Engineering Economics Unit IV MG 6863

1. S.Balamurugan AP/Mech AAA College of Engg. & Tech.3/19/2018 1

2. Replacement & maintenance analysis
• Equipment's should be continuously monitored for efficient functioning. Otherwise
quality of products will be poor.
• Over the period of time, the cost of their operation & maintenance would
increases.
• So it is necessary to maintain the equipment in good operating conditions with
economical cost.

3. Replacement & maintenance analysis
• Successful company has to take the following decisions,
– Whether to replace the old equipment or to retain it by
taking the cost of maintenance & operation into account.
Reasons for Replacement
Physical Damage – Changes in physical condition of
the machine. This would lead to decline in value of service
given, increased in operation & maintenance cost
Obsolescence – Becoming out of date, due to
improvement in technology.
Facilities can’t able to meet the demand
» Replacement of the existing equipment with a new one.
» Augmenting the existing one with an additional equipment.
Machines are to be periodically replaced, otherwise it will
become uneconomical for production

4. Types of Maintenance
Preventive Maintenance
• Proactive Approach
• The periodical inspection &
service activities which are
aimed to detect potential
failures & perform minor
adjustments or repairs.
• This will prevent major
operating problems in future.
• Preventive maintenance is
planned in such a way that it
will not disturb the normal
operations hence no down
time cost of equipment.
Breakdown Maintenance
• Reactive Approach
• The repair which is generally
done after the equipment has
attained down state.
• Emergency nature which will
associated penalty in terms of
high maintenance cost & down
time of an equipment.
Engine Oil : To prevent engine corrosion, it is very important that the engine
maintains the amount of oil it is supposed to receive. Negligence in this aspect will
lead to a deteriorating engine performance, resulting in less mileage.

5. Maintenance Costs

6. Types of Replacement Problem
A. Replacement of assets that deteriorate with
time.(Due to gradual failure, or wear & tear
of the components of the machine)
1. Determination of economic life of an asset.
2. Replacement of an existing asset with a new
asset.
B. Simple probabilistic model for assets which
fail completely (Due to sudden failure).
(Electronic Items)

8. Determination Economic life of an Asset
• A firm is considering replacement of an equipment, whose first cost is Rs.4,000 and the scrap value
is negligible at the end of any year. Based on experience, it was found that the maintenance cost is
zero during the first year and it increases by Rs.200 every year thereafter.
i.When should the equipment be replaced if i = 0%, ii. When should the equipment be replaced if i = 12%
i. When should the equipment be replaced if i = 0%
Economic Life of the equipment = 6 Years

9. ii. When should the equipment be replaced if i = 12 %
C
Economic Life of the equipment = 7 Years

10. Concept of Challenger & Defender
If an existing equipment is considered for replacement with a new equipment,
then the existing equipment is known as the defender and the new equipment is known
as challenger

11. Replacement of an existing asset with a new asset
• In this method, represents the concept of comparison of replacement of
an existing asset with a new asset.
• Step 1 – Annual Equivalent cost of each alternative should be computed
first.
• Step 2 – The alternative which has the least cost should be selected as
the best alternative.
Sunk Cost
• It is the past cost of an equipment / asset.
• Assume a Machine purchased for Rs.5,00,000 about 3 years back, now it
is considered for replacement.
• The supplier of the new equipment will take the old one for Rs.3,00,000.
this value of the equipment should be taken for the analysis

12. Concept of Challenger & Defender
Two years ago, a machine was purchased at a cost of Rs.2, 00,000 to be useful for
eight years. Its salvage at the end of its life is Rs.25,000. The annual maintenance cost
is Rs. 25,000. The market value of the present machine is Rs.1,20,000.Now, a new
machine to cater to the need of the present machine is available at Rs. 1, 50,000 to be
useful for six years. Its annual maintenance cost is RS. 14,000. The salvage value of
the new machine is RS. 20,000. Using an interest rate of 12%, find whether it is worth
replacing the present machine with the new machine.
Alternative 1 – Present Machine – Defender (Annual Equivalent Cost) (Cost Dominated)
1,20,000 A A A A A A
-2 -1 0 1 2 3 4 5 6
2,00,000 25,000
AE(12%) = ( P - F ) (A/P, i, n) + F × i + A = ( P - F ) (A/P, 12%, 6) + F × i + A
= ( 1,20,000 – 25,000) (0.2432) + 25,000 (0.12) + 25,000 = Rs. 51,104.
Alternative 2 – New Machine – Challenger (Annual Equivalent Cost) (Cost Dominated)
AE(12%) = ( P - F ) (A/P, i, n) + F × i + A = ( P - F ) (A/P, 12%, 6) + F × i + A
= ( 1,50,000 – 20,000) (0.2432) + 20,000 (0.12) + 14,000 = Rs. 48,016.
It is suggested that the present machine is replaced with new machine.
S.Balamurugan AP/Mech AAA College of Engg. & Tech.3/19/2018 12

