The document discusses various topics related to engineering economics and cost analysis. It begins by defining economics according to prominent economists like Robbins and Marshall. It then covers microeconomics, macroeconomics, economic goals and the flow in an economy. The document defines engineering economics and discusses principles, analysis procedures, and scope. It also covers concepts like cost elements, types of costs, cost estimating models, inflation, break even analysis, and value engineering. Finally, it discusses time value of money and interest calculations.

1. ENGINEERING
ECONOMICS AND COST
ANALYSIS
G.K.Manikandan
Professor, Department of Mechanical Engineering,
SSM College of Engineering
Komarapalayam, Namakkal District - 638183

2. ECONOMICS
 Prof. Lionel Robbins
 It studies human behaviour as a relationship between
ends and scarce means
 Alfred Marshall
 A study of man kind in the ordinary business of life

3. ECONOMICS
Economic Resources
5 M’s
Men, Money, Machine, Materials and Methods
 Land : (Water, Air, Mineral, Sunshine, Plants and Trees)
 Labour : (Highly , Semi & Unskilled)
 Capital : ( Real or Physical – M/c tools, Buildings)
(Financial , Human)

4. MICRO AND MACRO ECONOMICS
 Micro economics
 Study of particular firms, particular households,
individual wages, incomes, individual industries,
particular commodities
• Deals with Product Pricing, Factor pricing and
economic growth

5. MICRO AND MACRO ECONOMICS
 Macro economics
 Study of overall conditions like total production, income,
saving and investment in the economy
 Deals with production, consumption of goods , services and
the distribution of these for human welfare
• Deals with
• Theory of income, output & employment
• Theory of prices and economics of growth
• Macro theory of distribution

6. ECONOMICS
ECONOMIC GOALS
 High level of employment
 Price stability
 Efficiency
 Equitable distribution of income
 Growth

7. FLOW IN AN ECONOMY
BUSINESS TO HOUSE HOLDS
 Delivers goods, services
 Pay money for resources, rent, wages, salaries,
interest & profit
HOUSEHOLDS TO BUSINESS
 Pay the money for goods and services
 Provide Economic resources: (Land,
Labour,Capital)

8. ENGINEERING ECONOMICS
 DEFINITION:
 A set of principles, concepts , techniques & methods by which
alternatives within a project can be compared and evaluated for the
best
 PRINCIPLES OF ENGINEERING ECONOMICS
 Develop Alternatives
 Focus on differences
 Use Consistent view point
 Use a common unit of measure
 Consider all relevant criteria
 Make uncertainity explicit
 Revisit your decisions

9. ENGINEERING ECONOMICS
 ENGINEERING ECONOMICS ANALYSIS & PROCEDURE
 Problem Recognition, formulation & evaluation
 Development of alternatives
 Development of cash flow of each alternative
 Selection criteria
 Analysis and comparison of alternatives
 Selection of prepared alternatives
 Performance monetary & post evaluating results

10. ENGINEERING ECONOMICS
 SCOPE OF ENGINEERING ECONOMICS
 It is concerned with monetary consequences or financial
analysis of the products, project and processes that engineer’s
design
 It helps an engineer to evaluate and compare the overall cost of
available alternatives in the 3 P’s
 It helps to take the best decision among various alternatives in
an economical way

11. ENGINEERING ECONOMICS
 SCOPE OF ENGINEERING ECONOMICS
 Theses concepts are used in fields for improving production
reducing, human efforts increasing wealth by reducing the cost
 It helps in understanding the market condition, economic
environment in which the frirm is working
 It enacts as a basis for resource allocation

12. LAW OF SUPPLY & DEMAND
 Supply and Demand are independent
 They are functions of price
FACTORS INFLUENCING
 DEMAND
 Income
 Price of related goods
 Tastes of consumers
 SUPPLY
 Cost of inputs Technology Weather Price of related goods

13. EFFICIENCY
TECHNICAL O/I x 100
η B.thermal = (B.P/ Heat supplied) x 100
ECONOMIC ( PRODUCTIVITY)
(Worth/cost) x 100
Worth : Annual revenue generated from the
business
Cost: Total expenses incurred in carrying out the
business

