ENSC 2063
BY
FACULTY OF ENGINEERING SCIENCES DEPARTMENT, ESD 2020
Engr. Maricar B. Carreon
Engr. Babinezer D. Memoracion
Engr. Eduardo O. Dadivas
Engr. Jimmy L. Ocampo
Engr. Carmelita I. Durias
Engr. Ruben A. Pureza
Engr. Angela L. Israel
Engr. Roland C. Viray
ENGINEERING ECONOMY
INSTRUCTIONAL MATERIAL
THE OVERVIEW
This instructional material (IM) for Engineering Economy will give the students a good
understanding on what is the time value of money like the present worth and future relation and
how rate of interest affects their respective values. Likewise, it will also show the importance of
equation of value and its use to solve various problems in this subject. Similarly, it will also show
the different types of annuity and how depreciation changes the worth of a property due to passage
of time. It will also give the importance of break-even point in decision making whether a company
can make or break in its operation.
Several sample problems are presented as guide to solve the problems in the assessments at the
end of each module which eventually will give the student a chance to master the use of formulas
as presented in this instructional material.
THE LEARNING OBJECTIVES
This instructional material (IM) for the subject Engineering Economy will discuss the topics which
are commonly given in the Engineering Board Examination such as;
1. Simple Interest
2. Compound Interest
3. Present Worth (P) and Future Worth (F) Relations
4. Discount and Rate of Discount
5. Importance of Equation of Value
6. Annuity and Amortizations
7. Arithmetic and Geometric Gradients
8. Capitalized cost
9. Bond Value Calculation
10. Depreciation and Depletion
11. Break-Even Analysis and Profit Computation
COURSE MATERIALS:
1. Engineering Economy by de Garmo
2. Engineering Economy by Blank et.al.
3. Engineering Economy by Arreola
4. Engineering Economy by Sta Maria
5. Simplified Engineering Economy by Ocampo et.al.
6. Engineering Economy by Engr. Jimmy L. Ocampo at youtube.com
1
MODULE 1
LEARNING OBJECTIVES:
After the completion of Module 1, it is expected that the student have understand the
following:
1. Simple and Compound Interest
2. Present Worth (P) and Future Worth (F) relations
3. Discount and Rate of Discount
4. Importance of Equation of Value
ENGINEERING ECONOMY
- it deals with the use and application of economic principles in the analysis of
engineering decisions.
TOPIC 1 – MONEY AND INTEREST
MONEY – it is a measure of wealth
INTEREST – it is the amount paid for the use of borrowed capital
1. THE SIMPLE INTEREST, SI
- it is the interest earned by the principal alone over a given period of time usually
counted in number of days, months or in years.
a. Ordinary SI
Basis: 30 days / month
360 days / year
12 months / year
b. Exact SI
Basis: 365 days / year
366 days / leap year
NOTE: a year which is exactly divisible by four (4) is a leap year.
FORMULAS:
1. I = Pin
2. F = P (1 + in)
where,
I = interest, P
P = present worth or principal amount, P
i = rate of interest, reported as percent per unit time (yr) and used as
decimal in computation, % per unit time, e.q. 5% per year
n = no. of interest periods, the duration or time usually in years.
F = future worth or accumulated amount, P
NOTE: In formulas 1 and 2, the unit of time in i and n must be consistent.
2
EXAMPLES:
1. What is the interest of 8600P after 4 years at 12% simple interest rate?
Solution:
use, I = Pin
where,
P = 8600P
i = 12% “per year” = 0.12
n = 4 years
hence,
I = 8600(0.12)(4) = 4128P
2. 5000P will become how much after one year at simple interest of 15%?
Solution:
use, F = P (1 + in)
where,
P = 5000P
i = 15% “per year” = 0.15
n = 1 year
hence,
F = 5000 { 1 + (0.15) (1) }
F = 5750P
3. Find the present worth with a total interest of 5000P after 2 years at simple interest
rate of 6.25%.
Solution:
use, I = Pin
where,
I = 5000P
i = 6.25% “per year” = 0.0625
n = 2 years
hence,
5000 = P (0.0625) (2)
P = 40,000 P
4. In how many years will the investment to double its value at 5% simple interest?
Solution:
use, F = P (1 + in)
where,
F = 2P
i = 5% “per year” = 0.05
hence,
2P = P (1 + 0.05n)
n = 20 years
5. A man deposited 10,000P in a bank at 10% per annum for 3 years, 8 months and 25
days. Find the ordinary simple interest.
3
Solution:
use, I = Pin
where,
P = 10,000P
i = 10% “per year” = 0.10
n = 3 years + 8 months x
1𝑦𝑟
12 𝑚𝑜𝑠
+ 25 days x
1𝑦𝑟
360 𝑑𝑎𝑦𝑠
n = 3.74 years
hence,
I = 10000 (0.10) (3.74)
= 3740P
6. 10,000P was deposited in a bank at 10% per annum from Jan 15,2020 to Oct 25,2020.
Find the accumulated amount based on exact simple interest computation.
Solution:
use, F = P(1 + in)
where,
P = 10,000P
i = 10% per year = 0.1
n = ? year
Count the no of days covered by the deposit,
2020
4
= 505 “exact”
hence,
2020 is a Leap year and 366 days / Leap year
𝐽𝑎𝑛 15 − 31 = 16 "𝑒𝑥𝑐𝑙𝑢𝑑𝑒 𝐽𝑎𝑛 15"
𝐹𝑒𝑏𝑟𝑢𝑎𝑟𝑦 = 29 "𝑙𝑒𝑎𝑝 𝑦𝑒𝑎𝑟"
𝑀𝑎𝑟𝑐ℎ = 31
𝐴𝑝𝑟𝑖𝑙 = 30
𝑀𝑎𝑦 = 31
𝐽𝑢𝑛𝑒 = 30
𝐽𝑢𝑙𝑦 = 31
𝐴𝑢𝑔 = 31
𝑆𝑒𝑝𝑡 = 30
𝑂𝑐𝑡 1 − 25 = 25 "𝑖𝑛𝑐𝑙𝑢𝑑𝑒 𝑂𝑐𝑡 25"
n = 284 days
n=
284
366
= 0.776 yr
hence,
F = 10000 { 1 + (0.1)(0.776) } = 10776P
7. A price tag of 1500P is payable in 70 days but if paid in 35 days it will have a 5%
discount. Find the rate of interest.
Solution:
use, F = P (1 + in)
4
where,
F = 1500P
P = 1500 – 0.05 (1500)
P = 1425P
n = 35 days (
1𝑦𝑟
360 𝑑𝑎𝑦𝑠
) = 0.0972 yr
hence,
1500 = 1425 { 1 + i (0.0972) }
i = 0.5415 = 54.15% per yr
TOPIC 2 – COMPOUND INTEREST
- it is the interest on top of interest.
1. Nominal Rate of Interest ( j ) = the rate of interest that specifies the no of interest
periods in one year.
Ex:
j = 12% compounded quarterly (n1 = 4) --- i = 3% per quarter
FORMULA:
i=
𝒋
𝒏𝟏
where n1 = no. of interest periods in one year.
Common Methods of
Compounding
Values of n1
annually
semi-annually
quarterly
bi-monthly
monthly
daily
every 6 mos
every 3 mos
every 2 mos
1
2
4
6
12
365
2. Effective Rate of Interest ( ie ) = the actual rate of interest in one year.
FORMULAS:
a. ie = ( 1 + i )n1 - 1
b. ie = ( 1 +
𝒋
𝒏𝟏
)n1 – 1
Importance of ie
1. To identify which interest rate is higher
2. To convert an interest rate to other method of compounding.
NOTE: Two interest rates are equal if their effective rates are equal.
