UNIT I - Introduction to Economics PDF

UNIT - I Introduction to Economics:– Flow in an Economy, Law of supply and Demand, Concept of Engineering Economics – Engineering Efficiency, Economic Efficiency, Scope of Engineering Economics, Elements of costs, Marginal Cost, Marginal Revenue, Sunk cost, Opportunity cost, Break-Even Analysis, P/V...

UNIT - I Introduction to Economics:– Flow in an Economy, Law of supply and Demand, Concept of Engineering Economics – Engineering Efficiency, Economic Efficiency, Scope of Engineering Economics, Elements of costs, Marginal Cost, Marginal Revenue, Sunk cost, Opportunity cost, Break-Even Analysis, P/V ratio, Elementary Economics Analysis- Function, Aims, Value Engineering procedure, Interest Formulas and their Applications – Time Value of Money, Single Payment Compound Amount Factor, Single Payment Present Worth Factor, Equal Payment Series, Compound Amount Factor, Equal Payment Series Sinking Fund Factor, Equal Payment Series Present Worth Factor, Equal Payment Series Capital Recovery Factor, Uniform Gradient Series Annual Equivalent Factor, Effective Interest Rate, Examples in all the methods. Economics: Economics is the science that deals with the production and consumption of goods and services and the distribution and rendering of these for human welfare. The following are the economic goals;  A high level of employment   Price stability   Efficiency   An equitable distribution of income   Growth Flow in an Economy; The flow of goods, services, resources and money payments in a simple economy are shown in below diagram. Households and business are the two major entities in a simple economy. Business organizations use various economics resources like land, labour and capital which will be used by them. Business organizations make payment of money to the households for receiving various resources. The households in turn make payment of money to business organization for receiving consumer goods and services. This cycle shows the interdependence between the two major entities in a simple economy  Money payments for consumer goods and services Consumer goods, Services Business Households Provide goods and services to Consume final goods and services consumers. produced by business and services Use resources, inputs provided Provide productive inputs by households to businesses Money payments for resources, rents, wages, salaries, interest and profit Economics Resources; Land, Labour, Capital Law of supply and Demand An interesting aspect of the economy is that the demand and supply of a product are interdependent and they are sensitive with the respect to the price of that product. The interrelationships between them are shown in above diagram. And also it is clear that when there is a decrease in the price of a product, the demand for product increases and its supply decreases. Also, the product is more in demand for the product increases. At the same time, lowering of the price of the product makes the producers restrain from releasing more quantities of the product in the market. Hence the supply of the product is decreased. The point of intersection of the supply curve and the demand curve is known as the equilibrium point. At the price corresponding to this point, the quantity of supply is equal to the quantity of demand. Hence this point is called equilibrium point. Factors influencing demand The shape of the demand curve is influenced by the following factors; o Income of the people o Prices of related goods o Tastes of consumers If the income level of the people increases significantly, then their purchasing power will naturally improve. This would definitely shift the demand curve to the north east direction. A converse situation will shift the demand curve to the south west direction. If, for instance the price of television sets is lowered drastically its demand would naturally go up. As a result, the demand for its associated product, namely VCDs, would also increase. Hence the prices of related foods influence the demand of products. Over a period of time, the preferences of the people for a particular product may increase, which in turn, will affect the demand. For instance diabetic people prefer to have sugar free products. If the incidence of diabetes rises naturally there will be increased demand for sugar free products. Factors influencing supply The shape of the supply curve is affected by the following factors: 1) Cost of the inputs 2) Technology 3) Weather 4) Prices of related goods  If the cost of inputs increases, then naturally, the cost of the product will go up. In such a situation, at the prevailing price of the product the profit margin per unit will be less. The producers will then reduce the production quantity, which in turn will affect the supply of the product. For instance if the prices of fertilizers and cost of labor are increased significantly in agriculture the profit margin per bag of paddy will be reduced. So, the farmers will reduce the area of cultivation, and hence the quantity of supply of paddy will be reduced at the prevailing prices of the paddy. If there is advancement in technology used in manufacture of the product in the long run, there will be a reduction in the production cost per unit. This will enable the manufacturer to have a greater profit margin per unit at the prevailing price of the product. Hence, the producer will be tempted to supply more quantity to the market. Weather also has a direct bearing on the supply of products. For example demand for woolen products will increase during winter. This means the prices of woolen goods will be increased in winter. So, naturally, manufacturers will supply more volume of woolen goods during winter. Again take the case of television sets. If the price of TV sets is lowered significantly then its demand would naturally go up. As a result, the demand for associated products like VCDs would also go up. Over a period of time, this will lead to an increase in the price of VCDs, which would result in more supply of VCD‘s CONCEPT OF ENGINEERING ECONOMICS Science is a field of study where the basic principles of different physical systems are formulated and tested. Engineering is the application of science. It establishes varied applications systems based on different scientific principles. It is clear that price has a major role in deciding the demand and supply of the product. Hence from the organizations point of view, efficient and effective functioning of the organization would certainly help it to provide goods/services at a lowe cost which in turn will enable it to fix a lower price for its goods or services. The following discusses the different types of efficiency and their impact on the operation of businesses and the definition and scope of engineering economics. Types of Efficiency Efficiency of a system is generally defined as the ratio of its output to input. The efficiency can be classified into technical efficiency and economic efficiency.  Technical Efficiency It is the ratio of the output to input of a physical system. The physical system may be a diesel engine, a machine working in a shop floor, furnance etc, The technical efficiency of a diesel engine is as follows In practice technical efficiency can never be more than 100%. This is mainly due to frictional loss and incomplete combustion of fuel, which are considered to be unavoiadable phenomena in the working of a diesel engine. Economic efficiency Economic efficiency is the ratio of output to input of a business system. Worth is the annual revenue generated by way of operating the business and cost is the total annual expenses incurred in carrying out the business. For the survival and growth of any business the economic efficiency should be more than 100%. Economic efficiency is also called productivity. There are several wazs of improving productivity.  