193883462 03-part-3-pm-case-studies

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PREVENTIVE MAINTENANCE PLANNINGP2: The yearly PM programs information for six similar gas

turbines in a power station are as follows:

1- PM information:

Maintenance levels per gas turbine

PM Type Frequency Duration No. ofWorkers

Spare parts Cost

$1000Y– Level 1 Yearly 15 days 20 10S– Level 2 6 Monthly 10 days 20 83M– Level 3 3 Monthly 5 days 15 5M– Level 4 Monthly 2 days 10 2

2- Working conditions:

------------------------------------------------------------------------------------------Part 3: PM Case Studies - 1 -

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Gas turbine operating conditions: 24 hour/day Workers operating conditions: 300 day/year & 8 hour/day

3- CM information:

Average effort of CM = 380 man-day per gas turbine Average annual spare parts CM = $ 12000 per gas turbine Average CM downtime = 15 days/year per gas turbine Average downtime cost rate = $ 1000 per day

4- Cost rates:

Average labor cost rate = $ 10 per man-day Overhead cost = 25 % direct cost (spare parts & labor)


Required:

1) The size of maintenance labor force.2) Average system availability.3) Annual downtime cost losses.4) Annual maintenance cost.5) Annual PM plan.6) Maintenance resource profiles.7) Monthly PM plans.8) Maintenance work order

The size of maintenance labor force

PM Type

AnnualFrequency

Duration(day)

No. of Worker

Man-dayper PM type

Y 1 15 20 300 * 1= 300S 1 10 20 200 * 1 = 200

3M 2 5 15 75 * 2 = 150M 8 2 10 20 * 8 = 160

Annual PM man-day per gas turbine 810

Total PM annual man-day Required 810 * 6 = 4860

The size of PM labor force = 4860/300 =16.2 = 17 workers

The size of CM labor force = 380 * 6 / 300 = 8 workers

Total labor force = 17 + 8 = 25 workers

Crew check is ok (25 more than 20).


The average down time per year

PM Type AnnualFrequency

Duration(day)

PM Downtime(day)

Y1 15 15 * 1= 15

S 1 10 10 * 1 = 10

3M 2 5 5 * 2 = 10

M 8 2 2 * 8 = 16

PM downtime per gas turbine 51

Average down time = 51 + 15 = 66 day/year per gas turbine

Annual downtime cost losses = 66 * 6 * 1000 = $ 396000

Average equipment availability =Active operating time / Total time

= (364 – 66) / 364 = 82 %

System Reliability:Series or chain structure: Rs = R1 * R2 * R3 * … etc.Parallel structure: Rs = 1 –(1-R1)* (1-R2)* (1-R3) * .etc.

System time availability =Parallel structure: As = 1 – (1-A1)**6

= 1 – (1-0.82)**6 = 1 – (0.18)**6 = 99%


Annual maintenance cost

PM Type AnnualFrequency

Cost$1000

Spare parts PM Cost $1000

Y 1 10 10 * 1= 10S 1 8 8 * 1 = 8

3M 2 5 5 * 2 = 10M 8 2 2 * 8 = 16

Annual spare parts PM per gas turbine = 44Total annual spare parts PM cost = 44 * 6 = 264

The average annual spare parts CM cost =

$ 12000 * 6 = $ 72,000

Annual spare parts maintenance cost =

264000 + 72000 = $ 336,000

Annual labor cost = 25 workers * 300 day/year * $ 10 per man-day= $ 75,000

Annual direct maintenance cost = $ 336000 + $ 75000= $ 411000

Overhead cost = 25 % direct cost

Annual maintenance cost = $ 411000 * 1.25 = $ 513750

Annual maintenance cost = $ 513750Annual downtime cost losses = $ 396000


Basic Annual PM Plan

Eq.code

Month #1 2 3 4 5 6 7 8 9 10 11 12

G01 Y M M 3M M M S M M 3M M M

G02 M M Y M M 3M M M S M M 3M

G03 M 3M M M Y M M 3M M M S M

G04 S M M 3M M M Y M M 3M M M

G05 M M S M M 3M M M Y M M 3M

G06 M 3M M M S M M 3M M M Y M

Resource analysis:

Man-day 580 230 580 230 580 230 580 230 580 230 580 230

Day/month 24 24 24 24 24 24 24 24 24 24 24 24

Workers 24 10 24 10 24 10 24 10 24 10 24 10

SP cost 26 18 26 18 26 18 26 18 26 18 26 18

DT 33 18 33 18 33 18 33 18 33 18 33 18

Y= 300 S= 200 3M= 75 M= 20 man-dayY= 10 S= 8 3M= 5 M= 2 $1000Y= 15 S= 10 3M= 5 M= 2 day


Target Annual PM Plan # 1

Eq.code

Month #1 2 3 4 5 6 7 8 9 10 11 12

G01 Y M M 3M M M S M M 3M M M

G02 M M Y M M 3M M M S M M 3M

G03 M 3M M M Y M M 3M M M S M

G04 M S M M 3M M M Y M M 3M M

G05 3M M M S M M 3M M M Y M M

G06 M M 3M M M S M M 3M M M Y

Resource analysis:

