Repairable Systems
Non-repairables – such as light bulbs Three models of Repair Replacement model – new item replaces a failed item Maintenance model – preventive maintenance Repair model – repaired before returning to service. Examples: "socket" model, how many bulbs to retain or redundancy allocation problem Replacement (?) when to replace? failure? age? block? rd
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28. Maintenance Models
Preventive Maintenance - action taken before an item fails Examples: change oil in car, overhaul jet engine, lubricate a bearing Corrective Maintenance – action taken upon failure Examples: charging a dead car battery, repairing a failed printed circuit board in disc drive; fixing a flat. What is optimal time for preventive maintenance? Satellite repair is too costly to use preventive maintenance. rd
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29. Failure and Repair Process
Point Process Model X(t) = 0 if item failed at time t and = 1 if item is working at time t. x o x o x o x o X1 R1 X2 R2 X3 R3 X4 R4 Xi = ith time to failure and Ri the ith time to repair. x ~ failure times; o ~ time when repaired and returned to service
Poisson process rd
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30. Point Process
Poisson Process ~ when time to repair or replace is negligible compared to uptime. For example, when car is in shop an hour for an oil change (pit stop with no miles accrued) vs. an aircraft in a hangar for months having engines overhauled. Applicable to queue arrivals, earthquakes, crime, strikes, bankruptcies, wars, accidents. The s for an auto are many; for a light bulb is few. rd
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31. Non-repairable vs. Repairable
Non-repairable Repairable Item gets better with time Burn-in Improving Item gets worse with time Wear-out Deteriorating Stationary processes (Poisson) vs. independent increments rd
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32. Partitioning Availability
Time to detect failure Time to diagnose the problem Time to obtain parts and labor Repair time Testing time 0 1 1 rd
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33. Queuing Model
Breakdowns occur on average at 2 per day (Poisson) and each can be repaired at an average of 1/3 day (exponential). At a FCFS, what is the average number of items down at any time and the average waiting time before maintenance crew can begin service?  = 2/3 L = /(- 1 ) = (2/3)/(1/3) = 2 items Wq = /( - ) = 2/3(3-2) = 2/3 day (MM1 2 3) 
(MM1 2 3)
P =
W =
WQ =
L =
LQ =
UTILIZATION =
IDLE = rd
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34. Failure Rate
Failure Rate = number of failures = N operating time TT – Not-time 50 artificial heart valves were tested for 10,000 hours at a lab and 3 failed. a. What was the failure rate in %? b. number of failures per unit year? c. How many are expected to fail during a year from putting the valves into 100 patients? d. Find the MTBF. a) 3/50 = 6% b) 3/(10000*50 - 3*5,000) = /hr; assuming 3 failures occurred at average of 5,000 hours /yr c) * 100 = 5.42 d) MTBF = unit-yr/failure
1/5.42 = 18.46 rd
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35. Log-normal
Log-Normal Cumulative distribution function Erf (x) = (- (* 2 (phi (* x (sqrt 2)))) 1))
erf(x) = 2/sqrt(pi) * integral from 0 to x of exp(-t^2) dt. rd
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36. Maintenance Times
Mean Preventive Maintenance Time (MT) M-barpt  Median Active Corrective MT (Mct) Median Active Preventive MT (Mpt ) Mean Active MT M-bar Maximum Active Corrective MT (Mmax) Logistics Delay Time (LDT) Administrative Delay Time (ADT) (strikes, reassignments) Maintenance Downtime (MDT) total downtime rd
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37. Maintenance Labor Hours
MLH/OH MLH/cycle MLH/month MLH/MA rd
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38. Maintenance Frequency Factors
MTBM ~ mean time between maintenance a  b MTBM = 1 1/MTBMa + 1/MTBMb MTBR ~ mean time between replacement rd
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39. Maintenance Cost Factors
Cost per maintenance action ($/MA) Maintenance cost per system operating hour ($/OH) Maintenance cost per month ($/month) Maintenance cost per mission segment ($/mission) The ratio of maintenance cost to actual life cycle cost rd
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40. System Effectiveness (SE)
Probability that system performs as intended
Cost Effectiveness (CE) System benefits / life-cycle cost
System capacity / life-cycle cost
SE / life-cycle cost
Availability / life-cycle cost
Supportability / life-cycle cost
MTBM MTBR M-bar Mpt MLH/OH $/MA
Cost per maintenance hour labor hour / operating hour rd
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41. Maintainability Allocation
Suppose inherent availability probability = with MTBF = 450 hours; MLH/OH = 0.2; Mct = ? Mct = MTBF(1 – Ai) / Ai = 450 (1 – ) / = 0.5 hours  allocated to the components. continued  rd
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42. Maintainability Allocation
Item Q
Failure-rate Cf Cp
Mct
Ct = CfMct
Unit A 1
0.246
11%
0.9
0.221
Unit B
1.866
84%
0.4
0.746
Unit C
0.110 5%
1.0
Total
2.222
100%
1.077
Table13.7
Mct is Mean Corrective Maintenance time
Ct Contribution of total corrective M-time
Mct for System = / = hr (0.5 hr needed) Unit B's 0.4 hours is allocated to its components. Cc contribution of total corrective maintenance times rd
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43. Demonstration Method I
Failure is induced with only naturally occurring evidence
Technician called in to repair.
