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Build gas station First cost Rs. Annual property taxes Rs. Annual income Rs. Life of building years Salvage value Rs. Thus, building the gas station is the best alternative. Salvage value after 12 years It has received tenders from three different original manufacturers of annealing furnace. The details are as follows. Life years Annual operation and maintenance cost Rs. Solution Alternative 1—Manufacturer 1 First cost. Manufacturer 1 Initial cost Rs.
Future Worth Method 63 Annual operating and maintenance cost. The cash flow diagram of machine A is given in Fig. Krishna castings should download the annealing furnace from manufacturer 2. Use future worth method of comparison. Solution Machine A Initial cost of the machine. A company must decide whether to download machine A or Initial cost Useful life. The future worth function of Fig. The future worth function of Fig 5. Future Worth Method 65 2.
The cash flow diagram of the machine B is illustrated in Fig. Interest rate. Other comparative information are as follows: Diesel Vehicle cost Fuel cost per litre Mileage. A motorcycle is sold for Rs. The cars average A suburban taxi company is considering downloading taxis with diesel engines instead of petrol engines. A company must decide whether to download machine A or machine B: Machine A Initial cost Useful life. A Cost Uniform annual benefit Useful life years Rs.
The motorcycle dealer is willing to sell it on the following terms: Consider the following two mutually exclusive alternatives. The details of advertisements of the companies are as follows: Beta Finance Company is coming with a similar option of accepting Rs.
The policy will mature after 25 years. Alpha Finance Company is coming with an option of accepting Rs. Due to increasing awareness of customers. Use the future worth method of comparison. If a person survives till the end of the 25th year: An insurance company gives an endowment policy for a person aged 30 years. The yearly premium for an insured sum of Rs. 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 costbased 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.. P represents an initial investment.. Rj the net revenue at the end of the jth year. S R1 R2 R Annual Equivalent Method 69 In the second step.. The first step is to find the net present worth of the cash flow diagram using the following relation for a given interest rate.
The owner of the company is concerned about the increasing cost of petrol. If we have some non-standard cash flow diagram. The executive expects to drive an average of If he is offered similar service with the same quality on rental basis at Rs. If the rental car is preferred. His experience with his company car indicates that it averages 9 km per litre of petrol.
Such procedure is to be applied to all the alternatives and finally. In each of the cases presented in Sections 6. 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. What is the annual equivalent cost of fuel over this period of time?. Annual Equivalent Method 71 Therefore. The cash flow diagram for this situation is depicted in Fig.
Three original manufacturers have responded to its tender whose particulars are tabulated as follows: Manufacturer Down payment Rs. This amount is less than the annual rental value of Rs. Solution Alternative 1 Down payment.
Annual equal return Rs. The annual equivalent cost of manufacturer 3 is less than that of manufacturer 1 and manufacturer 2. The life of both alternatives is estimated to be 5 years with the following investments. Salvage value Rs. Annual Equivalent Method 73 Alternative 3 Down payment. Alternative A Investment Rs. Solution Alternative A Initial investment.
The annual equivalent revenue expression of the above cash flow diagram is as follows: Solution Machine X First cost. Base your answer on annual equivalent cost. The cash flow diagram of machine X is illustrated in Fig. Annual maintenance cost.
The cash flow diagram of machine Y is depicted in Fig. The following data are to be used in the analysis: Data on the routes are as follows: Annual Equivalent Method 77 The cash flow diagram for this alternative is shown in Fig. Other comparative details are as follows: Diesel Vehicle cost Rs. The monthly charge is Rs. At the end of the three-year period.
In either case. He finds. If the car could be sold for Rs. Comparison is done on common multiple lives of 12 years.
Annual Equivalent Method 79 Fuel cost. Ramu has decided to drive a low-priced automobile. The company will reimburse their salesman each month the fuel cost and maintenance cost. The monthly equivalent cost of alternative 1 is less than that of alternative 2. Machine B 6.
She estimates that it will have a five year useful life and no salvage value at the end of equipment life. The dealer. Solution In all the cases. Jothi Lakshimi should select alternative 2 for downloading the home equipment. The remaining money should be paid in 36 equal monthly installments of Rs. An automobile dealer has recently advertised for its new car.
