Download Engineering Economics By R. Panneerselvam – Designed as a text book for undergraduate students in various engineering disciplines – mechanical . This book on Engineering Economics is the outgrowth of my several years of teaching postgraduate courses in industrial engineering and production. Construction Management - II / Basics of Engineering Economics. Performance: Slidedoc. BUTE DCTM / Engineering Programs in English / Dr. Zoltán .

Author: | BELLA OLAUSEN |

Language: | English, Spanish, Dutch |

Country: | Colombia |

Genre: | Environment |

Pages: | 305 |

Published (Last): | 04.05.2016 |

ISBN: | 880-5-28123-154-2 |

PDF File Size: | 19.17 MB |

Distribution: | Free* [*Regsitration Required] |

Uploaded by: | FRANCES |

Engineering Economics. Cash Flow FERC. a1. Engineering Economics . What is the book value of the asset in the previous example after 3 years. vitecek.info Books will be door delivered after payment into AIR WALK engineering economics − Element of costs, Marginal cost, Marginal Revenue, Sunk. MG ENGINEERING ECONOMICS AND COST ANALYSIS. L T P C. 3 0 0 3 . The issues that are covered in this book are elementary economic analysis.

Save extra with 3 Offers. Designed as a textbook for undergraduate students in various engineering disciplinesMechanical, Civil, Industrial Engineering, Electronics Engineer-ing and Computer Scienceand for postgraduate students in Industrial Engineering and Water Resource Management, this comprehensive and well-organized book, now in its Second Edition, shows how complex economic decisions can be made from a number of given alternatives. It provides the managers not only a sound basis but also a clear-cut approach to making decisions. What is more, the book adequately illustrates the concepts with numerical problems and Indian cases. While retaining all the chapters of the previous edition, the book adds a number of topics to make it more comprehensive and more student friendly. Discusses different types of costs such as average cost, recurring cost, and life cycle cost.

Annual equivalent method 4. Rate of return method These methods are discussed in detail in Chapters 4—7. Explain the time value of money.

Give practical applications of various interest formulas. A person deposits a sum of Rs. Find the future amount of the deposited money at the time of admitting his son in the professional course. A person needs a sum of Rs. A person who is just 30 years old is planning for his retired life. A company is planning to expand its business after 5 years from now.

The expected money required for the expansion programme is Rs. The company can invest Rs. If not, find the difference in amounts for which the company should make some other arrangement after 5 years. A financial institution introduces a plan to pay a sum of Rs. Find the annual equivalent amount that a person should invest at the end of every year for the next 10 years to receive Rs. The money required for the expansion programme is Rs. A company wants to set-up a reserve which will help it to have an annual equivalent amount of Rs.

Find the single-payment that must be made as the reserve amount now. An automobile company recently advertised its car for a down payment of Rs. Alternatively, the car can be taken home by customers without making any payment, but they have to pay an equal yearly amount of Rs.

Suggest the best alternative to the customers. A company takes a loan of Rs. Find the equal installment amount that should be paid for the next 20 years. A bank gives loan to a company to purchase an equipment which is worth of Rs. This amount should be repaid in 25 yearly equal installments. A working woman is planning for her retired life. She has 20 more years of service. Find the total amount at the end of the 15th year of the above series. Consider the following cash flow diagram.

A person is planning for his retired life. Interest Formulas and Their Applications 41 A person invests a sum of Rs.

The compounding is monthly. Find the maturity amount of the deposit after 15 years. Then, depending on the type of decision, the best alternative will be selected by comparing the present worth amounts of the alternatives. The sign of various amounts at different points in time in a cash flow diagram is to be decided based on the type of the decision problem.

In a cost dominated cash flow diagram, the costs outflows will be assigned with positive sign and the profit, revenue, salvage value all inflows , etc. The costs outflows will be assigned with negative sign. In case the decision is to select the alternative with the minimum cost, then the alternative with the least present worth amount will be selected. On the other hand, if the decision is to select the alternative with the maximum profit, then the alternative with the maximum present worth will be selected.