5
EXAMPLES:
1. A bank charges 1.5% per month on credit card. Find (a) the nominal rate of interest
compounded monthly (b) the effective rate of interest (c) the equivalent nominal rate of
interest which is compounded quarterly.
Solution:
a)
i=
𝑗
𝑛1
where,
i = 1.5% per month
n1 = 12
j=?
hence,
1.5 =
𝑗
12
j = 18% compounded monthly
b)
ie = ( 1 + i )n1 - 1
hence,
ie = ( 1 + 0.015)12 – 1
ie = 0.1956
ie = 19.56% “per year”
c)
1.5% per month to ___% compounded quarterly
ie (quarterly) = ie (monthly)
n1 = 4
n1 = 12
j=?
i = 0.015
𝑗
( 1 + )4 – 1 = ( 1 + 0.015)12 – 1
4
solve for j,
j = 0.1827
j = 18.27% compounded quarterly
2. A bank advertises 9.5% account that yields 9.84% annually. Find how often is the
interest compounded?
Solution:
j = 9.5% compounded n1 = ?
ie = 9.84%
use,
ie = ( 1 +
0.0984 = ( 1 +
𝑗
)n1 – 1
𝑛1
0.095
𝑛1
)n1 – 1
by ES, Shift solve
n1 = 3.88 ≈ 4
hence,
9.5% is compounded quarterly
6
3. Find the nominal rate, which is converted quarterly could be used instead of 12%
compounded semi-annually.
Solution:
12% compounded semi-annually
to
__% compounded quarterly
ie (quarterly) = ie (semi-annually)
n1 = 4
n1 = 2
j=?
j = 0.12
𝑗
0.12
4
2
( 1 + )4 – 1 = ( 1 +
)2 – 1
j = 0.1183 = 11.83% compounded quarterly
TOPIC 3 – P AND F RELATION WITH COMPOUNDED INTEREST
CASH FLOW DIAGRAM, CFD
1
2
3
___ǀ___ǀ___ǀ___ _ _ n
i
F
FORMULAS:
1. F = P ( 1 + i )n
2. P = F ( 1 + i )-n
where,
( 1 + i )n = Single Payment Compound Amount Factor or
Future Value Factor (FVF)
( 1 + i )-n = Single Payment Present Value Factor (PVF)
EXAMPLES:
1. In 1906, an original painting of Picasso has a market price of 600P and in 1995 its
price has increased to 29,000,000P. What is the rate of interest of the painting?
Solution:
F = P(1 + i)n
where,
F = 29,000,000
P = 600P
n = 1995 – 1906 = 89 years
hence,
29,000,000 = 600(1 + i )89
i = 0.1288
i = 12.9%
7
2. If 10,000P is invested at 12% interest compounded monthly, find the 1st yr interest.
Solution:
j = 12% compounded monthly (n1 = 12)
i=
𝑗
𝑛1
=
12
12
i = 1% per month
I = F – P ------------------------ 1
where,
F = P (1 + i)n
P = 100,000P
i = 0.01
n = 1 yr = 12 mos
hence,
F = 100,000(1 + 0.01)12 = 112682.5P
subst. values to 1,
I = 112682.5 – 100,000
= 12682.50P
3. After how many years will an investment triple if invested at 10% per annum, net of
deduction, compounded quarterly?
Solution:
F = P(1 + i )n
where,
F = 3P
j = 10% compounded quarterly (n1 = 4)
i=
10
4
= 2.5% per quarter
hence,
3P = P(1 + 0.025)n
n = 44.5 quarters
in yrs n = 44.5 quarters (
1𝑦𝑟
4 𝑞𝑢𝑎𝑟𝑡𝑒𝑟𝑠
) = 11.12 yrs
TOPIC 4 – P and F Relation with Continuously Compounded Interest
*recall
The Exponential Law of Change in the Differential Equations
ln
𝑥
𝑥0
in monetary values
= kt or x = x0 ekt
x=F
x0 = P
k = j = continuously compounded interest
t = n, yrs
FORMULAS:
1. F = Pejn
2. ie = ej – 1
j – in decimal
n – in yrs
8
EXAMPLE:
1. Find the effective interest equivalent to 12% compounded continuously.
Solution:
use, ie = ej – 1 = e0.12 – 1
ie = 0.1275
= 12.75%
2. What is the future worth of 10,000P when invested at the rate of 12% compounded
continuously for 5 yrs?
Solution:
use, F = Pejn = 10000 e0.12(5)
F = 18221.19P
TOPIC 5 – DISCOUNT, D
- it is the difference between the future worth (F) and the present worth (P)
FORMULAS:
1. D = F – P
Rate of Discount, d = the discount on one unit in one unit of time.
P = 1(1+i)-1
CFD
________________ n = 1
0
F = 1P
hence,
2. d = 1 – (1 + i)-1 or =
3. i =
𝒅
or =
𝟏−𝒅
𝑫
𝑭
𝑫
𝑷
EXAMPLE:
1. What is the corresponding rate of interest for 18% simple discount rate?
Solution:
use, i =
𝑑
1−𝑑
where,
d = 18% = 0.18
hence,
i=
0.18
1 − 0.18
= 0.2195
i = 21.95%
9
TOPIC 6 – EQUATION OF VALUE, EV
- it is the resulting equation when comparing two sets of obligations at a certain
point of comparison called focal date.
EV at a focal date,
∑↑=∑↓
where,
∑ ↑ = sum of cash inflow
∑ ↓ = sum of cash outflow
NOTE: Use EV to solve unknown in a CFD.
EXAMPLE:
1. 12,000P is borrowed now at 12% interest. The 1st payment is 4000P and is made 3
years from now. Find the balance on the debt immediately after the 1st payment.
Solution:
draw CFD
12000
12000 (1.12)3
i = 12% per year
________________ 3 yrs
0
4000P
B = ? “balance”
set up EV at 3,
∑↑=∑↓
12000 (1.12)3 = 4000 + B
B = 12860P
10
2nd Solution:
12000P
i = 12% = 0.12
3
0
4000P
4000(1.12)-3
B
B(1.12)
-3
using zero as focal date
∑↑=∑↓
12000 = 4000(1.12)-3 + B(1.12)-3
B = 12860P
2. An investment pays 6000P at the end of the 1st year, 4000P at the end of the 2nd year
and 2000P at the end of the 3rd year. Compute the present value of the investment if a
10% rate of return is required.
Solution:
draw CFD
i = 10% “per yr”
P=?
1
2
3
yr
0 _____ǀ_____ǀ_____ǀ_
2000P
4000P
6000P
using zero as focal date, EV is
∑↑=∑↓
P = 2000(1.1)-3 + 4000(1.1)-2 + 6000(1.1)-1
P = 10262.96P
11
ASSESSMENT 1
1. A man wishes to accumulate 3722P after
5 yrs, 8 months and 28 days. How much
should be deposited by the man in a bank if
the ordinary simple interest is 15% per
annum?
2. A man deposited 2000P in a bank at the
rate of 15% per annum from March 21,1996
to October 25,1997. Find the exact simple
interest.
3. A bank charges 1.5% per month on a loan.
Find the equivalent nominal rate of interest.
4. A financing company charges 1.5% per
month on a loan. Find the equivalent effective
rate of interest.
5. A nominal rate of 12% compounded
monthly is equal to an effective rate of ____.
6. Convert 16% compounded semi-annually
to equivalent nominal rate which is
compounded daily.
is payable at once, if the bank gave him a
discount of 6%.
12. Find the cash price of a generator which
was bought in installment basis that requires
a down payment of 50,000P and payment of
30,000P after 1 yr, 40,000P after 2 yrs and a
final payment of 76,374.34P after 4 yrs at a
rate of 15% per annum.
13. A man made a loan of 100,000P at a rate
of 15% per annum and promise to pay it
according to the following manner, 30,000P
at the end of 1st yr, unknown payment at the
end of 2nd yr and a final payment of
76,374.38P at the end of 4th yr. Find the
unknown payment made by the man.
