2. Replacement Model
The problem of replacement arises when any one of the
components of productive resources, such as machinery, building
and men deteriorates due to time or usage. The examples are:
(a) A machine, which is purchased and installed in a production
system, due to usage some of its components wear out and its
efficiency is reduced.
(b) A building in which production activities are carried out, may
leave cracks in walls, roof etc, and needs repair.
3. The problem of replacement is experienced in systems
where machines, individuals or capital assets are the main
production or job performing units. The characteristics of
these units is that their level of performance or efficiency
decreases with time or usage and one has to formulate some
suitable replacement policy regarding these units to keep the
system up to some desired level of performance. The main
objective of replacement is to help the organization for
maximizing its profit or to minimize the cost.
4. FAILURE MECHANISIM OF ITEMS
The word failure has got a wider meaning in
industrial maintenance than what it has in our daily
life. We can categorize the failure in two classes.
They are (i) Gradual failure and (ii) Sudden failure.
Once again the sudden failure may be classified as:
(a) Progressive failure,
(b) Retrogressive failure and (c) Random failure.
5. Gradual failure
In this class as the life of the machine increases or due
continuous usage, due to wear and tear of components of the
facility, its efficiency deteriorates due to which the management can
experience:
(a) Progressive Increase in maintenance expenditure or operating
costs, (b)Decreased productivity of the equipment and (c) decrease
in the value of the equipment i.e. resale value of the
equipment/facility decreases.
Examples of this category are: Automobiles, Machine tools, etc.
6. Sudden failure
In this case, the items ultimately fail after a period of time. The
life of the equipment cannot be predicted and is some sort of
random variable. The period between installation and failure is not
constant for any particular type of equipment but will follow some
frequency distribution, which may be:
(a) Progressive failure: In this case probability of failure increases
with the increase in life of an item.
The best example is electrical bulbs and computer components
7. Retrogressive failure: Some items will have higher probability
of failure in the beginning of their life, and as the time passes
chances of failure becomes less. That is the ability of the item
to survive in the initial period of life increases its expected life.
The examples are newly installed machines in production
systems, new vehicles, and infant baby (The probability of
survival is very less in infant age, but once the baby get
accustomed to nature, the probability of failure decreases).
(c
8. Random failure:
In this class, constant probability of failure is
associated with items that fail from random
causes such as physical shocks, not related to
age. In such cases all items fail before aging
has any effect. Example is vacuum tubes.
9. Costs Associated with Maintenance
Our main aim in this chapter is to find optimal replacement period so as
to minimize the maintenance cost. Hence we are very much interested in
the various cost associated with maintenance. Various costs to be
discussed are:
Purchase cost or Capital cost: ( C )
This cost is independent of the age of the machine or usage of the
machine. This is incurred at the beginning of the life of the machine, i.e.
at the time of purchasing the machine or equipment. But the interest on
the invested money is an important factor to be considered.
10. (b) Salvage value / Scrap value / Resale value / Depreciation: (S)
As the age of the machine increases, the resale value decreases as its operating
efficiency decreases and the maintenance costs increases. It depends on the operating
conditions of the machine and life of the machine.
(c) Running costs including maintenance, Repair and Operating costs:
These costs are the functions of age of the machine and usage of the machine. As the
usage increases or the age increases, due to wear and tear, many components fail to
work and they are to be replaced. As the age increases, failures also increase and the
maintenance costs goes on increasing. At some period the maintenance costs are so
high, which will indicate that the replacement of the machine or equipment is essential.
12. Replacement of Items whose Maintenance Cost Increases with Time and the
Value of Money Remains Same During the Period
Case 1:
The total cost incurred on the item during period ‘y’ =
Capital cost + total maintenance cost in the period ‘y’ - Scrap value.
= C + M(y) - S
Hence average cost per unit of time incurred during the period ‘y’
on the item is given by:
u (y) = {C - S + M (y) } / y = G (y)
So, replace the item when the average annual cost reaches at the
minimum that will always occur at a time when the average cost
becomes equal to the current maintenance cost.
13. Case 2.
Here time ‘t’ is considered as a discrete variable
Average annual cost incurred during ‘y’ years is G (y) = {T (y) /
y} = {C + M (y) - S} / y
This show that do not replace, if the next years running cost is
less than the previous years average total cost but replace at the
end of ‘y’ years if the next year’s {i.e. (y + 1) th year} running
cost is more than the average cost of ‘y’ th year.
14. Points to remember
(a) If time is measured continuously, then the average
annual costs will be minimized by replacing the machine or
item, when the average cost to date becomes equal to the
current maintenance cost.
(b) If time is measured in discrete units, then the average
annual cost will be minimized by replacing the machine or
item when the next period’s maintenance cost becomes
greater than the current average cost
15. Problem
A firm is thinking of replacing a particular machine whose cost price is Rs.
