LECTURE GUIDE

AND
WORKBOOK

ENGINEERING ECONOMICS

ROMEO Q. TOLENTINO
CIVIL ENGINEER
MASTER OF MATHEMATICS
MASTER OF ENGINEERING SCIENCE

1
Engineering Economics - is the engineering
discipline mainly concerned with mathematical
analysis and evaluation of cost and benefits of
proposed ongoing business project and
ventures with the aim of making cost effective
decisions for such project and ventures , thus
ensuring the best of capital.

Principles of Engineering Economy

a. Deciding to implement a new project.
b. Selecting among alternative projects that
must be implemented within the capital budget
limit
c. Selecting between alternative technical
designs for component , product, structure,
system , service or process.
d. Analyzing the economic impact of
engineering improvements or modifications on
an existing product system of process.
e. Deciding on when to replace the existing
equipment.
f. Deciding whether to lease or purchase
equipment to improve existing operation.

2
Some Basic Concepts

Price – the amount of money or its equivalent

which can be given is exchange for a product or
a service .
Market - a place where sellers and buyers and
services come together to make transactions.
Local Market - limited locality
Domestic Market - whole country
World Market – International

Consumer Goods and Services – products or

services that are directly used by people to
satisfy their wants

Neccessities – products or services required to

support human life and activities.

Luxuries – products or services that are desired

by humans and will be purchased only if money
is available
and after necessities are obtained.

Demand – quantity of a certain commodity that

is bought at a certain price.

The law of demand - fundamental principle of

economics which states that at a higher price

3
consumers will demand a lower quantity of a
good.

Elastic demand is when price or other factors

have a big effect on the quantity consumers want
to buy.

Inelastic Demand – occurs when a change in

price causes a smaller percentage change in
demand.

Perfect Competition - the situation prevailing in

a market in which buyers and sellers are so
numerous and well informed that all elements of
monopoly are absent and the market price of a
commodity is beyond the control of individual
buyers and sellers.

Monopoly – opposite of perfect competition.

Oligoply – there are few suppliers of product or

services

Supply – quantity of certain commodity that is

offered for sale at a price and a given time.

Law of Supply

4
- all other factors being equal, as the price of a
good or service increases, the quantity of goods
or services that suppliers offer will increase, and
vice versa.

Law of Supply and Demand

Generally, low supply and high demand increase
price and vice versa.

5
Law of Diminishing Returns
-states that adding an additional factor of
production results in smaller increases in output

-economic law stating that if one input in the

production of a commodity is increased while all
other inputs are held fixed, a point will eventually
be reached at which additions of the input yield
progressively smaller, or diminishing, increases
in output.

6
 LESSON 2
SIMPLE INTEREST AND DISCOUNT
Interest - is the amount paid for the use of
borrowed capital. (borrowers viewpoint)
- is the amount produced by the money
which he has lent. (lenders viewpoint)
Simple Interest - interest on borrowed money is
directly proportional to the length of time the
amount or principal is borrowed.

Principal - amount of money borrowed.

Rate of Interest - amount earned by 1 unit of
principal during a unit of time.

Formula for simple interest

I = Pin F = P + Pin or F = P( 1 + in )

I = total interest earned by the principal

P = amount of the principal
i = rate of interest
n = number of interest periods.
F = total amount to be paid

Ordinary Simple Interest -

interest is computed on the basis of 1 bankers year
1 Banker’s Year = 12 months where 1 month =
30 days

7
Exact Simple Interest , interest is computed on the
exact number of days.
1 common year = 365 days 1 leap year = 366
days

Example 1
Determine the ordinary simple interest on P 20,000
for 9 months and 10 days if the rate of interest is
12%.
Solution:
9 months and 10 days = 9 x 30 + 10 = 280 days
P = 20,000

I = Pin
I = 20,000 ( 0.12)( 280/360) = P 1866.67

8
Example 2
Determine the ordinary and exact simple interests
on P 100,000 for the period
January 15 to June 20 2012 if interest is 15%.
Solution:
January 15 - 31 16 days ( excluding Jan 15 )
Feb 29 days (leap year )
March 31 days
April 30 days
May 31 days
June 20 (including June 20)
------------- Total = 157 days

Note: January 15 is excluded while June 20 is

included

Ordinary Simple Interest

I = Pin = 100,000 ( 0.15)( 157/360) = P 6541.67

Exact Simple Interest

I = Pin = 100,000(0.15)(157/366) = P 6434.43
Leap Year

Example 3 . If P 4000 is borrowed for 75 days at

16% per annum. How much will be due at the end
of 75 days?
Solution:
P = 4,000

9
i = 16%
n = 75 days = 75/360 year
F = P ( 1 + in )
F = 4,000 ( 1 + 0.16(75/360) )
F = 4,133.33

Example 4
How long will it take for a deposit of P 1, 500.00 to
earn P 186 if invested at the simple interest rate
of 7 1/3%?
Solution:
P = 1, 5000 I = 186 i = 0.07333 n = ?
I = Pin
186 = 1,500 (0.07333)n
n = 1.6909 years

Cash flow diagram – graphical representation of

cash flows drawn on a time scale.
Example 5
A loan of P 2,000 at simple interest rate of 5% will
become 2,000( 1 + 2(0.05) ) = 2,200 in 2 years.
Draw the cash flow diagram
a. from the viewpoint of the lender
b. from the viewpoint of the borrower.
Solution:

10
 Cash Flow Diagram from the viewpoint of
the lender

Cash Flow Diagram from the viewpoint of

the borrower

11
 Example 6
If you borrow money from your friend with simple
interest of 12%, find the present worth of P 20,000
at the end of 9 months.
Solution:
F = 20,000 i = 12% and n = 9/12 years
F = P( 1 + in )
20,000 = P ( 1 + 0.12(9/12) )
P = 18,348.62

Example 7
A deposit of P 110,000 was made for 31 days. The
net interest after deducting 20% withholding tax is P
890.36. Find the rate of return annually. PAST CE
BOARD
Solution:
P = 110,000 890.36 = 0.8 I I = 1,112.95 n
= 31/360
I = Pin
1,112.95 = 110,000 i(31/360)
i = 11.75%

12
Discount- is the difference between the worth in
the future and its present worth.
Discount = Future Value – Present Value
D= F–P
Kinds of Discounts
a. Trade Discount – discount offered by the seller to
induce trading.
b. Cash Discount - the reduction on the selling price
offered to a buyer to induce him to pay promptly.

Example 1
If a negotiable paper say a bond can be sold for P
100 seven months from now, but is sold for P 90
today, then P 10 is the discount.

Discount Rate (d) – is the discount on one unit of

principal per unit of time.
𝑖 𝐹−𝑃 𝑃 𝑃 𝑃
𝑑=𝐹= = 1 − 𝐹 = 1 − 𝑃(1+𝑖𝑛) Note: 𝑖 = 𝐹
𝐹
Let n = 1 period
then:
1 1+𝑖−1
𝑑 = 1 − 1+𝑖 or 𝑑= 1+𝑖
𝑖
𝑑=
𝑖+1
relationship between discount rate and interest
rate
𝑑
and 𝑖 = 1−𝑑

13
Example 8
Mr. T borrowed money from the bank. He receives
from the bank P 1, 340 and promised to pay P
1,500 at the end of 9 months. Compute:
a. Simple Interest Rate
b. Discount Rate
Solution:
F = 1500 P = 1340
n = 9/12 years i=? d=?
F = P ( 1 + in )
1500 = 1340 ( 1 + i(9/12) )
i = 15.92%
𝑖 0.1592
𝑑 = 𝑖+1 = 1.1592 = 13.73%

d = discount rate
𝐹−𝑃
𝑑= 𝐹
Fd = F - P
F – Fd = P
or P = F ( 1 – d ) for 1 year
P = F ( 1 – nd ) for n years

14
Example 8
Find the discount if P 2,000 is discounted for 6
months at 8% simple discount.
Solution:
P=? F = 2000 n = 6/12 years
P = F ( 1 – nd )
P = 2000( 1 – 6/12 (8% ) )
F = 1920
D = F – P = 2000 -1920 = 80

Example =9
Discount 1650 for 4 months at 6% simple interest.
What is the discount?
Solution:
F = P ( 1 + in )
F = 1650 P = ? n = 4/12 years
1650 = P( 1 + 0.06(4/12) )
P = 1617.65
Discount = F – P = 1650 – 1617.65 = 32.35

Example 10
On March 1 , 1996 Mr. X obtains a loan of P 1500
and signs a note promising to pay the principal and
the accumulated simple interest at the rate of 5% at
the end of 120 days. On May 15, 1996, Mr X
discounts the note at the bank whose discount rate
is 6%. What amount does he receive?

15
Solution:
n = 120/360 years P = 1500 i = 5% F = ?
F = P( 1 + in ) = 1500 ( 1 + 0.05(120/360) ) = 1525
This amount will be discounted by 6%
Days left = 45 days (March 1 to May 15 = 75 days)
120 – 75 = 45
P = F( 1 – nd )
P = 1525( 1 – 45/360 (6%) )
P = 1513.56 amount that Mr X received.

