value engg 2451 eeca

A Course Material on

Engineering Economics & Cost Analysis

By

Mr. R.Siby., M.E

ASSISTANT PROFESSOR

DEPARTMENT OF MECHANICAL ENGINEERING

SASURIE COLLEGE OF ENGINEERING

VIJAYAMANGALAM – 638 056

QUALITY CERTIFICATE

This is to certify that the e-course material

Subject Code : MG 2451

Subject : Engineering Economics & Cost Analysis

Class : IV Year Mech

Being prepared by me and it meets the knowledge requirement of the university curriculum.

Signature of the Author

Name : R. SIBY., M.E

Designation : ASSISTANT PROFESSOR

This is to certify that the course material being prepared by Mr. R.Siby., M.E is of adequate quality. Hehas referred more than five books among them minimum one is from abroad author.

Signature of HD

Name: E.R. SIVAKUMAR., M.E

SEAL

CONTENTS

SL NO TOPICS PAGE NO

UNIT I INTRODUCTION TO ECONOMICS

1.1 Introduction to economics 1

1.2 Flow in an economy 1

1.3 Law of supply and demand 2

1.4 Concept of engineering economics 4

1.5 Engineering efficiency 4

1.6 Economic efficiency 5

1.7 Scope of engineering economics 5

1.8 Marginal cost 7

1.9 Marginal revenue 7

1.10 Sunk cost 7

1.11 Opportunity cost 8

1.12 Break-even analysis 8

1.13 P\V ratio 10

1.14 Elementary economic analysis 10

1.15 Material selection for product design selection for a product 11

1.16 Design selection for a product 12

1.17 Process planning 14

UNIT-II VALUE ENGINEERING

2.1 Make or buy decision 18

2.2 Value engineering 18

2.3 Function 25

2.4 Aims 26

2.5 Value engineering procedure 26

2.6 Interest formulae and their applications 27

2.7 Time value of money 27

2.8 Single payment compound amount factor 28

2.9 Single payment present worth factor 28

2.10 Equal payment series sinking fund factor 29

2.11 Equal payment series payment present worth factor 29

2.12 Equal payment series capital recovery factor 30

2.13 Uniform gradient series annual equivalent factor 30

2.14 Effective interest rate 31

CONTENTS

UNIT III CASH FLOW

SL NO TOPICS PAGE NO

3.1 Methods of comparison of alternatives 32

3.1.1 Present worth method (Revenue dominated cash flow diagram), 32

3.1.2.1 Future worth method (Revenue dominated cash flow diagram) 33

3.1.2.2 Future worth method (Cost dominated cash flow diagram) 34

3.1.3.1 Annual equivalent method (Revenue dominated cash flow diagram) 35

3.1.3.2 Annual equivalent method (Cost dominated cash flow diagram) 35

3.1.4 Rate of return method 36

UNIT IV REPLACEMENT AND MAINTENANCE ANALYSIS

4.1 Replacement and maintenance analysis 46

4.2 Types of maintenance 47

4.3 Types of replacement problem 48

4.4 Determination of economic life of an asset 48

4.5 Replacement of an asset with a new asset 52

4.6 Capital recovery with return and concept of challenger and defender 53

4.7 Simple probabilistic model for items which fail completely 55

UNIT V DEPRECIATION

5.1 Depreciation- introduction 59

5.2.1 Straight line method of depreciation 59

5.2.2 Declining balance method of depreciation 61

5.2.3 Sum of the years digits method of depreciation 64

5.2.4 Sinking fund method of depreciation 66

5.2.5 Service output method of depreciation 69

5.3 Evaluation of public alternatives 70

5.4 Inflation adjusted decisions 70

5.5 Procedure to adjust inflation 70

5.6 Comparison of alternatives and determination of economic life of asset 70

* * * * *

6 Tutorial problems and Worked out examples 71

7 Question Bank 84

8 Previous year University questions 92

TEXT BOOKS:1. Panneer Selvam, R, “Engineering Economics”, Prentice Hall of India Ltd, New Delhi, 2001.

MG2451 ENGINEERING ECONOMICS AND COST ANALYSIS L T P C3 0 0 3

OBJECTIVES:To learn about the basics of economics and cost analysis related to engineering so as to take economically sounddecisions.

UNIT I INTRODUCTION TO ECONOMICS 8Introduction 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.

UNIT II VALUE ENGINEERING 10Make or buy decision, Value engineering – Function, aims, Value engineering procedure. Interest formulaeand their applications –Time value of money, Single payment compound amount factor, Single payment presentworth 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.

UNIT III CASH FLOW 9Methods of comparison of alternatives – present worth method (Revenue dominated cash flow diagram), Futureworth method (Revenue dominated cash flow diagram, cost dominated cash flow diagram), Annual equivalentmethod (Revenue dominated cash flow diagram, cost dominated cash flow diagram), rate of return method,Examples in all the methods.

UNIT IV REPLACEMENT AND MAINTENANCE ANALYSIS 9Replacement and Maintenance analysis – Types of maintenance, types of replacement problem, determination ofeconomic life of an asset, Replacement of an asset with a new asset – capital recovery with return and concept ofchallenger and defender, Simple probabilistic model for items which fail completely.

UNIT V DEPRECIATION 9Depreciation- Introduction, Straight line method of depreciation, declining balance method of depreciation-Sumof the years digits method of depreciation, sinking fund method of depreciation/ Annuity method ofdepreciation, service output method of depreciation-Evaluation of public alternatives- introduction, Examples,Inflation adjusted decisions – procedure to adjust inflation, Examples on comparison of alternatives anddetermination of economic life of asset.

TOTAL: 45 PERIODSTEXT BOOKS:1. Panneer Selvam, R, “Engineering Economics”, Prentice Hall of India Ltd, New Delhi,2001.2. Suma Damodaran, “ Managerial economics”, Oxford university press 2006.REFERENCES:1. Chan S.Park, “Contemporary Engineering Economics”, Prentice Hall of India, 2002.2. Donald.G. Newman, Jerome.P.Lavelle, “Engineering Economics and analysis” Engg.Press, Texas, 20023. Degarmo, E.P., Sullivan, W.G and Canada, J.R, “Engineering Economy”, Macmillan,New York, 19844. Grant.E.L., Ireson.W.G., and Leavenworth, R.S, “Principles of EngineeringEconomy”, Ronald Press, New York,1976.5. Smith, G.W., “Engineering Economy”, Lowa State Press, Iowa, 1973.6. Truett & Truett, “ Managerial economics- Analysis, problems & cases “ Wiley India 8Th edition 2004.7. Luke M Froeb / Brian T Mccann, “ Managerail Economics – A problem solvingapproach” Thomson learning 2007.

MG2451 ENGINEEERING ECONOMICS & COST ANALYSIS

SCE 1 Department of Mechanical Engineering

UNIT I - INTRODUCTION TO ECONOMICS

1.1 Introduction

Efficient functioning of any business organization would enable it to providegoods/services at a lower price.

In the process of managing organizations, the managers at different levels should takeappropriate economic decisions which will help in minimizing investment, operating andmaintenance expenditures besides increasing the revenue, savings and such other gains of theorganization.

These can be achieved through Engineering Economics which deals with the methodsthat enable one to make economic decisions towards minimizing costs and/or maximizingbenefits to business organizations.

This chapter discusses the elements of economics and the interaction between itsvarious components.

This is followed by an analysis of the need and scope of engineering economics. Later,elements of cost and break-even analysis are presented.

1.2 Economics - is a study of economic problems of the people concerning production,consumption, exchange and distribution of wealth.

Economics is the science that deals with the production and consumption of goods andservices 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

1.3 Flow in economy The flow of goods, services, resources and money payments in a simpleeconomy

Households and businesses are the two major entities in a simple economy.

Business organizations use various economic resources like land, labour and capitalwhich are provided by households to produce consumer goods and services whichwill be used by them.

Business organizations make payment of money to the households for receivingvarious resources.



The households in turn make payment of money to business organizations forreceiving consumer goods and services.

This cycle shows the interdependence between the two major entities ina simple economy.

Money payments for consumer goods and services

Consumer goods, services

Businesses

1. Provide goods and services toconsumers

2. Use resources, inputsprovided by households

Households

1. Consume final goodsand services produced

by businesses and services

2. Provide productiveinputs to businesses

Money payments for resources, rents, wages,salaries, interest and profit

Economic Resources: Land, labour, capital

Fig. Flow of goods, services, resources and money payments in a simple economy.

1.4 Laws of supply and demand

1.3.1 Laws of supply - states that the quantity of a commodity supplied varies directly withthe price, other determinants of supply remaining constant.

If the cost of inputs increases, then naturally, the cost of the product will go up. Insuch a situation, at the prevailing price of the product the profit margin per unitwill be less.

The producers will then reduce the production quantity, which in turn will affectthe supply of the product.

For instance, if the prices of fertilizers and cost of labour are increasedsignificantly, 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 ofsupply of paddy will be reduced at the prevailing prices of the paddy.

If there is an advancement in technology used in the manufacture of the product inthe 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 theprevailing price of the product. Hence, the producer will be tempted to supplymore quantity to the market.

Weather also has a direct bearing on the supply of products. For example, demandfor woollen products will increase during winter. This means the prices of woollengoods will be incresed in winter.

So, naturally, manufacturers will supply more volume of woollen goods duringwinter.



Pri

ce

Factors influencing supply

The shape of the supply curve is affected by the following factors:

Cost of the inputs Technology Weather Prices of related goods

1.3.2 Law of demand states that other things being equal demand when price falls andcontracts when price rises.

Market demand is the total quantity demanded by all the purchasers together.

Elasticity of Demand - Elasticity of demand may be defined as the degree ofresponsiveness of quantity demanded to a Change in price.

An interesting aspect of the economy is that the demand and supply of a productare interdependent and they are sensitive with respect to the price of thatproduct. .

From Fig. it is clear that when there is a decrease in the price of a product, thedemand for the product increases and its supply decreases.

Also, the product is more in demand and hence the demand of the productincreases.

At the same time, lowering of the price of the product makes the producers restrainfrom releasing more quantities of the product in the market.

Hence, the supply of the product is decreased. The point of intersection of thesupply 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 thequantity of demand. Hence, this point is called the equilibrium point.

Supply

Demand

Quantity

Fig. Demand and supply curve.



Factors influencing demand

The shape of the demand curve is influenced by the following factors:

Income of the people Prices of related goods Tastes of consumers

1.5 Concept of Engineering Economics

Science is a field of study where the basic principles of different physicalsystems are formulated and tested.

Engineering is the application of science. It establishes varied applicationsystems based on different scientific principles.

From the discussions in the previous section, it is clear that price has a major rolein deciding the demand and supply of a product.

Hence, from the organization’s point of view, efficient and effective functioning ofthe organization would certainly help it to provide goods/services at a lower costwhich in turn will enable it to fix a lower price for its goods or services.

The following section discusses the different types of efficiency and their impacton the operation of businesses and the definition and scope of engineeringeconomics.

.1.5 Types of Efficiency

Efficiency of a system is generally defined as the ratio of its output to input. Theefficiency can be classified into technical efficiency and economic efficiency.

1.Technical efficiency

It is the ratio of the output to input of a physical system. The physicalsystem may be a diesel engine, a machine working in a shop floor, a furnace,etc.

Technical efficiency (%) = Output × 100Input

The technical efficiency of a diesel engine is as follows:

Heat equivalent of mechanical

Technical efficiency (%) = energy produced × 100Heat equivalent of fuel used



In practice, technical efficiency can never be more than 100%.

This is mainly due to frictional loss and incomplete combustion of fuel, whichare considered to be unavoidable phenomena in the working of a diesel engine.

2. Economic efficiency

Economic efficiency is the ratio of output to input of a business system.

Economic efficiency (%) = Output × 100 = Worth × 100Input Cost

‘Worth’ is the annual revenue generated by way of operating the businessand ‘cost’ is the total annual expenses incurred in carrying out the business.

For the survival and growth of any business, the economic efficiencyshould be more than 100%.

Economic efficiency is also called ‘productivity’. There are several waysof improving productivity.

