Engineering Economics and Cost Analysis PDF

Engineering Economics and Cost Analysis S. Ramesh Babu Professor Department of Mechanical Engineering Sri Venkateswara College of Engineering  Syllabus UNIT I - INTRODUCTION TO ECONOMICS Introduction to Economics- Flow in an econom...

Engineering Economics and Cost Analysis S. Ramesh Babu Professor Department of Mechanical Engineering Sri Venkateswara College of Engineering  Syllabus UNIT I - INTRODUCTION TO ECONOMICS Introduction to Economics- Flow in an economy, Law of supply and demand, Concept of Engineering Economics – Engineering efficiency, Economic efficiency, Scope of engineering economics- Element of costs, Marginal cost, Marginal Revenue, Sunk cost, Opportunity cost, Break-even analysis- V ratio, Elementary economic Analysis – Material selection for product Design selection for a product, Process planning.  Syllabus UNIT II - VALUE ENGINEERING Make or buy decision, Value engineering – Function, aims, Value engineering procedure. Interest formulae and their applications –Time value of money, Single payment compound amount factor, Single payment present worth factor, Equal payment series sinking fund factor, Equal payment series payment Present worth factor- equal payment series capital recovery factor-Uniform gradient series annual equivalent factor, Effective interest rate, Examples in all the methods.  Syllabus UNIT III CASH FLOW Methods of comparison of alternatives – present worth method (Revenue dominated cash flow diagram), Future worth method (Revenue dominated cash flow diagram, cost dominated cash flow diagram), Annual equivalent method (Revenue dominated cash flow diagram, cost dominated cash flow diagram), rate of return method, Examples in all the methods.  Syllabus UNIT IV - REPLACEMENT AND MAINTENANCE ANALYSIS Replacement and Maintenance analysis – Types of maintenance, types of replacement problem, determination of economic life of an asset, Replacement of an asset with a new asset – capital recovery with return and concept of challenger and defender, Simple probabilistic model for items which fail completely.  Syllabus UNIT V - DEPRECIATION Depreciation- Introduction, Straight line method of depreciation, declining balance method of depreciation-Sum of the years digits method of depreciation, sinking fund method of depreciation/ Annuity method of depreciation, service output method of depreciation-Evaluation of public alternatives- introduction, Examples, Inflation adjusted decisions – procedure to adjust inflation, Examples on comparison of alternatives and determination of economic life of asset.  Contents 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 Break Even Analysis Elementary Economic Analysis Material selection for product Design selection for a product Process planning  Objective The key to success for any productive enterprise is taking the right and result oriented decisions – Objective Social problems- unemployment, inflation, poverty, population, taxes, international financial crises have economic roots.  What is economics about Economics is the social science concerned with the analysis of commercial activities and with how goods and services are produced. The field of economics studies how the things, people need and want are made and brought to them.  Goals of Economics A high level of employment Price stability Efficiency An equitable distribution of income Growth  Economics - definition “a social science concerned chiefly with the way, society chooses to employ its limited resource, which have alternative uses, to produce goods and services for present and future consumption”. 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  Goods Economic goods are those goods which possess money value or those for procuring which human effort is required. Economic value of a good depends upon the place under consideration. Economic goods have the following properties They satisfy human wants or have utility They are scarce or limited in supply. They are marketable i.e have money value and may be exchanged.  Want vs need A need is something you have to have, something you can't do without. A want is something you would like to have. It is not absolutely necessary, but it would be a good thing to have.  Functions of an Economic System To match supply to the effective demand for goods and services in an efficient manner To determine what goods and services are to be produced and in what quantities. To distribute scarce resources among the industries producing goods and services  Functions of an Economic System To distribute the products of industry among members of the community. To provide for maintenance and expansion of fixed capital investment. To fully utilise the resources of society.  Economic problems What goods and services should be produced? How should resources be organised for production? Who shall get the goods and services? How fast shall the economy grow?  Engineering Economics Engineering Economics is a science which deals with the application of economic theory in Engineering Practices. It is the study of allocation of resources available to a firm among its activities. Flow in an Economy  Flow in an Economy Households and businesses are the two major entities in a simple economy. Business organizations use various economic resources like land, labour and capital which are provided by households to produce consumer goods and services 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 organizations for receiving consumer goods and services.  Demand Demand is a relation showing the various amounts of a commodity that buyers would be willing and able to purchase at possible alternative prices during a given period of time, all other things remaining the same. The law of demand states that other things remaining the same, when the price of a commodity falls, its quantity demanded rises and when the price of a commodity rises, it quantity demanded falls. In other words, other things remaining the same, there is an inverse relationship between the price of a commodity and its quantity demanded.  Meaning of Demand Individual demand – Demand by one buyer for a commodity is called individual demand. – Demand for a commodity by an individual is the quantity of that commodity that the individual is willing to buy at a price over some period of time. Definition include – Quantity of a commodity that a buyer is willing to buy – The price of the commodity at which he is willing to buy that quantity, and – The time period during which he is willing to buy that quantity at the given price (The time period may be a day, a week, a month, a year or any other period)  Demand Demand for a commodity implies: – Desire to acquire it, – Willingness to pay for it, and – Ability to pay for it.  Demand The demand for mangoes by a consumer is 3kgs per day The demand for mangoes by a consumer is 3kgs per day when the price of mangoes is Rs. 10/- kg The demand for mangoes is 3 kgs, when the price of mangoes is Rs. 10/- kg  Market demand There are many buyers of a commodity. If we add the quantity of the commodity that each of its buyer is willing to buy at a price over a time period, we will get the market demand of the commodity. Thus market demand means the total quantity of a commodity that all its buyers are willing to buy at a given price over a time period.  Want or desire and demand Mere want or desire for a commodity by a person is not called his demand for that commodity. The want or desire will become a demand if we have the ability to buy it and we are willing to buy it. Thus demand is the want or desire for a good backed by the ability and willingness to pay for it.  