Man-day 455 355 455 355 455 355 355 455 355 455 355 455

Workers 19 15 19 15 19 15 15 19 15 19 15 19

SP cost 23 21 23 21 23 21 21 23 21 23 21 23

DT 28 23 28 23 28 23 23 28 23 28 23 28



Target Annual PM Plan # 2

Eq.code

Month #1 2 3 4 5 6 7 8 9 10 11 12

G01 Y M M 3M M M M S M 3M M M

G02 M M Y M M 3M M M M S M 3M

G03 M 3M M M Y M M 3M M M M S

G04 M S M 3M M M Y M M 3M M M

G05 M M M S M 3M M M Y M M 3M

G06 M 3M M M M S M 3M M M Y M

Resource analysis:

Man-day 400 410 400 410 400 410 400 410 400 410 400 410

Workers 17 17 17 17 17 17 17 17 17 17 17 17

SP cost 20 24 20 24 20 24 20 24 20 24 20 24

DT 25 26 25 26 25 26 25 26 25 26 25 26



Monthly Maintenance Plan: Month # 1Day G01 G02 G03 G04 G05 G06 PM worker

1. Y 202. Y 203. Y 204. Y 205. Y 206. Y 207. Y 208. Y 209. Y 2010. Y 2011. Y 2012. Y 2013. Y 2014. Y 2015. Y 2016. SB -17. M 1018. M 1019. SB -20. M 1021. M 1022. SB -23. M 1024. M 1025. SB -26. M 1027. M 1028. SB -29. M 1030. M 1031. SB -


MAINTENANCE WORK ORDER010120

Requester Section:Power Station PS03 - Gas Turbine G01 - Priority: AMaintenance type/level: Annual PM1- Check ….2- Clean …..3- Replace …..4- Adjust ……5- Repair …..

Eng. Attia GomaaPlanning Section:

Labor: 4 Mech. 2 Helper 5 days 5 Elec. 4 Helper 10 daysSpare parts: 2 valve xx1, 4 air filter yy3, .. etc.Special tools: xxx, yyyy, … etc,Expected down time (from 01/01 to 15/01/2004)Cost estimation ($ 10,000)Safety instructions:- Check … Eng. Aly Ahmed

Craft Feedback:1- Check ….2- Clean …..3- Replace …..4- Adjust ……5- Repair …..Labor: 3 Mech. 2 Helper 5 days 6 Elec. 3 Helper 11 days 1 Vib. 1 Helper 2 days Spare parts: 2 valve xx1, 4 air filter yy3, .. etc.Special tools: Vibrometer, … etc,Down time (01/01 to 17/01/2004) Actual Cost ($ 12,000)

Eng. Omer AlyCoding:


P3: The yearly PM programs information for four similar gas turbines (GT01 through GT04) in a power station are as follows:

Maintenance levels per gas turbinePM Level Frequency PM

TypeMaint.

durationNo. of

workersSpare parts

cost $1000Y – Level 1 Yearly Shutdown 14 day 20 10S – Level 2 6 Monthly Shutdown 7 day 15 83M – Level 3 3 Monthly Shutdown 4 day 10 5M– Level 4 Monthly Shutdown 2 day 8 2W – Level 5 Weekly Running 5 hours 2 --D – Level 6 Daily Running 1 hours 2 --

Gas turbine operating conditions: 24 hour/day Workers operating conditions: 300 day/year & 8 hour/day Average effort of CM = 380 man-day per gas turbine Average annual CM cost = $ 12000 per gas turbine Average CM downtime = 15 days/year per gas turbine Average downtime cost rate = $ 1000 per day Labor cost rate = $ 10 per man-day Overhead cost = 25 % direct cost (spare parts & labor)

Required:1. The size of maintenance labor force.2. Average system availability.3. Annual PM plan.4. Annual downtime cost losses.5. Annual maintenance cost.


P4:The yearly PM programs information for six similar fire tube

boilers (FB01 through FB06) in a textile company are as

follows:


Preventive Maintenance Program:Daily

(Running)Weekly

(Shutdown)Monthly

(Shutdown)S. Annually(Shutdown)

Annually(Shutdown)

Check water level

Check for tight closing of fuel valves

Inspect burnerClean low water cut-off(s)

Clean fireside surfaces

Blowdown boiler

Check fuel and air linkage

Analyze combustion

Check oil preheater Clean breeching

Blowdown water column

Check indicating lights and alarms Check cams Inspect

refractoryClean waterside surfaces

Check combustion visually

Check operating and limit controls

Inspect for flue gas leaks

Clean oil pump strainer and filter

Check oil storage tanks

Treat water according to the established program

Check safety and interlock controls

Inspect for hot spots

Clean air cleaner and air/oil separator

Check fluid levels on hydraulic valves

Record boiler operating pressure/ temperature

Check low water cutoff(s) operation

Review boiler blowdown procedures

Check pump coupling alignment

Check gauge glass

Record feed water pressure/ temperature

Check for leaks, noise, vibration, unusual conditions

Check combustion air supply

Reset combustion

Remove and recondition safety valves

Record flue gas temperature

Check operation of all motors

Check all filter elements

Inspect mercury switches

Check oil pumps

Record oil pressure and temperature

Check general burner operation

Check fuel systems Check boiler feed

pumps

Record gas pressure

Check lubricating oil levels

Check belt drives

Check condensate receivers

Record atomizing pressure

Check flame scanner assembly

Check lubrication requirements

Check chemical feed systems

Check general boiler/burner operation

Check packing glands

Tighten all electrical terminals


Maintenance levels per boilerPM Level Frequency Maintenance

duration (hour)No. of

workersSpare partscost $1000

Y – Level 1 Yearly 160 6 4S – Level 2 6 Monthly 60 3 2M– Level 4 Monthly 20 2 1W– Level 5 Weekly 10 2 0.5