Increased output for the same input

Decreased input for the same output

By a proportionate increase in the output which ismore than the proportionate increase in the input

By a proportionate decrease in the input which ismore than the proportionate decrease in the output

Through simultaneous increase in the output with decrease in theinput.

1.6 Definition and Scope of Engineering Economics

As stated earlier, efficient functioning of any business organization wouldenable it to provide goods/services at a lower price.

In the process of managing organizations, the managers at different levelsshould take appropriate economic decisions which will help in minimizinginvestment, operating and maintenance expenditures besides increasing therevenue, savings and other related gains of the organization.

1.6.1 Definition

Engineering economics deals with the methods that enable one to takeeconomic decisions towards minimizing costs and/or maximizing benefits tobusiness organizations.



1.6.2 Scope

The issues that are covered in this book are elementary economic analysis,interest formulae, bases for comparing alternatives, present worth method, futureworth method, annual equivalent method, rate of return method, replacementanalysis, depreciation, evaluation of public alternatives, inflation adjustedinvestment decisions, make or buy decisions, inventory control, projectmanagement, value engineering, and linear programming.

1.7 Elements Of Costs

Cost can be broadly classified 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 further classified into direct material cost, direct labourcost, and direct expenses. The overhead cost can be classified into factoryoverhead, administration overhead, selling overhead, and distributionoverhead.

Direct material costs are those costs of materials that are used to producethe product. Direct labour cost is the amount of wages paid to the direct labourinvolved in the production activities.

Direct expenses are those expenses that vary in relation to the productionvolume, other than the direct material costs and direct labour costs.

Overhead cost is the aggregate of indirect material costs, indirect labourcosts and indirect expenses. Administration overhead includes all the costs that areincurred in administering the business.

Selling overhead is the total expense that is incurred in the promotionalactivities and the expenses relating to sales force. Distribution overhead is the totalcost 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 + Directexpenses = Prime cost

(b) Prime cost + Factory overhead = Factory cost

(c) Factory cost + Office and administrative overhead= Costs of production

(d) Cost of production + Opening finished stock – Closing finishedstock

= Cost of goods sold



(e) Cost of goods sold + Selling and distribution overhead = Cost ofsales

(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 closingfinished stock, then the cost of production is equal to the cost of goods sold.

1..8 Other Costs/Revenues

The following are the costs/revenues other than the costs which are presented in theprevious section:

Marginal cost

Marginal revenue

Sunk cost

Opportunity cost

1. Marginal Cost

Marginal cost of a product is the cost of producing an additional unit of thatproduct. Let the cost of producing 20 units of a product be Rs. 10,000, and the cost ofproducing 21 units of the same product be Rs. 10,045. Then the marginal cost ofproducing the 21st unit is Rs. 45.

2. Marginal Revenue

Marginal revenue of a product is the incremental revenue of selling an additionalunit of that product. Let, the revenue of selling 20 units of a product be Rs. 15,000and the revenue of selling 21 units of the same product be Rs. 15,085. Then, themarginal revenue of selling the 21st unit is Rs. 85.

3. Sunk Cost

This is known as the past cost of an equipment/asset. Let us assume that anequipment has been purchased for Rs. 1,00,000 about three years back. If it is consideredfor replacement, then its present value is not Rs. 1,00,000. Instead, its present marketvalue 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.



4. 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 availablefor any other purpose. If the same money is invested in some other alternative (Y ), it mayfetch some return. Since the money is invested in the selected alternative (X ), one hasto forego the return from the other alternative (Y ).

The amount that is foregone by not investing in the other alternative (Y ) is knownas the opportunity cost of the selected alternative (X ). So the opportunity cost of analternative is the return that will be foregone by not investing the same money in anotheralternative.

1.9 Break-Even Analysis

The main objective of break-even analysis is to find the cut-off production volume fromwhere a firm will make profit.

Let

s = selling price per unitv = variable cost per unit

FC = fixed cost per periodQ = volume of production

The total sales revenue (S) of the firm is given by the following formula:

S = s Q

The total cost of the firm for a given production volume isgiven as

TC = Total variable cost + Fixedcost

= v Q + FC

The linear plots of the above two equations are shown in Fig. .

The intersection point of the total sales revenue line and the totalcost line is called the break-even point.

The corresponding volume of production on the X-axis is known asthe break-even sales quantity.

At the intersection point, the total cost is equal to the total revenue.This point is also called the no-loss or no-gain situation.

For any production quantity which is less than the break-evenquantity, the total cost is more than the total revenue.

Hence, the firm will be making loss.



Sales (S)

ProfitTotal cost (TC)

Break-evensales

Loss

Variable cost (VC)Fixed cost (FC)

BEP(Q*) Production quantity

Fig. Break-even chart.

For any production quantity which is more than the break-even quantity, thetotal revenue will be more than the total cost. Hence, the firm will be making profit.

Profit = Sales – (Fixed cost + Variable costs)= s Q – (FC + v Q)

The formulae to find the break-even quantity and break-even sales quantity

Break-even quantity =Fixed cost

Sellingprice/unit − Variable cost /unit

FC= s-v (in units)

Break-even sales = Fixed cost × Selling price/ unitSelling price/ unit − Variable cost / unit

=FC

× s (Rs.)s − v

The contribution is the difference between the sales and the variable costs. Themargin of safety (M.S.) is the sales over and above the break-even sales. Theformulae to compute these values are

Contribution = Sales – Variable costs

Contribution/unit = Selling price/unit – Variable cost/unit

M.S. = Actual sales – Break-even sales

=Profit

× salesContribution

M.S. as a per cent of sales = (M.S./Sales) 100



1.10 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:

P/V ratio =Contribution

SalesSales − Variable costs

= Sales

The relationship between BEP and P/V ratio is as follows:

BEP =Fixed cost

P/V ratio

1.11 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 decision making involved. One can managemany of these decision problems by using simple economic analysis.

For example, an industry can source its raw materials from a nearby place or from afar-off place. In this problem, the following factors will affect the decision:

Price of the raw material

Transportation cost of the raw material

`Availability of the raw material

Quality of the raw material

Consider the alternative of sourcing raw materials from a nearby placewith the following characteristics:

The raw material is more costly in the nearby area.

The availability of the raw material is not sufficient enough tosupport the operation of the industry throughout the year.

The raw material requires pre-processing before it is usedin the production process. This would certainly add cost to theproduct.

The cost of transportation is minimal under this alternative.

On the other hand, consider another alternative of sourcing theraw materials from a far-off place with the following characteristics:

The raw material is less costly at the far off place.

The cost of transportation is very high.

The availability of the raw material at this site is abundant and itcan support the plant throughout the year.

The raw material from this site does not require any pre-



processing before using it for production.

1.12 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 formsa major portion of the total cost of any product.

So, any attempt to find a suitable raw material will bring a reduction in the totalcost in any one or combinations of the following ways:

Reduced machining/process time

Enhanced durability of the product Cheaper raw material price

Therefore, the process of raw material selection/substitution will result in finding analternate raw material which will provide the necessary functions that are provided bythe raw material that is presently used.

In this process, if the new raw material provides any additional benefit, then itshould be treated as its welcoming feature. This concept is demonstrated with numericalproblem given below

Example

In the design of a jet engine part, the designer has a choice of specifying either analuminium alloy casting or a steel casting. Either material will provide equal service, butthe 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 perkg. 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 toincreased fuel consumption. Which material should be specified and what is theeconomic advantage of the selection per unit?

Solution (a) Cost of using aluminium metal for the jetengine part:

Weight of aluminium casting/unit = 1.2 kgCost of making aluminium casting = Rs. 80.00 per kgCost of machining aluminium casting per unit = Rs. 150.00Total cost of jet engine part made of aluminium/unit

= Cost of making aluminium casting/unit+ Cost of machining aluminium

casting/unit= 80 1.2 + 150 = 96 + 150= Rs. 246



(b) Cost of jet engine part made of steel/unit:Weight of steel casting/unit = 1.35 kgCost of making steel casting = Rs. 35.00 per kgCost of machining steel casting per unit = Rs. 170.00Penalty 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 steelcasting/unit

+ Penalty for excess weight of steelcasting

= 35 1.35 + 170 + 1,300(1.35 – 1.2)= Rs. 412.25

DECISION The total cost/unit of a jet engine part made ofaluminium is less than that for an engine made of steel. Hence,aluminium is suggested for making the jet engine part. Theeconomic advantage of using aluminium over steel/unit is Rs.412.25 – Rs. 246 = Rs. 166.25

1.12.1 Design Selection for a Product

a. The design modification of a product may result in reduced rawmaterial requirements, increased machinability of the materials andreduced labour.

b. Design is an important factor which decides the cost of the product for aspecified level of performance of that product.

Example

(Design selection for a process industry). The chief engineer of refineryoperations is not satisfied with the preliminary design for storage tanks to beused as part of a plant expansion programme. The engineer who submittedthe design was called in and asked to reconsider the overall dimensions in thelight of an article in the Chemical Engineer, entitled “How to size futureprocess vessels?”

The original design submitted called for 4 tanks 5.2 m in diameter and 7m in height. From a graph of the article, the engineer found that the presentratio of height to diameter of 1.35 is 111% of the minimum cost andthat the minimum cost for a tank was when the ratio of height to diameter was4 : 1. The cost for the tank design as originally submitted wasestimated to be Rs. 9,00,000. What are the optimum tank dimensions if thevolume remains the same as for the original design? What total savings may beexpected through the redesign?



Solution

(a) Original designNumber of tanks = 4Diameter of the tank = 5.2 mRadius of the tank = 2.6 mHeight of the tank = 7 mRatio of height to diameter = 7/5.2 = 1.35

Volume/tank = (22/7)r 2h = (22/7)(2.6)2 7

= 148.72 m3

(b) New designCost 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 : 14d = h

d = h/4r = h/8

Volume = (22/7)r2h = 148.72 (since, the volume remains the same)

(22/7)(h/8)2h = 148.72

h3 = 148.72

(22/7)

h = 14.47 m

64 = 3,028.48

r = h/8 = 14.47/8 = 1.81 m

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



1.13 Process Planning /Process Modification

While planning for a new component, a feasible sequence of operations withthe least cost of processing is to be considered.

The process sequence of a component which has been planned in the past isnot static.

It is always subject to modification with a view to minimize the cost ofmanufacturing the component.

So, the objective of process planning/process modification is to identify themost 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 isrequired.

2. Make recommendations to or consult with product engineerson product design changes.

3. List the basic operations required to produce the part tothe drawing or specifications.

4. Determine the most practical and economical manufacturingmethod and the form or tooling required for each operation.

5. Devise the best way to combine the operations and put themin sequence.

6. Specify the gauging required for the process.

Steps 3–5 aim to determine the most practical and economical sequence ofoperations to produce a component. This concept is demonstrated with anumerical problem.

Example

The process planning engineer of a firm listed the sequences of operations asshown in Table to produce a component.

Table Data for Example

SequenceProcess sequence

1 Turning – Milling – Shaping – Drilling2 Turning – Milling – Drilling3 All operations are performed with CNC machine

The details of processing times of the component for various operations andtheir machine hour rates are summarized in Table below



Operation Operation Time Machine Cost

(min) (hr) (Rs.) (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

No.(min) (hr)

hour rate(Rs.) (Rs.)

1 Turning 5 0.083 200 16.60

3 Drilling 3 0.050 300 15.00

Total: 124.80

Machine Hour Rates and Processing Times (minutes) for Example

Operation Machine hour Processsequence rate(Rs.) 1 2 3

Turning 200 5 5

– Milling 400 8 14

– Shaping 350 10 –

– Drilling 300 3 3

– CNC operations 1,000 – – 8

Find the most economical sequence of operations to manufacture thecomponent.

Solution (a) Cost of component using process sequence 1. The processsequence 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.

Table Workings for Process Sequence 1

No. hour rate

Total: 143.25

(b) Cost of component using process sequence 2. The process sequence 2of the component is as follows:

Turning – Milling – Drilling

The calculations for the cost of the above process sequence are given inTable



2 Milling 14 0.233 400 93.20




(min) (hr) (Rs.) (Rs.)

