Chapter 5 ESTIMATION OF MAINTENANCE COST PER HOUR USING AGE REPLACEMENT COST MODEL

Chapter 5 ESTIMATION OF MAINTENANCE COST PER HOUR USING AGE REPLACEMENT COST MODEL 87

ESTIMATION OF MAINTENANCE COST PER HOUR USING AGE REPLACEMENT COST MODEL 5.1 INTRODUCTION Maintenance is usually carried out to ensure the reliability of a system. Generally, there are two types of maintenance: preventive maintenance and corrective maintenance depending on whether the system is maintained before or after it breaks down. Corrective maintenance usually is costly, so it is necessary to carry out preventive maintenance. However, it is unwise to preventively maintain the system too frequently. From this point, many policies integrating preventive and corrective maintenance have been proposed such as age replacement policy, block replacement policy, and periodic replacement policy. 5.2 PREVENTIVE MAINTENANCE It is said to be preventive maintenance when planned and coordinated inspections, repairs, adjustments, and replacements are carried out to minimize the problems of breakdown maintenance. This is based on the premise that prevention is better than cure. This practice involves planning and scheduling the maintenance work without interruption in production schedule and thus improves the availability of equipment. Under preventive maintenance, a systematic inspection of each item of equipment or at least the critical parts will be carried out at predetermined times to unfold the conditions that lead to production stoppage and harmful depreciation. Planning and implementation of a preventive maintenance practice is a costly affair because it involves the replacement of all deteriorated parts / components during inspection. However, the higher cost of maintenance usually gets compensated by the prolonged operational life of the equipment. To avoid serious breakdowns, the preventive mode of maintenance is usually implemented in complex system. 5.2.1 Estimation of Preventive Maintenance Cost The preventive maintenance carried out after the 300 Hours working of diesel engine for compressor application, the preventive maintenance is calculated as per the preventive maintenance cost plus the down time cost. Cp=C +C 88

Where C p - Preventive replacement cost. C i - Cost preventive servicing. C d - Cost of down time. Diesel engine for compressor application is used for making the water tube well, for one ft drill rate is Rs 60 /- per ft. (sixty Rupees only). In one hour the target is maximum 90 ft. of drill made. According to (Annexure C) rate list, one hour charge is Rs 5400/- Rupees / Hour. Therefore Down time cost per hours = 90 60= Rs 5400/- Rupees / Hours. According to the service manual, the preventive maintenance is to be carried out after every 300 working Hours. i.e. 300, 600, 900 Hours. In the each preventive maintenance engine oil, fuel filter and ring, oil filter, Bypass filter elements, seal O ring and inhibitor corrosion is to be replaced, which cost Rs. 13587/- till 1200 hours. After each 1200 working Hours the air cleaner is to be replaced, which cost Rs. 6455.49 /-. That makes the cost of every fourth preventive maintenance cost Rs. 20,043/-. All other preventive maintenance cost in between remain as Rs. 13587 /- According to (Annexure A and B) preventive maintenance cost per intervals are shown in Table 5.1 Sr. No Table 5.1 Cost of preventive maintenance interval Time interval (Hours ) Preventive maintenance cost 1 000-300 Nil 2 301-600 13587 3 601-900 27174 4 901-1200 40761 5 1201-1500 60803.49 6 1501-1800 74390.49 7 1801-2100 87977.49 8 2101-2400 101564.49 9 2401-2700 121606.49 10 2701-3000 135193.98 89

As the total preventive maintenance cost (C p ) is equal to the down time cost (C d ) plus the preventive replacement cost (C i ). So Table 5.2 shows the diesel engine time to failure, down time cost and preventive replacement cost. Engine No Table 5.2 Preventive maintenance cost of diesel engine Time of Failure (hours) Cost of down time C d Preventive replacement cost C i 1 10 64800 0 64800 Total Preventive maintenance cost C p = C i + C d 2 41 64800 0 64800 3 628 43200 27174 70374 4 681 1555200 27174 1582374 5 720 43200 27174 70374 6 801 129600 27174 156774 7 886 64800 27174 91974 8 892 43200 27174 70374 9 915 64800 40761 105561 10 950 129600 40761 170361 11 1091 43200 40761 83961 12 1100 64800 40761 105561 13 1135 64800 40761 105561 14 1194 1166400 40761 1207161 15 1250 129600 60803.49 190403.49 16 1276 64800 60803.49 125603.49 17 1338 43200 60803.49 104003.49 18 1556 64800 74390.49 139190.49 19 1633 43200 74390.49 117590.49 20 1668 64800 74390.49 139190.49 21 1687 64800 74390.49 139190.49 22 1854 43200 87977.49 131177.49 23 1876 64800 87977.49 152777.49 Contd 90