True or False 1. My demand for milk is 10 litres 2. My demand for milk is 10 litres per month when the price of milk is Rs. 10 per litre 3. My demand for milk is 10 litres per month whatever may be the price of milk 4. Hari is a rich man and can buy a car, so Hari has a demand for a car. 5. Want for goods means demand for goods 6. Want of a consumer for goods becomes his demand for it when it is backed by ability and willingness of the consumer to pay for it.  Factors affecting law of demand Price of a commodity – Change in real income or purchasing power of the buyer of the commodity – Substitution of one commodity for other commodity Other factors which include Income of the buyer of the commodity Tastes and preferences of the buyer Prices of the related goods  Change in real income or purchasing power of the buyer of the commodity Purchasing power or real income means the quantity of goods and services that one can buy with the given money income. An increase in purchasing power means more can be bought with the same money income and a decrease in purchasing power means less can be bought with the same money income.  Change in real income or purchasing power of the buyer of the commodity Example – 15 litres of milk per month when the price is Rs. 8per litre – Price of the milk falls to Rs. 6 per litre – Price of the milk rises from Rs. 8 per litre to Rs. 10 per litre.  Substitution of one commodity for other commodity Two commodities are said to be substitutes of each other when one can be used in place of other. Example – Kerosene, LPG, electricity – Tea and coffee Other factors that affect the demand of a commodity Income of a buyer (Normal Vs Inferior Goods) Tastes and preferences of the buyer Prices of the related goods Substitute Goods Complementary Goods  Exceptions to the law of Demand Prestige goods Giffen goods Expectations  True or False The law of demand applies only on essential goods The law of demand states that other things remaining the same, the price of a commodity and its quantity demanded are inversely related Other things remaining same means the other factors affecting demand do not change. The law of demand also applies on goods that have prestige value. If the price of a commodity is rising and is expected to continue to rise in future, its quantity demanded will start falling. Price of a good is only one of the factors that affects the demand for a good. Demand curve slopes downward from left to right.  True or False Other things remaining the same, when the price of a good rises its demand also rises. An individual demand schedule shows the quantities demanded of a commodity at different prices. When the price of a good rises the purchasing power of its buyer also rises.  Fill in the blanks The demand for a good is also affected by price of ______ (all goods, related goods) In case of a ______good the increase in income of its buyer leads to a fall in its demand. (normal, inferior) The demand for a commodity __________ when the price of its substitute commodity rises. (decreases, increases) The demand for a commodity increases if the price of its complementary commodity ____( falls, rises)  Expansion of Demand and Increase in Demand Price of milk per Quantity Quantity litre demanded of demanded when (Rs) milk per day (in demand rises litres) 12 1 1.5 10 1.5 2.0 8 2 2.5 6 2.5 3.0 4 3 3.5  Expansion of Demand and Increase in Demand 14 12 10 Price of milk 8 6 4 2 0 0 0.5 1 1.5 2 2.5 3 3.5 4 Quanity demanded  Expansion of Demand and Increase in Demand The expansion of demand results in a downward movement along the demand curve Increase in demand results in a rightward shift of the demand curve.  Contraction of Demand and Decrease in Demand Price of milk per Quantity Quantity litre demanded of demanded (Rs) milk per day (in when demand litres) rises 12 1 0.5 10 1.5 1.0 8 2 1.5 6 2.5 2.0 4 3 2.5  Fill in the blanks When demand rises due to fall in price, it is called _____ of demand. A rightward shift in demand curve shows _____ in demand. A decrease in demand will result in a ____ shift of demand curve A upward movement along the same demand curve shows _____ of demand.  Elasticity of Demand The elasticity of demand measures the responsiveness of quantity demanded to a change in any one of the factors by keeping other factors constant. Types – Price elasticity of demand – Income elasticity of demand – Cross Elasticity of demand  Price Elasticity of Demand Price elasticity of demand is the degree of responsiveness of quantity demanded of a good to a change in its price. “The ratio of proportionate change in the quantity demanded of a good caused by a given proportionate change in price”  Income Elasticity of Demand Income is an important variable affecting the demand for a good. When there is a change in the level of income of a consumer, there is a change in the quantity demanded of a good, other factors remaining the same. The degree of change or responsiveness of quantity demanded of a good to a change in the income of a consumer is called income elasticity of demand.  Income Elasticity of Demand The ratio of percentage change in the quantity of a good purchased, per unit of time to a percentage change in the income of a consumer".  Cross Elasticity of Demand The concept of cross elasticity of demand is used for measuring the responsiveness of quantity demanded of a good to changes in the price of related goods. "The percentage change in the demand of one good as a result of the percentage change in the price of another good".  Classification of Elasticity of Demand 1. Price Elasticity of Demand a. Perfectly elastic demand b. Perfectly inelastic demand c. Relatively elastic demand d. Relatively inelastic demand e. Unitary elastic demand 2. Income Elasticity of Demand a. Zero income elasticity of demand b. Negative income elasticity of demand c. Unitary income elasticity of demand d. Income elasticity of demand is greater than one e. Income elasticity of demand is less than one 3. Cross Elasticity of Demand a. Cross elasticity of demand is positive b. Cross elasticity of demand is zero c. Cross elasticity of demand is negative  Classification of levels of Price Elasticity Perfectly Elastic Demand Perfectly Inelastic Demand Relatively More Elastic Demand Relatively less Inelastic demand Unitary Elastic Demand  Perfectly Elastic Demand When there is a small fall in the price of a product, it will result no change in the demand and in a small rise in the prices, it will lead to big contraction of demand, even to zero.  Perfectly Inelastic Demand When there is a big change in the price of certain commodities, there is no change in the demand of those commodities. E.g. Most necessity items like salt, rice.  Relatively More Elastic Demand When there is a small change in the price of any commodity, there is a greater change in its demand. Commodities fall under comforts and luxury will expand the demand when there is a small change in its price. The elasticity of demand is greater than one  Relatively Less Inelastic Demand When there is a large change in the price of any commodity, there is a smaller change in its quantity demand. Necessaries of life fall under this category. The elasticity of demand is less than one  Unitary Elastic Demand When there is a change in the price of any commodity, there is an equal change in its quantity demanded. The elasticity of demand is equal to one.  