1 CNC operations 8 0.133 1,000 133

(c) Cost of component using process sequence 3. The process sequence3 of the component is as follows:

Only CNC operations

The calculations for the cost of the above process sequence are summarizedin Table


No. hour rate

The process sequence 2 has the least cost. Therefore, it should be selected formanufacturing the component.




2.1 INTRODUCTION

Value analysis is one of the major techniques of cost reduction and costprevention. It is a disciplined approach that ensures necessary functions for minimumcost without sacrificing quality, reliability, performance, and appearance. Accordingto the Society of American Value Engineers (SAVE),

Value Analysis is the systematic application of recognized techniques whichidentify the function of a product or service, establish a monetary value for thefunction and provide the necessary function reliably at the lowest overall cost.

It is an organized approach to identify unnecessary costs associated with anyproduct, material part, component, system or service by analysing the function andeliminating such costs without impairing the quality, functional reliability, or thecapacity of the product to give service

2.1.1 WHEN TO APPLY VALUE ANALYSIS

One can definitely expect very good results by initiating a VA programme if oneor 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 thevolume of production.

6. Rate of return on investment has a falling trend.

7. Inability of the firm to meet its delivery commitments.

2.1.2 Value

The term ‘value’ is used in different ways and, consequently, has differentmeanings. The designer equates the value with reliability; a purchase person withprice paid for the item; a production person with what it costs to manufacture, and asales person with what the customer is willing to pay.

Example

Cost value. It is the summation of the labour, material, overhead and all otherelements 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 andfeatures of the product, which make the product possible of being traded for anotherproduct or for money.



Value analysis/value engineering

It is a special type of cost reduction technique. It critically investigates andanalyses the different aspects of materials, design, cost and production of each andevery component of the product in produce it economically without decreasing itsutility, function or reliability.

Applications

The various application areas of value engineering are machine toolindustries, industries making accessories for machine tools, auto industries, importsubstitutes, etc.

2.2 INTRODUCTION

In the process of carrying out business activities of an organization, acomponent/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 orbuy alternatives and then select the alternative which results in the lowest cost. This is animportant 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 acomponent/product may be economical today; but after some time, it may turn out to beuneconomical to make the same.

Thus, the make or buy decision should be reviewed periodically, say, every 1 to 3years. This is mainly to cope with the changes in the level of competition and variousother environmental factors.

Make or Buy Decisions - is a determination whether to produce a component partinternally or to buy it from an outside supplier. The Organization should evaluate thecosts and benefits of manufacturing a product or product component against purchasing itand then select the alternative which results in the lower cost.

2.2.1 CRITERIA FOR MAKE OR BUY

In this section the criteria for make or buy are discussed.

1. Criteria for make

The following are the criteria for make:

1. The finished product can be made cheaper by the firm than byoutside suppliers.



2. The finished product is being manufactured only by a limitednumber of outside firms which are unable to meet thedemand.

3. The part has an importance for the firm and requires extremelyclose quality control.

4. The part can be manufactured with the firm’s existingfacilities and similar to other items in which the company hasmanufacturing experience.

2. Criteria for buy

The following are the criteria for buy:

1. Requires high investments on facilities which are alreadyavailable at suppliers plant.

2. The company does not have facilities to make it and thereare more profitable opportunities for investing company’scapital.

3. Existing facilities of the company can be used moreeconomically to make other parts.

4. The skill of personnel employed by the company is not readilyadaptable 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.

2.3 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

1. Simple Cost Analysis

The concept is illustrated using an example problem.

EXAMPLE

A company has extra capacity that can be used to produce a sophisticatedfixture which it has been buying for Rs. 900 each. If the company makes thefixtures, it will incur materials cost of Rs. 300 per unit, labour costs of Rs. 250per unit, and variable overhead costs of Rs. 100 per unit. The annual fixed costassociated with the unused capacity is Rs. 10,00,000. Demand over the next yearis estimated at 5,000 units. Would it be profitable for the company to make thefixtures?



Solution

We assume that the unused capacity has alternative use.

Cost to makeVariable 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 associatedwith unused capacity + Rs. 10,00,000

Total cost = Rs. 42,50,000

Cost to buyPurchase cost = (5,000 units) (Rs. 900/unit)

= Rs. 45,00,000Add fixed cost associatedwith 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 fromoutside. Therefore, the organization should make the fixtures.



2. 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 aregiven in the following table:

Purchase model Manufacturing model

2CoD2CD

Q1 = CcQ2 = --------

Cc (1 − r /k)

TC = D P +DCo

Q1+

Q1 × Cc

2TC = D P +

DCo

Q2

+ Cc (k – r) Q2

2 * k

whereD = 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/yearQ1 = economic order size

Q2 = economic production size

TC = total cost per year



EXAMPLE

An item has a yearly demand of 2,000 units. The different costs inrespect of make and buy are as follows. Determine the best option.

Buy Make

Item cost/unit Rs. 8.00 Rs. 5.00

Procurement cost/order Rs. 120.00Set-up cost/set-up Rs. 60.00

Annual carrying cost/

item/year Rs. 1.60 Rs. 1.00Production rate/year 8,000 units

Solution

Buy option

D = 2,000 units/yearCo = Rs. 120/orderCc = Rs. 1.60/unit/year

Q1 =2CoD

=2 × 2,000 × 120

Cc 1.60

= 548 units (approx.)

TC = DP +DCo

Q1 +Q1Cc

2

= 2,000 8 +2,000 × 120

548+

548 × 1.60

2= Rs. 16,876.36

Make optionCo = Rs. 60/set-up

r = 2,000 units/year

Cc = Re 1/unit/year

k = 8,000 units/year

Q2 _-

2Cor

Cc[1 − (r /k)]

=2 × 60 × 2,000

1.0(1 − 2,000/8,000)= 566 units (approx.)



TC = DP +D × Co

Q2+ Cc (k – r)

Q2

2 × k

= 2,000 5.00 +

= Rs. 10,424.26

2,000 × 60

566+ 1.0 (8,000 – 2,000)

566

2 × 8,000

Result: The cost of making is less than the cost of buying.Therefore, the firm should go in for the making option.

3. Break-even Analysis

The break-even analysis chart is shown in Fig.

Sales (S)

Profit Total cost (TC)

A B

Loss

Variablecost (VC)

Fixedcost (FC)

TC = total costFC = fixed cost

CProduction quantity

Fig. 13.1 Break-even chart.

TC = FC + variable cost

B = the intersection of TC and sales (no loss or no gain situation)

A = break-even sales

C = break-even quantity/break-even point (BEP)

The formula for the break-even point (BEP) is

FC

BEP =Selling price/ unit − Variable cost /unit



EXAMPLE

There are three alternatives available to meet the demand of aparticular product. They are as follows:

(a) Manufacturing the product by using process

A (b) Manufacturing the product by using

process B (c) Buying the product

The details are as given in the following table:

Cost elements Manufacturing Manufacturing Buythe product by the product by

process A process B

Fixed cost/year (Rs.) 5,00,000 6,00,000

Variable/unit (Rs.)

Purchase price/unit (Rs.)

175 150

125

The annual demand of the product is 8,000 units. Should the company make theproduct using process A or process B or buy it?

SolutionAnnual cost of process A = FC + VC Volume

= 5,00,000 + 175 8,000= Rs. 19,00,000

Annual cost of process B = FC + VC Volume= 6,00,000 + 150 8,000

= Rs. 18,00,000

Annual cost of buy = Purchase price/unit Volume= 125 8,000= Rs. 10,00,000

Since the annual cost of buy option is the minimum among all the alternatives, the companyshould buy the product



2.4 Function

Function is the purpose for which the product is made. Identification of the basicfunctions and determination of the cost currently being spent on them are the two majorconsiderations of value analysis.

Function identifies the characteristics which make the product/component/part/item/device to work or sell. “Work functions” lend performance value while “sellfunctions” provide esteem value.

Verbs like “support”, “hold”, “transmit”, “prevent”, “protect”, “exhibits”, “control”,etc., are used to describe work functions, while “attract”, enhance”, “improve”, “create”,etc., are used to describe “sell” functions. For example, in a “bus driver cabin”, thefunctional analysis of some of the parts are given in Table

Functional Analysis of Some Parts of a Bus Driver Cabin

Component of study Functional analysis

Verb Noun Steering

wheel Control Direction Gear box

Change Speed Brake system Stop

Vehicle Wiper Clear Water

Horn Make Sound

Side mirror Show Side traffic

2.4.1Classification of the functions

Rarely do all functions assume equal importance. Usually, some functions are more important thanothers. Functions can be classified into the following three categories:

1. Primary function2. Secondary function3.Tertiary function

1. Primary functions are the basic functions for which the product is specially designed to achieve. Primaryfunctions, therefore, are the most essential functions whose non-performance would make the productworthless, e.g. a photo frame exhibits photographs, a chair supports weight, a fluorescent tube gives light.

2. Secondary functions are those which, if not in-built, would not prevent the device from performing itsprimary functions, e.g., arms of a chair provide support for hands. Secondary functions are usually related toconvenience. 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.

3. Tertiary functions are usually related to esteem appearance. For example, Sunmica top of a table givesesteem appearance for the table.



Examples

Let us consider a single example of painting a company bus to explain all the above three functions. Here, theprimary function of painting is to avoid corrosion.

The secondary function is to identify the company to which the bus belongs by the colour ofthe paint (e.g. bluecolour for Ashok Leyland Ltd.).

The tertiary function is to impart a very good appearance to the bus by using brilliant colours.

2.5 Aims The aims of value engineering are as follows: Simplify the product. Use (new) cheaper and better materials. Modify and improve product design. Use efficient processes. Reduce the product cost. Increase the utility of the product by economical means. Save money or increase the profits.

2.6 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 theproduct. First, identify the component for study. In future, any design change should addvalue and it should not make the product as obsolete one. Value engineering can be applied to a product as a whole or tosub-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 andmarketing details Latest development in related products

Step 3: Define different functions. Identify and define the primary, secondary and tertiary functions of theproduct 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 bedone 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.



Year end Interest(Rs.)

Compound amount(Rs.)

0 100.00

4 22.81 174.905 26.24 201.14

2.7 Interest Formulas Interest rate can be classified into simple interest rate and compoundinterest rate.

In simple interest, the interest is calculated, based on the initial deposit for everyinterest 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 theamount (principal plus interest up to the end of the previous period) at thebeginning 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 periodto/from the amount of deposit A1 at the end of period 1

2.8 Time Value of Money - . It represents the growth of capital per unit period. The periodmay be a month, a quarter, semiannual or a year.

An interest rate 15% compounded annually means that for every hundredrupees invested now, an amount of Rs. 15 will be added to the account at theend 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. 17.25will be added to the account.

Hence the total amount at the end of the second year will be Rs. 132.25. Theprocess will continue thus till the specified number of years.

If an investor invests a sum of Rs. 100 in a fixed deposit for five years with aninterest rate of 15% compounded annually, the accumulated amount at the end ofevery year will be as shown in Table

Compound Amounts

(amount of deposit = Rs. 100.00)

1 15.00 115.002 17.25 132.253 19.84 152.09



The formula to find the future worth in the third column is

F = P (1 + i)n

whereP = 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 theamount 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 assumesthat the inflation is at zero percentage.

2.9 Single-Payment Compound Amount - Here, the objective is to find the single future sum (F) ofthe initial payment (P) made at time 0 after n periods at an interest rate i compounded everyperiod. F

. .0 1 2 3 4 . . nP i %

Cash flow diagram of single-payment compound amount.

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.

2.10 Single-Payment Present Worth Amount - Here, the objective is to find the present worthamount (P) of a single future sum (F) which will be received after n periods at an interest rate ofi compounded at the end of every interest period.

F

. .0 1 2 3 4 . . n

P i %

Cash flow diagram of single-payment present worth amount.

The formula to obtain the present worth is

P = F

(1 + i)n

= F(P/F, i, n)



i %

. .

Where(P/F, i, n) is termed as single-payment present worth factor.

2.11 Equal-Payment Series Sinking Fund

In this type of investment mode, the objective is to find the equivalentamount (A) that should be deposited at the end of every interest period for n interestperiods to realize a future sum (F) at the end of the nth interest period at an interestrate of i.

n

0 1 2 3 4 . .

