2.3 : BASICS OF PROJECT MANAGEMENT: MAIN CONCEPTS: OTHER CONCEPTS:1.Project,2.Project network (cpm,pert etc),3.Activity,4.Node,5.An example of of CPM network,6.An eg of PERT network,

1.project crashing2. Project life cycle3. Project organisation4.The role of a project manager5.Project planning,scheduling and control techniques6.Project planning and control flow chart,7.0 Project versus plant 8.0 The 4 goals for successful project management9.Success factors in project management

1. PROJECT: Unique one-time operations designed to accomplish a specific set of objectives in a limited time frame.Egs: Construction of hospital, ship, R&D for launching a missile, new project, Design &Development of new product, Chandrayan, preparation for UPSC Exam…. The word “Project” conjures up a mental picture which is unique,big,multi-agency built, with huge no. of activities and a composite affair.eg :Visakhapatnam steel plant, recent space Mars Mission India etc., Exhibit: RINL,Visakhapatnam Steel Plant Exhibit :Mars mission(India)

In yester years, we had great projects, such as Egyptian pyramids,Tajmahal,Bhakra-Nangal

Dam,Kalpakkam Atomic Power Project etc.

Exhibit: Egyptian Pyramids Exhibit :TajMahal

Exhibit Bhakra Nangal Dam

KEY DECISIONS IN PROJECT MANAGEMENT

• Much of the success of projects depends on certain key managerial decisions. They are:1. Deciding which project to implement 2. Selecting the project manager3. Selecting the members of the project team4. Managing and controlling project resources5. Deciding whether to terminate a project and if so, when.

Project Network Techniques:For scheduling & monitoring,in the last 75 years, two techniques called;”PERT” and “CPM” were evolved and utilized.These are still used,though they are now a days computerized.Earlier, it was mentioned that a project should be completed “in time”.These networks are used for this purpose Exhibit Project Network

A project consists of many activities. For major categories like factories, dams etc., the no of activities run into thousands.

ACTIVITY :

An activity is a homogeneous work element , that consumes “Resources” and consumes (or takes) a definite amount of “TIME”.Also an activity is the smallest unit of productive work to be planned & scheduled. For a housing project, many activities are performed; like,foundation making,walls,roof,flooring,electrical wiring,plumbing,white washing are some of the activities.An Ex of an “Activity” : Site preparation,taking 10 days.Note : Activities can be in SERIES or PARALLEL or a COMBINATION.

NODE:

Nodes:Each activity will have 2 nodes(also called events),the starting node & Finishing node.(Eg : For the activity of “Site preparation”, the start instant is called Start node or Head node and the end instant is called End node or Tail node).These 2 nodes are shown as circles ,And the line connecting both is called the ACTIVITY.

SIMPLE ILLUSTRATION:Of preparation of Tea: Eqpt: 2 burner gas stove.There are 8 activities .Details of these and time taken are as shown.A project net work is to be drawn.

Activitycode

Activity description Time( mins)

A Arrange Vessels (2 Nos.),Tea powder , Sugar cubes , 2 Burner Stove & ??

3.0

B Pour water into Vessel#1, put on Burner #1, light the BURNER#1 & boil .

4.0

C On boiling , add tea powder & stir. 0.5D Add sugar, stir & lower down Vessel #1 & filter. using

Filter mesh. The output is tea decoction.2.0

E Pour milk in Vessel#2, put on Burner#1, light the BURNER#1 , boil & lower down Vessel #2.

6.0

F Add boiled milk to tea decoction in Vessel #1, put on the Burner#1 & do final boiling. The output is tea.

3.0

G Lower the Vessel#1, containing the tea 0.5H Pour into 4 cups . assemble on tray for use. 1.0

HOW MANY ACTIVITIRES ARE THERE : 8 ACTIVITIES.Each activity has ONE TIME ESTIMATE(EG:Activity B has 4.0 mins.What is the arithmetical total time ? “The arithmetical total time for the project is 20 Mins”

Let us a draw a PROJECT NETWORK.

(Re call):”The arithmetical total time for the project is 20 Mins”

(Recall) :”The arithmetical total time for the project is 20 Mins”

DEPENDENCY:OF ACTIVITIES.