This is months. The details are as follows: A company has three proposals for expanding its business operations. There are three alternatives of downloading the car which are explained below. Alternative 1 The customer can take delivery of a car after making a down payment of Rs.
It should be replaced four times during the month period. The annual equivalent cost of brand C is less than that of other brands. The download value of the milling machine is Rs. In Urban Bank. Suggest the most economical loan scheme for the company. There are two alternatives of replacing a machine. It has identified two banks for loan to download the milling machine. Option 1 Make a down payment of Rs.
Alternative 3 The customer can take delivery of the car by making full payment of Rs. Use the annual equivalent method. A company receives two options for downloading a copier machine for its office. The remaining money is to be paid in 24 equal monthly installments of Rs. In State Bank. The details of the alternatives are as follows: Alternative 1 download value of the new machine Life of the machine Salvage value of the new machine at the end of its life Annual operation and maintenance cost downloadback price of the existing machine Alternative 2 download value of the new machine Life of the machine Salvage value of the new machine at the end of its life Annual operation and maintenance cost downloadback price of the existing machine: A small-scale industry is in the process of downloading a milling machine.
Alternative Initial cost Rs.
Find the best alternative using the annual equivalent method of comparison. Life years Salvage value Rs. Annual receipt Rs. In this method of comparison.
In the above cash flow diagram. A generalized cash flow diagram to demonstrate the rate of return method of comparison is presented in Fig. It will be very difficult to find the exact value of i at which the present worth function reduces to zero. Then the alternative which has the highest rate of return is selected as the best alternative. In the figure.. In this type of analysis. The first step is to find the net present worth of the cash flow diagram using the following expression at a given interest rate.
The expected life of the business is five years. The initial outlay and cash flow pattern for the new business are as listed below. If so. Period Cash flow Rs. Find the rate of return for the new business. The annual net profit is 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 life of all the three alternatives is estimated to be five years with negligible salvage value. Alternative A1 Investment Annual net income Rs. The remaining two alternatives are qualified for consideration. The amounts are in rupees.
Among the alternatives A1 and A2. Alternative Rate of return A1 It has two alternatives for the expansion programme and the corresponding cash flows are tabulated below. Suggest the best alternative to the company.
Each alternative has a life of five years and a negligible salvage value. Initial investment Rs. The formula for the net present worth of alternative 2 is: The initial costs and the operating costs estimated for each system are now tabulated. Consider the following cash flow of a project: Year Cash flow 0 — The firm has reduced their choice to three different systems. A company is in the process of selecting the best alternative among the following three mutually exclusive alternatives: Find the rate of return of the business.
A shipping firm is considering the download of a materials handling system for unloading ships at a dock. At the end of the 10th year. The life of all three alternatives is estimated to be five years.
A firm has identified three mutually exclusive alternatives. If the firm must select one of the materials handling systems. Period of deposit years Maturity amount Rs. Annual incremental revenue Rs. Find the best alternative based on the rate of return method. Alternative Initial investment Rs. It has identified two different companies for the supply of the robot.
Life years Life-end salvage value Rs. A bank introduces two different investment schemes whose details are as follows: An automobile company is planning to download a robot for its forging unit.
Alpha Bank Deposit amount Rs. The details of cost and incremental revenue of using robots are summarized in the following table: Brand Speedex Initial cost Rs.
Suggest the best brand of robot to the company based on the rate of return method. If the company invests Rs. First cost Alternative 1 Alternative 2 Rs. A company is planning to expand its present business activity. Alternative 1 Alternative 2 4. Suggest whether the company should invest with the Gamma Bank for its expansion programme. Gamma Bank has recently introduced a scheme in this line. The expansion requires an equal sum of Rs. A company is planning for its expansion programme which will take place after five years.
Rate of Return Method 99 8. Consider the following table which summarizes data of two alternatives. It has two alternatives for the expansion programme and the corresponding cash flows are given in the following table. Obsolescence is due to improvement of the tools of production. Besides the quality of service of the facilities. There are two basic reasons for considering the replacement of an equipment—physical impairment of the various parts or obsolescence of the equipment.