Price Elasticity of Demand at a glance Numeral value of Term for elasticity of Descriptions elasticity demand Infinity Perfectly elastic Total revenue falls to zero Zero Perfectly Inelastic No change in the quantity demanded in response to the change in the price Greater than 1 Relatively elastic There is a small change in the price and there is a greater change in its demand Less than 1 Relatively inelastic There is a large change in the price and there is a small change in its demand One Unitary elastic Percentage change in quantity demanded is equal to the percentage change in its prices  Elasticity levels for Income elasticity of demand When the income of a person increases, his demand for goods also changes depending upon whether the good is a normal good or an inferior good. For normal goods, the value of elasticity is greater than zero but less than one. Goods with an income elasticity of less than 1 are called inferior goods. For example, people buy more food as their income rises but the % increase in its demand is less than the % increase in income. Elasticity levels for Cross elasticity of demand (i) Substitute Goods. When two goods are substitute of each other, such as coke and Pepsi, an increase in the price of one good will lead to an increase in demand for the other good. (ii) The numerical value of goods is positive. Elasticity levels for Cross elasticity of demand Complementary Goods. In case of complementary goods such as car and petrol, cricket bat and ball, a rise in the price of one good say cricket bat by 7% will bring a fall in the demand for the balls (say by 6%). The cross elasticity of demand which are complementary to each other is, therefore, 6% / 7% = 0.85 (negative). Elasticity levels for Cross elasticity of demand Unrelated Goods. The two goods which are unrelated to each other, say apples and pens, if the price of apple rises in the market, it is unlikely to result in a change in quantity demanded of pens. The elasticity is zero for unrelated goods.  Importance of elasticity of demand Fixing the price of a commodity If the demand of the commodity is inelastic, the company may rise the price to maximise the profits. If the demand of the commodity is elastic, the company may lower the price to maximise the demand.  Factors affecting the elasticity of demand Nature of the product Essential goods – inelastic / Comforts - Elastic Luxury goods – More Elastic Different use of a commodity / Extent use of the commodity Variety of uses – Elastic demand / Limited Use – Inelastic demand Range of substitutes Perfect substitute products – elastic demand Income level High income – Inelastic / Low income – elastic Proportion of Income spent on commodities Purchase frequency of a product – Elastic Habit and Fashion Deferred Consumption - Elastic  Law of supply The "law of supply" is a fundamental principal of economic theory which states that quantities respond in the same direction as price changes In other words, the law of supply states that (all other things unchanged) an increase in price results in an increase in quantity supplied. This means that producers are willing to offer more products for sale on the market at higher prices by increasing production as a way of increasing profits Law of supply Vs Law of Demand  Supply Schedule Price Per Slice of Slices Supplied Per Pizza (Rs) Day (#).50 100 1.00 150 1.50 200 2.00 250 2.50 300 3.00 350  S 3.00 2.50 2.00 1.50 1.00.50 0 0 100 150 200 250 300 350  Elasticity of Supply Elasticity of Supply: a measure of the way suppliers respond to a change in price Elastic: sensitive/responsive to change in price (greater than 1) Inelastic: not sensitive or not responsive to a change in price (less than 1) Unitary: Equal change in price to equal change in supply (= to 1)  Elasticity of Supply A supplier’s responsiveness to a price change is called _________________ Elasticity of Supply » (think like a supplier/seller) 3 Factors that will determine a product’s elasticity 1. Availability of resources required to make the product 2. Amount of time required to make the product 3. Skill level of the worker needed to make the product  Elastic Supply A product has elastic supply when a price change causes a significant change in the quantity supplied. 1. Abundance of resources required to make the product 2. Product can be made quickly 3. Low skill level of workers required  Slope of an Elastic supply curve Remember, if the price changes, the quantity supplied changes a lot. This creates a flatter curve. P P2 s P1 Q1 Q2 Q  Inelastic Supply A price change causes very little change in the quantity supplied. This happens because… The product requires scarce resources It takes a long time to make It requires a high skill level of workers – Hand crafted furniture – diamonds  Slope of an inelastic supply curve If the market price goes up but the supplier cannot increase production very much, then this creates a steeper curve. s P P2 P1 Q1 Q2 Q  Factors that Shift Supply Resource Prices Technology Prices of Related Goods and Services And Productivity Supply Number Expectations Of Of Producers Producers Price of Inputs (Resource Prices) When costs go up, profits go down, so that the incentive to supply also goes down.  Technology Advances in technology reduce the number of inputs needed to produce a given supply of goods. Costs go down, profits go up, leading to increased supply.  Expectations If suppliers expect prices to rise in the future, they may store today's supply to reap higher profits later.  Number of Suppliers As more people decide to supply a good the market supply increases (Rightward Shift).  Price of Related Goods or Services The opportunity cost of producing and selling any good is the foregone opportunity to produce another good. If the price of alternate good changes then the opportunity cost of producing changes too!  Taxes and Subsidies When taxes go up, costs go up, and profits go down, leading suppliers to reduce output. When government subsidies go up, costs go down, and profits go up, leading suppliers to increase output. Decrease in Supply Increase in Supply  Change in Supply vs. a Change in the Quantity Supplied  Concept of Engineering Economics Engineering is the application of science Price has a major role in deciding the demand and supply of a product. Hence for a organization, efficient and effective functioning would certainly help it to provide goods/ services at a lower cost which in turn will enable it to fix a lower price for its goods or services. – evolution of Engineering Economics  Concept of Engineering Economics Engineering Economics deals with the methods that enable one to take economic decisions towards minimizing costs and /or maximizing benefits to business organizations.  Scope of Engineering Economics Interest formulae Bases of comparing alternatives – Present worth method – Future worth method – Annual equivalent method – Rate of return method Replacement analysis Depreciation Evaluation of public alternatives Inflation adjusted investment decisions Make or buy decisions Inventory control Project management Value Engineering Linear Programming  Applications of Engineering Economics Selection of location and site for a new plant Production planning and control Selection of an equipment and their replacement analysis Selection of a material handling systems Determination of plant capacity Determination of wage structure of the workers It can be applied by a major corporation to analyse plans for a new manufacturing facility or a new research and development centre.  