14. Find the present worth of the following
payments, 5000P after 1 yr, 4000P after 2
yrs, 8000P after 4 yrs at a rate of 12% per
annum.
7. Find the accumulated amount of 1000P
after 5 yrs when deposited in a bank at a rate
of 16% compounded monthly.
15. Find the amount of the following
payments at the end of 5th yr, 3000P at the
end of 1st yr, 4500P at the end of 2nd yr and
6000P at the end of 4th yr if money worth 12%
per annum.
8. How long in yrs will a certain sum of money
to triple its amount when deposited at a rate
of 12% compounded annually?
16. How many yrs will it take for a certain sum
of money to triple its amount when deposited
at a rate of 12% compounded continuously?
9. How much should be deposited in a bank
at a rate of 12% compounded continuously
for 5 yrs if its accumulated amount is
9110.60P?
17. A bank is advertising 9.5% accounts that
yield 9.84% annually. How often is the
interest compounded?
10. An effective rate of interest, which is
12.75%, is equivalent to what percent if
compounded continuously.
18. A man borrows 2000P for 6 yrs at 8%. At
the end of 6 yrs, he renews the loan for the
amount due plus 2000P more for two yrs at
8%. What is the lump sum due?
11. How much is expected to be received by
a man that makes a loan of 851.06P, which
19. You deposit 1000P into a 9% account
today. At the end of two yrs, you will deposit
another 3000P. In five yrs, you plan a 4000P
12
purchase. How much is left in the account
one year after the purchase?
at the end of one year. Revenue of 150,000P
will be generated at the end of years 1 and 2.
What is the net present value of this business
if the effective annual interest rate is 10%?
20. Consider a business which involves the
investment of 100,000P now and 100,000P
ANSWERS TO ASSESSTMENT 1
1. 2000P
2. 480P
3.18%
4.19.56%
5. 12.68%
6.15.39%
7. 2210P
8. 9.2 yrs
9. 5000P
10. 12%
11. 800P
12. 150000P
13. 40000P
14. 12737P
15. 17763P
16. 9.16yrs
17. quarterly
18. 6035P
19. 1552P
20. 69422P
13
MODULE 2
LEARNING OBJECTIVES:
After the completion of Module 2, it is expected that the student have understand the
following:
1. Annuity and Amortization
2. Types of Annuity
3. Arithmetic and Geometric Gradients
4. Bond Value and Capitalized Cost
TOPIC 1 – ANNUITY AND AMORTIZATION
ANNUITY – it is a series of equal payments occurring at equal interval of time.
AMORTIZATION – a method of paying debt including the principal and interest which is
done in a series of equal payments occurring at equal interval of time.
TYPES OF ANNUITY
1. Ordinary Annuity – an annuity when payments or amortizations are made at
the end of each period.
Consider the CFD,
P
i
1
2
3
n
A
A
A
A = P / unit time
0 _______________ _ _
F
FORMULAS:
−𝒏
𝟏−(𝟏 + 𝒊)
}
𝒊
𝒏
(𝟏 + 𝒊) −𝟏
2. F = A {
}
𝒊
1. P = A {
where,
(1 + 𝑖)𝑛 −1
𝑖
𝐹
= ( 𝐴 , 𝑖%, 𝑛) – is called series of payments
Compound Amount Factor
1− (1 + 𝑖)−𝑛
𝑖
𝑃
= ( 𝐴 , 𝑖%, 𝑛) – is called series of payments
Present Value Factor
14
EXAMPLES:
1. The president of a growing engineering firm wishes to give each of 50 employees a
holiday bonus. How much is needed to invest monthly for a year at 12% nominal rate
compounded monthly, so that each employee will receive a 1000P bonus?
Solution:
j = 12% compounded monthly (n1 = 12) ------- i = 1% per month = 0.01
n = 1 yr = 12 mos
F = 50 (1000) = 50000P
𝑛
F=A{
(1 + 𝑖) −1
}
𝑖
50000 = A {
(1.01) −1
}
0.01
use,
12
A = 3942.44P / month
2. A young engineer borrowed 10000P at 12% interest and paid 2000P per annum for
the last 4 yrs. What does he have to pay at the end of 5th yr in order to pay his loan?
Solution:
draw CFD
10000P
i = 12% “per year”
0
PA
1
2
3
4
A
A
A
A
5
A = 2000P/yr
x=?
x(1.12)-5
set-up EV at zero
∑↑=∑↓
1000 = PA + x(1.12)-5 ------------------- 1
where,
PA = A {
1−(1+𝑖)
𝑖
PA = 2000 {
−𝑛
}
1−(1.12)
0.12
−4
} = 6074.70
subst. to 1
1000 = 6074.70 + x(1.12)-5
x = 6917.72P
3. An investment of 350,000P is made to be followed by revenue of 200,000P each year
for 3 years. What is most nearly the annual rate of return on investment for this project?
15
Solution:
draw CFD
PA
A
A
A = 200k P/yr
1
2
3
0
i=?
350kP
set-up EV at zero
∑↑=∑↓
PA = 350k
−𝑛
A{
1−(1+𝑖)
𝑖
200k {
1−(1+𝑖)
𝑖
} = 350k
−3
by ES,
} = 350k
i = 0.3268
= 33% per yr
4. To maintain a structure with a life of 20 yrs, it is necessary to provide the following for
repairs; 20000P at the end of 5th yr, 30000P at the end of 10th yr and 40000P at the end
of 15th yr. If money is worth 10% compounded annually, determine the equivalent uniform
annual maintenance cost for the 20 yr period.
Solution:
draw CFD
1.
P i = 10% per yr
5
10
15
0
20kP
30kP
40kP
2.
P=A{
𝟏−(𝟏.𝟏)−𝟐𝟎
𝟎.𝟏
0
} --------------- 1
20 yrs
A
A
A
A=?
16
at CFD #1 set up EV at zero
P = 20k(1.1)-5 + 30k(1.1)-10 + 40k(1.1)-15
P = 33560.41P,
subst. to 1 at CFD 2,
1−(1.1)
0.1
33560.41 = A {
−20
}
A = 3942P / yr
2. Deferred Annuity – type of annuity where payments are made several periods
after the annuity has started (late amortizations).
CFD
P
m
1
n
2
m
0
1
2
3
n
A
A
A
A
0'
“by EV at zero”
FORMULA:
−𝒏
P=A{
𝟏−(𝟏+𝒊)
𝒊
} (1 + i)-m
where,
m = deferred periods
n = ordinary annuity periods
EXAMPLES:
1. A man loans 187,400P from a bank with interest at 5% compounded annually. He
agrees to pay his obligations by paying 8 equal payments, the first being due at the end
of 10 years. Find the annual payments.
Solution:
draw CFD
P = 187400P
i = 5% per yr
m=9
1
2
n=8
9
01
0
10
1
2
3
n=8
A
A
A
A=?
use,
P=A{
1−(1 + 𝑖)
𝑖
−𝑛
} (1 + i)-m
subst. values,
187400 = A {
1−(1.05)
0.05
−8
} (1.05) -9
A = 44980.56P/yr
17
2. A student needs 4000P per year for four years to attend college. Her father invested
5000P in a 7% account for her education when she was born. If the student withdraws
4000P at the end of her 17th, 18th, 19th and 20th years, how much money will be left in the
account at the end of her 21st yr?