13. Simple probabilistic model for assets which fail
completely
• Electronic item fails all of a sudden.
• Failure may result in complete breakdown.
• To avoid the breakdown, we use the following
replacement policy.
– Individual Replacement policy - In this, an item
is replaced immediately after its failure
– Group Replacement policy - In this following
decisions are made
• At what equal intervals are all the items to be replaced
simultaneously with a provision to replace the items
individually which fail during a fixed group replacement
policy?

14. There are 10,000 bulbs in a decorative set. When any bulbs fail to be replaced, the cost of
replacing a bulb individually is Rs.Rs. 1 only. If all the bulbs are replaced at the same time, the
cost per bulb would be reduced to Rs.0.35. The percentage of bulbs surviving at the end of
month (t) i.e. S(t) and the probability of failures during the month (t) i.e. P(t) are given below.
Determine the optimal replacement policy.
t 0 1 2 3 4 5 6
S(t) 100 97 90 70 30 15 0
P(t) --- 0.03 0.07 0.20 0.40 0.15 0.15
Simple probabilistic model for assets which fail completely
Step 1 - Number of Bulbs replaced at the end of ith month (Ni) (i=0 to 6)
N0 = 10,000
N1 = N0 × P1 = 10,000 × 0.03 = 300
N2 = N0 × P2 + N1 × P1 = 10,000 × 0.07 + 300 × 0.03 = 709
N3 = N0 × P3 + N1 × P2 + N2 × P1 = 10,000 × 0.20 + 300 × 0.07 + 709 × 0.03 = 2042.27 = 2043
N4 = N0 × P4 + N1 × P3 + N2 × P2 + N3 × P1 = 10,000 × 0.40 + 300 × 0.20 + 709 × 0.07 + 2043 × 0.03 = 4170.92 = 4171
N5 = N0 × P5 + N1 × P4 + N2 × P3 + N3 × P2 + N4 × P1 = 10,000×0.15 + 300×0.40 + 709×0.20 + 2043×0.07 + 4171×0.03
= 2029.94 = 2030
N6 = N0 × P6 + N1 × P5 + N2 × P4 + N3 × P3 + N4 × P2 + N5 × P1 = 10,000×0.15 + 300×0.15 + 709×0.40 + 2043×0.20 +
4171×0.07 + 2030×0.03 = 2590.07 = 2591

15. Step 2 – Determination of Individual Replacement cost
Expected Life of Each Bulb = 𝑖 × Pi𝑛
𝑖=1 = 𝑖 × Pi6
𝑖=1
= 1 × 0.03 + 2 × 0.07 + 3 × 0.20 + 4 × 0.40 + 5 × 0.15 + 6 × 0.15
= 4.02 months
Average number of failures / month = 10,000 / 4.02 = 2487.56 = 2488 (Approximately)
Cost of Individual replacement = (No. of failures / month × Individual replacement cost / resistor)
= 2488 × 1 = Rs.2,488
Step 3 – Determination of Group Replacement cost
End of
month
A
Cost of replacing
10,000 bulbs at a
time (B) Rs.
Cost of replacing bulbs individually during
given replacement period
(C) Rs.
Total Cost
(D = B+C)
Rs.
Average
cost Rs.
(E= D/A)
1 10,000 × 0.35 = 3,500 300 × 1 = 300 3,800 3,800
2 3,500 (300+709) × 1 = 1,009 4,509 2254.5
3 3,500 (300+709+2043) × 1 = 3,052 6,552 2184
4 3,500 (300+709+2043+4171) × 1 = 7,223 10,723 2680.75
Average cost / Month is minimum for third month.
Minimum Group Replacement cost < Individual replacement cost
Group Replacement Policy is the Optimum replacement policy.