In addition to these facilities. In certain cases. This would lead to a decline in the value of the service rendered. Physical impairment refers only to changes in the physical condition of the machine itself.
If a firm wants to be in the same business competitively. Under such situation. All such facilities should be continuously monitored for their efficient functioning. 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. Preventive maintenance will reduce such cost up to a point. 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.
Beyond that point. Breakdown maintenance is the repair which is generally done after the equipment has attained down state. The level of maintenance corresponding to the minimum total cost is the optimal level of maintenance. The concepts are demonstrated in Fig. Replacement and Maintenance Analysis 8. The total cost. A typical shape of each of the above costs with respect to life of the machine is shown in Fig. Table 8.
The replacement alternatives can be evaluated based on the present worth criterion and annual equivalent criterion. This is summarized in column B of Table 8.
The basics of these criteria are already presented in Chapter 3. 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. Based on experience. If the interest rate is more than zero per cent. The point where the total cost is minimum is called the economic life of the machine.
From the beginning. Replacement and Maintenance Analysis From Fig. For this problem. Discount the maintenance costs to the beginning of year 1. Replacement and Maintenance Analysis 3. Identify the end of year for which the annual equivalent total cost is minimum.
Find the economic life of the machine assuming interest rate. Operation cost Maintenance at the end of cost at the year Rs. Find Column F by adding the first cost of Rs. Find the annual equivalent total cost through the years given. The details of machine B are summarized in Table 8. Assume that the scrap value of each of the machines is negligible at any year. Machine B. Determination of economic life and corresponding annual equivalent total cost of machine B.
The annual equivalent total cost corresponding to the economic life is Rs. The maintenance cost of machine B is estimated at Rs. Before discussing details. Then the alternative which has the least cost should be selected as the best alternative.
In this analysis. The comparison is made over the minimum common multiple of the lives of machine A and machine B. Selection of the best machine is based on the minimum annual equivalent total cost.
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. The annual maintenance cost is Rs.
Assume that an equipment has been downloadd about three years back for Rs. The market value of the present machine is Rs. The salvage value of the new machine is Rs. The download value of the existing equipment before three years is now known as sunk cost. Its annual maintenance cost is Rs. The supplier of the new equipment will take the old one for some money. Its salvage value at the end of its life is Rs. This should be treated as the present value of the existing equipment and it should be considered for all further economic analysis.
This improved version will have an estimated annual operating and maintenance cost of Rs. It has a present realizable market value of Rs. If kept. This engine can be replaced with an improved version costing 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 estimated that the annual maintenance cost of the reinforced bridge would exceed that of the concrete bridge by Rs. Such a bridge would have no salvage value. If the bridge is replaced by a new prestressed concrete bridge. What would you recommend? Solution There are two alternatives: Reinforcement would cost Rs. Reinforce the existing bridge.
The new prestressed concrete bridge would cost Rs. Replace the existing bridge by a new prestressed concrete bridge. If it is reinforced.
The cash flow diagram for alternative 2 is shown in Fig. Based on equal lives comparison over 40 years. Its useful life was estimated to be 10 years. The details of these motors are now tabulated. Due to the fast development of that locality.
Old 10 hp motor download cost P Rs. 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. Replacing the present 10 hp motor with a new 15 hp motor. Augmenting the present 10 hp motor with an additional 5 hp motor. Replacement and Maintenance Analysis Solution There are two alternatives to cope with the situation: Its life is six years and its salvage value at the end of its life is Rs.
Solution Old machine Let the comparative use value of the old machine be X. The company which is supplying the new machine is willing to take the old machine for Rs. The annual maintenance cost of the new machine is Rs. Its life is four years and its salvage value at the end of its life is Rs. 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?
Under this policy. The system may contain a collection of such items or just one item. The failure of the item may result in complete breakdown of the system. The following are the replacement policies which are applicable for this situation.
Find out the optimal replacement policy. Replacement and Maintenance Analysis There is a trade-off between the individual replacement policy and the group replacement policy. If all the transistors are replaced simultaneously. The failure rates of transistors in a computer are summarized Failure Rates of Transistors in Computers 1 0. If group replacement policy is optimal. This is explained with two numerical problems.