Solution:
B = ? (balance)
draw CFD
1
2
3
0
A
A
A
A
16
17
18
19
20
01
1
2
3
4
m = 16
21
n=4
5000P
i = 7% “per year”
a) using zero as focal date,
∑↑=∑↓
PD + B(1.07)-21 = 5000
−𝑛
A{
1−(1 + 𝑖)
𝑖
4000 {
1−(1.07)
0.07
} (1+i)-m + B(1.07)-21 = 5000
−4
} (1.07)-16 + B(1.07)-21 = 5000
B = 1700P
b) using 16 as focal date,
∑↑=∑↓
5000(1.07)16 = 4000 {
1−(1.07)
0.07
−4
} + B (1.07)-5
= 1700P
3. Perpetuity – type of annuity where payments are made indefinitely or forever.
CFD
P
1
2
3
4
n=∞
0
A
A
A
A
FORMULA:
P=
𝑨
𝒊
18
EXAMPLES:
1. What present sum would be needed to provide for annual end of year payments of
150,000 P each forever at an interest of 8%?
Solution:
𝐴
use, P =
𝑖
150000
P=
0.08
P = 1875000P
2. What amount of money deposited 50 yrs ago at 8% interest would now provide a
perpetual payment of 10000P per year?
Solution:
draw CFD
Pp= 𝟏𝟎𝟎𝟎𝟎
𝟎.𝟎𝟖
i = 8% “per yr”
A
A
A = 10000P/yr
1
2
3
∞
50
0
P(1.08)50
P
set-up EV at 50
∑↑=∑↓
10000
0.08
= P (1.08)50
P = 2665.15P
4. Annuity Due – type of annuity where payments started at the beginning of the annuity
periods
P
CFD
i
1
2
3
n
A
A
A
A
0
A
F
19
FORMULAS:
1. P = A {
2. F = A {
𝟏−(𝟏+𝒊)−(𝒏−𝟏)
𝒊
(𝟏+𝒊)(𝒏+𝟏) −𝟏
𝒊
+ 𝟏}
− 𝟏}
EXAMPLE:
1. A man borrows 100000P at 10% effective annual interest. He must pay back the loan
over 30 yrs with uniform monthly payments due on the first day of each month. What
amount does the man pay each month?
Solution:
ie = 10% ----------- i = ?
use, ie = (1 + i)n1 – 1
0.10 = (1 + i)12 – 1
i = 7.974x10-3
or
i = 0.7974% per month
P=A{
1−(1+𝑖)−(𝑛−1)
𝑖
+ 1}
n = 30 yrs = 360 mos
subst. values,
100000 = A {
1−(1+7.974 𝑥 10−3 )−359
7.974 𝑥 10−3
+ 1}
A = 839.18 P/month
5. Annuity with Continuously Compounded Interest – formulas are similar to Ordinary
Annuity but replace i by ej – 1
FORMULAS:
1. F = A {
2. P = A {
𝒆𝒋𝒏 − 𝟏
}
𝒆𝒋 − 𝟏
𝟏 − 𝒆−𝒋𝒏
𝒆𝒋 − 𝟏
}
where,
j = interest rate compounded continuously in decimal
n = no of years
EXAMPLE:
1. A man deposits 5000P each year into his savings account that pays 5% nominal
interest compounded continuously. How much will be the worth of the account at the end
of 5 years?
Solution:
F=A{
𝑒 𝑗𝑛 − 1
𝑒𝑗 − 1
}
20
where,
A = 5000P/yr
n = 5 yrs
j = 5% compounded continuously = 0.05
hence,
F = 5000 {
𝑒 0.05(5) − 1
𝑒 0.05 − 1
}
F = 27698.40P
TOPIC 2 – ARITHMETIC GRADIENT
- it is a series of payment with common difference and occurring at equal interval of time
CFD
P
1
2
3
n
0
A
A+G
A+2G
A+(n-1)G
where,
G = common difference
FORMULAS:
1. P = PA + PG
where, PA = A {
PG =
𝑮
𝒊
{
𝟏−(𝟏+𝒊)−𝒏
𝒊
𝟏−(𝟏+𝒊)−𝒏
}
− 𝒏(𝟏 + 𝒊)−𝒏 }
𝒊
2. F = FA + FG
where, FA = A{
FG =
𝑮
𝒊
{
(𝟏+𝒊)𝒏 −𝟏
𝒊
(𝟏+𝒊)𝒏 −𝟏
𝒊
}
− 𝒏}
EXAMPLE:
1. A farmer buys a tractor. There will be no maintenance cost the 1st yr as the tractor is
sold with one years free maintenance. The 2nd yr the maintenance is estimated at 2000P.
In the subsequent yrs the maintenance cost will increase by 2000P per year. How much
would need to be set aside now at 5% interest to pay the maintenance cost on the tractor
for the first 6 yrs of ownership?
21
Solution:
draw the CFD
1
2
3
4
5
6
0
0k
2k
4k
6k
8k
10k
A = 0P
G = 2k
P = PG = ?
n=6
i = 5% = 0.05
use,
P = PG =
𝐺
𝑖
{
1−(1+𝑖)−𝑛
𝑖
− 𝑛(1 + 𝑖)−𝑛 }
subst. values,
P=
2𝑘
0.05
{
1−(1.05)−6
0.05
− 6(1.05)−6 }
P = 23936P
TOPIC 3 – GEOMETRIC GRADIENT
- series of payment with common ratio and occurring at equal interval of time.
CFD
P
1
2
3
n
0
A
A(1+r)
A(1+r)2
A(1+r)n – 1
where,
r = % change in payments
1 + r = common ratio
FORMULAS:
1. P =
where,
𝑨
{
𝟏 − 𝒘𝒏
𝟏+𝒊 𝟏−𝒘
𝟏+𝒓
w=
𝟏+𝒊
}
------ use up to the 4th decimal place
22
2. if i = r
P=
𝑨𝒏
𝟏+𝒊
EXAMPLE:
1. A young man has decided to go into business at age 40. He wishes to accumulate
200000 P at that age. On his 25th birthday he deposits a certain amount and will increase
the deposit by 10% each year until the 40th yr. If the fund can be invested at 9.6%
compounded annually, how much should his initial investment be?
Solution:
draw CFD
200kP
i = 9.6%, r = 10%
25
0
26
1
27
2
40
15
x=?
w=
(𝟏+𝒓)
(𝟏+𝒊)
=
(𝟏.𝟏)
(𝟏.𝟎𝟗𝟔)
= 1.00365
A = x(1.1)
x(1.1)2
x(1.1)15
PGG
set-up EV at zero
∑↑=∑↓
x + PGG = 200k(1.096)-15
x+
𝑥(1.1)
1.096
{
1 − (1.00365)15
1 − 1.00365
} = 200k(1.096)-15
solve for x
x = 3074.85P
TOPIC 4 – BOND VALUE
- it is the present worth or cost of a bond.
CFD
C ------ if not given, C ≃ F
n = maturity period
I
I
I
I = Fr ------ if not given
F = face value or par- value
r = bond rate
n
I = dividend
0
i = yield of investment
P = bond value = ?
23
FORMULA:
set-up EV at zero
P=I{
𝟏 − (𝟏+𝒊)−𝒏
} + C (1+i)-n
𝒊
EXAMPLE:
1. A 1000P face value bond pays dividend of 110P at the end of each yr. If the bond
matures in 20 yrs, what is the approximate bond value at an interest of 12% per yr
compounded annually?
Solution:
use,
P=I{
1 − (1+𝑖)−𝑛
where, I = 110P/yr
n = 20 yrs
hence, P = 110 {
} + C (1+i)-n
𝑖
i = 12% per yr = 0.12
C ≃ F = 1000P
1 − (1.12)−20
0.12
} + 1000(1.12)-2 = 925.31P
TOPIC 5 – CAPITALIZED COST, CC
- it is the sum of the first cost (FC or C0) and the present worth of perpetual annual
maintenance and operational cost (MC), cost of repair (CR) at interval k yrs, and renewal
cost (RC) at the end of life L yrs.
