Engineering Economics for Civil Engineering.pdf
kiranmayuri
60 views
47 slides
Feb 20, 2025
Slide 1 of 47
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
About This Presentation
Engineering Economics
Slide Content
Unit no. 2
Engineering Economics
Mr. Kiran R. Patil
Assistant Professor,
Department of Civil Engineering,
D. Y. Patil College of Engineering & Technology, Kolhapur
Engineering Economics
Mr. Kiran R. Patil
Assistant Professor,
Department of Civil Engineering,
D. Y. Patil College of Engineering & Technology, Kolhapur
Introduction
•Engineering Economics is the economics applied to engineering problems. Engineering
Economics is a collection of mathematical techniques which simplifies economic
comparisons.
•Engineering Economics is based on economic analysis for evaluating various alternatives
using the economic concepts.
•Engineering Economics determines the economic factors and the economic criteria used
when one or two alternatives are considered for selection.
Importance of Engineering Economics
•In many cases when a specific task is to be performed there are alternative ways to
accomplish the task. The information for each alternative can be expressed quantitatively in
terms of rupee incomes and expenditures.
•When engineering alternatives involves capital investments for equipment, materials and
labour, the techniques of economic analysis may be used to assist in determining which is
the best alternative.
•Usually the monetary values are future estimates that would result if each alternative were
selected for implementation. These estimates are based on facts, experience, judgment and
comparison with similar projects.
•In many cases an engineer, rather than an economist, accountant, banker or tax expert,
performs the analysis, because the technical details are already known by the engineer.
• Economic analysis and engineering economy are extremely important for the engineer in
his professional, as well as personal life, for evaluating alternatives with respect to
investments, purchases, etc.
Mr. Kiran Patil 2
•Engineering Economics is the economics applied to engineering problems. Engineering
Economics is a collection of mathematical techniques which simplifies economic
comparisons.
•Engineering Economics is based on economic analysis for evaluating various alternatives
using the economic concepts.
•Engineering Economics determines the economic factors and the economic criteria used
when one or two alternatives are considered for selection.
Importance of Engineering Economics
•In many cases when a specific task is to be performed there are alternative ways to
accomplish the task. The information for each alternative can be expressed quantitatively in
terms of rupee incomes and expenditures.
•When engineering alternatives involves capital investments for equipment, materials and
labour, the techniques of economic analysis may be used to assist in determining which is
the best alternative.
•Usually the monetary values are future estimates that would result if each alternative were
selected for implementation. These estimates are based on facts, experience, judgment and
comparison with similar projects.
•In many cases an engineer, rather than an economist, accountant, banker or tax expert,
performs the analysis, because the technical details are already known by the engineer.
• Economic analysis and engineering economy are extremely important for the engineer in
his professional, as well as personal life, for evaluating alternatives with respect to
investments, purchases, etc.
Mr. Kiran Patil 2
•Engineering Economics deals with various important factors like site selection, planning,
equipment selection, labour payments, transportation systems, etc.
•Engineering Economics helps in understanding the general economic environment in which
the organization is working.
•Engineering Economics provides the basis for resource allocation.
•Engineering Economics is concerned with the decision making of engineering problems of
economic nature.
•Engineering Economics helps in controlling and reducing the costs.
• Engineering Economics helps in maximizing the profits.
•Engineering Economics helps in increasing production and productivity
Principles of Engineering Economics
•Economy studies are concerned with making comparisons between a number of alternative
ways of investing resources with a view to select the one which will give the optimal future
return of the investment.
•The alternatives usually involve purchase cost, the anticipated life of the asset, annual
maintenance and operating cost, the anticipated resale value (salvage value) and the rate of
return.
•In order to compare different alternatives, it is necessary to have an evaluation criterion that
can be used as a basis for judging the alternatives.
•In economic analysis, rupees are generally used as the basis for comparison.
Mr. Kiran Patil 3
equipment selection, labour payments, transportation systems, etc.
•Engineering Economics helps in understanding the general economic environment in which
the organization is working.
•Engineering Economics provides the basis for resource allocation.
•Engineering Economics is concerned with the decision making of engineering problems of
economic nature.
•Engineering Economics helps in controlling and reducing the costs.
• Engineering Economics helps in maximizing the profits.
•Engineering Economics helps in increasing production and productivity
Principles of Engineering Economics
•Economy studies are concerned with making comparisons between a number of alternative
ways of investing resources with a view to select the one which will give the optimal future
return of the investment.
•The alternatives usually involve purchase cost, the anticipated life of the asset, annual
maintenance and operating cost, the anticipated resale value (salvage value) and the rate of
return.
•In order to compare different alternatives, it is necessary to have an evaluation criterion that
can be used as a basis for judging the alternatives.
•In economic analysis, rupees are generally used as the basis for comparison.
Mr. Kiran Patil 3
Financial Analysis
•Financial Analysis is required to ascertain whether the proposed project will be financially
feasible and whether the project will satisfy the return expectations of those who provide the
capital.
• The aspects to be considered while doing financial analysis are,
1.Investment and cost of the project
2.Means of financing
3.Cost of capital
4.Projected profitability
5.Break-even point
6.Cash-flows
7.Level of risk
Time Value of Money
•Money has time value. Time-value is a relationship between the value of money today and
its value at some future date, considering the interest charged or paid for the use of money.
•If we invest money today in a bank, after certain period we will get more money than we
had invested. This change in the amount of money over a given time period is called time-
value of money. The increase in the time-value of money is due to interest. Interest is a
measure of the increase between the original sum borrowed or invested and the final amount
to be paid or obtained.
•If you invest money at some time in the past,
• Interest = (total amount accumulated - original investment)
Mr. Kiran Patil 4
•Financial Analysis is required to ascertain whether the proposed project will be financially
feasible and whether the project will satisfy the return expectations of those who provide the
capital.
• The aspects to be considered while doing financial analysis are,
1.Investment and cost of the project
2.Means of financing
3.Cost of capital
4.Projected profitability
5.Break-even point
6.Cash-flows
7.Level of risk
Time Value of Money
•Money has time value. Time-value is a relationship between the value of money today and
its value at some future date, considering the interest charged or paid for the use of money.
•If we invest money today in a bank, after certain period we will get more money than we
had invested. This change in the amount of money over a given time period is called time-
value of money. The increase in the time-value of money is due to interest. Interest is a
measure of the increase between the original sum borrowed or invested and the final amount
to be paid or obtained.
•If you invest money at some time in the past,
• Interest = (total amount accumulated - original investment)
Mr. Kiran Patil 4
•If you borrowed money at some time in the past,
•Interest = (present amount to be paid- original loan)
•The original investment or loan is called principal.
•Time-value of money is involved in the study to determine what is economical in the long
run.
•Simple Interest:
•When the total interest earned or charged is directly proportional to the principal amount
involved at the given interest rate at the period for which principle amount used that interest
is called simple interest.
•Simple interest is not used in commercial practice now-a-days.
I = P x N x i
Where,
P= Principle amount
N= no of years for principle amount is used
i= Rate of return per year
Compound Interest:
•When interest earn at the end of period becomes part of principle amount & itself earns
interest along with principle amount then it is called as compound interest.
Mr. Kiran Patil 5
•Interest = (present amount to be paid- original loan)
•The original investment or loan is called principal.
•Time-value of money is involved in the study to determine what is economical in the long
run.
•Simple Interest:
•When the total interest earned or charged is directly proportional to the principal amount
involved at the given interest rate at the period for which principle amount used that interest
is called simple interest.
•Simple interest is not used in commercial practice now-a-days.
I = P x N x i
Where,
P= Principle amount
N= no of years for principle amount is used
i= Rate of return per year
Compound Interest:
•When interest earn at the end of period becomes part of principle amount & itself earns
interest along with principle amount then it is called as compound interest.
Mr. Kiran Patil 5
Cash-Flow Diagrams
•Every company or person has receipts (income) and cash disbursements (costs) which occur
over a particular time span. The receipts and disbursements in a given time interval are
called cash flows.
•Positive cash-flows represent receipts.
•Negative cash-flows represent disbursements.
•Net cash flow = receipts – disbursements
•It is assumed that all cash-flows occur at the end of the interest period. This is known as
end-of-period convention.
•A cash-flow diagram is a graphical representation of cash flows drawn on a time scale. It
consists of two basic parts,
(i) The horizontal time line
(ii)The vertical cash-flow line
Time line (n years)
•The diagram should represent the statement of the problem. It should indicate what is given
and what is to be found. A vertical arrow pointing up indicates a positive cash flow. An
arrow pointing down indicate a negative cash flow
-ve cash-flow line
+ ve cash-flow line
Rate of interest i %
Mr. Kiran Patil 6
•Every company or person has receipts (income) and cash disbursements (costs) which occur
over a particular time span. The receipts and disbursements in a given time interval are
called cash flows.