Repairs in approved way
Data is collected on all operations to note delays, inadequacies, shortages, performance, test procedure.
The test is repeated on n simulated failures and an example of the allocation is shown in Table13.14. rd
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44. Demonstration 2
Given maintenance times, find M-bar, M-barct M-barpt Mmax Upper limit = M-barct + z/ (sqrt n) rd
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45. P13-27 Task Mcti Variance 1 58 16 2 72 100 3 32 900 … Total 3,105
= 3105/50 = 62.1 < 65
V( ) = 15,016/50 =
Mct = * 17.33/7.07
= 66.1 > 65 => Reject at 5%; OK at 20%
Specs: Mbar = 75 min
Mbar-ct = 65 min
Mbar-pt = 110 min
Mmax = 120 min
Producer's risk () = 5%
Task
Mcti
Variance 1 58 16 2 72
100 3 32
900 …
Total
3,105
15,016 rd
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46. P13-7
With availability set at and a calculated MTBF of 400 hours, find the mean corrective maintenance time Mct. Mct = MTBF(1 – Ai) / Ai = 400(1 – 0.99) / 0.99 = 4.04 hours rd
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47. P13-8
Determine Given Solutions-Answers Ai MTBM  = MTBF = 1/ = 250 hrs Aa MTBF TOT = 10K hrs Mbar = 50 – 30 = 20 hrs Ap M-bar MDT = 50 hr 10K * = 40 => 10 = PM Mbarct MTTRg TNoMA = = Mbarct *.004+6*.001/.005 M-barp = 6 hr Mbarct = 23.5 hrs Mean logistics + admin = 30 hr MTBM = 10000/ 50 = 200 hrs Ai = 250/( ) = Aa = 200/(200+20) = A0 = 200/(200+50) = 0.8
Ai ~ inherent availability See Page 436
Aa Achieved
Ao Operational availability
See
ToT = total operational time rd
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48. P13-11 Item Q  Contribution Cf = Q % Contribution Cp =Cf/Sum Mct *
(minutes) A 1
0.05
0.07 2
1.00 B
0.16
0.32
0.42
0.6
0.192 C
0.27
0.35
0.9
0.243 D
0.12
1.5
0.180
Total
0.76
0.715
Contribution = Cf * Mct
M-barct = 0.715/0.760 = 0.94 < 1 hr assigned to ABC => some relaxing
* Estimated times from inherent characteristics of design rd
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49. P13-25
(setf p13-25 '( )) (mu-std-err p13-25)  ( ) H0:   65 vs. H1:  < 65 with  = 10%, n = 50, s = X-bar = Z = 7.07(61.48 – 65)/16.25 = < p-value = <  = 10%, => Reject rd
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50. Unit B Allocation
Item Q
F-rate Cf Cp
Mct
Ci = CfMct 1
0.116 6%
0.5
0.058 2
1.550
83%
0.4
0.620 3
0.200
11%
0.3
0.060
Total
1.866
100%
0.738
Mct for Unit B = / = hr (0.4 hr needed) rd
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51. Component Selection and Application
Use standardized components (COTs); standard tools Built-in test (BITE) for quick, easy diagnostics for time consuming tests Ensure accessibility for remove and replace and modular independency Seek modular, physical and functional interchangeability of parts. Avoid preventive maintenance needs. Prioritize locations using time to repair and accessibility for high failure rate items or critical components Label components rd
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52. Maintainability Analysis
Tradeoff between reliability and maintainability -- Current need: 8 hrs daily operation, 360 days/yr for 10 years. Ai = 0.961, MTBF = 125 hrs, Mct = 5 hrs
New: Ai = 0.990, MTBF > 300, Mct < 5 12 10 8 6 4 2
Ai = 0.961
Mct
Ai = 0.990
Existing design
Configuration Ai
MTBF
Mct
Existing
0.961
125
5.0 A
0.991
450
4.0 B
0.990
375
3.5 C
320
2.8
Trade-off Area
Ai = 0.995
Page 446 of text for A, B, and C designs competing for contract
MTBF
Reliability MTBF, hr
Ai = MTBF/(MTBF + M-barct ) rd
