1
ABE 306: Engineering Economics
Credit unit = 2
This lecture note was prepared by Dr. K. O. Yusuf
Department Agricultural and Biosystems Engineering,
Faculty of Engineering and Technology, University of Ilorin, Ilorin, Nigeria
The lecture note is not for sale and should not be converted to textbook by any other person without
the consent of Dr. K. O. Yusuf
Course outline
1.
2.
3.
4.
The nature and scope of economics
Basic concepts of engineering economy
Interest formulae
Discounted cash flow, present worth, equivalent annual growth and rate of
return comparisons
5. Replacement analysis
6. Depreciation
7. Breakdown analysis
8. Benefit-cast analysis
9. Minimum acceptable rate of return
10. Judging attractiveness of proposed investment
Useful textbooks
1. Mao, K. (2006). Engineering Economics: overview and application in process Engineering
industry, 10.490 ICE
2. Newnan, D.G, Erchenbach, T.G and Larella, J.P (2004). Engineering Economics analysis, 9th
Edition, Oxford University press, New York.
3. Park, C.S (2007). Contemporary Engineering Economics 4th Edition, Pearson Prentice Hall,
Now Jersey.
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Introduction and definition of Engineering Economics
Before starting with Engineering Economics let us look at the definition of Economics so that the
knowledge can be applied to Engineering Economics. Economics is a social science which deals
with human behaviour as a relationship between ends and scarce means that have alternative uses.
Commodities or items are not readily available to meet the demand of man, human’s wants or needs
are unlimited and the resources or the money is insufficient to procure the items. Therefore, there is
need to prioritize our items according to the way they are needed and usefulness on a list called scale
of preference. With this brief definition of Economics and scope of Economics, we can now go to
Engineering Economics.
Engineering Economics was formerly called Engineering Economy, it deals with methods which are
used for justification and selection of project (s) from the available alternatives. Engineering
economics can be defined as the application of economic principles to evaluate alternatives by
computing the specific measure of value of money (cash flow) or project over a given period of time.
In the past, Engineers were concerned mainly with design and fabrication of machines, equipment
and tools. Today, Engineers have many challenges confronting them which make design and
fabrication difficult. This is so because natural resources and materials for construction are becoming
scarce and expensive. The demand for materials and natural resources for engineering work are
higher than the available resources and this normally makes cost of production of items higher. Not
only that, people are conscious of the negative effects of engineering innovations that normally cause
environmental degradation and pollution which could be hazardous to man.
Engineers must carry out analysis to know if a particular project is profitable, if there are
alternatives, cost of producing the project, choose a project that is more economical, profitable and
choose the best alternative for the project from the available alternatives.
The basis for selecting alternative(s) could be because of being economical simplicity and other
benefits. The focus of Engineering Economics is economic justification and selection of alternative
(s) with regards to the project (s), manufacturing, processing and investment. Engineers must carry
out analysis before taken decision.
Areas where engineering economics decision are needed are as follows:
(i)
Equipment and project selection
(ii)
Equipment replacement
(iii)
New product and product or project expansion
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(iv)
Cost of production
(v)
Public works
(vi)
Cost effectiveness of materials and production.
Engineering Economic Concepts
The basic concept of engineering economics is the principle of time value of money. This means that
time value of money is very important concept of engineering economics. Other concepts are:
(i)
Internal rate of return, interest rate, cash flow, equivalent worth of money and utility.
Economic concepts are usually qualitative in nature and not universal in application.
Utility: is the power of a goods or service to satisfy human needs (wants).
Value: This means the worth that a person attaches to an object or service. It is a measure or
appraisal of utility and is not the same as cost or price.
Utility goods: This could be as follows:
Consumer goods: goods and services that directly satisfy human wants (needs) e.g foods, clothing,
TV, houses, shoe and other items.
Producer goods: goods and services that satisfy human wants indirectly as part of the production or
construction process such as equipment, industrial chemicals and materials.
In Engineering Economics, we need to know the following because they are very important before a
decision is taken.
(i)
Time value of money
(ii)
Estimation of cash flows
(iii)
Quantitative measurements of profitability
(iv)
Comparison of alternatives before taken a decision
Time value of money
Time value of money is the relationship between interest and time. This means that time value of
money is characterized by the interest and time. For example
P
0
Present
P+I=F
1
2
3
4
5
6
7
8
9
Future
10 years
4
Where: P = present worth (value) of the money, I = interest on money after 10 years and F = Future
worth of the money after 10 years.
Money has time value because the purchasing power of money changes within short period of time
especially naira.
Interest: is the amount of money that is charged by financial institution or pay for the use of
borrowed capital to finance an enterprise or project over a given period of time.
Sources of income or capital for a project
The initial capital that can be used to start and finance a project or company can be obtained from the
following.
(i)
Loan from the bank, Loan from cooperative and Contribution
Purchasing power of money
Price of goods and services can go upward and downward. Therefore, purchasing power of money is
not stable or fixed but it changes with time.
i.
Price reduction: this occurs when there is increase in productivity and availability of
goods in the market for sale.
ii.
Price increase: this occurs due to government policies, price support scheme and deficit
financing.
Cash flow diagram
Cash flow diagrams are the means in form of a diagram for visualizing and simplifying the flow of
income (receipt) or expenditure (disbursement) for the acquisition and operation of items in an
enterprise.
Revenue: is the income or receipt and is represented by upward line arrow or upward pointed arrow.
The upward line arrow is called positive cash flow.
Disbursement: This is cash out flow and is normally negative cash flow. It is denoted by downward
pointed arrow. All disbursement and revenue are assumed to take place at the end of the year in
which they occur. This is known as the “End of year convection. Note that expenses incurred before
time t equals to zero (t = 0) are sunk costs and are not considered in the analysis.
Cash flow: is the sum of money recorded as receipts or disbursements in a project financial records.
N1120
N120 N120 N120 N120 N120
Expenses
Receipts
N1000
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From the cash flow diagram, the expenses are indicated by the arrows pointing downward while the
receipts (income) are denoted by the arrows pointing upward.
Time value of money changes and depreciates with time especially Nigerian currency (naira). This
means that time affect the worth or value of money. For instance, a TV set bought in 1983 at a rate of
N300 may be N30,000 in 2010.
Period of interest: is the time frame over which the interest or net return on investment is evaluated.
Interest formulae
Interest rate formula used is important in economic evaluation of engineering alternatives.
Types of Interest Formulae
(i)
Simple interest
(ii)
Compound interest
Simple interest = I
I=
𝑃𝑇𝑅
100
P = principal (limited) money, T = time in year, R = interest rate in %
I = interest charges at the end of the period
Example 1: Mr. Benson borrowed N100,000 at a simple interest of 17% per annum for 2years.
Calculate the simple interest on the money borrowed and total amount of money which Mr. Benson
will pay at the end of the 2 years.
I=
I=
𝑃𝑇𝑅
P = N100,000, T = 2 years, R = 17%
100
100000 𝑥 2 𝑥 17
100
= N34,000
At the end of 2nd year, Mr. Benson will pay N100,000+N34,000 = N134,000
Compound interest
Compound interest is the interest which banks normally use to determine their interest. It is based on
yearly interest.
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Use compound interest to solve example 1
Solution P = N100,000, R = 17% T = 2 but for 1 years interval
First year I =
100000𝑥1𝑥17
100
= N17,00
At the end of the first year, the debt = N117,000 and this will be the principal for the second year
2nd year: I =
117000 𝑥 1 𝑥 17
100
= 19,890, I = N19,890
Total debt = N117,000 +19,890 = N136,890
Derivation of a single payment compound interest formula
Notations
P = present worth=present value, F = future worth, i = interest rate (in fraction)
A = Annual worth or Annuity
Again, P is also principal
T = N = n = time = number of years
I = P n ἰ for the first year
First year: P x 1 x ἰ = Pἰ
F1 = P + Pἰ, F1 = P(1+ἰ), now P = F1
For the second year
y2 = I = P (1+ἰ) x 1 x i = P (1+ἰ) ἰ
∴F2 = P (1+ἰ) + P (1+ἰ) ἰ
F2 = Pἰ (1+ἰ) + P (1+ἰ)
(Pἰ+P)(1+ἰ) = P (1+ἰ)(1+ἰ) = P (1+ἰ)2
Similarly, F3 = P (1+ἰ)3 and F4 = (1+ἰ)4
∴Fn = P ( + ἰ)n
this is formula is for compound interest. This is simply as
F = (1+ ἰ)n
Fn means amount of money to be paid at the end of n-year considered
Example 2: Calculate the future worth of N100,000 borrowed from UBA which must be paid back
after 2 years with 17% interest rate using compound interest formula.
Solution P = N100,000, n = 2 and interest rate of 17%
17
ἰ =17% = 100 = 0.17
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F = P(1+ ἰ)n
F = 100,000 (1+0.17)2 = 100000(1.17)2v=136,890
F = N136,890
Class work: if the money is to be paid back at end of 5 years using the interest rate of 17% with P =
N100,000. Determine the future worth of the money using:
(i)
Single payment compound interest formula
(ii)
Normal compound interest formula (Ans = 219,244.80).
Example 3: If future worth of N100,000 is N136,890 after a certain period of time at interest rate of
17%. Calculate the number of years required to achieve it.
Fn = N136,890, P = N100,000, ἰ = 17%, n = ?
F = P (1+ ἰ)n
𝐹
𝑃
= (1+ ἰ)n take log of both sides
𝐹
𝑙𝑜𝑔 ( )
𝐹
𝑃
log (1+ ἰ)n =log (𝑃) n = log(1+ἰ)
𝑙𝑜𝑔
n=
10(
𝑙𝑜𝑔10
136,890
)
100,000
=
(1+0.17)
0.136371723
0.068185861
n = 2.00000015 = 2
Therefore, n = 2 years
Example 4: Dr. Taiye Johnson was given an appointment at the university of Ilorin in 1981 as a
lecturer I and his salary was N1350 (when N1 = USD1.50) and interest rate was 17%. Dr. Kehinde
Johnson was given the same appointment as a lecturer I in March, 2019 (38 years since 1981) at the
University of Ilorin and his salary was N320,000. Determine;
(i)
The future worth of salary of Dr. Taiye Johnson in 2019
(ii)
The difference in salary of Dr Taiye Johnson and Dr. Kehinde Johnson
(iii)
Who has better salary between Dr Taiye and Dr. Kehinde Johnson
(i)
Salary of Dr. Taiye Johnson in 1981 = N1350
P = N1350, ἰ =17% = 0.17, n = 38, F = ?
F = P (1+ ἰ)n
F = future worth of the money in 2019
F = 1350 (1+0.17)38 = 1350 (1.17)38 = 1350 (389.998)
F = 526,497.3
(ii)
= N526,497.30
526,497.30 -320,000.00 = N206,497.30
∴ The difference = N206,497.30
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Using USD = 1.5 x 1350 = USD 2025 in 1981
In 2019 when USD1 = N360
2025x360 = N729,000 in 2019
(iii)
Dr. Taiye Johnson had a better salary of N1350 in 1981 than Dr. Kehinde Johnson with
salary of N320,000 in 2019
Interest factors
Interest factor: this is a multiplication factor developed from the interest formula for a given interest
rate and period. The factor is used to convert cash flows occurring at different time to a common
time. The compound amount factor from the formula is (1 + ἰ)n but is normally represented by
(𝐹⁄𝑃 , ἰ%, n) in which (𝐹⁄𝑃 , ἰ, n) factor can be obtained from the table and then multiply by the
value of P to get the future worth (future value) which is denoted by F. (𝐹⁄𝑃 , ἰ, n) is called future
worth factor.
Again, if future worth sum is known or given at a given rate over a period of time, then, present
worth sum (P) could be calculated. The method of inverse of compounding from which present
worth sum is calculated when future worth sum is given is called discounting.
P = F (1+ ἰ)-n
P=F
∴P =
1
(1+ἰ𝑛 )
where
𝐹
1
(1+ἰ)𝑛
is the factor
P = F (𝑃⁄𝐹 , ἰ, n)
1+ἰ𝑛
from the table
Again, present worth could also be determined when annual worth sum called annuity (A) is given.
The formula is given as follow.
P=
𝐴 (1+ἰ)𝑛 −1
ἰ(1+ἰ)n
A=𝑃(
ἰ(1+ἰ)n
)
(1+ἰ)𝑛−1
From future worth sum when given annuity or annual worth sum (A)
(1+ἰ)𝑛 −1
F = A(
ἰ
) but (𝐹⁄𝐴, ἰ, n ) is called series compound amount factor
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ἰ
A = 𝐹(
(1+ἰ)𝑛 −1
) but (𝐴⁄𝐹, ἰ, n ) is called sinking fund factor which is the reciprocal of series
compound amount factor.
Summary of formulae for conversion of F to P and A
F = 𝑃(1 + ἰ)𝑛 − (1): 𝐹 = 𝑃 (𝐹⁄𝑃, ἰ, n )
P=
𝐹
(1+ἰ)𝑛
__(2): P = 𝐹 (𝑃⁄𝐹, ἰ, n )
from the table
from the table
(1+ἰ)𝑛 −1
P = A(
) _____(3): P = 𝐴 (𝑃⁄𝐴, ἰ, n )
ἰ(1+ἰ)𝑛
ἰ(1+ἰ)𝑛
A = P(
) _____(4): A = P(𝐴⁄𝑃, ἰ, n )
(1+ἰ)𝑛 −1
(1+ἰ)𝑛 −1
F = A(
)_____(5): F = A(𝐹⁄𝐴, ἰ, n )
ἰ
ἰ
A = F(
𝐴
)___(6): A = F(𝐹 , ἰ, 𝑛)
(1+ἰ)𝑛 −1
Capital recovery factor
It is important to be able to relate regular periodic payments to the present worth
A=P(
ἰ(1+ἰ)𝑛
𝐴
) but this could be obtained from the table as A = P (𝑃 , ἰ, 𝑛)
(1+ἰ)𝑛 −1
In this equation, the interest expression is called capital recovery factor. This is because the equation
defines or gives a regular income that is necessary to recover a capital from an investment but P
(𝑃⁄𝐴, ἰ, 𝑛)
series present worth factor.
Gradient.
It is occasionally necessary to treat a cash flow which regularly increases or decreases at each period.
Such pattern changes in cash flow is called gradient. The gradient may be in constant amount. This
means that the gradient could be linear (arithmetic progression) or geometric progression.
Arithmetic Gradient
A
A = Annuity = Annual worth
G = Arithmetic Gradient
A
G
A
2G
3G
4G
A
A
Formulae to find A when G is given and to find P when G is given
A
10
(1+ἰ)𝑛 −ἰ𝑛−1
A = G(
) - (7): A = G(𝑃⁄𝐺 , ἰ, 𝑛)
ἰ(1+ἰ)𝑛 −ἰ
 1  I n  in  1 
 - (8) : P = G (P/G, i, n)
P  G
n
2