Increased output for the same input   Decreased output for the same output   By a proporionate increase in the output which is more than the proportionate increase in the input 5   By a proportionate decrease in the input which is more than the proportionate decrease in the output   Through simultaneous increase in the output with decrease in the input. Increased output for the same input. In this strategy, the output is increased while keeping the input constant. Let us assume that in a steel plant, the layout of the existing facilities is not proper. By, slightly altering the location of the billet-making section., and bringing it closer to the furance which produces hot metal , the scale formation at the top of ladles will be considerably reduced. The molten metal is usually carried in ladles to the billet-making section. In the lng run, this would give more yield in terms of tonnes of billet produced. In this exercise, there is no extra cost involved. The only task is the relocation of the billet-making facility wich involves an insignficant cost. Decreased input for the same out put. In this strategy, the input is decreased to produce the same output. let us assume that there exists a substitue raw material to manfacture a product and it is available at a lower price. If we can identify such a material and use it for manfacturing the product then certainly it will reduce the input. In this excerise, the job of the purchase department is to identify an alternate subsitute material. The process of identification does not involve any extra cost. So, the productivity ratio will increase because of the decreased input by way of using cheaper raw materials to produce the same output. Less proportionate increase in output is more than that of the input. consider the example of introducing a new product into the existing product mix of an organization. Let us assume that the existing facilities are not fully utilized an the R&D wing of the company has identified a new product which has a very good market and which can be manfactured with the surplus facilities of the organization. If the new product is taken up for production, it will lead to  An increase in the revenus of the organiyation by way of selling the new product in addition to the existing product mix and   An increase in the material cost and operation and maintenance cost of machineries because of producing the new product. If we examine these these two increase closely, the proportionate increase in the revenue will be more than the proprotionate increase in the input cost. Hence, there will be a net increase in the productivity ratio. 6  When proprotionate decrease in input is more than that of the output.let us consider the converse of the pervious example, i.e dropping an uneconomical product from the existing product mix. This will result in the following:  A decrease in the revenue of the organization   A decrease in the material cost, and operation and maintenance cost of machinery If we closely examine these two decreases, we will se that the proportionate decrease in the input cost will more than the proportionate decrease in the revenue. Hence, there will be net increase in the productivity ratio. Simultaneous increase in output and decrease in input. let us asume that there are advanced automated technologies like robots and automated guided vechile system (AGVS, available in the market which can be empolyed in the organization we are interested in. If we employ these modern tools, then:  There will be drastic reduction in the operation cost, initiallly, the cost on equipment would be very high. But in the long run, the reduction in the operation cost would break-even the high intial investment and offer more savings on the input.   These advanced facilities would help in producing more products because they do not experience fatigue. The increased production will yield more revenue.   In this example in the long run, there is an increase in the revenue and a decrease in the input. hence, the productivity ratio will increase at a faster rate. 1.2.2 Definition and scope of engineering economics As stated earlier, efficient functioning of any business organization would enable it to provide goods/services at a lower price. In the process of managing organizations, the managers at different levels should take appropriate economic decisions which will help in minimizing investment, operating and maintenance expenditures besides increasing the revenue, savings and other related gains of the organization. Definition Engineering economics deals with the methods that enable one to take economic decisions towards minimizing costs and /or maximizing benefits to business organizations. Scope The issues that covered in this book are elementary economic analysis, intrest formulae, bases for comparing alternatives, present worth method, future worth method, annual equilant method, rate of return method,replacement analysis, depreciation, evaluationof public alternatives, inflation adjusted investment decisions, make or buy decisions, inventory control, project management, value engineering and linear programming 1.3 ELEMENTS OF COSTS Cost can be broadly classfied into variable cost and overhead cost. Variable cost varies with the volume of production while overhead cost is fixed, irrespective of the production volume. Variable cost can be futher classfied into direct material cost, direct labour cost, and direct expenses. The overhead cost can be classfied into factory overhead, adminstration overhead, selling overhead, and distribution overhead. Direct material costs are those costs of materials that are used to produce the product. Direct labour cost is the amount of wages paid to the direct labour involved in the production activities. Direct expenses are those expenses that vary in relation to the production volume, other than the direct material costs and direct labour costs. Overall cost is the aggregate of indirect material costs, indirect labour costs and indirect expenses. Administration overhead includes all the costs that are incurred in administering the business. Selling overhead is the total expense that is incurred in the promotional activities and the expenses relating to sales force. Distribution overhead is the total cost of shipping the items from the factory site to the customer sites The selling price of a product is derived as shown below: a) Direct material costs + Direct labour costs+ Direct expenses= Prime cost b) Prime cost+ Factory overhead = Factory cost. c) Factory cost+ office and administrative overhead = cost of production. d) cost of production +opening finished stock- Closing finished stock = cost of goods sold. e) cost of goods sold + selling and distribution overhead = cost of sales f) cost of sales + profit = sales g) sales/ Quantity sold = selling price per unit  in the above calculations, if the opening finished stock is equal to the closing finished stock, then the cost of production is equal to the cost of goods sold. 1.4 OTHER COSTS/ REVENUES The following are the costs/revenue other than the costs which are presented in the previous