Boiler operating conditions: 24 hour/day Workers operating conditions: 300 day/year & 8 hour/day

Average effort of CM = 250 man-hour per boiler Average annual CM cost = $ 2000 per boiler Average CM downtime = 120 hours/year per boiler

Average downtime cost rate = $ 1000 per day Labor cost rate = $ 10 per man-day Overhead cost = 25 % direct cost (spare parts & labor)

Required:

1. The size of maintenance labor force.

2. Average system availability.

3. Annual PM plan.

4. Annual downtime cost losses.

5. Annual maintenance cost.


P5: The yearly PM programs information for some machines in a company are as follows:

M/cType

Numberof M/C

Maintenance level for each machine

Weekly Monthly YearlyMan Hour Man Hour Man Hour

Turning 6 3 3 - - 5 24Grinding 3 2 3 4 12 - -

M/cType

Maintenance Spare Parts Cost for each machine

Weekly level Monthly level Yearly levelTurning 400 - 3000Grinding 500 1600 -

(Maint. working conditions: 52 week/year, 6 day/week, 8 hr/day)

Required:

1. The size of PM labor force.

2. Annual PM cost.

3. Annual PM plan.

4. Average down time per month.

5. Average system availability.


P6: The yearly Reliability and PM programs information for

some machines in a company are as follows:

M/cType

Numberof

M/cs

Maintenance level10 days 6 Monthly Yearly

W D O W D O W D OLath 4 3 1 10 4 2 20 5 4 20

Drilling 2 2 1 10 4 2 20 - - -

Grinding 1 - - - 3 2 20 3 3 20

W: Number of workers D: Maint. duration (day) O: Oil (liter)

- Working conditions:

Operating: 52 week/year, 6 day/week, two shifts

Manpower: 45 week/year, 6 day/week, one shift


T0.95

T0.9

D0.95

D0.8

T0.9

T0.8

G0.9

Annual CM analysis per machine Lath Drilling Grinding

Average Manpower (workers) 2 2 2Average oil quantity (liter) 100 60 80Average Spar parts costs ($/year) 2000 1000 1000Average Down time (day/year) 20 12 10

- Oil cost is $ 5 /liter- Average labor rate is $ 10 / man-hour- Average Indirect costs (or Overhead) are 40 % direct

maintenance cost- Average down time cost rate of the PM is $ 100 / machine-

hour, whereas that of CM is $ 300 / machine-hour.

Required:1. The size of maintenance labor force.2. Annual oil quantity.3. Annual maintenance budget.4. Annual maintenance plan with the best manpower

profile.5. Annual down time for each machine.6. Average availability for each machine.7. Annual down time cost for each type. 8. System availability. 9. System reliability.


P7: The yearly PM programs information for a production

line in a shop are as follows:

M/c Type

Numberof

machines

PM Maintenance levels per machine10 days 6 Monthly Yearly

Man Day Man day Man dayTurning 3 3 1 4 2 5 4Milling 2 2 1 - - 4 3Drilling 1 2 1 4 2 - -Grinding 1 - - 3 2 3 3

M/c Type PM Spar parts and materials cost (L.E.)

10 days 6 Monthly YearlyTurning 400 1500 3000Milling 600 - 4000Drilling 500 1600 -Grinding - 1000 2000

- Working conditions:Production & maintenance 52 week/year, 6 day/week, two

shifts.Manpower 45 week/year, 6 day/week, one shift.


T Standby

T T M

M Standby

D G

- Average labor rate is 10 L.E./ man-hour

- Average other costs are 40 % direct maintenance cost

- Average CM Spare parts costs are 15 % PM spar parts

costs.

- Average CM manpower are 30 % PM manpower.

- Average down time cost rate is 100 L.E./machine-hour.- Average CM down times per machine (day /year) are as follows:

Machine Turning Milling Drilling Grinding

Down time 15 10 12 8

Required:

1. The size of maintenance labor force.

2. Annual maintenance cost.

3. Annual maintenance plan with the best manpower

profile.

