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
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
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
X1
of
Size
X2
of
Size
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)
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
/
cost
Variable
unit
price/
Selling
cost
Fixed
unit
/
cost
Variable
unit
price/
Selling
cost
Fixed
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
sales
x
on
contributi
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
on
Contributi
Sales
costs
Variable
-
Sales
ratio
P/V
cost
Fixed
ratio
P/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
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
)
i
1
(
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
)
i
1
( 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
)
i
1
( 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
)
i
1
(
i
1
)
i
1
(
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
)
i
1
(
)
i
1
(
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
)
i
1
(
i
1
)
i
1
(
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
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
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
)
i
1
(
i
1
)
i
1
(
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
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
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
)
i
1
(
)
i
1
(
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
)
i
1
(
)
i
1
(
i
n
n
1
)
i
1
(
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