section:  Marginal cost   Marginal revenue   Sunk cost   Opportunity cost Marginal cost: Marginal cost of a product is the cost of producing an additional unit of that product. Let the cost of producing 20 units of a product be Rs.10,000, and the cost of producing 21 units of the same product be Rs. 10045. Then the marginal cost of producing the 21 units is Rs.45. Marginal revenue: Marginal revenue of a product is the incremental revenue of selling an additional unit of that product. Let the revenue of selling 20 units of a product be Rs.15,000 and the revenue selling 21 units of the same product be Rs.15085. then, the marginal revenue of selling the 21st unit is Rs. 85. Sunk Cost: This is known as the past cost of an equipment/asset. Let us assume that an equipment has been purchased for Rs 1,00,000 about three years back. If it is considered for replacement, then its present value is not Rs. 100000. Instead, its present market value should be taken as the present value of the equipment for further analysis. So, the purchase value of the equipment in the past is known as its sunk cost. The sunk cost should not be considered for any analysis done from nowonwards. Opportunity Cost In practice, if an alternative (X) is selected from a set of competing alternatives(X, Y), then the corresponding investment in the selected alternative is not available for any other purpose. If the same money is invested in some other alternative (Y), it may fetch some return. Since the money is invested in the selected alternative(X), one has to forego the return form the other alternative (Y). The amount that is foregone by not investing in the other alternative(Y) is known as opportunity cost of the selected alternative(X). So the opportunity cost of an alternative is the return that will be foregone by not investing the same money in another alternative. Consider that a person has invested a sum of Rs 50, 000 in shares. Let the expected annual return by this alternative be rs 7500. If the same amount is invested in fixed deposit, a bank will pay a return of 18%. Then, the corresponding total return per year for the investment in the bank is rs 9000. This return is greater than the return from shares. The foregone excess return of rs 1500 by way of not investing in the bank is the opportunity cost of investing in shares. BREAK EVEN ANALYSIS The main objective of break even analysis is to find the cutt off production volume from where a firm will make profit. Let , s = selling price per unit , v = variable cost per unit FC = fixed cost per period Q = volume of production The total sales revenue (s) of the firm is given by the following formula; S = sxQ The total cost of the firm for a given production volume is given as TC = Total variable cost + Fixed cost = v x Q + FC The linear plots of the above two equations are shows in below diagram. The intersection point of the total sales revenue line and the total cost line is called the break even point. The corresponding volume of production on the X axis us known as break even sales quantity. At the intersection point, the total cost is equal to the total revenue. This point is also called as no loss or no gain situation. For any production quantity which is less than the break even quantity, the total cost is more than the total revenue. Hence, the firm will be making loss. For any production quantity which is more thatn the break even quantity, the total revenue will be more than the total cost. Hence, the firm will be making profit. Profit = Sales - (Fixed cost + Variable cost) = s x Q - (FC + v x Q ) The formulae to find the break even quantity and break even sales quantity The contribution is the difference between the sales and the variable costs. The margin of safty is the sales over and above the break even sales. The formulae to compute these values are Contribution = Sales – Variable costs Contribution /unit = Selling Price/ unit - Variable cost/unit M. S = Actual Sales – Break even sales M.S as a per cent of sales = (M.S/ Sales) x 100 EXAMPLE 1.1 Alpha Associates has the following details: Fixed cost = Rs. 20,00,000; Variable cost per unit = Rs. 100; Selling price per unit = Rs. 200 Find (a) The break-even sales quantity, (b) The break-even sales (c) If the actual production quantity is 60,000, find (i) contribution; and (ii) margin of safety by all methods. Solution Fixed cost (FC) = Rs. 20,00,000 Variable cost per unit (v) = Rs. 100 Selling price per unit (s) = Rs. 200. PROFIT/VOLUME RATIO (P/V RATIO) P/V ratio is a valid ratio which is useful for further analysis. The different formulae for the P/V ratio are as follows: The relationship between BEP and P/V ratio is as follows: The following formula helps us find the M.S. using the P/V ratio: EXAMPLE 1.2 Consider the following data of a company for the year 1997: Sales = Rs. 1,20,000 Fixed cost = Rs. 25,000 Variable cost = Rs. 45,000 Find the following: (a) Contribution (b) Profit (c) BEP (d) M.S. Solution  EXAMPLE 1.3 Consider the following data of a company for the year 1998: Sales = Rs. 80,000; Fixed cost = Rs. 15,000; Variable cost = 35,000 Find the following: (a) Contribution (b) Profit (c) BEP (d) M.S. Solution; ELEMENTARY ECONOMIC ANALYSIS Whether it is a business situation or a day to day event in somebody‘s personal life, there are a large number of economic decisions making involved. One can manage many of these decision problem by sing simple economic analysis. For example an industry can source its raw materials from a nearby place or from a far off place. In this problem the following factors will affect the decisions.   Price of the raw materials     Transportation cost of the raw materials    Availability of the raw materials   Quality of the raw materials Consider the alternative of sourcing raw materials from a nearby place with the following characteristics;   The raw material is more costly in the nearby area  The availability of the raw material is not sufficient enough to support the  operation of the industry throughout the year  The raw material requires pre-processing before it is used in the production  process. This would certainly add cost to the product  The cost of transportation is minimal under this alternative On the other hand, consider another alternative of sourcing the raw materials form a far off place with the following characteristics;   The raw materials is less costly at the far off the place     The cost of transportation is very high   raw material at this site is abundant and it can support the plant The availability of the throughout the year.   from this site does not require any pre processing before using it The raw material for production EXAMPLES FOR SIMPLE ECONOMIC ANALYSIS In this section, the concept of simple economic analysis is illustrated using suitable examples in the following areas:  Material selection for a product   Design selection for a product   Design selection for a process industry   Building material selection for construction activities   Process planning/Process modification Material Selection for a Product/Substitution of Raw Material The cost of a product can be reduced greatly by substitution of the raw materials. Among various elements of cost, raw material cost is most significant and it forms a major portion of the total cost of any product. So, any attempt to find a suitable raw material will bring a reduction in the total cost in any one or combinations of the following ways:  Cheaper raw material price   Reduced machining/process