Advantages of Engineering Economics Better decision making on the part of Engineers Efficient use of resources result in better output and economic advancement. Cost of production can be reduced. Alternative courses of action using economic principles may result in reduction of prices of goods and services. Elimination of waste can result in application of engineering economics More capital will be made available for investment and growth Improves the standard of living with the result of better products, more wages and salaries, more output, etc from the firm applying engineering economics  Types of Efficiency Technical Efficiency Economic Efficiency Technical Efficiency is the ratio of the output to input of a physical system. The physical system may be diesel engine, a machine working in a shop floor, a furnace, etc. Economic efficiency is the ratio of output to input of a business system or ratio of worth to cost of a business firm.  Types of Efficiency 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. Several ways to improve economic efficiency Increased output for the same input Decreased input for the same output. By a proportionate increase in the output which is more than the proportionate increase in the input. 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.  Elements of Cost Classification of Cost: 1. Variable Cost Variable cost varies with the production volume. Classification of Variable Cost: 1. Direct Material Cost: cost of materials that are used to produce the product. 2. Direct Labour Cost: amount of wages paid to the labour, who involved in the production activities. 3. Direct Expenses: expenses that vary in relation to the production volume, other than the direct material and direct labour costs.  Elements of Cost Classification of Cost: 2. Overhead Cost Overhead cost is fixed, irrespective of production volume. Classification of Overhead Cost: 1. Factory Overhead: indirect expenses incurred in factory right from work-order receipt till goods-despatch. 2. Administrative Overhead: all the costs incurred in administering the business. 3. Selling Overhead: total expenses incurred in promotional activities & expenses relating to sales force. 4. Distribution Overhead: total cost of shipping the items from factory site to customer site.  Selling Price of a product 1. Prime Cost = = DMC + DLC + DE 2. Factory cost = Prime cost + Factory Overhead 3. Production Cost = Factory cost + Administrative overhead 4. Cost of goods sold = Production Cost + Opening stock – Closing stock 5. Cost of Sales = Cost of goods sold + Selling and distribution overhead 6. Selling Price = Cost of sales + Profit 7. Selling price per unit = Sales ÷ Quantity sold  Marginal Cost  It 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. Let the cost of producing 21 units of the same product be Rs. 10,045. Then the marginal cost of producing the 21st unit is Rs. 45.  Marginal Revenue  It is the incremental revenue of selling an additional unit of a product.  Let, the revenue of selling 20 units of a product be Rs. 15,000. Let the revenue of selling 21 units of the same product be Rs. 15,085. Then, the marginal revenue of selling the 21st unit is Rs. 85.  Sunk Cost  Past cost of an equipment / asset.  An equipment was purchased for Rs. 1,00,000 about 3 years back.  If it is considered for replacement, then its present value is not Rs. 1,00,000.  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 now onwards.  Opportunity Cost A set of alternatives (X &Y) is available for investment. Let us assume an alternative (X) is selected and you get return from it. If the same money is invested in alternative (Y), it may fetch some more return. Since the money is invested in the selected alternative (X), one has to forego the return from the other alternative (Y).  Opportunity Cost The amount that is foregone by not investing in the other alternative (Y) is known as the 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.  Opportunity Cost - Example Consider You invested a sum of Rs. 50,000 in shares. Let the expected annual return be Rs. 7,500. If the same amount is invested in a fixed deposit, a bank will pay a return of 18%. Then, the corresponding total return per year is Rs. 9,000. Return from Bank is greater than the return from shares. The foregone excess return of Rs. 1,500 by way of not investing in the bank is the opportunity cost of investing in shares.  Break-even Point Break Even Point: – Total sales revenue equals Total expenses – Point at which no profit, no loss occurs. In Graphical representation, the intersection point of Total sales revenue line & Total cost line Breakeven Chart  Break-even Point If the production qty < the break-even qty: Total expense is more than total revenue. Hence, the firm will be making LOSS. If the prodn. qty is more than the break-even qty: Total revenue is more than total expenses. Hence, the firm will be making PROFIT.  Break-even Analysis Let s = selling price per unit v = variable cost per unit FC = fixed cost per period Q = volume of production 1. Total sales revenue (S) = sxQ 2. Total cost (TC) = Total variable cost + Fixed Cost = (v x Q) + FC 3. Profit = Total Sales Revenue – Total Cost = s x Q – (FC + v x Q) Fixed cost (FC) 4. Break-even quantity (BEQ) = ----------------------------------------------------- Selling price/unit (s) − Variable cost/unit (v) 5. Break-even sales = BEQ x Selling price/unit 6. Contribution = Sales – Variable costs 7. Margin of Safety = Actual sales – Break-even sales Profit = ------------------------------- × sales Contribution  Break Even Analysis: Problem 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.  Break Even Analysis: Solution for Q1 FC 20,000,00 (a) Break-even quantity = ---------- = ----------------- s –v 200 – 100 = 20,00,000/100 = 20,000 units FC 20,000,00 (b) Break-even sales = -------- x s = ---------------- x Rs.200 s –v 200 – 100 = 40,000,000  Break Even Analysis: Solution for Q1 (c) (i) Contribution = Sales – Variable cost = (s x Q) – (v x Q) = (200 x 60,000) – (100 x 60,000) = 1,20,00,000 – 60,00,000 = Rs. 60,00,000 (c) (ii) Margin of = Sales – Break-even sales Safety = (200 x 60,000) – (200 x 20,000) = 1,20,00,000 – 40,00,000 = Rs. 80,00,000  Contribution The contribution margin is a concept used with breakeven point or in breakeven analysis. The contribution margin is the amount of money a company has to cover its fixed costs after it pays all of its variable expenses. It is also the amount, after covering fixed costs, that contributes to the net operating profit or net operating loss of the business firm  Contribution Contribution Margin = Sales Revenue – Variable expenses On a per unit basis, contribution is calculated as: Contribution Margin per unit of sales = Sales Revenue per unit – Variable expenses per unit Contribution Margin – Fixed Costs = Net Profit or Loss. Contribution - Example  Contribution Break even point = 50,000 units The contribution margin in this case is Rs. 1.20 per unit. The company has to produce and sell 50,000 units of their products in order to cover their total expenses, fixed and variable. At this level of sales, they will make no profit but will just break even with a contribution margin towards the fixed costs of Rs. 1.20 per unit sold or Rs. 60,000  Profit / Volume (P/V) Ratio It is the ratio of contribution to sales. Contribution (Sales - Variable costs) P/V ratio = --------------------- = --------------------------------- Sales Sales Contribution is the difference between sales revenue and variable cost i.e. (Sales Revenue – Variable Cost). A high P/V Ratio: indicates that higher profits. A low P/V ratio: indicates that low profitability.  Calculating Profit-Volume Ratio 1. look up the price of the product. 2. Calculate the cost involved to produce a single unit of the product. 3. Subtract that number from the initial price number, and this is profit. 4. Divide that profit figure by the price point, and this is profit-volume ratio. 