INFORMATION REQUIRED FOR DRAWING OF (PROJECT) NETWORKS :

(1) List of all the activities (2) Activity interdependencies (3) Time Estimates

(A) PROJECT NETWORK :METHOD : CRITICAL PATH METHOD (CPM)POWER PLANT PROJECT EXAMPLE :

Activity Activity Code (start &end nodes)

Activity Description

Activity Dependency on

Time estimate (most likely) ,” m”

A 1-2 Designing of plant Nil 12

B 2-3 Selection of site A 8

C 2-4 Select ion of vendor A 4

D 2-6 Select Employees A 3

E 3-5 Site preparation B 12

F 4-5 Manufacturing Generator by vendor C 18

G 4-6 Preparation of opern & maintenance manuals

C 5

H 5-7 Installation of Generator E,F 4

I 6-7 Train the employees D,G 9

J 7-8 Obtain License for operation H,I 6

81

Note: Column (4) gives the dependency relationships of activities. ie, if an activity is to commence after completion of another activity (Serial), or 2 (or more) activities can start at the same time(Parallel) ,or an activity can start independently (Eg : Starting activities).In the eg above, activity (1-2) which is A, does not depend on start or end of any activity.So the project network is drawn starting with A It is seen earlier that an activity has a starting and an ending node (2 nodes).This is as shown .Similarly, Activity(7-8) which is J should appear at the end .Systematically , proceeding, with these concepts, the Net work is evolved.Note : It is seen that , there will always be one start node (initial one) at the beginning end and a finish node (Final one) at final end.”The arithmetical total time for the project is 81 Days”

(Re call) :”The arithmetical total time for the project is 81 Days”

(B) PROJECT NETWORK : METHOD : PROGRAM EVALUATION and REVIEW TECHNIQUE (PERT)POWER PLANT PROJECT EXAMPLE :

• PERT TYPE PROJECT NETWORK EXAMPLE• Suppose, Project planners have sought the judgement of engineers, foremen & vendors &

developed 3 time estimates, as shown. Determination of• (a) the critical path & time• (b) the probability the project will be finished within 4 years and• (c)what is the probability that it will take more than 55 months is required..• In this eg, there are 3 time estimates are available for each activity, the project network may be

called ”PERT”.

Activity Activity Code (start &end nodes)

Activity Description

Activity Dependency

Time estimate (Optimistic) ,” a”

Time estimate (most likely) ,” m”

Time estimate (pessimistic) ,” b”

A 1-2 Designing of plant Nil 10 12 16

B 2-3 Selection of site A 2 8 36

C 2-4 Select ion of vendor A 1 4 5

D 2-6 Select Employees A 2 3 4

E 3-5 Site preparation B 8 12 20

F 4-5 Manufacturing Generator by vendor

C 15 18 30

G 4-6 Preparation of operations& maintenance manuals

C 3 5 8

H 5-7 Installion of Generator E,F 2 4 8

I 6-7 Train the employees D,G 6 9 12

J 7-8 Obtain License for operation

H,I 4 6 14

“ Total of a = 53Days ,Total of b = 81Days & Total of c = 153 Days “

PERT EXAMPLE (a) Drawing critical path & obtaining total project completion time: STEP 1

”Activity wise” computation of (i) mean time from the 3 time estimates & (ii) ”Activity wise” computation of Varience are made, as shown.

Activity Activity Description

Dependency

Time est, a (weeks)

Time est, m (weeks)

Time est, b (weeks)

Te=(a+4m+b)/6 ( as per Formula) (weeks)

Varience ( as per Formula)