Example 11
A man borrowed P 5,000 from a bank and agreed
to pay the loan at the end of 9 months. The bank
discounted the loan and gave him P 4,000 in cash.
a. What was the rate of discount? (for 9 mo )
b. What was the rate of interest? ( for 9 mo )
c. What was the rate of interest for one year?
Solution:
a. d for 9 months = discount ÷ principal
= 1000÷ 5,000 = 20%
𝑑 20%
b. 𝑖 = 1−𝑑 = 1−20% = 25% i for 9 months
𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 1000
or 𝑖 = 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑊𝑜𝑟𝑡ℎ = 4000 = 25%
c. Using Ratio and Proportion:
12 12
𝑖1 𝑦𝑒𝑎𝑟 = 𝑖9 𝑚𝑜𝑛𝑡ℎ𝑠 = (25%) = 33.33%
9 9

16
 PROBLEM SET 1-A
Name: ________________________________
Score______
Subject and Section: _____

1. Calculated the exact interest on an investment

of P2 000.00 for a period from January 30 to
September 15, 2001 if the rate of interest is 10%.
(Ans. P124.93)

2. What amount will be available in eight months if

P15 000.00 is invested now at 10% simple interest
per year? (Ans. P16 000.00)

3. A deposit of P 110,000 was made for 31 days.

The net interest after deducting 20% withholding tax

17
is P 890.36. Find the rate of return annually. Ans.
11.75%

4. Ann buys a television set who ask for P 1300

at the end of 60 days. Ann wishes to pay
immediately and the merchant offers to compute the
cash price on the assumption that money is worth
8% simple interest

5. If you borrow money from your friend with simple

interest of 12%, find the present worth of P20
000.00 which is due at the end of nine months?
(Ans. P18 348.60)

6. Find the amount due at the end of 15 months

whose present value is P2, 000 at 5% simple
discount.
Ans. F = P2, 133.33 (amount due at the end of 15
months)

18
7. A man buys an electric fan from a merchant that
charges P1500.00 at the end of 90 days. The man
wishes to pay cash. What is the cash price if money
is worth 10% simple interest?(Ans. P1463.41)

8. P1000.00 becomes P1500.00 in three years.

Find the simple interest rate.
(Ans. 16.67%)

9. An engineer borrowed a sum of money under

the following terms: P650 000.00 if paid in 90 days,
or P600 000.00 if paid in 30 days. What is the
equivalent annual rate of simple interest? (Ans.
50%)

19
10. Mr. Flores made a money market placement of
P1, 000.000 for 30 days at 7.5% per year. If the
withholding tax is 20%, what is the net interest that
Mr. Flores will receive at the end of the month?
Ans. Net Interest = P5, 000

20
 PROBLEM SET 2 DISOUNT
Name: __________________Score: ________
Subject and Section: ______________

1. Discount P 5,000 for 4 months at 6% simple

interest. What is the discount?
Answer: Discount = 98.03

2. Compute the discount if P2, 000 is discounted

for 6 months at 8% simple interest.
Answer: Discount = P76.92

3. What simple interest rate is equivalent the

simple discount rate of 6% in discounting an

21
amount P1, 000 due at the end of 3 months?
Answer: r = 0.0609 = 6.09%

4. A man borrowed P2, 000 from a bank and

promise to pay the amount for one year. He
received Only the amount of P1, 920 after the bank
collected an advance interest of P80.00. What was
the rate of discount and the rate of interest that the
bank collected in advance? Answer: Rate of
discount = 4% Rate of interest = 4.17%

22
5. X borrowed P2,000 from a bank and promise to
pay the amount for one year. He received only the
amount of P1, 920 after the bank collected an
advance interest of P80.00. What was the rate of
discount and the rate of interest that the bank
collected in advance?
Answer:
Rate of discount = 4% Rate of interest = 4.17%

6 Find the amount due at the end of 15 months

whose present value is P 2, 000 at 5% simple
discount? Ans. 2133.33

23
7. What simple interest rate is equivalent to simple
discount rate of 6% in discounting an amount of
1000 due at the end of 3 months?
Ans. 0.0609 or 6.19% Simple Interest Rate

8. Find the discount if P 2,000 is discounted for

6 months at 8% simple discount.
Ans. 80

24
9. Discount 1650 for 4 months at 6% simple
interest. What is the discount?
Ans. 32.353

10. Compute the discount if P 2,000 for 6 months

at 8% simple interest.
Ans. 76.92

25
 LESSON 2
COMPOUND INTEREST

Compound Interest – interest of principal is based

not only on the original amount of the principal but
the amount of the principal plus accumulated
interest.
Formula: 𝐹 = 𝑃( 1 + 𝑖 )𝑛
i = interest per period
n = number of interest periods
(1 + i)n = single payment compound amount
factor
1
(1+𝑖)𝑛
= single payment present worth factor

12% compounded quarterly is 12%/4 = 3%

interest per quarter
12% compounded semiannually is 12%/2 = 6%
interest per half year
12% compounded monthly = 12%/12 = 1% interest
per month

26
Example 1
The amount of P 20,000 was deposited in a bank
earning an interest rate of 6.5% per annum.
a. Determine the total amount at the end of 7 years
if the principal and interest were not withdrawn
during this period.
Solution:
P = 20,000 i = 6.5% n = 7 periods
𝑛
𝐹 = 𝑃 (1 + 𝑖 )
𝐹 = 20,000(1 + 6.5%)7
F = P 31,079.73

Example 2
A man expects to receive P 25,000 in 8 years. How
much is that money worth now considering interest
at 8% compounded quarterly?
Solution:
F = 25,000 P = ?
i = 8% compounded quarterly = 8%/4 = 2% per
quarter
n = 8 years = 8 x 4 = 32 quarters
F = P( 1 + i)n
25,000 = P ( 1 + 2%)32
P = 13,265.83

27
Example 3
How many years will P 100,000 earn a
compounded interest of P 50,000 if interest is 9%
compounded quarterly?
Solution:
F = 100,000 + 50,000 = 150,000 P = 100,000 i
= 9%/4 = 0.0225 per quarter
𝐹 = 𝑃 ( 1 + 𝑖 )𝑛
150,000 = 100,000 ( 1 + 0.0225)n
1.0225n = 1.5
n ln 1.0225 = ln 1.5
n = 18.22 quarters
number of years = 18.22/4 = 4.56 years

Nominal and Effective Rates of Interest

Rate of Interest – amount earned by a unit principal
per unit time.
Nominal Rate of Interest – basic annual rate of
interest.
Example 8% compounded quarterly.

Effective Rate of Interest – exact rate of interest

earned on the principal during a one year
period.

Conversion of Interest rates:

28
Example 4
Find the effective rate of interest corresponding to
8% compounded quarterly.
Solution:
Assume P = 1 Let x = effective rate
1( 1 + 8%/4)4 = 1 ( 1 + x )
x = 8.243%

Example 5
Find the nominal rate, which if converted
quarterly could be used instead of 12%
compounded semiannually?
Solution:
Let x = rate compounded quarterly
1( 1 + x/4)4 = 1 ( 1 + 12%/2)2
4
𝑥 = ( √( 1.06)2 -1)4 = 11.825%

29
Example 6
If money is worth 5% compounded quarterly, find
the equated time for paying a loan of P 150,000
due in one year and P 280,000 in 2 years.

150,000 280,000
𝑃1 = 5% 1×4
= 142,278.64 𝑃2 = 5% 2×4
=
(1+ ) (1+ )
4 4
253,511.57
150,000 + 280,00
𝑃3 = 142,278.64 + 253,511.57 =
5% 4𝑛
(1 + 4 )
430,000
=
1.01254𝑛
n = 1.64 years

30
Example 7
Five years ago, you paid P 340,000 for a lot. Today
you sold it at P500,000. What is the annual rate of
appreciation?
Solution:
F = 500,000 P = 340,000 n = 5
𝐹 = 𝑃 ( 1 + 𝑖 )𝑛
500,000 = 340,000 ( 1 + i)n
500,000 27
(1 + 𝑖 )5 = =
340,000 17
5 25
(1 + 𝑖 ) = √
17
5 25
𝑖 = √17 -1 = 8%

31
 PROBLEM SET 3
Name: _____________Score: ________
Subject and Section: ____________

1. John borrowed P50, 000.00 from the bank at

30% compounded semi-annually. What is the
equivalent effective rate of interest? Answer:

2. Find the present worth of a future payment of

350, 000 to be made 5 years with an interest rate of
8% per annum.

32
3. How long will it take money to double itself if
invested at 8% compounded annually?
Answer:

4. The amount of 70, 000 was deposit in the bank

earning an interest of 8.5% per annum. Determine
the total amount at the end of 5 years, if the principal
and interest were not withdrawn during the period?

5. What is the corresponding effective interest rate

of 11% compounded semi-quarterly?

33
6. Find the present worth of a future payment of
P120, 000 to be made in 10 years with an interest
of 12% compounded quarterly.

7. In how many years is required for P2, 000 to

increase by P3, 000 if interest at 19% compounded
semi-annually?

8. Compute the equivalent rate of 8% compounded

semi-annually to a rate compounded quarterly.

34
9. If P5, 000.00 shall accumulate for 15 years at
8% compounded quarterly. Find the compounded
interest at the end of 10 years.

10. A sum of P1, 000, 000.00 is invested now and

left for eight years, at which time the principal is
withdrawn. The interest has accrued is left for
another eight years. If the effective annual interest
rate is 5%, what will be the withdrawal amount at
the end of the 16th year?

11. P1, 500.00 was deposited in a bank account

26 years ago. Today it is worth
P3, 000.00. Interest is paid semi-annually.
Determine the interest rate paid on this account?

35
12. A merchant puts in his P8,000.00 to a small
business for a period of six years with a given
interest rate on the investment of 15% per year,
compounded annually, how much will he collect at
the end of the sixth year?

13. By the condition of a will, the sum of P20, 000

is left to a girl to be held in trust fund by her
guardian until it amounts to P90, 000, when will the
girl received the money if the fund is invested at 8%
compounded quarterly?

14. The amount of P12, 800 in 4 years at 8%

compounded quarterly is ________.

36
15.A man expects to receive P45, 000 in 8 years.
How much is that money worth now considering
interest at 8% compounded quarterly?

16. At an interest rate of 11% compounded

annually, how much will a deposit of P1, 500 be in
15 years?

17. How many years will P120, 000 earned a

compound interest of P50, 000 if the interest rate is
9% compounded quarterly?

18. The effective rate of 17% compounded semi-

annually is?

37
 LESSON 3
CONTINUOUS COMPOUNDING INTEREST
For continuous compounding:
𝑑𝑃
= 𝑖𝑃 n = number of years and ic = interest
𝑑𝑛
compounded continuously .
𝑑𝑃
= 𝑖𝑐 𝑑𝑛
𝑃
The solution of this ODE is
𝐹 = 𝑃𝑒 𝑖𝑐 𝑛
𝑒 𝑖𝐶 (𝑛) = 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑖𝑛𝑔 𝑎𝑚𝑜𝑢𝑛𝑡 𝑓𝑎𝑐𝑡𝑜𝑟
1
= present worth of continuous compounding
𝑒 𝑖𝑐 (𝑛)
factor

Example 1
P 100,000 is deposited in a bank that earns 5%
compounded continuously . What will be the
amount after 10 years?
Solution:
P = 100,000 F = ? ic = 5%
𝐹 = 𝑃𝑒 𝑖𝑐 (𝑛) = 100,000𝑒 0.05(10) = 16,487.21

Example 2
Money is deposited in a certain account for which
interest is compounded continuously. If the balance
doubles in 6 years, what is the annual percentage
rate?

