ASSET MANAGEMENT FOR FLEETS OF MINING EQUIPMENT IN OPEN-PIT OPERATIONS Víctor Barrientos Boccardo Skype: victor.barrientos.boccardo, Chile Abstract: The results obtained from three studies are presented in this article. The first study was conducted on the performance of final drives in 100-ton trucks, the results of which indicate that the repaired final drives have a cost of US$3.42 per hour, in comparison with US$6.48 and US$8.71 per hour of new and exchanged components, respectively. The second study was conducted based on the optimum rebuild strategy for a pool of components in a fleet of drills, where the results estimate that the optimum strategy will generate a cost of US$19.34 per hour and approximately 546 annual machine shutdown hours of fleet for 16 selected components. The last study refers to the methodology CAE (Equivalent Annual Cost) to define the optimum time to replace 100-ton trucks, based on equipment investment, component and spares cost. The decision point is when major components such as diesel engine, transmission, differential, etc. need to be replaced. Keywords: Repairable system; Multi-component system; Age replacement policy1 INTRODUCTIONAs assets accumulate hours of operation, they become deteriorated by usage. Production capacity of the asset which isgradually affected by the increase of costs is preserved by means of preventive maintenance, structural repairs, and majorand minor components change strategy.In capital-intensive industries, maintenance costs are a significant portion of operating costs and production costs. Instudies conducted on open-pit mining operations in Chile, maintenance costs have been estimated at 44% of totalproduction cost of the mine. Likewise, it is estimated that the repair cost for major systems accounts for 9 to 18% of totaloperating cost (Lugtigheid, Banjevic & Jardine, 2008) , (Topal & Ramazan, 2010).In 2005, mine purchased new equipment for drilling, loading and hauling processes, as well as an ancillary equipmentfleet. The versions of such equipment were new and, therefore, there was no reliable information about the useful life oftheir components at disposal. The only information available was that which was provided by the manufacturer.In five years of operation, the history information about the useful life of new, exchanged and repaired components isanalyzed. By assessing the cost of the repaired components and overall operation and maintenance (O&M) costs, threelines of study were developed with the purpose of reducing maintenance costs, increasing availability of the equipment,and reducing the workload at maintenance workshops. The lines of study were:• Strategy to change the final drives of a fleet of 100-ton trucks.• Strategy to change a pool of components for a fleet of hydraulic drills.• Strategy to replace 100-ton mining trucks in open-pit operations.
2 DESCRIPTION OF METHODOLOGIESThe assessment methodologies for the three lines of study are shown below:2.1 Strategy to change final drives of a fleet of 100-ton trucksThis methodology consists in determining the relevant repair cost and useful life for different types of components (new,repaired, and exchanged). With the previous, one can determine different costs per hour in order to compare and identifythe rebuild modality that represents the most inexpensive alternative under a condition-based maintenance.LN : Life of an original new component.LR : Life of own component repaired by the dealer.LI : Life of an exchange component purchased from the dealer.QN : Cost of new component, transport and acquisition management.QR : Cost of repaired component, transport, inventory, and acquisition management.QI : Cost of the exchanged component, transport and acquisition management. Minimum = Figure 1 Function to define the minimum cost per hourFor this assessment, the useful life of a component was considered such as its hours of operation until the time itevidenced loss in its optimum performance, following monitoring of the condition (sample of oil, magnetic plugs,magnetic grids, etc.) where the component needs to be restored.For the purpose of the assessment, data corresponding to changes of components based on hours of operation and repairsmade by alternative suppliers other than the Dealer was not considered.The cost of a component includes its value as a new, repaired or exchanged, in addition to transport expense, inventorycost and expenses related to the acquisition management. The costs related to loss of profit caused by unavailability havenot been estimated in this study.2.2 Strategy to change a pool of components for a fleet of hydraulic drillsThis methodology