5. Multiply the result by 100 for a percentage. Contribution (Sales - Variable costs) P/V ratio = --------------------- = --------------------------------- Sales Sales  Improving Profit-Volume Ratio By the following ways, an improvement in P/V ratio can be achieved: 1. The selling price increase; but the risk that the volume of sales might be affected. 2. By purchasing the latest machinery, a reduction in the variable cost per unit can be achieved, thereby cutting the production hours. 3. By concentrating on the products by which highest contribution can be achieved. For doing business analysis, in the hands of management, the P/V ratio is an invaluable tool.  Advantages of P/V Ratio 1. This ratio determines profitability of a line of product & also overall profitability. 2. This ratio compares the profitability of different lines of products, sales, companies, factories etc. 3. This ratio calculates break-even sales, profit at different levels of output, turnover which may be required for a desired profit or to offset reduction in price or to meet increased expenditure.  Problem No. 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 for Problem No. 2 (a) Contribution = Sales – Variable costs = Rs. 1,20,000 – Rs. 45,000 = Rs. 75,000 (b) Profit = Contribution – Fixed cost = Rs. 75,000 – Rs. 25,000 = Rs. 50,000 Solution for Problem No. 2  Problem No. 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 for Problem No. 3 (a) Contribution = Sales – Variable costs = Rs. 80,000 – Rs. 35,000 = Rs. 45,000 (b) Profit = Contribution – Fixed cost = Rs. 45,000 – Rs. 15,000 = Rs. 30,000 Solution for Problem No. 3  Problem No. 4 A company produces a single article. About its product, the following cost data has been given: Selling price per unit Rs. 40 Marginal cost per unit Rs. 24 Fixed cost per annum Rs. 1600 Calculate: (a) P/V ratio, (b) Break-even sales, (c) Sales to earn a profit of Rs. 200, (d) Profit at sales of Rs. 12000, (e) If sales price is reduced by 10%, then a new break- even sales.  Solution for Problem No. 4 We know that Sales – Variable cost = Fixed cost + Profit By multiplying & dividing left hand side by Sales, Sales x (Sales –Variable Cost) -------------------------------------------- = Fixed cost + Profit Sales i.e Sales x P/V ratio = Contribution (a) P/V ratio = Contribution / Sales * 100 = [(40-24)/40] * 100 = 16/40 * 100 = 40%  Solution for Problem No. 4 (b) Fixed cost Break even Sales = ----------------- P/V ratio Sales = 1600/40 = Rs. 4000 (c) Sales to earn a profit of Rs. 200: Sales * P/V ratio = Fixed cost + Profit Sales * 40% = 1600 + 200 Sales = 1800 / 40% Sales = Rs. 4500  Solution for Problem No. 4 (d) Profit at sales of Rs. 12000: Sales * P/V ratio = Fixed cost + Profit 12000 * 40% = 1600 + Profit Profit = Rs. 3200 (e) New break-even sales, if sales price is reduced by 10%: New Sales price = Rs. 40 – Rs. 4 = Rs. 36 Marginal cost = Rs. 24 Contribution = Rs. 36 – Rs. 24 = Rs. 12 P/V ratio = Contribution / Sales = (12/36) *100 = 33.33% B.E.S * P/V ratio = Fixed Cost (at B.E.P, contribution is equal to fixed cost) Or, B.E.S = 1600/33.33% Or, B.E.S = Rs. 4800  Elementary Economic Analysis Introduction: economic decision making involved in day-to- day events. Example: purchasing of raw materials from a nearby or far-off place. Factors affect from a nearby place from far-off place taking decision Price more costly Less cost Transportation cost Minimum Very high not sufficient enough to support Abundant; can support Availability the operation throughout the throughout the year year Does not require pre- requires pre-processing before processing before it is Quality it is used in the production used in the production process process  Elementary Economic Analysis The procurement of the raw material should be decided in such a way that the overall cost is minimized.  Elementary Economic Analysis EXAMPLES FOR SIMPLE ECONOMIC ANALYSIS: 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  Process planning  Material selection for a Product Among various elements of cost, raw material cost is most significant and it forms a major portion of the total cost of any product. Objective :- To find a suitable raw material that 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 Note If the new raw material provides any additional benefit, then it should be treated as its welcoming feature.  Material selection for a Product 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.  Material selection for a Product 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?  Material selection for a Product (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  Material selection for a Product 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/-  Material selection for a Product (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  Material selection for a Product 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  Material selection for a Product: Problem 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: ----------------------------------------------------------------------------------------------- Description of item Quantity Cost ----------------------------------------------------------------------------------------------- Wood for frame and legs 0.1 m3 Rs. 12,000/m3 Table top with sunmica finish 1 Rs. 3,000 Leg bushes 4 Rs. 10/bush Nails 100 g Rs. 300/kg Total labour 15 hr Rs. 50/hr 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. 132  Material selection for a Product: Problem 2 The materials and labour requirements along with cost details to manufacture a table with granite top are given below: ----------------------------------------------------------------------------------------------- Description of item Quantity Cost ----------------------------------------------------------------------------------------------- Wood for frame and legs 0.15 m3 Rs. 12,000/m3 Granite table top 1.62 m2 Rs. 800/m2 Leg bushes 4 Rs. 25/bush Nails 50 g Rs. 300/kg Total labour 8 hr Rs. 50/hr ----------------------------------------------------------------------------------------------- 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. 133  Material selection for a Product: Problem 2 (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. 5,020 134  Material selection for a Product: Problem 2 (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 x 1.62 = Rs. 1,296 Cost of bushes = 25 x 4 = Rs. 100 Cost of nails = 300 x (50/1,000) = Rs. 15 Cost of labour = 50 x 8 = Rs. 400 Total = Rs. 3,611 135  Material selection for a Product: Problem 2 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. 1,409 136  Design selection for a Product The design modification of a product may result in reduced raw material requirements, increased machinability of the materials, reduced labour, etc. Design is an important factor. It decides the cost of the product. Economic analysis applied to the selection of design for a product. 137  Design selection for a Product – Problem 1 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. 1. Design A will require 16 hours of lathe time and 4.5 hours of grinder time per 1,000 units. 2. Design B will require 7 hours of lathe time and 12 hours of grinder time per 1,000 units. 3. The operating cost of the lathe including labour is Rs. 200 per hour. 4. 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? 138  Design selection for a Product – Problem 1 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 = 16 200 + 4.5 x 150 = Rs. 3,875 Total cost of design A/1,00,000 units = 3,875 x1,00,000/1,000 = Rs. 3,87,500 139  Design selection for a Product – Problem 1 (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 x150 = Rs. 3,200 Total cost of design B/1,00,000 units = 3,200 x 1,00,000/1,000 = Rs. 3,20,000 140  Design selection for a Product – Problem 1 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. 