A,1-2 Designing of plant

Nil 10 12 16 12.33 Square of (16-10)/6=1 X1=1.0

B,2-3 Selection of site

A 2 8 36 11.67 Sq of (36-2)/6=5.6 X 5.6=32.11

C,2-4 Selection of vendor

A 1 4 5 3.67 Sq of(5-1)/6=0.67X0.67=0.44

D,2-6 Select Employees

A 2 3 4 3.00 Sq of (4-2)/6=0.33 X0.33=0.11

E,3-5 Site preparation

B 8 12 20 12.67 Sq of (20-8)/6=2 X2=4.0

F,4-5 Manufacturing Generator by vendor

C 15 18 30 19.50 Sq of (30-15)/6=2.5 X2.5=6.25

G,4-6 Preparation of operations& maintenance manuals

C 3 5 8 5.17 Sq of(8-3)/6=0.83 sX 0.83=0.69

H,5-7 Installion of Generator

E,F 2 4 8 4.33 Sq of(8-2)/6= 1.0 X 1=1

I,6-7 Train the employees

D,G 6 9 12 9.000 Sq of (12-6)/6=1 sX 1=1.0

J,7-8 Obtain License for operation

H,I 4 6 14 7.00 Sq of (14-4)/6=1.67 X 1.67=2.78

PERT EXAMPLE : STEP II : Project network is drawn. (Here in called “PERT” Net work).

PERT EXAMPLE : STEP III : Identify various “all possible”paths from start to finish.

Path Times (in weeks)

Path (i) ;1-2-3-5-7-8(critical) 12.33+11.67+12.67+4.33+7.00=48.00

(ii);1-2-4-5-7-8 46.83

(iii);1-2-4-6-7-8 37.17

(iv):1-2-6-7-8 31.33

Comment :Various paths from start to finish:

There are 4 such paths.The critical path (which is the Longest duration route from beginning to end) is 1-2-3-5-7-8(critical), with times as 12. 33+11.67+12.67+4.33+7.00=48.00 weeks

(b) The probability the project will be finished within 4 years ? Te= 48 months; The probability (normal distribution) = 0.50

(c) what is the probability that it will take more than 55 months ? To determine probabilities of any other completion time , it is required to calculate the variances of the distribution of completion times along THE ACTIVITIES ON THE CRITICAL PATH, sum them and estimate Sigma. Sigma= square root of(sigma varience)=sq rt of (1.00+32.11+4.00+1.00+2.78)=6.4 months. Z= (Tx-TE)/Sigma=(55.00-48.00)/6.4=1.09 From Z graph, for normal distribution,P(X>Tx)=0.5000-0.3621=0.1379 Hence, the Probability=0.14

PERT NETWORK-a note• This probability computation is of utmost importance to top management to intervene, if

needed. • This is how PERT Network is found more useful than CPM Network. • How ever, for simple projects or projects which are repetitive(like building construction

projects),CPM will serve the purpose.

OTHER CONCEPTS:

1.CRASHING PROJECT NETWORKS

• Crashing Project Networks: To reduce the normal project duration to the desired shorter duration, the activities on the critical path are crashed(reduction of activity time duration), by the use of additional resources/expenses.

Crashing CPM Networks-eg.,A net work has 4 following activities.The minimum feasible time and for achieving this, the costs to be incurred/day to crash project duration are as given.If fixed project costs are Rs 90/Day, what is the lowest cost- time schedule?,

Activity Normal time,days

MinimumFeasible time,days

Direct Costs of Time reduction(Rs)

1-2 5 2 40(each day)

1-3 4 2 35(1st day),80(2nd day)

2-4 4 4 None possible

3-4 6 3 45(1st day),110(other days)

STEP 1 : ALT ( 1)……… Reference cost for 10 days=Rs 90*10= Rs 900 (*)

STEP 2: :ALT (2)

Present Reference cost for 9 days=Rs90*9= Rs 810 (**) Crash cost for decreasing , time of activity 1 – 3, by 1 day, from 4 to 3 days = minimum cost slope selected = Rs 35COST differential = Previous step cost – present reference cost for 9 days – cost slope selected =

( Rs900)(*) - ( Rs 90*9 +Rs 35) =Rs 900 - 845 = Rs 55

Hence, Cost differential = 55 ( Positive number)

STEP 3: :ALT (3)

Present Reference cost for 8 days=Rs90*8= Rs 720

Crash cost for decreasing , time of activity 1 – 2, by 1 day, from 5 to 4 days & also time of activity 3 – 4, by 1 day, from 6 to 5 days = the 2minimum cost slopes selected = Rs 40 + Rs 45 = Rs 85COST differential = Previous step cost – present reference cost for 8 days – cost slopes selected =

( Rs 810)(**) - ( Rs 90*8 +Rs 85) =Rs 810 - 805 = Rs 5

Hence, Cost differential = 5

STEP 4: :ALT (4)

Any further reduction of time on any critical path/s, will render the Cost differential as a Negative number. Hence, no more reduction of time is “PROFITABLE”.Crashing by only 2 days is POSSIBLE. i.E, THE PROJECT CAN BE COMPLETED IN 8 DAYS ,INSTEAD OF 10 DAYS,FOR WHICH EXTRA COSTS (CALLED, CRASHING COSTS) ARE TO BE INCURRED.