 i 1  i 

Example 1: If N5000 is invested at 12% per annum compound interest in January, 1998, determine
the future worth sum of the money in January, 2008
N5000
N =10
0
1
𝑛
F = P(1 + ἰ) = ἰ = 12% = 0.12
3
2
4
5 6
F = 5000 (1 + 0.12)10
7
8
9
10
ἰ =12% = 0.12
10
F = 5000 (1.12) = 5000 (3.10584208) = 15,529.4
F=?
F = 15, 529.24 = N15,529.24
Using the Table to determine the factor
F = P(𝐹⁄𝑃 , ἰ, 𝑛) = 5000 (𝐹⁄𝑃 , 12% 10)
= 5000 (3.1058)
F = 15,529 = N15,529
Example 2 Use the value of future worth sum of N15,529.24 with the same interest rate of 12% and
10years to find the present worth sum (P)
Solution. This is to move from right to left.
P=?
0
1
2
3
4
5 6
7
8
9
10
ἰ =12%
P=
𝐹
(1+ἰ)𝑛
=
15529.24
(1+0.12)10
= 0.321973236
P = N4999.9997 ≈ N5,000
From the Table
P = F(𝑃⁄𝐹 , ἰ, 𝑛) = 15529.24 (𝑃⁄𝐹 , 12%, 10)
F =N15,529.24
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P = 15529.24 (0.3220) = 5000.42
P = N5000.42≈N5000
Example 3: Calculate the interest rate that is required on N5000 put in a bank to yield N15,000 in 10
years solution F = N15,000 P = N5000 n = 10 years
F = P (𝐹⁄𝑃 , ἰ, 𝑛) you can use future worth Table to solve this problem
𝐹
𝑝
=
15000
5000
= 3.0000
Now check
𝐹
𝑝
from the Table at 10 years period where you can get 3.0000 from the Table, 3.0000 is
not available.
Then, choose 2 values that are close to it and interpolate.
Recall that by interpolation, the actual interest rate required would be determined
From the Table interpolation
12% =3.1058
ἰ−11
12−11
=
3.000−2.8394
3.1058−2.8391
ἰr =3.0000
11% = 2.8394
ἰ-11 =
ἰ−11
1
=
0.1606
0.2664
0.1606𝑥1
0.2664
0.1606𝑥1
ἰ-11 = 0.2664 =0.602852852
ἰ =11+ 0.602852852 =11.602852852
Alternatively,
F  P1  i  ,
15000 50001  i 
10
n
1
 15000 10
1 i  
 =1.1161, i = 1.1161 – 1 = 0.1161 x 100. Therefore, i = 11.6%.
 5000 
now, let us check if we can get N15,000
F = P (1+ ἰ)n
, P = 5,000, n =10
F = 5000 (1 +
ἰ = value obtained
11.602852852 10
100
)
F=5000 (1.116029)10 =5000 (2.9975)
F = 14,987.50
F = N14,987.50
What is the difference between the actual figure of the money and the calculated figure.
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15,000 - 14,987.50 = N12.50
The error is due to approximation of values from the table
Class-work: Calculate the interest rate required on N5,000 that is put in a bank to achieve
N15,529.24 in 10 years.
Example 4: To calculate present worth sum (P) if the annual worth sum or annuity is N315.45 at
interest rate of 10% and in 4 years
Solution
Recall that
A=(
ἰ(1+ἰ)𝑛
(1+ἰ)𝑛 −1
) but P =A ( ἰ(1+ἰ)𝑛 )
(1+ἰ)𝑛 −1
(1+ἰ)𝑛 −1
∴P = A (
ἰ(1+ἰ)𝑛
) = A (P/A, i, n)
(1+0.10)4 −1
P =315.45 (
N315.45
N315.45
N315.45 N315.45
ἰ(1+ἰ)𝑛
10%=0.10,
)
1.4641−1
0.4641
) = 315.45 (0.14641)
0.1(1.4641)
P = 315.45 (
P =315.45 (3.169865446) = 999.9341
∴ P = N999.93≈N1000
Similarly, if P = N1000, ἰ =10% and n=4 determine the annual worth sum (annuity)
P =N1000
4 years ἰ =10%
A =P (
A
A
A
ἰ(1+ἰ)𝑛
(1+ἰ)𝑛 −1
)
0.14641
A= 1000 (
A
0.4641
) =1000 (0.315470803)
A = N315.47
Uniform Gradient series or Arithmetic Gradient
P
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To find P when G is given
ἰ(1+ἰ)𝑛 −1−ἰ𝑛
P=G(
ἰ2 (1+ἰ)𝑛
)
P =G (𝑃⁄𝐺 , ἰ%, 𝑛)
Note that for uniform gradient series, there is an increment every year with a constant value. To
convert a uniform gradient to present worth sum, you need to calculate annuity by converting the
annuity to present worth sum as P, and P2 for gradient to present worth sum.
Example 1: The yearly payment at the end of 1st, 2nd, 3rd, 4th and 5th year are N1000, N1500, N2000,
N2500 and N3000, respectively if the interest rate is 8%, calculate the equivalent present worth of
the payment and annual worth of the payment.
P
O
A
N1000
N1500
N2000
N2500
N3000
14
P1
P2
+
O
O
N1000
o
N1000
A
G=500
500
1000
1500
A to P
G to P
2000
𝑃
P2 = G( ⁄𝐺, 8%, 5) from the table
P1 = A(𝑃⁄𝐴, 8%, 5)
(1+ἰ)𝑛 −1
P1=A(
ἰ(1+ἰ)𝑛
 1  i n  1  in 
 using formula
P2  G
n
2