Contd 24 1895 64800 87977.49 152777.49 25 2000 64800 87977.49 152777.49 26 2106 583200 101564.49 684764.49 27 2109 43200 101564.49 144764.49 28 2117 64800 101564.49 166364.49 29 2174 194400 101564.49 295964.49 30 2346 64800 101564.49 166364.49 31 2434 64800 121606.98 186406.98 32 2440 64800 121606.98 186406.98 33 2470 64800 121606.98 186406.98 34 2547 64800 121606.98 186406.98 35 2570 64800 121606.98 186406.98 36 2607 64800 121606.98 186406.98 37 2628 64800 121606.98 186406.98 38 2646 43200 121606.98 164806.98 39 2904 43200 135193.98 178393.98 40 2961 129600 135193.98 264793.98 5.3 CORRECTIVE MAINTENANCE The practice of preventive maintenance brings out the nature of repetitive failures of a certain part of the equipment. When such repetitive types of failures are observed, corrective maintenance can be applied so that reoccurrence of such failures can be avoided. These types of failures can be reported to the manufacturer to suggest modifications to the equipment. Corrective maintenance can be defined as the practice carried out to restore the full performance of the equipment that has stopped working to acceptable standards. 5.3.1 Estimation of corrective maintenance cost Corrective maintenance cost is equal to replacement cost of failed component and spare part cost plus down time cost. C =C +C 91

Where C f - Corrective (Failure) cost C r - Cost of replacement system and components C d - Cost of down time Table 5.3 Corrective maintenance cost of diesel engine Engine No Time of Failure (hours) Cost of down time C d Cost of replacement system /components & spare cost C r Total Corrective maintenance cost C f = C r + C d 1 10 64800 3119.31 67919.31 2 41 64800 2363.01 67163.01 3 628 43200 40237.05 83437.05 4 681 1555200 21607.52 1576807.52 5 720 43200 20166.86 63366.86 6 801 129600 20579.85 150179.85 7 886 64800 611.51 65411.51 8 892 43200 20265.01 63465.01 9 915 64800 14145.03 78945.03 10 950 129600 87052 216652 11 1091 43200 6467.88 49667.88 12 1100 64800 11058.57 75858.57 13 1135 64800 5980.53 70780.53 14 1194 1166400 15071.81 1181471.81 15 1250 129600 5314.7 134914.7 16 1276 64800 6368.53 71168.53 17 1338 43200 10.76 43210.76 18 1556 64800 9089.22 73889.22 19 1633 43200 20551.34 63751.34 20 1668 64800 4820.75 69620.75 21 1687 64800 20502.96 85302.96 22 1854 43200 9710.06 52910.06 Contd 92

Contd 23 1876 64800 6275.22 71075.22 24 1895 64800 5052.43 69852.43 25 2000 64800 1296.6 66096.6 26 2106 583200 22159.92 605359.92 27 2109 43200 4695.61 47895.61 28 2117 64800 6275.32 71075.32 29 2174 194400 20260.22 214660.22 30 2346 64800 5814.06 70614.06 31 2434 64800 580.16 65380.16 32 2440 64800 20548.88 85348.88 33 2470 64800 13475.04 78275.04 34 2547 64800 6275.31 71075.31 35 2570 64800 8792.35 73592.35 36 2607 64800 7031.96 71831.96 37 2628 64800 4759.8 69559.8 38 2646 43200 20562.96 63762.96 39 2904 43200 973.41 44173.41 40 2961 129600 50000 179600 5. 4 AGE REPLACEMENT COST MODEL The classical policy used in maintenance application is called failure replacement policy, or age replacement policy (ARP). Under a preventive maintenance policy, the replacement of the component is either made after a specified time interval or in the case of component failure before the next scheduled time for replacement. The idea of this maintenance strategy is to replace the component with a new one (i.e. maximal repair) when it fails or when it has been in operation for T time units, whichever comes first. The expected maintenance cost per unit time, C (T), using Age replacement model can be written as C T = C F t + C R t R t dt 5.1 93