38
Solution:
𝐹 = 𝑃𝑒 𝑖𝐶(𝑛)
F = 2 P = 1 N = 6 NR = ?
2 = 1𝑒 𝑖𝐶(6)
𝑒 𝑖𝑐 (6) = 2
6𝑖𝐶 = ln 2
ic = 0.1155 OR 11.55%

Example 3
A man wishes to have P 40,000 in a certain fund at
the end of 8 years. How much should he invest in a
fund that will pay 6% compounded continuously?
Solution:
F = Pe0.06n
F = 40,000 P = ? n = 8
40,000 = Pe0.06(8)
P = 24, 751.34

Example 4
If the effective annual interest rate is 4%, compute
the equivalent nominal interest compounded
continuously.
er -1 = 4%
er = 1.04
r = ln(1.04) = 0.03922
r = 3.924%

39
Example 5
What is the nominal rate of interest compounded
continuously for 10 years if the compound amount
factor is 1.34986?
𝑒 𝑖𝑐 (𝑁) = 1.34986
𝑒 𝑖𝑐 𝑛 = 1.34986 = 𝑒 𝑖𝑐 (10)
ic =( ln 1.34986 )/10 = 3%

Example 6
Money is deposited in a certain account which is
compounded continuously. If the balance doubles in
6 years, what is the annual percentage rate?
Solution:
𝐹 = 𝑃𝑒 𝑖𝑐 𝑛
2 = 1𝑒 𝑖𝑐 (6)
𝒍𝒏(𝟐)
𝒊𝒄 = = 𝟏𝟏. 𝟓𝟓%
𝟔

40
 PROBLEM SET 4
Name: ____________________________
Score: ________
Subject and Section:

1. What is the nominal rate of interest compounded

continuously for 12 years if the compound amount
factor is equal to 1.34986?

2. City Bank charges 1 1/3% interest per month,

compounded continuously on the unpaid balance
purchases made on this credit card. Compute the
effective rate of interest.

41
3 If the nominal interest is 10% compounded
continuously, compute the effective annual rate of
interest.

4. If the effective interest rate is 28%, what nominal

rate of interest is charged for a continuously
compounded loan?Answer:

5. Compute the difference in the future amount of

P800 compounded annually at nominal rate of 5%
and if it is compounded continuously for 5 years at
the same rate.

42
6. Deposits of P35 000.00, P48 000.00, and P25
000.00 were made in a savings account eight
years, five years, and two years ago, respectively.
Determine the accumulate amount in the account
today if a withdrawal of P55 000.00 was made four
years ago. The applied interest rate is 12%
compounded continuously.

7 A man wishes to have P 40,000 in a certain fund

at the end of 8 years. How much should he invest
in a fund that will pay 9% compounded
continuously?

8. If the effective annual interest rate is 8%,

compute the equivalent nominal interest
compounded continuously.

43
 LESSON 4
ORDINARY ANNUITY

Annuity – series of equal payments occurring at

equal interval of time.
Ordinary Annuity – payments are made at the
end of each period.

Sum of Ordinary Annuities Derivations

Transfer all A’s to zero

𝐴 𝐴 𝐴 𝐴
𝑃= + + + … . .
1 + 𝑖 ( 1 + 𝑖 )2 ( 1 + 𝑖 )3 ( 1 + 𝑖 )𝑛

This is a geometric series with common ratio 𝑟 =

1 𝐴
and 1st term is .
1+𝑖 1+𝑖
The sum of the geometric series is:
𝑎(1−𝑟𝑛 )
𝑆𝑛 = where a is the 1st term, n is the
1−𝑟
number of terms and r is the common ratio.

44
 𝐴 1 𝑛
(1 − (1 + 𝑖) ) = 𝐴[(1 + 𝑖 )𝑛 − 1]
𝑃 = 1+𝑖
1 [(1 + 𝑖 ) − 1](1 + 𝑖 )𝑛 ]
1−
1+𝑖
𝐴[(1 + 𝑖 )𝑛 − 1]
=
𝑖 (1 + 𝑖 )𝑛
𝐴[(1+𝑖)𝑛−1] (1−(1+𝑖)−𝑛
P= =𝐴 )
𝑖 (1+𝑖)𝑛 𝑖
(1−(1+𝑖)−𝑛
) = is called uniform series present
𝑖
worth factor.
(USPWF )

To get F , transfer P to n.
𝐴[(1+𝑖)𝑛 −1]
𝐹 = 𝑃 (1 + 𝑖 )𝑛 = (1 + 𝑖 )𝑛
𝑖 (1+𝑖)𝑛
𝐴[(1+𝑖)𝑛−1]
𝐹= (USCAF)
𝑖
(1+𝑖)𝑛 −1
= is called uniform series compound
𝑖
amount factor.

45
Example 1
Find the annual payment to extinguish a debt of P
100,000 payable for 6 years at 12% interest
annually.
Solution:

P = 100,000 i = 12% A=?

(1 − (1 + 𝑖 )−𝑛
𝑃=𝐴 )
𝑖
P = 100,000 i= 12% n = 6 years

A = P 24,322.57

46
Example 2
What annuity is required over 12 years to equate to
a future amount of P 200,000?
i = 8% effective

Solution:

F = 200,000 A = ? i = 8%
𝑛
𝐴[(1 + 𝑖 ) − 1]
𝐹=
𝑖
𝐴(1.0812 − 1)
200,000 =
0.08
A = 10,539.00

47
Example 3
A man paid 10% downpayment of P 200,000 for
a house and lot and agreed to pay the 90% balance
on monthly installment for 60 months at an interest
rate of 15% compounded monthly. Compute the
amount of monthly payment.
Solution:
Let x = amount of the house and lot
Then 0.1x = 200,000
x = 2,000,000
Amount to be amortized = 2,000,000 – 200,000 =
1,800,000

(1 − (1 + 𝑖 )−𝑛
𝑃=𝐴 )
𝑖
sub i= 0.15/12 n = 60 and P = 1,800,000
A = 42,821.87

48
Example 4. Mr Y bought a house and lot for
P 2,800,000 with a downpayment of 300,000.
Interest is 5% compounded monthly to be paid
for 30 years on a monthly basis. Construct and
Amortization Schedule.
Solution:

The loan to be amortized = 2,800,000 – 300,000 =

2,500,000

(1 − (1 + 𝑖 )−𝑛
𝑃=𝐴
𝑖
sub P = 2,500,000 i= 5%/12 n = 30 x 12 = 360

A = 13,420.54 monthly payment

49
 Amortization Schedule
Payment Amount Principal Interest Balance
1 13,420.54 3,003.87 10,416.67 2,496,996.13
2 13,420.54 3,016.38 10,404.15 2,499,979.74
3 13,420.54 3,028.96 10,391.58 2,490,950.78
4 13,420.54 3,041.58 10,378.96 2,487,909.21
5 13,420.54 3,054.25 10,366.29 2,484,854.95

Discussion:
For Payment 1: Interest = 2,500,000 x 5%/12 =
10,416.67
Principal = 13,420.54 - 10,416.67= 3,003.57
Balance = 2,500,000 - 3,003.57 = 2,496,966.13

For Payment 2: Interest = 2,496,996.13 x 5%/12 =

10,404.15
Principal = 13,420.54 - 10,404.15 = 3016.38
Balance = 2,496,996.13 - 3,016.38 = 2,499,979.74

50
 PROBLEM SET 5
Name: ____________________________
Score: ________
Subject and Section:

1. A man paid 16% down payment of P200, 000 for

a house and lot and agreed to pay the 90% balance
on monthly installments for 60 months at an interest
rate of 15% compounded monthly. Compute the
amount of the monthly payment.

2. What is the present worth of a 3 years annuity

paying P8, 000? At the end of each year, with
interest at 8% compounded annually?

51
3. How much must be deposited at 6% each year
beginning on January 1, year 1, in order to
accumulate P5, 000 on the date of the last deposit,
January 1, year 6?

4. An instructor plans to retire in one year and want

an account that will pay him P28% a year for the
next 15 years. Assuming a 6% annual effective
interest rate, what is the amount he would need to
deposit now? (The fund will be depleted after 15
years).

52
5. A piece of machinery can be bought for P200,
000 cash, or for P20, 000 down and payments of
P20,000 per year for 15 years. What is the annual
interest rate of the time payments?

6. Money borrowed today is to be paid in 6 equal

payments at the end of 6 quarters. If the interest is
10% compounded quarterly, how much was initially
borrowed if quarterly payments are P2, 000?

7. What is the accumulated amount of the five year

annuity paying P8, 000 at the end of each year,
with interest at 15% compounded annually?

53
8. A debt of P10, 000 with 18% interest
compounded semi-annually is to be amortized by
semi-annual payments over the next 7 years. The
first due is 6 months. Determine the semi-annual
payments.

9. A man purchased on monthly installment a P200,

000 worth of land. The interest rate is 12% nominal
and payable in twenty years. What is the monthly
amortization?

10. Mr. ALVIN RENZ FLORES borrowed P150,

000 . The terms of the loan are 10% interest for 10
years with uniform payments. What is the yearly
amortization?

54
11. Mr. Flores plans a deposit of P5000 at the end
of each month for 10 years at 11% annual interest,
compounded monthly. The amount that will be
available in two years is.

12. A machine costs P20, 000.00 today and has an

estimated scrap value of P2, 000.00 after 8 years.
Inflation is 2% per year. The effective annual
interest rate earned on money invested is 8%. How
much money needs to be set aside each year to
replace the machine with an identical model 8 years
from now?

13. A man inherited a regular endowment of P100,

000 every of 3 months for 10 years. However, he
may choose to get a single lump sum payment at
55
the end of 4 years. How much is this lump sum if
the cost of money is 14% compounded quarterly?
Answer: Total lump sum = P3, 702,939.73

14. A service car whose cash price was P540,

000 was bought with a down payment of P163, 000
and monthly installment of P 10,345.67 for 5 years.
What was the rate of interest if compounded
monthly?