•Positive cash-flows represent receipts.
•Negative cash-flows represent disbursements.
•Net cash flow = receipts – disbursements
•It is assumed that all cash-flows occur at the end of the interest period. This is known as
end-of-period convention.
•A cash-flow diagram is a graphical representation of cash flows drawn on a time scale. It
consists of two basic parts,
(i) The horizontal time line
(ii)The vertical cash-flow line
Time line (n years)
•The diagram should represent the statement of the problem. It should indicate what is given
and what is to be found. A vertical arrow pointing up indicates a positive cash flow. An
arrow pointing down indicate a negative cash flow
-ve cash-flow line
+ ve cash-flow line
Rate of interest i %
Mr. Kiran Patil 6
Equivalence
•The concept of equivalence is the combination of time-value of money and interest rate.
Equivalence means the different sums of money at different times are equal in economic
value.
• Example:
•If the interest rate is 10 % per year, Rs. 1000 today would be equivalent to Rs.1100 one year
from today
•[1000 + (0.10) 1000] = 1100
•Therefore, if someone offers you a gift of Rs.1000 today or Rs.1100 one year from today, it
would make no difference. The two sums of money are equivalent to each other when the
interest rate is 10 % per year.
•The same principle can be applied to determine equivalence for the past years. If you have
Rs.1000 now, it is equivalent to last year.
Interest Formulae
•P = Present Value/ Present Worth of money
•F = Future Value/ Future Worth of money
•n = Number of interest periods in years
•i = Rate of interest per year
•A = Uniform series compound amount (occurs at the end of each year)
Mr. Kiran Patil 7
•The concept of equivalence is the combination of time-value of money and interest rate.
Equivalence means the different sums of money at different times are equal in economic
value.
• Example:
•If the interest rate is 10 % per year, Rs. 1000 today would be equivalent to Rs.1100 one year
from today
•[1000 + (0.10) 1000] = 1100
•Therefore, if someone offers you a gift of Rs.1000 today or Rs.1100 one year from today, it
would make no difference. The two sums of money are equivalent to each other when the
interest rate is 10 % per year.
•The same principle can be applied to determine equivalence for the past years. If you have
Rs.1000 now, it is equivalent to last year.
Interest Formulae
•P = Present Value/ Present Worth of money
•F = Future Value/ Future Worth of money
•n = Number of interest periods in years
•i = Rate of interest per year
•A = Uniform series compound amount (occurs at the end of each year)
Mr. Kiran Patil 7
•For single cash-flow
1. Single Payment Compound Amount Factor (SPCAF)
• A single payment (P) is multiplied to find its compound amount (F) at a specified time in
future.
i n
•Ex. If Rs. 5000 are invested now at 8 % per annum, how much will it accumulate in 10
years?
i=8% n=10
SPCAF = (F/P, i, n)
??????=??????1+??????
�
??????=50001+0.08
10
F = Rs.10794.60
P= Given
F= ?
SPCAF = (F/P, i, n)
??????=????????????+??????
??????
P= 5000
F= ?
Mr. Kiran Patil 8
1. Single Payment Compound Amount Factor (SPCAF)
• A single payment (P) is multiplied to find its compound amount (F) at a specified time in
future.
i n
•Ex. If Rs. 5000 are invested now at 8 % per annum, how much will it accumulate in 10
years?
i=8% n=10
SPCAF = (F/P, i, n)
??????=??????1+??????
�
??????=50001+0.08
10
F = Rs.10794.60
P= Given
F= ?
SPCAF = (F/P, i, n)
??????=????????????+??????
??????
P= 5000
F= ?
Mr. Kiran Patil 8
2. Single Payment Present Worth factor (SPPWF)
•When a single future payment (F) is multiplied by this factor, the present worth (P) is
obtained.
•A person desires to have Rs. 10795 after 10 years from now. What amount he should
deposit now to provide for it? Interest rate is 8 % p.a.
P=?
F= Given
i n
SPPWF = (P/F, i, n)
??????=??????
??????
??????+??????
??????
F= 10795
P=?
i = 8% n = 10
SPPWF = (P/F, i, n)
??????=??????
1
1+??????
�
??????=10795
1
1+0.08
10
P = Rs. 5000
Mr. Kiran Patil 9
•When a single future payment (F) is multiplied by this factor, the present worth (P) is
obtained.
•A person desires to have Rs. 10795 after 10 years from now. What amount he should
deposit now to provide for it? Interest rate is 8 % p.a.
P=?
F= Given
i n
SPPWF = (P/F, i, n)
??????=??????
??????
??????+??????
??????
F= 10795
P=?
i = 8% n = 10
SPPWF = (P/F, i, n)
??????=??????
1
1+??????
�
??????=10795
1
1+0.08
10
P = Rs. 5000
Mr. Kiran Patil 9
•For Uniform Series of Payments (Annuities)
3. Uniform Series Compound Amount Factor (USCAF)
•This factor converts a uniform series payment (A) to its compound amount (F), as shown
in the cash-flow diagram.
•Example: If eight annual deposits of Rs. 5000 each are placed in the account at 8 %
interest per year, how much money will be accumulated immediately after the last
deposit?
•USCAF = (F/A, i %, n)
F = A
1+??????
??????
− 1
??????
F = 5000
1+0.08
8
− 1
0.08
F = Rs. 53183
F= ? A= Given USCAF = (F/A, i %, n)
F = A
??????+??????
??????
− ??????
??????
0 1 2 3 4 5 6 7 8
A= 5000
F= ?
i n
i = 8%
n= 8 Years
Mr. Kiran Patil 10
3. Uniform Series Compound Amount Factor (USCAF)
•This factor converts a uniform series payment (A) to its compound amount (F), as shown
in the cash-flow diagram.
•Example: If eight annual deposits of Rs. 5000 each are placed in the account at 8 %
interest per year, how much money will be accumulated immediately after the last
deposit?
•USCAF = (F/A, i %, n)
F = A
1+??????
??????
− 1
??????
F = 5000
1+0.08
8
− 1
0.08
F = Rs. 53183
F= ? A= Given USCAF = (F/A, i %, n)
F = A
??????+??????
??????
− ??????
??????
0 1 2 3 4 5 6 7 8
A= 5000
F= ?
i n
i = 8%
n= 8 Years
Mr. Kiran Patil 10
4. Uniform Series Present Worth Factor (USPWF)
•This factor is used to determine the present worth (P) for a uniform series payment of A,
as shown in the cash-flow diagram.
•Example: How much should be deposited in a fund to provide for eight end-of-year
withdrawals of Rs. 5000 each? Interest rate is 8 % p.a.
•USPWF = (P/A, i %, n)
P = A
1+??????
??????
− 1
??????1+??????
??????
P=5000
1+0.08
8
− 1
0.081+0.08
8
P = Rs. 28733
A= Given
USPWF = (P/A, i %, n)
P = A
??????+??????
??????
− ??????
????????????+??????
??????
0 1 2 3 4 5 6 7 8
A= 5000
P= ?
P= ?
i n
i = 8%
n= 8 Years
Mr. Kiran Patil 11
•This factor is used to determine the present worth (P) for a uniform series payment of A,
as shown in the cash-flow diagram.
•Example: How much should be deposited in a fund to provide for eight end-of-year
withdrawals of Rs. 5000 each? Interest rate is 8 % p.a.
•USPWF = (P/A, i %, n)
P = A
1+??????
??????
− 1
??????1+??????
??????
P=5000
1+0.08
8
− 1
0.081+0.08
8
P = Rs. 28733
A= Given
USPWF = (P/A, i %, n)
P = A
??????+??????
??????
− ??????
????????????+??????
??????
0 1 2 3 4 5 6 7 8
A= 5000
P= ?
P= ?
i n
i = 8%
n= 8 Years
Mr. Kiran Patil 11
5. Sinking Fund Factor (SFF)
•It is the factor by which a future sum (F) is multiplied to find a uniform sum (A) that
should be set kept aside regularly, such that the final value of the funds kept aside is F.
•Example: What uniform annual amount should be deposited each year at interest rate 8
% p.a. to accumulate Rs. 50,000 at the time of the eighth annual deposit?
•SFF = (A/F, i %, n)
•A = F
??????
1+??????
??????
− 1
•A = 50000
0.08
1+0.08
8
− 1
•A = Rs. 4700
A= ?
SFF = (A/F, i %, n)
A = F
??????
??????+??????
??????
− ??????
0 1 2 3 4 5 6 7 8
A= ?
F= 50000
F= Given
i n
i = 8%
n= 8 Years
Mr. Kiran Patil 12
•It is the factor by which a future sum (F) is multiplied to find a uniform sum (A) that
should be set kept aside regularly, such that the final value of the funds kept aside is F.
•Example: What uniform annual amount should be deposited each year at interest rate 8
% p.a. to accumulate Rs. 50,000 at the time of the eighth annual deposit?