S R1 R2 R3. The interest rate is i, compounded annually. S is the salvage value at the end of the nth year. If we have some more alternatives which are to be compared with this alternative, then the corresponding present worth amounts are to be computed and compared. Finally, the alternative with the maximum present worth amount should be selected as the best alternative.

Finally, the alternative with the minimum present worth amount should be selected as the best alternative. It has identified three different technologies for meeting the goal. The initial outlay and annual revenues with respect to each of the technologies are summarized in Table 4.

Table 4. Therefore, technology 2 is suggested for implementation to expand the production. The details of the bids for the elevators are as follows: Alpha Elevator Inc. Solution Bid 1: The cash flow diagram of bid 1 is shown in Fig. The present worth of the above cash flow diagram is computed as follows: Beta Elevator Inc. Hence, bid 1 is to be selected for implementation.

That is, the elevator from Alpha Elevator Inc. Proposal End of years 0 1 2 3 4 A Rs. Which proposal should be selected? The cash flow diagram of proposal A is shown in Fig. The cash flow diagram of the proposal B is shown in Fig. Therefore, select proposal B. If it is purchased under down payment, the cost of the machine is Rs. Solution There are two alternatives available for the company: Down payment of Rs.

The cash flow diagram of the second alternative is shown in Fig. Hence, the company should select the second alternative to buy the fully automated granite cutting machine. In plan 1, the company pays Rs. In plan 2, for every Rs. Solution Plan 1. The cash flow diagram for plan 1 is illustrated in Fig.

The cash flow diagram for plan 2 is shown in Fig. Innovative Investment Ltd. Which is the best investment alternative? At the end of the life of the business, the salvage value is zero. A project involves an initial outlay of Rs. The salvage value at the end of the life of the project after five years is Rs.

End Maintenance and Revenue of year operating expense Rs. Find the present worth of the following cash flow series. Consider the following cash flow series over a year period. End of year Cash flow Rs. The cost of erecting an oil well is Rs. The annual equivalent yield from the oil well is Rs. The salvage value after its useful life of 10 years is Rs. The details of the feasibility report of a project are as shown below. You are asked to advise the best alternative for the customers based on the present worth method of comparison.

The cash flows of two project proposals are as given below. Each of the project has an expected life of 10 years. Initial Annual Salvage outlay equivalent value after revenue 10 years Rs. Project 1 —7,50, 2,00, 50, Project 2 —9,50, 2,25, 1,00, 9. A company has two alternatives for satisfying its daily travel requirements of its employees for the next five years: Renting a vehicle at a cost of Rs.

Alternative 2: Buying a vehicle for Rs. The salvage value of the vehicle after five years is Rs. She would like to have an annual equivalent amount of Rs. Find the single amount that should be deposited now so that she receives the above mentioned annual equivalent amount at the end of every year for 20 years after her retirement.

Then, the alternative with the maximum future worth of net revenue or with the minimum future worth of net cost will be selected as the best alternative for implementation. Finally, the alternative with the maximum future worth amount should be selected as the best alternative. If we have some more alternatives which are to be compared with this alternative, then the corresponding future worth amounts are to be computed and compared. Finally, the alternative with the minimum future worth amount should be selected as the best alternative.

End of year Alternative 0 1 2 3 4 A Rs. Thus, alternative A should be selected. He must decide which of the several alternatives to select in trying to obtain a desirable return on his investment. After much study and calculation, he decides that the two best alternatives are as given in the following table: Build Build soft gas station ice-cream stand First cost Rs. Thus, building the gas station is the best alternative.

Therefore, alternative 2 must be selected. Thus, none of the two alternatives should be selected. It has received tenders from three different original manufacturers of annealing furnace. The details are as follows. Manufacturer 1 2 3 Initial cost Rs. Krishna castings should buy the annealing furnace from manufacturer 2.

Use future worth method of comparison. The cash flow diagram of machine A is given in Fig. The future worth function of Fig. The cash flow diagram of the machine B is illustrated in Fig. The future worth function of Fig 5. Therefore, machine A should be selected.