A A A Ai %

A = equal amount to be deposited at the end of each interest periodn = No. of interest periodsi = rate of interestF = single future amount at the end of the nth period

The formula to get F is

Where

A = Fi

(1 + i)n − 1= F(A/F, i, n)

(A/F, i, n) is called as equal-payment series sinking fund factor.

2.12 Equal-Payment Series Present Worth Amount The objective of this mode of investment is tofind the present worth of an equal payment made at the end of every interest period for ninterest 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 worthA = annual equivalent paymenti = interest raten = No. of interest periods

The formula to compute P is

(1 + i)n − 1P = A = A(P/A, i, n)

i(1 + i)n

Where(P/A, i, n) is called equal-payment series present worth factor



. .

.

2.13 Equal-Payment Series capital recovery The objective of this mode of investment isto find the annual equivalent amount (A) which is to be recovered at the end of everyinterest period for n interest periods for a loan (P) which is sanctioned now at an interestrate of i compounded at the end of every interest period

P

i %

0 1 2 3 4 . . n

A A A A A

Cash flow diagram of equal-payment series capital recovery amount.

P = present worth (loan amount)A = annual equivalent payment (recovery amount)i = interest raten = No. of interest periods

The formula to compute P is as follows:

i(1 + i)n

A = P = P(A/P, i, n)(1 + i)n − 1

Where,(A/P, i, n) is called equal-payment series capital recovery factor.

2.14 Uniform Gradient series annual equivalent The objective of this mode of investment is tofind the annual equivalent amount of a series with an amount A1 at the end of the first yearand with an equal increment (G) at the end of each of the following n – 1 years with aninterest rate i compounded annually.

The corresponding cash flow diagram is shown in Fig

0 1 2 3 4 . . 10.

A1A1+G A1+2G

A1+3G

A1+ (n – 1)G

Cash flow diagram of uniform gradient series annual equivalent amount.

The formula to compute A under this situation is

(1 + i)n − in − 1A = A1 + G

i(1 + i)n − i= A1 + G (A/G, i, n)

hereWhere

(A/G, i, n) is called uniform gradient series factor.



2.15 Effective Interest rate Let i be the nominal interest rate compounded annually. But, inpractice, the compounding may occur less than a year. For example, compounding maybe monthly, quarterly, or semi-annually. Compounding monthly means that the interest iscomputed at the end of every month. There are 12 interest periods in a year if theinterest is compounded monthly. Under such situations, the formula to compute theeffective interest rate, which is compounded annually, is Effective interest rate, R = 1 +

i/CC − 1

where,i = the nominal interest rateC = the number of interest periods in a year.



UNIT III CASH FLOW

3.1 Introduction

In this method of comparison, the cash flows of each alternative will be reduced to timezero by assuming an interest rate i. Then, depending on the type of decision, the bestalternative will be selected by comparing the present worth amounts of the alternatives.

The sign of various amounts at different points in time in a cash flow diagram is to bedecided based on the type of the decision problem.

In a cost dominated cash flow diagram, the costs (outflows) will be assigned with positivesign and the profit, revenue, salvage value (all inflows), etc. will be assigned with negativesign.

In a revenue/profit-dominated cash flow diagram, the profit, revenue, salvage value (allinflows to an organization) will be assigned with positive sign. The costs (outflows) will beassigned with negative sign.

In case the decision is to select the alternative with the minimum cost, then the alternativewith the least present worth amount will be selected. On the other hand, if the decision is toselect the alternative with the maximum profit, then the alternative with the maximum presentworth will be selected.

3.2 BASES FOR COMPARISON OF ALTERNATIVES

In most of the practical decision environments, executives will be forced to select thebest alternative from a set of competing alternatives.

Let us assume that an organization has a huge sum of money for potential investment andthere are three different projects whose initial outlay and annual revenues during their livesare known. The executive has to select the best alternative among these three competingprojects.

There are several bases for comparing the worthiness of the projects. These bases are:

1. Present worth method2. Future worth method3. Annual equivalent method4. Rate of return method

3.2.1 PRESENT WORTH METHOD

In this method of comparison, the cash flows of each alternative will be reduced totime zero by assuming an interest rate i.

Then, depending on the type of decision, the best alternative will be selected bycomparing the present worth amounts of the alternatives.

In a cost dominated cash flow diagram, the costs (outflows) will be assigned with



positive sign and the profit, revenue, salvage value (all inflows), etc. will be assignedwith negative sign.

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(outflows) will be assigned with negative sign.

3.1.1.1 Revenue-Dominated Cash Flow Diagram

A generalized revenue-dominated cash flow diagram to demonstrate thepresent worth method of comparison is presented in Fig.

S

R1 R2 R3.

0 1 2 3

Rj Rn

j n

To find the present worth of the above cash flow diagram for a giveninterest rate, the formula is

PW(i) = – P + R1[1/(1 + i)1] + R2[1/(1 + i)2] + ...

+ Rj [1/(1 + i) j] + Rn[1/(1 + i)n] + S[1/(1 + i)n]

3.1.1.2 Cost-Dominated Cash Flow Diagram

A generalized cost-dominated cash flow diagram to demonstrate the present worthmethod of comparison is presented in Fig.

S

0 1 2 . j

C1 C2 . Cj Cn

P

To compute the present worth amount of the above cash flow diagram for a given interestrate i, we have the formula

PW(i) = P + C1[1/(1 + i)1] + C2[1/(1 + i)2] + ... + Cj[1/(1 + i) j]

+ Cn[1/(1 + i)n] – S[1/(1 + i)n]



Initial outlay(Rs.)

Annual revenue(Rs.)

Life(years)

Technology 1 12,00,000 4,00,000 10

Technology 3 18,00,000 5,00,000 10

EXAMPLE

Alpha Industry is planning to expand its production operation. It has identified three differenttechnologies for meeting the goal. The initial outlay and annual revenues with respect to each of thetechnologies are summarized in Table 1. Suggest the best technology which is to be implementedbased on the present worth method of comparison assuming 20% interest rate, compoundedannually.

Table 1

Technology 2 20,00,000 6,00,000 10

SolutionIn all the technologies, the initial outlay is assigned a negative sign and

the annual revenues are assigned a positive sign.

TECHNOLOGY 1

Initial outlay, P = Rs. 12,00,000Annual revenue, A = Rs. 4,00,000Interest rate, i = 20%, compounded annuallyLife of this technology, n = 10 years

The cash flow diagram of this technology is as shown in Fig. 4.3.

4,00,000 4,00,000 4,00,000 . . 4,00,000

0 1 2 3.

i=20%

j 10

12,00,000

Fig. Cash flow diagram for technology 1.

The present worth expression for this technology is

PW(20%)1 = –12,00,000 + 4,00,000 (P/A, 20%, 10)

= –12,00,000 + 4,00,000 (4.1925)

= –12,00,000 + 16,77,000

= Rs. 4,77,000



TECHNOLOGY 2


The cash flow diagram of this technology is shown in Fig. 4.4.

6,00,000 6,00,000 6,00,000

.0 1 2

i = 20%

.10

20,00,000



PW(20%)2 = – 20,00,000 + 6,00,000 (P/A, 20%, 10)

= – 20,00,000 + 6,00,000 (4.1925)

= – 20,00,000 + 25,15,500

= Rs. 5,15,500

TECHNOLOGY 3


The cash flow diagram of this technology is shown in Fig. 4.5.

5,00,000 5,00,000 5,00,000

0 1 2.

i=20%

.10

18,00,000



PW(20%)3 = –18,00,000 + 5,00,000 (P/A, 20%, 10)

= –18,00,000 + 5,00,000 (4.1925)

= –18,00,000 + 20,96,250

= Rs. 2,96,250

From the above calculations, it is clear that the present worth of technology 2is the highest among all the technologies. Therefore, technology 2 is suggestedfor implementation to expand the production.



3.1.2 FUTURE WORTHMETHOD

In the future worth method of comparison of alternatives, the future worth of various alternativeswill be computed.

Then, the alternative with the maximum future worth of net revenue or with the minimum futureworth of net cost will be selected as the best alternative for implementation.


A generalized revenue-dominated cash flow diagram to demonstrate the future worthmethod of comparison is presented in Fig.

S

R1 R2 R3 Rj Rn

0 1 2 3 . j n

P

Fig Revenue-dominated cash flow diagram.

In Fig. P represents an initial investment, Rj the net-revenue at the end of the jth year, and Sthe salvage value at the end of the nth year.

The formula for the future worth of the above cash flow diagram for a given interest rate, i is

FW(i) = –P(1 + i)n + R1(1 + i)n–1 + R2(1 + i)n–2 + ...

+ R j(1 + i)n–j + ... + Rn + S

In the above formula, the expenditure is assigned with negative sign and the revenues areassigned with positive sign.




A generalized cost-dominated cash flow diagram to demonstrate the future worth methodof comparison is given in Fig.

0 1 2

C1 C2

S

j

Cj Cn

P

Fig. Cost-dominated cash flow diagram.

In Fig. 5.2, P represents an initial investment, Cj the net cost of operation andmaintenance at the end of the j th year, and S the salvage value at the end ofthe nth year.

The formula for the future worth of the above cash flow diagramfor a given interest rate, i is

FW(i) = P(1 + i)n + C1(1 + i )n–1 + C2(1 +

i)n–2 + ...

+ Cj(1 + i)n–j + ... + Cn – S



EXAMPLE

Consider the following two mutually exclusive alternatives:

End of year

Alternative 0 1 2 3 4

A (Rs.) – 50,00,000 20,00,000 20,00,000 20,00,000 20,00,000

B (Rs.) – 45,00,000 18,00,000 18,00,000 18,00,000 18,00,000

At i = 18%, select the best alternative based on futureworth method of comparison.

Solution Alternative A

Initial investment, P = Rs. 50,00,000Annual equivalent revenue, A = Rs. 20,00,000Interest rate, i = 18%, compounded annuallyLife of alternative A = 4 years

The cash flow diagram of alternative A is shown in Fig.

20,00,000 20,00,000 20,00,000 20,00,000

0 1 2 3 4

i=18%

50,00,000

Fig. Cash flow diagram for alternative A.

The future worth amount of alternative B is computed as

FWA(18%) = –50,00,000(F/P, 18%, 4) + 20,00,000(F/A, 18%, 4)

= –50,00,000(1.939) + 20,00,000(5.215)

= Rs. 7,35,000



Alternative B

Initial investment, P = Rs. 45,00,000Annual equivalent revenue, A = Rs. 18,00,000Interest rate, i = 18%, compounded annuallyLife of alternative B = 4 years

The cash flow diagram of alternative B is illustrated in Fig..

18,00,000 18,00,000 18,00,000 18,00,000

0 1 2 3 4

i=18%

45,00,000Fig. Cash flow diagram for alternative B.

The future worth amount of alternative B is computed as

FWB(18%) = – 45,00,000(F/P, 18%, 4) + 18,00,000 (F/A, 18%, 4)

= – 45,00,000(1.939) + 18,00,000(5.215)

= Rs. 6,61,500



3.1.3 ANNUAL EQUIVALENT METHOD

In the annual equivalent method of comparison, first the annual equivalent cost orthe revenue of each alternative will be computed.

Then the alternative with the maximum annual equivalent revenue in the case ofrevenue-based comparison or with the minimum annual equivalent cost in thecase of cost- based comparison will be selected as the best alternative.


A generalized revenue-dominated cash flow diagram to demonstrate the annualequivalent method of comparison is presented in Fig.

S

R1 R2 R3 Rj Rn

0 1 2 3 . j n

P

Fig. Revenue-dominated cash flow diagram.

In Fig. P represents an initial investment, Rj the net revenue at the end of thej th year, and S the salvage value at the end of the nth year.

The first step is to find the net present worth of the cash flow diagram using thefollowing expression for a given interest rate, i:

PW(i) = –P + R1/(1 + i)1 + R2/(1 + i)2 + ...

+ Rj/(1 + i) j + ... + Rn/(1 + i)n + S/(1 + i)n

In the above formula, the expenditure is assigned with a negative sign and therevenues are assigned with a positive sign.