•SFF = (A/F, i %, n)
•A = F
??????
1+??????
??????
− 1
•A = 50000
0.08
1+0.08
8
− 1
•A = Rs. 4700
A= ?
SFF = (A/F, i %, n)
A = F
??????
??????+??????
??????
− ??????
0 1 2 3 4 5 6 7 8
A= ?
F= 50000
F= Given
i n
i = 8%
n= 8 Years
Mr. Kiran Patil 12
6. Capital Recovery Factor (CRF)
•This factor can be used to find the uniform payments of ‘A’ to exactly recover a present
capital sum (P) with interest.
•Example: What is the amount of eight equal payments to repay a loan of Rs. 5, 00,000 at
interest rate of 10 % per year? The first payment is due 1 year after receiving the loan.
•CRF = (A/P, i %, n)
•A = P
??????1+??????
??????
1+??????
??????
− 1
•A = 500000
0.101+0.10
8
1+0.10
8
− 1
•A = Rs. 93, 722
A= ?
CRF = (A/P, i %, n)
A = P
??????1+??????
??????
1+??????
??????
− 1
0 1 2 3 4 5 6 7 8
A= ?
i n
i = 10%
n= 8 Years
P= Given
P= 500000
Mr. Kiran Patil 13
•This factor can be used to find the uniform payments of ‘A’ to exactly recover a present
capital sum (P) with interest.
•Example: What is the amount of eight equal payments to repay a loan of Rs. 5, 00,000 at
interest rate of 10 % per year? The first payment is due 1 year after receiving the loan.
•CRF = (A/P, i %, n)
•A = P
??????1+??????
??????
1+??????
??????
− 1
•A = 500000
0.101+0.10
8
1+0.10
8
− 1
•A = Rs. 93, 722
A= ?
CRF = (A/P, i %, n)
A = P
??????1+??????
??????
1+??????
??????
− 1
0 1 2 3 4 5 6 7 8
A= ?
i n
i = 10%
n= 8 Years
P= Given
P= 500000
Mr. Kiran Patil 13
•For Single cash-flow
•For Uniform series (Annuities)
To
Find
Given Formula Factor Name Symbol
F P ??????=????????????+??????
??????
Single Payment Compound Amount Factor
(SPCAF)
(F/P, i %, n)
P F
??????=??????
??????
??????+??????
??????
Single Payment Present Worth Factor (SPPWF) (P/F, i %, n)
To
Find
Given Formula Factor Name Symbol
F A
F = A
??????+??????
??????
− ??????
??????
Uniform Series Compound Amount Factor
(USCAF)
(F/A, i %, n)
P A
P = A
??????+??????
??????
− ??????
????????????+??????
??????
Uniform Series Present Worth Factor
(USPWF)
(P/A, i %, n)
A F
A = F
??????
??????+??????
??????
− ??????
Sinking Fund Factor (SFF)
(A/F, i %, n)
A P
A = P
??????1+??????
�
1+??????
�
− 1
Capital Recovery Factor
(CRF)
(A/P, i %, n)
Mr. Kiran Patil 14
•For Uniform series (Annuities)
To
Find
Given Formula Factor Name Symbol
F P ??????=????????????+??????
??????
Single Payment Compound Amount Factor
(SPCAF)
(F/P, i %, n)
P F
??????=??????
??????
??????+??????
??????
Single Payment Present Worth Factor (SPPWF) (P/F, i %, n)
To
Find
Given Formula Factor Name Symbol
F A
F = A
??????+??????
??????
− ??????
??????
Uniform Series Compound Amount Factor
(USCAF)
(F/A, i %, n)
P A
P = A
??????+??????
??????
− ??????
????????????+??????
??????
Uniform Series Present Worth Factor
(USPWF)
(P/A, i %, n)
A F
A = F
??????
??????+??????
??????
− ??????
Sinking Fund Factor (SFF)
(A/F, i %, n)
A P
A = P
??????1+??????
�
1+??????
�
− 1
Capital Recovery Factor
(CRF)
(A/P, i %, n)
Mr. Kiran Patil 14
Economic Comparisons
1. Equivalent Present Worth Method ( PW Method)
•For evaluating the alternatives, the present worth (PW) of both the alternatives are found
out. The alternative with the maximum PW is the most economical alternative.
•For cost-dominated cash flow diagrams, the alternative with the lowest present worth is
chosen.
•In case of cash flow diagrams involving both, cost and revenues, the difference of PW of
revenues and costs are found. This is called net present worth or net present value (NPV).
•Problem No. 1:
The details of two construction machines are given below. Make a comparison if the rate of
interest is 12% per year. Use present work method.
Machine A Machine B
Initial cost (Rs) 1,00,000 1,70,000
Annual operating & Maintainece cost (Rs) 60,000 50,000
Salvage value(Rs) 10,000 20,000
Life in years 10 10
Mr. Kiran Patil 15
1. Equivalent Present Worth Method ( PW Method)
•For evaluating the alternatives, the present worth (PW) of both the alternatives are found
out. The alternative with the maximum PW is the most economical alternative.
•For cost-dominated cash flow diagrams, the alternative with the lowest present worth is
chosen.
•In case of cash flow diagrams involving both, cost and revenues, the difference of PW of
revenues and costs are found. This is called net present worth or net present value (NPV).
•Problem No. 1:
The details of two construction machines are given below. Make a comparison if the rate of
interest is 12% per year. Use present work method.
Machine A Machine B
Initial cost (Rs) 1,00,000 1,70,000
Annual operating & Maintainece cost (Rs) 60,000 50,000
Salvage value(Rs) 10,000 20,000
Life in years 10 10
Mr. Kiran Patil 15
Cash flow for equipment A
i= 12%
0 1 2 3 4 5 6 7 8 9 10
a)Present worth for initial cost= Rs. -1,00,000
b)Present worth for O & M cost (P/A, i, n)
Uniform Series Present Worth Factor (USPWF)
P = A
1+??????
??????
− 1
??????1+??????
??????
P=60000
1+0.12
10
− 1
0.121+0.12
10
P= Rs. - 3,42,162.16
c)Present worth of salvage value (P/F, i, n)
Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
??????
??????=10000
1
1+0.12
10
P = Rs. 3219.73
P= 100000
F= 10000
d) Total present worth of equipment A= -1,00,000 -
3,42,162.16 + 3219.73 = Rs - 4,38,942.43
A= 60000
Mr. Kiran Patil 16
i= 12%
0 1 2 3 4 5 6 7 8 9 10
a)Present worth for initial cost= Rs. -1,00,000
b)Present worth for O & M cost (P/A, i, n)
Uniform Series Present Worth Factor (USPWF)
P = A
1+??????
??????
− 1
??????1+??????
??????
P=60000
1+0.12
10
− 1
0.121+0.12
10
P= Rs. - 3,42,162.16
c)Present worth of salvage value (P/F, i, n)
Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
??????
??????=10000
1
1+0.12
10
P = Rs. 3219.73
P= 100000
F= 10000
d) Total present worth of equipment A= -1,00,000 -
3,42,162.16 + 3219.73 = Rs - 4,38,942.43
A= 60000
Mr. Kiran Patil 16
Cash flow for equipment B
i= 12%
0 1 2 3 4 5 6 7 8 9 10
a)Present worth for initial cost= Rs. -1,70,000
b)Present worth for O & M cost (P/A, i, n)
Uniform Series Present Worth Factor (USPWF)
P = A
1+??????
??????
− 1
??????1+??????
??????
P=50000
1+0.12
10
− 1
0.121+0.12
10
P= Rs. – 2,85,135.14
c)Present worth of salvage value (P/F, i, n)
Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
??????
??????=20000
1
1+0.12
10
P = Rs. 6,439.46
P= 170000
F= 20000
d) Total present worth of equipment B= -1,70,000 -
2,85,135.14 + 6,439.46 = Rs – 4,48,695.68
A= 50000
Mr. Kiran Patil 17
i= 12%
0 1 2 3 4 5 6 7 8 9 10
a)Present worth for initial cost= Rs. -1,70,000
b)Present worth for O & M cost (P/A, i, n)
Uniform Series Present Worth Factor (USPWF)
P = A
1+??????
??????
− 1
??????1+??????
??????
P=50000
1+0.12
10
− 1
0.121+0.12
10
P= Rs. – 2,85,135.14
c)Present worth of salvage value (P/F, i, n)
Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
??????
??????=20000
1
1+0.12
10
P = Rs. 6,439.46
P= 170000
F= 20000
d) Total present worth of equipment B= -1,70,000 -
2,85,135.14 + 6,439.46 = Rs – 4,48,695.68
A= 50000
Mr. Kiran Patil 17
•Total present work of equipment A= Rs - 4,38,942.43
•Total present work of equipment A= Rs – 4,48,695.68
•For cost dominated alternatives, the alternative with the lowest present worth of cost (less
–ve) is selected.
