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

Transcription

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

2 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 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

3 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 /- till 1200 hours. After each 1200 working Hours the air cleaner is to be replaced, which cost Rs /-. That makes the cost of every fourth preventive maintenance cost Rs. 20,043/-. All other preventive maintenance cost in between remain as Rs /- 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 Nil

4 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 Total Preventive maintenance cost C p = C i + C d Contd 90

5 Contd 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 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

6 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 Contd 92

7 Contd 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

8 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 = and η = F (t) =1 -.. = = 1 -. = = 0.86 R (t) = 1-F (t) = =

9 C f = = C p = = C (T) =.... = 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) Contd 95

10 Contd Contd 96

11 Contd 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

12 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 ) 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

13 800 C(T) y = 0.064x R² = Linear (C(T)) 700 y = 25.64e 0.000x R² = Expon. (C(T)) 600 y = 37.66ln(x) R² = Log. (C(T)) 500 y = 2E-05x x R² = Poly. (C(T)) Cost Per Hours y = 3.239x R² = Power (C(T)) 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 Polynomial Y = 2E -05x x Logarithmic Y = 37.66ln(x) Exponential Y = 25.61e 0.000x Power Y =

14 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 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 = = Rs /- Depreciation for 2000 = = Rs /- Depreciation for 3000 = = Rs /- 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

15 Table 5.6 Depreciation cost interval of diesel engine Sr. No. Working time Depreciation cost (Hours) According to data from Table 5.6 the depreciation cost trend shown in Graph 5.2 Depreciation cost Depereciation cost per hours y = x + 2E+06 R² = Working Hour Depreciation Graph 5.2 Depreciation cost interval of diesel engine per hour 101

16 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 Contd 102

17 Contd

18 According to data from Table 5.7 the depreciation cost per time to failure is shown in graph 5.3 Depreciation Cost Deperciation cost per hours y = x + 2E+06 R² = Working hours Depreciation Cost Graph 5.3 Depreciation cost of diesel engine per hour Ω