4. Average down time for each machine type.

5. Average availability for each machine type.

6. Average system availability

7. Average down time cost for each machine type.


MAINTENANCE FORECASTING

Y = a + b X

Sum Y = n . a + b Sum X , Sum XY = a Sum X + b Sum X2

Measuring the accuracy of forecasting:1- Coefficient of correlation

r =( n Sum XY - Sum X Sum Y) / ( n Sum X2 – (Sum X) 2 ) 0.5 ( n Sum Y2 – (Sum Y) 2 ) 0.5

2- Mean Absolute Deviation

MAD = sum | A – F | / (n -1)

3- Mean Squared Error

MSE = sum (A – F)2 / (n -1)

Controlling the forecast:

CLs = 0 ± Z S

S = Standard deviation of the distribution of errors = MSE

Z = 3 99.7% of the values can be expected.Z = 2 95.5% of the values can be expected.Z = 1 68.5% of the values can be expected.


0

P8: Maintenance spare parts cost ($):

Year1999

Year2000

Year2001

Year2002

Exp.2003

Forecasting limits2003

1450 1300 1200 1000 ? ?

X 1 2 3 4 5Y 1450 1300 1200 1000 ?

XY 1450 2600 3600 4000

n = 4Sum X = 10 Sum X2 = 30Sum Y = 4950 Sum XY = 11650

Sum Y = n . a + b Sum X , Sum XY = a Sum X + b Sum X2

4950 = 4 a + 10 b 11650 = 10 a + 30 b14850 = 12 a + 30 b

a = 1600 b = - 145

Y = 1600 – 145 X Y5 = 1600 – 145 (5) = 875

X 1 2 3 4 5A 1450 1300 1200 1000 -F 1445 1310 1165 1020 875

(A-F) 5 10 35 20(A-F)2 25 100 1225 400

MSE = 1750 / (4 -1) = 583S = 24 Z = 2

CLs = 0 ± Z S = 0 ± 48


P9: An casting company has the following yearly production and overhead

cost data:

Year 96 97 98 99 00 01 02 03Production quantity(1000 ton)

5.2 5.5 6.0 6.4 6.5 7.0 7.5 7.9

Maintenance cost (M L.E.)

1.3 1.5 1.8 2.0 2.1 2.2 2.5 3.0

Required:1) Plot the maintenance cost vs. production quantities?

2) Estimate and plot a trend for the maintenance cost and

production quantities?

3) Compute the maintenance cost forecast for year 2004, if the

production plan is 8,500 tons?

4) Find the unit maintenance fixed cost and unit maintenance

variable cost, for the year 2003?

5) Measure and discuss the accuracy of the maintenance cost

forecasts using both the correlation coefficient and mean

absolute deviation?

6) Develop and discuss a forecasting control chart with 99.7%

and 95.5% confidence levels (probability of expected

values)?


P10 :

Uncertain spare parts cost

Spare parts cost

$ 100,000

Probability

%

9 20

10 50

11 20

12 7

13 3

Estimate the spare parts budget based on the following:

1- Average method ($ 1,100,000)

2- Probability method ($ 1,023,000)

3- PERT method ($ 1,033,000)


MONTHLY M. PLANNING USING CPM

P11: The monthly PM programs information for a machining shop are as follows:

Machine Code

T01 D01 M01 T02 M02

Machine Description

Turning Drilling Milling Turning Milling

Predecessors - - - T01 M01Duration (day)

8 5 6 8 20

Worker/day 5 8 7 5 5Spare Parts cost $ 1000

5 4 3 6 12

Required:

1. If the maximum worker limit is 12 worker/day,

construct the corresponding monthly maintenance

plan.

2. Calculate the monthly spare parts cost.

3. Construct the monthly spare parts cost profile.


P12: The monthly PM programs information for a machining shop are as follows:

Machine Code

T01 D01 M01 T02 D02 M02 M03

Machine Description

Turning Drilling Milling Turning Drilling Milling Milling

Predecessor - - - T01 D01 M01 M02Duration (day)

8 4 6 8 5 8 12

Worker/day 5 6 7 5 6 5 5Spare Parts cost $ 1000

2 1.5 3 4 2 4 5

Required:

1. If the maximum worker limit is 12 worker/day,

construct the corresponding monthly

maintenance plan.

2. Calculate the monthly spare parts cost.

3. Construct the monthly spare parts cost profile.


CAPITAL MAINTENANCE (OVERHAUL) PLANNING USING CPM

P13: A Bus company has about 1500 bus. The capital maintenance and engine overhaul is carrying out after about 250,000 to 300,000 km (about two years) and its duration is about 10 day. The overhaul information for each bus engine are as follows:

Activity ID A B C D E F G HPredecessors - - - - B & D C F E & GDuration (day) 2 1 1 1 2 1 2 2Worker/day 4 2 3 2 4 3 1 1

Average spare parts cost = 10,000 L.E./Bus.Average labor cost rate = 20 L.E./man-dayIndirect cost rate = 50 L.E./dayDown time cost rate = 300 L.E./Bus-day

Required:1. Construct the bus overhaul plan, If the maximum

worker limit for each overhaul is 6 worker/day.2. Construct the Annual maintenance plan for all

buses.3. Calculate the size of maintenance labor force.4. Comparison between the current and developed

maintenance planning systems.