•Since PW of cost of machine A is less than PW of cost of machine B, So select machine
A.
Mr. Kiran Patil 18
•Total present work of equipment A= Rs – 4,48,695.68
•For cost dominated alternatives, the alternative with the lowest present worth of cost (less
–ve) is selected.
•Since PW of cost of machine A is less than PW of cost of machine B, So select machine
A.
Mr. Kiran Patil 18
•Problem No. 2:
Using present worth method compare the following equipments & suggest which should be
purchase if the rate of interest is 15% per year.
Solution:
1.Find LCM ( Least common multiply)
• 4/2= 2 so 6 x 2= 12
• 6/2= 3 so 4 x 3= 12
2. Present worth of equipment A
0 1 2 3 4 5 6 7 8 9 10 11 12
Machine A Machine B
Initial cost (Rs) 2,50,000 3,50,000
Annual operating & Maintainece cost (Rs) 90,000 70,000
Salvage value(Rs) 20,000 20,000
Life in years 4 6
P=2,50,000 P=2,50,000
A=90,000
F=20,000
F=20,000 F=20,000
Mr. Kiran Patil 19
Using present worth method compare the following equipments & suggest which should be
purchase if the rate of interest is 15% per year.
Solution:
1.Find LCM ( Least common multiply)
• 4/2= 2 so 6 x 2= 12
• 6/2= 3 so 4 x 3= 12
2. Present worth of equipment A
0 1 2 3 4 5 6 7 8 9 10 11 12
Machine A Machine B
Initial cost (Rs) 2,50,000 3,50,000
Annual operating & Maintainece cost (Rs) 90,000 70,000
Salvage value(Rs) 20,000 20,000
Life in years 4 6
P=2,50,000 P=2,50,000
A=90,000
F=20,000
F=20,000 F=20,000
Mr. Kiran Patil 19
a)Present worth of initial cost
•Present worth of initial cost = Rs. - 2,50,000
•Present worth at 1
st
replacement at 4 year (P/F, i, n)
Use Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
�
??????=250000
1
1+0.15
4
P = Rs. – 1,42,938.31
•Present worth at 2nd replacement at 8 year (P/F, i, n)
Use Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
�
??????=250000
1
1+0.15
8
P = Rs. – 81,725.44
•Total present worth of initial cost Rs= -2,50,000 -1,42,938.31- 81,725.44
Rs = - 4,74,663.75
b) Present worth of O & M cost
(P/A,i,n)
Uniform Series Present Worth Factor (USPWF)
P = A
1+??????
??????
− 1
??????1+??????
??????
=90000
1+0.15
12
− 1
0.151+0.15
12
=Rs. – 4,89,375
Mr. Kiran Patil 20
•Present worth of initial cost = Rs. - 2,50,000
•Present worth at 1
st
replacement at 4 year (P/F, i, n)
Use Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
�
??????=250000
1
1+0.15
4
P = Rs. – 1,42,938.31
•Present worth at 2nd replacement at 8 year (P/F, i, n)
Use Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
�
??????=250000
1
1+0.15
8
P = Rs. – 81,725.44
•Total present worth of initial cost Rs= -2,50,000 -1,42,938.31- 81,725.44
Rs = - 4,74,663.75
b) Present worth of O & M cost
(P/A,i,n)
Uniform Series Present Worth Factor (USPWF)
P = A
1+??????
??????
− 1
??????1+??????
??????
=90000
1+0.15
12
− 1
0.151+0.15
12
=Rs. – 4,89,375
Mr. Kiran Patil 20
c) Present worth of salvage value
•Present worth at 1
st
replacement at 4 year (P/F, i, n)
Use Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
�
??????=20000
1
1+0.15
4
P = Rs. 11,435.06
•Present worth at 2nd replacement at 8 year (P/F, i, n)
??????=??????
1
1+??????
�
??????=20000
1
1+0.15
8
P = Rs. 6,538.04
•Present worth at 3
rd
replacement at 12 year (P/F, i, n)
??????=??????
1
1+??????
�
??????=20000
1
1+0.15
12
P = Rs. 3738.14
•Total of present worth of salvage value Rs= 11,435.06+ 6,538.04+ 3738.14= 21,711.24
•So, Total present worth of equipment A = - 4,74,663.75 – 4,89,375 + 21,711.24
Rs = - 9,42,327.51
Mr. Kiran Patil 21
•Present worth at 1
st
replacement at 4 year (P/F, i, n)
Use Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
�
??????=20000
1
1+0.15
4
P = Rs. 11,435.06
•Present worth at 2nd replacement at 8 year (P/F, i, n)
??????=??????
1
1+??????
�
??????=20000
1
1+0.15
8
P = Rs. 6,538.04
•Present worth at 3
rd
replacement at 12 year (P/F, i, n)
??????=??????
1
1+??????
�
??????=20000
1
1+0.15
12
P = Rs. 3738.14
•Total of present worth of salvage value Rs= 11,435.06+ 6,538.04+ 3738.14= 21,711.24
•So, Total present worth of equipment A = - 4,74,663.75 – 4,89,375 + 21,711.24
Rs = - 9,42,327.51
Mr. Kiran Patil 21
3. Present worth of equipment B
0 1 2 3 4 5 6 7 8 9 10 11 12
a)Present worth of initial cost
•Present worth of initial cost = Rs. - 3,50,000
•Present worth at 1
st
replacement at 6 year (P/F, i, n)
Use Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
�
??????=350000
1
1+0.15
6
P = Rs. – 1,51, 314.66
•Total present worth Rs= -3,50,000 -1,51,314.66
Rs = - 5,01,314.66
P=3,50,000
A=70,000
F=20,000
F=20,000
P=3,50,000
Mr. Kiran Patil 22
0 1 2 3 4 5 6 7 8 9 10 11 12
a)Present worth of initial cost
•Present worth of initial cost = Rs. - 3,50,000
•Present worth at 1
st
replacement at 6 year (P/F, i, n)
Use Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
�
??????=350000
1
1+0.15
6
P = Rs. – 1,51, 314.66
•Total present worth Rs= -3,50,000 -1,51,314.66
Rs = - 5,01,314.66
P=3,50,000
A=70,000
F=20,000
F=20,000
P=3,50,000
Mr. Kiran Patil 22
b) Present worth of O & M cost
(P/A,i,n)
Uniform Series Present Worth Factor (USPWF)
P = A
1+??????
??????
− 1
??????1+??????
??????
P= 70000
1+0.15
12
− 1
0.151+0.15
12
P= Rs. – 3,80,625
c) Present worth of salvage value
•Present worth at 1
st
replacement at 6 year (P/F, i, n)
Use Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
�
??????=20000
1
1+0.15
6
P = Rs. 8,646.55
•Present worth at 2nd replacement at 12 year (P/F, i, n)
??????=??????
1
1+??????
�
??????=20000
1
1+0.15
12
P = Rs. 3,738.14
•Total of present worth of salvage value Rs= 8,646.55+ 3,738.14 = 12,384.69
•So, Total present worth of equipment B = - 5,01,314.66 - 3,80,625 + 12,384.69
Rs = - 8,69,554.97
Mr. Kiran Patil 23
(P/A,i,n)
Uniform Series Present Worth Factor (USPWF)
P = A
1+??????
??????
− 1
??????1+??????
??????
P= 70000
1+0.15
12
− 1
0.151+0.15
12
P= Rs. – 3,80,625
c) Present worth of salvage value
•Present worth at 1
st
replacement at 6 year (P/F, i, n)
Use Single Payment Present Worth Factor (SPPWF)
??????=??????
1
1+??????
�
??????=20000
1
1+0.15
6
P = Rs. 8,646.55
•Present worth at 2nd replacement at 12 year (P/F, i, n)
??????=??????
1
1+??????
�
??????=20000
1
1+0.15
12
P = Rs. 3,738.14
•Total of present worth of salvage value Rs= 8,646.55+ 3,738.14 = 12,384.69
•So, Total present worth of equipment B = - 5,01,314.66 - 3,80,625 + 12,384.69
Rs = - 8,69,554.97
Mr. Kiran Patil 23
•Total present work of equipment A= Rs - 9,42,327.51
•Total present work of equipment A= Rs – 8,69,554.97
•For cost dominated alternatives, the alternative with the lowest present worth of cost
(less –ve) is selected.
•Since PW of cost of machine B is less than PW of cost of machine A, So select
machine B.
Mr. Kiran Patil 24
•Total present work of equipment A= Rs – 8,69,554.97
•For cost dominated alternatives, the alternative with the lowest present worth of cost
(less –ve) is selected.
•Since PW of cost of machine B is less than PW of cost of machine A, So select
machine B.
Mr. Kiran Patil 24
Equivalent Uniform Annual Cost (EUAC) Method
•This method is widely used for comparing alternatives. In this method, all payments and
disbursements are converted into an annualized cost series.