A generalized cost-dominated cash flow diagram to demonstrate the annualequivalent method of comparison is illustrated in Fig.

S

0 1 2 .

C1 C2 Cj Cn

P

In Fig, P represents an initial investment, Cj the net cost of operation and maintenance atthe end of the jth year, and S the salvage value at the end of the nth year.

The first step is to find the net present worth of the cash flow diagram using the followingrelation for a given interest rate, i.

PW(i) = P + C1/(1 + i)1 + C2/(1 + i)2 + ...

+ Cj/(1 + i) j + ... + Cn/(1 + i)n – S/(1 + i)n



EXAMPLE

A company provides a car to its chief executive. The owner of thecompany is concerned about the increasing cost of petrol. The cost per litreof petrol for the first year of operation is Rs. 21. He feels that the cost of petrolwill be increasing by Re.1 every year. His experience with his company carindicates that it averages 9 km per litre of petrol. The executive expects todrive an average of 20,000 km each year for the next four years. What is theannual equivalent cost of fuel over this period of time?. If he is offeredsimilar service with the same quality on rental basis at Rs. 60,000 per year,should the owner continue to provide company car for his executive oralternatively provide a rental car to his executive? Assume i = 18%. If therental car is preferred, then the company car will find some other use withinthe company.

Solution

Average number of km run/year = 20,000 kmNumber of km/litre of petrol = 9 km

Therefore,

Petrol consumption/year = 20,000/9 = 2222.2 litreCost/litre of petrol for the 1st year = Rs. 21Cost/litre of petrol for the 2nd year = Rs. 21.00 + Re. 1.00

= Rs. 22.00Cost/litre of petrol for the 3rd year = Rs. 22.00 + Re. 1.00

= Rs. 23.00Cost/litre of petrol for the 4th year = Rs. 23.00 + Re. 1.00

= Rs. 24.00Fuel expenditure for 1st year = 2222.2 21 = Rs. 46,666.20Fuel expenditure for 2nd year = 2222.2 22 = Rs. 48,888.40Fuel expenditure for 3rd year = 2222.2 23 = Rs. 51,110.60Fuel expenditure for 4th year = 2222.2 24 = Rs. 53,332.80

The annual equal increment of the above expenditures is Rs. 2,222.20

(G). The cash flow diagram for this situation is depicted in Fig.

0 1 2 3 4

A1

A1+GA1+2G

A1+3G

Fig. Uniform gradient series cash flow diagram.



In Fig., A1 = Rs. 46,666.20 and G = Rs. 2,222.20

A = A1 + G(A/G, 18%, 4)= 46,666.20 + 2222.2(1.2947)

= Rs. 49,543.28

The proposal of using the company car by spending for petrol by thecompany will cost an annual equivalent amount of Rs. 49,543.28 for fouryears. This amount is less than the annual rental value of Rs. 60,000.Therefore, the company should continue to provide its own car to itsexecutive.



3.1.4 RATE OF RETURNMETHOD

The rate of return of a cash flow pattern is the interest rate at which the present worth of thatcash flow pattern reduces to zero.

In this method of comparison, the rate of return for each alternative is computed. Then thealternative which has the highest rate of return is selected as the best alternative.

A generalized cash flow diagram to demonstrate the rate of return method of comparison ispresented in Fig

S

R1 R2 R3 . Rj Rn

0 1 2 3 . j nP

Fig. Generalized cash flow diagram.

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

The first step is to find the net present worth of the cash flow diagram using the followingexpression at a given interest rate, i.

PW(i) = – P + R1/(1 + i)1 + R2/(1 + i)2 + ...

+ Rj/(1 + i) j + ... + Rn/(1 + i)n + S/(1 + i)n



EXAMPLE

A person is planning a new business. The initial outlay and cashflow pattern for the new business are as listed below. The expectedlife of the business is five years. Find the rate of return for thenew business.

Period 0 1 2 3 4 5

Cash flow –1,00,000 30,000 30,000 30,000 30,000 30,000(Rs.)

Solution

Initial investment = Rs. 1,00,000Annual equal revenue = Rs. 30,000Life = 5 yearsThe cash flow diagram for this situation is illustrated in Fig.

30,000 30,000 30,000 30,000 30,000

0 1 2 3 4 5

1,00,000

Fig. Cash flow diagram.

The present worth function for the business is

PW(i) = –1,00,000 + 30,000(P/A, i, 5)



When i = 10%,

PW(10%) = –1,00,000 + 30,000(P/A, 10%, 5)

= –1,00,000 + 30,000(3.7908)

= Rs. 13,724.

When i = 15%,

PW(15%) = –1,00,000 + 30,000(P/A, 15%, 5)

= –1,00,000 + 30,000(3.3522)

= Rs. 566.

When i = 18%,

PW(18%) = –1,00,000 + 30,000(P/A, 18%, 5)

= –1,00,000 + 30,000(3.1272)

= Rs. – 6,184

i = 15% + 566 − 0

566 − (− 6184)

= 15% + 0.252%

= 15.252%

(3%)

Therefore, the rate of return for the new business is 15.252%.



UNIT IV REPLACEMENT AND MAINTENANCE ANALYSIS

4.1 Introduction

Organizations providing goods/services use several facilities like equipment andmachinery which are directly required in their operations. In addition to these facilities,there are several other items which are necessary to facilitate the functioning oforganizations.

All such facilities should be continuously monitored for their efficient functioning;otherwise, the quality of service will be poor. Besides the quality of service of thefacilities, the cost of their operation and maintenance would increase with the passage oftime.

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

If a firm wants to be in the same business competitively, it has to take decision onwhether to replace the old equipment or to retain it by taking the cost of maintenance andoperation into account.

There are two basic reasons for considering the replacement of an equipment physicalimpairment of the various parts or obsolescence of the equipment.

Physical impairment refers only to changes in the physical condition of the machineitself. This would lead to a decline in the value of the service rendered, increasedoperating cost, increased maintenance cost or a combination of these.

Obsolescence is due to improvement of the tools of production, mainly improvementin technology.

So, it would be uneconomical to continue production with the same machineunder any of the above situations. Hence, the machines are to be periodically replaced.

Sometimes, the capacity of existing facilities may be inadequate to meet the currentdemand. Under such situation, the following alternatives will be considered.

Replacement of the existing equipment with a new one.

Augmenting the existing one with an additional equipment.



4.2 Types of Maintenance

Maintenance activity can be classified into two types:

Preventive maintenance and Breakdown maintenance.

Preventive maintenance (PM) is the periodical inspection and service activities whichare aimed to detect potential failures and perform minor adjustments or repairs which willprevent major operating problems in future.

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

Preventive maintenance will reduce such cost up to a point. Beyond that point, the costof preventive maintenance will be more when compared to the breakdown maintenancecost.

The total cost, which is the sum of the preventive maintenance cost and the breakdownmaintenance cost, will go on decreasing with an increase in the level of maintenance up toa point.

Beyond that point, the total cost will start increasing. The level of maintenancecorresponding to the minimum total cost is the optimal level of maintenance. Theconcepts are demonstrated in Fig.



4.3 Types of Replacement Problem

Replacement study can be classified into two categories:

(a) Replacement of assets that deteriorate with time(Replacement due to gradual failure, or wear and tear of thecomponents of the machines).

This can be further classified into the following types:

(i) Determination of economic life of an asset.

(ii) Replacement of an existing asset with a new asset.

(b) Simple probabilistic model for assets which fail completely(replacement due to sudden failure).

4.4 Determination of Economic Life of an Asset

Any asset will have the following cost components:

Capital recovery cost (average first cost), computed from the first cost(purchase price) of the machine.

Average operating and maintenance cost (O & M cost)

Total cost which is the sum of capital recovery cost (average first cost)and average maintenance cost.



EXAMPLE

A firm is considering replacement of an equipment, whose first cost is Rs.4,000 and the scrap value is negligible at the end of any year. Based on experience,it was found that the maintenance cost is zero during the first year and it increasesby Rs. 200 every year thereafter.

(a) When should the equipment be replaced if i = 0%?

(b) When should the equipment be replaced if i = 12%?

(a) When i = 0%. In this problem

(i) First cost = Rs. 4,000

(ii) Maintenance cost is Rs. 0 during the first year and it increasesby Rs. 200 every year thereafter.

This is summarized in column B of Table



=n

+n

the given period

Column C summarizes the summation of maintenance costs for eachreplacement period. The value corresponding to any end of year in this columnrepresents the total maintenance cost of using the equipment till the end of thatparticular year.

Average total cost = First cost (FC) + Summation of maintenance costReplacement period

= Average first cost

for+ Average maintenance

cost for the givenperiod

Column F = Column E + Column D

The value corresponding to any end of year (n) in Column F representsthe average total cost of using the equipment till the end of that particularyear.

For this problem, the average total cost decreases till the end of year 6and then it increases. Therefore, the optimal replacement period is sixyears, i.e. economic life of the equipment is six years.

(b) When interest rate, i = 12%. When the interest rate is more than0%, the steps to be taken for getting the economic life are summarized withreference to Table

Table Calculations to Determine Economic Life (First cost = Rs. 4,000,Interest = 12%)

End ofyear

(n)

Maintenancecost at end of

year

P/F, 12%,n

Present worth as ofbeginning of year 1

of maintenancecosts

Summation ofpresent worth

of maintenancecosts through year

given

Present worth ofcumulative

maintenance cost& first cost

A/P, 12%,n

Annualequivalent

totalcost throughyear given

(B C) D E + Rs. 4,000 F G

A B (Rs) C D (Rs.) E (Rs.) F (Rs.) G H (Rs.)

1 0 0.8929 0.00 0.00 4,000.00 1.1200 4,480.00

2 200 0.7972 159.44 159.44 4,159.44 0.5917 2,461.14

3 400 0.7118 284.72 444.16 4,444.16 0.4163 1,850.10

4 600 0.6355 381.30 825.46 4,825.46 0.3292 1,588.54

5 800 0.5674 453.92 1,279.38 5,279.38 0.2774 1,464.50

6 1,000 0.5066 506.60 1,785.98 5,785.98 0.2432 1,407.15

7 1,200 0.4524 542.88 2,328.86 6,328.86 0.2191 1,386.65*

8 1,400 0.4039 565.46 2,894.32 6,894.32 0.2013 1,387.83

9 1,600 0.3606 576.96 3,471.28 7,471.28 0.1877 1,402.36

10 1,800 0.3220 579.60 4,050.88 8,050.88 0.1770 1,425.00

*Economic life of the machine = 7 years



The steps are summarized now:

1. Discount the maintenance costs to the beginning of year 1.

Column D = Column B1

(1 + i)n

= Column B (P/F, i, n) = Column B Column C.

2. Find the summation of present worth of maintenance coststhrough the year given (Column E = Column D).

3. Find Column F by adding the first cost of Rs. 4,000 toColumn E.

4. Find the annual equivalent total cost through the years given.

i(1 + i)n

Column H = Column F (1 + i)n − 1

= Column F (A/P, 12%, n) = Column F Column G

5. Identify the end of year for which the annual equivalenttotal cost is minimum.

For this problem, the annual equivalent total cost is minimum at the end ofyear 7. Therefore, the economic life of the equipment is seven years.



REPLACEMENT OF EXISTING ASSET WITH A NEW ASSET

In this section, the concept of comparison of replacement of an existing assetwith a new asset is presented. In this analysis, the annual equivalent cost of eachalternative should be computed first.

Then the alternative which has the least cost should be selected as the bestalternative. Before discussing details, some preliminary concepts which are essentialfor this type of replacement analysis are presented.

Capital Recovery with Return

Consider the following data of a machine.

Let

P = purchase price of the machine,

F = salvage value of the machine at the end of machine life,

n = life of the machine in years, and

i = interest rate, compounded annually


The equation for the annual equivalent amount for the abovecash flow diagram is

AE(i) = (P – F ) (A/P, i, n) + F i

This equation represents the capital recovery with return.