Assume that a transistors that fail during a week are replaced just before the end of the week. Solution Assume that there are transistors in use. Any one of the following two options can be followed to replace the transistors: The per cent surviving.
If all the resistors are replaced at the same time. S i at the end of month i is tabulated as follows: Solution Let pi be the probability of failure during the month i. The cost of replacing a resistor individually is Rs. We assume that the resistors failing during a month are accounted at the end of the month. When any resistor fails. Replacement and Maintenance Analysis Table 8. From Table 8. List and explain the different types of maintenance. Discuss the reasons for replacement. A firm is considering replacement of an equipment.
The following table gives the operation cost. Distinguish between breakdown maintenance and preventive maintenance. Replacement and Maintenance Analysis From Table 8. The maintenance costs of the machine B are estimated at Rs. The details of these motors are given in the following table.
Three years back. A manufacturer is offered two machines A and B. Machine B which has the same capacity is priced at Rs. A steel highway bridge must either be reinforced or replaced. A new machine to cater to the need of the present machine is available at Rs. Life in years n Salvage value at the end of machine life Rs. Due to rapid development of that locality. Its salvage value at the end of its estimated life is Rs. A is priced at Rs. Its annual operation and maintenance cost is Rs. If the group replacement policy is optimal.
An electronic equipment contains 1. Find out which is the optimal replacement policy. S i at the end of month i is tabulated now. Replacement and Maintenance Analysis The current market value of the 10 hp motor is Rs.
The failure rates of transistors in a computer are summarized in the following table. End of week Probability of failure to date 1 0. These are as follows: The replacement of the equipment at the end of its life involves money.
This must be internally generated from the earnings of the equipment. This may be due to wear and tear of the equipment or obsolescence of technology. Declining balance method of depreciation 3. 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.
Sum of the years—digits method of depreciation 4. Straight line method of depreciation 2. Service output method of depreciation These are now discussed in detail. Sinking-fund method of depreciation 5. The estimated salvage value of the equipment at the end of its lifetime is Rs. The formulae for depreciation and book value are as follows: Determine the depreciation charge and book value at the end of various years using the straight line method of depreciation.
Table 9. In this approach. The calculations pertaining to Bt for different values of t are summarized in Table 9.
This approach is a more realistic approach. If this rate is used. 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.
For any year. The calculations of Dt and Bt for different values of t are summarized in Table 9. The rates of depreciation for the years 1—8.
If the asset has a life of eight years. Depreciation 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 some situations, it may not be realistic to compute depreciation based on time period. 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 downloadd a capsulating machine for Rs. The plant engineer estimates that the machine has a useful. Compute the depreciation schedule for the machine by each of the following depreciation methods: A company has recently downloadd 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 downloadd.
An automobile company has downloadd 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 downloadd 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-yearsdigits method of depreciation: A company has downloadd 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. It is expected that a total of eight years will be required to. The firm will download Rs. During the preparation of the job cost estimate, the following utilization schedule was computed for the special equipment: At the end of the job, it is estimated that the equipment can be sold at auction for Rs.
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. The benefits may occur at different time periods of the public activity. 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.
These are to be converted to a common time base as done in the equivalent benefits. If this ratio is at least one, the public activity is justified; otherwise, it is not justified. This results in excessive travel time and increased fuel cost. So, the state government is planning to construct a bridge across the river. The estimated initial investment for constructing the bridge is Rs. The estimated life of the bridge is 15 years. The annual operation and maintenance cost is Rs.
The value of fuel savings due to the construction of the bridge is Rs. Total present worth of fuel savings BP: Since the BC ratio is more than 1, the construction of the bridge across the river is justified. In addition to the production of electric power, this project will provide flood control, irrigation and recreation benefits.
The estimated benefits and costs that are expected to be derived from this project are as follows: There is no salvage value associated with either of the projects. The initial outlay for the project A2 is Rs. Using the benefit cost ratio. Project A1 requires an initial outlay of Rs. Within the next 20 years. The right of way.