•Annual cost is the cost pattern of each alternative converted into an equivalent uniform
series of annual cost at a given interest rate. The alternative with the least annual cost is
chosen.
•Types of economic Comparisons
1. Alternative with equal lives
2.Alternative with different lives
Mr. Kiran Patil 25
•This method is widely used for comparing alternatives. In this method, all payments and
disbursements are converted into an annualized cost series.
•Annual cost is the cost pattern of each alternative converted into an equivalent uniform
series of annual cost at a given interest rate. The alternative with the least annual cost is
chosen.
•Types of economic Comparisons
1. Alternative with equal lives
2.Alternative with different lives
Mr. Kiran Patil 25
•Problem No. 1:
A manager is trying to decide between the following equipments. Suggest which equipment
he should purchase. If the rate of interest is 12%. Consider following data & use EUAC
method.
Solution:
A.EUAC for equipment A:-
0 1 2 3 4 5 6 7 8 9 10 11 12
Machine A Machine B
Initial cost (Rs) 60,000 1,30,000
Annual operating & Maintainece cost (Rs) 8,000 6,000
Salvage value(Rs) 10,000 25,000
Life in years 12 12
P= 60000
A= 8000
F= 10000
Mr. Kiran Patil 26
A manager is trying to decide between the following equipments. Suggest which equipment
he should purchase. If the rate of interest is 12%. Consider following data & use EUAC
method.
Solution:
A.EUAC for equipment A:-
0 1 2 3 4 5 6 7 8 9 10 11 12
Machine A Machine B
Initial cost (Rs) 60,000 1,30,000
Annual operating & Maintainece cost (Rs) 8,000 6,000
Salvage value(Rs) 10,000 25,000
Life in years 12 12
P= 60000
A= 8000
F= 10000
Mr. Kiran Patil 26
a)EUAC for initial cost (A/P, i, n)
Use capital recovery factor
A = P
??????1+??????
??????
1+??????
??????
− 1
A = 60000
0.121+0.12
12
1+0.12
12
− 1
A = Rs. 9517.24
b) EUAC for annual O & M cost= Rs 8000
c) EUAC for salvage value (A/F, i, n)
Use Sinking fund factor
A = F
??????
1+??????
??????
− 1
A = 10000
0.12
1+0.12
12
− 1
A = Rs. 414.37
d) Total EUAC for equipment A Rs= - 9,517.24 - 8,000 +414.37
Rs = - 17,102.63
Mr. Kiran Patil 27
Use capital recovery factor
A = P
??????1+??????
??????
1+??????
??????
− 1
A = 60000
0.121+0.12
12
1+0.12
12
− 1
A = Rs. 9517.24
b) EUAC for annual O & M cost= Rs 8000
c) EUAC for salvage value (A/F, i, n)
Use Sinking fund factor
A = F
??????
1+??????
??????
− 1
A = 10000
0.12
1+0.12
12
− 1
A = Rs. 414.37
d) Total EUAC for equipment A Rs= - 9,517.24 - 8,000 +414.37
Rs = - 17,102.63
Mr. Kiran Patil 27
B. EUAC for equipment B:-
0 1 2 3 4 5 6 7 8 9 10 11 12
P= 1,30,000
A= 6000
F= 25000
a)EUAC for initial cost (A/P, i, n)
Use capital recovery factor
A = P
??????1+??????
??????
1+??????
??????
− 1
A = 130000
0.121+0.12
12
1+0.12
12
− 1
A = Rs. 20,620.69
b) EUAC for annual O & M cost= Rs 6000
c) EUAC for salvage value (A/F, i, n)
Use Sinking fund factor
A = F
??????
1+??????
??????
− 1
A = 25000
0.12
1+0.12
12
− 1
A = Rs. 1034.48
d) Total EUAC for equipment B Rs= - 20,620.69 - 6,000 +1034.48
Rs = - 25,586.21
Mr. Kiran Patil 28
0 1 2 3 4 5 6 7 8 9 10 11 12
P= 1,30,000
A= 6000
F= 25000
a)EUAC for initial cost (A/P, i, n)
Use capital recovery factor
A = P
??????1+??????
??????
1+??????
??????
− 1
A = 130000
0.121+0.12
12
1+0.12
12
− 1
A = Rs. 20,620.69
b) EUAC for annual O & M cost= Rs 6000
c) EUAC for salvage value (A/F, i, n)
Use Sinking fund factor
A = F
??????
1+??????
??????
− 1
A = 25000
0.12
1+0.12
12
− 1
A = Rs. 1034.48
d) Total EUAC for equipment B Rs= - 20,620.69 - 6,000 +1034.48
Rs = - 25,586.21
Mr. Kiran Patil 28
•Total EUAC of equipment A= Rs – 17,102.63
•Total EUAC of equipment B= Rs – 25,586.21
•For cost dominated alternatives, the alternative with the lowest EUAC of cost (less –ve) is
selected.
•Since EUAC of cost of machine A is less than EUAC of cost of machine B, So select
machine A.
Mr. Kiran Patil 29
•Total EUAC of equipment B= Rs – 25,586.21
•For cost dominated alternatives, the alternative with the lowest EUAC of cost (less –ve) is
selected.
•Since EUAC of cost of machine A is less than EUAC of cost of machine B, So select
machine A.
Mr. Kiran Patil 29
•Problem No. 1:
Using EUAC method suggest which machine should be purchased if rate of interest is 15%.
Consider following data & use EUAC method.
Solution:
A.EUAC for equipment A:-
0 1 2 3 4 5 6
Machine A Machine B
Initial cost (Rs) 2,60,000 3,60,000
Annual operating & Maintainece cost (Rs) 8,000 3,000
Annual labour cost (Rs) 1,10,000 70,000
Extra Annual income tax (Rs) --- 26,000
Salvage value(Rs) 20,000 30,000
Life in years 6 10
P=2,60,000
A= 8000 A= 1,10,000
F= 20,000
Mr. Kiran Patil 30
Using EUAC method suggest which machine should be purchased if rate of interest is 15%.
Consider following data & use EUAC method.
Solution:
A.EUAC for equipment A:-
0 1 2 3 4 5 6
Machine A Machine B
Initial cost (Rs) 2,60,000 3,60,000
Annual operating & Maintainece cost (Rs) 8,000 3,000
Annual labour cost (Rs) 1,10,000 70,000
Extra Annual income tax (Rs) --- 26,000
Salvage value(Rs) 20,000 30,000
Life in years 6 10
P=2,60,000
A= 8000 A= 1,10,000
F= 20,000
Mr. Kiran Patil 30
a)EUAC for initial cost (A/P, i, n)
Use capital recovery factor
A = P
??????1+??????
??????
1+??????
??????
− 1
A = 2,60,000
0.151+0.15
6
1+0.15
6
− 1
A = Rs. 34,236.64
b) EUAC for annual O & M cost= Rs 8000
c) EUAC for Labor cost= Rs 1,10,000
d) EUAC for salvage value (A/F, i, n)
Use Sinking fund factor
A = F
??????
1+??????
??????
− 1
A = 20000
0.15
1+0.15
6
− 1
A = Rs. 2290.07
d) Total EUAC for equipment A Rs= - 34,236.64 – 8,000 – 1,10,000 + 2290.07
Rs = - 1,49,946.57
Mr. Kiran Patil 31
Use capital recovery factor
A = P
??????1+??????
??????
1+??????
??????
− 1
A = 2,60,000
0.151+0.15
6
1+0.15
6
− 1
A = Rs. 34,236.64
b) EUAC for annual O & M cost= Rs 8000
c) EUAC for Labor cost= Rs 1,10,000
d) EUAC for salvage value (A/F, i, n)
Use Sinking fund factor
A = F
??????
1+??????
??????
− 1
A = 20000
0.15
1+0.15
6
− 1
A = Rs. 2290.07
d) Total EUAC for equipment A Rs= - 34,236.64 – 8,000 – 1,10,000 + 2290.07
Rs = - 1,49,946.57
Mr. Kiran Patil 31
B. EUAC for equipment B:-
0 1 2 3 4 5 6 7 8 9 10
P= 3,60,0000
A= 3000
F= 30000
a) EUAC for initial cost (A/P, i, n)
Use capital recovery factor
A = P
??????1+??????
??????
1+??????
??????
− 1
= 360000
0.151+0.15
10
1+0.15
10
− 1
A = Rs. 71,764.71
b) EUAC for annual O & M cost= Rs 3000
c) EUAC for labor cost= Rs. 70000
d) EUAC for extra annual tax = Rs. 26000
c) EUAC for salvage value (A/F, i, n)
Use Sinking fund factor
A = F
??????
1+??????
??????