FORMULA:
CC = FC +
𝑴𝑪
𝒊
+
𝑪𝑹
(𝟏+𝒊)𝒌 −𝟏
𝑹𝑪
+
(𝟏+𝒊)𝑳 −𝟏
NOTE:
1. if Life L yrs is given and RC is not given,
use, RC ≃ FC – CR – SV
where, SV or CL = salvage value
2. k is a factor of L
EXAMPLES:
1. A machine costs 80000P and with a salvage value of 20000P at the end of useful life of
20 yrs. The annual operating costs is 18000P. Find the capitalized cost of the machine at
an interest rate of 10% per annum.
Solution:
use, CC = FC +
𝑀𝐶
𝑖
+
𝐶𝑅
(1+𝑖)𝑘 −1
+
𝑅𝐶
(1+𝑖)𝐿 −1
where,
FC = 80kP
SV = 20kP
L = 20 yrs
i = 10%
MC = 18kS/yr CR = 0
RC = ?
RC = FC – SV – CR = 80 – 20 – 0 = 60kP
subst. value,
CC = 80k +
18𝑘
0.10
+
60𝑘
(1.1)20 −1
= 270475.77P
24
2. A dam was constructed for 200kP. The annual maintenance cost is 5kP. Find the
capitalized cost of the dam at an interest rate of 5% per annum.
Solution:
use, CC = FC +
𝑀𝐶
𝑖
+
𝐶𝑅
(1+𝑖)𝑘 −1
+
𝑅𝐶
(1+𝑖)𝐿 −1
where,
FC = 200kP
MC = 5kP/yr
CR = 0
RC = 0
i = 5%
hence,
CC = 200kP +
5𝑘
0.05
= 300000P
25
ASSESSMENT 2
1. Find the accumulated amount of the
ordinary annuity paying an amortization of
1000P per month at a rate of 12%
compounded monthly for 5 years.
2. What present sum is equivalent to a series
of 1000P annual end-of-year payments, if a
total of 20 payments are made and interest is
12%?
3. A man made ten annual-end-of year
purchases of 1000P common stock. At the
end of 10th year he sold all the stock for
12000P. What interest rate did he obtain on
his investment?
4. A piece of property is purchased for
10000P and yields a 1000P yearly profit. If
the property is sold after 5 years, what is the
maximum price to break-even if the interest
is 6% per annum?
5. A condominium unit can be bought at a
down payment of 150000P and a monthly
payment of 10000P for 10 years starting at
the end of 5th year from the date of purchase.
If money is worth 12% compounded monthly,
what is the cash price of the condominium
unit?
6. The owner of the quarry signs a contract to
sell his stone on the following basis. The
purchaser is to remove the stone from the
certain portion of the pit according to a fixed
schedule of volume, price and time. The
contract is to run 18 years as follows. Eight
years excavating a total of 20,000 m per year
at 10P per meter, the remaining ten years,
excavating a total of 50,000 m per year at
15P per meter. On the basis of equal yearend payments during each period by the
purchaser, what is the present worth of the pit
to the owner on the basis of 15% interest?
7. A wealthy man donated a certain amount
of money to provide scholarship grants to
deserving students. The fund will grant
10,000P per year for the first 10 years and
20,000P per year on the years thereafter.
The scholarship grants started one year after
the money was donated. How much was
donated by the man if the fund earns 12%
interest.
8. What amount of money deposited 40 years
ago at 12% interest would now provide a
perpetual payment of 10,000P per annum?
9. A company rent a building for 50,000P per
month for a period of 10 years. Find the
accumulated amount of the rentals if the
rental for each month is being paid at the start
of each month and money is worth 12%
compounded monthly.
10. The amount of the perspective investor
pay for a bond if he desires an 8% return on
his investment and the bond will return 1000P
per year for 20 years and 20,000P after 20
years is
11. A machine costs 50,000P. Find the
capitalized cost if the annual maintenance
and operational cost is 5000P and money
worth 15% per annum.
12. A machine cost 50,000P. Find the
capitalized cost if the annual maintenance
cost is 5000P and cost of repair is 4000P
every 4 years and money worth 12% per
annum.
13. A building cost 10 million and the salvage
value is 150,000P after 25 years. The annual
maintenance cost is 60,000P costs of repair
is 200,000P every 5 years. Find the
capitalized cost if money worth 15% per
annum.
26
14. A salesman earns 1000P on the 1st
month, 1500P on the 2nd month, 2000P on
the 3rd month and so on. Find the
accumulated amount of his income at the 10th
month if money worth 12% compounded
monthly.
15. A man wishes to accumulate a total of
500,000P at the age of 30. On his 20th
birthday, he deposited a certain amount of
money at a rate of 12% per annum. If he
increases his deposit by 10% each year until
the 30th birthday, how much should his initial
deposit be?
16. If 2000P is deposited in a savings
account at the beginning of each of 15 years
and the account draws interest at 7% per
year, compounded annually. Find the value
of the account at the end of 15 years.
17. A man deposits 1000P every year for 10
years in a bank. He makes no deposit during
the subsequent 5 years. If the bank pays 8%
interest, find the amount of the account at the
end of 15 years.
18. Twenty-five thousand pesos is deposited
in a savings account that pays 5% interest,
compounded semi-annually. Equal annual
withdrawals are to be made from the account,
beginning one year from now and continuing
forever. Find the maximum amount of the
equal annual withdrawal.
19. What amount of money deposited 50
years ago at 8% interest would now provide
a perpetual payment of 10000P per year?
20. A man buys a motor cycle. There will be
no maintenance cost the first year as the
motor cycle is sold with one year free
maintenance. The 2nd year the maintenance
is estimated at 2000P. In subsequent years
the maintenance cost will increase by 2000P
per year. How much would need to be set
aside now at 5% interest to pay the
maintenance costs of the motor cycle for the
first 6 years of ownership?
ANSWERS TO ASSESSMENT 2
1. 81670P
2. 7470P
3. 4%
4. 7745P
5. 539171P
6. 2127948P
7. 110165P
8. 896P
9. 11616954P
10. 14109P
11. 83334P
12. 98641P
13. 10.9MP
14. 33573P
15. 15987P
16. 53776P
17. 21286P
18. 1265P/yr
19. 2665P
20. 23936P
27
MODULE 3
LEARNING OBJECTIVES:
After the completion of Module 3, it is expected that the student have understand the
following:
1. Depreciation and Depletion
2. Common Methods to Calculate depreciation
3. Methods of Evaluating Depletion
4. Hoskold’s Formula for Valuation
TOPIC 1 – COMMON METHODS OF EVALUATING DEPRECIATION
DEPRECIATION – it is the decrease in worth or value of a property due to passage of time.
a. Straight Line Method, SLM
▪ the simplest method
▪ depreciation charge per year (d) is constant
FORMULAS:
1. d =
𝑪𝟎 − 𝑪𝑳
𝑳
2. Dn = nd =
𝒏
𝑳
(C0 – CL)
3. Cn = C0 – Dn
where,
C0 = original cost
CL = salvage value
DN = total depreciation after n years
CN = book value after n years
EXAMPLES:
1. A machine costing 1.8M P has a life of 8 yrs. Using SLM, the total depreciation at the
end of 4th year is 800kP. Determine the salvage value of the machine.
Solution:
use, DN =
𝑛
𝐿
(C0 – CL)
for n = 4,
D4 =
4
8
(C0 – CL)
subst. values,
800k =
1
2
(1.8M – CL)
CL = 200000P
2. A drill press is purchased for 10000P and has an estimated life of 12 yrs. The salvage
value at the end of 12 yrs is estimated to be 1300P. Using SLM, compute the book value
of the drill press at the end of 8 yrs.