It is proposed to broaden a river flowing from the state to the seaport at a cost of Rs. The state will have to pay to each of them a welfare cheque of Rs. The reduction in the income from the taxes on the railroad will be compensated by the taxes on the barges. There would be some side effects of the change-over as follows. The average transport charge is Rs. The railroad would be bankrupt and be sold for no salvage value. This will make the river navigable to barges and will reduce the transport cost to Rs.
The annual goods transported is 1. The comparison is made on a year period which is the minimum common multiple of the lives of alternatives 1 and 2.
What is the benefit-cost ratio based on the next 20 years of operation? The estimated benefits and costs that are expected from the three alternatives under consideration are given in the following table. Since the BC ratio is more than 1. In addition to the production of electric power. Since A is the only eligible alternative. Initial cost P Benefit-cost ratio From the last row of Table List the different benefits and costs related to this alternative. This results in excessive time of travel and increased fuel cost.
Discuss the difference in evaluating alternatives of private and public organizations. Consider the evaluation of the alternative of constructing a bridge across a river. Annual equivalent of the initial cost 1. The estimated benefits and costs that are expected to be derived from this project are listed below. In a particular locality of a state. A state government is planning a hydroelectric project for a river basin.
Two mutually exclusive projects are being considered for investment. A government is planning a hydroelectric project for a river basin. An inland state is presently connected to a seaport by means of a railroad system.
The estimated benefits and costs expected from the three alternatives under consideration are listed in the following table: Within the next 25 years. Using the BC ratio. What is the BC ratio based on the next 25 years of operation? The annual goods transported amount to 1. It is proposed to improve a river flowing from the state to the seaport at a cost of Rs.
Besides the production of electric power. For practical decision making. Inflation is the rate of increase in the prices of goods per period. If economic decisions are taken without considering the effect of inflation into account. The same is true for succeeding years and hence the rate of inflation is compounded in the same manner that an interest rate is compounded.
But due to various reasons. As per our requirement. But there is always difficulty in determining the rate of inflation. If the rate of inflation is very high. Step 2. What equal amount should he save each year until he retires so that he can make withdrawals at the end of each year commencing from the end of the 21st year from now that will allow him to live as comfortably as he desires for 10 years beyond his retirement?
Solution Step 1. The overall cash flow diagram for the savings and withdrawal in terms of future rupees is shown in Fig. He plans to retire at the age of 60 and estimates that he can live comfortably on Rs. The formula which is given below is used to get future equivalent of Rs. Modification of the costs estimated in step 1 is summarized in Table The method of computing the present equivalent of the withdrawals is as follows: A machine which is downloadd today cannot be used forever.
The annual equivalent amount A. This kind of analysis is called replacement analysis. The productivity of any organization is a function of many factors. It is largely affected by efficient and effective use of machinery and equipment. It has a definite economic lifetime. After the economic life. The person has to invest an amount of Rs. Inflation Adjusted Decisions The sum of the present equivalents of the year end withdrawals from the year 21 to 30 is computed by assuming the end of the year 20 as the base time zero and it is shown at the end of the year 20 in Fig.
Annual operation and maintenance cost 3. The year with the minimum total cost is called as the economic life of the machine. Initial cost 2. Salvage value. download cost initial cost 2. But the capital recovery with return decreases with the life of the machine. Operation and maintenance cost 3. Salvage value at the end of every year. The total cost of the machine goes on decreasing initially but it starts increasing after some years.
But the increase in the operation and maintenance cost due to inflation is not considered. Their resale value is going up year after year. In replacement analysis. This may be partly due to inflation and partly due to good quality of the engine parts. The internal combustion engines R. Its annual operation cost during the first year is Rs. The maintenance cost during the first year is Rs.
To highlight this particular fact on salvage value. The corresponding steps are explained in Table The resale value of the machine is Rs. From the Table From the table it is clear that the total annual equivalent cost is minimum if the machine is used for 14 years. In the existing model. At the end. First the concept of replacement analysis is demonstrated without taking the inflation into account.
Lister which were made in England during pre-independence of India are still functioning well.