i
1

i


)
(1+0.08)5 −1
P1 =1000 (
=1000 (
1000 (
0.08(1.08)5
)
 1  0.085  1  0.08  5 

P2  500
2
5





0
.
08
1

0
.
08


1.4633− 1
0.11755
)
0.46933
)
0.11755
1.46933
P2 = 500(
(0.08)5 −1−0.008𝑥5
0.06933
1000(3.9926)
500 (
P1=3992.60
500(7.3724)
P1=N3992.60
0.0064(1.08)5
0.009404
)
)
P1 = N3,686.20
P= P1+ P2 = N3992.60 + 3,686.20 = 7678.8
∴ Present worth sum = N7678.80k
Similarly from the Table
P1=A(𝑃⁄𝐴 8%, 5) +P2=G (𝑃⁄𝐺 8%, 5)
1000 (3.9927) +500 (7.37123)
P =3992.70 +3685.65 = N7678.35k
∴The equivalent present worth of the payment = N7678.80k (from calculation without having table
to check value of conversion factor) or N7678.35k (from Table).
15
To convert uniform gradient to annual worth use the data provided in example 1 and calculate the
annual worth sum of the money and then convert the annual worth sum to present worth sum.
By calculation
Annual worth sum or annuity (A)
A = 1000 +G (𝐴⁄𝐺 , ἰ% 𝑛)
ἰ(1+ἰ)𝑛 ἰ𝑛−1
A = 1000 +500 (
ἰ(1+ἰ)𝑛 −1
G = N500
)
By calculation without table
(1+ἰ)𝑛 ἰ𝑛−1
A= G (
ἰ(1+ἰ)𝑛 −1
)
(1.08)5 −0.08𝑥5−1
A=1000+500 (
A=1000+500
A=1000+500
0.08(1.08)5 −0.08)
)
1.46933−0.4−1
0.08(1.46933)−0.08
0.06933
0.03755
=1000+500 (1.8463)
A=N1923.15k
∴ The annual worth sum for 5year = N1923.15
From table or using Table
A = P (𝐴⁄𝐺 , ἰ% 𝑛) = if P = N7678.23
ἰ = 8%, n = 5 years
A=7678.28(0.2505) = N1923.40k
Similarly, the present worth sum of the annual worth could be converted to PW (present worth) or
net present worth (NPW)
By calculation
P = A(𝑃⁄𝐴 8%, 5)

(1+ἰ)𝑛 −1
A(
ἰ(1+ἰ)𝑛
P=1923.15(
from the previous calculation
) A=N1923.15K now convert this A to NPV or NPW
(1.08)5 −1
)
0.08(1.08)5
P = 1923.15 (
0.46933
0.11755
)
1923.15 (
1.46933−1
0.08(1.46933)
= N7678.77K
)
16
Basic methods for making decision
The basic methods that are commonly used to analyze economic alternatives in order to make
decision in engineering economic are as follows:
(1) Present worth (PW or PV) method
(2) Future worth (PW) method
(3) Annual worth (AW) method
(4) Internal rate of Return (IRR) method
(5) External rate of Return (ERR) method
(6) Explicit Reimbursement Rate of Return (ERRR)
(7) Payback period
Example 3: Determine the annual amount of money that must be set aside in order to overcome
anticipated expenses in the following table if the interest rate is 8%.
End of year
2
3
5
8
Expenses (N)
12,000
6,000
20,000
13,000
P=?
0
2
3
4
1
N12,000
5
6
7
N6000
N20,000
N13,000
Solution
Step 1: Calculate the present worth sum
Step 2: convert the present worth sum to annual worth sum (A)


P = F P F ,8%,2 + F(𝑃⁄𝐹, 8%, 3) + F(𝑃⁄𝐹, 8%, 5) + F(𝑃⁄𝐹, 8%, 8)
From the Table
P = 12000(0.8573) + 6000(7938) + 2000 (0.6606) + 13000 (0.5403)
P = N35,686.30
7
By calculating the factor using formula is a little bit cumbersome but also accurate.
P = F(𝑃⁄𝐹 ἰ%, 𝑛)
17
P = F(1 + ἰ)−𝑛 =
P=
12000
(1.08)2
𝐹
(1+ἰ)𝑛
6000
20000
+ (1.08)3 + (1.08)5 +
13000
(1.08)8
10,288.07+4,762.99+13,611.66+7,023.50=35686.22
From calculation
PW = N35,686.22K
Now convert this PW to annual worth
Solution
A= P (𝐴⁄𝑃, 8%, 8) = 35686.22 (0.1740) A = N6209.40K
Or, A = P (
ἰ(1+ἰ)𝑛
),
(1+ἰ)𝑛 −1
0.1480744168
35656.22 (
0.85093021
A = 3586.22
0.08(1.08)8
(1+0.08)8 −1
, 35686.22 (
0.08(1.85093021)
18.5093021−1
)
) = 35686.22 (0.17401476) = N6,209.93
∴ sum of N6,209.93k should be set aside every year to take care of the expenses.
Example 2: Mr. Benson is planning to embark on a project with an initial capital of N10,000 which
is expected to yield annual revenue of N5,310 for 5 years. The project has a salvage value of N2000.
Annual disbursement is N3000 for the operation and maintenance of the project. The interest rate on
the capital project is 10%. Show whether this project is a desirable investment.
Solution
Salvage value = SV is the market value of worth of an item or asset at the end of its useful life. SV =
future worth at the end of 5 years = F = future worth
Method to use = annual worth sum (A) for this method, you convert initial capital and salvage value
to annual worth sum (A) for comparison.
N5310
N5310
N5310
N5310
N5310
+5V of N2000
N10,000
0
1
2
3
4
N3000
N3000
N3000
N3000
5
N3000
18
Note that SV is still part of revenue which Mr. Benson can sell at the end of the 5 th year. Net annual
worth (NAW) or Net annual value (NAV)
The methods that can be used and the equations to use are:
(1) Total annual revenue – total disbursement = NAV or
(2) AW = P (𝐴⁄𝑃 ἰ%, 𝑛) - SV (𝐴⁄𝐹 ἰ%, 𝑛) − − − (1)
If Net Annual Worth (NAW) ≥ 0 the project is good and profitable but
If NAW < 0, the project is not profitable
First method:
Total disbursement => but convert initial capital to annual worth
P = N10,000 => initial capital +
Annual disbursement on operation and maintenance of N3000, ἰ =10%, n = 5years
𝐴
AW of initial capital =P(𝑃 , 10%, 5)
From the table
AW=1000 (02638) =N2638
If table is not available and the calculation is done using formula
𝐴
ἰ (1+ἰ)n
𝑃
(1+ἰ)𝑛 −1
AW=P ( , ἰ%, 𝑛) = 𝑃 (
0.1 (ἰ+1)5
AW =1000 ( (ἰ+1)5 −1 ) =1000
AW =
0161051
1.61051
)
01(1.61051)
1.61051−1
=10000 (0.26379748)
To 4d.p
AW = 10000 (02638) = N2638, Therefore, N10,000 present worth = N2,638 AW
Total disbursement = 2638 + 3000 = N5638.00
Total revenue =5310 + salvage value
SV = F, this is so because at end of 5th year the item would be sold
A = F(𝐴⁄𝐹 , ἰ%, n), A = SV (𝐴⁄𝑆𝑉 , ἰ%, n) from table 0.1638 at ἰ =10% n=5 ,
A = 2000 (
0.1
0.1
) = 2000 (0.61051)
(1+1)5 −1
A = 2000 (0.16379748) =2000 (0.1638) = 327.6
A = N327.60k
Total annual revenue = 5310 + 327.60 = 5637.60
NAW = Revenue – Disbursement
NAW = 5637 – 5638.00 = -N0.4 = 40k = 40 kobo (100k = N1)
19
Second method by converting o present worth
Annual disbursement = AD
(𝑙+𝑖)𝑛 −1
(1+0.1)5 −1
𝑖(𝑙+𝑖)
0.1 (1.1) 5)
=
𝑛
=
0.61051
0.161051
ADP = P + 3000 (𝑃⁄𝐴, 10%, 5), 11000+ 3000 (3.7908)
10,000+ 11372. 40 = #21,372.40
Annual revenue to present worth
ARP= 5310 (𝑃⁄𝐴, 10% , 5) + 2000 (𝑃⁄𝐹, 10% , 5)
3.7908 also
0.6209 from the table
ARP = 5310 (3.7908) + 2000 (0.6209)
20,129.15 +1,241.80 = N21,370.95
NPW = Revenue – Disbursement