14. WAYS OF IMRPOVING PRODUCTIVITY
 Increased o/p for the same i/p
 Altering the lay out
 Decrease i/p for the same o/p
 Alternate raw material
 Proportionate increase in o/p which is more than the
increase in i/p
 Introducing a new product
 Proportionate decrease in i/p which is more than o/p
 Dropping a product from the list
 Simultaneous increase in o/p with decrease in i/p
 Using advance technologies like AGV, Robot etc.,

15. WAYS OF IMRPOVING PRODUCTIVITY
 What does engineering economics deals with?
 Methods that enable one to take economic decision
towards
 Minimizing costs
OR
 Maximizing benefits

16. ELEMENTS OF COST
MATERIAL LABOUR EXPENSE
Direct
Material
In Direct
Material
Direct
Labour
In Direct
Labour
Direct
Expense
In Direct
Expense
DIRECT INDIRECT
MATERIAL MATERIAL
LABOUR PRIME LABOUR OVERHEAD
EXPENSE EXPENSE
Production Administration Selling &
Distribution

17. ELEMENTS OF COST
 Variable
 Varies with volume of production
 Overhead cost
 It is fixed and does not depend on volume of production
Variable Overhead cost
Direct material Indirect material
Direct labour Indirect labour
Direct Expenses Indirect Expenses

18. ELEMENTS OF COST
Direct
Material
Direct
labour
Direct
Expense
Indirect
Material
Indirect
Labour
InDirect
Expense
Raw
Material
(Shaft, Pin,
bush etc.,)
Machinist
Turner
Welder
Hire
charges,
Design,
Special
pattern
Grease,
coolant,
Cotton
Waste
Store
keeper,
Security
Rent, Power,
Advertising
PRIME COST = D.M + D.L + D.E
FACTORY COST = PRIME + FACTORY OVER HEAD
PRODUCTION COST = FACTORY COST + OFF & ADMIN. EXP
COST OF GOODS SOLD = PRODUCTION COST +( Opening finished
stock – Closing finished stock)
SALES COST = COST OF GOODS SOLD + SELLING &
DISTRIBUTION OVER HEAD
SALES = COST OF SALES + PROFIT
SELLING PRICE /UNIT = SALES/QTY SOLD

19. PRIME COST = D.M + D.L + D.E
FACTORY COST = PRIME + FACTORY OVER HEAD
PRODUCTION COST = FACTORY COST + OFF & ADMIN. EXP
COST OF GOODS SOLD = PRODUCTION COST +(Opening
finished stock – Closing finished stock)
SALES COST = COST OF GOODS SOLD + SELLING &
DISTRIBUTION OVER HEAD
SALES = SALES COST + PROFIT
SELLING PRICE /UNIT = SALES/QTY SOLD

20. FACTORY OVERHEAD / WORKS COST/ MANUFACTURING COST
 Indirect material
 Indirect Wages
 Factory rent
 Fac. Lighting and heating
 Power & fuel
 Repairs & maintenance
 Research & Experiment cost
 Depriciation of factory plant
 Stationery
 Insurance
 Managers salary

21. OFFICE AND ADMINISTRATIVE OVERHEAD
 Office salaries
 Office rent
 Lighting & heating
 Cleaning
 Telephone & postages
 Printing & Stationery
 Depreciation of office furniture
 Depreciation of office equipment
 Insurance
 Legal expenses

22. SELLING AND DISTRIBUTION OVER HEADS
 Advertising
 Salesmen salaries
 Samples & free gifts
 Sales office rent
 Sales promotion expenses
 Packing & demonstration
 Shoe room rent
 Commission
 Travelling Expenses
 Ware house rent
 Repair & maintenance of delivery van
 Carriage freight outwards etc.,

23. OTHER COSTS
 MARGINAL COST /MARGINAL REVENUE:
Cost of producing an extra unit/ Revenue by selling an extra unit
 AVERAGE COST
 Total cost of producing a given volume/ volume of production
 SUNK COST
 Purchase value of an equipment in the past (Ex; bike)
 OPPORTUNITY COST
The return that will be fore gone by not investing the money in another altternative
 RECURRING COST
 Periodically repeated cost (Salary, rent, E.B, Maintenance, employees Welfare etc.,)
 NON RECURRING
 Initial outlay, expansion / modernization of plant etc.,

