Msa University - Maintenance Planning And Control Final Exam 2009

MSA tsB424 Faculty of Engineering Maintenance Planning & Control Final Term -Spring 2009 Time: 3 Hours Answer all of the following questions: Ouestion I (Total Mark: 5) A maintenancejob has been studied for 40 observations.The actual mean time is 7.3 minutes and the standarddeviation of the time is estimatedto be 2.5 minutes. How many additional observationsshould be taken for 98.8% confidence that the actual mean time hasbeendetermined withtn l2yo.

Ouestion2 (a) What arethe two main typesof preventivemaintenance?

(Total Mark 5)

(b) what are the two types of condition-basedpreventive maintenance? (c) What are the two types of statistically and reliability-basedpreventive maintenance? (d) Discussin your own words the relationship betweenthe three maintenancefunctions of load forecasting,capacity planning and scheduling. (e) what is maintenanceload composedof? Is it certain or uncertain? (f) Define "preventive maintenance"in your own words. Question 3 (Total Mark: g) Consider thefollowingyearlydatafor maintenance loadin labor-hours for theiast6 vears. Year

I

2

a J

4

5

6

Maintenance Load

700

900

8s0

l 100

I 150

I 300

(a) Find the maintenance load forecast for years 7, 8 and 9 using the three-period moving averageand the linear regressionmodel. (b) Determine which method is more accurateusing the mean square error (MSE) basedon the years4,5 and 6. ouestion 4 (Totar Mark: g) Assume we have four time periods with maintenanceloads 300, 150,200 and 250 respectively.The in-houseregularcapacityis 120, 250,200 and 150 for periods 1,2,3 and 4 respectively.The overtime is at most2}Yo of the in-house capacity. Subcontracting has no limit. The cost of performing one in-house man-hour ii taken to be I unit, overtime per man-hour costs 60% more than regular time (i.e. 1.6 units) and subcontractingcosts2.2 wits. Backlogging of one man-hour costs n : 0.2 per period. (a) Use the least cost heuristic metlod to allocate the maintenance load in the four time periods to the available types of resources. (b) Evaluatethe total cost of the plan developedin part (a).

Pagelof 2

Ouestion 5

(Total Mark: 7)

An analysthas observeda maintenancejob long enoughto becomefamiliar with it and has divided it into five elements.The elementtimes (in minutes)for the first 5 cyclesare shown in the following table,along with the performancerating for eachelement.

Cycle

Element Number

Performance Rating (%)

1.5

1.3

1.5

1.6

r.4

80

2.5

2.0

z.)

2.4

1 a

z.J

110

1.8

2 .0

3.0

1.8

1.9

9s

1.3

1.1

1.4

r.2

1.2

90

t.7

1.5

1.6

1.7

1.8

100

(a) Computean estimatedbasictime for thejob on the basisof the availabledata. (b) On basisof the mean and standarddeviationof the availableobserveddata of the job as a whole (not broken down to elements),how many additionalcycles should be observedfor 96.6ohconfidencethat the actualmean time has been determined within 5%. Question6

(Total Mark: 7)

Considerthe following quarterlydata for the maintenanceload in labor-hoursfor the last 5 years.

Quarter II

m

215

190

170

250

230

220

185

270

2s0

225

2t0

28s

255

245

230

295

280

260

240

320

Year

(a) Determine an appropriateseasonalindex for each quarter. (b) Deseasonalize the dataand fit them to a linear regressionmodel. (c) Predict the quarter values for 6 and 7 years. Good Luck!

2c5\-\ t

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,

n}t' -[]'J

[rf {t)at tll

,=(Z"Po\'

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MSE =

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A= ; - 6 i

STANDIRD STATISTICAL 1. Areaq, under

the

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TABLES

Norroal Distribution

Ibe talle gives tb mlative pmbability up to tie stailhdis€d lonrl valuez i"e. I

P l l < ? l

Ilrrp(+s:) dl P[z
0.03

0.00

0.01

nffi

0.0 0.1 0.2 0.3 0.{

0.5000 0.50{0 0,5080 0.5120 0.5159 0.5199 0.5239 A.5219 0,5319 0.5359 0,5398 0.5{38 0.5178 0.5517 0.5557 0.5596 0.5635 0,5675 0.571t 0.5753 0.5793 0.5832 0,58?1 0.5910 0.59{8 0.598? 0.5025 0,6061 0.6103 0.51.{1 0.6179 0.6217 0.6255 0.6291 0.6331 0.5358 0.6106 0.6{{3 0.6{S0 0.651? 0.6551 0.6591 0.5528 0,6561 0.6700 0i6736 0.67?2 0.5808 0;68d4 0.58?9