NPW = 21370.95 - 21372.40 = -1.45
From the calculation above, NPW< 0
Now let us make or take decision based on the calculation
Our decision from the analysis is that the project is not profitable
Therefore, don’t embark on the project.
Example 3: Use the information provided in the table to carry out the analysis using annual worth
method and choose the best oil extracting machine that is economical for a particular company. The
minimum required rate of return is 8%
S/No
Cost
Machine A (N)
Machine B (N)
1
Initial cost
26,000
36,000
2
Annual maintenance cost
800
300
3
Annual labour cost
11,000
7,000
4
Annual expenditure
-
2,600
5
Salvage value
2,000
3,000
6
Expected useful life in year
6
10
Machine A
Annual maintenance + labour cost + expenditure,
SV=N2000
A=?
6yr
N800
N26,000
800 + 11,000 + 0 = N11,800
N11,000
20
Note that salvage value is a revenue SV = F = N2000
Total annual cost of operating machine A
AC= P(𝐴⁄𝑃 , 8%, 6) + 11,800 – SV (𝐴⁄𝐹 , 8%, 6)
AC=26000 (0.2163) +11800 – 2000 (0.1363) = N17,151.20
Machine B has a SV of N3,000
Annual cost of maintenance + labour + expenditure = 300 + 700 + 2600 = N9,900
AC = AW = P (A/P, 8%, 10) + 9,900 – 3000 (A/F, 8%, 10)
AC = 36000 (0.1490) + 9,900 – 3000 (0.0690)
5364 + 9900 – 207 = N15,057
Cost of running machine A is N17,151.20k and for machine B is N15,057.00. Therefore, machine B
is economically better for the company than machine B. In the computation to take the decision,
present worth method was not used but annual cost (AC) because of unequal useful life in years of
machine A and machine B.
Third Method: Internal Rate of Return (IRR)
Internal rate of return: This is also known as investor method. IRR is the interest rate that makes the
net present value (NPV) of all cash flow equal to zero. With this method, the cash flow of the project
is converted to present worth sum and net present worth (NPW) or NPV is determined.
NPV = PVin – PVout must be equal to zero for good IRR
Example 1: A lathe machine is purchased at a rate of N100,000 for producing machine tools. The
lathe machine has a useful life of 10 years with salvage value of N20,000. The machine can generate
N40,000 as annual revenue while N22,000 is needed as annual disbursement for the operating cost
and maintenance of the machine. If the minimum acceptable rate of return (MARR) is 13%, calculate
the required internal rate of return. Cash flow diagram
A=N40,000
1
2
P=N100,000
3
4
5
+SV=N20,000
6
7
8
9
10
21
Note that IRR method is based on trial and error or iteration until a reasonable value is obtained.
With this method there may be need for interpolation to get the required value of internal rate of
return (ἱ).
Note that
Arrows pointing upward =Revenue
Arrows pointing downward =Expenses or disbursement
Assumption: let the initial IRR be =12%
Now, convert all cash flows to PW (PV)
PWin = A (𝑃⁄𝐴, 12% , 10) + 5𝑉 (𝑃⁄𝐹, 12% , 10)
5.6502from the Table 0.3220 from Table
PWin = 40,000 (5.6502) + 20000 (0.3220)
PWin = N232,448.00
PWout = P + A (P/A, 12%, 10)
PWout = 100,00 +22,000 (5.6502)=
100,000+124,304.40 =N224,304.40
Net present worth or net present value
NPV = PWin – Pwout
NP1 = 232,448.00 - 224,304.40 = N8,143.60
NPV = N8143.6 which is greater than zero
Therefore, choose another value of IRR >12%
Let IRR = 13% => ἱ = 13% for new iteration
PWin = 40,000 (𝑃⁄𝐴, 13% , 10) + 𝑆𝑉 (𝑃⁄𝐹, 13% , 10)
= 40000 (5.4262) +20,000 (0.2946) = PWin = N222,940
PWout =100,000 +22,000 (𝑃⁄𝐴, 13% , 10)
=100,000 +22,000 (5.4262) = 219376.4
NPV2 = 222,940 - 219,376.4 = N3563.6 > 0
NPV1 and NPV2 > 0
Then, try IRR = 14% to check if is close to zero
If IRR = 14%
PWin = PVin = 40,000 (𝑃⁄𝐴, 14% , 10) + 𝑆𝑉 (𝑃⁄𝐹, 14% , 10)
PWin = 40,000 (5.2161) +20000 (0.2697) = 214,038
22
PWout =100,000 + 22000 (𝑃⁄𝐴, 14% , 10)
100,000+22,000 (5.2161) = N214,754.20
NPV = PVin – Pvout
= 214,038 - 214,754.20 = - N716.20 which is closer to zero than 12%
The required IRR is between 13% and 14%
13% =-716.20 the required IRR is expected to give 0.00
IRR? =0.00
14% =3,563.60
Let us do the interpolation
𝐼𝑅𝑅 − 13
0.00 − 3,563.60
=
14 − 13
− 716.20 − 3,563.60
𝐼𝑅𝑅 − 13 −3,563.60
=
1
−4,279.80
IRR – 13 = 0.83266
IRR = 13 + 0.8327 = 13.8327%
∴ IRR = 13.8327% but to 2 d.p =13.83%
Let us check the NPV or NPW if is =zero
IRR =13.83%
PWin = 40,000 (𝑃⁄𝐴, 13.83% , 10) + 20,000 (𝑃⁄𝐹, 13.83% , 10)
= 40,000 (
40,000 (
(1+ἱ)𝑛 −1
ἱ`(1+ἱ)𝑛
1
) + 20,000 ((1+ἱ)𝑛 )
(1.1383)10 −1
1
) + 20,000 ((1.1383)10)
0.1383(1.138310 )
3.65231−1
40000 (
0.1383(3.65231)
2.65231
40,000 (
0.50511
) + 20000 (0.2738)
) + 20000 (0.2738)
40,000 (5.2510) +20,000 (0.2738) =N215,516
PWin = PVin = N215,516
(𝑃⁄𝐹, 13.83% , 10)
PWout = 100,000+22,000 (5.2510), 100,000 +115,522 = 215,522
PWout = N215,522
NPV=PVin – Pvout = 215,516 -215522
NPV = -N4
23
IRR =13.83%, the NPV is – N4
For 13.8%
(1.138)10 −1
1
PWin = 40,000(0.138(1.138)10 ) + 20,000 ((1.138)10 )
2.6427
40,000 (0.5027) + 20000 (0.2745)
(𝑃⁄𝐴, 13.8% , 10)
210, 280.49 + 5,490 = N215,770.49
PWout= 100,000 +22000 (𝑃⁄𝐹, 13.8% , 10)
100,000+22000 (
2.6427
0.5027
) =100000 +115,654.27
PWout =100,000+115,654.27 = N215,654.27
NPV = PWin -PWout =215,770.49-215,654.27
NPV = N116.22
IRR =13.8% = N116.22
IRR =13.83% = - N4.00 and when IRR is 13.8%, NPV = - N6
A computer programme will give the best IRR
NPV =4 0,000 (
(1+ἱ)10 −1
ἱ`(1+ἱ)10
1
) + 20,000 ((1+ἱ)10)
- [100,000 + 22,000 (
(1+ἱ)10
ἱ(1+ἱ)10
)]
NPV = 0 by putting the appropriate value of ἱ
4th method: future worth (FW)
Use future worth method to show whether the lathe machine in example is economically good for a
company or not.
Data given annual revenue = N40,000, Salvage value of the machine = N20,000
Annual disbursement = N22,000, Initial cost of lathe machine = N100,000
MARR = ἱ = 13%
A=N40,000
+SV=N20,000
10
P= N100,000
Year
= N22,000
SV=F=already in future worth at the end of 10 year
th
24
All the cash flow are converted to future worth l =13% n =10
FWin = A (F/A, 13%, 10) + 20,000 or
 i  i n  1 
 + 20000 = FW in 40000 i.1310  1  + 20000 = N756,789.97
FW in A