Let







AE(i) = (P – F ) (A/P, i, n) + F i









Let







AE(i) = (P – F ) (A/P, i, n) + F i




Concept of Challenger and Defender

o If an existing equipment is considered for replacement with a newequipment, then the existing equipment is known as the defender and thenew equipment is known as challenger.

o Assume that an equipment has been purchased about three years back forRs. 5,00,000 and it is considered for replacement with a new equipment.The supplier of the new equipment will take the old one for somemoney, say, Rs. 3,00,000.

o This should be treated as the present value of the existing equipment andit should be considered for all further economic analysis.

o The purchase value of the existing equipment before three years is nowknown as sunk cost, and it should not be considered for further analysis.

EXAMPLE

Two years ago, a machine was purchased at a cost of Rs. 2,00,000 to beuseful for eight years. Its salvage value at the end of its life is Rs. 25,000. Theannual maintenance cost is Rs. 25,000.

The market value of the present machine is Rs. 1,20,000. Now, a newmachine to cater to the need of the present machine is available at Rs. 1,50,000 tobe useful for six years. Its annual maintenance cost is Rs. 14,000. The salvagevalue of the new machine is Rs. 20,000.

Using an interest rate of 12%, find whether it is worth replacing the presentmachine with the new machine.

Solution

Alternative 1—

Present machine

Purchase price = Rs. 2,00,000Present value (P) = Rs. 1,20,000Salvage value (F) = Rs. 25,000Annual maintenance cost (A) = Rs. 25,000Remaining life = 6 yearsInterest rate = 12%

The cash flow diagram of the present machine is illustrated in Fig.



2,20,000

25,000

–2

2,00,000

–1 0

1,20,000

1 2 3 4 5 6

25,000 25,000 25,000 25,000 25,000 25,000

Fig. Cash flow diagram for alternative 1.

annual maintenance cost for the preceding periods are not shownin this figure. The annual equivalent cost is computed as

AE(12%) = (P – F)(A/P, 12%, 6) + F i + A

= (1,20,000 – 25,000)(0.2432) + 25,000 0.12 + 25,000

= Rs. 51,104

Alternative 2 —

New machine

Purchase price (P) = Rs. 1,50,000Salvage value (F) = Rs. 20,000Annual maintenance cost (A) = Rs. 14,000Life = 6 yearsInterest rate = 12%

The cash flow diagram of the new machine is depicted in Fig.

20,000

0 1 2 3 4 5 6

14,000 14,000 14,000 14,000 14,000 14,000

1,50,000

Fig. Cash flow diagram for alternative 2.



The formula for the annual equivalent cost is

AE(12%) = (P – F)(A/P, 12%, 6) + F i + A

= (1,50,000 – 20,000)(0.2432) + 20,000 0.12 + 14,000

= Rs. 48,016

Since the annual equivalent cost of the new machine is less than that of thepresent machine, it is suggested that the present machine be replaced with thenew machine.

Simple Probabilistic Model For Items Which Fail Completely

Electronic items like transistors, resistors, tubelights, bulbs, etc. could failall of a sudden, instead of gradual deterioration. The failure of the item mayresult in complete breakdown of the system. The system may contain acollection of such items or just one item, say a tubelight.

Therefore, we use some replacement policy for such items which wouldavoid the possibility of a complete breakdown.

The following are the replacement policies which are applicable for thissituation.

(i) Individual replacement policy. Under this policy, an item is replacedimmediately after its failure.

(ii) Group replacement policy. Under this policy, the following decision ismade:

At what equal intervals are all the items to be replaced simultaneously witha provision to replace the items individually which fail during a fixed groupreplacement period?

There is a trade-off between the individual replacement policy and thegroup replacement policy. Hence, for a given problem, each of the replacementpolicies is evaluated and the most economical policy is selected forimplementation. This is explained with two numerical problems.



EXAMPLE

The failure rates of transistors in a computer are summarized inTable .

Table Failure Rates of Transistors in Computers

End of week 1 2 3 4 5 6 7

Probability offailure to date

0.07 0.18 0.30 0.48 0.69 0.89 1.00

The cost of replacing an individual failed transistor is Rs. 9. Ifall the transistors are replaced simultaneously, it would cost Rs.3.00 per transistor. Any one of the following two options can befollowed to replace the transistors:

(a) Replace the transistors individually when they fail(individual replacement policy).

(b) Replace all the transistors simultaneously at fixed intervalsand replace the individual transistors as they fail in serviceduring the fixed interval (group replacement policy).

Find out the optimal replacement policy, i.e. individualreplacement policy or group replacement policy. If groupreplacement policy is optimal, then find at what equal intervalsshould all the transistors be replaced.

Solution

Assume that there are 100 transistors in use.

Let,

pi be the probability that a transistor which was new whenplaced in position for use, fails during the ith week of its life.Hence,

p1 = 0.07, p2 = 0.11, p3 = 0.12, p4 = 0.18,

p5 = 0.21, p6 = 0.20, p7 = 0.11

Since the sum of pis is equal to 1 at the end of the 7th week, thetransistors are sure to fail during the seventh week.

Assume that(a) transistors that fail during a week are replaced just before

the end of the week, and(b) the actual percentage of failures during a week for a sub-

group of transistors with the same age is same as the



expected percentage of failures during the week for that sub-group oftransistors.

LetNi = the number of transistors replaced at the end of the ith week

N0 = number of transistors replaced at the end of the week 0 (orat the beginning of the first week).

= 100

N1 = number of transistors replaced at the end of the 1st week= N0 p1 = 100 0.07 = 7

N2 = number of transistors replaced at the end of the 2nd week= N0 p2 + N1 p1= 100 0.11 + 7 0.07 = 12

N3 = N0 p3 + N1 p2 + N2 p1= 100 0.12 + 7 0.11 + 12 0.07= 14

N4 = N0 p4 + N1 p3 + N2 p2 + N3 p1= 100 0.18 + 7 0.12 + 12 0.11 + 14 0.07= 21

N5 = N0 p5 + N1 p4 + N2 p3 + N3 p2 + N4 p1= 100 0.21 + 7 0.18 + 12 0.12 + 14 0.11 + 21 0.07= 27

N6 = N0 p6 + N1 p5 + N2 p4 + N3 p3 + N4 p2 + N5 p1= 100 0.2 + 7 0.21 + 12 0.18 + 14 0.12 + 21 0.11 + 27 0.07= 30

N7 = N0 p7 + N1 p6 + N2 p5 + N3 p4 + N4 p3 + N5 p2+ N6 p1

= 100 0.11 + 7 0.2 + 12 0.21 + 14 0.18 + 21 0.12+ 27 0.11 + 30 0.07

= 25

Calculation of individual replacement cost

7

Expected life of each transistor = ∑i=1

i pi

= 1 0.07 + 2 0.11 + 3 0.12 + 4 0.18

+ 5 0.21 + 6 0.2 + 7 0.11

= 4.39 weeks

Average No. of failures/week = 100/4.39 = 23

(approx.) Therefore,

Cost of individual replacement

= (No. of failures/week Individual replacement cost/transistor)

= 23 9 = Rs. 207.



Determination of group replacement cost

Cost of transistor when replaced simultaneously = Rs. 3

Cost of transistor when replaced individually = Rs. 9

The costs of group replacement policy for several replacementperiods are summarized in Table.

Table Calculations of Cost for Preventive Maintenance

End ofweek

A

Cost of replacing100 transistors

at a timeB (Rs.)

Cost of replacing transistorsindividually during given

replacement periodC (Rs.)

Total cost

(B + C)D (Rs.)

Averagecost/week

(D/A)E (Rs.)

1 300 7 9 = 63 363 363.00

2 300 (7 + 12) 9 = 171 471 235.50

3 300 (7 + 12 + 14) 9 = 297 597 199.00

4 300 (7 + 12 + 14 + 21) 9 = 486 786 196.50*

5 300 (7 + 12 + 14 + 21 + 27) 9 = 729 1,029 205.80

6 300 (7 + 12 + 14 + 21 + 27 + 30) 9 = 999 1,299 216.50

7 300 (7 + 12 + 14 + 21 + 27 + 30 + 25) 9 = 1,224 1,524 217.71

*Indicates the minimum average cost/week.

From Table it is clear that the average cost/week is minimumfor the fourth week. Hence, the group replacement period is fourweeks.

Individual replacement cost/week = Rs. 207Minimum group replacement cost/week = Rs. 196.50

Since the minimum group replacement cost/week is less thanthe individual replacement cost/week, the group replacementpolicy is the best, and hence all the transistors should be replacedonce in four weeks and the transistors which fail during this four-week period are to be replaced individually.



UNIT V DEPRECIATION

5.1 Introduction

Any equipment which is purchased today will not work for ever. This may bedue to wear and tear of the equipment or obsolescence of technology.

Hence, it is to be replaced at the proper time for continuance of anybusiness.

The replacement of the equipment at the end of its life involves money. Thismust be internally generated from the earnings of the equipment.

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

Thus the word “depreciation” means decrease in value of any physical assetwith the passage of time.

5.2 Methods of Depreciation

There are several methods of accounting depreciation fund. These areas follows:

1. Straight line method of depreciation

2. Declining balance method of depreciation

3. Sum of the years—digits method of depreciation

4. Sinking-fund method of depreciation

5. Service output method of depreciation

1. Straight Line Method of Depreciation

In this method of depreciation, a fixed sum is charged as thedepreciation amount throughout the lifetime of an asset such that the accumulatedsum at the end of the life of the asset is exactly equal to the purchase value of theasset.Here, we make an important assumption that inflation is absent.

LetP = first cost of the asset,F = salvage value of the assetn = life of the asset,

Bt = book value of the asset at the end of the period t,Dt = depreciation amount for the period t.



The formulae for depreciation and book value are as follows:

Dt = (P – F)/n

Bt = Bt–1 – Dt = P – t [(P – F)/n]

EXAMPLE

A company has purchased an equipment whose first cost is Rs. 1,00,000 with anestimated life of eight years. The estimated salvage value of the equipment at the end ofits lifetime is Rs. 20,000. Determine the depreciation charge and book value at the end ofvarious years using the straight line method of depreciation.

Solution

P = Rs. 1,00,000F = Rs. 20,000n = 8 years

Dt = (P – F)/n= (1,00,000 – 20,000)/8= Rs. 10,000

In this method of depreciation, the value of Dt is the same for all the years.The calculations pertaining to Bt for different values of t are summarizedin Table .

Table Dt and Bt Values under Straight line Method of Depreciation

End of year Depreciation Book value(t) (Dt) (Bt = Bt–1 – Dt)

0 1,00,000

1 10,000 90,000

2 10,000 80,000

3 10,000 70,000

4 10,000 60,000

5 10,000 50,000

6 10,000 40,000

7 10,000 30,000

8 10,000 20,000

If we are interested in computing Dt and Bt for a specific period (t),the formulae can be used. In this approach, it should be noted that thedepreciation is the same for all the periods.



EXAMPLE

Consider Example and compute the depreciation and the book value for period 5.

P = Rs. 1,00,000

F = Rs. 20,000

n = 8 years

D5 = (P – F)/n

= (1,00,000 – 20,000)/8

= Rs. 10,000 (This is independent of the time period.)

Bt = P – t (P – F)/n

B5 = 1,00,000 – 5 (1,00,000 – 20,000)/8

= Rs. 50,000

2. Declining Balance Method of Depreciation

In this method of depreciation, a constant percentage of the book valueof the previous period of the asset will be charged as the depreciationamount for the current period.

This approach is a more realistic approach, since the depreciationcharge decreases with the life of the asset which matches with theearning potential of the asset.

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

Let

P = first cost of the asset,

F = salvage value of the asset,

n = life of the asset,

Bt = book value of the asset at the end of the period t,

K = a fixed percentage, and

Dt = depreciation amount at the end of the period t.

The formulae for depreciation and book value are as follows:Dt = K Bt-1

Bt = Bt–1 – Dt = Bt–1 – K Bt–1

= (1 – K) Bt–1

.



The formulae for depreciation and book value in terms of P areas follows:

Dt = K(1 – K)t–1

P Bt = (1 – K)t

P

While availing income-tax exception for the depreciationamount paid in each year, the rate K is limited to at the most 2/n. Ifthis rate is used, then the corresponding approach is called thedouble declining balance method of depreciation.