56
 LESSON 6
ANNUITY DUE, DEFERRED ANNUITY AND
PERPETUITY
Annuity Due – annuity whose 1st payment occurs
immediately. The 1st payment occurs at the
beginning of each term and ends one payment
interval after the last payment.

Deferred Annuity - annuity whose term does not

begin until the expiration of a specified time.

Perpetuity - an annuity where the payment

period extends forever, which means that the
periodic payments continue indefinitely.

57
Formula:
𝐴
𝑃= A = Annuity i = rate of interest
𝑖

Example 1
If money is worth 4% compounded semiannually,
find the present amount of an annuity due paying P
5,000 semiannually for a term of 3.5 years.
Solution:

A = 5,000 i = 4% compounded semiannually

n = 3.5 x 2 = 7

4% 6
5,000[(1+ ) −1]
2
𝑃= 4% 4% 6
+ 5,000 = P 33,007.15
(1+ )
2 2

58
Note: All the A’s are added from 1 to 6 using the
Ordinary Annuity Formula then A is added.

Example 2
A man agrees to make equal payments at the
beginning of each 6 months for 10 years to
discharge a debt of P 50,000 due now. If money is
worth 8% compounded semiannually, find the
semiannual payment.
Solution:
1−(1+𝑖)−𝑛
𝑃=𝐴 Ordinary Annuity Formula
𝑖

8% 19
𝐴[(1+ ) −1] 1− 1.04−19
2
50,000 = 𝐴 + 8% 19
= 𝐴+ 𝐴
8%
(1+ ) 0.04
2 2
50,000 = 14.1339394 A
A= P 3,537.58
59
Example 3
To accumulate a fund of P 80,000 at the end of
10 years, a man will make equal annual deposits in
the fund at the beginning of each year. How much
should he deposit if the fund is invested at 5%
compounded annually?
Solution:

Use the sum of ordinary annuities ( Future Worth )

𝐴[(1+𝑖)𝑛−1]
𝐹= 𝑖
𝐴[(1+5%)11 −1]
80,000 = −𝐴
5%

80,000 = 13.20678A
A = 6,057.49

60
Example 4 Deferred Annuity
The present value of an annuity of R pesos
payable annually for 8 years, with the 1st payment
at the end of 10 years is P 187,481.25. Find the
value of R if money if money is worth 5%.
Solution:

Transfer P to year 9.
P9 = 187,481.25(1+5%)9 = 290,844.9531
𝐴[(1+𝑖)𝑛 −1] 1−(1+𝑖)−𝑛
Then Use: P = =𝐴
𝑖 (1+𝑖)𝑛 𝑖
𝑅[ (1+5%)8 −1 1−(1.05)−8
290,844.9531 = =𝑅
5%(1+5%)8 0.05
R = 45,000.06

61
Example 5
A parent on the day that child is born wishes to
determine what lump sum would have to be paid
into an account bearing interest of 5% compounded
annually, in order to withdraw P 20,000 each on
the child’s 18th, 19th , 20th and 21th birthdays?
Solution:

20,000 [(1+5%)4 −1] 1−1.05−4

𝑃(1 + 5%)17 = =20,000
5%(1+5%)4 0.05
P = P 30,941.73

Example 6
Find the present value of a perpetuity of P 15,000
payable semiannually if money is worth 8%
compounded quarterly.
Solution:
Convert 8% compounded quarterly to x%
compounded semiannually.
8% 4 𝑥% 2
(1 + ) = (1+ )
4 2

62
x = 8.08% compounded semiannually
𝐴
𝑃=
𝑖
15,000
𝑃= = 𝑃 371,287
8.08%
2

Example 6
If money is worth 8%,determine the present value
of a perpetuity of P 1,000 payable annually with the
1st payment due at the end of 5 years.
Solution:

𝐴 1,000
𝑃(1.08)4 = =
𝑖 8%
P = 9,187.87

63
 PROBLEM SET 6
Name: __________________________________
Score: ________
Subject and Section: _________

1. A man bought a machine costing P 35,000

payable in 12 semiannual payments payable at the
beginning of each period. interest = 26%
compounded semiannually. Determine the amount
of installment.

3. A man invest P 50,000 now for the college

education of his 2 year old son. Money is 14%
effective. How much will the son get each year
starting from his 18th to 22nd birthday?
Ans. P 103,598.20

64
4. A person buys a property for P 100,000
downpayment and 10 deferrred semiannnual
payments of P 9,000 each starting three years from
now. What is the present value of the investment?
interest = 12% compounded semiannually.

5. A house cost P 400,000.00. Mr X will pay

P 96,000 cash, 62,000 at the end of 2 years and a
sequence of 6 equal annual payments starting with
the 1st at the end of 4 years to pay the house.
Interest is 7% compounded annually. Find the
annual payment for the 6 years.

6. Compute the present and future worth of

payments of
a. 6,000 at the beginning of each month for 12
months.

65
b. Future worth of 12 payments of 8,000 each at
the beginning of each month ( at the end of the 12
months )
Interest = 1% per month.

7. Robert wants to deposit 3000 into a fund at the

beginning of each month. If he can earn 10%
compounded interest monthly, how much amount
will be there in the fund at the end of 8 years?

66
8. The monthly rent on an apartment is 9500 per
month payable at the beginning of each month. If the
current interest is 12% compounded monthly, what
single payment 12 months in advance would be
equal to a year’s rent?

9. Find the present worth of a perpetuity of 2,000

annual payments with interest of 6% .

10. Find the present value of a perpetuity of 1,000

every quarter with interest of 6% effective.

11. Find the present value of a perpetuity of 1000

monthly payments if interest if 10% compounded
annually.

67
12. Find the present value of a perpetuity that
starts monthly payment of 1200 after 6 months with
interest of 13% compounded quarterly.

68
 LESSON 7
UNIFORM GRADIENT SERIES ANNUITY

Some Problems in Engineering Economy involves

receipts or disbursements that may increase or
decrease as the interest period progresses. An
example is the maintenance of an engineering
equipment. If the increase is assumed to be uniform,
the periodic increase or decrease is called a uniform
gradient.

Formulas for Arithmetic Gradient

Future Worth
(1+𝑖)𝑛−1 𝐺 (1+𝑖)𝑛−1
𝐹 = 𝐴[ ]+ [ − 𝑛]
𝑖 𝑖 𝑖
F = future worth of the gradient
A is the 1st payment at period 1.
G is the uniform increase starting at period 2.

Present Worth Formula:

𝐴[1 − (1 + 𝑖 )−𝑛 ] 𝐺 1 − (1 + 𝑖 )−𝑛 𝑛
𝑃= + [ − ]
𝑖 𝑖 𝑖 ( 1 + 𝑖 )𝑛
G = G is the increase or decrease starting at
period 2
A is the 1st payment at period 1.

69
Example 1
An individual makes 5 deposits that increase
uniformly by
P 300 every month in a savings account that earns
12% compounded monthly. If the initial deposit is
P 4500, determine the accumulated amount in the
account just after the last deposit.
Solution: A = 4500 G = 300 i = 12%/12 = 1%

(1 + 𝑖 )𝑛 − 1 𝐺 (1 + 𝑖 )𝑛 − 1
𝐹 = 𝐴[ ]+ [ − 𝑛]
𝑖 𝑖 𝑖
(1 + 0.01)5 − 1 300 (1 + 0.01)5 − 1
𝐹 = 4500 [ ]+ [ − 5]
0.01 0.01 0.01
= 25,984.67

70
Example 2
An amortization of a debt is in a form of a gradient
series. What is the equivalent present worth of the
debt if interest is 5%. Determine also the future
amount of amortization as well as the equivalent
uniform periodic payment.

Solution:
𝐴[(1 + 𝑖 )𝑛 − 1] 𝐺 1 − (1 + 𝑖 )𝑛 𝑛
𝑃= + [ − ]
𝑖 (1 + 𝑖 ) 𝑛 𝑖 𝑖 (1 + 𝑖 )𝑛
A = 5,000 G = -500 i= 5%
5000[(1.05)4 − 1]
𝑃=
0.05(1.05)4
−500 1 − (1.05)−4 4
+ [ − ]
0.05 0.05 1.054
P = 15,178.34

To get the Future Worth:

71
F = P( 1 + i)n
F = 15,178.34 ( 1+0.05) 4 = 18,449.37
To get the equivalent Annual Cost

𝐴[(1+𝑖)𝑛 −1]
P= 𝑖(1+𝑖)𝑛
𝐴(1.054 −1)
15,178.34 = A = P 4,280.47
0.05(1.05)4

Geometric Gradient Formulas

𝐺 1−𝑥 𝑛 1+𝑟
𝑃 = 1+𝑖 ( 1−𝑥 ) where 𝑥 = 1+𝑖
𝐺= 1stterm of a Geometric Gradient
1 + r = fixed percentage
i = interest rate
n = number of interest periods.

72
Example 3
What is the present worth of the given cash flow
diagram if interest rate is 10%. Determine also the
equivalent future worth.

Solution:
𝐺 1−𝑥 𝑛 1+𝑟
𝑃 = 1+𝑖 ( 1−𝑥 ) where 𝑥 = 1+𝑖
1 + r = 540/500 = 583.2/540 = 1.08
r = 0.08
1+𝑟 1.08 54
𝑥= = =
1 + 𝑖 1 + 10% 55
Then:
54 5
𝐺 1−𝑥 𝑛
500 1 − ( )
𝑃= ( )= ( 55 ) = 2,191.57
1+𝑖 1−𝑥 1.1 54
1 − 55
And F = P( 1 + i)n = 2,191.57(1 + 10%)5 =
3529.54

73
 PROBLEM SET 7
Name: __________________________
Score: ________
Subject and Section:
1. Given the series of payment at the end of each
year as follows:
Year 1 P 1000
Year 2 1500
Year 3 2000
Up to Year 10
If i = 5% effective
a. Find the 10th payment.
b. Find the present worth of all payments.
c. Find the future worth of all payments (Year 10 )
d. Find the equivalent annual payments.

74
2. Find the future worth of all payments. i =
10%. Also find the equivalent annual payments.
End of
Year 1 5000
Year 2 6000
Year 3 7,000
…..
Year 20 24,000
Answers. 659,124.99 63,983.95

75
3. Given the monthly payment as shown with
interest as 1 % per month.
Month
1 100
2 110
3 121
4 133.1
Up to Month 12.
Find the 12th payment, the present worth of all
payments , the future worth of all payments and the
equivalent monthly payment.