Age replacement model is more useful in practical application for the determination of maintenance cost (preventive and corrective) and estimation of maintenance cost per hour. To determine the cost function C(T), using Weibull distribution model is shown in equation. (5.2). C T = C 1 + C e 5.2 C =C +C (5.3) C =C +C (5.4) Where: C f - Corrective (Failure) cost C p - Preventive replacement cost C r - Cost of replacement system and components C d - Cost of down time C i - Cost preventive servicing F (t) - Cumulative distribution function R (t) - Reliability function η Scale parameters (characteristic life), β- Shape parameters (variation of the failure rate) Estimation the maintenance cost of diesel engine for compressor application using the Age replacement cost model t = 3000 hours, β = 2.198 and η = 1929.46 F (t) =1 -.. = 1 -.. = 1 -. = 1-0.14 = 0.86 R (t) = 1-F (t) = 1-0.86 = 0.14 94

C f = 4759.8 + 64800 = 69559.8 C p = 121606.98 +64800 = 186406.98 C (T) =.... = 204.56 Cost / hour Table 5.4 shows that, the diesel engines time to failure, total corrective maintenance cost, total preventive cost and the maintenance cost per hours according to age replacement cost model. Engine No Time of Failure (Hours) Table 5.4 Estimation of maintenance cost per hour Total Corrective maintenance cost C f Total Preventive maintenance cost C p Cost per Hour C(T) 1 10 67919.31 64800 21.82868 2 41 67163.01 64800 21.82614 3 628 83437.05 70374 25.87647 4 681 1576807.52 1582374 585.8583 5 720 63366.86 70374 26.6679 6 801 150179.85 156774 60.4073 7 886 65411.51 91974 34.81111 8 892 63465.01 70374 27.79095 9 915 78945.03 105561 40.96346 10 950 216652 170361 73.72687 11 1091 49667.88 83961 33.50566 12 1100 75858.57 105561 43.61573 13 1135 70780.53 105561 43.91336 Contd 95

Contd 14 1194 1181471.81 1207161 571.1687 15 1250 134914.7 190403.49 84.63097 16 1276 71168.53 125603.49 53.55221 17 1338 43210.76 104003.49 42.76986 18 1556 73889.22 139190.49 67.37771 19 1633 63751.34 117590.49 59.61391 20 1668 69620.75 139190.49 71.53767 21 1687 85302.96 139190.49 77.20069 22 1854 52910.06 131177.49 70.18088 23 1876 71075.22 152777.49 87.98213 24 1895 69852.43 152777.49 88.91573 25 2000 66096.6 152777.49 95.65787 26 2106 605359.92 684764.49 699.0904 27 2109 47895.61 144764.49 85.50696 28 2117 71075.32 166364.49 113.4588 29 2174 214660.22 295964.49 292.1138 30 2346 70614.06 166364.49 138.9079 31 2434 65380.16 186406.98 155.0444 32 2440 85348.88 186406.98 147.4846 33 2470 78275.04 186406.98 189.5244 Contd 96

Contd 34 2547 71075.31 186406.98 171.7456 35 2570 73592.35 186406.98 201.1435 36 2607 71831.96 186406.98 197.8183 37 2628 69559.8 186406.98 204.5676 38 2646 63762.96 164806.98 197.1762 39 2904 44173.41 178393.98 208.3454 40 2961 179600 264793.98 380.5549 5.5 REGRESSION ANALYSIS Linear regression analysis is statistical technique used to predict one variable (the dependent or output variable) from another (the independent or input variable) the methods of linear regression analysis are to describe the relationship between two variables by identifying an equation. Regression analysis is used to identify the line or curve which provides the best fit through a set of data points. This curve can be useful to identify a trend in the data, whether it is linear, parabolic, or of some other form. The result of linear regression analysis is regression line, a straight line that passes through the points that form the scatter diagram. The regression line is the basis for the equation that summarizes the relationship between the variables. The regression line also is used for describing the location of the values of the dependent variable. In addition to fitting a curve to given data, regression analysis can be used in combination with statistical techniques to determine the validity of data points within a data set. Deviation of regression line requires reference to the basic rules of algebra and properties of straight line, the equation for straight line is calculated by equation Y=mX+C Where In equation C is the Y axis intercept. That is C is that point on the Y - axis where the straight line intercepts (crosses) the Y- axis, or it is the point on Y axis 97