P14: Monthly Maintenance Plan for Wire Production Line

Project Name : MMPW Project start: 1 Jan. 2004Planning unit : Day 6 DAYS /WEEK

1- Activity List

Activity ID Duration (day) Predece

ssors

Relations(SS, FS, FF, and SF)

1 Preparation PRP 2 - -2 Mech. maintenance # 01 MM1 7 PRP -3 Elec. maintenance # 01 EM1 9 M

M1

SS 3

4 Mech. maintenance # 02 MM2 6 PRP -5 Elec. maintenance # 02 EM2 8 MM2 SS 26 Mech. maintenance # 03 MM3 5 PRP -7 Elec. maintenance # 03 EM3 7 MM3 SS 28 Setup STP 1 EM1

EM2EM3

-

2- Resource List

ResourceCode

Resources description Unit Limits/day PriceLE/unitNorm. Max.

L01 Mechanical worker md 3 6 40L02 Electrical worker md 4 8 60SPS Spare parts & supplies cost - - 1000


W01 9 W02 6 W03 3 2

3- Resource Allocation

Activity IDResource

L01/day

L02/day

SPS(Total)

1 Preparation PRP 2 1 12 Mech. maintenance # 01 MM1 4 - 33 Elec. maintenance # 01 EM1 - 5 44 Mech. maintenance # 02 MM2 3 - 25 Elec. maintenance # 02 EM2 - 4 36 Mech. maintenance # 03 MM3 2 - 27 Elec. maintenance # 03 EM3 - 3 38 Setup STP 2 2 1

4- Base Calendar (Working periods)Saturday Sunday Monday Tuesday Wednesday Thursday Friday

X X X X X X1/01/04

Holidays: 20 to 21 Jan. 2004Required:

1. Draw the project network (logic diagram)?2. Draw the corresponding Gantt chart?3. Construct the corresponding smoothed worker

loading?4. Construct the corresponding worker leveling?5. Construct the target action plan?.6. Construct the cost profile & S-curve?7. Construct the target master plan?


P15: Annual Maintenance Plan for AUC-IT Labs.

Project Name : AMIT Project start: 1 Jan. 2004Planning unit : Day 6 DAYS /WEEK

1- Activity List

Activity ID Duration (day)

Predecessors

Relations

1 Preparation PRP 1 - -2 Server maintenance SRM 3 PRP -3 Hardware maintenance

Lab #01HM1 4 SRM -

4 Software maintenance Lab #01

SM1 5 HM1 SS 2

5 Hardware maintenance Lab #02

HM2 3 SRM -


SM2 4 HM2 SS 1


HM3 3 SRM -


SM3 4 HM3 SS 1

9 Setup STP 1 SM1SM2SM3

-

2- Resource ListResourceCode

Resources description Unit Limits/day PriceLE/unitNorm. Max.

L01 Hardware Engineer md 3 6 120L02 Software Engineer md 4 8 100SPS Spare parts & supplies cost - - 1000


3- Resource Allocation

Activity IDResource

L01/day

L02/day

SPS(Total)

1 Preparation PRP 2 1 12 Server maintenance SRM 1 1 13 Hardware maintenance

Lab #01HM1 4 - 2


SM1 - 5 3


HM2 3 - 1


SM2 - 4 2


HM3 2 - 1


SM3 - 3 2

9 Setup STP 2 2 1

4- Base Calendar (Working periods)Saturday Sunday Monday Tuesday Wednesday Thursday Friday

X X X X X X 1/01/04Holidays: 20 to 21 Jan. 2004

Required:1. Draw the project network (logic diagram)?2. Draw the corresponding Gantt chart?3. Construct the corresponding smoothed worker loading?4. Construct the corresponding worker leveling?5. Construct the target action plan?.6. Construct the cost profile & S-curve?7. Construct the target master plan?


MATERIALS REQUIREMENTS PLANNING (MRP) FOR MAINTENANCE

P16:A monthly maintenance plan for 50 similar equipment

to replace the gear box for these equipment. The gear box

structure is shown below.

Component A B C D E F G

Lead time (week) 1 2 1 1 2 3 2

On-Hand 10 15 20 10 10 5 0

Required:1. Time-phased for the gear box structure2. Gross requirements plan for 50 gear box3. Net material requirements plan for 50 gear box.


A

B(2) C(2)

D(2) E(2) E(2) F(2)

G(1) D(2)

CM CAPACITY PLANNING USING WAITING LINES MODELS

P17: A maintenance service facility has Poisson arrival

rates, negative exponential service times, and operates on a

first-come first-served queue discipline. Breakdowns occur

on an average of = three per day with a range of zero to

eight. The maintenance crew can service an average of =

six machines per day with a range of from zero to seven.

Find the following:

1) Utilization factor of the service facility.