24. OTHER COSTS
 CASH COST
Cost paid by the business using cash/cheque and not credit
 BOOK COST
The money spent in buying an equipment or other facilities
(Ex: purchase of share 100 share X Rs 10 per share = Rs 1000)
 LIFE CYCLE COST
The economic cost of purchase of raw materials, procurement of components /
sub assemblies, production and assembly of components & sub assemblies,
maintenance, waste management etc.,

25. COST ESTIMATING MODELS
 PER UNIT MODEL
Estimated cost per unit X Volume of production
 SEGMENTING MODEL
Ex: Bike
 COST INDEXES
Cost of product at current point of time (P2) = P1 X (C2/C1)
Where C is the cost index value at the given time
 LEARNING CURVE MODEL
The time taken for the Nth unit of task or product
Time for completing N units, TN = T1 X Nk

26. COST ESTIMATING MODELS
 POWER SIZE MODEL
Cost of product X2 = Cost of X1
Ex: Cost of X2 = cost of 20 H.P engine
Cost of X1 = cost of 10 H.P engine
Size of X1 = 10 H.P and Size of X2 = 20 H.P
K= exponent which indicates the economy of scale
K= 1 No economy ; K> 1 diseconomy
 PRICE INDEX NUMBER
It represents the change in value of a set of related variables from one
time period to the other time period
K
X1ofSize
X2ofSize







27. INFLATION
 The rise in the price of goods and services in a given period of a
Country. It is based on CPI( consumer price index) or WPI (Whole sale price
index)
 CPI: It represents the changes in the price of a select set of goods & services
for a given consumer class. CPI has an effect on the purchasing power of
people.
 WPI: It represents the monthly average of goods sold in large quantities
 The price of goods & services change due to several reasons
Such as
 Population growth
 Excessive spending by govt., public
 Decline in industrial activities etc.,

28. BREAK EVEN ANALYSIS
Objective:
To find the cut-off production volume fro where a
firm will make profit
ie.,Total sales revenue = Total cost incurred ( At B.E pt)
S = TC
s x Q = v x Q + FC
s = S.P /unit
v = Variable cost/unit
FC = Fixed cost
Q = Volume of production
Profit = Sales – Total cost
= Sales – (Fixed + variable)
= s x Q – (v x Q + FC)

29. BREAK EVEN ANALYSIS
Profit
Loss
Total cost (TC)
Fixed Cost
Variable Cost
Sales (S)
BEP (Q*)
Production quantity
Break- even
sales

30. BREAK EVEN ANALYSIS
At Break even point
Sales = Total cost
Sales = (Fixed + variable)
s x Q = (v x Q + FC)
(B.E.Qty) Q =
Q =
B.E. Sales s = Break even Qty X Selling price per unit
x S.P/unit
S = x s (Rs)
unit/costVariableunitprice/Selling
costFixed

unit/costVariableunitprice/Selling
costFixed

v-s
FC
v-s
FC

31. BREAK EVEN ANALYSIS
Contribution = Sales – variable cost
Contribution/unit = Sales/unit – variable cost/unit
Margin of safety M.S = Actual sales – break-even sales
M.S =
M.S as a percent of sales = (M.S / Sales) X 100
salesx
oncontributi
Profit

32. PROFIT – VOLUME RATIO
It expresses the relationship of contribution to sales
P/V ratio = =
Relation b/w BEP and P/V ratio
BEP =
Relation b/w BEP and P/V ratio
M.S =
Sales
onContributi
Sales
costsVariable-Sales
ratioP/V
costFixed
ratioP/V
Profit

33. ELEMENTARY ECONOMIC ANALYSIS
 MATERIAL SELECTION FOR A PRODUCT
Cheaper raw material price
Reduced machining/process time
Enhanced durability of the product
 DESIGN SELECTION FOR A PRODUCT
Two alternatives
 DESIGN SELECTION FOR A PROCESS INDUSTRY
Ex: Storage Tank in a chemical industry
 BUILDING MATERIAL SELECTION FOR CONSTRUCTION
ACTIVITIES
Sourcing of raw material
 PROCESS PLANNING/PROCESS MODIFICATION
Sequence of operation with least cost