A suburban taxi company is considering buying taxis with diesel engines instead of petrol engines. The cars average 50, km a year, with a useful life of three years for the taxi with the petrol engine and four years for the diesel taxi. Other comparative information are as follows: Diesel Petrol Vehicle cost Rs. A motorcycle is sold for Rs.

The motorcycle dealer is willing to sell it on the following terms: Consider the following two mutually exclusive alternatives. A B Cost Rs. A company must decide whether to buy machine A or machine B: Use the future worth method of comparison. Due to increasing awareness of customers, two different television manufacturing companies started a marketing war.

The details of advertisements of the companies are as follows: Alpha Finance Company is coming with an option of accepting Rs. Beta Finance Company is coming with a similar option of accepting Rs. An insurance company gives an endowment policy for a person aged 30 years.

The yearly premium for an insured sum of Rs. The policy will mature after 25 years. Also, the person is entitled for a bonus of Rs. If a person survives till the end of the 25th year: Then the alternative with the maximum annual equivalent revenue in the case of revenue-based comparison or with the minimum annual equivalent cost in the case of cost- based comparison will be selected as the best alternative.

The first step is to find the net present worth of the cash flow diagram using the following expression for a given interest rate, i: If we have some more alternatives which are to be compared with this alternative, then the corresponding annual equivalent revenues are to be computed and compared. Finally, the alternative with the maximum annual equivalent revenue should be selected as the best alternative.

The first step is to find the net present worth of the cash flow diagram using the following relation for a given interest rate, i. Then, in the second step, the annual equivalent cost is computed using the following equation: Finally, the alternative with the minimum annual equivalent cost should be selected as the best alternative.

Such procedure is to be applied to all the alternatives and finally, the best alternative is to be selected. In each of the cases presented in Sections 6.

The owner of the company is concerned about the increasing cost of petrol. The cost per litre of petrol for the first year of operation is Rs.

He feels that the cost of petrol will be increasing by Re. His experience with his company car indicates that it averages 9 km per litre of petrol. The executive expects to drive an average of 20, km each year for the next four years. What is the annual equivalent cost of fuel over this period of time?. If he is offered similar service with the same quality on rental basis at Rs. If the rental car is preferred, then the company car will find some other use within the company.

The cash flow diagram for this situation is depicted in Fig. This amount is less than the annual rental value of Rs. Therefore, the company should continue to provide its own car to its executive. Three original manufacturers have responded to its tender whose particulars are tabulated as follows: Manufacturer Down payment Yearly equal No.

The annual equivalent cost of manufacturer 3 is less than that of manufacturer 1 and manufacturer 2. Therefore, the company should buy the advanced machine centre from manufacturer 3.

The life of both alternatives is estimated to be 5 years with the following investments, annual returns and salvage values. Alternative A B Investment Rs.

The annual equivalent revenue expression of the above cash flow diagram is as follows: Thus, the company should select alternative B. The following data are to be used in the analysis: Base your answer on annual equivalent cost. The cash flow diagram of machine X is illustrated in Fig. The cash flow diagram of machine Y is depicted in Fig. So, machine X is the more cost effective machine.

Data on the routes are as follows: Around the lake Under the lake Length 15 km 5 km First cost Rs. Therefore, select the route around the lake for laying the power line. The cars average 60, km a year with a useful life of three years for the petrol taxi and four years for the diesel taxi. Other comparative details are as follows: Therefore, the taxi company should buy cars with diesel engine.

Comparison is done on common multiple lives of 12 years. The company will reimburse their salesman each month the fuel cost and maintenance cost. Ramu has decided to drive a low-priced automobile.

He finds, however, that there are two different ways of obtaining the automobile. In either case, the fuel cost and maintenance cost are borne by the company. The monthly charge is Rs. At the end of the three-year period, the car is returned to the leasing company.

If the car could be sold for Rs. The monthly equivalent cost of alternative 1 is less than that of alternative 2. Hence, the salesman should purchase the car for cash.