consists in determining the relevant costs of repair and useful life for every new and repairedcomponent, as well as the hours of shutdown required to replace the component. Those values allow the determination ofthe different costs per hour and machine shutdown hours, in order to compare and assess the rebuild modality thatrepresents the most relevant alternative.The following nomenclature is defined for the proposed methodology:Ln : Useful life of an original new componentLr : Useful life of a component repaired in workshopsCn : Cost of new component, transport and acquisition managementCr : Cost of repaired component, transport, inventory and acquisition managementHM : Machine shutdown hours required to change the componentHMaqYear : Annual machine hours of operation for the equipment fleet
d : Total costs per hour for a decision vectorf : Total machine shutdown hours for a decision vectorPoint (d,f) : Pair of costs per hour and machine shutdown hours for a decision vectorA rebuild strategy or replacement strategy may be defined for each component or classes of components – i -, which maybe a new one ( xj = 0) or a repaired one ( xj = 1). Each rebuild strategy represents a sequence of component changes thatdefine a cost per hour and machine shutdown hours.A similar definition is used in other studies (Aissani, Chateauneuf & Laggoune, 2010).When defining a particular rebuild strategy, this may be represented as a decision vector X= [x1; x2; xi;xi+1….; xp ],which defines an ordered pair Point (d,f) and where the values of “d” and “f” may me calculated as shown in Figure N°2. New or repaired component cost Useful life of new or repaired component Point ( d , f ) Decision vector Machine shutdown hours Useful life of new or repaired component Figure 2 Calculation of Point (d,f) for a decision vector2.3 Strategy to replace 100-ton mining trucks in open-pit operationsThis methodology consists in determining the optimum time to replace equipment for a new one. This assessment usesthe equivalent annual cost (CAE) calculation methodology that establishes the minimization of the current value of futurecosts as the optimum point, considering the following variables:CAE n : Equivalent annual cost in life year i= 1, 2, 3 …. nA : Value of the initial investment or acquisition cost in year zeroTn : Residual or trading value of the asset in year n of lifeC1 : Operation and maintenance (O&M) cost in year 1 of lifeCi : Operation and maintenance (O&M) cost in years i= 2, 3….n of lifen : Years of life for equipment replacementr : Agreed discount rateThe mathematical expression that needs to be resolved is shown in Figure N° 3.
Figure 3 Equivalent Annual Cost (CAE)The objective of this expression is to annualize two different behavior terms: on the one hand, the total costs of operationthat increase as time passes by; and, on the other hand, the net equipment investment that tends to decrease whenconsidering more years in its annualization. The combination of both criteria may lead to a point where the equivalentannual cost (CAE) has a minimum value. The number of years for which the equivalent annual cost (CAE) is minimumaccounts for the economic life of the equipment being analyzed.The assessment methodology consisted of resolving the Equivalent Annual Cost (CAE) function mathematically shownin Figure N° 3. The optimum replacement point corresponds to the minimum value of the function, where the conditionshown in Figure N° 4 must be met. Figure 4 Criterion to define the optimum replacement point3 RESULTSThe results obtained according to the assessment methodologies for the three lines of study are shown below:3.1 Results for final drives of a fleet of 100-ton trucks, durations and costs per hourOnce the history data of useful life and the costs per hour for final drives of 100-ton trucks have been collected andtabulated, it is concluded that:• The time of useful life for new components is 2.72 times (18,485 hours vs. 6,786 hours) higher than that of repaired ones.• The value of exchange components is 2.61 times (US$60,667 vs. US$23,201) higher than that of repaired ones.• In 85% of the cases, the hour cost of a new component is higher than that of a repaired one.• In 100% of the three existing cases, the hour cost of an exchange component is higher than that of the repaired one.When comparing the rebuild modalities, they show that the repaired components have a cost of US$3.42 per hour, incomparison with US$6.48 and US$8.71 per hour for the new and exchange components, respectively.A “repaired component” is defined as one owned by the operation and subjected to successive repairs; therefore, itshistory is known.