time   Enhanced durability of the product Therefore, the process of raw material selection/substitution will result in finding an alternate raw material which will provide the necessary functions that are provided by the raw material that is presently used. In this process, if the new raw material provides any additional benefit, then it should be treated as its welcoming feature EXAMPLE 2.1 In the design of a 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 1.2 kg as compared with 1.35 kg for the steel casting. The aluminium can be cast for Rs. 80.00 per kg. and the steel one for Rs. 35.00 per kg. The cost of machining per unit is Rs. 150.00 for aluminium and Rs. 170.00 for steel. Every kilogram of excess weight is associated with a penalty of Rs. 1,300 due to increased fuel consumption. Which material should be specified and what is the economic advantage of the selection per unit? Solution (a) Cost of using aluminium metal for the jet engine part: Weight of aluminium casting/unit = 1.2 kg Cost of making aluminium casting = Rs. 80.00 per kg Cost of machining aluminium casting per unit = Rs. 150.00 Total cost of jet engine part made of aluminium/unit = Cost of making aluminium casting/unit + Cost of machining aluminium casting/unit = 80 x 1.2 + 150 = 96 + 150 = Rs. 246 (b) Cost of jet engine part made of steel/unit: Weight of steel casting/unit = 1.35 kg Cost of making steel casting = Rs. 35.00 per kg Cost of machining steel casting per unit = Rs. 170.00 Penalty of excess weight of steel casting = Rs. 1,300 per kg Total cost of jet engine part made of steel/unit = Cost of making steel casting/unit + Cost of machining steel casting/unit + Penalty for excess weight of steel casting = 35 x 1.35 + 170 + 1,300(1.35 – 1.2) = Rs. 412.25 DECISION The total cost/unit of a jet engine part made of aluminium is less than that for an engine made of steel. Hence, aluminium is suggested for making the jet engine part. The economic advantage of using aluminium over steel/unit is Rs. 412.25 – Rs. 246 = Rs. 166.25 EXAMPLE 2.2 A company manufactures dining tables which mainly consist of a wooden frame and a table top. The different materials used to manufacture the tables and their costs are given in Table 2.1. In view of the growing awareness towards deforestation and environmental conservation, the company feels that the use of wood should be minimal. The wooden top therefore could be replaced with a granite top. This would require additional wood for the frame and legs to take the extra weight of the granite top. The materials and labour requirements along with cost details to manufacture a table with granite top are given in Table 2.2. If the cost of the dining table with a granite top works out to be lesser than that of the table with wooden top, the company is willing to manufacture dining tables with granite tops. Compute the cost of manufacture of the table under each of the alternatives described above and suggest the best alternative. Also, find the economic advantage of the best alternative. Solution (a) Cost of table with wooden top Cost of wood for frame and legs = 12,000 x 0.1 = Rs. 1,200 Cost of wooden top = Rs. 3,000  Cost of bushes = 10 x 4 = Rs. 40 Cost of nails = 300 x (100/1,000) = Rs. 30 Cost of labour = 50 x 15 = Rs. 750 Total = Rs. 5,020 (b) Cost of table with granite top Cost of wood for frame and legs = 12,000 x 0.15 = Rs. 1,800 Cost of granite top = 800 x1.62 = Rs. 1,296 Cost of bushes = 25 x 4 = Rs. 100 Cost of nails = 300x (50/1,000) = Rs. 15 Cost of labour = 50x 8 = Rs. 400 Total = Rs. 3,611 The cost of a table with granite top works out to be less than that of a table with a wooden top. Hence, the table with granite top should be selected by the manufacturer. (c) Economic advantage Cost of a table with wooden top = Rs. 5,020 Cost of a table with granite top = Rs. 3,611 Economic advantage of table with granite top = Rs. 1,409 Design Selection for a Product The design modification of a product may result in reduced raw material requirements, increased machinability of the materials and reduced labour. Design is an important factor which decides the cost of the product for a specified level of performance of that product. The elementary economic analysis applied to the selection of design for a product is illustrated with two example problems. EXAMPLE 2.3 Two alternatives are under consideration for a tapered fastening pin. Either design will serve the purpose and will involve the same material and manufacturing cost except for the lathe and grinder operations. Design A will require 16 hours of lathe time and 4.5 hours of grinder time per 1,000 units. Design B will require 7 hours of lathe time and 12 hours of grinder time per 1,000 units. The operating cost of the lathe including labour is Rs. 200 per hour. The operating cost of the grinder including labour is Rs. 150 per hour. Which design should be adopted if 1,00,000 units are required per year and what is the economic advantage of the best alternative? Solution Operating cost of lathe including labour = Rs. 200 per hr Operating cost of grinder including labour = Rs. 150 per hr (a) Cost of design A No. of hours of lathe time per 1,000 units = 16 hr No. of hours of grinder time per 1,000 units = 4.5 hr Total cost of design A/1,000 units = Cost of lathe operation per 1,000 units + Cost of grinder operation per 1,000 units = 16 x 200 + 4.5 x 150 = Rs. 3,875 Total cost of design A/1,00,000 units = 3,875 x 1,00,000/1,000 = Rs. 3,87,500 (b) Cost of design B No. of hours of lathe time per 1,000 units = 7 hr No. of hours of grinder time per 1,000 units = 12 hr Total cost of design B/1,000 units = Cost of lathe operation/1,000 units + Cost of grinder operation/1,000 units = 7 x 200 + 12 x 150 = Rs. 3,200 DECISION The total cost/1,00,000 units of design B is less than that of design A. Hence, design B is recommended for making the tapered fastening pin. Economic advantage of the design B over design A per 1,00,000 units = Rs. 3,87,500 – Rs. 3,20,000 = Rs. 67,500. Building material selection for construction activities The sourcing of raw materials will have a significant effect on the cost of any product. Hence it is assumed that the price of raw materials is location dependent. While sourcing a raw material, the cost of transportation is to be considered in conjunction with the price of raw material. EXAMPLE 2.5 In the design of buildings to be constructed in Alpha State, the designer is considering the type of window frame to specify. Either steel oraluminium 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 purchased in Beta State and transported for a distance of 2,500 km to the site. The price of window frames of the type required is Rs. 1,000 each for steel frames and Rs. 1,500 each for aluminium frames. 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 100 km. Which designshould be specified and what is the economic advantage of the selection? Solution Distance between Alpha State and Beta State = 2,500 km Transportation cost = Re 1/kg/100 km (a) Steel window frame Price of steel window frame/unit = Rs 1,000 Weight of steel window frame/unit = 75 kg Total cost of steel window frame/unit = Price of steel window frame/unit + Transportation cost of steel window frame/unit = 1,000 + (75 x2,500 x 1)/100 = Rs. 2,875 (b) Aluminium window frame Price of aluminium window frame/unit = Rs. 1,500 Weight of aluminium window frame/unit = 28 kg Total cost of aluminium window frame/unit = Price of aluminium window  frame/unit+Transportation cost of aluminium window frame/unit = 1,500 + (28 x 2,500x 1)/100 = Rs. 2,200 DECISION The total cost/unit of the aluminium window frame is less than that of steel window frame. Hence, aluminium window frame is recommended. The economic advantage/unit of the aluminium window frame over thesteel window frame = Rs. 2,875 – 2,200 = Rs. 675 Process Planning /Process Modification While planning for a new component, a feasible sequence of operations with the least cost of processing is to be considered. 