She estimates that it will have a five year useful life and no salvage value at the end of equipment life. The dealer, who is a friend has offered Jothi Lakshimi two alternative ways to pay for the equipment. Hence, Jothi Lakshimi should select alternative 2 for purchasing the home equipment. Brand Tyre warranty Price per tyre months Rs. This is months. The annual equivalent cost of brand C is less than that of other brands.

Hence, it should be used in the vehicles of the trucking company. It should be replaced four times during the month period. A company has three proposals for expanding its business operations. The details are as follows: Alternative Initial cost Annual revenue Life Rs. An automobile dealer has recently advertised for its new car. There are three alternatives of purchasing the car which are explained below.

Alternative 1 The customer can take delivery of a car after making a down payment of Rs. The remaining money should be paid in 36 equal monthly installments of Rs. Alternative 3 The customer can take delivery of the car by making full payment of Rs. Use the annual equivalent method. A small-scale industry is in the process of buying a milling machine.

The purchase value of the milling machine is Rs. It has identified two banks for loan to purchase the milling machine. In Urban Bank, the loan is to be repaid in 60 equal monthly installments of Rs.

In State Bank, the loan is to be repaid in 40 equal monthly installments of Rs. Suggest the most economical loan scheme for the company, based on the annual equivalent method of comparison. There are two alternatives of replacing a machine. The details of the alternatives are as follows: Alternative 1 Purchase value of the new machine: A company receives two options for purchasing a copier machine for its office. Option 1 Make a down payment of Rs.

The remaining money is to be paid in 24 equal monthly installments of Rs. Find the best alternative using the annual equivalent method of comparison. In this method of comparison, the rate of return for each alternative is computed. Then the alternative which has the highest rate of return is selected as the best alternative. A generalized cash flow diagram to demonstrate the rate of return method of comparison is presented in Fig.

Rj Rn 0. In the above cash flow diagram, P represents an initial investment, Rj the net revenue at the end of the jth year, and S the salvage value at the end of the nth year.

The first step is to find the net present worth of the cash flow diagram using the following expression at a given interest rate, i. In the figure, the present worth goes on decreasing when the interest rate is increased. It will be very difficult to find the exact value of i at which the present worth function reduces to zero.

So, one has to start with an intuitive value of i and check whether the present worth function is positive.

If so, increase the value of i until PW i becomes negative. Then, the rate of return is determined by interpolation method in the range of values of i for which the sign of the present worth function changes from positive to negative.

The initial outlay and cash flow pattern for the new business are as listed below. The expected life of the business is five years. Find the rate of return for the new business.

Period 0 1 2 3 4 5 Cash flow —1,00, 30, 30, 30, 30, 30, Rs. The life of the project is 10 years with no salvage value at the end of its life. The initial outlay of the project is Rs. The annual net profit is Rs. The life of all the three alternatives is estimated to be five years with negligible salvage value. Alternative A1 A2 A3 Rate of return So, it should not be considered for comparison. The remaining two alternatives are qualified for consideration.

Among the alternatives A1 and A2, the rate of return of alternative A1 is greater than that of alternative A2. Hence, alternative A1 should be selected. The amounts are in rupees. It has two alternatives for the expansion programme and the corresponding cash flows are tabulated below. Each alternative has a life of five years and a negligible salvage value.

Suggest the best alternative to the company. Initial investment Yearly revenue Rs. The formula for the net present worth of alternative 2 is: Consider the following cash flow of a project: Year 0 1 2 3 4 5 Cash flow —10, 4, 4, 5, 5, 6, Find the rate of return of the project.

At the end of the 10th year, the salvage value of the business is Rs. Find the rate of return of the business. A company is in the process of selecting the best alternative among the following three mutually exclusive alternatives: Alternative Initial Annual revenue Life investment Rs. A shipping firm is considering the purchase of a materials handling system for unloading ships at a dock. The firm has reduced their choice to three different systems, all of which are expected to provide the same unloading speed.