An “exchange component” is defined as a component of unknown history and acquired from dealer, giving a “core” aspartial payment. Hours of Operation vs. Costs per Hours Costs per hour [US$ / Hrs] Hours of Operation (hrs) New Exchange Repaired Poly. (Exchange) Poly. (New) Poly. (Repaired) Figure 5 Costs per hour by type of rebuild.It is noted that eight of the components restored as repair and exchange have shown loss in their optimum cost. Theamount of information about the useful life of these rebuild alternatives is not sufficient in order to arrive at conclusiveresults about the cost in dollars per hour. This information provides us with the first signs and trends of the repair costsand useful life of the repaired and exchange final drives.3.2 Results for pool of 16 components of hydraulic drills, costs and machine shutdown hoursEach one of the decision vectors X = [x1; x2; xi;xi+1….; xP ] generates an ordered pair Points (d, f) indicating the costsper hour and the machine shutdown hours for this decision vector. The total decision vectors or ordered pairs Points (d,f) is of 2p, where - for the case study - the vectors are a total of 216 = 66,536 possible combinations. Some of them areshown in Figure N° 6. Rebuild Strategies represented by Points (d,f) f = Machine shutdown hours [Hrs] Total Points (d,f) of 66,536 d = Cost per hour [US$/hrs] Figure 6 Zone of result for vectors Points (d,f)Each one of Point (d,f) in Figure N° 6 represents a specific rebuild strategy for the 16 classes of components included inthis assessment. Table N° 1 shows its decision vector and the name of the rebuild strategy for every relevant point inFigure N° 6.
Table 1 Decision vectors shown in Figure N° 6 Machine shutdown Costs per hour Points Restoration Strategy Description Decision vector hours [Hrs] [US$/hrs] Current strategy Strategy with less machine hours New strategy (proposal) Strategy with less costs per hourAs an initial approach to find the New Rebuild Strategy (Point 2), we are looking for the strategy that meets the criteriato reduce the machine shutdown hours and components expense reduction, which in this case corresponds to Point B. Itis estimated that the rebuild strategy represented by Point B will have a cost of US$18.12 per hour and about 586machine shutdown hours for the fleet.It is considered that the New Rebuild Strategy for the drill fleet – which is referred to as Point 2 – is located near Point B,because Point B corresponds to a mathematically based decision vector that needs to be reviewed to make sure that it isoperatively executable. Some considerations that need to be taken into account when selecting the New Rebuild Strategy(Point 2) are:• Accessibility of components in equipment• Components that should be changed as a whole according to technical recommendations• Availability and lead time of new and repaired componentsWhen applying the foregoing criteria, the New Rebuild Strategy is obtained for the fleet of drills shown in Point 2. It isestimated that the cost will be US$19.34 per hour and about 546 machine shutdown hours for the drill fleet. Table N° 2shows the details of the New Rebuild Strategy for the pool of 16 components under study. Table 2 New Rebuild Strategy New Rebuild N° Component Classes Strategy Point 2 (d**, f**) 1 Alternator Repaired 2 Starter motor Repaired 3 Turbo Repaired 4 Double pump New 5 Pulldown pump New 6 Rotation pump New 7 Fixed flow motor New 8 Variable flow motor New 9 Indexing motor New 10 Carrousel cylinder Repaired 11 Tower lift cylinder Repaired 12 Breakout wrench cylinder Repaired 13 Leveling cylinder Repaired 14 Pulldown cylinder New 15 Pump drive New 16 Rotational head Repaired
3.2 Results obtained when assessing the replacement of 100-ton mining trucksThe analysis compares operation and maintenance (O&M),t and investment costs for new equipment vs. existing truckswith a residual value according to equipment remaining life.The results from the various simulations of Equivalent Annual Cost (CAE) for different asset investment costs and(O&M) cost profiles show an optimal replacement point. This point is mainly related to the cost associated to replacemajor components such as diesel engine, drive and differential and cost of a new equipment and residual value.In addition a key variable for mining trucks is the structural state of the chassis, which usually conditions its operationalprojection in terms of extend equipment life.Example of the methodology applied to a 100 ton haul truck fleet.Figure N° 7 shows the results of the