where the line would cross the vertical if it were extended and m is the regression line s slope. Correlation Regression analysis can be used to determine the equation of the line that passes through a collection of ordered points that from a scatter diagram. The next logical step is to try to quantify how strong the linear relationship is between the two variables. To accomplish this, the correlation, a method for determining what proportion of the variability in one set of scores can be predicated by the variability in another set of scores. Coefficient of Correlation: The strength of the evidence of a linear relationship between two variables is described; it is denoted by R Coefficient of determination: determine the proportion of variation in the dependent variable that can be explained by variation in the independent variable. The coefficient determination is calculated by squaring Coefficient of Correlation (R 2 ) 5.5.1 Estimation of appropriate maintenance cost model for diesel Engine To estimate the appropriate maintenance cost model for diesel engine using different regression models, some researchers has suggested following regression models. Linear model =Y=a+bx 5.5 Polynomial = Y=a+bx+cx 5.6 Logarithmic=Y=a+lnbx (5.7) Exponential= Y=ae 5.8 Power= Y=ax 5.9 Microsoft MS Excel software is used to perform regression analysis of data given in Table 5.4 for estimating maintenance cost model of diesel engine. According to the data the scatter plot is constructed, the trend lines are shows the nature of failure engine and related maintenance cost. Graph 5.1 shows the regression equations with coefficient correlation value, using this different regression model for the maintenance cost of diesel engine is obtained. The regression equations and coefficient of determination is shown in Table 5.5 98

800 C(T) y = 0.064x + 38.24 R² = 0.101 Linear (C(T)) 700 y = 25.64e 0.000x R² = 0.445 Expon. (C(T)) 600 y = 37.66ln(x) -124.4 R² = 0.064 Log. (C(T)) 500 y = 2E-05x 2-0.004x + 79.42 R² = 0.108 Poly. (C(T)) Cost Per Hours 400 300 y = 3.239x 0.470 R² = 0.301 Power (C(T)) 200 100 0 0 500 1000 1500 2000 2500 3000 3500-100 Time to failure Graph 5.1 Maintenance cost Regression models Table 5.5 Maintenance cost regression models Sr. No Regression Models Regression Equations R 2 1 Linear Y = 0.064X+ 38.24 0.101 2 Polynomial Y = 2E -05x 2-0.004x +81.82 0.108 3 Logarithmic Y = 37.66ln(x) 124.4 0.064 4 Exponential Y = 25.61e 0.000x 0.445 5 Power Y = 3.239.438 0.301 99

5.6 ESTIMATION OF DEPRECIATION COST OF DIESEL ENGINE FOR COMPRESSOR APPLICATION The cost of using long-term operating assets, also known as depreciation, is recorded gradually as the assets are used. Depreciation is an allocation of the cost that is calculated using an asset s acquisition cost, it s expected life, and expected residual value. Straight-line (Machine hour) Method: The straight-line method (SLM), also known as the fixed installment method, allocates an equal amount of an asset s cost to each year of its expected useful life. This allocation assumes that an equal amount of an asset s potential is consumed in each period of its life. However, this may not be true under all circumstances. The repairs and maintenance cost will be lower in earlier years of use but will gradually be higher as the asset becomes old. Moreover, the asset might have different capacities over different years of its life. The amount of depreciation for each period is computed by deducting the asset s expected residual value from its acquisition cost, and dividing the result by the assets expected economical and useful life. An engine is purchased Rupees 1500000 Lakhs and estimated working hours are 40,000 Hours. hours Engine hour rate of depreciation = Cost / Life in terms of effective working = 15, 00,000 / 40,000 = Rs 37.5 Cost / Hours Engine hour rate of depreciation = Rs 37.5 per hour Depreciation for 1000 = 37.5 1000 = Rs 37500 /- Depreciation for 2000 = 37.5 2000 = Rs 75000 /- Depreciation for 3000 = 37.5 3000 = Rs 112500 /- The engine is working per year approximately 3000 hour, Accordingly company manual mentioned that, diesel engine is work positively upto 40,000 Hours and more without any major breakdown. The depreciation of diesel engine per 3000 Hours is given in Table 5.6. 100