i
 0.13 


FWout  1000001.13
10
+ A (F/A, 13%, 10)
100000 (3.3940) + 22000 (18.4197) = N744,693.40
NFW = FWin – Fwout. NFW = 756,789.97 – 744,693.40 = N12,096.57
Total annual revenue= 5310 + 327.60 = 5637.60
Since NFW is greater than zero, the project good and profitable because revenue is greater than the
disbursement.
Now, NAW = Revenue - Disbursement
= 5637.60 - 5638.00 = -#.4k = 40k
NAW< 0: the project is not profitable second method
Calculate the total disbursement
AW = P (A/P, i%, n) –SV (𝐴⁄𝐹 i%, n), = 10,000(0.2638) - 2000 (0.1638) = 2638- 327.6
AW = 2310.40: art of disbursement
Total disbursement = 3000 + 2310.40 = N5310.40
NAW = Revenue - Disbursement
NAW = 5310 - 5310.40 = - 40k
NAW < 0: The project is not profitable
Alliteratively, the annual cost could be converted to present worth sum.
Annual revenue = N5310
Salvage value = N2000
Initial cost or capital= N10,000
Annual maintenance cost = N3000
531
5310
5310
5310
5310
#3000
#3000
#3000
#3000
#3000
FWin = A (𝐹⁄𝐴, 13. % , 10) + 20,000
25
= 40,000 (
(1+ἱ)𝑛 −1
ἱ`
(1.13)10 −1
40,000 (
0.13
) + 20,000
) + 20,000 = N756789.97
FWin = N756,789.97
P(𝐹⁄𝑃 , ἱ, 𝑛)
FWout =100000 (1+ἱ)1+(𝐹⁄𝐴, 13. % , 10)
(1.13)10 −1
100,000 (3.3946) +22000 (
0.13
)
339,460 +22000 (18.4197) = N744693.40
FWout = N744,693.40K
NFW = FWin – FWout
NFW =756,790 -744,693.40 = N12,096.60K
NFW > 0 The project is good and profitable because revenue > Disbursement
5th Method: payback period
Payback period: is the expected number of years required to recover the initial capital investment.
This means that the number of years required for the cash in flow of a project to return to the initial
cash investment of the project. Payback period as a tool of analysis is often used because it is easy to
apply and easy to understand for most individuals regardless of academic training or field of
endevour. Actually, payback period does not properly account for the time value of money, risk and
opportunity cost. In short, the short coming of this method are:
(1) Time value of money is not considered
(2) Not all the project cash inflows are considered
Formula for computing payback period (PBP) is given in the following expression.
PBP = year before recovery +
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 𝑁𝐶𝐹 𝑖𝑛 𝑡ℎ𝑎𝑡 𝑦𝑒𝑎𝑟
𝑐𝑎𝑠ℎ 𝑓𝑙𝑜𝑤 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑦𝑒𝑎𝑟
NCF = Net cash flow
Example 1: Calculate the payback period of a project with an initial capital of N1000. Annual
revenue from the project for the 1st, 2nd, 3rd, 4th, and 5th year are N500, N400, N200, N200 and N100,
respectively.
26
Solution
Let C.F = cash in flow
COF = cash out flow => first cost of project
NCF = C.F + COF
Year
Cash in flow
Cash out flow
Cumulative
NCF = CF + COF
0
-
-1000
-1000
1
500
-
-500
2
400
-
-100
3
200
-
+100
4
200
-
+300
5
100
-
+400
Payback period or break-even point occurs during the third year but not up to third year because at
the end of third year initial capital N1000 has been exceeded by N100.
∴ PBP = Last year with a negative NCF +
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 𝑁𝐶𝐹 𝑖𝑛 𝑡ℎ𝑎𝑡 𝑦𝑒𝑎𝑟
𝑐𝑎𝑠ℎ 𝑖𝑛 𝑓𝑙𝑜𝑤 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑦𝑒𝑎𝑟
100
PBP = 2 + 200 = 2,5 years. Therefore, payback period = 2.5 years
Example 2: Mr. John is planning to embark on a project with an initial cost of N10,000 and cash
inflow for four years are N1000, N3000, N4000 and N6000 as an Engr. advise Mr. John on his plan
whether to go ahead or not.
Solution
N1000
N10,000
N3000
N4000
N6000
27
Year
Cash in flow
Cash out flow
Cumulative
(N)
(N)
NCF = CIF + COF
0
-
-10,000
-10,000
1
1000
-
-9,000
2
3000
-
-6,000
3
4000
-
-2,000, ignore –ve
4
6000
-
+4000
PBP = Last year with a negative NCF +
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑣𝑎𝑙𝑢𝑒 𝑁𝐶𝐹 𝑖𝑛 𝑡ℎ𝑎𝑡 𝑦𝑒𝑎𝑟
𝑐𝑎𝑠ℎ 𝑖𝑛 𝑓𝑙𝑜𝑤 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑦𝑒𝑎𝑟
Payback period is between the 3rd and 4th year
PBP = 3 +
2000
6000
= 3 + 0.33 = 3.33 years
Therefore, payback period = 3.33years. Mr. John is advised to start the project.
Depreciation
Depreciation: means gradual loss or reduction in the value of an asset (project) over a given period of
time. The book value of an asset decreases as the economic or useful life of the asset reduces over a
given period of time. Depreciation is influenced by deterioration (worn-out) and obsolete (outdated)
over a period of years. Life span or useful life of an asset or project is the anticipated period of time
in years. Which an asset or machine could be used for producing item before the machine is
replaced. Examples of depreciable items includes machine, vehicle, equipment, TV, etc.
Salvage value: is the market or trade value of an asset or a machine at the end of its useful life. Note
that land is not commonly regarded as a depreciable item because it does not wear-out or become
absolute.
Methods for computing depreciation
1. Straight line depreciation method (SL)
2. Sum of year digit method (SYD)
3. Declining balance method (DB)
4. Double declining balance method (DDB) the commonly used methods are the SL and SYD
straight line method (SL) the following formula of SL is as given in the following expression
Dt =
𝑃−𝑆𝑉
𝑛
Where Dt = annual depreciation
28
P = initial cost of the machine or item
N = useful life or recovery period
SV = salvage value.
1
The rate of depreciation is 𝑛 for SL, it is constant.
Disadvantage: the writing off is not accelerated. The process of charging the depreciation is called
writing off the asset book value (BVt) = P – Dt x t
:. BVt = P – t D t where t = number of year t = 1,2,3,4,…
Example1: A Toyota carina car purchased in 1990 at a rate of N 80,000 and have a salvage value of
N20,000 in 1995. Calculate (i) the annual depreciation (ii) compute the book value of car after each
year (iii) plot the book value against the time. Use straight line (SL) method to solve the example.
Data given P= N 80,000 SV= N 20, 000 and n = 5 years
Solution
Dt = annual depreciation Dt =
D=
80,000−20,000
5
=
60,000
5
𝑃−𝑆𝑉
𝑛
= N 12,000
Therefore, the annual depreciation is N 12, 000,
Book value of the car at the end of each year
When t = 0, 1, 2, 3, 4 and 5 years. At t = 0
BVt = P – t Dt
BV0 = 80,000 – 0 x 12000 = N 80, 000. But when t = 1
BV1 = 80 000 – 1x12000 = 68, 000. When t = 2
BV2 = 80 000 – 2x12000 = N 56,000 when t =3
BV3 = 80 000 – 3x 12000 = N 44,000
BV4 = 80 000 – 4x12000 = N 32, 000
BV5 = 80 000 – 5x12000 = N 20, 000
Note that at the end of the 5th year, book value is equal to the salvage value of the car
BV5 = SV = N 20, 000
Student should not be surprised that a Toyota carina car was purchased a rate of N80, 000 in 1990
but the value of the same car after 29 years (1990- 2019) if the interest is 12%, calculate the current
value of N 80, 000 in 2019.
: FW = P (1 + ί)n
P = N80, 000 ί = 12% = 0.12 n = 29
FW = 80, 000 (1.12)29 = N 2,139,994.44k
The Toyota carina car now worth N2,139,994.44k when ί = 12%. The graph is a straight line.
29
90,000
80,000
70,000
Book value (N)
60,000
50,000
40,000
30,000
20,000
10,000
0
0
1
2
3
4
5
6
Time (year)
Graph of book value versus time using SL
Sum of the year digit method (SYD)
Formula for computing depreciation by SYD is given as
𝑃−𝑆𝑉
Dt = (n – t + 1) (
𝑆
)=(
𝑛−𝑡+1
𝑆
)(𝑃 − 𝑆𝑉)
𝑛−𝑡+1
𝑆
= is the rate of depreciation and it varies
according time t.
The book value of a particular project for different years(s) could be calculated using the following
formula
BVt = P – t (
Where S =
𝑛−0.5𝑡+0.5
5
) (𝑃 − 𝑆𝑉)
𝑛 (𝑛+𝐼)
2
BVt = book value of the project in a given time
n = useful life or life span of the project in year
t = time of the year
p = initial cost
SV = salvage value
Dt = depreciation rate
30
Example1: use SYD method to solve the example previously solved using SL to determine the
depreciation at every year and compute the book value. Also plot the graph of the book value versus
time. The information is given on page 44.
Solution
Let us calculate the depreciation using SYD and then calculate the book value at every year of the
useful life.
𝑛−𝑡+𝐼
Dt = (
𝑆
)(𝑃 − 𝑆𝑉) but S =
𝑛 (𝑛+1)
2
Data given P = N 80, 000, SV = N 20, 000
n = 5years
S=
5 (5+1)
=
2
5 (6)
2
= 15 :. S = 15
The value of S = 15 will remain constant when t=1 but n= 5
5−1+1
D1 = (
15
) (80,000 − 20000)
5
D1 = (15) (60,000) = N 20, 000
t = 2 for second year
5−2+1
D2 = (
15
4
) (60,000) = 15 (60,000) = N 16, 000
t=3
5−3+1
D3 = (
15
5−4+1
D4 = (
15
N
15
2
) (60,000) = 15 (60,000) = N 8, 000
5−5+1
D5 = (
3
) (60,000) = 15 (60,000) = N 12, 000
1
) (60,000) = 5 (60,000) = N 4, 000
Depreciation (N)
Cumulative(N)
Book value (N)
depreciation
80000 – 0 = 80,000
0
-----
------
1
20, 000
20,000
80000 – 20000 = 60,000
2
16,000
36,000
80,000 – 36000 = 44,000
3
12,000
48,000
80,000 – 48000 = 32,000
4
8,000
56,000
80000 – 56000 = 24,000
5
4,000
60,000
80000 – 60000 = 20,000
31
Alternative, by book value could be determined using the following formula for SYD method
BVt = P - t [(n-0.5t +0.5)(
𝑃−𝑆𝑉
5
Let us determine S
S=
𝑛 (𝑛+1)
2
=
)]
n=5
5 (5+1)
2
= 15
Now when t = 0 what is the book value of the project (machine) P = N 80, 000
BV0 = 80000- [(5-0.5x0x0.5)x SV = N 20, 000 n = 5
80 000−20000
(
15
)
t = 0, 1, 2, 3, 4 & 5
BV0 = 80000 – 0 (5.0) (4,000) = 80000 – 0
BV0 = N 80, 000 = book value is still the same 80000-20000 = 60, 000= N 50
BV1 = 80000 – 1 [5-0.5x1+ 0.5) (
60,000 60000
15
)
15
= 4000
80,000 -1 (5) (4000) = N 60, 000
BV2 = 80000 -2 (5-0.5x2 + 0.5) (4000) = N 44, 000
BV3 =80, 000 -3 (5-0.5x3 +0.5) (4000) = 80,000 -3 (4) (4000) = N 32, 000
BV4 = 80,000 -4 (5-0.5x4+0.5) (4000) =80, 000 -4 (3.5 (4000) = N 24, 000
BV5 = 80000 -5 (5-0.5 x 5 + 0.5) (4000) 80000 -5 (3) (4000) = N 20, 000
Book value at end of 5th year = SV = N 20, 000
Both SL and SYD gave the same book value at the end of the useful life of the project = N 20, 000
90,000
80,000
70,000
Book value (N)
60,000
50,000
40,000
30,000
Salvage
value
20,000
10,000
0
0
1
2
3
4
Time (year)
Graph of book value against time using SYD
5
6
32
Assignment/Home work
An airline company bought an aircraft equipment for N50,000. The salvage value after 8 years useful
life is N2000,using the sum of the year digit depreciation method, determine;
(i) The depreciation at the end of each year to the end of useful life of the equipment.
(ii) The book value of the equipment at the end of each year.
(iii) Plot the graph of book versus time.
(iv)
Also use straight line to solve i,ii and iii as given above.
Declining Balance (DB) method and Double Declining Balance (DDB) depreciation method
Declining balance (DB) method: this method is also called diminishing balance method. DB, DDB
and SYD are accelerated depreciation method in which large part of the depreciation occur at the
beginning of the life of the fixed asset and become smaller towards the end of the useful life of the
asset. DB could never depreciation the asset to zero because the salvage value in the formula must be
positive. Common type of declining balance method is the double declining balance depreciation
method (DDB). Note that for straight line depreciation method, depreciation occur uniformly over
the useful life the fixed asset
Example 1
A Toyota carina car purchased in 1990 at a rate of ₦20,000 in 1995 Use declining balance (DB)
method to compute TV annual depreciation and the book value of the assets. Present your answer in
tabular form. Also plot the graph of book value against time. (hint :(
𝑛−𝑡+𝑖
𝑠
)(P-SV)
Formula for computing DB
1


SV
 
n 
Dt  BVt ` 1  
 
  P  


n=5
P = ₦80,000
SV = N20, 000
When t = 1 BV1-1 = BV0 = book value at the beginning of first year
BVt = first price = initial cost = ₦80,000
1


  20000 5 
D1 = (80000) [1 – (
] Dt  8000001  
  = (80000) (1-0.757858203)
80000
  80000 


20000
33
D1 = (80000) (0.242141716) = ₦ 19, 371.34
D1 = ₦ 19, 371.34
BV1 = BV0 – D1 = P – D, = 80000 – 19371.34
𝑆𝑉
BV1 = ₦ 60,628.66 note that (1-( 𝑃 )1⁄𝑛) is constant throughout which is 0.242141716
D2 = (BV1) (0.242141716)
D2 = (60628.66) (0.242141716) = ₦ 14680.73
BV2 = BV1 – D2 = 60628.66 – 14680.73 = 45947.93
BV2 = ₦ 45947.93
D3 = (BV2) (0.242141716)
D3 = (45947.93) (0.242141716) = 11, 125. 91
D3 = ₦ 11,125.91
BV3 = 45,947.93 – 11,125.91 = ₦ 34,822.02
D4 = (BV3) (0.242141716)
= (34822.02) (0.242141716) = ₦8,431.86
BV4 = BV3- D4 = 34822.02-8,431.86
BV4 = ₦ 26,390.16
D5 = (BV4) (0.242141716)
= (26390.16) (0.242141716) = ₦ 6,390.16
BV5 = BV4-D5= 26390.16-6390.16
BV5 = ₦ 20,000
N
First price ₦
Annual
Book value₦
depreciation₦
Cumulative
depreciation
0
80,000
________
80,000
________
1
80,000
19,371.34
60,628.66
19371.34
2
80,000
14,680.73
45,947.93
34,052.07
3
80,000
11,125.91
34,822.02
45,177.98
4
80,000
8,431.86
26,390.66
53,609.84
5
80,000
6,390.16
20,000
60,000.00
34
90,000
80,000
70,000
Book value (N)
60,000
50,000
DB
40,000
30,000
20,000
10,000
0
0
1
2
3
4
5
6
Time (year)
4th method
Double declining balance method (DDB).
The formula for DDB for computing annual depreciation is given in the following expression.
 2  n  2 
Dt  P 