34. VALUE ENGINEERING

35. VALUE ENGG OR VALUE ANALYSIS
 Systematic identification & elimination of unessential or
unnecessary costs
 Why value analysis has to be carried out?
- Competitive market
 Value engg is a very effective tool of Material
Management & cost reduction
 Value analysis investigates as
 How the value of the product can be improved OR
 If a part can be replaced (or) eliminated by any other part
of the same value / lesser cost

36. VALUE ENGG OR VALUE ANALYSIS
 The main aim is to study the relation between design
functionality & cost of a part
 DEFINITION:
 The Process of objectively studying every item purchased or
manufactured to eliminate every cost factor which does not
contribute to usefulness or utility
 A technique that yields value improvement by the
determination of the essential function of a product/item and
the accomplishment of this function at minimum cost without
degradation in its quality

37. VALUE ENGG OR VALUE ANALYSIS
VALUE RATIO
Value =
TYPES OF VALUES
1. Esteem value ( Imported car)
2. Use value
3. Cost value
4. Exchange value (Gold ornaments)
OBJECTIVES OF VALUE ANALYSIS
 Reduce the cost of product
 Improve the quality of product & profit
 To modify/improve design
Cost
Function

38. VALUE ENGG OR VALUE ANALYSIS
 To ensure greater returns
 To simplify the product
 To develop logical & analytical approach to solve problems
 To promote creativity and quality awareness amongst workers
PHASES OF VALUE ANALYSIS PROCEDURE
1. Orientation
2. Information
3. Functional
4. Creation
5. Evaluation
6. Investigation
7. Recommendation & implementation

39. 5. Evaluation phase
Select the most promising ideas
Identify the ideas that are most suitable
5. Investigation phase
Find out the feasibility and limitation
Prepare a work plan to convert the ideas into proposals
Consult experts & vendors
5. Recommendation & Implementation phase

40. PHASES OF VALUE ANALYSIS PROCEDURE
1. Orientation
Select the project for study (Ex: A car/clutch/fuel injection system)
Form a team from design, sales, purchase & accounts etc.,
2. Information phase
• Collect facts (Drawings, parts, suppliers, manufacturing methods etc.,)
• Determine cost
• Fixation of cost (Segregate the specification & actual requirements)
3. Function Phase (Relates Value & function)
• Identify the part & the function it performs
• Relate its functions with cost & worth of providing them

41. PHASES OF VALUE ANALYSIS PROCEDURE
4. Creation Phase
Create ideas for alternate ways
Ask & Answer questions
What is Achieved?
Why is it essential?
How is it achieved?& Why that way?
Where does it take place & Why there?
When is it done & Why then?
Who does it & Why that man
Simple steps in creative thinking
Identification of the problem
Determination of the facts
Idea determination
Determination of solution

42. TIME VALUE OF MONEY
A rupee today is more valuable than a year ago
Capital should be employed productively to
generate greater returns
An investment of 1 rupee today would grow to
1+r after a year
To find the future worth of money
F = P x (1+i)n
F= Future amount
P = Principal amount invested at time 0
i = interest rate compounded annually
n = Period of deposit

43. INTEREST
1. SIMPLE
2. COMPOUND
NOTATIONS USED
1. P = Principal amount
2. n= No. of interest periods
3. i = interest rate ( compounded monthly, quarterly, semiannually or
annually)
4. A = Equal amount deposited at the end of interest period
5. G= Uniform amount which will be added or subtracted to/from the
amount of deposit A1 at the end of period 1

44. INTEREST
1. Single payment compound amount
F=P(1+i)n F
0 1 2 3 4 ……………………. n
P
If I deposit 20,000, what will I get say after 20 years?
2. Single payment present worth amount
If I need 2,00,000 after 10 years, how much should I deposit
today?
P = = F (P / F, i, n)n
)i1(
F


45. INTEREST
3. Equal payment series compound amount
F=A F
0 1 2 3 4 n
P Ex.,20,000 20,000 20,000 20,000
If iam paying an equal monthly investment, what
will I get after say 20 years
i
1)i1( n


46. INTEREST
4. Equal payment series sinking fund
F
0 1 2 3 4 n
A A A A
A
If i need say 5,00,000 after 10 years, how much
should i pay every equal monthly investment?
2. F = A i
1)i1( n


47. INTEREST
5. Equal payment series present worth amount
P
0 1 2 3 4 n
A A A A A
What is the single payment that a company should
reserve so that its reserve can grow annually to
meet the employees welfare measures?
P = A = A(P / A, i, n)
n
n
)i1(i
1)i1(