The initial costs and the operating costs estimated for each system are now tabulated. If the firm must select one of the materials handling systems, which one is the most desirable?. A firm has identified three mutually exclusive alternatives. The life of all three alternatives is estimated to be five years.

Find the best alternative based on the rate of return method. An automobile company is planning to buy a robot for its forging unit. It has identified two different companies for the supply of the robot.

The details of cost and incremental revenue of using robots are summarized in the following table: Brand Speedex Giant Initial cost Rs. Suggest the best brand of robot to the company based on the rate of return method. A bank introduces two different investment schemes whose details are as follows: A company is planning for its expansion programme which will take place after five years.

The expansion requires an equal sum of Rs. Gamma Bank has recently introduced a scheme in this line. If the company invests Rs.

Suggest whether the company should invest with the Gamma Bank for its expansion programme. Consider the following table which summarizes data of two alternatives. First cost Annual return Life Alternative 1 Rs.

A company is planning to expand its present business activity. It has two alternatives for the expansion programme and the corresponding cash flows are given in the following table. In addition to these facilities, there are several other items which are necessary to facilitate the functioning of organizations. All such facilities should be continuously monitored for their efficient functioning; otherwise, the quality of service will be poor. Besides the quality of service of the facilities, the cost of their operation and maintenance would increase with the passage of time.

Hence, it is an absolute necessity to maintain the equipment in good operating conditions with economical cost. Thus, we need an integrated approach to minimize the cost of maintenance. In certain cases, the equipment will be obsolete over a period of time.

If a firm wants to be in the same business competitively, it has to take decision on whether to replace the old equipment or to retain it by taking the cost of maintenance and operation into account. There are two basic reasons for considering the replacement of an equipment—physical impairment of the various parts or obsolescence of the equipment.

Physical impairment refers only to changes in the physical condition of the machine itself. This would lead to a decline in the value of the service rendered, increased operating cost, increased maintenance cost or a combination of these. Obsolescence is due to improvement of the tools of production, mainly improvement in technology.

So, it would be uneconomical to continue production with the same machine under any of the above situations. Hence, the machines are to be periodically replaced. Sometimes, the capacity of existing facilities may be inadequate to meet the current demand. Under such situation, the following alternatives will be considered. Preventive maintenance PM is the periodical inspection and service activities which are aimed to detect potential failures and perform minor adjustments or repairs which will prevent major operating problems in future.

Breakdown maintenance is the repair which is generally done after the equipment has attained down state. It is often of an emergency nature which will have associated penalty in terms of expediting cost of maintenance and down time cost of equipment.

Preventive maintenance will reduce such cost up to a point. Beyond that point, the cost of preventive maintenance will be more when compared to the breakdown maintenance cost. The total cost, which is the sum of the preventive maintenance cost and the breakdown maintenance cost, will go on decreasing with an increase in the level of maintenance up to a point. Beyond that point, the total cost will start increasing. The level of maintenance corresponding to the minimum total cost is the optimal level of maintenance.

The concepts are demonstrated in Fig. A typical shape of each of the above costs with respect to life of the machine is shown in Fig. Replacement and Maintenance Analysis From Fig. From the beginning, the total cost continues to decrease up to a particular life and then it starts increasing.

The point where the total cost is minimum is called the economic life of the machine. If the interest rate is more than zero per cent, then we use interest formulas to determine the economic life. The replacement alternatives can be evaluated based on the present worth criterion and annual equivalent criterion. The basics of these criteria are already presented in Chapter 3.

Based on experience, it was found that the maintenance cost is zero during the first year and it increases by Rs. This is summarized in column B of Table 8. Table 8. The value corresponding to any end of year in this column represents the total maintenance cost of using the equipment till the end of that particular year. For this problem, the average total cost decreases till the end of year 6 and then it increases. Therefore, the optimal replacement period is six years, i.

Discount the maintenance costs to the beginning of year 1. Replacement and Maintenance Analysis 3. Find Column F by adding the first cost of Rs. Find the annual equivalent total cost through the years given. Identify the end of year for which the annual equivalent total cost is minimum. For this problem, the annual equivalent total cost is minimum at the end of year 7. Therefore, the economic life of the equipment is seven years.