0 1,00,000.00

8 4,194.30 16,777.22

EXAMPLEConsider Example 9.1 and demonstrate the calculations of the

declining balance method of depreciation by assuming 0.2 for K.

Solution

P = Rs. 1,00,000

F = Rs. 20,000

n = 8 years

K = 0.2

The calculations pertaining to Dt and Bt for different values of tare summarized in Table 9.2 using the following formulae:

Dt = K Bt–1

Bt = Bt–1 – Dt

Table Dt and Bt according to Declining Balance Method of Depreciation

End of year Depreciation Book value(n) (Dt) (Bt)

1 20,000.00 80,000.002 16,000.00 64,000.003 12,800.00 51,200.004 10,240.00 40,960.005 8,192.00 32,768.006 6,553.60 26,214.407 5,242.88 20,971.52

If we are interested in computing Dt and Bt for a specificperiod t, the respective formulae can be used.

EXAMPLE Consider Example 9.1 and calculate the depreciation and thebook value for period 5 using the declining balance method of depreciation byassuming 0.2 for K.

SolutionP = Rs. 1,00,000

F = Rs. 20,000

n = 8 years

K = 0.2

Dt = K(1 – K)t –1 P

D5 = 0.2(1 – 0.2)4 1,00,000

= Rs. 8,192



Bt = (1 – K)t P

B5 = (1 – 0.2)5 1,00,000

= Rs. 32,768

3. Sum-of-the-Years-Digits Method of Depreciation

In this method of depreciation also, it is assumed that the book value of theasset decreases at a decreasing rate. If the asset has a life of eight years, first thesum of the years is computed as

Sum of the years = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8= 36 = n(n + 1)/2

The rate of depreciation charge for the first year is assumed as the highest and then itdecreases. The rates of depreciation for the years 1–8, respectively are as follows:8/36, 7/36, 6/36, 5/36, 4/36, 3/36, 2/36, and 1/36.

For any year, the depreciation is calculated bymultiplying the corresponding rate of depreciation with (P – F).

Dt = Rate (P – F)

Bt = Bt–1 – Dt

The formulae for Dt and Bt for a specific year t areas follows:

Dt =n − t + 1

n(n + 1)/2(P – F)

Bt = (P – F)(n − t)

n

(n − t + 1)+ F

(n + 1)

EXAMPLEConsider Example 9.1 and demonstrate the calculations of the

sum-of-the-years-digits method of depreciation.

SolutionP = Rs.1,00,000

F = Rs.20,000

n = 8 years

Sum = n (n + 1)/2 = 8 9/2 = 36

The rates for years 1–8, are respectively 8/36, 7/36, 6/36, 5/36, 4/36,3/36,2/36 and 1/36.

The calculations of Dt and Bt for different values of t are summarized in



0 1,00,000.00

7 4,444.44 22,222.268 2,222.22 20,000.04

t

Table using the following formulae:

Dt = Rate (P – F)

Bt = Bt–1 – Dt

Table Dt and Bt under Sum-of-the-years-digits Method of Depreciation

End of year Depreciation Book value(n) (Dt) (Bt)

1 17,777.77 82,222.232 15,555.55 66,666.683 13,333.33 53,333.354 11,111.11 42,222.245 8,888.88 33,333.366 6,666.66 26,666.70

If we are interested in calculating Dt and Bt for a specific t, then the usageof the formulae would be better.

EXAMPLE 9.6 Consider Example 9.1 and find the depreciation and bookvalue for the 5th year using the sum-of-the-years-digits method of depreciation.

Solution

P = Rs. 1,00,000

F = Rs. 20,000

n = 8 years

Dt =n − t + 1

n(n + 1)/2(P – F)

D5 =8 − 5 + 1

8(8 + 1)/2(1,00,000 – 20,000)

= Rs. 8,888.88

B = (P – F)n − t

n

n − t + 1+ F

n + 1

B5 = (1,00,000 – 20,000)8 − 5

8

8 − 5 + 1

8 + 1+ 20,000

= 80,000 (3/8) (4/9) + 20,000

= Rs. 33,333.33



4. Sinking Fund Method of Depreciation

In this method of depreciation, the book value decreases at increasing rates withrespect to the life of the asset

Let

P = first cost of the asset,F = salvage value of the asset,n = life of the asset,

i = rate of return compounded annually,

A = the annual equivalent amount,

Bt = the book value of the asset at the end of the period t, and

Dt = the depreciation amount at the end of the period t.

The loss in value of the asset (P – F) is made available an the form of cumulativedepreciation amount at the end of the life of the asset by setting up an equaldepreciation amount (A) at the end of each period during the lifetime of the asset.

A = (P – F) [A/F, i, n]

The fixed sum depreciated at the end of every time period earns an interest atthe rate of i% compounded annually, and hence the actual depreciation amountwill be in the increasing manner with respect to the time period. A generalizedformula for Dt is

Dt = (P – F) (A/F, i, n) (F/P, i, t – 1)

The formula to calculate the book value at the end of period t is

Bt = P – (P – F) (A/F, i, n) (F/A, i, t)

The above two formulae are very useful if we have to calculate Dt andBt for any specific period.

If we calculate Dt and Bt for all the periods, then the tabular approachwould be better.



EXAMPLE

Consider Example and give the calculations regarding the sinking fundmethod of depreciation with an interest rate of 12%, compounded annually.

SolutionP = Rs.1,00,000F = Rs.20,000n = 8 yearsi = 12%

A = (P – F) [A/F, 12%, 8]= (1,00,000 – 20,000) 0.0813= Rs. 6,504

In this method of depreciation, a fixed amount of Rs. 6,504 will bedepreciated at the end of every year from the earning of the asset.

The depreciated amount will earn interest for the remaining period of lifeof the asset at an interest rate of 12%, compounded annually.

For example, the calculations of net depreciation for some periods are asfollows:

Depreciation at the end of year 1 (D1) = Rs. 6,504.

Depreciation at the end of year 2 (D2) = 6,504 + 6,504 0.12

= Rs. 7,284.48



End of yeart

Fixeddepreciation

Net depreciationDt

Book valueBt

(Rs.) (Rs.) (Rs.)

0 6,504 – 1,00,000.00

7 6,504 12,837.74 34,381.078 6,504 14,378.27 20,002.80

Depreciation at the end of the year 3 (D3)

= 6,504 + (6,504 + 7,284.48) .12

= Rs. 8,158.62

Depreciation at the end of year 4 (D4)

= 6,504 + (6,504 + 7,284.48 + 8,158.62) 0.12

= Rs. 9,137.65

These calculations along with book values are summarized in Table .

Table Dt and Bt according to Sinking Fund Method of Depreciation

1 6,504 6,504.00 93,496.002 6,504 7,284.48 86,211.523 6,504 8,158.62 78,052.904 6,504 9,137.65 68,915.255 6,504 10,234.17 58,681.086 6,504 11,462.27 47,218.81

Bt = Bt–1 – Dt

EXAMPLE

Consider Example and compute D5 and B7 using thesinking fund method of depreciation with an interest rate of 12%,compounded annually.


F = Rs. 20,000

n = 8 yearsi = 12%

Dt = (P – F) (A/F, i, n) (F/P, i, t – 1)

D5 = (P – F) (A/F, 12%, 8) (F/P, 12%, 4)

= (1,00,000 – 20,000) 0.0813 1.574

= Rs. 10,237.30

This is almost the same as the corresponding value given inthe table. The minor difference is due to truncation error.

Bt = P – (P – F) (A/F, i, n) (F/A, i, t)

B7 = P – (P – F) (A/F, 12%, 8) (F/A, 12%, 7)

= 1,00,000 – (1,00,000 – 20,000) 0.0813 10.089

= 34,381.10



5. Service Output Method of Depreciation

In some situations, it may not be realistic to compute depreciationbased on time period. In such cases, the depreciation is computedbased on service rendered by an asset. Let

P = first cost of the asset

F = salvage value of the asset

X = maximum capacity of service of the asset during its lifetime

x = quantity of service rendered in a period.

Then, the depreciation is defined per unit of service

rendered: Depreciation/unit of service = (P – F)/X

Depreciation for x units of service in a period = P − F (x)X

EXAMPLE

The first coat of a road laying machine is Rs. 80,00,000. Its salvage valueafter five years is Rs. 50,000. The length of road that can be laid by themachine during its lifetime is 75,000 km. In its third year of operation, thelength of road laid is 2,000 km. Find the depreciation of the equipment for thatyear.


F = Rs. 50,000

X = 75,000 km

x = 2,000 km

Depreciation for x units of service in a period =

P − Fx

X

Depreciation for year 3 = (80,00,000 − 50,000)

75,000

= Rs. 2,12,000



5.3 Evaluation Of Public AlternativesIn evaluating alternatives of private organizations, the criterion is to select thealternative with the maximum profit.

The profit maximization is the main goal of private organizations whileproviding goods/services as per specifications to their customers.

But the same criterion cannot be used while evaluating public alternatives.

Examples of some public alternatives are

constructing bridges,

roads, dams,

establishing public utilities, etc.

The main objective of any public alternative is to provide goods/services to thepublic at the minimum cost. In this process, one should see whether the benefits of thepublic activity are at least equal to its costs.

5.4 Inflation Adjusted Decisions

Inflation is the rate of increase in the prices of goods per period. So, it has acompounding effect. Thus, prices that are inflated at a rate of 7% per year will increase7% in the first year, and for the next year the expected increase will be 7% of these newprices.

If economic decisions are taken without considering the effect of inflation intoaccount, most of them would become meaningless and as a result the organizationswould end up with unpredictable return.

5.5 Procedure To Adjust Inflation

A procedure to deal with this situation is summarizednow.

1. Estimate all the costs/returns associated with an investment proposal in terms oftoday’s rupees.

2. Modify the costs/returns estimated in step 1 using an assumed inflation rate so thatat each future date they represent the costs/returns at that date in terms of therupees that must be expended/received at that time, respectively.

3. As per our requirement, calculate either the annual equivalent amount or futureamount or present amount of the cash flow resulting from step 2 by consideringthe time value of money.



WORKEDOUT PROBLEMS

Single-Payment Compound Amount

A person deposits a sum of Rs. 20,000 at the interest rate of 18% compoundedannually for 10 years. Find the maturity value after 10 years.

Solution

P = Rs. 20,000i = 18% compounded annuallyn = 10 years

F = P(1 + i)n = P(F/P, i, n)= 20,000 (F/P, 18%, 10)= 20,000 5.234 = Rs. 1,04,680

The maturity value of Rs. 20,000 invested now at 18% compounded yearly is equalto Rs. 1,04,680 after 10 years.

Single-Payment Present Worth Amount

A person wishes to have a future sum of Rs. 1,00,000 for his son’s education after 10years from now. What is the single-payment that he should deposit now so that he gets thedesired amount after 10 years? The bank gives 15% interest rate compounded annually.

Solution

F = Rs. 1,00,000i = 15%, compounded annuallyn = 10 years

P = F/(1 + i)n = F(P/F, i, n)= 1,00,000 (P/F, 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.



. .

Equal-Payment Series Sinking Fund

A company has to replace a present facility after 15 years at an outlay of Rs. 5,00,000. Itplans to deposit an equal amount at the end of every year for the next 15 years at an interestrate of 18% compounded annually. Find the equivalent amount that must be deposited at theend of every year for the next 15 years.

Solution

F = Rs. 5,00,000n = 15 yearsi = 18%A = ?

The corresponding cash flow diagram is shown in Fig.

5,00,000

i = 18%

0 1 2 3 4 . . 15

A A A A AFig. Cash flow diagram of equal-payment series sinking fund.

A = Fi

(1 + i)n −1

= F(A/F, i, n)

= 5,00,000(A/F, 18%, 15)

= 5,00,000 0.0164

= Rs. 8,200

The annual equal amount which must be deposited for 15 years is Rs. 8,200.



1,00,00,000

.

Equal-Payment Series Present Worth Amount

A company wants to set up a reserve which will help the company to have an annualequivalent amount of Rs. 10,00,000 for the next 20 years towards its employees welfaremeasures. 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. 10,00,000i = 15%n = 20 yearsP = ?