− 1
A = 30000
0.15
1+0.15
10
− 1
A = Rs. 1475.41
d) Total EUAC for equipment B Rs= - 71,764.71 – 3000 – 70000- 26000 +1475.41
= -1,69,289.30
A= 70,000 A= 26,000
Mr. Kiran Patil 32
0 1 2 3 4 5 6 7 8 9 10
P= 3,60,0000
A= 3000
F= 30000
a) EUAC for initial cost (A/P, i, n)
Use capital recovery factor
A = P
??????1+??????
??????
1+??????
??????
− 1
= 360000
0.151+0.15
10
1+0.15
10
− 1
A = Rs. 71,764.71
b) EUAC for annual O & M cost= Rs 3000
c) EUAC for labor cost= Rs. 70000
d) EUAC for extra annual tax = Rs. 26000
c) EUAC for salvage value (A/F, i, n)
Use Sinking fund factor
A = F
??????
1+??????
??????
− 1
A = 30000
0.15
1+0.15
10
− 1
A = Rs. 1475.41
d) Total EUAC for equipment B Rs= - 71,764.71 – 3000 – 70000- 26000 +1475.41
= -1,69,289.30
A= 70,000 A= 26,000
Mr. Kiran Patil 32
•Total EUAC of equipment A= Rs - 1,49,946.57
•Total EUAC of equipment B= Rs -1,69,289.30
•For cost dominated alternatives, the alternative with the lowest EUAC of cost (less –ve) is
selected.
•Since EUAC of cost of machine A is less than EUAC of cost of machine B, So select
machine A.
Mr. Kiran Patil 33
•Total EUAC of equipment B= Rs -1,69,289.30
•For cost dominated alternatives, the alternative with the lowest EUAC of cost (less –ve) is
selected.
•Since EUAC of cost of machine A is less than EUAC of cost of machine B, So select
machine A.
Mr. Kiran Patil 33
3. Capitalized cost method:
•Capitalized cost (Cc) refers to the present worth of a project that is assumed to least
forever.
•Some public works project such as dams, irrigation systems, railway etc are evaluationg
using capitalized cost method.
•In general, the procedure followed in calculating the capitalized cost of an infinite square
of cash flows as follows:
Step 1: Draw a cash flow dia. Showing all non-recurring costs (one time) & at least two
cycles of all non recurring cost & receipts.
Step 2: Find present worth of all non recurring amount.
Step 3: Find equivalent uniform annual worth (AW) through one life cycle of all recurring
amounts & add this to all other uniform amounts occurring in year 1 through infinity (∞).
Step 4: Divide the AW obtained in step 3 by interest rate (i) to get capitalized cost.
Step 5: Add the value obtained in step 2 to the value obtained in step 4.
Capitalized cost =
????????????
??????
????????????
????????????
??????
or
??????
??????
We know (P/A,i,n) factor when n= ∞. The equation for PW using factor
P = A
1+??????
??????
− 1
??????1+??????
??????
Divide numerator & denominator by 1+??????
�
P = A
1−1/1+??????
�
??????
Mr. Kiran Patil 34
•Capitalized cost (Cc) refers to the present worth of a project that is assumed to least
forever.
•Some public works project such as dams, irrigation systems, railway etc are evaluationg
using capitalized cost method.
•In general, the procedure followed in calculating the capitalized cost of an infinite square
of cash flows as follows:
Step 1: Draw a cash flow dia. Showing all non-recurring costs (one time) & at least two
cycles of all non recurring cost & receipts.
Step 2: Find present worth of all non recurring amount.
Step 3: Find equivalent uniform annual worth (AW) through one life cycle of all recurring
amounts & add this to all other uniform amounts occurring in year 1 through infinity (∞).
Step 4: Divide the AW obtained in step 3 by interest rate (i) to get capitalized cost.
Step 5: Add the value obtained in step 2 to the value obtained in step 4.
Capitalized cost =
????????????
??????
????????????
????????????
??????
or
??????
??????
We know (P/A,i,n) factor when n= ∞. The equation for PW using factor
P = A
1+??????
??????
− 1
??????1+??????
??????
Divide numerator & denominator by 1+??????
�
P = A
1−1/1+??????
�
??????
Mr. Kiran Patil 34
Problem : Calculate the capitalised cost of a project that has an initial cost of Rs. 1,50,000 &
an additional investment cost Rs. 50,000 after 10 years. The annual operating cast will be Rs.
5,000 for the first 4 years & Rs. 8,000 thereafter. In addition, there is expected to be a recurring
major rework cost of Rs. 15,000 after 13 years. Assume that i= 15% per year.
Solution:
1)Cash flow diagram:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
2) Find present worth of non-recurring amount
•P1= -1,50,000
•For find P2 (P/F,i,n)
• Use Single Payment Present Worth Factor (SPPWF)
•P2=??????
1
1+??????
??????
=5000
1
1+0.15
10
= 12,360/-
• Total of PW of Non- recurring amount= P1+P2= -1,50,000- 12,360= -1,62,360/-
3) Convert the recurring cost of Rs. 15000 every 13 years into an annual worth A for 1
st
13
years
We know (A/F, i,n)
1,50,000 50,000 15,000 15,000
5000
8000
Mr. Kiran Patil 35
an additional investment cost Rs. 50,000 after 10 years. The annual operating cast will be Rs.
5,000 for the first 4 years & Rs. 8,000 thereafter. In addition, there is expected to be a recurring
major rework cost of Rs. 15,000 after 13 years. Assume that i= 15% per year.
Solution:
1)Cash flow diagram:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
2) Find present worth of non-recurring amount
•P1= -1,50,000
•For find P2 (P/F,i,n)
• Use Single Payment Present Worth Factor (SPPWF)
•P2=??????
1
1+??????
??????
=5000
1
1+0.15
10
= 12,360/-
• Total of PW of Non- recurring amount= P1+P2= -1,50,000- 12,360= -1,62,360/-
3) Convert the recurring cost of Rs. 15000 every 13 years into an annual worth A for 1
st
13
years
We know (A/F, i,n)
1,50,000 50,000 15,000 15,000
5000
8000
Mr. Kiran Patil 35
•Use Sinking fund factor
A1 = F
??????
1+??????
??????
− 1
= 15000
0.15
1+0.15
13
− 1
= 437/-
4. The capitalised cost for the Rs 5,000 annual cost series may be computed by either 2 ways
a)Consider a series of Rs. 5,000/- from now to infinity & find present worth of (8000-
5000=3000) from years 5 on.
OR
b) Find the present worth of Rs. 5000 for 4 years & present worth of Rs. 8000 from 5 years to
infinity.
We use 1
st
option
•A2= -5000
•P =
�
??????
=
3000
0.15
= -20,000
•For initial 4 years for A=5000 we require total Rs. 20,000 (F)
• Use Single Payment Present Worth Factor (SPPWF)
•P2=??????
1
1+??????
??????
=20000
1
1+0.15
4
= 11,435/-
5. The two annual costs are converted into a capitalised cost (P)
P3=
�1+ A2
??????
=
−437−5000
0.15
=??????�.−36,247
6. Total Capitalised cost= P1+P2+P3
= -1,62,360-11,435-,36,247= -2,10,043/-
Mr. Kiran Patil 36
A1 = F
??????
1+??????
??????
− 1
= 15000
0.15
1+0.15
13
− 1
= 437/-
4. The capitalised cost for the Rs 5,000 annual cost series may be computed by either 2 ways
a)Consider a series of Rs. 5,000/- from now to infinity & find present worth of (8000-
5000=3000) from years 5 on.
OR
b) Find the present worth of Rs. 5000 for 4 years & present worth of Rs. 8000 from 5 years to
infinity.
We use 1
st
option
•A2= -5000
•P =
�
??????
=
3000
0.15
= -20,000
•For initial 4 years for A=5000 we require total Rs. 20,000 (F)
• Use Single Payment Present Worth Factor (SPPWF)
•P2=??????
1
1+??????
??????
=20000
1
1+0.15
4
= 11,435/-
5. The two annual costs are converted into a capitalised cost (P)
P3=
�1+ A2
??????
=
−437−5000
0.15
=??????�.−36,247
6. Total Capitalised cost= P1+P2+P3
= -1,62,360-11,435-,36,247= -2,10,043/-
Mr. Kiran Patil 36
•Net Present Value (NPV)
•The NPV of a project is equal to the sum of the present values of all the cash flows associated
with it.
•??????????????????=�??????�
1
1+??????
0
+ �??????1
1
1+??????
1
+ �??????2
1
1+??????
2
+ …………… . + �??????�
1
1+??????
??????
•∴ ??????????????????=
�??????�
1+??????
0
+
�??????1
1+??????
1
+
�??????2
1+??????
2
+ …………… . +
�??????�
1+??????
??????
•Accept the project if its NPV is +ve and reject it if the NPV is –ve.
•NPV of zero indicates that the benefits of the project are just enough to, recover the
capital invested, and earn the required return on the capital invested.
•A +ve NPV implies that the project earns an excess return.