28
Solution:
Cn = C0 – DN
𝑛
Dn =
(C0 – CL)
𝐿
Cn = C0 -
𝑛
(C0 – CL)
𝐿
for
n=8
C0 = 10kP
subst. values
L = 12
CL = 1.3kP
8
C8 = 10k –
( 10k – 1.3k)
12
C8 = 4200P
b. Sinking Fund Method, SFM
▪ d = constant per year
▪ interest rate i is considered in the computations
CFD
C0 - CL
i
0
Dn
1
2
3
n
d
d
d
d
L
d
using,
F=A{
(1+𝑖)𝑛 −1
𝑖
}
FORMULAS:
1. d =
where,
(𝑪𝟎 − 𝑪𝑳 ) 𝒊
(𝟏 + 𝒊)𝑳 −𝟏
𝒊
= Sinking Fund Factor, SFF
(𝟏 + 𝒊)𝑳 −𝟏
(𝟏+𝒊)𝒏 −𝟏
2. Dn = d {
}
Dn =
(C0 – CL)
𝒊
(𝟏+𝒊)𝒏 −𝟏
(𝟏+𝒊)𝑳 −𝟏
or
3. Cn = C0 – Dn
EXAMPLE:
1. An equipment cost 10kP with a salvage value of 500P at the end of 10yrs. Calculate
the annual depreciation cost by sinking fund method if interest rate is 4%.
Solution:
use,
d=
(𝑪𝟎 − 𝑪𝑳 ) 𝒊
(𝟏 + 𝒊)𝑳 −𝟏
29
where,
C0 = 10kP
CL = 500P
L = 10 yrs
i = 4% = 0.04
hence,
d=
(10𝑘 − 500)(0.04)
(1.04)10 −1
d = 791.26P
c. Sum of the Years Digit Method, SYDM
▪ depreciation charges varies from yr to yr
▪ evaluated by the principle of Arithmetic Progression
FORMULAS:
1. dn =
2. Dn =
𝟐(𝑳 − 𝒏 + 𝟏)
(C0 – CL)
𝑳(𝑳 + 𝟏)
---------Ocampo’s Formula
𝒏(𝟐𝑳 − 𝒏 + 𝟏)
𝑳(𝑳 + 𝟏)
(C0 – CL) -----------
3. Cn = C0 - Dn
EXAMPLES:
1. An asset is purchased for 120kP, its estimated life is 10 yrs, after which it will be sold
for 12k P. Find the depreciation for the 2nd yr using sum of the years digit method.
Solution:
use, dn =
for
2(𝐿 − 𝑛 + 1)
𝐿(𝐿 + 1)
n=2
L = 10
(C0 – CL)
C0 = 120kP
CL = 12kP
hence,
d2 =
2(10 − 2 + 1)
10(11)
(120k – 12k) = 17672.73P
2. What is the book value of equipment purchased 3 yrs ago for 15kP if it is depreciated
using sum of the years digit method and the expected life is 5 yrs?
Solution:
Cn = C0 – Dn
Dn =
for
𝑛(2𝐿 − 𝑛 + 1)
𝐿(𝐿 + 1)
n=3
L=5
(C0 – CL)
C0 = 15kP
CL = 0
hence,
D3 =
3{2(5) − 3 + 1}
5(6)
(15k – 0) = 12kP
C3 = C0 – D3 = 15k – 12k = 3000P
30
d. Declining Balance Method, DBM
or Matheson’s Formula or Constant Percentage Method
▪ d varies from yr to yr
▪ N/A if CL = 0
▪ by Principle of Geometric Progression
FORMULAS:
1. Constant percentage
𝑳
k=1- √
𝑪𝑳
𝑪𝟎
2. dn = C0k (1 – k)
𝒏
=1- √
𝑪𝒏
𝑪𝟎
n–1
𝒏
𝑪𝑳 𝑳
3. Cn = C0 (1 – k)n = C0 ( )
𝑪𝟎
4. CL = C0 (1 – k)L
5. Dn = C0 { 1 – (1 - k)n }
EXAMPLE:
1. A radio service panel truck initially costs 560kP. Its resale value at the end of 5 th yr is
estimated at 150kP. Find the depreciation charge on the second year by Declining
Balance Method.
Solution:
use, dn = C0k (1 – k)n – 1
𝐿
k=1- √
for
𝐶𝐿
𝐶0
L = 5yrs
C0 = 560kP
CL = 150kP
5
k=1- √
150𝑘
560𝑘
= 0.2316
hence, for n = 2
d2 = (560k) (0.2316) (1-0.2316)1
d2 = 99658.41P
e. Double Declining Balance Method, DDBM
▪
same formulas as in DBM, simply replace k by
2
𝐿
.
FORMULAS:
1. dn =
𝟐𝑪𝟎
𝟐 𝒏−𝟏
(𝟏 − 𝑳)
𝑳
2. Cn = C0 (𝟏 −
3. CL = C0 (𝟏 −
𝟐 𝒏
)
𝑳
𝟐 𝑳
𝑳
)
--- etc --31
EXAMPLE:
1. A machine costs 100kP and the useful life is 10 yrs. Find the depreciation charge at
the 3rd yr Double Declining Balance Method.
Solution:
2𝐶0
dn =
for
𝐿
2 𝑛−1
(1 − )
𝐿
n=3
L = 10
C0 = 100kP
d3 =
2(100𝑘)
10
2 2
(1 − 10)
d3 = 12800P
f. Service-Output Method
1. Per Hour Basis
FORMULA:
𝒅
𝒉𝒓
=
𝑪𝟎 − 𝑪𝑳
𝑯
𝒅
or Dn = (
𝒉𝒓
) Hn
or
𝑪𝟎 − 𝑪𝑳
Dn = (
𝑯
) Hn
where,
H = total generating hours within the service life
Hn = no. of hours used during n period
Dn = total depreciation within the n period
2. Per Unit Basis
FORMULA:
𝒅
𝒖𝒏𝒊𝒕
=
𝑪𝟎 − 𝑪𝑳
or Dn = (
𝑯
𝒅
) Tn
𝒉𝒓
or
𝑪𝟎 − 𝑪𝑳
Dn = (
𝑻
) Tn
where,
T = total no of units produced within the service life
Tn = no of units produced within n period
EXAMPLE:
1. An asphalt and aggregate mixing plant having a capacity of 50 m3/hr costs 2.5M P. It
is estimated to process 800k m3. If its scrap value is 100kP, determine (a) the
depreciation chargeable per batch of m3. (b) the depreciation chargeable per batch of
50m3. (c) the total depreciation during a certain yr if it processed 60k m3.
32
Solution:
a.
b.
c.
𝑑
=
𝑚3
𝐶0 − 𝐶𝐿
=
𝑇
𝑃
(2.5𝑀−100𝑘)𝑃
800𝑘 𝑚3
50𝑚3
= 3 P/m3
𝑑
= (3 3 ) (
) = 150 P/batch
𝑚3
𝑚
𝑏𝑎𝑡𝑐ℎ
𝐶0 − 𝐶𝐿
2.5𝑀−100𝑘
Dn = (
)Tn = (
𝑇
800𝑘
) (60k) = 180000P
TOPIC 2 – SUNK COST, SC
- it is the cost which can not be recovered due to poor estimate of book value.
FORMULA:
SC = Cn ---- Resale Value or Trade-In Value
where,
Cn = book value after n yrs when replacement occur.
EXAMPLE:
1. A machine was purchased 5 yrs ago at a cost of 120k P. Its estimated salvage value
at the end of 10 yrs is 10k P. If it is sold now for 30k P, what is the sunk cost if the
depreciation method used is straight line method?
Solution:
SC = Cn --- Resale Value
for
n=5
Resale Value = 30k P
SC = c5 – 30k -------- 1
by SLM,
Cn = C0 – Dn
Dn =
for
𝑛
𝐿
(C0 – CL)
n=5
L = 10
D5 =
5
10
C0 = 120k P
CL = 10k P
(120k – 10k)
D5 = 55k P
C5 = C0 – D5
C5 = 120k – 55k = 65k P
subst. values to 1
SC = 65k – 30k
SC = 35000 P
TOPIC 3 – DEPRECIATION TAX SHIELD, DTS
- it is the present sum of money needed for the payments of the depreciation taxes.