 n  n 
t 1
Example: use DDB for computing annual depreciation of the previous example. Data in the example
are: P: N80,000, 5v = N20,000, n = 5
 2  n  2 
Dt  P 

 n  n 
t 1
when t =1
Student plot the graph of book value versus time as assignment /home work.
2
5−2 t-1
5
5
D1 = 80,000 ( ) (
D1 = ₦32,000
)
= (80,000) (0.4) (0.6)0 = 80,000 (0.4) (1)
35
D2 = (80,000) (0.4) 2-1
when t = 2
= (80,000) (0.4) (0.6)1
D2 = ₦19.200
D3 = (80,000) (0.4) 2 = ₦11,520
Book value:
Bvt = P – Dt
Bv1 = 80,000 – 32,000 = ₦48,000
Bv2 = 80,000 – (32,000 + 19,200 ) = ₦28,800
The remain value to salvage value = ₦8,800
i.e 28,800 – 20,000 = ₦8,800
But from calculate
Bv3 = 80,000 – (51,200 + 11,520) = ₦17,280
₦17,280 is less than the salvage value.
This means that n depreciation again from the question because salvage value is ₦20,000 the useful
life
n
2
𝑛−2
First price
Dt = P (𝑛) ( 𝑛 ) t-1
Cum.
(₦)
(₦)
Depreciation
Book value
Bvt = P – cum. Dt
(₦)
(₦)
0
80,000
-------
------
80,000
1
80,000
32,000
32,000
48,000
2
80,000
19,200
51,200
28,800
3
80,000
11,520 = 8,800
62,720
20,000
4
80,000
No Depreciation
60,000
20,000
5
80,000
No Depreciation
60,000
20,000
Assignment /home work
A mazida car is purchased at a rate of N421,775 and has a useful life of 5years and salvage value of
the car at the end of the 5th year is N50,000. Use DDB method to compute the annual depreciation
and book value of the car. Present your computation (answer) in tabular form.
 2  n  2 
Hint: Dt  P 

 n  n 
t 1
36
Replacement analysis
Replacement analysis: this is the aspect of engineering economics that deals with economic
comparison of two assets and choose the beat out of the alternative that is economical and profitable.
An asset is usually considered for replacement prior to the anticipated life spam of the assets because
of the physical deterioration and obsolescence (old fashion). In replacement analysis, an assets or a
project that is currently being owned by a company is called defender while the new project to be
purchased for the placement is called challenger.
In comparing the two assets the current, market value of the defender is normally considered as the
first cost of the assets and all other associated costs such as salvage value and annual operating cost
of the machine are considered.
Example 1: A lathe machine purchased 3 years ago N200,000 with an estimated lifespan of 9 years
and salvage value of N60,000 annual operation cost of the old lathe machine is N50,000 and the
market value for the old lathe machine is N170,000 A challenger has been old lathe machine at a cost
of N220,000. The life span of the proposed for the replacement of the old lathe machine is 9years
with salvage value of N70,000 and annual operating cost of N40,000. Should the machine be
replaced or not if the minimum acceptable rate of return (MARR) is 10%.
Solution
Let
AOC = Annual operating cost
EAC = Equivalent annual cost
D = defender
C = Challenger
SV = salvage value =future value of the assets (f)
In solving this problem, convert all necessary cost(s) to annual worth for uniformity and easy
comparison and easy comparison
Let
EACD = Equivalent annual cost of defender
EACC = Equivalent annual cost of challenge
Let P= current cost of the lathe machine = N170,000
SV = salvage value = N60,000
Annual operating cost = AOCD = N50,000
EACD = P(A/P, ἱ %, n) – SV (A/F , ἱ %, 𝑛 )
37
EACD = 170,000(A/P, 10%, 9) – 60,000 (A/F 10%,9) + 50,000