48. INTEREST
6. Equal payment series capital recovery amount
P
0 1 2 3 4 n
A A A A A
If a bank sanctions me a loan for machine
purchase, How much equal installment should I pay
every year?
A= P = P(A / P, i, n)
1)i1(
)i1(i
n
n



49. INTEREST
7. Uniform gradient series annual equivalent amount
A person wishes to save 20% of his salary every year . In the
subsequent years his annual increase is also deposited. How
much will he get at the of say 25 years?
0 1 2 3 4 n
A1 + 2G A1 + 3G
A = A1 + G = A1 + G(A/G,i,n)
i)i1(i
1)i1(i
n
n

 in
A1 + (n-1)G

50. SELECTING THE BEST ALTERNATIVE
 PRESENT WORTH METHOD
 FUTURE WORTH METHOD
 ANNUAL EQIVALENT METHOD
 RATE OF FRETURN METHOD

51. PRESENT WORTH METHOD

52. PRESENT WORTH METHOD
 The present worth of all the alternatives is determined
 To take a decision in selecting the best alternatives,
the cash flow diagram can be adapted
 Cost dominated Cash flow diagram
Cost/Expenditure : +ve Sign
Profit, Salvage, Revenue: -ve Sign
 Revenue/profit –dominated Cash flow diagram

53. To select an alternative with minimum cost, then the
alternative with least present amount will be selected
If the decision is to select the alternative with
Maximum profit, then the alternative with Maximum
present amount will be selected

54. COST DOMINATED CASH FLOW DIAGRAM
EXPENDITURE : +VE
REVENUE : − VE
Present Worth Cash Flow
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]
0 1 2 . j : : n
C1 C2 . Cj : : Cn
P
S

55. REVEVENUE DOMINATED CASH FLOW DIAGRAM
Present Worth Cash Flow
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]
P
S
0 1 2 3 . j n
R1 R2 R3 . Rj Rn

56. PROBLEM
 An industry wants to expands its production. It has identified
three technologies. Suggest the best one for an interest rate
of 20%
 Technology 1
P = Rs. 12,00,000
A = Rs. 4,00,000
i = 20%
n = 10
Technology Initial outlay
(Rs)
Annual
Revenue(Rs)
Life (Yrs)
1 12,00,000 4,00,000 10
2 20,00,000 6,00,000 10
3 18,00,000 5,00,000 10

57. SOLUTION
 Since the revenue is given use Revenue dominated cash flow
diagram
 The best technology is the one that yields the maximum present
worth of the technology
PW ( 20%) = − 12,00,000 + 4,00,000 x ( P/A, 20%,10)
= − 12,00,000 + 4,00,000 x (4.1925)
= RS. 4,77,000
P/A = = A(P / A, i, n)
Similarly calculate for the other two technologies and select the one with
maximum present worth
Answer: Technology 2 Rs. 5,15,500
n
n
)i1(i
1)i1(



58. PROBLEM
 An engineer has two bids for an elevator to be installed in a
new building
Determine which bid should be accepted based on present worth of
comparison assuming 15% rate compounded annually.
 Solution
Alpha elevator : PW (15%) = 4,50,000 + 27,000 ( P/A, 15%, 15)
= 4,50,000 + 27,000 (5.8474)
= Rs. 6,07,879.80
Bid Initial cost
(Rs.)
Service
Life
(Year)
Annual operations &
Maintenance cost
(Rs.)
Alpha Elevator
Inc.
4,50,000 15 27,000
Beta Elevator
Inc.
5,40,000 15 28,500

59. PROBLEM
 Investment Proposals A & B have the net cash as follows
Compare the present worth of A & B at i= 18%
 Solution
Cash flow
Proposal
End of Years
0 1 2 3 4
A (Rs) -10,000 3,000 3,000 7,000 6,000
B(Rs) -10,000 6,000 6,000 3,000 3,000
3,000 3,000 7,000 6,000
0 1 2 3 4
10,000