2) Percent idle time % I

3) Mean time in the system Ts

4) Mean number in the system Ns

5) Mean Waiting time Tw

6) Mean number in the queue Nq

7) Probability of finding n=2 machines in the system




Solution

1) Utilization factor of the service facility =

= Mean arrival rate / Mean service rate

= / = 3 / 6 = 50 %

2) Percent idle time % I =

% I = (Total) - (Utilization factor)

= 100 – 50 = 50 %

3) Mean time in the system Ts =

Ts = 1 / (Mean service rate – Mean arrival rate)

= 1 / ( - ) = 1 / (6 – 3) = 1/3 day

4) Mean number in the system Ns =

Ns = (Mean time in system) (Mean arrival rate)

= / ( - )= 3 / (6 – 3) = 1 machine

5) Mean Waiting time Tw------------------------------------------------------------------------------------------Part 3: PM Case Studies - 33 -

Maintenance service facility

Mean arrival

rate

= 3, {0,8}

Mean service

rate = 6, {0,7}

Tw = Total time in system – Service time

= {1/( - )} – { 1/ }

= {1/(6 - 3)} – { 1/6 } = 1/6 day

6) Mean number in the queue Nq =

= (Mean number in system)-(Mean number being served)

= { / ( - )} – {/}

= {3 / (6 – 3)} - {3/6} = 1/2 machine


Pn = (1 - /) (/)n = (1 - 3/6) (3/6)2 = 0.125


Pn = (1 - /) (/)n = (1 - 3/6) (3/6)4 = 0.03125


Pn = (1 - /) (/)n = (1 - 3/6) (3/6)6 = 0.0078


P18: A maintenance service facility has Poisson arrival

rates, negative exponential service times, and operates on a

first-come first-served queue discipline. Breakdowns occur

on an average of = five per day with a range of zero to

nine. The maintenance crew can service an average of =

six machines per day with a range of from zero to eight.

Find the following:

1. Utilization factor of the service facility.

2. Percent idle time % I

3. Mean time in the system Ts

4. Mean number in the system Ns

5. Mean Waiting time Tw

6. Mean number in the queue Nq

7. Probability of finding n=2 machines in the system

8. Probability of finding n=4 machines in the system

9. Probability of finding n=6 machines in the

system


P19:

The CM information for a production department are as

follows:

1. Labor maintenance cost rate = $10/hr

2. Downtime cost rate = $50/hr per machine.

3. The mean number of failures is four/hr, and distributed

according to a Poisson distribution.

4. The mean number of repairs is six repairs per worker-

hour, and distributed exponentially

What is the best maintenance crew size?


Solution

Crew

Size

Ns =

/ ( -

)

Labor

cost

$/hr

Downtime

cost

$/hr

Total

cost

$/hr

1 6 2.00 10 100.00 110

2 12 0.50 20 25.00 45

3 18 0.29 30 14.29 44.29

4 24 0.20 40 10.00 50.00

5 30 0.15 50 7.69 57.69

The best maintenance crew size is 3 workers


Maintenance service facility

Mean arrival rate = 4, Poisson

Labor cost = $10/hr

Downtime cost = $50/hr/machine

Mean service rate

= 6 /worker, Exp.

Crew Size?

P20:

The CM information for a production department are as

follows:

5. Labor maintenance cost rate = $20/hr

6. Downtime cost rate = $150/hr per machine.

7. The mean number of failures is five/hr, and distributed


8. The mean number of repairs is five repairs per worker-

hour, and distributed exponentially

What is the best maintenance crew size?


P21:

The CM information for a production department is as

follows:

Labor maintenance cost rate = $20/hr

Downtime cost rate = $150/hr per machine.

The mean number of failures is five/hr, and distributed


The mean number of repairs is five repairs per worker-hour,

and distributed exponentially

Required:

1. Determine the best maintenance crew size?

2. Calculate the utilization factor of the maintenance

crew?

3. Determine the mean time in the system?

4. Calculate the probability of finding two machines in the

system?


P22:A factory has 200 machines and the maintenance

engineer supervises the repair crews who repair

malfunctioning machines. The maintenance policy is to

repair the broken down machine and bring back in

production within 2 hours on the average. If average

breakdown rate is 3.5 machines/hour and each repair

crew can repair 0.25 machine per hour on the average.

How many repair crews are required ?

The formula for average repair rate () is

1ts = ---------- or = + 1/ ts -

Where = repair rate

= arrival rate of malfunctioning machines

ts = average time arrivals in the system

Required average repair rate

= 3.5 + 1 / 2 = 4 machines / hour

No. of Crews =

¸ machines/hour a crew can repair

=

4 ¸ 0.25 = 16 repair crews required


MAINTENANCE WORK CONTROL USING CONTROL CHARTS

P23: The weekly backlog data for the mechanical section of

the maintenance department are given as follows in terms of

labor-hours for the last four months:

Number Month Week 1 Week 2 Week 3 Week 4

1

2

3

4

March

April

May

June

360

340

420

390

320

350

412

370

300

320

452

350

380

380

380

410

Determine and plot the control chart for the weekly backlog. (For n = 4, d2 = 2.059.)


SPARE PARTS COST CONTROL

P24: The monthly plan and the actual maintenance spare parts in ABC Company are as follows:

Spare part #

Plan (Jan. 2001) Actual (Jan. 2001)Planned quantity

(unit)

Standard cost

(L.E./unit)

Actual quantity

(unit)

Actual cost(L.E./unit)

A11 40 1000 40 1100A12 30 1200 20 1200A13 50 900 40 1000A14 20 850 10 800A15 20 950 20 900

Based on these data, determine the different performance

indicators.