0.5 0.6 0,7 0;8 0.9

0.6915 0.6950 0.6985 0.7019 0.705t 0.7088 0.7123 0.715? 0.7190 0.7224 0,7257 0.7291 0.732{ 0,7357 0.n$ 0,71n 0,7{5{ 0.7486 0.?517 0.7519 017580 0.7611 0.'16t2 0.?573 0.7701 O"nU ,0.??61 0.?791 0.?823 0.?854 0.7841 ,0;7t10 0.7939 A,79670.7995 0180?30.8051 0,8078 0.8105 0.8133 0.8159 0.81s6 0.8212 0.8238 0.E26{ 0.8289 0;8315 0.8310 0.8355 0.8389

t,0 l.l 1.2 1.3 l.{

0.8113 0.s{38 0.8151 0.8{85 0.S50S0.S531 0.855{ 0.S5?7 0.S599 A.S62L 0,86{3 0.8665 0.8685 q.S70S0,8729 0,87{9 0.8770 0.8?90 O'.sS0t0.ss30 0.8819 0.8869 0.8888 0.8907 0.3925 0189110.8962 0.89S0 0.899? 0.9015 0.9032 0.90{9 0.S66 0.9082 0.9099 0.9115 0,9131 0.91{? 0.e162 0.917? 0.9192 0.9207 0.92n 0.9236 0.981 0.9265 0,9279 0,9292 0.9306 0.9319

1.5 1.5 l.? 1.8 1.9

0.9332 0.93{5 0.e{52 0.9{63 0.95st 0.956t 0.9511 0.96{9 0.9713 0.9n9

2.0

0.97n

0.9357 0.93?0 0.9382 0;9391 0.9{06 0.9{18 0.9129 0.9{t1 0.9{7{ 0.9{8t 0.9t95 0.9505 0.9515 0,9525 0.9535 0.95{5 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 0.9656 0.966{ 0.9671 0.95?8 0.9685 0.9693 0.9699 0.9?06 0.9726 0.n32 0.9?38 0.97{{ 0.9750 0.9?55 0.9761 0.9'16't

0.9778 0.983

0.9788 019793 0.9798 0.9803 0.9808 0.9812 0.98t7

2-t 0.9821 0.9826 0.9830 0.983{ 0,9838 0r98t2 0.9816 0.9850 0.985{ 0.9857 z;r--0-it6t-0.988--T3Ffl 0,9E7r 0,98?t 0.9878 0.9881 0.98E1 0.9887 0.9890 2.3 0.9893 0.9995 0.9898 0.9901 0,990{ 0;9906 0.9909 0.9911 0.9913 0.9916 2.1 0.9918 0.9920 0.99n 0.9924 0.992? 0.9929 0.9931 0.9932 0.993t 0.9936 2..5 2:6 2,7 2.8 2.9

0.9938 9.9910 0.9941 0.99{3 0.99{5 0i9916 0.9918 0.9919 0.9951 0.9952 0.9953 0.9t55 0.9955 0.9957 0.9959 0.9950 0.9961 0.9962 9:9953 0.996{ 0,9965 0.9965 0.9967 0.9968 0.9969 0.99?0 o.gyn 0,9972 0.9973 0;997{ 0.997{ 0.99?5 0.e9?6 0.9977 0.9977 0.99?8 0.99?9 0.e980 0.9980 0.9981 0.9e81 0,9982 0.9982 0.9983 0.998{ 0.99St 0.9985 0,9985 0.9986 0.9985

r 3.00 3.10 3.20 3.30 3.{0 3.50 3.50 3.?0 3.80 3.90 P 0.9986 0.9990 0.9993 0.9995 0.999? 0.9998 0.9998 0:9999 0.9999 1.0000

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