•The NPV method is based on the assumption that the intermediate cash inflows of the
project are re-invested at the rate of return equal to the cost of capital.
•NPV takes into account the time-value of money.
Ex. Calculate the NPV of the given project and state whether the project is acceptable if the
rate of interest is 10 %.
Year Cash flow
0 - 1 00 000
1 + 2 00 000
2 + 2 00 000
3 + 3 00 000
4 + 3 00 000
5 + 5 50 000 Mr. Kiran Patil 37
•The NPV of a project is equal to the sum of the present values of all the cash flows associated
with it.
•??????????????????=�??????�
1
1+??????
0
+ �??????1
1
1+??????
1
+ �??????2
1
1+??????
2
+ …………… . + �??????�
1
1+??????
??????
•∴ ??????????????????=
�??????�
1+??????
0
+
�??????1
1+??????
1
+
�??????2
1+??????
2
+ …………… . +
�??????�
1+??????
??????
•Accept the project if its NPV is +ve and reject it if the NPV is –ve.
•NPV of zero indicates that the benefits of the project are just enough to, recover the
capital invested, and earn the required return on the capital invested.
•A +ve NPV implies that the project earns an excess return.
•The NPV method is based on the assumption that the intermediate cash inflows of the
project are re-invested at the rate of return equal to the cost of capital.
•NPV takes into account the time-value of money.
Ex. Calculate the NPV of the given project and state whether the project is acceptable if the
rate of interest is 10 %.
Year Cash flow
0 - 1 00 000
1 + 2 00 000
2 + 2 00 000
3 + 3 00 000
4 + 3 00 000
5 + 5 50 000 Mr. Kiran Patil 37
•Solution:
•??????????????????=
�??????�
1+??????
0
+
�??????1
1+??????
1
+
�??????2
1+??????
2
+
�??????3
1+??????
3
+
�??????4
1+??????
4
+
�??????5
1+??????
5
• =
− 1 00 000
1+0.10
0
+
2 00 000
1+0.10
1
+
2 00 000
1+0.10
2
+
3 00 000
1+0.10
3
+
3 00 000
1+0.10
4
+
5 50 000
1+0.10
5
• = −100000 + 181818.18 + 165289.25 + 225394.44+ 204904.03 +
341507.73
• = + 1018913.63
•Since the NPV is +ve, the project is acceptable.
Mr. Kiran Patil 38
•??????????????????=
�??????�
1+??????
0
+
�??????1
1+??????
1
+
�??????2
1+??????
2
+
�??????3
1+??????
3
+
�??????4
1+??????
4
+
�??????5
1+??????
5
• =
− 1 00 000
1+0.10
0
+
2 00 000
1+0.10
1
+
2 00 000
1+0.10
2
+
3 00 000
1+0.10
3
+
3 00 000
1+0.10
4
+
5 50 000
1+0.10
5
• = −100000 + 181818.18 + 165289.25 + 225394.44+ 204904.03 +
341507.73
• = + 1018913.63
•Since the NPV is +ve, the project is acceptable.
Mr. Kiran Patil 38
•Internal Rate of Return (IRR)
•The IRR of a project is the discount rate which makes its NPV zero.
•0=
�??????�
1+�
0
+
�??????1
1+�
1
+
�??????2
1+�
2
+ ……………. +
�??????�
1+�
??????
•Where r = discount rate
•In the NPV calculation, we assume that the discount rate (cost of capital) is known and
determine the NPV of the project.
•In the IRR calculation, we set the NPV equal to zero and determine the discount rate
(IRR) which satisfies the condition.
Ex. Calculate the IRR of the given project.
•IRR
•150000=
35000
1+�
1
+
40000
1+�
2
+
45000
1+�
3
+
50000
1+�
4
•Trial 1 Assume i= 4%
Year Cash flow
0 1 50 000
1 35 000
2 40 000
3 45 000
4 50 000
Mr. Kiran Patil 39
•The IRR of a project is the discount rate which makes its NPV zero.
•0=
�??????�
1+�
0
+
�??????1
1+�
1
+
�??????2
1+�
2
+ ……………. +
�??????�
1+�
??????
•Where r = discount rate
•In the NPV calculation, we assume that the discount rate (cost of capital) is known and
determine the NPV of the project.
•In the IRR calculation, we set the NPV equal to zero and determine the discount rate
(IRR) which satisfies the condition.
Ex. Calculate the IRR of the given project.
•IRR
•150000=
35000
1+�
1
+
40000
1+�
2
+
45000
1+�
3
+
50000
1+�
4
•Trial 1 Assume i= 4%
Year Cash flow
0 1 50 000
1 35 000
2 40 000
3 45 000
4 50 000
Mr. Kiran Patil 39
•150000=
35000
1+0.04
1
+
40000
1+0.04
2
+
45000
1+0.04
3
+
50000
1+0.04
4
•150000 = 33653.85 + 36982.25 + 40004.84 + 42740.21
•150000 = 153381.14
•LHS < RHS
•Trial 2 i= 5 %
•150000=
35000
1+0.05
1
+
40000
1+0.05
2
+
45000
1+0.05
3
+
50000
1+0.05
4
•150000 = 33333.33+ 36281.18 + 38872.69+ 41135.12
•150000 = 149622.32
•LHS > RHS
•It shows that i is lies in between 4% to 5%
•To determine closet rate of return
•If i= 4% -------3381.15 (153381.14- 150000)
•If i= 5% -------377.68 (150000- 149622.32)
•Find sum of absolute values = 3381.15+377.68= 3758.83/-
•Calculate ratio=
377.68
3758.83
= 0.10
•Subtract the ratio obtained into discount rate= 5-0.10= 4.90%
Mr. Kiran Patil 40
35000
1+0.04
1
+
40000
1+0.04
2
+
45000
1+0.04
3
+
50000
1+0.04
4
•150000 = 33653.85 + 36982.25 + 40004.84 + 42740.21
•150000 = 153381.14
•LHS < RHS
•Trial 2 i= 5 %
•150000=
35000
1+0.05
1
+
40000
1+0.05
2
+
45000
1+0.05
3
+
50000
1+0.05
4
•150000 = 33333.33+ 36281.18 + 38872.69+ 41135.12
•150000 = 149622.32
•LHS > RHS
•It shows that i is lies in between 4% to 5%
•To determine closet rate of return
•If i= 4% -------3381.15 (153381.14- 150000)
•If i= 5% -------377.68 (150000- 149622.32)
•Find sum of absolute values = 3381.15+377.68= 3758.83/-
•Calculate ratio=
377.68
3758.83
= 0.10
•Subtract the ratio obtained into discount rate= 5-0.10= 4.90%
Mr. Kiran Patil 40
•Benefit-Cost Ratio (B.C.R.)
•This method is used for economic analysis of public projects because profit is almost
never involved. Such projects have multiple benefits which cannot be measured in terms
of rupees.
•�.�.??????.=
??????������ ????????????��� �� �����??????��
??????�??????�????????????� ??????���������
=
????????????�
??????
•Present value of benefits = ????????????�=
�����??????� ??????� � �ℎ ??????�??????�
1+??????
??????
•If B.C.R. > 1, the project is acceptable.
•If B.C.R. = 1, the project is indifferent.
•If B.C.R. < 1, the project is not acceptable.
•Example: A project is having initial investment of Rs. 10 00 000. The rate of interest is 12
%. The benefits are as below,
•Year 1: Rs. 2 50 000
•Year 2: Rs. 4 00 000
•Year 3: Rs. 4 00 000
•Year 4: Rs. 5 00 000
•State whether the project is acceptable.
Mr. Kiran Patil 41
•This method is used for economic analysis of public projects because profit is almost
never involved. Such projects have multiple benefits which cannot be measured in terms
of rupees.
•�.�.??????.=
??????������ ????????????��� �� �����??????��
??????�??????�????????????� ??????���������
=
????????????�
??????
•Present value of benefits = ????????????�=
�����??????� ??????� � �ℎ ??????�??????�
1+??????
??????
•If B.C.R. > 1, the project is acceptable.
•If B.C.R. = 1, the project is indifferent.
•If B.C.R. < 1, the project is not acceptable.
•Example: A project is having initial investment of Rs. 10 00 000. The rate of interest is 12
%. The benefits are as below,
•Year 1: Rs. 2 50 000
•Year 2: Rs. 4 00 000
•Year 3: Rs. 4 00 000
•Year 4: Rs. 5 00 000
•State whether the project is acceptable.
Mr. Kiran Patil 41
•Solution: �.�.??????.=
????????????�
??????
=
??????�??????��????????????
1+??????
??????
??????
•B.C.R. =
2 50 000
1+0.12
1
+
4 00 000
1+0.12
2
+
4 00 000
1+0.12
3
+
5 00 000
1+0.12
4
10 00 000
=
11 44 563
10 00 000
= 1.14
•B.C.R. > 1, the project is acceptable.