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. So, the objective of process planning/process modification is to identify the most economical sequence of operations to produce a component. The steps in process planning are as follows; 1. Analyze the part drawing to get an overall picture of what is required 2. Make recommendations to or consult with product engineers on product design changes 3. List the basic operations required to produce the part to the drawings or specifications 4. Determine the most practical and economical manufacturing method and the form or tooling required for each operations 5. Devise the best way to combine the operations and put them in sequence. 6. Specify the gauging required for the process. EXAMPLE 2.6 The process planning engineer of a firm listed the sequences of operations as shown in Table 2.3 to produce a component. The details of processing times of the component for various operations and their machine hour rates are summarized in Table 2.4. Find the most economical sequence of operations to manufacture the component. Solution (a) Cost of component using process sequence 1. The process sequence 1 of the component is as follows: Turning – Milling – Shaping – Drilling The calculations for the cost of the above process sequence are summarized in Table 2.5. Table 2.5 Workings for Process Sequence 1 (b) Cost of component using process sequence 2. The process sequence 2 of the component is as follows: Turning – Milling – Drilling The calculations for the cost of the above process sequence are given in Table 2.6. (c) Cost of component using process sequence 3. 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 2.7 The process sequence 2 has the least cost. Therefore, it should be selected for manufacturing the component. Make or Buy Decision; In the process of carrying out business activities of an organization, a component/product can be made within the organization or bought from a subcontractor. Each decision involves its own costs. So, in a given situation, the organization should evaluate each of the above make or buy alternatives and then select the alternative which results in the lowest cost. This is an important decision since it affects the productivity of the organization. In the long run, the make or buy decision is not static. The make option of a component/product may be economical today; but after some time, it may turn out to be uneconomical to make the same. CRITERIA FOR MAKE OR BUY Criteria for make The following are the criteria for make: 1. The finished product can be made cheaper by the firm than by outside suppliers. 2. The finished product is being manufactured only by a limited number of outside firms which are unable to meet the demand. 3. The part has an importance for the firm and requires extremely close quality control. 4. The part can be manufactured with the firm‘s existing facilities and similar to other items in which the company has manufacturing experience. Criteria for buy The following are the criteria for buy: 1. Requires high investments on facilities which are already available at suppliers plant. 2. The company does not have facilities to make it and there are more profitable opportunities for investing company‘s capital. 3. Existing facilities of the company can be used more economically to make other parts. 4. The skill of personnel employed by the company is not readily adaptable to make the part. 5. Patent or other legal barriers prevent the company for making the part. 6. Demand for the part is either temporary or seasonal. APPROACHES FOR MAKE OR BUY DECISION Types of analysis followed in make or buy decision are as follows: 1. Simple cost analysis 2. Economic analysis 3. Break-even analysis Simple Cost Analysis EXAMPLE 13.1 A company has extra capacity that can be used to produce a sophisticated fixture which it has been buying for Rs. 900 each. If the company makes the fixtures, it will incur materials cost of Rs. 300 per unit, labour costs of Rs. 250 per unit, and variable overhead costs of Rs. 100 per unit. The annual fixed cost associated with the unused capacity is Rs. 10,00,000.Demand over the next year is estimated at 5,000 units. Would it be profitable forthe company to make the fixtures? We assume that the unused capacity has alternative use. Cost to make Variable cost/unit = Material + labour + overheads = Rs. 300 + Rs. 250 + Rs. 100 = Rs. 650 Total variable cost = (5,000 units) (Rs. 650/unit) = Rs. 32,50,000 Add fixed cost associated with unused capacity + Rs. 10,00,000 Total cost = Rs. 42,50,000 Cost to buy Purchase cost = (5,000 units) (Rs. 900/unit) = Rs. 45,00,000 Add fixed cost associated with unused capacity + Rs. 10,00,000 Total cost = Rs. 55, 00,000 The cost of making fixtures is less than the cost of buying fixtures from outside. Therefore, the organization should make the fixtures. Economic Analysis The following inventory models are considered to illustrate this concept:  Purchase model  Manufacturing model The formulae for EOQ and total cost (TC) for each model are given in the following table: Where D = demand/year P = purchase price/unit Cc = carrying cost/unit/year Co = ordering cost/order or set-up cost/set-up k = production rate (No. of units/year) r = demand/year Q1 = economic order size Q2 = economic production size TC = total cost per year EXAMPLE 13.2 An item has a yearly demand of 2,000 units. The different costs in respect of make and buy are as follows. Determine the best option. Buy option D = 2,000 units/year Co = Rs. 120/order Cc = Rs. 1.60/unit/year Make option Co = Rs. 60/set-up r = 2,000 units/year Cc = Re 1/unit/year k = 8,000 units/year Result: The cost of making is less than the cost of buying. Therefore, the firm should go in for the making option. VALUE ANALYSIS/ VALUE ENGINEERING Value Analysis is the systematic application of recognized techniques which identify the function of a product or service, establish a monetary value for the function and provide the necessary function reliably at the lowest overall cost. WHEN TO APPLY VALUE ANALYSIS One can definitely expect very good results by initiating a VA programme if one or more of the following symptoms are present: 1. Company‘s products show decline in sales. 2. Company‘s prices are higher than those of its competitors. 3. Raw materials cost has grown disproportionate to the volume of production. 4. New designs are being introduced. 5. The cost of manufacture is rising disproportionate to the volume of production. 6. Rate of return on investment has a falling trend. 