141  Design selection for a process industry 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?” 142  Design selection for a process industry The original design submitted called for 4 tanks 5.2 m in diameter and 7 m in height. From a graph of the article, the engineer found that the present ratio of height to diameter of 1.35 is 111% of the minimum cost and that the minimum cost for a tank was when the ratio of height to diameter was 4 : 1. The cost for the tank design as originally submitted was estimated to be Rs. 9,00,000. 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? 143  Design selection for a process industry (a) Original design Number of tanks = 4 Diameter of the tank = 5.2 m Radius of the tank = 2.6 m Height of the tank = 7 m Ratio of height to diameter = 7/5.2 = 1.35 Volume/tank = (22/7)r2h = (22/7)(2.6)2 * 7 = 148.72 m3 144  Design selection for a process industry (a) New design Cost of the old design = 111% of the cost of the new design (optimal design) Optimal ratio of the height to diameter = 4:1 h:d=4:1 4d = h d = h/4 r = h/8 145 Design selection for a process industry 146  Design selection for a process industry (b)New design Therefore, Diameter of the new design = 1.81* 2 = 3.62 m Cost of the new design = 9,00,000 (100/111) = Rs. 8,10,810.81 Expected savings by the redesign = Rs. 9,00,000 – Rs. 8,10,810.81 = Rs. 89,189.19 147  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.  Process Planning /Process Modification The objective of process planning/process modification is to identify the most economical sequence of operations to produce a component.  Process Planning /Process Modification The steps in process planning are as follows: 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.  Example for Process planning The process planning engineer of a firm listed the sequences of operations as shown in Table. 1 to produce a component. Sequence Process Sequence 1 Turning – Milling – Shaping - Drilling 2 Turning – Milling - Drilling 3 All operations are performed with CNC machine  Example for Process planning The details of processing times of the component for various operations and their machine hour rates are summarized in Table 2. Find the most economical sequence of operations to manufacture the component Operation Machine Process sequence hour rate 1 2 3 (Rs.) Turning 200 5 5 - Milling 400 8 14 - Shaping 350 10 - - Drilling 300 3 3 - CNC 1000 - - 8 operations  Example for Process planning (a) Cost of component using process sequence 1. The process sequence 1 of the component is as follows Turning – Milling – Shaping – Drilling Operation Operation Time Machine Cost No Min hr Hour Rs. Rate Rs. 1 Turning 5 0.083 200 16.60 2 Milling 8 0.133 400 53.20 3 Shaping 10 0.167 350 58.45 4 Drilling 3 0.050 300 15.00 Total 143.25  Example for Process planning (b) Cost of component using process sequence 2. The process sequence 2 of the component is as follows Turning – Milling – Drilling Operation Operation Time Machine Cost No Min hr Hour Rs. Rate Rs. 1 Turning 5 0.083 200 16.60 2 Milling 14 0.233 400 93.20 3 Drilling 3 0.050 300 15.00 Total 124.80  Example for Process planning (b) Cost of component using process sequence 3. The process sequence 2 of the component is as follows Only CNC operations Operation Operation Time Machine Cost No Min hr Hour Rs. Rate Rs. 1 CNC 8 0.133 1,000 133 Operations The Process Sequence 2 has the least cost. Therefore it should be selected for manufacturing the component.  UNIT II Interest Formulae  Interest Rate Interest rate is the time value of money It represents the growth of capital per unit period The period may be a month, a quarter, semi annual or a year.  Interest Rate – Types Simple Interest Compound Interest  Simple Interest Simple interest is defined as a fixed percentage of the principal multiplied by the life the load. I=nxixP I = Total amount of simple interest n = Life of the load i = interest rate P = Principal  Simple Interest Rate Fund accumulated at the end of First year = P + iP = P (1+i) Fund accumulated at the end of second year = P + iP + iP = P + 2iP = P(1+2i) IIIly Fund accumulated at the end if nth year = = P(1+ni)  Compound Interest Rate When the interest rate is compounded, the total time period is divided into several interest periods. Interest is credited at the end of each period and is allowed to accumulate from one interest period to the next.  Compound Interest Rate During a given interest period, the current interest is determined as the percentage of the total amount owned (i.e the principle plus the previously accumulated interest.  Compound Interest Rate For the first period, the interest is determined as I1 = iP Fund accumulated at the end of First year F1 = P + I1 = P+iP = P (1+i) For the second period, the interest is determined as I2 = iF1 Fund accumulated at the end of second year F2 = P + I1 + I2 = P + iP + i P (1+i) = P + iP + iP + i2 P = P( 1 + 2i + i2) = P (1 + i)2 IIIly Fund accumulated at the end if nth year = = P(1+i)n  Time value of money 1. Single Payment Compound Amount 2. Single Payment Present Worth Amount 3. Equal Payment Series Compound Amount 4. Equal Payment Series Sinking Fund 5. Equal Payment Series Present Worth Amount 6. Equal Payment Series Capital Recovery Amount 7. Uniform Gradient Series Annual Equivalent Amount  Single payment compound amount 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.  Single payment compound amount The cash flow diagram is: F i=% 0 1 2 3 n P P is known, F has to be determined n 𝐹 F=P 1 + i = 𝑃( , 𝑖, 𝑛) 𝑃 𝐹 Here , 𝑖, 𝑛 is called compound amount 𝑃 factor  Single Payment Present Worth Amount Objective is to find the present worth(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  Single Payment Present Worth The cash flow diagram is: F i=% 0 1 2 3 n P F is known, P has to be determined 𝐹 𝑃 Present worth P= = 𝐹( , 𝑖, 𝑛) (1+𝑖)𝑛 𝐹 𝑃 Here ( , 𝑖, 𝑛) is called single payment 𝐹 present worth factor  Equal Payment Series Compound Amount Objective is to find the future worth of ‘n’ equal payments which are made at the end of every interval period till the end of 𝑛𝑡ℎ Interest period at an interest rate of ‘i’ compounded at the end of each interest period.  Equal Payment Series Compound Amount The cash flow diagram is: F i=% 0 1 2 3 n A A A A A is known, F has to be determined 𝐹 Here ( , 𝑖, 𝑛) is called Equal Payment 𝐴 Series Compound Amount Factor  Equal Payment Series Compound Amount The cash flow diagram is: F i=% 0 1 2 3 n A A A A A = Equal Amount Deposited at the end of each interest period n = Number of interest periods i = Rate of interest F = Single Future Amount  Equal Payment Series Compound Amount The cash flow diagram is: F i=% 3 n 1 2 0 A A A A 2 Future amount F = A + A(1+i) +A 1 + 𝑖 +.