•Payback Period
•The payback period is the length of time required to recover the initial investment on the
project. e.g. If a project involves an initial investment of Rs. 5,00,000 and generates cash
inflows of Rs. 1,50,000, Rs. 1,50,000 and Rs. 2,00,000 in the first, second and third year
respectively, its payback period is three years because the sum of cash inflows during three
years is equal to the initial investment.
•∴ ??????�??????���?????? ���??????�� =
??????�??????�????????????� ??????���������
����??????� ????????????�ℎ ??????�����
=
�����??????�
1+??????
??????
Merits:
• It is simple and does not involve tedious calculations.
•It favours projects which generate substantial cash inflows in earlier years.
Demerits:
•The method does not consider time-value of money. Cash inflows are simply added
without suitable discounting.
•It overlooks cash flows beyond the payback period. This leads to injustice against the
projects which generate substantial cash inflows in later years.
•Payback period considers return of investment and not return on investment
Mr. Kiran Patil 42
????????????�
??????
=
??????�??????��????????????
1+??????
??????
??????
•B.C.R. =
2 50 000
1+0.12
1
+
4 00 000
1+0.12
2
+
4 00 000
1+0.12
3
+
5 00 000
1+0.12
4
10 00 000
=
11 44 563
10 00 000
= 1.14
•B.C.R. > 1, the project is acceptable.
•Payback Period
•The payback period is the length of time required to recover the initial investment on the
project. e.g. If a project involves an initial investment of Rs. 5,00,000 and generates cash
inflows of Rs. 1,50,000, Rs. 1,50,000 and Rs. 2,00,000 in the first, second and third year
respectively, its payback period is three years because the sum of cash inflows during three
years is equal to the initial investment.
•∴ ??????�??????���?????? ���??????�� =
??????�??????�????????????� ??????���������
����??????� ????????????�ℎ ??????�����
=
�����??????�
1+??????
??????
Merits:
• It is simple and does not involve tedious calculations.
•It favours projects which generate substantial cash inflows in earlier years.
Demerits:
•The method does not consider time-value of money. Cash inflows are simply added
without suitable discounting.
•It overlooks cash flows beyond the payback period. This leads to injustice against the
projects which generate substantial cash inflows in later years.
•Payback period considers return of investment and not return on investment
Mr. Kiran Patil 42
Ex: Using payback period method suggest weather following project the expected if expected
pay back period is 4 years.
1.Initial investment= 2,00,000
2.Benefit for 1
st
year= 50,000
3.Benefit for 2
nd
year= 80,000
4.Benefit for 3
rd
year= 1,00,000
5.Benefit for 4
th
year= 1,50,000
If rate of interest is 10%.
Solution: Given data
1)P= 2,00,000 2) i=10% 3) F1= 50,000 4) F2= 80,000 5) F3= 1,00,000 6) F4=
1,50,000
Payback period:
Year Cash flow Discount factor Discounted cash flow Net Investment
0 -2,00,000 1.00 -2,00,000 -2,00,000
1 50,000 0.91 45,500 -1,54,500
2 80,000 0.83 66,400 -88,100
3 1,00,000 0.75 75,000 -13,100
4 1,50,000 0.68 1,02,000 88,900
Mr. Kiran Patil 43
pay back period is 4 years.
1.Initial investment= 2,00,000
2.Benefit for 1
st
year= 50,000
3.Benefit for 2
nd
year= 80,000
4.Benefit for 3
rd
year= 1,00,000
5.Benefit for 4
th
year= 1,50,000
If rate of interest is 10%.
Solution: Given data
1)P= 2,00,000 2) i=10% 3) F1= 50,000 4) F2= 80,000 5) F3= 1,00,000 6) F4=
1,50,000
Payback period:
Year Cash flow Discount factor Discounted cash flow Net Investment
0 -2,00,000 1.00 -2,00,000 -2,00,000
1 50,000 0.91 45,500 -1,54,500
2 80,000 0.83 66,400 -88,100
3 1,00,000 0.75 75,000 -13,100
4 1,50,000 0.68 1,02,000 88,900
Mr. Kiran Patil 43
•Discount Factor
For 0 years=
1
1+??????
??????
=
1
1+0.1
0
=1
For 1 year=
1
1+0.1
1
= 0.91
For 2 year=
1
1+0.1
2
= 0.83
For 1 year=
1
1+0.1
3
= 0.75
For 1 year=
1
1+0.1
4
= 0.68
Payback period of this project is within 4 years so accept the project.
Mr. Kiran Patil 44
For 0 years=
1
1+??????
??????
=
1
1+0.1
0
=1
For 1 year=
1
1+0.1
1
= 0.91
For 2 year=
1
1+0.1
2
= 0.83
For 1 year=
1
1+0.1
3
= 0.75
For 1 year=
1
1+0.1
4
= 0.68
Payback period of this project is within 4 years so accept the project.
Mr. Kiran Patil 44
Break Even Point:
•Break even point is that level of output or sales at which no profit or loss achieved. A
business is said to have break even when its income equals its expenditure.
•When production excess BEP the business makes profit & when it is below the BEP the
business makes loss.
Mr. Kiran Patil 45
•Break even point is that level of output or sales at which no profit or loss achieved. A
business is said to have break even when its income equals its expenditure.
•When production excess BEP the business makes profit & when it is below the BEP the
business makes loss.
Mr. Kiran Patil 45
•Sales as expenditure in Rs. Is represented on Y-axis & output is represented on X-axis.
•Line (A) represented the fixed cost, line (B) represented total cost & line (C) represented
sales & revenue cost & indicates income at various levels of output.
•The point where the line B & C intersect at each other is called as Break even point (BEP).
•The space between line B & C to the right of BEP represent profit where as to the left of
BEP represent loss. The amount of loss or profit can be measured in vertical scale.
Basic elements of break even analysis:
1.Fixed cost:
Fixed costs are those which do not vary with the volume of production. Such costs are usually
independent of volume of production. They include depreciation, insurance, salaries of clerical
& technical staff, indirect labor cost etc.
2. Variable cost:
Variable costs are those which vary directly with the valume of production. Such costs are
dependent on volume of production. They include costs direct materials, direct labours, power
& supply etc.
3. Total costs:
Total cost is the sum of fixed & variable cost.
4. Break even point:
It is the situation where there is no profit or no loss ie revenues & costs are equal.
5. Angle of incidence:
It is an angle at which total revenue line intersect total cost line. The magnitude of this angle Mr. Kiran Patil 46
•Line (A) represented the fixed cost, line (B) represented total cost & line (C) represented
sales & revenue cost & indicates income at various levels of output.
•The point where the line B & C intersect at each other is called as Break even point (BEP).
•The space between line B & C to the right of BEP represent profit where as to the left of
BEP represent loss. The amount of loss or profit can be measured in vertical scale.
Basic elements of break even analysis:
1.Fixed cost:
Fixed costs are those which do not vary with the volume of production. Such costs are usually
independent of volume of production. They include depreciation, insurance, salaries of clerical
& technical staff, indirect labor cost etc.
2. Variable cost:
Variable costs are those which vary directly with the valume of production. Such costs are
dependent on volume of production. They include costs direct materials, direct labours, power
& supply etc.
3. Total costs:
Total cost is the sum of fixed & variable cost.
4. Break even point:
It is the situation where there is no profit or no loss ie revenues & costs are equal.
5. Angle of incidence:
It is an angle at which total revenue line intersect total cost line. The magnitude of this angle Mr. Kiran Patil 46
indicates the level of loss or profit.
Larger the angle of incidence, higher will be the profit & vice-versa.
6. Margin of incidence:
It is excess of budgeted sales or actual sales over the break even sales volume i.e.
Margin of safety= Budget sales- sales at BEP.
Mr. Kiran Patil 47
Larger the angle of incidence, higher will be the profit & vice-versa.
6. Margin of incidence:
It is excess of budgeted sales or actual sales over the break even sales volume i.e.
Margin of safety= Budget sales- sales at BEP.
Mr. Kiran Patil 47
Categories
GeneralDownload
Download Slideshow Get the original presentation fileQuick Actions
Statistics
- Views 60
- Slides 47
- Age 294 days
Related Slideshows
22
Pray For The Peace Of Jerusalem and You Will Prosper
RodolfoMoralesMarcuc
36 views
26
Don_t_Waste_Your_Life_God.....powerpoint
chalobrido8
40 views
31
VILLASUR_FACTORS_TO_CONSIDER_IN_PLATING_SALAD_10-13.pdf
JaiJai148317
35 views
14
Fertility awareness methods for women in the society
Isaiah47
32 views
35
Chapter 5 Arithmetic Functions Computer Organisation and Architecture
RitikSharma297999
31 views
5
syakira bhasa inggris (1) (1).pptx.......
ourcommunity56
32 views