7. Inability of the firm to meet its delivery commitments. Value Analysis vs. Value Engineering Often the terms value analysis and value engineering are used synonymously. Though the philosophy underlying the two is same, i.e. identification of unnecessary cost, yet they are different. The difference lies in the time and the stage at which the techniques are applied. Value analysis is the application of a set of techniques to an existing product with a view to improve its value. It is thus a remedial process. Value engineering is the application of exactly the same set of techniques to a new product at the design stage, project concept or preliminary design when no hardware exists to ensure that bad features are not added. Value engineering, therefore, is a preventive process. Value The term ‗value‘ is used in different ways and, consequently, has different meanings. The designer equates the value with reliability; a purchase person with price paid for the item; a production person with what it costs to manufacture, and a sales person with what the customer is willing to pay. Value, in value investigation, refers to ―economic value‖, which itself can be divided into four types: cost value, exchange value, use value, and esteem value. These are now briefly described. Cost value. It is the summation of the labour, material, overhead and all other elements of cost required to produce an item or provide a service compared to a base. Exchange value. It is the measure of all the properties, qualities and features of the product, which make the product possible of being traded for another product or for money. In a conventional sense, exchange value refers to the price that a purchaser will offer for the product, the price being dependent upon satisfaction (value) which he derives from the product. Value derived from the product consists of two parts ―use value‖ and ―esteem value‖, which are now described. Use value. It is known as the function value. The use value is equal to the value of the functions performed. Therefore, it is the price paid by the buyer (buyer‘s view), or the cost incurred by the manufacturer (manufacturer‘s view) in order to ensure that the product performs its intended functions efficiently. The use value is the fundamental form of economic value. An item without ―use value‖ can have neither ―exchange value‖ nor ―esteem value‖ Esteem value. It involves the qualities and appearance of a product (like a TV set), which attract persons and create in them a desire to possess the product. Therefore, esteem value is the price paid by the buyer or the cost incurred by the manufacturer beyond the use value. PERFORMANCE; The performance of a product is the measure of functional features and properties that make it suitable for a specific purpose. Appropriate performance requires that (a) the product reliably accomplish the intended use of work or service requirement (functional requirements), (b) the product provide protection against accident, harmful effects on body and danger to human life (safety requirements), (c) the product give trouble-free service cover during its specified life span (reliability requirements), (d) service and maintenance work can be carried out on the product with ease and with simple tools (maintainability requirements), and (e) appearance of the product creates an impression on the buyer and induces in him or her the desire to own the product (appearance requirements). Performance and cost must be interwoven. Desired performance at the least cost should be achieved by selecting appropriate materials and manufacturing operations, which is the measure of value. Therefore, the value of the product is the ratio of performance (utility) to cost. Thus, Value can be increased by increasing the utility for the same cost or by decreasing the cost for the same utility. Satisfactory performance at lesser cost through identification and development of low cost alternatives is the philosophy of Value analysis. FUNCTION Function is the purpose for which the product is made. Identification of the basic functions and determination of the cost currently being spent on them are the two major considerations of value analysis. Functional Analysis of Some Parts of a Bus Driver Cabin Classification of the functions Rarely do all functions assume equal importance. Usually, some functions are more important than others. Functions can be classified into the following three categories: 1. Primary function 2. Secondary function 3. Tertiary function Primary functions are the basic functions for which the product is specially designed to achieve. Primary functions, therefore, are the most essential functions whose non- performance would make the product worthless, e.g. a photo frame exhibits photographs, a chair supports weight, a fluorescent tube gives light. Secondary functions are those which, if not in-built, would not prevent the device from performing its primary functions, e.g., arms of a chair provide support for hands. Secondary functions are usually related to convenience. The product can still work and fulfill its intended objective even if these functions are not in-built and yet they may be necessary to sell the product. Tertiary functions are usually related to esteem appearance. For example, Sunmica top of a table gives esteem appearance for the table. Let us consider a single example of painting a company bus to explain all the above three functions. Here, the primary function of painting is to avoid corrosion. The secondary function is to identify the company to which the bus belongs by the colour of the paint (e.g. blue colour for Ashok Leyland Ltd.). The tertiary function is to impart a very good appearance to the bus by using brilliant colours. AIMS The aims of value engineering are as follows: 1. Simplify the product. 2. Use (new) cheaper and better materials. 3. Modify and improve product design. 4. Use efficient processes. 5. Reduce the product cost. 6. Increase the utility of the product by economical means. 7. Save money or increase the profits.  The value content of each piece of a product is assessed using the following questions: 1. Does its use contribute to value? 2. Is its cost proportionate to its usefulness? 3. Does it need all its features? These three questions pertain to the function of the part which may decide the elimination of parts.  Is there anything better for the intended use?   Can company or vendor standard be used?   Can a usable part be made by a lower-cost method?   Is it made with the proper tooling, considering volume?   Does the part yield suitable profit?   Can another vendor furnish the same at a lower cost? VALUE ENGINEERING PROCEDURE The basic steps of value engineering are as follows: (a) Blast (i) Identify the product. (ii) Collect relevant information. (iii) Define different functions. (b) Create (iv) Different alternatives. (v) Critically evaluate the alternatives. (c) Refine (vi) Develop the best alternative. (vii) Implement the alternative. Step 1: Identify the product First, identify the component for study. In future, any design change should add value and it should not make the product as obsolete one. Value engineering can be applied to a product as a whole or to sub-units. Step 2: Collect relevant information. Information relevant to the following must be collected: _ Technical specifications with drawings _ Production processes, machine layout and instruction sheet _ Time study details and manufacturing capacity _ Complete cost data and marketing details _ Latest development in related products Step 3: Define different functions Identify and define the primary, secondary and tertiary functions of the product or parts of interest. Also, specify the value content of each function and identify the high cost areas. Step 4: Different alternatives Knowing the functions of each component part and its manufacturing details, generate the ideas and create different