…+A(1 +  Equal Payment Series Compound Amount Subtract ❶ from ❷, 𝐹(1+𝑖) 𝐹 - = (1+ i)n – 1 𝐴 𝐴 𝐹 𝐹 𝐹 + *i- = (1+ i)n – 1 𝐴 𝐴 𝐴 𝐹 (1+ i)n – 1 = 𝐴 𝑖 A ∗( (1+ i)n – 1) F = 𝑖 𝐹 A( , 𝑖, 𝑛) is termed as equal payment series, 𝐴 compound amount factor  Equal Payment Series Sinking Fund 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 on nth interest period at an interest rate of ‘i’ Equal Payment Series Sinking Fund The cash flow diagram is: F i=% 0 1 2 3 n A A A A F is known, A has to be determined 𝐴 Here ( , 𝑖, 𝑛) is called Equal Payment 𝐹 Series Sinking Fund Factor i 𝐴 A = Fx = F ( , 𝑖, 𝑛) 1+𝑖 −1 𝑛 𝐹 Equal Payment Series Present Worth Amount Objective 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. Equal Payment Series Present Worth Amount The cash flow diagram is: P i=% 0 1 2 3 n A A A A A = Equal Amount Deposited at the end of each interest period n = Number of interest periods i = Rate of interest P = Present Worth Equal Payment Series Present Worth Amount 0 1 2 3 4 n A A A A A P 𝐴 𝐴 𝐴 𝐴 P= + + + ……. + (1+𝑖) (1+𝑖)2 (1+𝑖)3 (1+𝑖)𝑛 1 1 1 1 P = A[ + + + ……. + ]–① (1+𝑖) (1+𝑖)2 (1+𝑖)3 (1+𝑖)𝑛 1 Multiply ① by & subtract ① - ② (1+𝑖)  Equal Payment Series Present Worth 𝑃 1 1 Amount 1 1 = A[ + + ……. + ] (1+𝑖) (1+𝑖)2 (1+𝑖)3 (1+𝑖)4 (1+𝑖)𝑛+1 -② 1 1 1 1 P =A[ + + + ……. + ] (1+𝑖) (1+𝑖)2 (1+𝑖)3 (1+𝑖)𝑛 –① ②-① 𝑃−𝑃 −𝑖𝑃 1 1 =A[ - ] (1+𝑖) (1+𝑖)𝑛+1 (1+𝑖) 1 −𝑖𝑃 =A[ - 1 ] (1+𝑖)𝑛 (1+𝑖)𝑛 −1 𝑃=A[ ] 𝑖(1+𝑖)𝑛 Equal Payment Series Present Worth Amount 𝑃 𝑃 𝐹 1 = x = 𝑛 x 𝐴 𝐹 𝐴 1+𝑖 1+𝑖 𝑛 −1 𝑖 1+𝑖 𝑛 −1 = 𝑖 1+𝑖 𝑛 1+𝑖 𝑛 −1 𝑃 P= Ax = A( , 𝑖, 𝑛) 𝑖 1+𝑖 𝑛 𝐴 𝑃 Here ( , 𝑖, 𝑛) is called equal payment 𝐴 series present worth amount.  Equal Payment Series Capital Recovery Amount Objective 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 year.  Equal Payment Series Capital Recovery Amount The cash flow diagram is: P i=% 0 1 2 3 n A A A A A = Equal Amount Deposited at the end of each interest period n = Number of interest periods i = Rate of interest P = Present Worth / Loan sanctioned at initial period  Equal Payment Series Capital Recovery Amount The cash flow diagram is: P i=% 0 1 2 3 n A A A A Here P is given and the objective is to determine A We know that from Equal Payment Series 1+𝑖 𝑛 −1 Present Worth Amount P = A x 𝑛 𝑖 1+𝑖  Equal Payment Series Capital Recovery Amount The cash flow diagram is: P i=% 0 1 2 3 n A A A A 𝑃[𝑖 1+𝑖 𝑛 ] 𝐴 A= 𝑛 −1 = P( , 𝑖, 𝑛) 1+𝑖 𝑃 𝐴 Here ( , 𝑖, 𝑛)is called equal payment series 𝑃 capital recovery factor.  Uniform Gradient Series Annual Equivalent Amount Objective is to find the annual equivalent amount of a series with an amount A at the end of the final year and with an equal increment (G) at the end of each of the following ‘n-1’ years with an interest rate of ‘i’ compounded annually.  Uniform Gradient Series Annual Equivalent Amount Cash Flow Diagram 0 i= 1 2 3 4 n % A1 A1 + G A1 + 2G A1 + 3G A1 + (n- 1)G  Uniform Gradient Series- Present Worth Amount 0 1 2 3 4 n G 2G 3G (n-1)G 𝑃 𝑃 𝑃 P=G( , 𝑖,2 ) + 2G ( , 𝑖, 3 ) + 3G ( , 𝑖, 4) 𝐹 𝐹 𝐹 +………..… 𝑃 𝑃 + G(n-2) ( , 𝑖,n-1 +G(n-1) ( , 𝑖,n ) 𝐹 𝐹 P= G [ 1 (1+𝑖)2 + 2 (1+𝑖)3 + 3 (1+𝑖)4 ……. + 𝑛−2 (1+𝑖)𝑛−1 + 𝑛−1 (1+𝑖)𝑛 ]- ① Uniform Gradient Series- Present Worth Amount Multiply ① x ( i + 1) & subtract ① from ② 1 2 3 𝑛−2 P(i +1 )=G[ + + + ……. + (1+𝑖)1 (1+𝑖)2 (1+𝑖)3 (1+𝑖)𝑛−2 𝑛−1 + (1+𝑖)𝑛−1 ] -② 1 2 3 𝑛−2 𝑛−1 P=G [ + ++ ……. ] -- + (1+𝑖)2 (1+𝑖)3 (1+𝑖)4 (1+𝑖)𝑛−1 (1+𝑖)𝑛 ① _____________________________________________ _ 1 1 1 1 𝑛−1 iP = G [ (1+𝑖)1 + (1+𝑖)2 + (1+𝑖)3 ……. + - (1+𝑖)𝑛−1 (1+𝑖)𝑛 ] 1 1 1 1 𝑛 iP = G [ (1+𝑖)1 + (1+𝑖)2 + (1+𝑖)3 ……. + (1+𝑖)𝑛 ] – G [ (1+𝑖)𝑛 ] Uniform Gradient Series- Present Worth Amount Refer Equal Payment Series – Present worth Amount (1+𝑖)𝑛 −1 𝑛 iP = G [ ]–G[ ] 𝑖(1+𝑖)𝑛 (1+𝑖)𝑛 G (1+𝑖)𝑛 −1 𝑛 3 P= [ - ] ………… 𝑖 𝑖(1+𝑖)𝑛 (1+𝑖)𝑛 Eqn. 3 is the general relation to convert an arithmetic gradient ‘G’ for ‘n’ years into a present worth at year ‘0’ Uniform Gradient Series- Present Worth Amount The conversion of cash flow diagram 0 1 2 3 4 n G 2G 3G (n-1)G P 1 2 3 4 n- n 1  Uniform Gradient Series- Annual equal Amount The equivalent Uniform annual series (A value) for an arithmetic gradient ‘g’ is found by multiplying the present worth in eqn. 3 by the (A/P) factor expression Ag A P = x G P G Ag 1 (1+𝑖)𝑛 −1 𝑛 𝑖(1+𝑖)𝑛 G = [ 𝑖 𝑖(1+𝑖)𝑛 - (1+𝑖)𝑛 ] x [(1+𝑖)𝑛−1 ] (1+i)n (1+i)n (1+i)n n(1+i)n Ag = G [ - - ] i(1+i)n ( 1+i n −1) i(1+i)n ( 1+i n −1) (1+i)n ( 1+i n −1)  Uniform Gradient Series- Annual equal Amount (1+𝑖)𝑛 1 𝑛 Ag = G [ - - ] 𝑖( 1+𝑖 𝑛 −1) 𝑖( 1+𝑖 𝑛 −1) ( 1+𝑖 𝑛 −1) (1+𝑖)𝑛 −1 𝑛 Ag = G [ - ] 𝑖( 1+𝑖 𝑛 −1) ( 1+𝑖 𝑛 −1) 1 𝑛 Ag = G [ - ] 𝑖 1+𝑖 𝑛 −1 1+𝑖 𝑛 −𝑖𝑛−1 Ag = G [ ] 𝑖 1+𝑖 𝑛 −1 Uniform Gradient Series- Annual equal Amount 𝐺 1+𝑖 𝑛 −in−1 A=A+ 𝑖 1+𝑖 𝑛 −1 𝐴𝑔 = A + AG( , 𝑖, 𝑛) 𝐺 𝐴𝑔 Here ( , 𝑖, 𝑛) is called uniform gradient 𝐺 series factor.  Uniform Gradient Series NOTE: The total present worth 𝑃𝑇 for a gradient series must consider the base and the gradient separately. Thus, for cash flow series involving conventional gradient: 1. The base amount is the uniform-series amount ‘A’ that begins in year 1 and extends through year ‘n’. It’s present worth is represented by 𝑃𝐴  Uniform Gradient Series 2. For an increasing gradient, the gradient amount must be added to the uniform series amount. The present worth is 𝑃𝐺. 3. For a decreasing gradient, the gradient amount must be subtracted from the uniform series amount. The present worth is 𝑃𝐺. The general equations for calculating total present worth 𝑃𝑇 of conventional arithmetic gradients are: 1. 𝑃𝑇 = 𝑃𝐴 + 𝑃𝐺 2. 𝑃𝑇 = 𝑃𝐴 - 𝑃𝐺  Problem -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. Given Data: P = Rs. 20,000 i = 18% compounded annually n = 10 years  Solution for problem - 1 The cash flow diagram is: F i = 18% 0 1 2 3 1 0 P= 20000 n 18 10 F= P 1+i = 20000 ∗ 1 + 100 = 20000 * 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.  Problem -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. Given Data: F = Rs. 1,00,000 i = 15% compounded annually n = 10 years  Solution for problem - 2 The cash flow diagram is: F= 1,00,000 i= 0 1 2 15% 3 1 0 P 𝐹 1 Present worth P= = 1,00,000 ∗ (1+𝑖)𝑛 1+0.15 10 = 1,00,000 * 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.  Problem -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 for the 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. Given Data A = Rs. 10,000 n = 25 years i = 20% F=?  Solution for Problem - 3 The cash flow diagram is: i= F 0 1 20% 2 3 25 A= A= A= A= 10000 10000 10000 10000 A ∗ ((1+ i)n – 1) F = 𝑖 10,000 ∗ ( 1+0.20 25 −1) = 0.20 = 10, 000 * 471.981 = Rs. 47, 19, 810/- The future sum of the annual equal payments after 25 years is equal toRs. 47,19,810.  Problem -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. Given Data F = Rs. 5,00,000 n = 15 years i = 18% A=?  Solution for Problem - 4 5,00,00 The cash flow diagram is: 0 i = 18% 0 1 2 3 1 5 A A A A i A = Fx 1+𝑖 𝑛 −1 0.18 = 5,00,000* ( 15 ) 1+0.18 −1 = 5,00,000 * 0.0164 = Rs. 8,200/- The annual equal amount which must be deposited for 15 years is Rs. 8,200.  Problem -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. Given Data A = Rs. 10,00,000 n = 20 years i = 15% P=?  Solution for Problem - 5 The cash flow diagram is: P i = 15% 0 1 2 3 2 0 10,00,00 10,00,00 10,00,00 10,00,00 0 01+𝑖 𝑛 −1 0 0 P = Ax 𝑖 1+𝑖 𝑛 1+0.15 20 −1 = 10,00,000* ( 20 ) 0.15 ∗ 1+0.15 = 10,00,000 * 6.2593 = Rs. 6 59, 300/- The amount of reserve which must be set-up now is equal to Rs. 62,59,300.  Problem -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. Given Data P = Rs. 10,00,000 n = 15 years i = 18% A=?  Solution for Problem - 6 The cash flow diagram is: 10,00,00 i = 18% 0 1 1 0 2 3 5 A A A A 𝑃[𝑖 1+𝑖 𝑛 ] A = 1+𝑖 𝑛 −1 0.18 ∗ 1+0.18 15 = 10,00,000* ( 15 ) 1+0.18 −1 = 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.  Problem -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. Given Data A = Rs. 4,000 n = 10 years i = 15% G = Rs. 500? F= ?  Solution for Problem - 7 Cash Flow Diagram 0 i= 1 2 3 4 1 15% 0 400 0 4000+50 0 4000+100 0 4000+150 0 4000+450 0  Solution for Problem - 7 1+𝑖 𝑛 −1−𝑖𝑛 𝐴𝑔 = G[ ] 𝑖 1+𝑖 𝑛 −𝑖 1+0.15 10 −1−0.15 𝑋 10 = 500 [ 10 ] 0.15 1+0.15 −0.15 1.5456 = 500 [ ] 0.4568 = 500 x 3.3835 𝐴𝑔 = Rs.1,691.75  Solution for Problem - 7 A = A1 + A G = 4,000 + 1,691.75 = Rs.5,691.75 This is equivalent to paying an equivalent amount of Rs.5,691.75 at the end of every year for the next 10 years.  Solution for Problem - 7 The future worth sum of this revised series at the end of the 10th year is: 𝐴( 1+𝑖 𝑛 −1) F = 𝑖 5695.75( 1+0.15 10 −1) = 0.15 = Rs.1,15,563.68 At the end of the 10th year, the compound amount of all his payments will be Rs. 1,15,563.68.  