End of year Operation cost Maintenance Salvage value n at the end of cost at the at the end of year Rs. Therefore, the economic life of the machine is five years. The annual equivalent total cost corresponding to the economic life is Rs. Now, the manufacturer of machine B has approached the company. Machine B, which has the same capacity as that of machine A, is priced at Rs.

The maintenance cost of machine B is estimated at Rs. Assume that the scrap value of each of the machines is negligible at any year. Determination of economic life and corresponding annual equivalent total cost of machine B. The details of machine B are summarized in Table 8.

The first cost of machine B is equal to Rs.

Hence the economic life of machine B is 8 years and the corresponding annual equivalent total cost is Rs. Selection of the best machine is based on the minimum annual equivalent total cost.

The comparison is made over the minimum common multiple of the lives of machine A and machine B, i. In this analysis, the annual equivalent cost of each alternative should be computed first.

Then the alternative which has the least cost should be selected as the best alternative. Before discussing details, some preliminary concepts which are essential for this type of replacement analysis are presented. P Fig. Assume that an equipment has been purchased about three years back for Rs. The supplier of the new equipment will take the old one for some money, say, Rs. This should be treated as the present value of the existing equipment and it should be considered for all further economic analysis.

The purchase value of the existing equipment before three years is now known as sunk cost, and it should not be considered for further analysis. Its salvage value at the end of its life is Rs. The annual maintenance cost is Rs.

The market value of the present machine is Rs. Now, a new machine to cater to the need of the present machine is available at Rs. Its annual maintenance cost is Rs. The salvage value of the new machine is Rs. It has a present realizable market value of Rs. If kept, it can be expected to last five years more, with operating and maintenance cost of Rs. This engine can be replaced with an improved version costing Rs.

This improved version will have an estimated annual operating and maintenance cost of Rs. Equal lives are nothing but the least common multiple of the lives of the alternatives.

Since the annual equivalent cost of the old diesel engine is less than that of the new diesel engine, it is suggested to keep the old diesel engine. Here, an important assumption is that the old engine will be replaced four times during the 20 years period of comparison.

Reinforcement would cost Rs. If it is reinforced, it is estimated that its net salvage value would be Rs. The new prestressed concrete bridge would cost Rs. Such a bridge would have no salvage value.

It is estimated that the annual maintenance cost of the reinforced bridge would exceed that of the concrete bridge by Rs. If the bridge is replaced by a new prestressed concrete bridge, the scrap value of the steel would exceed the demolition cost by Rs. What would you recommend? Solution There are two alternatives: Reinforce the existing bridge. Replace the existing bridge by a new prestressed concrete bridge.

The cash flow diagram for alternative 2 is shown in Fig. Based on equal lives comparison over 40 years, alternative 2 is selected as the best alternative. Thus, it is suggested to go in for prestressed concrete bridge. Its useful life was estimated to be 10 years.

Due to the fast development of that locality, the municipality is unable to meet the current demand for water with the existing motor.

The municipality can cope with the situation either by augmenting an additional 5 hp motor or replacing the existing 10 hp motor with a new 15 hp motor. The details of these motors are now tabulated.

Replacement and Maintenance Analysis Solution There are two alternatives to cope with the situation: Augmenting the present 10 hp motor with an additional 5 hp motor. Replacing the present 10 hp motor with a new 15 hp motor.

Therefore, it is suggested that the present 10 hp motor be augmented with a new 5 hp motor. Its life is six years and its salvage value at the end of its life is Rs.

Now, a company is offering a new machine at a cost of Rs. Its life is four years and its salvage value at the end of its life is Rs. The annual maintenance cost of the new machine is Rs. The company which is supplying the new machine is willing to take the old machine for Rs.

Solution Old machine Let the comparative use value of the old machine be X. X — 1, 0. Therefore, it is advisable to replace the old machine with the new one. The failure of the item may result in complete breakdown of the system. The system may contain a collection of such items or just one item, say a tubelight. Therefore, we use some replacement policy for such items which would avoid the possibility of a complete breakdown.