TOTAL MAINTENANCE CONTROL

P25:Monthly production information on Foundry Shop FS510 was as follows:Item January

2004February2004

Working days 31 28Standard production rate (ton/hr) 8 8Average daily time (hr/day) 24 24Average down time (hr/day) 6 4Average standby (hr/day) 3 3Average target quantity (ton/day) 120 136Average actual quantity (ton/day) 80 105Average sound quantity (ton/day) 70 98Average defect quantity (ton/day) 10 7Average energy consumption (1000 kwh/day) 49 67Material cost (1000 L.E/day) 100 130

Required:1. Equipment avaliability.2. Equipment performance efficiency.3. Equipment utilization factor.4. Equipment quality rate5. Equipment uptime ratio.6. Energy productivity.7. Material productivity8. Overall equipment effectiveness (OEE).9. Total effective equipment productivity (TEEP).

10. Net equipment effectiveness (NEE).


Basic dataItem January

2004February2004

Feb. / Jan.

Production rate (ton/hr) 8 8 100 %

Total time (hr/day) 24 24 100 %

Average down time(hr/day)

6 4 67 %

Average available time (hr)

18 20 111 %

Average standby (hr/day)

3 3 100 %

Average used time(hr/day)

15 17 113 %

Average target quantity (ton/day)

120 136 113 %

Average actual quantity (ton/day)

80 105 125 %

Average sound quantity (ton/day)

70 98 129 %

Average defect quantity (ton/day)

10 (14%)

7(7%)

64 %

Energy productivity (kwh/ton)

700 684 98%

Material productivity (1000 L.E/ton)

1429 1326 92%


Performance EvaluationIndicator January

2004February2004

Feb. / Jan.

Availability 18/24= 75 % 20/24= 83 % 111 %

Performance efficiency

80/120= 67 % 105/136= 77 % 115 %

Quality rate 70/80= 88 % 98/105= 93 % 106 %

Utilization ratio 15/18= 83 % 17/20= 85 % 102 %

Uptime (hr/day) 70/8= 8.75 98/8= 12.25 140 %

Uptime ratio 8.75/15= 49% 12.25/17=72 % 147 %

OEE 44 % 60 % 136 %

TEEP 37 % 51 % 138 %

NEE 29 % 52 % 179 %

Energy productivity (kwh/ton)

700 684 98%

Material productivity (1000 L.E/ton)

1429 1326 92%


P26:The six-monthly maintenance costs ($1000) for a productive system are as follows:

Target Costs:

Cost item Month #Jan Feb Mar Apr May Jun Jly

PM Cost: Spar parts Labor

10050

10050

10050

10050

10050

10050

10050

CM Cost: Spar parts Labor

200150

200150

200150

200150

200150

200150

200150

DT Cost 300 300 300 300 300 300 300

Actual Costs:



2332

3865

4996

5694

6894

6590

5472


231503

213370

181293

185164

199201

196193

157142

DT Cost 407 397 320 290 330 320 362

Based on these data, determine the different performance evaluation indicators for the maintenance system.


Target:

Cost item Month #Jan Feb Mar Apr May Jun Jly Total

PM Cost 150 150 150 150 150 150 150 1050

CM Cost 350 350 350 350 350 350 350 2450

TM Cost 800 800 800 800 800 800 800 5600

DT Cost 300 300 300 300 300 300 300 2100

TM+DT 1100 1100 1100 1100 1100 1100 1100 7700

PM/TM 0.14 0.14 0.14 0.14 0.14 0.14 0.14 0.955

CM/PM 2.33 2.33 2.33 2.33 2.33 2.33 2.33 16.33

Actual:


PM Cost 55 103 145 150 162 155 126 896CM Cost 734 583 474 349 400 369 299 320

8TM Cost 1196 1083 939 789 892 864 787 655

0DT Cost 407 397 320 290 330 320 362 242

6TM+DT 1603 1480 1259 1079 1222 1184 1149 897

6PM/TM 0.05 0.10 0.15 0.19 0.18 0.18 0.16 1.00

7CM/PM 13.35 5.66 3.27 2.33 2.47 2.38 2.37 31.8

2


Change %:


PM Cost CM CostTM CostDT CostTM+DTPM/TMCM/PM


P27:

The yearly PM programs information for six similar gas turbines in a power station are as follows:

Target work performed:

Item PM CM TotalTotal labor force (worker) 18 7 25Annual spare parts cost ($1000) 264 72 336Annual labor cost ($1000) -- -- 75Overhead cost ($1000) -- -- 514Average down time(day/year per gas turbine)

51 15 66

Average downtime cost rate = $ 1000 per day

Actual work performed:

Item PM CM TotalTotal labor force (worker) 20 10 30Annual spare parts cost ($1000) 300 100 400Annual labor cost ($1000) -- -- 80Overhead cost ($1000) -- -- 520Average down time(day/year per gas turbine)

45 5 50



Performance Evaluation Sheet:

Item Target Actual Change %

Total labor force (worker) 25 30 + 20

Annual s. parts cost

($1000)336 400 + 19

Annual labor cost ($1000) 75 80 + 6.6

Overhead cost ($1000) 514 520 + 1.2

Total m. cost ($1000) 925 1000 + 8.1

Average down time 66 50 - 24.3

Down time cost ($1000) 66 50 - 24.3

TMC + DTC 991 1050 + 6.0

Availability % 81.9 86.3 + 5.3

CM/PM % (labor force) 7/18 =38.9

10/20 =50

+ 28.5

CM/PM % (Spare parts) 72/264 =27.3

100/300 =33.3

+ 22.0

Overhead % 514/411=1.25

520/480=1.08

- 13.6

Labor productivity %(worker/gas turbine)