 i 1  i n 
i

 - 60000 
EACD  170000
n
n
 1  i   1  + 50000



1

i

1




=170,000 (
0.1(1.1)9
(1.1)9−1
) - 60,000 (
0.1
) +50,000
(1.1)9−1
=170,000 (0.1736) - 60,000 (0.0736) + 50,000
EACD + N75,096
For challenger,
P = N220,000
SV = N70,000 AOCC = N40,000
EACC = P (A/P, ἱ %, n) - SV (A/F , ἱ %, 𝑛 ) + AOCC
=220,000 (A/P, 10%, 9) – 70,000 ((A/F , 10 %, 9 ) + 40,000
=220,000 (0.1736) – 70,000 (0.0736) + 40,000
EACC = N73,040
From the above calculation or analysis
EACC = N73,040
EACD = 75,096
EACC < EACD: This lathe machine) is higher than that of challenger. Therefore replacement is
necessary is necessary to buy new lathe machine.
Example 2: When annual maintenance cost is not constant but varies linearly: sAn electric oven in a
bakery is being considered for replacement. Its salvage value and maintenance costs are give in the
following table. A new oven cost N80,000 ad this price includes a complete guarantee of the
maintenance of the new oven are shown in the table. Should the oven be replaced this year if the
MARR is 10%. SV at the beginning = first price (P) = N20,000. (see page 60aand 60b).
Year
SV(₦)
Maintenance cost (₦)
SV (₦)
Maintenance cost
(₦)
0
20,000
-------
80,000
----
1
17,000
9,500
75,000
0
2
14,000
9,600
70,000
0
3
11,000
9,700
66,000
1000
4
7,000
9,800
62,000
3000
Old oven (Defender)
38
Let the current market price of the oven = SV and SV for old oven = N20,000 if the oven is dispose.
The question is on page 60a and 60b .
Let EACD = equivalent annual cost for defender AMC = Annual maintenance cost.
SV for year 0 = N20,000
From the table, annual maintenance cost is not Constance, therefore, the calculation should be done
yearly (each year).
Note that maintenance cost must be converted to annual maintenance cost. Maintenance cost occur
as present worth but should be converted to future worth for a common value and them to annual
worth cost.
17,000
14,000
11,000
7000
P=N20,000
G=N100
For the first year n=1
Defender
EACD = (P-5) (A/P , ἱ %, 𝑛 ) + SV (ἱ %) + AMC --------(i)
EACD = P(A/P , ἱ %, 𝑛 ) + SV (A/F , ἱ %, 𝑛 ) + AMC --------(2)
From Equation
EACD1 for the first year D1, maintenance cost of N9,500 is annual cost because it occurs for the first
year, there is no need to regard it as present cost or convert it to further worth and later to annual
worth sum (A)
EACD1 = (20,000-17,000) (A/P , 10 %, 1) + 17,000 (0.1) + 9,500
3000 (1.1)+ 17,000 +9500 = N14,500
n=1
39
OR using Equation (2)
EACD1 = 20,000 (
P = N20,000
ἱ(1+ἱ)𝑛
(1+ἱ)𝑛−1
=20,000 (
=20,000 (
(1+ἱ)𝑛−1
0.1(1.1)1
0.1
(1.1)1−1
(1.1)1−1
0.11
0.1
ἱ
) - 17,000 (
) + 17,000 (
) +9500
) + 9500
0.1
) - 17,000 (0.1) + 9500
=20,000 (1. 1) - 17,000 (1) + 9500 =N14,500
:- EACD1 = N14,500
Both equations(1) and (2) gave the same answer as EACD1 = N14,500
For second year when n=2
Maintenance cost of 1st and 2nd year must be combined together as the total cost and converted to
annual worth (i.e N9,500 t future worth, add it to that of 2nd year –N9600 and then convert the total
or sum to annual worth sum).
EACD2 = (P-SV) (A/P , 10 %, 2 ) + SV (102) + (9500(F/P,10%,1) +9600) (A/F, 10%,2)
= (20,000-14,000) (0.5762) + 14,000 (0.1)
= (6000) (0.5752) + 14000 + (9500(F/P, 10%, 1)+ 9600 (A/F, 10%,2)).
Let us do or determine the future value of N9500 that occurs in the first and add it to N9600 for the
second year as cumulative money or cost spent on maintenance and then convert it to annual worth.
9500 (1.1) =10,450=this is future worth of N9500 in second year 10450 + 9600 = N20,050.
Therefore cost maintenance at the end of second year = N20,050 but we need to convert it annual
worth (A).
20,050 (A/F, 10%,2) = 20,050 (
0.1
ἱ
(1+ἱ)𝑛−1
) or from table
0.1
20,050 ((1.1)2−1) 20,050 (1.21−1)
20,050 (
0.1
) = 20,050 (0.4762) you can get this from table
0.21
AMCD1= N9,547.81
Total cost = EACD2
back to previous calculation
EACD2 = 6000 (0.5762) + 1400 +9547.81
=3457.2+1400+9547.81 =N14,405.01
= N14,405.01
The little difference in answer of EACD2 that wasN14,405.01 instead of N14,404.82 is due to round
off error which was due approximation.
40
20,050
(
0.1
) 20,050 (0.476190476)
0.21
= 9,547.619048
From the table
6000 (0.5762) + 1400 +9547.62+N14,404.82
:- EACD3 = N14,405.01 = N14,404.82 Error of 0.19 due to approximation but this is not a problem
just for you to understand that round-off error can affect your final answer with little magnitude.
Now Let s calculate EACD3 for the 3rd Year
EACD3 =(P-SV) (A/P , 10 %, 3 ) + SV (ἱ %) +AMC3
9000 (0.4021)+ 1100 (0.1) = 3 618.9+1100
Let us calculate the AMC3 separately
AMC3 =9500 (F/P, 10%,2) + 9600 (F/P, 10%,1) + 9700
=9500 (1.210) + (9600(1.1) + 9700
11,495 + 10,560 +9700 = N31,755
This N31,755 is future worth at the end of 3rd year but it must be converted to annual worth
31,755 (A/F, 10%,3) = 31,755 (0.3021)
AMC3 =N95993.19
EACD3 3618.9 +1100+9593.19 = N14,312.09
:- EACD3 = N14,312.09
EACD4
EACD4 = (P-SV) (A/P , 10 %, 4 ) + SV (ἱ %) +AMC4
= (20,000 -7000) (0.3155) + 7000 (0.1)+AMC4
Let us calculate AMC4
Different in 4-1= 3 :- 9500 to 9700 =n=3 not 4years
AMC4 = 9500 (F/P , 10 %, 3 ) + 9600 (F/P , 10 %, 2) + 9700 (F/P , 10 %, 1)+ 9800
9500 (1.331) + 9600 (1.210) + 9700 (1.100) + 9800)
12,644.5 + 11,616.0+10,670+9800
FW of maintenance = FMC =44,730.5
FMC4 to AMC4 = 44730.5 (A/P , 10 %, 4)
AMC4 = 44,730.5 (0.2155)
AMC4 = N9639.42
:- EACD4 = (20,000 -7000) (0.3155) + 7000(0.1) + 09639.42
(13000) (0.3155) +700 + 9639.42
4,101.5 + 700 + 9639.42 = N14,440.9
41
:- EACD4 14,440.94 from the text EACD4 = N14,439.60
14,440.94 – 14,439.60 = N1.32 different due round off error
EACD1 = N14,500.00
EACD2 = N14,405.01
All these are for the defender
EACD3= N14,312.09
EACD4 = N14,440.94
For the challenger
EACC1 = (P-SV) (A/P , ἱ %, 𝑛 ) + SV (ἱ %) +AMC
= (80,000 – 75,000)
(A/P , 10 %, 1) + 75,000 (0.1) +0 as AMC =0 for the first year
(5,000) (1.1)+7,500 = N13,000
EACC2 (80,000 -70,000) (A/P , 10 %, 2) + 7000 (0.1) at the 2nd year AMC =0
= (10,000) (0.5762) + 7000 + 0 = N12,762 for the 3rdyear
EACC3 (80,000 - 66,000) (A/P , 10 %, 3) + 66,000 (0.1)+AMC3
=(14,000) (0.4021) +6,600 + AMC3 …N1000 is future worth compare to year 1 and 2.
AMC3 = 1000 (A/P , 10 %, 3) = 1000 (0.3021)
AMC3 = 302.1 = N302.10
Now back to the original calculation
EACC3 = (14,000) (0.4021) + 6,600 + 302.10=12531.5
EACC3 = N12,531.50
For the 4th year
n=4
EACC4 = (80,000-62,000) (A/P , 10 %, 4) – 62,000 (0.1) + (1000 (F/P , 10 %, 1)+3000) (A/P , 10 %, 4)
AMC4 = 1000 (F/P , 10 %, 1 ) + 3000) (A/P , ἱ10%, 4 ) 1000 (1.100)+3000) (0.2155)
(1100+3000) (0.2155) = (4100) (0.2155)
AMC4 (18,000) (0.3155)+6,200+883.55
EACC4 = 5679+6200 +883.55 = N12,762.55
Summary of the cost of running the oven
Year
Defender₦
Challenger ₦
0
20,000
80,000
1
14,500.00
13,000.00
2
14,405.01
12,762.00
3
14,312.09
12,531.50
4
14,440.94
12,762.55
42
The annual operating cost at the end of the 3rd year was very small both for defender and challenge
but that of defender is greater than that of challenge.
₦14.312.09 > ₦12,531.50
Therefore, replace the defender with the new electric oven at the end of the 3rd year.
This is done to reduce coast of production and make more profit.
43
Break-Even Analysis
Break-Even: is the point at which total revenue received is exactly equal to the cost of production.
At this point no profit is made and no lost incurred. Break-even point (BEP) can expressed in terms
of unit sales or unit indicates the level of sales that is required to cover the total cost of incurred for
the production. Sales above this point will definitely give profit but sales below this point result in to
a loss. This means that break even analysis aims at finding the value of individual variables at which
the project NPV is zero
Break even is required to quantify the level of production that is needed for a new business.
Break even point (BEP) unit = BEPU
BEPU =
BEPu =
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑎𝑛𝑛𝑢𝑎𝑙 𝑓𝑖𝑥𝑒𝑑 𝑐𝑜𝑠𝑡
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑒𝑙𝑙𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 − 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡
𝐹.𝐶
(unit e.g kg or bags )
𝑆𝑃−𝑉.𝐶
Break even sales (BEP) in naira (₦)
BEPS =
𝐹.𝐶
1−(𝑉𝐶 ÷𝑆𝑃)
Profit =selling price x quantity - (F.C + V.C x quantity )
Profit = SP x Q – (F.C + V.C x Q)
Example 1: A fertilizer company located in Kaduna normally produces fertilizer for retailers sale of
₦500 per bag . the average variable cost per bag is ₦200 and the annual fixed cost is ₦6,000,000 .
determine
(i) Break even unit
(ii)
Break even sales
Solution
Break even unit
BEPU =
=
𝐹.𝐶
𝑆𝑃−𝑉.𝐶
6,000,000
500 −280
BEPu = 27,273.73
=
6,000,000
220
≈ 27, 273 bags
44
(ii) Break even sale
BEPS = BES =
𝐹.𝐶
𝐼−(𝑉𝐶÷5𝑃)
6,000,000
BEPS = 1−(280÷500 )
BEPS =
6,000,000
0.44
=
6,000,000
1−0.56
= ₦13,636,363.64
Alternatively, break even sales could be determined using the following formula
BEPS = BEPU x sales in naira per bag
BEPS = 27,273 X 500 = ₦13,636,500 or as if it not approximated to whole naira
27,272.73 x 500 = ₦13,636,365
Break-even point as it was defined before as the production (number of units produced) without
profit loss
Cost and revenue
BEP
E
DD
B
F
C
A
Units
Break even graph
Let F.C = fixed cost, V.C = variable cost.
BEP = break-even point, AB = variable cost, CF = fixed cost and CD = F.C + V.C
Note that AB should be parallel to CD
Note the following expressions
Total cost = F.C + n (V.C)
45
Again, for simplicity, let F = fixed point (₦), V = variable cost (₦), n = unit of production at BEP
R = selling price per unit (₦)
Total cost = F + n V
Total revenue = n R
F + n V = nR
F = nR – nV
n=
𝐹
𝑅−𝑉
or as BEPU =
F.C
SP−V.C
Where
SP = selling price
V.C = variable cost
F.C = fixed point
BEPu = break-even point of unit production in term of kg or bags
Example 2: Nigerian society o engineers (NSE) is planning to have a one day seminar (workshop)
for its members. The following cost data are necessary for NSE to have the program conducted.
Fixed cost : Brochure printing = ₦12,600
Mailing brochure = ₦19000
Room charges =
₦15000
Speaker gift =
₦8000
Total fixed cost =
₦54,600
Variable costs: The following costs are for each participant for the seminar
Lunch
= ₦650
Coffee
= ₦ 375
Folder
= ₦ 250
Total variable cost = ₦ 1275
Calculate
(i) The break-even point for the seminar if the cost of attending the seminar is
₦5000 per person, (ii) Profit to be made if 42 people attended the seminar
Solution
Total fixed cost F.C = ₦54,600
V.C = ₦1275
46
R = ₦5000
Let F.C = F, n = BEPU
R = selling price per people or unit, = variable cost n = number of unit at BEP
n=
n=
F
R−V
,
54600
3725
n=
54600
5000 − 1275
= 14.6577 units
Q = 42 people attended it
BEP = N = 14.66 units (participants)
Profit = R x Q – (F + V X Q)
Profit = RQ – (F + VQ)
= 5000 x 42 – (54600 + 1275 X 42)
= 210,000 – 108, 150
= 101,850
Profit that could be made from the seminar = ₦101,850
Benefit-cost ratio
Profitability index =
Benefit
cost
Profitability index is also called benefit -cost ratio (B-C ratio).
Benefit -cost ratio: is the ratio of the sum of present value of future benefits to the sum of the future
capital expenditure and cost.
Profitability index = Benefit -cost ratio.
If benefit -cost ratio > 1, the project is good but if it is < 1, the project is not good.
Example 1: Use the data provided in the last example in which NSE is planning to conduct a one
day seminar for its members.
From data given FC = ₦54,600
VC (per person) = ₦1275
Number of participants = 42
V.C for 42 participants =₦1275 x 42 = ₦53,550
Total cost = F.C + V.C = ₦54600 + ₦53550 = ₦108,150
Expected selling price or revenue from 42 people =₦ 5000 x 42 = ₦210,000
Benefit – cost ratio =
Profitability index =
Benefit
cost
Benefit
cost
=
210,000
108,50
= 1.942, B.C ratio > 1
That means that the seminar is profitable because the benefit – cost ratio is greater than 1.0
47
ABE 306 OBJECTIVE TEST QUESTIONS AND SOLUTIONS
(1) Calculate the future worth of ₦100,000 borrowed from UBA which must be payback after
3 years at 15% interest rate
Solution F = P (1 + ἱ)n F = 100,000 (1.15)3 = ₦ 152,087.5
A. ₦136,890
B. ₦200,000
C. ₦152,087.5
D. ₦152,000
(2) Dr. Taiye Johnson put ₦ X (Naira X) in UBA in 1980 at interest rate of 18% and he intended to
collect the money 40 years later (in 2020) for his first child to start a business when it would be
₦1,000,000 . Calculate the present worth of the money (₦ X) put in the bank
Solution F = P(1 + ἱ)N
A.₦1350.00
(3
𝑛
P = (1+ἱ)n
B. ₦1332.66
=
C. ₦1500
1,000,000
(1.18)40
= 1,332.66
D. ₦2,000
If the future of ₦100,000 after a certain period of time at interest rate of 15% is ₦152, 087.5,
calculate the years required to achieve the repayment.
Solution F = P (1+ ἱ) n
F
= (1 + ἱ) n
P
 152087.5 
F
log10 
log10  