The following are the replacement policies which are applicable for this situation. Under this policy, an item is replaced immediately after its failure. Under this policy, the following decision is 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 period?

Replacement and Maintenance Analysis There is a trade-off between the individual replacement policy and the group replacement policy. Hence, for a given problem, each of the replacement policies is evaluated and the most economical policy is selected for implementation.

This is explained with two numerical problems. If all the transistors are replaced simultaneously, it would cost Rs. Any one of the following two options can be followed to replace the transistors: Find out the optimal replacement policy, i.

If group replacement policy is optimal, then find at what equal intervals should all the transistors be replaced. Solution Assume that there are transistors in use. Let, pi be the probability that a transistor which was new when placed in position for use, fails during the ith week of its life. Assume that a transistors that fail during a week are replaced just before the end of the week, and b the actual percentage of failures during a week for a sub-group of transistors with the same age is same as the expected percentage of failures during the week for that sub-group of transistors.

Replacement and Maintenance Analysis Table 8. From Table 8. Hence, the group replacement period is four weeks. When any resistor fails, it is replaced.

The cost of replacing a resistor individually is Rs. If all the resistors are replaced at the same time, the cost per resistor is Rs. The per cent surviving, S i at the end of month i is tabulated as follows: Solution Let pi be the probability of failure during the month i.

Hence, a resistor which has survived for five months would certainly fail during the sixth month. We assume that the resistors failing during a month are accounted at the end of the month. Replacement and Maintenance Analysis From Table 8.

Thus, the group replacement period is three months. List and explain the different types of maintenance. Discuss the reasons for replacement. Distinguish between breakdown maintenance and preventive maintenance.

A firm is considering replacement of an equipment, whose first cost is Rs. The following table gives the operation cost, maintenance cost and salvage value at the end of every year of a machine whose purchase value is Rs.

A manufacturer is offered two machines A and B. A is priced at Rs. Machine B which has the same capacity is priced at Rs. The maintenance costs of the machine B are estimated at Rs. Three years back, a machine was purchased at a cost of Rs. Its salvage value at the end of its estimated life is Rs. A new machine to cater to the need of the present machine is available at Rs. A steel highway bridge must either be reinforced or replaced.

Three years back, a municipality purchased a 10 hp motor for pumping drinking water. Its annual operation and maintenance cost is Rs. Due to rapid development of that locality, the municipality is unable to meet the current demand for water with the existing motor. The details of these motors are given in the following table. The failure rates of transistors in a computer are summarized in the following table.

End of week 1 2 3 4 5 6 7 Probability of 0. Find out which is the optimal replacement policy, i.

If the group replacement policy is optimal, then find at what equal intervals should all the transistors be replaced. An electronic equipment contains 1, resistors. The per cent surviving, S i at the end of month i is tabulated now. This may be due to wear and tear of the equipment or obsolescence of technology. Hence, it is to be replaced at the proper time for continuance of any business. The replacement of the equipment at the end of its life involves money. This must be internally generated from the earnings of the equipment.

The recovery of money from the earnings of an equipment for its replacement purpose is called depreciation fund since we make an assumption that the value of the equipment decreases with the passage of time.

These are as follows: Straight line method of depreciation 2. Declining balance method of depreciation 3. Sum of the years—digits method of depreciation 4. Sinking-fund method of depreciation 5. Service output method of depreciation These are now discussed in detail.

Here, we make an important assumption that inflation is absent. The formulae for depreciation and book value are as follows: The estimated salvage value of the equipment at the end of its lifetime is Rs. Determine the depreciation charge and book value at the end of various years using the straight line method of depreciation.

The calculations pertaining to Bt for different values of t are summarized in Table 9. Table 9. In this approach, it should be noted that the depreciation is the same for all the periods. This approach is a more realistic approach, since the depreciation charge decreases with the life of the asset which matches with the earning potential of the asset.