76
4. Given the series of payments at the end of
each year. i = 6%
Year 1 90,000
Year 2 85,000
Year 3 80,000
a. In what year will there be no payment?
b. What is the present worth of all payment?
c. What is the equivalent annual payment?
Ans. 19, P 597,699.71 , P 53,566.36

77
5. Given Deferred Monthly Deposits i =
2%/month
Month Deposits
1 None
2 None
3 None
4 None
5 1200
6 1700
7 2200
….
..
n 11,200

n =?
Present Worth of All Monthly Deposits=?
Future worth of all monthly deposits =?

78
 LESSON 8
BONDS

Bond – financial security note issued by business or

corporations and by the government as a means of
borrowing long term fund.

Some Terms Related to Bonds

Selling Price – actual purchase price of the bond.

Face Value - Price of redemption upon maturity.

Bond Discount – difference between the face value

and the selling price of the bond which is sold now
below its face value.

Bond Premium - difference between the selling

price and the face value of the bond when it sold
above its face value.

Coupon- is the amount of interest paid on the bond.

Normal Cycle of a Bond

1. Borrower Sold Bond to lender.
2. Lender gets certificate from borrower.
3. Lender receives periodic interest.

79
4. Borrower redeems bond after n years, pays
principal and gets back certificate.

Formula:
Vn = value of bond n years before redemption
Z = par value of the bond
r = rate of interest on the bond per period
C = redemption price of bond
i = interest rate per period
n = number of years before redemption

1 − (1 + 𝑖 )−𝑛 𝐶
𝑉𝑛 = 𝑍𝑟 +
𝑖 (1 + 𝑖 )𝑛

80
Example 1
A P 1,000 ,000 issue of 3%, 15 year bond was sold
at 95%. What is the rate of interest of this
investment?
Solution:
𝑍𝑟[(1 + 𝑖 )𝑛 − 1] 𝐶
𝑉𝑛 = +
(𝑖 )(1 + 𝑖 ) 𝑛 (1 + 𝑖 )𝑛
𝑉𝑛 = 0.95(1,000,000) = 950,000
C = 1,000,000
Zr = 1,000,000 x 0.03 = 30,000
Then:
30,000((1+𝑖)15−1] 1,000,000
950,000 = + (1+𝑖)15
𝑖 (1+𝑖)15
i = 0.03432

Example 2

A bond with a par value of P 1,000 and with a bond

rate of 9% payable annually is to be redeemed at
1,050 at the end of 6 years from now. If it is sold now,
what should be the selling price to yield 8%?

1 − (1 + 𝑖 )−𝑛 𝐶
𝑉𝑛 = 𝑍𝑟 +
𝑖 (1 + 𝑖 )𝑛
Zr = 1000 x 0.09 = 90 C = 1050 i= 0.08 n = 6

Vn = P1, 077.73

81
 Example 3
Mr. X purchase a bond at P 5,100. The bond pays
P 200 per year. It is redeemable for P 5,050 at the
end of 10 years. What is the net rate of interest on
your investment?
Solution:
𝑍𝑟[(1 + 𝑖 )𝑛 − 1] 𝐶
𝑉𝑛 = +
(𝑖 )(1 + 𝑖 ) 𝑛 (1 + 𝑖 )𝑛
Vn = 5,100 Zr = 200 C = 5,050 n = 10
𝑍𝑟[(1 + 𝑖 )𝑛 − 1] 𝐶
𝑉𝑛 = +
(𝑖 )(1 + 𝑖 ) 𝑛 (1 + 𝑖 )𝑛
200[(1+𝑖)10 −1] 5,050
5,100 = + (1+𝑖)10
𝑖 (1+𝑖)10
i = 3.85%

Example 4. A man purchased a share of stock for

P 135 six years ago and has received a dividend of
P 6 at the end of each year. If he sold the share for
P 120 today, at what rate of return could he
considered being paid? Assume that the reinvested
interest earned at 4% compounded annually.

82
Solution:

1 − (1 + 𝑖 )−𝑛 𝐶
𝑉𝑛 = 𝑍𝑟 +
𝑖 (1 + 𝑖 )𝑛
Vn = 135 Zr = 6 C = 120

1 − (1 + 𝑖 )−6 120
135 = 6 +
𝑖 (1 + 𝑖 )6
i= 2.714%

83
 PROBLEM SET 8
Name: ________________________________
Score: ________
Subject and Section:
1. A man was offered a landbank certificate with a
face value of P 120,000 which is bearing an interest
rate of 8% per year payable semiannually and due
in 6 years. If he wants to earn 6% semiannually, how
much must he pay the certificate?

2. To secure a return of 4%, at what price should a

bond be purchased if it is redeemable at P 1,200 in
10 years and plays annual dividends of P 33.00?

84
3. You purchased a P 520 bond for P 5,100.00.
The bond pays 200 per year. It is redeemable for P
5,050 after 10 years. What is the net rate of interest
on your investment?

4. A bond with a par value of P 1,000 and with a bond

rate of 10% payable annually is sold now for P 1,070.
If the yield is to be 12%, how much should be the
redemption price at the end of 8 years?

5. A new company developed a program in which

the employees will be allowed to purchase share of
stocks of the company at the end of its 5th year of
operation, when the company thought to have
gained stability already, at par value of P 100 per

85
share. Believing in a good potential of the company,
an employee decided to save in a bank the amount
f P 8,000 for him at 9% interest compounded yearly.
How much share of stocks will he be able to
purchase at the end of the 5th year of his yearly
deposits?

6. A company issued 50 bonds of P 1,000 face

value each redeemable at par at the end of 15 years
to accumulate the funds required for redemption.
The firm established a sinking fund consisting of
annual deposits, the interest rate of the fund being
4%. What was the principal in the fund at the end of
the 12th year?

86
 LESSON 9
STRAIGHT LINE DEPRECIATION
Depreciation – gradual decrease in value of a
material property due to physical or economic
reasons.
A given property is subject to depreciation if it
meets the following requirements.
1. It must be used in business.
2. It must have a determinable useful life that is
longer than 1 year.
3. It is not product inventory, stocks, bonds or other
asset of investment.
4. It must be a property that loses its value due to
deterioration, obsolescence and other natural
causes.

Terms in Depreciation Problems

1. Book Value – worth of a property at a given
time. It is the acquisition cost of the property minus
the accumulated depreciation charges for a given
period.
2. Salvage Value – estimated worth of property at
the end of its productive life.
3. Scrap Value – price of the property if it sold as
junk.
4. Economic Life - properties depreciable life.
5. Yearly Depreciation Charge – depreciation
charge for a given year of a property.

87
6. Accumulated Depreciation – total depreciation
expense charged to a property after n years of
service.
7. First Cost - purchase of the property plus any
expenses ( shipping,
installation , repair ) prior to initial service or
operation of the property.

Depreciation Methods
The Straight Line Method - assumes that the
loss of value is directly proportional to the age of
the property.
Equations:
𝐹𝐶−𝑆𝑉
𝑑= 𝐵𝑉𝑛 = 𝐹𝐶 − 𝑛𝑑 𝐷𝑛 = 𝑛𝑑
𝐿
FC = first cost
SV = salvage value
L = economic life
d = annual depreciation
BVn = book value after n years
Dn = accumulated depreciation after n years

Example 1
1. A machine has an initial cost of P 50,000 and a
salvage value of P 10,000 after 10 years.
Using Straight Line Method of Depreciation
a. What is the annual depreciation?
b. What is the book value after 5 years?

88
c. What is the total depreciation after 3 years?
d. Construct a depreciation Table
Solution:
FC = 50,000 SV = 10,000 L = 10
𝐹𝐶−𝑆𝑉 50,000−10,000
a. 𝑑 = = = 4,000
𝐿 10

b. 𝐵𝑉5 = 50,000 − 5(4,000) = 30,000

c. D3 = 3(4,000) = 12,000
DEPRECIATION TABLE
Depreciation Accumulated Book
Year
Charge Depreciation Value
1 4000 4000 46000
2 4000 8000 42000
3 4000 12000 38000
4 4000 16000 34000
5 4000 20000 30000
6 4000 24000 26000
7 4000 28000 22000
8 4000 32000 18000
9 4000 36000 14000
10 4000 40000 10000

89
Example 2 An Engineer bought an equipment for P
500,000. He spent an additional amount of P 30,000
for installation and other expenses. The salvage
value is 10% of the 1st cost. If the book value at the
end of 5 years is P 291,500 using straight line
depreciation, compute the life of the equipment in
years.

Solution:
FC = 500,000 + 30,000 = 530,000
SV = 10% ( 530,000) = 53,000
BV at year 5 = 530,000 - AD ( 5)
530,000−53,000
291,500 =530,000- 5 𝐿
L = 10 years

Example 3
A machine which cost P 10,000 was sold as scrap
after being used for 10 years. The scrap value is P
500. Determine the total depreciation at the end of
5 years.
Solution:
FC = 10,000
L = 10 years
SV = 500
𝐹𝐶−𝑆𝑉 10,000−500
AD = = = 950
10 10
Total Depreciation after 5 years = 5 x 950 = 4750

90
 PROBLEM SET 9
Name: ______________________________
Score: ________
Subject and Section: ____________

1. Erectors Co owns an equipment that cost P

120,000.00. After 8 years it will have an estimated
value of P 17,000. Compute the accumulated
depreciation charge after 2 years and the book
value after 5 years using straight line method.

2. The 1st cost of a certain machine is P

950,000.00 . After 12 years, it will have a salvage
value of P 180,000.00. Compute the book value at
the end of 8 years.

91
3. A machine cost P 74,500 and has a life of 9
years with a salvage value of P 3500 at the end of
8 years. Determine the book value at the end of 6
years using straight line method.

4. An engineer bought an equipment for P

600,000.00. He spent an additional amount of P
30,000 for installation and other expenses. The
salvage value is 12% of the initial first cost. Life =
16 years.
Compute the following.
a. Annual Depreciation.
b. Book Value after 6 years.
c. Total depreciation after 10 years

92
5. An asset was purchased at P580 000.00. It has
an estimated life of 10 years, and will be sold at P50
000.00 at that time. What is the depreciation for the
first year? USE SL method.