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53. Prediction of Mct Table 13.10
Maintainability Prediction Worksheet (Assembly i)
Maintenance Times (Hr)
Part  N N Loc Iso Acc Ali Cck Int Mct NMct A B
C …
Total 
Underlined constitute a maintenance cycle time
Int ~ interchange
Locate isolate access align check interchange
N ~ number of parts rd
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54. Maintainability Analysis
Maintainability Prediction Reliability-Centered Maintenance (RCM) Level-of-Repair Analysis (LORA) – repair or replace Maintenance Task Analysis (MTA) – what's needed Total Productive Maintenance (TPM) Maintainability Demonstration – ORI Maintainability Assessment – of MTBM. MDT, MLH/OR rd
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55. Downtime
Logistics Delay Time (LDT) Administrative Delay Time (ADT) Maintenance Downtime (MDT) rd
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56. Mean Corrective Maintenance Time
= MTTR = rd
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57. Mean Preventive Maintenance Time pt=
fpi is the frequency of the ith preventive maintenance action per system operating hour
Mpt is elapsed time needed for the ith action to exclude logistic delay and administrative delay times. rd
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58. Median Active Corrective Maintenance Time
= antilog rd
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59. Median Active Preventive Maintenance Timerd
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60. Mean Active Maintenance Time
fpt is the preventive maintenance rate rd
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61. Maximum Active Corrective Maintenance Time (Mmax)
Mean of the logarithms of Mct; Z ~ to standard deviation, sample logarithms of average repair times rd
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62. Total Productive Maintenance (TPM)
Integrated life-cycle approach to maintaining a manufacturing plant Overall equipment effectiveness (OEE) OEE = availability (A) * performance rate (P) * quality rate (Q) Availability = loading time – downtime loading time Loading time – total time available to make products P = output * actual cycle time * ideal cycle time loading time – downtime actual cycle time Q = [input – (quality defects + startup defects + rework)] / input rd
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63. Summary
Maintainability Analysis Interrelationship of Maintainability with Reliability Systems Life-Cycle Approach rd
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64. Problem 13-5
((41 . 2) (39 . 3) (47 . 2) (35 . 5) ( ) ( ) (33 . 6) ( ) ( ) (37 . 4) ( ) (36 . 5) (31 . 7) (13 . 3) (11 . 2) (15 . 8) (29 . 8)(21 14)
( )
( )
Range = 47 – 11 = 36
(setf P13-5 (flatten (mapcar #' list-of '( ) '( )))) 
( )
(Range P13-5)  36; rd
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65. 13 in (21/2 31/2) (* * * * * * * * * * * * *)
(Histogram P )
Problem 13-5 histogram N = 126
13 in (21/2 31/2) (* * * * * * * * * * * * *)
24 in (31/2 41/2) (* * * * * * * * * * * * * * * * * * * * * * * *)
37 in (41/2 51/2) (* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *)
18 in (51/2 61/2) (* * * * * * * * * * * * * * * * * *)
18 in (61/2 71/2) (* * * * * * * * * * * * * * * * * *)
12 in (71/2 81/2) (* * * * * * * * * * * *)
2 in (81/2 91/2) (* *)
2 in (91/2 101/2) (* *)
0 in (101/2 111/2)
Log Normal Distribution (skewed right)
(mu-std-err p13-5)  ( )
(median p13-5)  23
(mu-std-err (mapcar #' log p13-5))  ( ) * Using Log Base e