FORMULA:
DTS = d {
𝟏 − (𝟏+𝒊)−𝑳
𝒊
} (TR)
33
where,
d = constant depreciation charge per yr which
can be evaluate by SLM or SFM.
TR = tax rate
EXAMPLE:
1. A company purchase 200kP of equipment in year zero. It decides to use SLM of
depreciation over the expected 20 yr life of the equipment. The interest rate is 14%. If the
overall tax rate is 40%, what is the present worth of the depreciation tax shield?
Solution:
DTS = d {
1 − (1+𝑖)−𝐿
𝑖
} (TR) -------- 1
by SLM,
d=
d=
𝐶0 − 𝐶𝐿
𝐿
200𝑘 − 0
20
= 10kP
subst. values to 1
DTS = 10k {
1 − (1.14)−20
0.14
} (0.40)
DTS = 26492.52P
TOPIC 4 – DEPLETION
- the decrease in worth of a natural resources (e.q. mine)
due to gradual extraction of each content.
METHODS OF CALCULATION
1. Per Unit or Factor Method
FORMULA:
𝑪𝟎 − 𝑪𝑳
dn = (
𝑻
) Sn
where,
C0 = orig. cost
CL = salvage value
T = total no. of units available
Sn = no. of units extracted and sold at n period
EXAMPLE:
1. A mining company invested 25M P to develop an oil well which us estimated to contain
1M barrels of oil. During a certain yr, 200k barrels were produced from this well. Compute
the depletion charge during the year.
Solution:
𝐶0 − 𝐶𝐿
dn = (
𝑇
) Sn
34
where,
C0 = 25M P
CL = 0
T = 1M barrels
Sn = 200k barrels
hence,
25𝑀 −0
dn = (
1𝑀
) (200k)
dn = 5MP
2. Percentage Allowance Method
a. Based on Gross Income, GI
𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑎𝑙𝑙𝑜𝑤𝑎𝑛𝑐𝑒
dn = (
) (𝐺𝐼 )
𝑓𝑜𝑟 𝑡ℎ𝑒 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑟𝑒𝑠𝑜𝑢𝑟𝑐𝑒𝑠
from Table
b. Based on Net Income, NI
NI = GI – Expenses Excluding Depletion
dn = 50% NI
NOTE: Compare results of a and b and report whichever is less
EXAMPLE:
1. The total gross income of an oil company is 30MP. The taxable income after deducting
all expenses excluding depletion is 11.8MP. Determine the allowable depletion allowance
for the year. The percentage allowance for oil is 22% of the gross income.
Solution:
a. Based on Gross Income
dn = 0.22 (30M) = 6.6MP
b. Based on net Income
dn = 0.5 (11.8M) = 5.9MP
* by comparison of results of a and b, dn = 5.9MP
TOPIC 5 – VALUATION
- the process of estimating the cost of a natural resources which is done by authorized
individual called appraiser.
FORMULA:
A = Pr +
(𝑷−𝑪𝑳 )𝒊
(𝟏+𝒊)𝑳 −𝟏
------------ Hoskold’s Formula
where,
A = net annual income
P = estimated cost of mine
r = rate of return
CL = salvage value
i = interest on the sinking fund
L = life, yrs
35
EXAMPLE:
1. A timber tract will yield an annual income of 1MP for 10 yrs after which the timber tract
will be exhausted. The land can be sold for 120kP. If a prospected buyer wishes to earn
12% on his investment and can deposit money in a sinking fund at 6%, determine the
maximum price he could pay for the tract.
Solution:
A = Pr +
(𝑃−𝐶𝐿 )𝑖
(1+𝑖)𝐿 −1
where,
A = 1MP
L = 10 yrs
r = 12%
i = 6%
CL = 120kP
1M = P(0.12) +
solve for P,
(𝑃−120𝑘)(0.06)
(1.06)10 −1
P = 5151961.37P
36
ASSESSMENT 3
1. A brand new car costs 500,000P and the
salvage value is 10% of the original cost after
20 years. Find the book value of this car after
5 years by straight line method.
9. A tractor costs 800,000P and whose
salvage value is 40,000P after 10 years. Find
the total depreciation after 6 years by
declining balance method.
2. A generator set costs 500,000P and the
salvage value is 10% of the original cost after
20 years, find the depreciation charge per
year if money worth 12%.
10. A generator costs 500,000P and whose
salvage value is 10,000P after 20 years. Find
the book value after 12 years by sum of the
year’s digit method.
3. A car costs 500,000P and the salvage
value is 10% of the original cost after 20
years. Find the depreciation charge at the 8th
year by sum of the year’s digit method
(SYDM).
11. A personal computer costs 60,000P and
the salvage value is 5000P after 10 years.
Find the book value after 6 years if money
worth’s 12% per annum.
4. A lathe machine cost 650,000P and the
salvage value is 65,000P after 20 years, find
the depreciation charge at the 5th year by
Matheson’s Formula, (Declining Balance
Method, DBM).
5. A pump costs 50,000P and the useful life
is 10 years. Find the book value after 8 years
by double declining balance method
(DDBM).
6. A concrete hollow blocks (CHB) machine
cost 40,000P and the salvage value is 4000P
after 6 years. If it can make 108,000 pieces
of hollow blocks within the useful life, find the
depreciation charge at the year 1999 if it
made 15,000 pieces only.
7. A car costs 500,000P and the salvage
value is 50,000P after 20 years. Find the
book value after 5 years if money worth 12%
per annum.
8. A motor costs 60,000P and the salvage
value is 6000P after 10 years. If it was used
for 28,800 hours within the useful life, find its
depreciation at the year 1998 if its total
operating hours were 2520.
12. A car costs 800,000P 4 years ago and the
salvage value is 50,000P 6 years from now.
If it is to be replaced by a new one and the
trade in value is 450,000P find the sunk cost
if money worth’s 12%.
13. To develop a timberland containing
2,000,000 trees required an initial investment
of 30,000,000P. In a certain year, 400,000
trees were cut off. Find the depletion charge
during the year.
14. A mining company has a gross income of
32,000,000P per month from the production
of iron core. All expenses, excluding
depletion expenses, amount of 26,000,000P
per month. If the fixed depletion rate of iron
core is 15%, what is the monthly depletion
allowance?
15. Ten hectares of timberland will yield an
annual profit of 1,000,000P for 10 years, after
which the timber will be exhausted. The land
can be sold for 12,000P per hectare. If the
prospective buyer wishes to earn 15% on his
investment and can deposit money in a
sinking fund at 8%, determine the maximum
price he could pay for the timberland.
37
16. A machine cost 100,000P and with a
useful life of 25 years. Find the book value
after 3 years by using double declining
balance method.
17. A machine cost 80,000P and the salvage
value is 20,000P after 20 years. Find the
book value after 2 years by using sum of the
years digit method.
18. A machine cost 80,000P and the salvage
value is 20,000P after 20 years. Find the
constant percentage in the declining book
value.
19. A machine cost 80,000P and the salvage
value is 20,000P after 20 years. Find the
sinking fund factor if interest rate is 8% per
year.
20. A machine cost 80,000P and the salvage
value is 20,000P after 20 years. Find the
depreciation charge by sinking fund method
if interest rate is 8% per year.