6. A machine has an initial cost of P60, 000.00 and

a salvage value of P10 ,000.00 after 12 years.
Find/What is the book value after five years using
straight-line depreciation?

8. An asset waspurchased for P700 000.00. The

salvage value after 25 years is P100 000.00. What
is the total depreciation of the first years using
straight-line method?

93
9. A heavy-duty copying machine was procured for
P120 000.00 with an estimated salvage value of P10
000.00 after 10 years. What is the book value after
five years?

10. A unit of welding machine costs P45 000.00 with

an estimated life of five years. Its salvage value is
P2500.00. Find its depreciation rate by straight-line
method.

94
 LESSON 10
DEPRECIATION DECLINING BALANCE,
DOUBLE DECLINING BALANCE METHOD

Declining Balance Method

– assumes that the depreciation charge for a given
year is a fixed percentage of the book value at the
start of the year.
Formulas:
𝐿 𝑆𝑉
𝑘 =1− √
𝐹𝐶
𝑑𝑛 = 𝐹𝐶 (1 − 𝑘 )𝑛−1 𝑘
depreciation at year n
𝐵𝑉𝑛 = 𝐹𝐶 (1 − 𝑘 )𝑛
k is called the Matheson’s Constant

Example 1 A machine costing P 720,000 is

estimated to have a book value of P 40, 545.73
when retired at the end of 10 years. Depreciation
cost is computed using a constant percentage of
the declining value.
a. What is the annual rate of depreciation?
b. What is the book value after 3 years?
c. What is the depreciation charge after 4 years?

95
d. What is the total depreciation after 6 years?
e. Construct a Depreciation Table

Solution:
FC = 720,000 SV = P 40,545.73 L = 10
𝐿 𝑆𝑉 10 40,545.73
a. 𝑘 = 1 − √𝐹𝐶 = 1 − √ 720,000 = 0.25
𝐵𝑉𝑛 = 𝐹𝐶 (1 − 𝑘 )𝑛
b. 𝐵𝑉3 = 720,000 ( 1 – 0.25)3 = 303,750
c. 𝑑𝑛 = 𝐹𝐶 (1 − 𝑘 )𝑛−1 𝑘
𝑑4 = 720,000(1 − 0.25)4−1 (0.25) = 75,937.5
d. Total depreciation after 6 years:
𝐵𝑉6 = 720,000(1 − 0.25)6 = 128,144.53
𝐷6 = 720,000 − 128,144.53 = 591,855.47

Depreciation Table

96
 Double Declining Balance Method
In this method , the depreciation is computed as :
2 𝑛−1 2
𝑑𝑛 = 𝐹𝐶 (1 − )
𝐿 𝐿
2 𝑛
𝐵𝑉𝑛 = 𝐹𝐶 (1 − )
𝐿
where dn is the depreciation charge after n years
and BVn is the book value after n years.

Example 3:
A machine has a 1st cost of P 140,000 and a life of
8 years with a salvage value of P 10,000 at the end
of its useful life. Using double declining
balance method
a. What is the Book Value on the 3rd year?
b. What is the depreciation charge on the 4th
year?
Solution:
2 𝑛
a. 𝐵𝑉𝑛 = 𝐹𝐶 (1 − 𝐿)
2 3
𝐵𝑉3 = 140,000 − ) = 59,062.5
(1
8
2 𝑛−1 2
b. 𝑑𝑛 = 𝐹𝐶 (1 − )
𝐿 𝐿
2 4−1 2
𝑑4 = 140,000 (1 − 8) (8) = 14,765.63

97
 PROBLEM SET 10
Name: __________________________________
Score: ________
Subject and Section:

DECLINING BALANCE METHOD

1. A machine costing P900,000 is estimated to
have a book value of P 70,000 when retired at the
end of 10 years. Depreciation cost is computed
using a constant percentage of the declining value.
PAST CE BOARD EXAM
a. What is the annual rate of depreciation?
b. What is the book value after 3 years?
c. What is the depreciation charge after 4 years?
d. What is the total depreciation after 6 years?

98
2. The cost of a certain asset is P 300,000. Its life
is 8 years with scrap value of P 50,000. Using
declining balance method , find the annual rate of
depreciation under a constant percentage method
and construct a depreciation table.

99
3. A machine having a certain 1st cost has a life of
10 years and a salvage value of 7.633% of the 1st
cost of 10 years. If it has a book value of P 68,914
after 6 years, how much is the 1st cost of the
machine using Matheson’s Method?
(Declining Balance Method )

4. A machine has a current price of P 400,000. If

its selling price is expected to decline at the rate of
12% per annum, what will be the selling price after
5 years?

100
 DOUBLE DECLINING BALANCE METHOD
5. A machine has a 1st cost of P 350,000 and a
life of 10 years with a salvage value of P 12,000 at
the end of its useful life. Using double declining
balance method
a. What is the Book Value on the 3rd year? b. b.
What is the depreciation charge in the 4th year?

6. An equipment costs P 600,000 and has a

salvage value of P 25,000 after its 35 years of
useful life. Using Double Declining Balance Method,
what will be the book value after 8 years?

101
7. A machine costing P 55,000 is estimated to have a
salvage value of P 4,350 when retired at the end of 9 years.
Using declining balance method
a. Find the Matheson’s Constant
b. Find the book value after 3 years.

8. An equipment cost P 66,000. Its resale value at

the end of the 5th year is P 15,000,00. Using
Declining Balance Method, determine the
depreciation charge for the 2nd year.

8. An equipment cost P 600,000 and has a

salvage value of P 25,000 after its 25 years of
useful life. Using double declining balance method,
what will be the value of the equipment at the end
of 8 years?

Using Double declining balance method, find the book value

at the end of 5 years. Ans. P 21,375

102
 LESSON 11
DEPRECIATION USING SUM OF THE YEARS
DIGIT METHOD (SOYD)
Given FC (First Cost) , SV (Salvage Value ) a
nd L (economic life )
Procedure:
𝐿(𝐿 + 1)
𝑌 = 1 + 2 + 3 + ⋯.𝐿 =
2
𝐿
d1 = 𝑌 (𝐹𝐶 − 𝑆𝑉) 𝐵𝑉1 = 𝐹𝐶 − 𝑑1
𝐿−1
𝑑2 = (𝐹𝐶 − 𝑆𝑉) 𝐵𝑉2 = 𝐹𝐶 − (𝑑1 + 𝑑2 )
𝑌
𝐿−2
𝑑3 = (𝐹𝐶 − 𝑆𝑉) 𝐵𝑉3 = 𝐹𝐶 − (𝑑1 + 𝑑2 +
𝑌
𝑑3 )
…….. …..

Example 4. An asset is purchased for P 9000. Its

estimated life is 10 years, after which is will be sold
for P 1,000. Using SOYD
a. Find the book value during the 3rd year.
b. Find the depreciation during the 2nd year.
c. Find the total depreciation after 4 years.
d. Construct a Depreciation Table

Solution:
10(11)
𝑌= = 55
2
𝐵𝑉3 = 𝐹𝐶 − (𝑑1 + 𝑑2 + 𝑑3 )

103
 10 10
𝑑1 = (𝐹𝐶 − 𝑆𝑉 ) = ( 9000 − 1000) = 1454.55
55 55
9 9
𝑑2 = (𝐹𝐶 − 𝑆𝑉 ) = (9000 − 1000) = 1309.09
55 55
8 8
𝑑3 = (𝐹𝐶 − 𝑆𝑉 ) = (9000 − 1000) = 1163.64
55 55
𝐵𝑉3 = 9000 − (1454.55 + 1309.09 + 1163.64)
= 5072.72
= book value during the 3rd year.

𝑑2 = 1309.09 = depreciation during the 2nd year

7 7
𝑑4 = (𝐹𝐶 − 𝑆𝑉 ) = (9000 − 1000) = 1018.18
55 55

𝐷4 = 𝑑1 + 𝑑2 + 𝑑3 + 𝑑4
= 1454.55 + 1309.09 + 1163.64
+ 1018.18
= 4945.45 = total depreciation after 4 years

104
DEPRECIATION TABLE

105
 PROBLEM SET 11
Name:
_______________________________Score: __
Subject and Section:

SUM OF YEARS DIGIT METHOD

1. An asset is purchased for P 300,000. Its
estimated life is 15 years, after which is will be sold
for P70,000 . Using SOYD
a. Find the book value during the 3rd year.
Find the depreciation during the 2 nd year. c. Find
the total depreciation after 4 years.
d. Construct a depreciation and Book Value Table.

106
2. A certain corporation makes it a policy that for
any/every- new equipment purchased, the annual
depreciation cost should not exceed 30% of the first
cost at any time with no/without- salvage value.
Determine the length of the service life if the
depreciation used is SYD method.

107
3. Q purchased a Bulk Milk Cooler for P780 000.00.
Shipping, tax, and installation costs amounted to
P25,000.00, P20,00.00 and 15,000.00 years,
determine the book value after four years and the
depreciation charge on its last year of service by
preparing an SYD depreciation table.

108
4. A telephone company purchased microwave radio
equipment for P 9 million, freight and installation
charges amounted to 4% of the purchased price. If
the equipment will be depreciated over a period of
12 years with a salvage value of 8%, determine the
depreciation cost during 5th year using SYD.

5.An asset is purchased for P160,000.00. Its

estimated life is 12 years, after which it will be sold
for P12,000.00. Find the depreciation for the first
year using the sum-of-the-year’s digit method.

109
6. An asset is purchased for P8000.00. Its estimated
life is 14 years, after which it will be sold for P
1000.00 find the book value during the second year
if sum-of-the year’s digit (SYOD) depreciation is
used.

7. A company purchases an asset for P10 000.00

and plans to keep it for 24 years. If the salvage value
is zero at the end of the 20th year, what is the
depreciation in the third year? Use sum-of-the-year’s
digits depreciation.

110
8. An equipment costing P700 000.00 has a life
expectancy of 8 years. Using some-of-the-year’s
digit method of depreciation, what must be its
salvage value such that its depreciation charge for
the first year is P100,000.00?