60. PROBLEM
 Present worth
PWA (18%) = - 10,000 + 3,000(P/F, 18%,1) + 7,000(P/F, 18%,2) + 6,000(P/F,
18%,3) + 3,000(P/F, 18%,4)
F=P(1+i)n P/F =
= - 10,000 + 3,000(0.8475) + 7,000(0.7182) + 6,000(0.6086) + 3,000(0.5158)
= Rs. 2,052.10
n
i)(1
1


61. FUTURE WORTH METHOD

62. CASH DOMINATED CASH FLOW DIAGRAM
EXPENDITURE : +VE
REVENUE : − VE
Present Worth Cash Flow
FW(i) = P(1+i)n + C1(1+i) n-1 + C1(1+i) n-2.......
Cj(1+i) n-j + ………. + Cn − S
0 1 2 . j : : n
C1 C2 . Cj : : Cn
S
P

63. REVEVENUE DOMINATED CASH FLOW DIAGRAM
Future worth Cash Flow
FW(i) = − P( 1+i)n + R1(1+i) n-1 + R2(1+i) n-2 + .......
Rj(1+i) n-j +……….+ Rn+ S
P
S
0 1 2 3 . j n
R1 R2 R3 . Rj Rn

64. PROBLEM
 Select the best alternative, if i = 18%
 Alternative A
Initial investment P = Rs. 50,00,000
Annual equal revenue A = Rs. 20,00,000
Life of alternative = 4 years
FW (18%)A = − 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
FW (18%)B = Rs. 6,61,500
End of year
Alternative 0 1 2 3 4
A (Rs.) - 50 L 20 L 20 L 20 L 20 L
B (Rs.) - 45 L 18 L 18 L 18 L 18 L

65. ANNUAL EQIVALENT
METHOD

66. REVEVENUE DOMINATED CASH FLOW DIAGRAM
Step 1: Find the Present worth of Cash Flow
PW(i) = − P + R1/(1+i) 1 + R2 / (1+i) 2 + .......
Rj / (1+i) j +……….+ Rn / (1+i) n + S / (1+i) n
P
S
0 1 2 3 . j n
R1 R2 R3 . Rj Rn

67.  Step 2: Next calculate the annual equivalent revenue
A = P = P(A / P, i, n)
(Equal payment series capital recovery factor)
1)i1(
)i1(i
n
n



68. CASH DOMINATED CASH FLOW DIAGRAM
EXPENDITURE : +VE
REVENUE : − VE
Step 1: Find the Present worth of Cash Flow
PW(i) = P + C1/(1+i) 1 + C2 / (1+i) 2 + ....... Cj / (1+i) j +……….+
Cn / (1+i) n - S / (1+i) n
0 1 2 . j : : n
S
P
C1 C2 . Cj : Cn

69.  Step 2: Next calculate the annual equivalent revenue
A = P = P(A / P, i, n)
(Equal payment series capital recovery factor)
ALTERNATE APPROACH
Step 1: Find the future worth
A = F = F(A / F, i, n)
Step 2: Calculate the annual equivalent cost/ revenue
1)i1(
)i1(i
n
n


1)i1(
i
n


70. PROBLEM
 Two possible routes for laying a power line are under
study. If 15% interest is used, should the power line be
routed around the lake?
Solution: Around the Lake
First cost = 1,50,000 x 15 = Rs 22,50,000
Maintenance cost per year = 6,000 x 15 = Rs. 90,000
Power loss/Yr = 15,000 x 15 = Rs 2,25,000
Around the lake Under the lake
Length 15 Km 5 Km
First cost (Rs.) 1,50,000/km 7,50,000/km
Useful life (Yrs) 6,000/km/yr 12,000 km/yr
Salvage value (Rs) 90,000/km 1,50,000/km
Yearly power loss 15,000/km 15,000/km

71. PROBLEM
Maintenance cost & power loss/yr = Rs 90,000 + Rs 2,25,000
= Rs. 3,15,000
Salvage value = 90,000 x 15 = Rs 13,50,000
Cash flow diagram
13,50,000
i=15%
3,15,000 3,15,000 3,15,000 ……… 3,15,000
0 1 2 3 15
22,50,000

72. PROBLEM
The annual equivalent cost
AE1(15%) = 22,50,000(A/P, 15%,15) + 3,15,000 – 13,50,000 (A/F, 15%, 15)
= 22,50,000 (0.1710) + 3,15,000 – 13,50,000 (0.0210)
= Rs 6,71,,400