SUMMARY OF STEPS IN NETWORK CRASHING:

Step 1 : Determine the time-cost ratio for each activity in the network. This ratio represents the increase in cost for a unit decrease in time.

• Time-cost ratio =Increase in cost/Decrease in time;• = (Crash cost-Normal cost)/(Normal time-Crash time)

Step 2 : Identify the activities on the critical path and select that activity on the critical path which has the smallest time-cost ratio and crash that activity to the extent possible.Step 3 : Observe whether there is any change in the critical path i.e., any other path becomes critical. If so, calculate the time-cost ratio for the activities in the new critical path.Step 4 : Repeat step 2 and 3, till the activities are crashed to reduce the project duration to the desired time period.Note : Reource Levelling / Slacks or Floats.

2. PROJECT LIFE CYCLE: The life cycle through which a project passes. Its various phases are 1. The concept phase, 2. Feasibility phase, 3. Detailed planning phase, 4. Organisation phase, 5. Execution phase and 6. Termination phase.

3. PROJECT ORGANISATION: Organisation developed to ensure the successful completion of the project ,Its VARIETIES: Functional /Project coordinator /Project matrix

4.THE ROLE OF A PROJECT MANAGERSix basic functions that project management must address are

1. Manage the project's scope to define the goals and work to be done, in sufficient detail to facilitate understanding and correct performance by participants.

2. Manage the human resources involved in the project.3. Manage communications to see that the appropriate parties are informed.4. Manage time by planning and meeting a schedule.5. Manage quality so that the project's results are satisfactory6. Manage costs so that the project is performed at the minimum practical cost and within budget if

possible.

5.PROJECT PLANNING,SCHEDULING AND CONTROL TECHNIQUES

PROJECT PLANNING: All activities that result in a course of action for a project. PROJECT SCHEDULING: Establishes times and sequences of the various phases of the project. PROJECT CONTROL: Activities designed to measure the status of component activities of a

project, transmit data to a control centre where it is compared with the performance standard and initiate corrective action when required.

6.PROJECT PLANNING AND CONTROL FLOW CHART,

7.0 PROJECT VS PLANT :The project upon its completion is called “Plant or operational unit or entity”,which functions “repetitively” or “continuously” to seve society’s needs..

During project stage, huge resources are deployed/spent,and no production and revenue occurs.Once project is completed, regular production starts, the plant becomes “revenue” i.e. it earns revenue.Eg: a 2000 MW NTPC Project, generally takes 4 years for its planning and execution.During this period, only investments/expenditures are incurred. But on its completion, it produces power, earns revenue and meets society’s needs. In the first 4 years , it is called “NTPC Project” & after that it is called”NTPC Plant”.

8.0 THE 4 GOALS FOR SUCCESSFUL PROJECT MANAGEMENT:

Exhibit Goals

Successful project management implies 4 goals.These are8.1.Project must get”Completed.8.2.Get completed ”within Budget”8.3.Get completed ” within Allocated Time”8.4.On completion(of project) subsequently the“Plant”“must perform to all the stake holders’ satisfaction”.

9..SUCCESS FACTORS IN PROJECT MANAGEMENT: Exhibit Success Factors in Project management

1.A thorough Feasibility study2. A well planned Detailed Project Report(DPR)3. A Capable Project Manager4. An Adequate Organisation Structure.5. A Quick Conflict Resolution Mechanism6 .An Adequate & Timely financing.7. Proper Monitoring & Control system

2.4(IV)MAINTENANCE MANAGEMENT

MAINTENANCE OBJECTIVE : Adding value through Maximisation of Availability of FACILITY.