111
 LESSON 12
DEPRECIATION USING SINKING FUND
METHOD

Sinking Fund Method- assumes that a sinking fund

is created for replacement purposes. The fund is
established by depositing equal amount at the end
of each year for L years at r% interest per year.
Let FC = first cost SV = salvage value and L =
life in years

𝐿−1
(1 + 𝑖 ) 𝐿 − 1
𝐹𝐶 − 𝑆𝑉 = ∑ 𝑑 (1 + 𝑖 )𝑥 = 𝑑
𝑖
0
(𝐹𝐶−𝑆𝑉)𝑖
Then: 𝑑 = (1+𝑖 )𝐿 −1

𝑛−1
(1 + 𝑖 )𝑛 − 1
𝐷𝑛 = ∑ 𝑑 (1 + 𝑖 )𝑥 = 𝑑( )
𝑖
0

112
𝐵𝑉𝑛 = 𝐹𝐶 − 𝐷𝑛

Example 1
Jade company purchased a computer for P 45,000.
If the salvage value is estimated to be P 7,000 after
5 years. Prepare a depreciation table using the
sinking fund method at 4% interest.
Solution:
FC = 45,000 SV = 7,0000 L = 5 years
(𝐹𝐶−𝑆𝑉)𝑖 (45,000−7,000)(4%)
𝑑= (1+𝑖)𝐿 −1
= (1+4%)5−1
= 𝑃 7,015.83

(1 + 𝑖 )𝑛 − 1
𝐷𝑛 = 𝑑( )
𝑖
1.041−1
𝐷1 = 7,015.83 ( ) = 7015.83
0.04
2
1.04 − 1
𝐷2 = 7,015.83( ) = 14,312.29
0.04
1.043 − 1
𝐷3 = 7,015.83 ( ) = 21,900.61
0.04

113
 DEPRECIATION TABLE

Example 2
Given FC = 100,000
SV = 10,000
L = 10 years
i = 5%
Compute
a. Annual Depreciation d.
b. Book Value after 3 years.
c. Book Value after 8 years.

Solution:
(𝐹𝐶 − 𝑆𝑉 )𝑖
𝑑=
(1+𝑖 )𝐿 − 1
(100,000−10,000)5%
a. 𝑑 = (1+5%)10−1
= 7155.41
b. Book value after 3 years
𝐵𝑉𝑛 = 𝐹𝐶 − 𝐷𝑛
(1+𝑖)𝑛−1
BV3 = 100,000 - 𝑑( )
𝑖

114
 1.053−1
= 100,000 – 7155.41 = 𝑃 77,442.56
0.05
(1+𝑖)𝑛 −1
BV8 = 100,000 - 𝑑( ) = 100,000 −
𝑖
1.058 −1
7155.41 =
0.05
= P 31,672.21

115
 PROBLEM SET 12
Name: _________________________________
Score: ________
Subject and Section:
1. An equipment cost P 500,000 with a salvage
value of 10,000 at the end of 15 years.
Using Sinking Fund Method with interest rate= 8
%.
a. Compute the annual depreciation cost.
b. Find the book values at years 5 to 10.

116
2. A plant erected to manufacture socks with a first
cost of P 10,000,000 with an estimated salvage
value of P 200,000 at the end of 25 years. Find the
appraised value to the nearest 100 by sinking fund
method at 7% interest rate at the end of
a. 10 years
b. 20 years

117
3. A factory is constructed at a 1st cost of
P 8,000,000 and with an estimated salvage value of
P 200,000 at the end of 25 years. Find its
appraised value to the nearest 100 at the end of 10
years by using sinking fund of depreciation
assuming an interest of 5%. Ans. P 5,944,400

118
4. Given FC = 200,000 SV = 50,000 L = 10
years i= 7%
a. Annual depreciation d = ? Ans. 10, 856.62
b. BV3 = ? 165,097.04
c. Total depreciation after 6 years? P 77,660.6
d. Construct a depreciation table.

119
5. A four-stroke motorbike costs P75,000.00. It will
have a salvage value of P10,000.00 when worn out
at the end of eight years. Determine the annual
replacement deposit, and present a depreciation
schedule using the SFF at 8%.

120
6. The original cost of a certain piece of equipment
is P 150,000.00 and its depreciated by a 12% sinking
fund method. Determine the annual depreciation
charge if the book value of the equipment after 10
years is the same as if it had been depreciated at
P14 000.00 each year by straight-line method.

121
 LESSON 13
CAPITALIZED COST

Capitalized Cost – sum of its 1st cost and the

present worth of all its costs for replacement,
operation and maintenance for a long time.
= First cost + cost of perpetual maintenance
It is the sum of the first cost and the present worth
of all future payments and replacements which is
assumed to continue forever.

Example 1 The 1st cost of a certain equipment is P

324,000 and a salvage value of P 50,000 at the end
of its life of 4 years. If money is worth 6%
compounded annually, find the Capitalized cost.
Solution:
Compute annual depreciation:
(𝐹𝐶 − 𝑆𝑉 )𝑖
𝑑=
(1+𝑖 )𝐿 − 1

(324,000 − 50,000)0.06
𝑑=
1.064 − 1
d = 62,634.07

62,634.07
Capitalized cost = FC + = 1,367,091.14
0.06
where FC = 324,000

122
Example 2
Find the capitalized cost of a bridge whose cost is
P 250 M and life is 20 years. If the bridge must be
partially rebuilt at a cost of 100 M at the end of each
20 years. i = 6%.
Solution.
Convert interest rate of 6% every 20 years.
( 1 + 0.06)20 -1 = 2.207135 or 220.7135%
Capitalized Cost = First cost + cost of perpetual
maintenance
100 𝑀
250 M + 2.207135 = 295.3076 M

123
Example 3
What is the total capitalized cost of a structure
requiring &150,000 for initial construction, $10,000
every year for the first 8 years and then $14,000
each year thereafter for upkeep and replacements of
$100,000 at the end of every 10 years. Assume 12%
interest and that upkeep expenses occur at the end
of each year.

1 − (1.12)−8
𝐶𝐴𝑃 𝐶𝑂𝑆𝑇 = 150,000 + 10,000
0.12
100,000 14.000
+ + (1.12−8 )
1.1210 − 1 0.12

= $294,282.91

1−(1.12)−8
10,000 = Present worth of 10,000 for 8
0.12
years
14,000/0.12 ( 1.12)-8 Present worth of perpetuity
of 14,000 starting at year 9.
1.1210 -1 12% converted every 10 years

124
Example 4
A machine cost P 150,000 and will have a scrap
value of P 10,000 when retired at the end of 15
years. If money is worth 4%, find the annual
investment and the capitalized cost of the machine.
Solution:
Annual Investment = First Cost x i + Annual
Depreciation
FC = 150,000 SV = 10,000
(𝐹𝐶−𝑆𝑉)
𝑑 = 1.0415 −1 (0.04) = 6,991.75
Then Annual Investment =
150,000 (0.04) + 6,991.75 = 12,991.75
Capitalized Cost = 12,991.75/ 0.04 =
P 324,793.85

125
 PROBLEM SET 13
Name: ____________________________
Score: ________
Subject and Section:

1. A machine cost P 400,000 and will have a scrap

value of P 50,000 when retired at the end of 15
years. If money is worth 5%, what is the capitalized
cost?

2. A bridge that was constructed at a cost of P 7.5

M is expected to last 30 years at the end of which
time its renewal cost will be P 5 M. Annual repairs
and maintenance is P 300,000. What is the
capitalized cost of the bridge at an interest of 6%?

126
3. An item is purchased for P 100,000. Annual cost
is P 18,000. Using i = 8%, what is the capitalized
cost of perpetual service? Ans. P 325,000

4. An asset was purchased for P 100,000 and retired

at the end of 15 years with a salvage value of P
4,000. Determine the capitalized cost of the asset
base on interest rate of 8%. Ans. P 369,195

5. A machine cost P 600,000 with a salvage value

of 50,000 after 15 years. What is the capitalized
cost if i = 6%.

127
6. Compute the capitalized cost of a project that cost
P 35 M now and requires 2 M maintenance annually.

7. Find the capitalized cost of a bridge whose cost is

P 4 M with an annual maintenance of 200,000 with
i = 20%.

128
8. At 6%, what is the capitalized cost of a bridge
whose cost is 200 M and life of 20 years if the bridge
must be partially rebuilt at a cost of 100 M at the end
of each 25 years?

9. What are the capitalized cost of the following?

a. An asphalt pavement costing P 100,000 that

would last for 5 years with negligible repairs. At the
end of 5 years, P 5,000 would be spent to remove
the old surface before P 100,000 is spent again for
the new surface. i = 8%.

129
b. A thick concrete pavement costing P 250,000
which would last indefinitely with a cost P 20,000
for minor repairs at the end of every 3 years. Use i
= 8%

130
 LESSON 14
BREAK EVEN ANALYSIS
Break-Even Analysis - A technique for analyzing
how revenue, expenses and profit vary with
changes in sales volume .
Assumptions:
1. All units are sold at a constant price per unit.
2. There is no income other than that of operations.
3. Variable costs are directly proportional to
production rate from zero to 100% capacity.
4. Fixed Cost are constant regardless of the
number of units produced.

Example 1
A company has a production capacity of 200 units
per month and its fixed cost is P 20,000 per month.
The variable cost per unit is P 300 and the unit can
be sold for P 450. Economy measures are
instituted to reduce the fixed cost by 10% and the
variable cost by 20%.
a. Compute the old and new break even point.
b. Draw a break even chart.
Solution:
Let x = number of units to be produced and sold to
break even.

131
Then Total Cost = Total income to break even
20,000 + 300x = 450x
x = 133.33
Production must be 133.33/200 = 66.67 % to break
even.

If the fixed cost is reduced by 10% and the variable

cost by 20%
Then:
0.9(20,000) + 0.8 (300x) = 450x
x = 85.7
Production is 85.7/200 = 42.9%

Graph:
Graph y = 20,000 + 300x and y = 450x

132
 Break Even Chart (old )

Break Even Chart ( Reduced Cost )

133
Example 2
A firm has the capacity to produce 1,000,000 units
of product per year. At present, it is able to produce
and sell only 600,000 units yearly at a total income
of P 720,000. Annual fixed costs are P 250,000
and the variable cost per unit are 0.70.
a. Compute the annual firms profit or loss for the
present production.
b. How many units must be sold annually to break
even.
c. Draw a break even chart.