12,200. The scrap value of the machine is Rs. 200/-. The maintenance
costs are found to be as follows:
Year 1 2 3 4 5 6 7 8
Maintenance
Cost in Rs. 220 500 800 1200 1800 2500 3200 4000
Determine when the firm should get the machine replaced.
16. Problem
The maintenance cost and resale value per year of a machine whose
purchase price is Rs. 7000/ is given below:
Year: 1 2 3 4 5 6 7 8
Maintenance
cost in Rs.: 900 1200 1600 2100 2800 3700 4700 5900
Resale value in Rs.: 4000 2000 1200 600 500 400 400 400
When should the machine be replaced?
17. CPM/PERT or Network Analysis
These are two techniques that are widely
used in planning and scheduling the large
projects. A project is a combination of various
activities. For example, Construction of a house
can be considered as a project. Similarly,
conducting a public meeting may also be
considered as a project.
18. • In CPM activities are shown as a network of precedence
relationships using activity-on-node network construction
– Single estimate of activity time
– Deterministic activity times
• USED IN: Production management - for the jobs of
repetitive in nature where the activity time estimates can
be predicted with considerable certainty due to the
existence of past experience.
19. • In PERT activities are shown as a network of precedence
relationships using activity-on-arrow network
construction
– Multiple time estimates
– Probabilistic activity times
• USED IN: Project management - for non-repetitive jobs
(research and development work), where the time and
cost estimates tend to be quite uncertain. This technique
uses probabilistic time estimates.
20. Applications of CPM / PERT
• Construction of a dam or a canal system in a region
• Construction of a building or highway
• Maintenance or overhaul of airplanes or oil refinery
• Space flight
• Cost control of a project using PERT / COST
• Designing a prototype of a machine
• Development of supersonic planes
21. Basic Steps in PERT / CPM
Project scheduling by PERT / CPM consists of four
main steps
• Planning
• Scheduling
• Allocation of resources
• Controlling
25. Advantages
•A PERT/CPM chart explicitly defines and makes visible dependencies (precedence
relationships) between the elements,
•PERT/CPM facilitates identification of the critical path and makes this visible,
•PERT/CPM facilitates identification of early start, late start, and slack for each activity,
•PERT/CPM provides for potentially reduced project duration due to better
understanding of dependencies leading to improved overlapping of activities and tasks
where feasible.
26. Disadvantages
•There can be potentially hundreds or thousands of activities and individual
dependency relationships,
•The network charts tend to be large and unwieldy requiring several pages to print and
requiring special size paper,
•The lack of a timeframe on most PERT/CPM charts makes it harder to show status
although colours can help (e.g., specific colour for completed nodes),
•When the PERT/CPM charts become unwieldy, they are no longer used to manage the
project.
27. Examples on CPM
Determine the early start and late start in respect of all node
points and identify critical path for the following network.
28. Solution
From the table, there are two possible critical paths
1 → 2 → 5 → 8 → 10
1 → 2 → 5 → 7 → 10
29. Project Evaluation and Review Technique (PERT)
The main objective in the analysis through PERT is to find out the completion
for a particular event within specified date. The PERT approach takes into
account the uncertainties. The three time values are associated with each
activity.
30. Optimistic time – It is the shortest possible time in which the activity can be
finished. It assumes that every thing goes very well. This is denoted by t0.
Most likely time – It is the estimate of the normal time the activity would take.
This assumes normal delays. If a graph is plotted in the time of completion and the
frequency of completion in that time period, then most likely time will represent
the highest frequency of occurrence. This is denoted by tm.
Pessimistic time – It represents the longest time the activity could take if
everything goes wrong. As in optimistic estimate, this value may be such that only
one in hundred or one in twenty will take time longer than this value. This is
denoted by tp.
31. In PERT calculation, all values are used to obtain the percent expected
value.
Expected time – It is the average time an activity will take if it were
to be repeated on large number of times and is based on the assumption
that the activity time follows Beta distribution, this is given by
Expected Time or Average Time = tE = (tO + 4tL + tP) / 6
The variance for the activity is given by
σ2 = [(tp – to) / 6] 2
Standard deviation = (tP - tO)/6= σ , tP - tO is known as range.
32. Example on PERT
A project has the following characteristics. Construct a PERT network. Find
the critical path and variance for each event.
Activity
Most optimistic time
(a)
Most pessimistic time
(b)
Most likely time
(m)
(1 – 2)
(2 – 3)
(2 – 4)
(3 – 5)
(4 – 5)
(4 – 6)
(5 – 7)
(6 – 7)
(7 – 8)
(7 – 9)
(8 – 10)
(9 – 10)
1
1
1
3
2
3
4
6
2
5
1
3
5
3
5
5
4
7
6
8
6
8
3
7
1.5
2
3
4
3
5
5
7
4
6
2
5