ANSWERS TO ASSESSMENT 3
1. 387500P
2. 6245P
3. 27857P
4. 44590P
5. 8389P
6. 5000P
7. 460324P
8. 4725P
9. 667459P
10. 94000P
11. 34566P
12. 145741P
13. 6MP
14. 3MP
15. 4603415P
16. 77869P
17. 68857P
18. 6.69%
19. 0.0219
20. 1311P
38
MODULE 4
LEARNING OBJECTIVES:
After the completion of Module 4, it is expected that the student have understand the
following:
1. Evaluation of Break Even Point
2. Calculation of Profit
TOPIC 1 – BREAK-EVEN ANALYSIS
1. Break-Even Point (BEP) – a point in economic study where the sales volume is just
enough pay the costs of production. Hence no loss, no gain.
FORMULA:
S(x) = C + V(x) ---------- x = no. of units needed
at BEP
where,
𝑆𝑒𝑙𝑙𝑖𝑛𝑔 𝑃𝑟𝑖𝑐𝑒
S(x) = sales function = (
) (𝑥 )
𝑃𝑒𝑟 𝑈𝑛𝑖𝑡
C = Fixed cost
V(x) = variable cost function
𝑀𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝐶𝑜𝑠𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡
+ 𝐿𝑎𝑏𝑜𝑟 𝐶𝑜𝑠𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡
={
} (𝑥)
+ 𝑂𝑡ℎ𝑒𝑟 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝐶𝑜𝑠𝑡
𝑝𝑒𝑟 𝑢𝑛𝑖𝑡
EXAMPLES:
1. A steel drum manufacturer incurs a yearly fixed operating cost of 200kP. Each drum
manufactured costs 160P to produce and sells for 200P. What is the manufacturer’s breakeven sales volume in drums per year?
Solution:
let x = no of drums needed per year at BEP
use, S(x) = C + V(x)
200x = 200k + 160x
Solve for x,
x = 5000
2. A company manufactures bookcases that it sells for 65P each. It costs 35kP per yr to
operate its plant. This sum includes rent, depreciation charges on equipment and salary
payments. If the additional cost to produce one bookcase is 50P, how many cases must
be sold each year for the company to avoid taking a loss?
Solution:
let x = no of bookcases needed per year at BEP
S(x) = C + V(x)
65x = 35k + 50x
x = 2334
3. A piece of property is purchased for 10kP and yields 1kP yearly profit. The property is
sold after 5 years. At 6% interest, what is the minimum price to break-even?
39
Solution:
draw CFD
S(1.06)-5
S = ? “selling price”
PA
0
A
A
A
A
A = 1KP/yr
1
2
3
4
5 (yrs)
i = 6% “per year”
10kP
set-up EV at zero
∑↑=∑↓
PA + S(1.06)-5 = 10k
1 − (1.06)−5
1k {
0.06
} + S(1.06)-5 = 10k
S = 7745.16P
4. Project A requires 100kP now. Project B requires an 80kP investment now and an
additional 40kP investment later. At 8% interest, what is the BEP on the timing of the
additional 40k later?
Solution:
at BEP, the present worth of projects A and B must be equivalent.
CFD for Project B
PB
n=?
0
40kP
80kP
where,
PA = 100kP
PB = 80k + 40k (1.08)-n
EV at zero
hence,
100k = 80k + 40k (1.08)-n
n = 9 yrs
2. Unhealthy Point - it is a point in economic study where the sales volume is just enough
to pay the dividends.
FORMULA:
S(x) = C + V(x) + D
40
where,
D = payment of dividends usually P/yr
NOTE:
be consistent with time unit
3. Profit Calculation
FORMULA:
Sales = Fixed cost + Variable cost + Dividend + Profit
Capital
or
Profit = Sales – Capital
EXAMPLE:
1. A company assembling small radio produced and sold 100 units per month. It costs
800P to produce a unit which is sold at 1200P. If the company has a fixed cost of 20kP per
month and pays 10% on its 10k shares with a par value of 200P/share dividends, calculate
the profit or loss of the company.
Solution:
Profit = Sales – Expenses or Capital ---------- 1
in 1month,
1200𝑃
Sales = (
Expenses =
Fixed cost =
Production cost =
Dividend =
=
𝑢𝑛𝑖𝑡
) (100 units)
= 120kP
Fixed cost + Production cost + Dividend
20kP
800𝑃
( 𝑢𝑛𝑖𝑡 ) (100 units)
= 80kP
{ (0.10)(10k shares) } (
200𝑃
1𝑦𝑟
) (12 𝑚𝑜𝑠 )
𝑠ℎ𝑎𝑟𝑒 𝑦𝑟
16666.67 P/yr
hence,
Expenses = 116666.67P
subst. values to 1,
Profit = 3333.33P
2. An investor is considering a stock portfolio that costs 55P. If he invests in the portfolio,
there is a 0.5 probability that he will receive a total revenue of 20P. If that event does not
occur, he will receive a total revenue of 100P, what will be the investors expected profit if
he decides to invest?
Solution:
Profit = Income or Sales – Capital or Expenses
where,
Income = 0.5(20) + 0.5(100) = 60P
Capital = 55P
hence,
Profit = 60 – 55 = 5P
41
ASSESSMENT 4
1. A concrete hollow blocks (CHB)plant has
an overhead cost of 150,000P per month.
The material cost is 3.75 per unit and labor
cost 2.25 per unit. How many units should be
made per month to break-even if the selling
price is 7.50 per unit.
2. A high voltage gloves manufacturer
produces a pair of gloves at labor cost of 15
and a material cost of 40 a pair. The fixed
charges on the business are 90,000P a
month and the variable cost are 15 a pair. If
the gloves sell for 160 a pair, how many pairs
must be produced per month by the
manufacturer to break-even?
3. A steel drum manufacturer incurs a yearly
fixed operating cost of 2,000,000P. Each
drum manufacturer cost 160P to produce and
sells for 200P. What is the manufacturers
break-even sales volume in drums per year?
4. The direct labor cost and direct material
cost of certain product are 300P and 400P
per unit, respectively. Fixed charges are
100,000P per month and other variable costs
are 100P per unit. If the product is sold for
1200P per unit, how many units must be
produced and sold to break-even?
5. A local factory assembling calculators
produces 400 units per month and sells them
at 1800P each. Dividends are 8% on the
8000 shares with par value of 250P each.
The fixed operating cost per month is
25000P. Other costs are 1000P per unit.
Determine the number of units needed to be
produced per month at unhealthy point.
6. A certain firm has a capacity to produce
650,000P units of a certain product per year.
At present, it is operating at 62% capacity.
The firm’s annual income is 4,160,000P.
Annual fixed cost is 1,920,000P and the
variable costs are equal to 3.56 per unit.
What is the annual profit or loss?
7. A shoe manufacturer produces a pair of
shoes at a labor cost of 90P a pair and
materials cost of 80P a pair. The fixed
charges of the business are 90,000P a month
and the variable cost is 40P a pair. If the
shoes sell for 300P a pair, how many pairs
must be produced each month by the
manufacturer to break-even?
8. A plant has capacity of producing 8000
units per month of a product, which it sells for
1.50P per unit regardless of output. The
monthly fixed costs are 2800P and a variable
cost of 4800P at 75% capacity. What is the
fixed cost per unit at the break-even point?
9. A local company assembling stereo radio
cassettes produces 300 units per month at a
cost of 800P per unit. Each cassette sells for
1200P. If the firm makes a profit of 10% on
its 10,000P shares with a par value of 200P
per share, and the total fixed cost per month
is 20,000P what is the break-even point?
10. Company A and B manufactures the
same article. Company A, relying mostly on
machines has fixed expenses of 12000P per
month and direct cost of 8 per unit. Company
B, using more hand work, has fixed expenses
of 4000P and direct cost of 20 per unit. At
what monthly production rate will the total
cost per unit is the same for the two
companies?
42
ANSWERS TO ASSESSMENT 4
1. 100000
2. 1000
3. 50000P
4. 250
5. 48
6. 805320P
7. 1000
8. 0.70
9. 92
10. 667
43