Solution:
Cost per product = 720,000 / 600,000 = 1.2
Expenses = 250,000 + 0.7x

134
Income = 1.2x
To break even:
Expenses = Income
250,000 + 0.7x = 1.2x
x = 500,000 units

Profit = Income – Expenses for x = 600,000

units
= 1.2(600,000) - [ 250,000 + 0.7(600,000) ]
= P 50,000
Break Even Chart

135
136
 PROBLEM SET 14
Name: __________________________________
Score: ________
Subject and Section: ______________________

1. A manufacturer producer certain items at a labor

cost per unit of P 315, material cost per unit is P100,
variable cost of P3.00 each. If the item has a selling
price of P995, how many units must be
manufactured each month for the manufacturer to
breakeven if the monthly overhead is P461, 600?
Draw the breakeven chart.

2. The annual maintenance cost of a machine shop

is P69, 994. If the cost of making a forging is P56
per unit and its selling price is P135 per forge unit,
find the number of units to be forged to break-even.
If the number of units sold is 1000, what is the
loss/profit? Draw the break even chart.

137
3. A manufacturer produces certain item at a labor
cost of P115 each, material cost of P76 each and
variable cost of P2.32 each. If the item has a unit
price of P600, how many units must be
manufactured each month for the manufacturer to
break even if the monthly overhead is P428, 000?
Draw the breakeven chart.

138
4. Steel drum manufacturer incurs a yearly fixed
operating cost of $ 200,000. Each drum
manufactured cost $ 160 to produce and sells $ 200.
What is the manufacturer’s break-even sales volume
in drums per year? Draw the break even chart.

5. XYZ Corporation manufactures bookcases that

sell for P 65.00 each. It costs XYZ Corporation P35,
000 per year to operate its plant. This sum includes
rent, depreciation charges on equipment, and
salary payments. If the cost to produce one
bookcase is P50.00 how many cases must be sold
each year for XYZ to avoid taking a loss?

139
6 A company which manufactures electric motors
has a production capacity of 200 motors a month.
The variable costs are P150.00 per motor. The
average selling price of the motors is P275.00 Fixed
cost of the company amount to P20,000 per month
which includes taxes. Find the number of motor that
must be sold each month to breakeven.
If the company sold 135 motors, find the profit or
loss.
Draw the break even chart.

140
7. The annual maintenance cost of a machine is
P70, 000. If cost of making a forging is P56 and its
selling price is P 125 per forged unit. Find the
number of units to be forged to break even. Ans.
1,015 units

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 LESSON 15
COMPARISON OF ALTERNATIVES

Fundamental Principle for Comparison of

Alternatives
Methods of Comparison
Present Worth Analysis – involves conversion of
future cash flows to its equivalent present amount. In
the analysis, include present worth of all operation
and maintenance expenses, replacement and other
out of pocket cost. The alternative with the least
present cost is the most logical choice.

Future Worth Analysis – It is based on the concept

of finding equivalent compound amount of receipts
and disbursements at the end of the investment
period.

Rate of Return on Additional Investment Method

– assumes that unlimited capital is available and
therefore an alternative requiring a bigger
investment may be provided the rate of return on the
additional investment justifies the bigger outlay of
capital.
𝑛𝑒𝑡 𝑎𝑛𝑛𝑢𝑎𝑙 𝑝𝑟𝑜𝑓𝑖𝑡
Rate of Return = 𝑐𝑎𝑝𝑖𝑡𝑎𝑙 𝑖𝑛𝑣𝑒𝑠𝑡𝑒𝑑

142
Minimum Annual Cost Method - This method
determine the annual cost of each of the several
alternatives and the alternative with the least annual
cost is chosen.

Capitalized Cost - widely used in comparing

alternatives where structures have long lives.
Capitalized Cost of any asset is the sum of its 1st cost
and the present worth of all cost for replacement.

Example 1
An investment of P 270,000 can be made in a project
that will produce a uniform annual revenue of P
185,400 and then have a salvage value of 10% of
the initial investment.
Out of Pocket Cost for Operations and Maintenance
= P 81,000 per year
Taxes and Insurance = 4% of 1st cost per year.
The company expects capital to earn not less than
25% before income taxes.
Is this a desirable investment? Use Rate of Return
Method

143
Solution 1:
Rate of Return Method
Annual Revenue P 185,400 for
5 years
Annual Depreciation:
270,000−10%(270,000)
∑30 1.25𝑥
= 29,608.76
Taxes and Insurance
270,000 (0.04) = 10,800
Operation and Maintenance P 81,000
Annual Cost = 121,408.76
Profit = 185,400 - 121,408.76 = 63,991.24

63,991.24
Rate of Return = = 23.7%
270,000
Rate of Return < 25% Investment not justified.

Solution 2: Using Annual Worth Method

Annual Revenue P 185,400
Annual Cost :
Annual Depreciation: 29,608.76
Operation and Maintenance 81,000
Taxes and Insurance 10,800
Interest on Capital 0.25(270,000) = P 67,500
Total Annual Cost P 188,909.76 >
Annual Revenue
Investment Not Justified

144
Solution 3 : Using Present Worth Method
Present Worth of Cash Inflows =
185,400 27000
∑51 + = P 507,439.87
1.25𝑥 1.255
Annual Costs ( Excluding Depreciation ) =
81,000 + 270,000(0.04) = P 91,800
Present Worth of Cash Ouflows =
91800
∑51 + 270,000 = P 516,875.90
1.25𝑥
Present Worth of Cash Ouflow = P 516,875.90 >
Present Worth of Cash Inflows
> P 507,439.87
Investment Not Justified

145
Example 2
A new piece of equipment has been propose to
increase production. The equipment will cost
P 55,000 will last for seven years and will have a
salvage value of P 5,000. With this investment
income it is projected to increase by
P 12,000 each year for the 1st four years and P
14,000 each year for the remaining 3 years. Using
present worth analysis and rate of return of 14%,
determine if equipment purchase is justified.

(1+𝑖)𝑛−1
Note: 𝑃 = 𝐴 𝑖 (1+𝑖)𝑛

12,000 14,000 5000

Present Worth = ∑41 ( + ∑75 + 1.147 =
1.14)𝑥 1.14𝑥
56,207.03
=
)4
(1.14 − 1 1.143 − 1
12,000 [ ] + 14,000 [ ] (1.14)−4
0.14(1.14)4 0.14(1.14)3
5,000
+
1.147
146
 = P56,207.03 < 55,000
The equipment purchase is justified.

Example 3
A certain dam costs P 4,000,000 and will serve the
purpose for 20 years. At the end of that time, it must
be enlarged at a cost of P 2,000,000. If the dam had
been built larger than originally, the cost would have
been P 5,00,000. Bond pay interest at the rate of 5%
per annum and the interest paid annually. Compare
the 2 alternatives and compute the savings or loss.
Solution:
Solution 1: Present Worth Method
The present worth of building the dam in case 1 is:
2,000,000
P1 = 4,000,000 + = 4,753,778.97
1.0520
P2 = 5,000,0000
Difference in initial cost = 5,000,0000 -
4,753,778.97 = 246, 221.03
(The 1st method is better by P 246,221.03 )

Solution 2: Future Worth Method

The difference in initial cost is 5,000,000 –
4,000,000 = 1,000,000
If this difference is invested at 5% per annum for 20
years.
F = P( 1 + i)n = 1,000,000 ( 1.0520) = 2,653,297.71
> P 2,000,000

147
The 1st alternative is better and the savings will be
2,653,297.71 - 2,000,000 = P 653,397.71

Example 4
A new equipment that will increase the revenue by
P 94, 500 a year requires an investment of P
430,000. It is estimated to have a net salvage value
of P 50,000 at the end of 12 years with annual
expenses for repairs and maintenance totaling P
15,000. Determine if investment is justified using
Future worth analysis and 18% rate of return.
Solution:

Find the future worth of all costs.

Future Worth =
∑11 𝑥
0 (94,500 − 15,000)(1.18) + 50,000 =
P 2,827,020.08
Future Worth of P 430,000 is
430,000( 1.18)12= 3,133,664.83

148
Investment is not justified since
3,133,664.83 > 2,827,020.08
(This means if you invest P 430,000 at the rate of
18% , it will earn 3,133,664.83 greater than investing
in the equipment. )

149
 PROBLEM SET 15
Name: _____________________________
Score: ________
Subject and Section:

1. The data below are estimated for a project study.

i = 10%
Plan A
Initial Investment P 35,000
Annual Operating Cost P 6,450
Life 4 years
Salvage Value none
Annual Revenue 19,000

Plan B
Initial Investment P 50,000
Annual Revenue P 25,000
Annual Disbursement P 13830
Life 8 years
Salvage Value none

Which plan would you recommend? Use Present

Worth Method and Future Worth Method and 8
years of study period.

150
2 Two different type of machines are being
considered for a power plant. i = 7%
The data are as follows:
Machine A:
First Cost P 120,000
Salvage Value P 15,000
Life 6 years
Annual Maintenance Cost P 9,000

Machine B
First Cost P 136,000
Life 8 years
Salvage Value None
Annual Operating Cost P 7,000

Use analysis period of 24 years.

a. Find the present worth of Machine A and B.

b. Which machine would you recommend?

151
4. A company is considering two alternatives of
buying new machine or retaining an old one.
Data:
Alternative A: Buying a new one
Cost P 135,000
Life 5 years
Salvage Value 0
Annual Operating Cost P 150,000

Alternative B Retaining the old one

Present Value P 7,500
Remaining Life 5 years
Salvage Value none
Annual Operating Cost P 187,500

Using Sinking Fund Deprecation Method (i = 10%)

and Annual Cost Analysis, which is profitable
alternative.

152
5. Which machine will you recommend using
Annual Cost Method and sinking fund depreciation?
New Machine
First Cost P 1,200,000
Life 16 years
Overhead Cost 5,000 Annually
Salvage Value P 120,000
i = 10%

Old Machine
Cost P 600,000
Overhead Cost P 10,000 annually
Life 6 years
Salvage value P 60,000
i = 10%

153
6. A machine costs P 500,000 and has an expected
life of 10 years. Salvage value is P 20,000 and
annual operating cost of P 10,000. What is the
return on investment if the annual revenue is P
100,000?
Note: Annual Cost = Annual Revenue

154