Problem -8 An engineer is planning for a 15 year retirement. In order to supplement his pension and offset the anticipated effects of inflation, he intends to withdraw Rs.5,000 at the end of the first year and to increase the withdrawal by Rs.1,000 at the end of each successive year. How much money must he have in his savings account at the start of his retirement, if money earns 6% per year, compounded annually.  Solution for Problem - 8 Given Data n = 15years A1 = Rs.5,000 G = Rs.1,000 i = 6/100 = 0.06 To find: A=? P=?  Solution for Problem - 8 1+𝑖 𝑛 −1−𝑖𝑛 𝐴𝑔 = G[ 𝑛 ] 𝑖 1+𝑖 −𝑖 1+0.06 15 −1−0.06 𝑋 15 𝐴𝑔 = 1000 [ ] 0.06 1+0.06 15 −0.06 0.4 9 6 5 5 8 2 𝐴𝑔 = 1000 [ ] 0.0837935 𝐴𝑔 = Rs.5,926 A = 5,000+ 5,926 𝐴 = Rs.10,926  Solution for Problem - 8 The money that he must have in his savings account at the start of his retirement 1+𝑖 𝑛 − 1 P = A[ 𝑛 ] 𝑖 1+𝑖 1+0.06 15 − 1 P = 10926 [ 15 ] 0.06 1+0.06 P = Rs.1,06,116.03  Problem -9 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 there afterwards he wishes to deposit the amount with an annual decrease 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 for Problem - 8 Given Data: A = Rs.8,500 G = -Rs.500 I = 15% n = 10 years To find: A = ? F = ?  Solution for Problem - 9 Cash Flow Diagram 0 i= 1 2 3 4 1 15% 0 8500- 8500- 4500 8500- 1500 8500- 1000 500 850 0  Solution for Problem - 9 1+𝑖 𝑛 −𝑖𝑛−1 𝐴𝐺 = G[ 𝑛 ] 𝑖 1+𝑖 −𝑖 1+0.15 10 −0.15 𝑋 10 −1 𝐴𝐺 = 500 [ 10 ] 0.15 1+0.15 −0.15 𝐴𝐺 = 500 * 3.3832 = Rs.1,691.60 A = 8,500 - 1,691.60 A = Rs.6,808.40  Solution for Problem - 9 The future worth sum of this revised series at the end of the 10th year is : 1+𝑖 𝑛 − 1 F = A( ) 𝑖 1+0.15 10 − 1 F = 6808.4 ( ) 0.15 F = Rs.1,38, 235.84  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  Effective Interest Rate If the interest is compounded monthly, then the formula to compute the effective interest rate which is compounded annually is: 𝑖 𝑁 R = 1+ −1 𝐶 Where, i = nominal interest rate C = The number of interest periods in a year N = Total number of interest periods in a year  Problem - 10 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. Given Data: P = Rs. 5000 Nominal Interest rate per year =12% Interest rate is compounding quarterly Number of years = 10  Solution for Problem - 10 Number of interest periods per year = 4 Number of interest periods in 10 years=10x4=40 Hence, N=40 Interest rate per quarter= r= 12/4 = 3 % compounded quarterly 𝑛 40 F= P 1+𝑟 =5000 1 + 0.03 F = Rs.16,310.19  Solution for Problem - 10 Solution 2: Number on interest periods in a year, C = 4 𝑖 𝑛 Effective interest rate , R = 1 + −1 𝐶 12 4 R= 1+ −1 4 R =12.55% compounded annually 𝑛 F=P 1+𝑅 10 F = 5000 1 + 0.1255 F =Rs.16,308.91  Problem - 11 How much money must initially be deposited in a savings account paying 5% per year, compounded annually to provide for ten annual withdrawals that start at Rs.6000 and decreases by Rs.500 each year. Diagram  Solution for Problem - 11 1+𝑖 𝑛 −𝑖𝑛−1 𝐴𝐺 = G[ 𝑛 ] 𝑖 1+𝑖 −𝑖 1+0.05 10 −0.05 𝑋 10 −1 𝐴𝐺 = 500 [ ] 0.05 1+0.05 10 −0.05 𝐴𝐺 = Rs.2,049.50 A = 𝐴1 - 𝐴𝐺 = 6000 – 2049.5 A = Rs.3950.50  Solution for Problem - 11 1+𝑖 𝑛 − 1 Present worth = P = A [ 𝑛 ] 𝑖 1+𝑖 1+0.05 10 − 1 P = 3950.5[ 10 ] 0.05 1+0.05 P = Rs.30,504.19  Unit III Comparison of Alternatives  Bases for comparison of Alternatives In most of the practical decision environments , executives will be forced to select the best alternative from a set of competing alternatives.  Bases for Comparing the Worthiness of the Projects. 1.Present Worth Method 2.Future Worth Method 3.Annual Equivalent Method 4.Rate of Return Method  Present Worth Method of Comparison In this method the Cash flows of each alternative will be reduced to time Zero by assuming an interest rate i. Depending on the type of decision , the best alternative will be selected by comparing the present Worth amounts of the alternatives.  Present Worth Method of Comparison The sign of various amounts at different points in time in a cash flow diagram is to be decided based on the type of the decision problem. Cash dominated Cash flow diagram Revenue/Profit dominated Cash flow diagram  Present Worth Method of Comparison In a Cash dominated Cash flow diagram , the cost (out flows) will be assigned with positive sign and the profit , revenue , salvage value (all inflows) etc. will be assigned with negative sign.  Present Worth Method of Comparison In a Revenue/profit dominated Cash flow diagram , the profit , revenue , salvage value(all inflows to an organization) will be assigned with positive sign. The costs (outflow) will be assigned with negative sign.  Cost Dominated Cash flow diagram S 0 1 2 3 J n C1 C2 C3 Cj Cn P P represents the initial investment. C j the net Cost of operation and maintenance at the end of the Jth year. S is the salvage Value at the end of the nth year.  Cost Dominated Cash flow diagram S 0 1 2 3 J n C1 C2 C3 Cj Cn P To calculate the present worth amount of the above Cash flow diagram for a given interest i , we have 𝟏 𝟏 PW(i)= P + C1 ( ) + C2 ( 𝟐 ) + ….. + Cj 𝟏+𝒊 𝟏+𝒊 𝟏 𝟏 𝟏 ( 𝟏+𝒊 𝐣 ) + Cn ( 𝟏+𝒊 𝐧 ) -S( 𝟏+𝒊 𝐧 )  Cost Dominated Cash flow diagram If we have some more alternatives which are to be compared with this alternative , then the Corresponding present worth amount are to be computed and compared. Finally, the alternative with the minimum present worth amount should be selected as the best alternative.  Revenue Dominated Cash flow diagram S R1 R2 R3 R4 R5 Rj Rn 0 1 2 3 4 5 J n P P represents the initial investment. R j the net revenue at the end of the Jth year. The interest rate i is compounded annually. S is the salvage Value at the end of the nth year.  Revenue Dominated Cash flow 𝟏 diagram 𝟏 𝟐 𝟏 PW(i)= -P + R1 ( 𝟏+𝒊 ) + R2 ( 𝟏+𝒊 ) + ….. + Rn ( 𝟏+𝒊 ) + S 𝒏 𝟏 ( ) 𝟏+𝒊 𝒏 Expenditure is assigned a negative sign and revenue a positive sign. If we have some more alternatives which are to be compared with this alternative , then the Corresponding present worth amount are to be computed and compared. Finally, the alternative with the maximum present worth amount should be selected as the best alternative.  Problems related to present worth comparison ⑴ Alpha industry is planning to expand its production operation.It has identified three different technologies for meeting the goal. The initial outlay and annual revenues with respect to each of the technologies are summarized in table. Suggest the best technology which is to be implemented based on the present worth method of comparison assuming 20% interest rate compounded annually. Initial Outlay Annual Life(Years) (Rs) Income (Rs) Technology 1 12,00,000 4,00,000 10 Technology 2 20,00,000 6,00,000 10 Technology 3 18,00,000 4,00,000 10  Solution:4,00,000 4,00,000 4,00,000 i=20% 1 0 Technology-1 12,00,00 0 Present Worth for this technology is P PW(20%)1 = -12,00,000 + 4,00,000 X (A , 20%, 10) (1+i)n −1 = -12,00,000 + 4,00,000 X ( i(1+i)n ) (1+0.2)10 −1 = -12,00,000 + 4,00,000 X ( ) 0.2(1+0.2)10 = Rs 4,76,988/=  6,00,000 6,00,000 6,00,000 i=20% 1 Technology-2 0 20,00,000 Present Worth for this technology is P PW(20%)2 = -20,00,000 + 6,00,000 X ( , 20%, 10) A = -20,00,000 + 6,00,000 X (1+0.2)10 −1 ( ) 0.2(1+0.2)10 = Rs 5,15,483.3/=  5,00,000 5,00,000 5,00,000 Technology-3 i=20% 10 18,00,000 Present Worth for this technology is P PW(20%)3 = -18,00,000 + 6,00,000 X ( , 20%, 10) A (1+0.2)10 −1 = -18,00,000 + 5,00,000 X (0.2(1+0.2)10 ) = Rs 2,96,236/= It is clear that the present worth of technology 2 is the highest. Among all the technologies ∴ Technology 2 is suggested for implementation to expand production. ⑵ An Engineer