MAIN CONCEPTS: OTHER CONCEPTS:1.Definition2.Objectives3.SSH,MAH & %MAH,MUH&5MUH4.Categories5.Maintenance policies6.TPM

1.Replacement policies2.Reliability3.Measurement of Maintenance Hours,MTBF4.MIS5. Organisation chart for Maintenance management

7.PROBLEMS

Ex 1: 1 Shift Operation: (1 year old Motor Cycle’s) Use by College student V/s (1 year old Motor Cycle’s) Use by Medical Sales Representative

AND

Ex 2:1 Shift operation:(4 years old Car’s) Use by Co.Executive V/s (4 years old Car’s(Cab) Use by Cab operator.

Note : In the diagrams shown:Ellipse…………represents “BREAKDOWN & REPAIR” period.&Rectangle……… represents “UTILIZATION” period.

Ex 3: 3 Shifts operation: 1 year old Cab’s Use Vs 4 years old Cab’s Use

S.No

Example SSH=Stnd

Scheuled Hours

BDH=Break

Down/Mntnce

Hours

MAH=Mc

Available Hours

MUH=McUtilized Hours

% A= %

Availability

%U=%

Utilization

% U/

% A

Comments

c1 (c2) (c3) (c4) (c5) (c6) (c7) (c8) (c9) (c10)

1 Student,1 yr old M/Cy(1Shift oprn)

8 Hours=

(480mins)100%(Base)

0.083Hours

=(5 mins)

8-0.083=7.917 Hours=(480-

5=475mins)

0.5 Hours=(30 mins)

(480-5)X100/480=98.95%

30X100/480=

6.25%

(30)X100/475=

6.315%

2 Med Repr,1 yr old M/Cy(1Shift oprn)

480m= 100%

30 min 450min 180min (480-30)X100/480=93.75%

180X100/480=

37.5%

(180)X100/450=

6.315%

3 Exec,4 yrs old M/Car(1Shift oprn)

480m= 100%

15 min 465min 90min (480-15)X100/480=96.87%

90X100/480=

18.75%

(90)X100/465=

19.35%

4 Cab Oprtr4 yrs old M/Car(1Shift oprn)

480m= 100%

60 min 420min 240min (480-60)X100/480=87.5%

240X100/480=

50.0%

(240)X100/420=

57.15%

5 Cab oprtr,1 yr old M/Car,(3 ShiftsOprn)

3X480=

1440 m= 100%

90 min 1440-90=1350

min

(1480-90)X100/1440=93.75%

Mc. Available time= 1350 mins, No. of Br.Downs=

3Nos.Hence, MTBF= 1350/3=450

mins(7.5 hrs). ie, on the average, once in 7.5 hours,there is a Break

down.

6 Cab oprtr,5 ysr old M/Car,(3 ShiftsOprn)

480m= 100%

210 min 1440-210=123

0 min

(1440-210)X100/1440=85.42%

Mc. Available time= 1230 mins, No. of Br.Downs=

6Nos.Hence, MTBF= 1230/6=450

mins(7.5 hrs). ie, on the average, once in 3 hrs 25 mins ,there is a

Break down.

COMPUTATIONS OF SSH,MAH ,MUH & MUH / MAH ,at a Glance

S.No

Machine

User

Schedule

Shifts/Day

SSH(SchdShiftHours

=(SchdShiftMins)

BDH (BreakDown Hours)

=(BreakDown Mins

MAH(MachineAvailable Hours)

 =(MachineAvailable Mins)

% Availability=%A

MUH(MachineUtilised Hours)

 =(Machine Utilised Mins)

% Utilisation= %U

For top Mgt(% U /%A)

c1 c2 c3

 c4 c5 c6= c5X8

c7=c6X60

c8 c9=c8X60

C10=c6-c8

c11=c10X60

c12=c10X100/c6

c13 c14=c13X60

c15=c13X100/c6

c16=c15X100/c12

                               1.a Mo

bike,1Yr

Student

Day 

1 8 480 0.083 5 7.917=8-.083

475 98.96 0.50 30 6.25 6.3

          7.917*100/8

  0.5X100/8

 6.25 X 100 / 98.9

               