alternatives so as to increase the value of the product. Value engineering should be done after a brain storming session. All feasible or non- feasible suggestions are recorded without any criticism; rather, persons are encouraged to express their views freely. Step 5: Critically evaluate the alternatives. Different ideas recorded under step 4 are compared, evaluated and critically assessed for their virtues, validity and feasibility as regards their financial and technical  requirements. The ideas technically found and involving lower costs are further developed. Step 6: Develop the best alternative. Detailed development plans are made for those ideas which emerged during step 5 and appear most suitable and promising. Development plans comprise drawing the sketches, building of models, conducting discussions with the purchase section, finance section, marketing division, etc. Step 7: Implement the alternative. The best alternative is converted into a proto-type manufacturing model which ultimately goes into operation and itsresults are recorded. ADVANTAGES AND APPLICATION AREAS Advantages. The advantages of value engineering are as follows:   It is a much faster cost reduction technique.   It is a less expensive technique.  It reduces production costs and adds value to sales income of the product. Applications The various application areas of value engineering are machine tool industries, industries making accessories for machine tools, auto industries, import substitutes, etc. INTEREST FORMULAS AND THEIR APPLICATIONS INTRODUCTION Interest rate is the rental value of money. It represents the growth of capital per unit period. The period may be a month, a quarter, semiannual or a year. An interest rate 15% compounded annually means that for every hundred rupees invested now, an amount of Rs. 15 will be added to the account at the end of the first year. So, the total amount at the end of the first year will be Rs. 115. At the end of the second year, again 15% of Rs. 115, i.e. Rs. 17.25 will be added to the account. Hence the total amount at the end of the second year will be Rs. 132.25. The process will continue thus till the specified number of years. TIME VALUE OF MONEY If an investor invests a sum of Rs. 100 in a fixed deposit for five years with an interest rate of 15% compounded annually, the accumulated amount at the end of every year will be as shown in Table 3.1. The formula to find the future worth in the third column is F = P X (1 + i)n Where P = principal amount invested at time 0, F = future amount i = interest rate compounded annually n = period of deposit. The maturity value at the end of the fifth year is Rs. 201.14. This means that the amount Rs. 201.14 at the end of the fifth year is equivalent to Rs. 100.00 at time 0 (i.e. at present). This is diagrammatically shown in Fig. 3.1. This explanation assumes that the inflation is at zero percentage. Alternatively, the above concept may be discussed as follows: If we want Rs. 100.00 at the end of the nth year, what is the amount that we should deposit now at a given interest rate, say 15%? A detailed working is shown in Table 3.2. The formula to find the present worth in the second column is From Table 3.2, it is clear that if we want Rs. 100 at the end of the fifth year, we should now deposit an amount of Rs. 49.72. Similarly, if we want Rs. 100.00 at the end of the 10th year, we should now deposit an amount of Rs. 24.72. INTEREST FORMULAS While making investment decisions, computations will be done in many ways. To simplify all these computations, it is extremely important to know how to use interest formulas more effectively. Before discussing the effective application of the interest formulas for investment-decision making, the various interest formulas are presented first. Interest rate can be classified into simple interest rate and compound interest rate. In simple interest, the interest is calculated, based on the initial deposit for every interest period. In this case, calculation of interest on interest is not applicable. In compound interest, the interest for the current period is computed based on the amount (principal plus interest up to the end of the previous period) at the beginning of the current period. The notations which are used in various interest formulae are as follows: P = principal amount n = No. of interest periods i = interest rate (It may be compounded monthly, quarterly, semiannually or annually) F = future amount at the end of year n A = equal amount deposited at the end of every interest period G = uniform amount which will be added/subtracted period after period to/ from the amount of deposit A1 at the end of period 1 Single-Payment Compound Amount Here, the objective is to find the single future sum (F) of the initial payment (P) made at time 0 after n periods at an interest rate i compounded every period. The cash flow diagram of this situation is shown in Fig. 3.2 The formula to obtain the single-payment compound amount is F = P(1 + i)n = P(F/P, i, n) where (F/P, i, n) is called as single-payment compound amount factor. EXAMPLE 3.1 A person deposits a sum of Rs. 20,000 at the interest rate of 18% compounded annually for 10 years. Find the maturity value after 10 years. Solution P = Rs. 20,000 i = 18% compounded annually n = 10 years F = P(1 + i)n = P(F/P, i, n) = 20,000 (F/P, 18%, 10) = 20,000 x 5.234 = Rs. 1,04,680 The maturity value of Rs. 20,000 invested now at 18% compounded yearly is equal to Rs. 1,04,680 after 10 years. Single-Payment Present Worth Amount Here, the objective is to find the present worth amount (P) of a single future sum (F) which will be received after n periods at an interest rate of i compounded at the end of every interest period. The corresponding cash flow diagram is shown in Fig. 3.3. Cash flow diagram of single-payment present worth amount. The formula to obtain the present worth is Where (P/F, i, n) is termed as single-payment present worth factor. EXAMPLE 3.2 A person wishes to have a future sum of Rs. 1,00,000 for his son‘s education after 10 years from now. What is the single-payment that he should deposit now so that he gets the desired amount after 10 years? The bank gives 15% interest rate compounded annually. Solution F = Rs. 1,00,000 i = 15%, compounded annually n = 10 years P = F/(1 + i)n = F(P/F, i, n) = 1,00,000 (P/F, 15%, 10) = 1,00,000 x 0.2472 = Rs. 24,720 The person has to invest Rs. 24,720 now so that he will get a sum of Rs. 1,00,000 after 10 years at 15% interest rate compounded annually. Equal-Payment Series Compound Amount In this type of investment mode, the objective is to find the future worth of n equal payments which are made at the end of every interest period till the end of the nth interest period at an interest rate of i compounded at the end of each interest period. The corresponding cash flow diagram is shown in Fig. 3.4. Fig. 3.4 Cash flow diagram of equal-payment series compound amount. In Fig. 3.4, A = equal amount deposited at the end of each interest period n = No. of interest periods i = rate of interest F = single future amount The formula to get F is where (F/A, i, n) is termed as equal-payment series compound amount factor. EXAMPLE 3.3 A person who is now 35 years old is planning for his retired life. He plans to invest an equal sum of Rs. 10,000 at the end of every year forthe next 25 years starting from the end of the next year. The bank gives 20% interest rate, compounded annually. Find the maturity value of his account when