(geometric-mean P13-5) 
(exp (mu (mapcar #' log p13-5)))  = Mct = Geometric Mean
(std-err (mapcar #' log p13-5 (list-of )))  * Using Log base 10
Mmax = antilog ( * 0.138) where 1.38 is (antilog and = antilog is (inv-phi 90) and
= 36 minutes is antilog rd
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66. Problem 13-6
(setf P13-6 (flatten (mapcar #' list-of '( ) '( ))) 
( )
The boundaries of the class intervals are:
6 in (17/2 27/2) (* * * * * *) NORMAL
10 in (27/2 37/2) (* * * * * * * * * *) Appearing
35 in (37/2 47/2) (* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *)
22 in (47/2 57/2) (* * * * * * * * * * * * * * * * * * * * * *)
17 in (57/2 67/2) (* * * * * * * * * * * * * * * * *)
3 in (67/2 77/2) (* * *)
(range p13-6)  28; (mu-std-err p13-6)  ( ) rd
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67. Problem 13-7
With a specified inherent availability of 0.99 and a calculated MTBF of 400 hours, find the mean corrective maintenance time. Mct = MTBF(1 - Ai)/Ai = 400(1 – 0.99)/0.99 = 4.04 hours rd
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68. Problem 13-8 Given  = 0.004, total operation time = 10,000 hours,
Mean downtime = 50 hours, Total number of maintenance actions = 50 and mean preventive maintenance time = 6 hours, and mean logistics plus administrative = 30, find
MTBF = 1/ = 1/0.004 = 250 hours
M-bar = 50 – 30 = 20 hours (assumption) rd
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69. Problem 13-9
Given Mct = 0.5 hour; MTBMa = 2 hours Mpt = 2 hours; MTBMs = 1000 hours; compute the achieved availability. 1 MTBM = 1/MTBMa + 1/MTBMs = 1/(1/2 + 1/1000) = hours  = 0.5 and frequency of preventive fpt = M-bar = (0.5 * * 2)/( ) [See 13.8] = hours Aa = MTBM/(MTBM + M-bar) = 1.966/( ) = 0.796
a is unscheduled maintenance; s is scheduled maintenance rd
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70. Problem 13-12
System operates 40 hours per week for 50 weeks a year for 15 years. MTBF is 400 hours and mean corrective time is 2 hours. Two techs are assigned for the duration of each maintenance action. Find the MLH/OH. (40 * 50 * 15)/400 = 30,000/400 = 75 maintenance actions. 75 * 2 hours * 2 techs = 300 MLH/30,000 OH = 0.01 MLH/OH rd
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71. P13-26
Mpt = 120 minutes required. (setf p13-26 '( )) (mu-std-err p13-26)  ( ) Z = H0:   120 vs. H1:  < 120 with  = 10%, n = 50, s = 26.68, X-bar = => Xcritical = * 26.68/7.07 = ; (inv-phi 0.10)  Z = 7.071( – 120)/26.68 = < => p-value = => Reject * / 7.07 = < 120 => Passed Test (depict p13-26)
(U-normal 120 (/ (square 26.68) 50) )  % rd
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72. P13-27 Task Mcti Variance 1 58 16 2 72 100 3 32 900 … Total 3,105
= 3105/50 = 62.1 < 65
V( ) = 15,016/50 =
Mct = * 17.33/7.07
= 66.1 > 65 => Reject at 5%; OK at 20%
Specs: Mbar = 75 min
Mbar-ct = 65 min
Mbar-pt = 110 min
Mmax = 120 min
Producer's risk () = 5%
Task
Mcti
Variance 1 58 16 2 72
100 3 32
900 …
Total
3,105
15,016 rd
5/17/2019

73. Reliability of a System
Compute reliability of system. Compute reliability if system redundancy is used. Ans ,
0.95
0.98
0.90 rd
5/17/2019

74. Preventive Maintenance Policy
Find expected number of breakdowns (bds). Each bd cost $280. For a cost of $150/month, a firm guarantees 1 bd/month. Is this a good deal? Σ # of breakdowns # months (freq) probability 1/12 1/3 5/12 1/8 1/24 E(bds) = 1.71 => costing 1.71 * $280  $479 (wo contract) = $430 (w contract) $ 49/month (EV '(1/12 1/3 5/12 1/8 1/24) (upt0 4))  rd
5/17/2019