Table 5.6 Depreciation cost interval of diesel engine Sr. No. Working time Depreciation cost (Hours) 1 0 1500000 2 3000 1387500 3 6000 1275000 4 9000 1162500 5 12000 1050000 6 15000 937500 7 18000 825000 8 21000 712500 9 24000 600000 10 27000 487500 11 30000 375000 12 33000 262000 13 36000 150000 14 39000 37500 15 40000 0 According to data from Table 5.6 the depreciation cost trend shown in Graph 5.2 Depreciation cost 2000000 Depereciation cost per hours 1500000 1000000 500000 0-500000 y = -37.50x + 2E+06 R² = 1 0 10000 20000 30000 40000 50000 Working Hour Depreciation Graph 5.2 Depreciation cost interval of diesel engine per hour 101

In Table 5.7 represents the time to failure of diesel engines with actual depreciation cost Table 5.7 Depreciation cost of Diesel engine per hour Engine No Time of Failure (Hours) Cost per hour depreciation Total cost Project Cost Actual cost after depreciation 1 10 37.5 375 1500000 1499625 2 41 37.5 1537.5 1500000 1498462.5 3 628 37.5 23550 1500000 1476450 4 681 37.5 25537.5 1500000 1474462.5 5 720 37.5 27000 1500000 1473000 6 801 37.5 30037.5 1500000 1469962.5 7 886 37.5 33225 1500000 1466775 8 892 37.5 33450 1500000 1466550 9 915 37.5 34312.5 1500000 1465687.5 10 950 37.5 35625 1500000 1464375 11 1091 37.5 40912.5 1500000 1459087.5 12 1100 37.5 41250 1500000 1458750 13 1135 37.5 42562.5 1500000 1457437.5 14 1194 37.5 44775 1500000 1455225 15 1250 37.5 46875 1500000 1453125 16 1276 37.5 47850 1500000 1452150 17 1338 37.5 50175 1500000 1449825 18 1556 37.5 58350 1500000 1441650 19 1633 37.5 61237.5 1500000 1438762.5 20 1668 37.5 62550 1500000 1437450 21 1687 37.5 63262.5 1500000 1436737.5 22 1854 37.5 69525 1500000 1430475 23 1876 37.5 70350 1500000 1429650 24 1895 37.5 71062.5 1500000 1428937.5 25 2000 37.5 75000 1500000 1425000 26 2106 37.5 78975 1500000 1421025 Contd 102

Contd 27 2109 37.5 79087.5 1500000 1420912.5 28 2117 37.5 79387.5 1500000 1420612.5 29 2174 37.5 81525 1500000 1418475 30 2346 37.5 87975 1500000 1412025 31 2434 37.5 91275 1500000 1408725 32 2440 37.5 91500 1500000 1408500 33 2470 37.5 92625 1500000 1407375 34 2547 37.5 95512.5 1500000 1404487.5 35 2570 37.5 96375 1500000 1403625 36 2607 37.5 97762.5 1500000 1402237.5 37 2628 37.5 98550 1500000 1401450 38 2646 37.5 99225 1500000 1400775 39 2904 37.5 108900 1500000 1391100 40 2961 37.5 111037.5 1500000 1388962.5 41 3000 37.5 112500 1500000 1387500 42 6000 37.5 225000 1500000 1275000 43 9000 37.5 337500 1500000 1162500 44 12000 37.5 450000 1500000 1050000 45 15000 37.5 562500 1500000 937500 46 18000 37.5 675000 1500000 825000 47 21000 37.5 787500 1500000 712500 48 24000 37.5 900000 1500000 600000 49 27000 37.5 1012500 1500000 487500 50 30000 37.5 1125000 1500000 375000 51 33000 37.5 1237500 1500000 262000 52 36000 37.5 1350000 1500000 150000 53 39000 37.5 1462500 1500000 37500 54 40000 37.5 1500000 1500000 0 103

According to data from Table 5.7 the depreciation cost per time to failure is shown in graph 5.3 Depreciation Cost 2000000 Deperciation cost per hours 1500000 1000000 500000 0-500000 y = -37.50x + 2E+06 R² = 1 0 10000 20000 30000 40000 50000 Working hours Depreciation Cost Graph 5.3 Depreciation cost of diesel engine per hour ----------------Ω---------------- --- 104