2.a Mobike,1Yr

M/Repr

Day

1 8 480 0.500 30 7.500=8-.0.5

450 93.75 3.00 180 37.50 40.0

          7.5x100/8

  3X100/8

 37.50 X 100 / 93.75

2.b Mobike,1Yr

M/Repr

Week(6d)

6 48 2880 3.000 180 45.000=48.0-.3.0

2700 93.75 18 180 37.50 40.0

2.c Mobike,1Yr

M/Repr

Month(25d)

25 200 12000

12.500

750 187.500=200-12.5

11250 93.75 75.00 180 37.50 40.0

2.d Mobike,1Yr

M/Repr

Year(300d)

300 2400 144000

150.000

9000 2250.000=2400-150

135000 93.75 900.00 180 37.50 40.0

     

     3.a

car,4Yrs

Exec

Day

1 8 480 0.250 15 7.750 465 96.88 1.50 90 18.75 19.4

          7.75*100/8

  1.5*100/8

 18.75 X 100 / 96.8

     

     4.a

cab,4Yrs

Cab optr

Day

1 8 480 1.000 60 7.000 420 87.50 4.00 240 50.00 57.1

          7x100/8

  4 x 100/8

 50.0X 100 / 87.50

          

5.a cab,4Yrs

Cab optr

Day

3 24 1440 1.500 90 22.500 1350 93.75        

          22.5X100/24

       

          

6.a Cab,5Yrs

Cab optr

Day

3 24 1440 3.500 210 20.500 1230 85.42        

          20.5X100/24

       

COMMENTS: More the work (or UTILIZATION) for EQUIPMENT (Stress), the more will be the BREAK DOWNS.

So, if more emphasis is given to have REGULAR PREVENTIVE MAINTENANCE , the lesser will be the BREAK DOWNS.

Ex4:MTBF Concept:Dates of Breakdown Time Interval(Days)1 Jan 0 (Ref)24 Jan 2310 Feb 1716 March 3430 April 456 May 6Total 125Avg Interval between consecutive failures 125/5=25 DaysMTBF 25 Days

Comments:If MTBF is less,What does it indicate? If MTBF is more,What does it indicate? Which alternative is preferred?What should be MTBF for (i) Aeroplane (ii) Nuclear Reactor ,(iii) Motor cycle(iv)Ambulance,108 ,Power systems for Medical O.Th,ICU etc.

MAINTENANCE &MAINTENANCE MANAGEMENT Maintenance: The function of production management, that is concerned with day -to-day problem

of keeping the physical plant in good working condition. Maintenance management: Concerned with planning, organising and directing the resources in

order to control the availability and performance of the industrial plants to some specified level.OBJECTIVES OF MAINTENANCE MANAGEMENT

1. Minimizing the loss of productive time because of equipment failure (i.e., minimizing idle time of equipment due to break down).

2. Minimizing the repair time and repair cost.3. Minimizing the loss due to production stoppages.4. Efficient use of maintenance personnel and equipments.5. To keep all productive assets in good working condition.6. Prolonging the life of capital assets by minimizing the rate of wear and tear.7. To maximize efficiency and economy in production through optimum use of facilities.8. To minimize accidents through regular inspection and repair of safety devices.9. To minimize the total maintenance cost which includes the cost of repair, cost of preventive

maintenance and inventory carrying costs due to spare parts inventory.10. To improve the quality of products and to improve productivity.

AREAS & TYPES OF MAINTENANCE

AREAS OF MAINTENANCE • Civil Maintenance• Mechanical Maintenance• Electrical Maintenance

TYPES OF MAINTENANCE• Break down maintenance or corrective maintenance,• Preventive maintenance,• Predictive maintenance,• Routine maintenance,• Planned maintenance•

Types of Maintenance

BREAK-DOWN-MAINTENANCE : Corrective or remedial maintenance that occurs when or after equipments or machines fail and must be repaired on an urgent basis.