he is 60 years old. Solution A = Rs. 10,000 n = 25 years i = 20% F=? = A(F/A, i, n) = 10,000(F/A, 20%, 25) = 10,000 x 471.981 = Rs. 47,19,810 The future sum of the annual equal payments after 25 years is equal to Rs. 47,19,810. Equal-Payment Series Sinking Fund In this type of investment mode, the objective is to find the equivalent amount (A) that should be deposited at the end of every interest period for n interest periods to realize a future sum (F) at the end of the nth interest period at an interest rate of i. The corresponding cash flow diagram is shown in Fig. 3.6. In Fig. 3.6, A = equal amount to be deposited at the end of each interest period n = No. of interest periods i = rate of interest F = single future amount at the end of the nth period where (A/F, i, n) is called as equal-payment series sinking fund factor. EXAMPLE 3.4 A company has to replace a present facility after 15 years at an outlay of Rs. 5,00,000. It plans to deposit an equal amount at the end of every year for the next 15 years at an interest rate of 18% compounded annually. Find the equivalent amount that must be deposited at the end of every year for the next 15 years. Solution F = Rs. 5,00,000, n = 15 years i = 18% A = ? The corresponding cash flow diagram is shown in Fig. 3.7. = 5,00,000(A/F, 18%, 15) = 5,00,000 x 0.0164 = Rs. 8,200 The annual equal amount which must be deposited for 15 years is Rs. 8,200 Equal-Payment Series Present Worth Amount The objective of this mode of investment is to find the present worth of an equal payment made at the end of every interest period for n interest periods at an interest rate of i compounded at the end of every interest period. The corresponding cash flow diagram is shown in Fig. 3.8. Here P = present worth A = annual equivalent payment i = interest rate n = No. of interest periods The formula to compute P is EXAMPLE 3.5 A company wants to set up a reserve which will help the company to have an annual equivalent amount of Rs. 10,00,000 for the next 20 years towards its employees welfare measures. The reserve is assumed to grow at the rate of 15% annually. Find the single-payment that must be made now as the reserve amount. Solution A = Rs. 10,00,000, i = 15%, n = 20 years P=? The corresponding cash flow diagram is illustrated in Fig. 3.9. = 10,00,000 x (P/A, 15%, 20) = 10,00,000 x 6.2593 = Rs. 62,59,300 The amount of reserve which must be set-up now is equal to Rs. 62,59,300. Equal-Payment Series Capital Recovery Amount The objective of this mode of investment is to find the annual equivalent amount (A) which is to be recovered at the end of every interest period for n interest periods for a loan (P) which is sanctioned now at an interest rate of i compounded at the end of every interest period (see Fig. 3.10). Cash flow diagram of equal-payment series capital recovery amount In Fig. 3.10, P = present worth (loan amount) A = annual equivalent payment (recovery amount) i = interest rate n = No. of interest periods The formula to compute P is as follows: where, (A/P, i, n) is called equal-payment series capital recovery factor. EXAMPLE 3.6 A bank gives a loan to a company to purchase an equipment worth Rs. 10,00,000 at an interest rate of 18% compounded annually. This amount should be repaid in 15 yearly equal installments. Find the installment amount that the company has to pay to the bank. Solution P = Rs. 10,00,000, i = 18% n = 15 years A = ? The corresponding cash flow diagram is shown in Fig. 3.11. Fig. 3.11 Cash flow diagram of equal-payment series capital recovery amount. = 10,00,000 _ (A/P, 18%, 15) = 10,00,000 _ (0.1964) = Rs. 1,96,400 The annual equivalent installment to be paid by the company to the bank is Rs. 1,96,400. Uniform Gradient Series Annual Equivalent Amount The objective of this mode of investment is to find the annual equivalent amount of a series with an amount A1 at the end of the first year and with an equal increment (G) at the end of each of the following n – 1 years with an interest rate i compounded annually. The corresponding cash flow diagram is shown in Fig. 3.12. Fig. 3.12 Cash flow diagram of uniform gradient series annual equivalent amount. The formula to compute A under this situation is where (A/G, i, n) is called uniform gradient series factor. EXAMPLE 3.7 A person is planning for his retired life. He has 10 more years of service. He would like to deposit 20% of his salary, which is Rs. 4,000, at the end of the first year, and thereafter he wishes to deposit the amount with an annual increase of Rs. 500 for the next 9 years with an interest rate of 15%. Find the total amount at the end of the 10th year of the above series. Solution Here, A1 = Rs. 4,000 G = Rs. 500 i = 15% n = 10 years A = ? & F = ? The cash flow diagram is shown in Fig. 3.13. Cash flow diagram of uniform gradient series annual equivalent amount. = 4,000 + 500(A/G, 15%, 10) = 4,000 + 500 x 3.3832 = Rs. 5,691.60 This is equivalent to paying an equivalent amount of Rs. 5,691.60 at the end of every year for the next 10 years. The future worth sum of this revised series at the end of the 10th year is obtained as follows: F = A(F/A, i, n) = A(F/A, 15%, 10) = 5,691.60(20.304) = Rs. 1,15,562.25 At the end of the 10th year, the compound amount of all his payments will be Rs. 1,15,562.25. EXAMPLE 3.8 A person is planning for his retired life. He has 10 more years of service. He would like to deposit Rs. 8,500 at the end of the first year and thereafter he wishes to deposit the amount with an annual decrease of Rs. 500for the next 9 years with an interest rate of 15%. Find the total amount at the end of the 10th year of the above series. Solution Here, A1 = Rs. 8,500, G = –Rs. 500, i = 15% , n = 10 years, A = ? & F = ? The cash flow diagram is shown in Fig. 3.14. Fig. 3.14 Cash flow diagram of uniform gradient series annual equivalent amount. This is equivalent to paying an equivalent amount of Rs. 6,808.40 at the end of every year for the next 10 years. The future worth sum of this revised series at the end of the 10th year is obtained as follows: F = A(F/A, i, n) = A(F/A, 15%, 10) = 6,808.40(20.304) = Rs. 1,38,237.75 At the end of the 10th year, the compound amount of all his payments is Rs. 1,38,237.75. 3.3.8 Effective Interest Rate Let i be the nominal interest rate compounded annually. But, in practice, the compounding may occur less than a year. For example, compounding may be monthly, quarterly, or semi-annually. Compounding monthly means that the interest is computed at the end of every month. There are 12 interest periods in a year if the interest is compounded monthly. Under such situations, the formula to compute the effective interest rate, which is compounded annually, is where, i = the nominal interest rate C = the number of interest periods in a year. EXAMPLE 3.9 A person invests a sum of Rs. 5,000 in a bank at a nominal interest rate of 12% for 10 years. The compounding is quarterly. Find the maturity amount of the deposit after 10 years. Solution P = Rs. 5,000, n = 10 years, i = 12% (Nominal interest rate), F = ? METHOD 1 No. of interest periods per year = 4 No. of interest periods in 10 years = 10 _ 4 = 40 Revised No. of periods (No. of quarters), N = 40 Interest rate per quarter, r = 12%/4 = 3%, compounded quarterly. F = P(1 + r)N = 5,000(1 + 0.03)40 = Rs. 16,310.19 METHOD 2 No. of interest periods per year, C = 4 Effective interest rate, R = (1 + i/C)C – 1 = (1 + 12%/4)4 – 1 = 12.55%, compounded annually. F = P(1 + R)n = 5,000(1 + 0.1255)10 = Rs. 16,308.91