The book value at the end of the life of the asset may not be exactly equal to the salvage value of the asset. This is a major limitation of this approach.

If this rate is used, then the corresponding approach is called the double declining balance method of depreciation. The rates of depreciation for the years 1—8, respectively are as follows: For any year, the depreciation is calculated by multiplying the corresponding rate of depreciation with P — F.

The calculations of Dt and Bt for different values of t are summarized in Table 9. The loss in value of the asset P — F is made available an the form of cumulative depreciation amount at the end of the life of the asset by setting up an equal depreciation amount A at the end of each period during the lifetime of the asset. If we calculate Dt and Bt for all the periods, then the tabular approach would be better.

For example, the calculations of net depreciation for some periods are as follows: The minor difference is due to truncation error. In such cases, the depreciation is computed based on service rendered by an asset. Then, the depreciation is defined per unit of service rendered: Its salvage value after five years is Rs. The length of road that can be laid by the machine during its lifetime is 75, km.

In its third year of operation, the length of road laid is 2, km. Find the depreciation of the equipment for that year. Define the following: Distinguish between declining balance method of depreciation and double declining balance method of depreciation. The Alpha Drug Company has just purchased a capsulating machine for Rs. Compute the depreciation schedule for the machine by each of the following depreciation methods: A company has recently purchased an overhead travelling crane for Rs.

Its expected life is seven years and the salvage value at the end of the life of the overhead travelling crane is Rs. Using the straight line method of depreciation, find the depreciation and the book value at the end of third and fourth year after the crane is purchased. An automobile company has purchased a wheel alignment device for Rs. The device can be used for 15 years. Find the following using the double declining balance method of depreciation: A company has purchased a bus for its officers for Rs.

The expected life of the bus is eight years. The salvage value of the bus at the end of its life is Rs. Find the following using the sinking fund method of depreciation: Consider Problem 4 and find the following using the sum-of-the-years- digits method of depreciation: A company has purchased a Xerox machine for Rs. The salvage value of the machine at the end of its useful life would be insignificant. The maximum number of copies that can be taken during its lifetime is 1,00,00, During the fourth year of its operation, the number of copies taken is 9,00, Find the depreciation for the fourth year of operation of the Xerox machine using the service output method of depreciation.

A heavy construction firm has been awarded a contract to build a large concrete dam. The firm will buy Rs. During the preparation of the job cost estimate, the following utilization schedule was computed for the special equipment: Prepare the depreciation schedule for all the years of operation of the equipment using the service output method of depreciation.

But the same criterion cannot be used while evaluating public alternatives. Examples of some public alternatives are constructing bridges, roads, dams, establishing public utilities, etc.

In this process, one should see whether the benefits of the public activity are at least equal to its costs. If yes, then the public activity can be undertaken for implementation. Otherwise, it can be cancelled. For the purpose of comparison, these are to be converted into a common time base present worth or future worth or annual equivalent. Similarly, the costs consist of initial investment and yearly operation and maintenance cost.

It provides the managers not only a sound basis but also a clear-cut approach to decision making. What is more, the book adequately illustrates these approaches with numerical problems and Indian cases. A distinguishing feature of the book is that it has an Appendix on interest tables for a wide range of interest rates 0. Visitor Kindly Note: EasyEngineering team try to Helping the students and others who cannot afford buying books is our aim. Thank you. Kindly Note: Thank you for visiting my thread.

Hope this post is helpful to you. Have a great day! Kindly share this post with your friends to make this exclusive release more useful. Notify me of follow-up comments by email. Notify me of new posts by email. Leave this field empty. Welcome to EasyEngineering, One of the trusted educational blog.

Check your Email after Joining and Confirm your mail id to get updates alerts. Panneerselvam Book Free Download. Other Useful Links. Your Comments About This Post. Is our service is satisfied, Anything want to say? Cancel reply. Please enter your comment! Please enter your name here. You have entered an incorrect email address!

Get New Updates Email Alerts Enter your email address to subscribe this blog and receive notifications of new posts by email. Join With us. Today Updates.