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Replacement & Maintenance Analysis Unit IV MG 6863

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Engineering Economics Unit IV MG 6863

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Replacement & Maintenance Analysis Unit IV MG 6863

  1. 1. S.Balamurugan AP/Mech AAA College of Engg. & Tech.3/19/2018 1
  2. 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. 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This would lead to decline in value of service given, increased in operation & maintenance cost Obsolescence – Becoming out of date, due to improvement in technology.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. 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. 5. Maintenance Costs
  6. 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)
  7. 7. 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
  8. 8. ii. When should the equipment be replaced if i = 12 % C Economic Life of the equipment = 7 Years
  9. 9. 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
  10. 10. 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
  11. 11. 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
  12. 12. 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?
  13. 13. 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
  14. 14. 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.

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