100000 
P


n
Take log of both sides, Log10 (F/p) = log10 (1+ ἱ) , n 
, n
 3 years
log10 1  i 
log10 1.15
A.
3 years
B. 3.5years
C. 5years
D. 2years
(4) Calculate the interest rate to 4 decimal places that is require on N5000 put in a bank to yield
N15000 in 10years.
Solution F = P (1+ ἱ) n ,
(1+i)10 =
15000
15000 = 5000 (1+i)10
, 1+ i = 310 , 1+ i = 1.116123174, i – 1.116123174 – 1 = 0.1151232 x 100
1
5000
i = 11.6123%
A. 11.61232%
B. 11.0000%
C. 12.0000% D. 10.500%
(5) Calculate the interest rate to 4 decimal places that is require on ₦1,000,000 Put in bank to yield
₦3,500,000 in 16 years. Solution F = P (1+ ἱ) n ,
𝐹
𝑃
= (1 + ἱ) n
1  i 16  3500000 , 1  i  3.516 , i = 3..50.0625  1 , 1.081444541 – 1,
1
1000000
i = 0.0814445 x 100 = 8.1445%
A. 8.1445%
B. 18.445%
C. 10.08%
D. 20.08%
(6) The following assets depreciate over a given period of time except
A. US dollar
B. Lathe machine C. money D. land
48
(7) In replacement analysis, the old machine that is currently being used for production is called?
A. Replacement
B. Defender
C. Asset D. Challenger
(8) Calculate the annual worth if the present worth of money is ₦250,000 at 10% interest rate over a
given period of 10 years.
i1  i 
Hint A = P (𝐴⁄𝑃 , 10%, n) is A = P
to 4 decimal places
1  i n  1
n
0.11.1
10
Solution: A = P
A. 25,000
1.1  1
10
B. 50,000
P
0.2594
, BUT P = 250,000, A = 250000 (01628) = 40,700
1.5937
C. 40,000
D. 40,700
(9) The yearly payment at the end of 1st , 2nd, 3rd, 4th and 5th years are ₦10,000 , ₦10,500 , ₦11,000,
₦11,500 and ₦12,000, respectively. If the interest rate is 8%, calculate equivalent present worth of
the payment. Hint: (𝑃⁄𝐴 , 8%, 5 = 3.9926), (P/G, 8%, 5 = 7.3713)
Solution: P = A (𝑃⁄𝐴 , 8%, 5) + 𝐺(𝑃⁄𝐺 , 8%, 5), 10,000(3.9926) + 500(7.3713) = 43,611.65
A. ₦43,611.65
(10)
B. ₦39,926
C. ₦3,685,65
D. ₦50,000
The interest rate that makes the net present value of all cash flows equal to zero is called?
A. Simple interest rate B. Payback period C. Internal rate of return D. Interest rate
(11)
The number of year required to recover the initial capital investment is called?
A. Payback period
(12)
B. Internal rate return
C. Minimum acceptable rate of return D. Break even
Gradual loss in the value of an asset over a given period of time is called?
A. Book value B. Diminishing in value of asset C. Appreciation D. Depreciation
(13)
The market value of an asset at the end of its useful life is called?
A. Depreciation
(14)
B. Salvage value C. Future worth D. Break-even
Determine the book value of a lathe machine at the end of the 3rd year if the initial cost of the
lathe machine is 80,000 and the salvage value of the machine is 20,000 and useful life is 5years (use
straight line method).
Solution: D5 
P  SV 80000  20000

 12000 , Book value (BV3) = 80000 – 12000 x 3 = 44000
n
5
A. ₦12,000 B. ₦44,000 C. ₦56,000 D. ₦20,000
(15)
Use sum of year digit (SYD) method to calculate the depreciation at the end of first year if
the initial cost of a pumping machine is ₦50,000 salvage value is 2000 and its useful life is 8 years.
Hint (
𝑛−𝑡+1
𝑛(𝑛+1)
𝑠
2
) (P - SV) and S =
Solution Using SYD: S 
useful life = n = 8, SV = N2000, P = N50,000, t = 1 year
8  1  150000  2000   10,666 .67
88  1
 36 , D1 
2
36
49
A. ₦10,666.67
(16)
B. ₦9,333.33 C. ₦48,000 D. ₦20,000
Determine the book value at the end of third year (BV3) using sum of year digit if the first
cost of lathe machine is 80,000 and salvage value of the machine is 15000 and useful life of the
lathe machine is 10 years
Solution n = 10 , P = ₦80,000 , t = 3 but S =
𝑃−𝑆𝑉
BVT = P – t [(n − 0.5t + 0.5)(
S=
10(10−1)
2
𝑆
𝑛(𝑛+1)
2
)]
= 5(11) = 55
BV3 = 80,000 – 3(10 – 0.5 x 3 + 0.5) (
₦80,000−₦15,000
55
) = 80,000 - 3 (9)(65,000)
= 80,000 – 31,909.09 = 48,090.91
A. ₦48,090.91
(17)
B. ₦31.909.09
C. ₦30,000 D. ₦35,000
Determine the depreciation of the end of the first year using declining balance method. If the
initially cost of motorcycle is 80,000 has a salvage value of 20,000 and useful life of the
1


  SV  n 
motor cycle is 10years. Hint; Dt  BVt 1 1  
 
  P  


1


  2000  10 
Solution D1  800001  
  = 80000 (0.129449436)0.1 = 10,355,95
  80000 


A. ₦19,371.34
(18)
B. ₦10,355.95
C. ₦25,000
D. ₦69,644.05
Calculate the book value of motorcycle at the end of second year using SYD if the initial cost
is ₦90,000, salvage value is ₦16,000 and useful life of the motor cycle is 7 years
Solution: t = 2, n = 7, P = N90,000, SV = N16,000
S=
n(n+1)
2
=
7(7+1)
2
 P  SV 
= 7 x 4 = 28, BVt  P  t n  0,5t  0.5

 S 
 90000 16000
BVt  90000 27  0,5  2  0.5
  55,642.86
28


A. ₦55,642.86
B. ₦34,357.14 C. ₦74000
D. ₦2, 642.86
(19) The aspect of engineering economics that deals with economic comparison of two assets and
choose the best out of the alternatives that is economical and profitable is called?
A. Break even analysis
(20)
B. Engineering economy C. replacement analysis
D. Benefit cost ratio
Calculate the profitability index if the benefit is ₦500,000 and cost is ₦400,000
50
Solution: Profitability index =
A. ₦100,000
(21)
B. 1.25%
Benefit
cost
500,000
==
C. 1.25
400,000
= 1.25
D. 1.00
Use the straight line method to calculate the depreciation if the initial cost of a Toyota car is
₦940,000 and the salvage value is ₦150,000 and the useful life of the Toyota is 10 years
Solution P = ₦940,000
D=
940,000 – 150,000
n = ₦10years
= ₦79,000
10
A. ₦79,000
(22)
SV = ₦150,000
B. ₦100,000
C. ₦250,000
D. ₦50,000
The amount of money that is charged by financial institution for the use of borrowed capital
over a given period of time is called
A. Tax B. Liquidity ratio
(23)
C. Revenue D. Interest
Calculate the annual worth of the present worth money of ₦100,000 over a given period of
8years and interest rate is 12%. Hint : A = P(
A = 100,000 (
A. ₦20,130
(24)
A
, i , n) is
P
p=(
i(1+i)n
(1+i)n−1
0.12(1.12)8
0.12(2.4760
0.2971
(1.12)8−1
1.476
1.4760
) = 100,000 (
B. ₦25,000
C. ₦30,000
) = 100,000 (
) Solution
) = 100,000 (0.2013) = ₦20,130
D. ₦15,000
Calculate the present worth of money if the future worth at the of 10years is ₦250,000 and
the interest rate is 10%
Solution F  P1  i  , P 
n
A. ₦648,435.62
(25)
F
1  i 
n
B. ₦96,385.82
, P
250000
 96,385.82
1.110
C. ₦120,000 B. ₦100,000
Determine the interest rate to 1 decimal place that is needed on ₦250,000 to yield
₦3,500,000 at end of 12 years
1
1
 35000,000 12
 F n
n
Solution F  P1  i  , i     1 , i  
  1  0.246  24.6 %
P
 250,000 
A. 8.3%
(26)
B. 24.6%
C. 15.0%
D. 25%
If the future worth on ₦250,000 after certain period of time at interest rate of 24.6% is
₦3,500,000. Calculate the year required to achieve it to 1 decimal place
Solution: F  P1  i  ,
n
F
F
n
n
 1  i  , Take log of both sides, log10    log10 1  i 
P
P
 
 3500000
log10 

250000 

n
 11.999  12.0 years
log10 1.246
A. 12.0
B. 10.0
C. 15.0
D. 20.0
51
(27)
The fixed cost for organizing a workshop by NSE ₦54,600, variable cost per unit is 1,275
and cost per participant is ₦5,000. Calculate the number of participants at break-even point.
V = ₦1275, n 
Solution FC = ₦54,600 R = ₦5000
BEPU = n =
54600
5000−1275
A. 14.66 B. 10.92
(28)
=
54600
3725
C. 42.82
FC
FC
or as BEPU 
R V
SP  V .C
= 14.6577 = 14.66
D. 18.00
The yearly payment at the end of 1st, 2nd, 3rd, 14th, and 5th year are ₦1000, ₦1500, ₦2,000,
₦2500, and ₦3000, respectively. Calculate the annual worth of the payment if the interest rate
1  i n  in  1 , G = N500 and A = N1000
n
i1  i   1
(1+ ἱ)𝑛−&𝑛−1
A
is10%. (hint: G( /P, % n) is
ἱ(1+ ἱ) 𝑛−1
1.15  0.1  5  1
 0.11051
A= 1000 + 500
, 1000 + 500 
 = 1000+500 (1.797) = 1,899.5
5
0.11.1  0.1
 0.0615 
A ₦2000
B ₦15000
C ₦1,898.5
D ₦4,500
(29) calculate the net annual worth (NAW) if the annual revenue and disbursement are ₦785 and
₦715, respectively. Solution: NAW = Revenue – Disbursement = 785 – 715 = N70
A ₦1500 B ₦750
D ₦70
C 500
(30) The initial cost of a Yahama motor cycle is ₦50,000 and its salvage value at the end of 8 years
useful life is ₦12,000 calculate the book value at end the of 5th year using straight line method.
D 
50000−12000
P  SV
=
= D8 = N4,750, Book value = P – t x Dt = 5 x 4750 = 26250
8
n
A ₦26,250
B ₦4750
C ₦7600
D ₦15,000
(31) A fertilizer company producing fertilizer for retail sales at rate of ₦2000 per bag. The average
variable cost per bag is ₦1,200 and annual fixed cost ₦5,000,000. Calculate the break -ven point
sales. Solution BEPS 
A ₦12,500,000
FC
,
 VC 
1 

 SP 
BEPS 
B ₦10,000.000
5,000,000 5,000,000

 N12,500,000
1  0.6
 1200
1 

 2000
C ₦2,5000
D ₦4,166.67
(32) What is the full meaning of MARR
A. Maximum average rate of return,
B. Minimize acceptable rate of return
C. Minimum average rate of return,
D. Marginal acceptable rate of return
52
(32) Profitability index is called
A Profit after tax,
B Benefit cost ratio
C Marginal profit
D Break-even point
(34) in replacement analysis, the new equipment to procure for replacement of old equipment is
called. A Asset
B Defender C Challenger
D Depreciation
(35) The expression P (A/P, ἱ &, n) means
A Conversion of present worth to annual worth
B Conversion of annual worth to present
C Conversion of present worth to future worth
D Conversion of annual worth to gradient worth