simulation for the acquisition price for trucks in 2005 (Equip Price 1) and theestimated investment cost at 2011 price list (Equip Price 2). The trend is similar in both cases, in the case of higherequipment investment (Equip Price 2) CAE after 36.000 operating hours is almost flat, and reflects mainly the impact ofequipment purchase, no major variations after 42.000 hrs.The first major component change is around 24.000 hrs, period associated with an increase in CAE from 612 KUS$/yearthat continues up to 42.000 hrs which is around 700 KUS$/year, afterwards CAE is relatively similar. Equivalent Annual Cost (CAE) 1,100,000 1,000,000 900,000 CAE ( US$ ) 800,000 Equip Price 2 700,000 Equip Price 1 600,000 500,000 0 6 12 18 24 30 36 42 48 54 60 66 72 78 84 90 96 Equipment Operating Hours ( x 1.000 hrs ) Figure 7 Equivalent Annual Cost (CAE) for different Equipment Price4 CONCLUSIONSCase 1 : Final drives in 100-ton truckThe limitation of this methodology is that it does not include overall costs such as unavailability.When comparing rebuild strategies, they do not indicate that the repaired final drives have a cost of US$3.42 per hour, incomparison with US$6.48 and US$8.71 per hour of new and exchange final drives, respectively. Thus, the strategyadopted to have own components repaired as the preferred alternative is reaffirmed.
Case 2 : Drill fleet componentsThis methodology requires a significant amount of reliable information related to the costs and useful life. As morecomponents are included in the assessment, an exponential growth of rebuild strategy options is evidenced.Conclusively, for the study of the pool of 16 components of the drill fleet, it is established that a cost of US$19.34 perhour and 546 machine shutdown hours a year are possible to obtain. An annual expense reduction of about US$60,000 ayear and an availability increase of 1.5% to 2.0% is estimated.Case 3 : Replacement of 100-ton trucksThis methodology has various limitations, such as: It does not consider availability and increased productivity in case of replacement for new equipment. It requires an estimation of operation and maintenance (O&M) costs, which may be particularly inaccurate in the case of new equipment that lack consolidated design and cost. The estimation of asset sales value may be an inaccurate variable due to the variables that impact the value. The sales value will depend on aspects such as the number of equipment to be replaced, the conditions of the rental and sales market – and in general terms, on the world economy and local activity level, and the geographical location.Once it is determined – by means of the equivalent annual cost (CAE) method – that the optimum replacement time for apiece of equipment is today and not tomorrow, it is important to calculate the NPV (Net Present Value) and IRR (InternalRate of Return) indicators in order to make this type of projects comparable with others of similar characteristics.5 GENERAL COMMENT The amount of information on repair costs and useful life of components for mining equipment may be scarce in the case of fleets composed of few equipment, as in these three cases. The models and serial numbers of the mining equipment change every eight to ten years. This fact introduces an additional variable of technical obsolescence when some major components have been discontinued. Considering the importance of expenses in components in comparison with total maintenance, it is important to conduct similar studies, both on major and minor components, taking into account the benefits derived from keeping rigorous record of the changes and repairs made, as well as an adequate control of information if maintenance is conducted with own, external or combined resources. Such information will serve as a data base for more accurate analysis of maintenance processes. An extended version of these three lines of study was presented at the “6° Encuentro Internacional de Mantenedores de Equipos Mina” (www.mantemin.cl) held in Chile in September, 2011.6 REFERENCESAissani D., Chateauneuf A., Laggoune R. (2010) Impact of few failure data on the opportunistic replacement policy for multi-component systems. Reliability Engineering and System Safety 95 108-119Lugtigheid D.; Banjevic D.; Jardine A. (2008) System repairs: When to perform and what to do. Reliability Engineering and System Safety, pp. 604-615Topal E.; Ramazan S. (2010) A New MIP Model for Mine Equipment Scheduling by Minimizing Maintenance Cost. European Journal of Operational Research, pp. 1065-1071.