The corresponding cash flow diagram is illustrated in Fig.

P

i = 15%

0 1 2 3 4 . 20.

10,00,000 10000001000000 100000

Fig. Cash flow diagram of equal-payment series present worth amount.

The formula to compute P is

(1 + i)n − 1P = A = A(P/A, i, n)

i(1 + i)n

= 10,00,000 (P/A, 15%, 20)

= 10,00,000 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

A bank gives a loan to a company to purchase an equipment worth Rs. 10,00,000 at aninterest rate of 18% compounded annually. This amount should be repaid in 15 yearly equalinstallments. Find the installment amount that the company has to pay to the bank.

Solution

P = Rs. 10,00,000i = 18%n = 15 yearsA = ?

The corresponding cash flow diagram is shown in Fig..

10,00,000

i = 18%

0 1 2 3 4

.. 15.

A A A A AFig. Cash flow diagram of equal-payment series capital recovery amount.

i(1 + i)n

A = P = P(A/P, i, n)(1 + i)n − 1

= 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

A person is planning for his retired life. He has 10 more years of service. He would like todeposit 20% of his salary, which is Rs. 4,000, at the end of the first year, and thereafterhe wishes to deposit the amount with an annual increase of Rs. 500 for the next 9 yearswith an interest rate of 15%. Find the total amount at the end of the 10th year of the aboveseries.

SolutionHere,

A1 = Rs. 4,000G = Rs. 500i = 15%n = 10 yearsA = ? & F = ?

The cash flow diagram is shown in Fig.

i = 15%

0 1 2 3 4

4,0004,000 + 500

4,000 + 1,0004,000 + 1,500

4,000 + 4,500

Fig. Cash flow diagram of uniform gradient series annual equivalent amount.

A = A1 + G(1 + i)n − in − 1

i(1 + i)n − i

= A1 + G(A/G, i, n)

= 4,000 + 500(A/G, 15%, 10)

= 4,000 + 500 3.3832

= Rs. 5,691.60

This is equivalent to paying an equivalent amount of Rs. 5,691.60 at the end of everyyear for the next 10 years. The future worth sum of this revised series at the end of the10th 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.



Effective Interest Rate

A person invests a sum of Rs. 5,000 in a bank at a nominal interest rate of 12% for 10years. The compounding is quarterly. Find the maturity amount of the deposit after 10years.

SolutionP = Rs. 5,000n = 10 yearsi = 12% (Nominal interest rate)

F = ?

METHOD 1No. of interest periods per year = 4No. of interest periods in 10 years = 10 4 = 40Revised No. of periods (No. of quarters), N = 40Interest 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 2No. 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



UNIT I INTRODUCTION TO ECONOMICS

PART – A (2 MARKS)

1. What is elasticity of Demand?Elasticity of demand may be defined as the degree of responsiveness of quantity

demanded to a Change in price.

2. Define the term `cost’?Cost may be defined as a total of all expenses incurred, whether paid of

outstanding in the manufacture and sale of a product.

3. What is opportunity cost?Opportunity cost may be defined as the potential benefit that is given up as you

seek an alternative course of action. In other words, the expected return or benefit forgone in rejecting one course of action for another.

4. What do you mean by marginal cost?The institute of cost & works Accountants of India defined marginal cost as, “the

amount at any given volume of output by which aggregate cost are changed, if thevolume of output is increased or decreased by one unit.

5. Explain marginal costing?Marginal costing is defined by the ICWA as, “the ascertainment by

differentiating between fixed costs, of marginal costs and of the effect on profit of changesin volume or type of output”.

6. What is meant by marginal revenue?The revenue that can be obtained from selling one more unit of product is called

marginal revenue.

7. Give a short note on sunk cost?A cost which was incurred or sunk in the past and is not relevant to the particular

decision making is a sunk cost or sunk loss. It may be variable or fixed or both.

8. List out the elements of cost?The elements of cost are:

V Materials Labor cost Expenses

9. Define the term costing?Institute of costs and Management Accountants, (I.C.M.A) London has defines

costing as the ascertainment of costs. “it refers to the techniques and process ofascertaining costs and studies the principles and rules concerning the determination ofcosts of products and services” .



10. What is Break-even point?

The Break-even point is, therefore, the volume of output at which neither a profitis made nor a loss is incurred. It is a point where the total sales are equal to total cost.

11. Define P/V ratio.

Profit-Volume ratio expressed as a percentage indicates the relative profitability ofdifferent products

PART: B (16 MARKS)

1. Explain in detail about flow in an economy?2. Explain the concept of law of supply and demand with suitable example?3. Briefly explain about element of cost and its classification?4. Explain the concept of break even analysis with clear diagram?5. Briefly explain about process planning and its various types?



12. What do you mean by `Make or Buy Decisions? Make or buy decision is a determination whether to produce a component

part internally or to buy it from an outside supplier. The Organization should evaluate the costs and benefits of

manufacturing a product or product component against purchasing it andthen select the alternative which results in the lower cost.

13. What are the different approaches followed in make or buy decision?The following are the approaches followed in make or buy decision. Simple cost analysis Economic analysis Break-even analysis

14. What is mean by value analysis/value engineering? Value analysis is a special type of cost reduction technique. It critically investigates and analyses the different aspects of materials,

design, cost and production of each and every component of the product inproduce it economically without decreasing its utility, function orreliability.

15. What do you mean by value of a product?Value differs from both price and cost in the sense that it is the cost proportionate

to the function. We can express value mathematically as Value =function or utility/ cost



16. Explain `function’.Function specifies the purpose of the product or what the product does, what is its

utility etc.

17. What are the different types of values? Cost value Exchange value Use value Esteem value

18. What are the various functions of a product?

Functions can be classified into the following three categories:

Primary functions Secondary functions Tertiary functions

19. Write any four objectives of value analysis. Reduce the cost of the product Simplify the product Use (new) cheaper and better materials Modify and improve product design so as to make it acceptable to

consumer

20. List any four advantage of value engineering. Value engineering/analysis identifies and reduces the product cost. It modifies and improve the product design It increases the performance/utility of the product by economical means. It helps to generate new ideas.

PART :B (16 MARKS)

1. Explain in details about criteria for make or buy decision and its approaches? (Also seeproblems)

2. Problems in single –payment compound amount method?3. Problems in single payment present worth factor?4. Problems in equal payment series sinking fund factor method?5. Problems in equal payment series present worth factor method?6. Problems in equal payment series capital recovery factor method?



UNIT-III CASH FLOW


21. What is revenue dominated cash flow?The profit/revenue, salvage value of all inflows to an organization will be assigned

with positive sign and the cost outflows will be assigned with negative sign is calledrevenue dominated cash flow.

22. What is cost of dominated cash flow? The cost outflow will be assigned with positive sign and profit, revenue

salvage value all inflows etc,. Will be assigned with negative sign is called cost dominated cash flow.

23. Mention the various rate of return method. Internal rate of return(IRR) Average rate of return(ARR) Net present value method (NPV) Pay-back period (PBP)

24. What is rate of return?Rate of return is the break-even interest rate, I, which equates the present worth of

a project’s cash outflows to the present worth its cash inflow

25. What is present worth method? The present worth measures the surplus in an investments project at time zero (0). The present worth of all cash inflows is computed the present worth of all cash

outflows associated with an investment of project is called present worth method.

PART: B (16 MARKS)

1. Problems in present worth method (Revenue dominated cash flow diagram)2. Problems in future worth method (Revenue dominated cash flow diagram)3. Problems in Annual equivalent method (Revenue dominated cash flow diagram)4. Problems in Annual equivalent method (cost dominated cash flow diagram)5. Problems in rate of return method



UNIT-IV REPLACEMENT AND MAINTENANCE ANALYSIS


26. What is future worth analysis? Net future worth measures the surplus at time period other then 0. Future worth analysis is particularly useful in an investment situation where we

need to compute the equivalent worth of a project at the end of its investmentperiod.

27. What is annual equivalent method?The criterion provide a basis for measuring investment worth by determining equal

payments on an annual basis is called annual equivalent method.

28. What is Replacement analysis? Replacement analysis involves the Replacement of existing obsolete or worn-out

assets in order to avoid failure in operations. The problems often faced by management of various industries are whether to

replace equipment or to continue to use existing equipment, and when existingequipment should be replaced with efficient equipment.

This class of decision analysis is known as replacement analysis.

29. What is mean by gradual failure? Gradual failure is progressive in nature. It can be depicted in machine equipment, which is gradually depreciating or

deteriorating with the time resulting in very high operating and maintenance costand/or decreased residual value.

It is easier to predict such type of failure and taking the corrective measures byproviding a replacement policy for such machine equipment.

30. Define economic service life of an asset?The economic service life of an asset is defined to be the period of useful life that

minimizes the annual equivalent cost of owning and operating the asset.

31. What are the types of replacement problem?

(a) Replacement of assets that deteriorate with time (replacement due to gradualfailure, or were and tear of the components of the machine).

This can be further classified in to the following types:1. Determination of economic life an asset2. Replacement of an asset with a new asset.

(b)Simple probabilistic model for assets which fails completely. (Replacement dueto sudden failure).

32. Explain annual equivalent total cost.

Annual equivalent total cost of owning and operating an asset is asummation of the capital recovery cost (average first cost) and the annual equivalentoperating cost of the asset.



33. Name the types of maintenance. Corrective or Breakdown maintenance. Scheduled maintenance Preventive maintenance and Predictive maintenance.

34. State the main causes of breakdown. Failure to replace worn out parts Lack of lubrication Indifference towards minor faults

35. State any two disadvantages of break down maintenance.

1. Delays in production2. Faster plant deterioration

36. Explain Predictive maintenance. It is comparatively a newer maintenance technique. Equipment conditions are measured periodically or on a continuous basis

and this enable maintenance men to take a timely action such asequipment adjustments repair or overhaul.

UNIT-V DEPRECIATION


37. Define the term “Depreciation”.

Depreciation is the process of allocating the acquisition cost of the tangible assetsless salvage value, if any, in a systematic and a rational manner over the estimated life ofthe asset.

38. Mention the various method used in depreciation calculation.

The various methods used in depreciation calculation are:1. Straight line method2. Declining method3. Sum of the years digits method4. Sinking fund or annuity method5. Service output method

39. What is service output method of depreciation?

Service output method of depreciation is a type of computing depreciation base onservice rendered by an asset



40. What are the causes of depreciation?

The causes of depreciation are: Wear and tear Depletion Obsolescence Lapse of time

41. Write five reasons for providing depreciation.

The reasons for providing depreciations are: To know the correct profits To show correct financial position. To make provision for replacement of asset To compute tax liability To decide for show much to buy or sell the assets in the second hand

market

42. What is evaluation of public alternatives?

Evaluation of public alternatives is nothing but the selecting of best alternativefrom the available alternatives.

43. Define the term inflation? Inflation may be defined as a sustained in the general price level. It is an economic condition where there is a rise in prices resulting in the

fall in the purchasing power of money

44. What is sinking fund? A depreciation fund equal to be actual loss in the value of the asset is

estimated for each year. This amount is invested outside the business in a separate account called

sinking fund.

45. What is amortization? Amortization is a routine decrease in value of an intangible asset, or the

process of paying off a debt over time through regular payments. Amortization refers to the expensing of intangible capital assets (intellectual

property: patents, trademarks, copyrights. Etc,) in order to show theirdecrease in value as a result of use or passage of time.

PART – B (16 MARKS)

1. Problems in different types of depreciation methods (16)

2. Problems in inflation adjusted decision (16)

3. Problems in finding the economic life of an asset (16)



4. (a) (i) How to adjust inflation in evaluating public alternatives? Explain the procedure. (8)(ii)Find the depreciation annuity by annuity method after three years, when the initial cost

of the machine is Rs 8,00,000 an salvage value at the end of three years is Rs 4,00,000.Rate ofinterest 10 % (8)

5. (i) What is economic life of an asset? How to determine it? Explain(ii) The cost of a machine is Rs 1,60,000 and its scrap value is Rs 40,000 .Estimate life 5

years .Using sum of years digits method ,determine depreciation charges for each year. (16)

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