The annual cost of operation and maintenance of the machine will increase with the age of the machine due to decline in efficiency of the machine. Inflation Adjusted Decisions In the existing model to deal with this type of replacement analysis. The operating cost of the grinder including labour is Rs. Which design should be adopted if 1,00, units are required per year and what is the economic advantage of the best alternative?
Hence, design B is recommended for making the tapered fastening pin. The chief engineer of refinery operations is not satisfied with the preliminary design for storage tanks to be used as part of a plant expansion programme.
The engineer who submitted the design was called in and asked to reconsider the overall dimensions in the light of an article in the Chemical Engineer, entitled How to size future process vessels? The original design submitted called for 4 tanks 5. From a graph of the article, the engineer found that the present ratio of height to diameter of 1. The cost for the tank design as originally submitted was estimated to be Rs. What are the optimum tank dimensions if the volume remains the same as for the original design?
What total savings may be expected through the redesign? Hence, it is assumed that the price of raw material is location dependent. While sourcing a raw material, the cost of transportation is to be considered in conjunction with the price of the raw material. This concept is demonstrated with a numerical example. Either steel or aluminium window frames will satisfy the design criteria. Because of the remote location of the building site and lack of building materials in Alpha State, the window frames will be downloadd in Beta State and transported for a distance of 2, km to the site.
The price of window frames of the type required is Rs. The weight of steel window frames is 75 kg each and that of aluminium window frame is 28 kg each. The shipping rate is Re 1 per kg per km. Which design should be specified and what is the economic advantage of the selection? Hence, aluminium window frame is recommended. The process sequence of a component which has been planned in the past is not static.
It is always subject to modification with a view to minimize the cost of manufacturing the component. The steps in process planning are as follows: 1. Analyze the part drawing to get an overall picture of what is required. Make recommendations to or consult with product engineers on product design changes. List the basic operations required to produce the part to the drawing or specifications. Determine the most practical and economical manufacturing method and the form or tooling required for each operation.
Devise the best way to combine the operations and put them in sequence. Specify the gauging required for the process. Steps 35 aim to determine the most practical and economical sequence of operations to produce a component.
This concept is demonstrated with a numerical problem. Solution a Cost of component using process sequence 1. The process sequence 3 of the component is as follows: Only CNC operations The calculations for the cost of the above process sequence are summarized in Table 2.
Therefore, it should be selected for manufacturing the component. List and explain the different situations deserving elementary economic analysis. Explain the steps in the process planning. In the design of an aircraft jet engine part, the designer has a choice of specifying either an aluminium alloy casting or a steel casting.
Either material will provide equal service, but the aluminium casting will weigh 5 kg as compared with 7 kg for the steel casting. The aluminium part can be cast for Rs. Two alternatives are under consideration for a hexagonal bolt fastening pin.
Either design will serve equally well and will involve the same material and manufacturing cost except for the lathe and grinder operations. Elementary Economic Analysis 25 Design A will require 20 hours of lathe time and 8 hours of grinder time per 10, units. Design B will require 10 hours of lathe time and 22 hours of grinder time per 10, units. Which design should be adopted if 10,00, units are required per year and what is the economic advantage of the best alternative?
A building contractor can source door frames from either a nearby shop or a far-off forest area. The cost details are as summarized in the following table. The total requirement of wood for the construction work is 75 tons. Also find the economic advantage of the best decision. Consider Example 2. Rework this example if the ratio of the height to diameter corresponding to the minimum cost is instead of The process planning engineer of a firm listed down the sequences of operations, as shown in the following table to produce a component: Sequence 1 2 3 Process sequence Turning Milling Shaping Drilling Turning Milling Drilling All operations are performed with CNC machine The details of process time for the components for various operations and their machine hour rates are tabulated now.
Operation Machine hour rate Rs. It represents the growth of capital per unit period. The period may be a month, a quarter, semiannual or a year. So, the total amount at the end of the first year will be Rs. Hence the total amount at the end of the second year will be Rs. The process will continue thus till the specified number of years.
Table 3. The maturity value at the end of the fifth year is Rs. This means that the amount Rs. This is diagrammatically shown in Fig. This explanation assumes that the inflation is at zero percentage. Alternatively, the above concept may be discussed as follows: If we want Rs.