25/6=4.17

30/6=5.00

- 16.6


P28:The six-monthly maintenance costs ($1000) for a productive system are as follows:

Target Costs:



100 50

100 50

100 50

100 50

100 50

100 50

100 50


200150

200150

200150

200150

200150

200150

200150

DT Cost 300 300 300 300 300 300 300

Actual Costs:



2332

3865

4996

5694

6894

6590

5472


231503

213370

181293

185164

199201

196193

157142

DT Cost 407 397 320 290 330 320 362



Target:

Cost item Month #

Jan Feb Mar Apr May Jun Jly

PM Cost 150 150 150 150 150 150 150CM Cost 350 350 350 350 350 350 350DT Cost 300 300 300 300 300 300 300TM Cost 800 800 800 800 800 800 800

Actual:

Cost itemMonth #

Jan Feb Mar Apr May Jun Jly

PM Cost 55 103 145 150 162 155 126CM Cost 734 583 474 349 400 369 299DT Cost 407 397 320 290 330 320 362TM Cost 1196 1083 939 789 892 864 787


P29: The monthly budget and the actual maintenance cost for XXX Company are as follows:

Cost itemStandard Actual

Quantity Rate Quantity Rate

Spare parts 20,00

ton

6000

L.E.

15,00

ton

6400

L.E.

Direct labor 40,000

hour

15

L.E.

45,000

hour

18

L.E.

Indirect materials 150,000 L.E. 180,000 L.E.

Indirect labor 100,000 L.E. 80,000 L.E.

Fixed overhead 300,000 L.E. 350,000 L.E.


indicators.


P30:

The annual budget and the actual maintenance cost for XXX production department are as follows:

Cost item (L.E.)

Budget Actual

Current

month

Year to

date

Current

month

Year to

date

Direct labor 6000 60000 5000 65000

Indirect labor 5936 53834 4111 58834

Spare parts 2000 12000 700 10000

Indirect materials 1000 10000 500 5000

Other overheads 1000 10000 1200 12000


indicators.


P31:Maintenance Planning for Spare Parts Work Shop

1- Equipment List

Plant Systems

Location Equipment Type Number of Machines

Productive systems

Machining shop

Turning Milling DrillingGrindingPress

42221

Foundry shop

Induction furnacesMolding machines

25

Welding shop

Arc Welding 1

Supportive systems

Material handling

Fork lift 4

Air room Compressor 2Water room Pump – 50 HP

Pump – 100 HP22

Power room

Diesel generator 2


2- Layout of machining shop

3- Equipment reliability

4- Equipment PM Program for machining shop

4-1- PM Manpower & Downtime:

Eq. Type Weekly Monthly 6 Monthly YearlyMan Hr Man Hr Man Hr Man Hr

Turning 2 5 3 16 - - 5 24Milling 2 5 3 12 - - 5 16Drilling 1 3 - - 2 10 3 12Grinding 2 3 - - 3 12 3 16Press 1 2 - - 2 6 3 12


T T

T T

M

M

D

D

G

G

T0.9

T0.9

M0.8

M0.8

T0.9

T0.9

D0.8

D0.8

G0.9

G0.9

4-2- PM Spare Parts & Supplies:

Eq. Type Weekly Monthly 6 Monthly YearlyB O B O B O B O

Turning - 3 - 5 - - 4 12Milling - 2 - 4 - - 4 10Drilling - 1 - - 4 6 4 8Grinding - 1 - - 4 8 4 8Press - 1 - - 4 6 4 8

B : Bearing O: Oil (liter)4-3- PM Cost:

Average labor rate $ 10 / hour

Spare Parts & Supplies cost:Bearing $ 50 /unit & Oil $ 5 /liter

Other costs (Indirect or Overhead) 40 % PM

5- Failure Analysis for the last year (CM)

Average per equipment

Turning Milling Drilling Grinding Press

Man-hour 100 60 40 50 30Down time hr 50 40 30 40 30Spare parts cost $

1000 800 500 600 500

Oil (liter) 20 15 12 16 106- Working conditions:Machine Working conditions:------------------------------------------------------------------------------------------Part 3: PM Case Studies - 57 -

2 shifts/day, 8 hr./shift, 6 day/week, 45 week/year.

Manpower working conditions: 8 hr./day, 6 day/week, 45 week/year.

II- Maintenance Planning ElementsRequired:1. Equipment coding.2. Equipment priorities.3. Maintenance labor force.4. Spare parts & supplies requirement planning.5. Annual maintenance cost.6. Equipment average down time per month.7. Equipment average time availability.8. System average down time per month.9. System average time availability10. System reliability11. Annual PM plans.12. Annual man-hours & manpower profile 13. Monthly PM plans.14. Maintenance Work orders

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193883462 03-part-3-pm-case-studies

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