PREVENTIVE MAINTENANCE : Maintenance activities under taken before the machines or equipments fail. Aims to minimise the possibility of unanticipated or major break-downs.(Inspection, servicing,lubricating,greasing, mnt of buildings etc.,

PREDICTIVE MAINTENANCE : Modern approach to preventive maintenance using sensitive instruments to predict anticipated failure of machines and equipments.(vibration, temperature, pressures etc.,)

ROUTINE MAINTENANCE : This includes activities such as periodic inspection, cleaning, lubrication, and repairs.(Running maintenance, when m/c is working, Shut down maintenance)

PLANNED MAINTENANCE : Maintenance activities that are carried out according to a predetermined schedule. Also known as scheduled maintenance or productive maintenance.(BDs do not occur in a planned manner, but maintenance work can be planned in advance , which REDUCES sudden BDs needing emergency maintenance).

MAINTENANCE POLICY& ECONOMIES OF MAINTENANCE

MAINTENANCE POLICY:1. On size of the maintenance department.2. On extent of maintenance activities and

ECONOMIES OF MAINTENANCE At the optimal level of maintenance preventive maintenance cost equals breakdown maintenance cost.

OPTIMAL LEVEL OF MAINTENANCE,

CREW SIZE COSTS:

TOTAL PRODUCTIVE MAINTENANCE(TPM)

TPM Combines total quality management concept with strategic view of maintenance from process and equipment design to preventive maintenance.

TPM means(i)total employee involvement (ii)Total equipment effectiveness(zero downtime, Zero breakdown) & (iii) total maintenance delivery system.

SPECIFIC ACTIONS OF TPM REQUIRE THE FOLLOWING; Restoring equipment to a like-new condition, having equipment operators involved in maintenance ,improving maintenance effectiveness, training labour pool to improve job skills & effective use of predictive & preventive maintenance technology. IT IS THE MOST MODERN CONCEPT.

PROBLEMS ON:MAINTENANCE MANAGEMENT : PROBLEM-1,

No of Break Downs(B/Ds) (x)

Freq of B/Ds in days, F(X)

Probability Distribution of B/Ds, (Px)

Expected value of B/Ds= x *Px

0 40 40/300=0.133 Nil

1 150 0.500 0.5000

2 70 0.233 0.466

3 30 0.100 0.300

4 10 .033 0.132

Total 300 1.000 1.400

• Each BD costs Rs650, and is considering having PM at Rs200/day, to limit BDs to 1/day. Comment?

• Average BDs/day-1.40, Cost of BDs/day=1.4*650=910, (Alt) Cost of PM & 1 BD/day=200+650=850, Saving=60/day, 60*300=Rs18000/year

PROBLEM-2• Requst for maintenance service from a centralized maintenance facility ,simulated for 8 hours data

is given. Labour charges for maintenance crew is Rs40/hr, working or idle. The waiting time of operators & machinery under breakdown is Rs70/hour.

• A) idle time cost of maintenance facility?B)Waiting time of operators & machinery(not including repair)?C)total facility idle time & machinery waiting time cost?

• D)Assuming that for an additional cost of Rs10/hour, the maintenance centre could add another crew & decrease the repair time by one third, would the additional cost be justified?

Request arrival time

Repair time reqd, with one crew(mins)

Repair time begins (clock time)

Repair time ends (clock time)

M/c down time with one crew(idle time +repair time)=Total time

01:30 60 01:30 02:30 Nil+1.00=1.00

02:00 20 02:30 02:50 0.5+0.33=0.83

04:15 45 04:15 05:00 Nil+0.75=0.75

04:30 120 05:00 07:00 0.5+2.00=2.50

05:30 30 07:00 07:30 1.5+0.50=2.00

07:00 10 07:30 07:40 0.5+0.166=0.666

Total 3.00+4.746=7.75

•• A) idle time cost of maintenance facility?

idle time=Repair =(8.00-4.75) hours.

cost= 3.25*40=Rs130/

• B)Waiting time of operators & machinery(not including repair)? =3.00 *70=Rs 210/

• C)total facility idle time &machinery waiting time cost? =Rs130+Rs210=Rs340/

•• D)Assuming that for an additional cost of Rs10/hour,the maintenance centre could add

another crew & decrease the repair time by one third, would the additional cost be justified?

• One more crew’s cost= Rs10*8=Rs 80/• Decrease in repair time= 1/3 rd of original• =1/3 *4.75=1.582 hrs• Decrease in idle time cost, for machinery=1.582*70=Rs110.74,• It is justified to have one more maintenance crew.

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