UNIT II.

THE TIME VALUE OF MONEY

Overview
This unit presents the concept of interest and cash flow diagrams as the first and
basic concept in analysis of engineering economy problems. This will help you identify .and
analyze which problems will use the different formula of interest.
Learning Objectives
At the end of the unit, I am able to:

1. to solve problems involving interest;

2. to draw and label cash flow diagrams from a given problem; and
3. to differentiate which problems, need the specific formula to the kind of interest
used.
Topics
1.1 Simple Interest
1.2 Compound Interest
1.3 Cash Flow Diagrams
1.4 Rates of Interest
1.5 Continuous compounding
1.6 Discount
Setting up

Name: Date:
Course/Year/Section:

Directions: Solve the following questions.

1. Catherine Espino was granted a loan of P80,000 by her employer ORE Industries with
an interest of 6% for 180 days on the principal collected in advance. The company
would accept a promissory note for P80,000 non-interest for 180 days. If discounted at
once, find the proceeds on the note.

2. In buying a refrigerator, Mr. Alcantara was offered the options of paying P27,500 cash
at the end of 30 days or P29,700 at the end of 120 days. At what rate is Mr. Alcantara
paying simple interest if he agrees to pay at the end of 120 days?
3. Mr. Martin applied for a car loan from ORE Finance Corporation which charges a rate of
interest of 18%. Interest in this type of transaction is to be deducted from the loan at
the time the money is released. At the end of one year the borrower will have to pay the
same amount that is stated in his application for the loan. What is the actual interest did
the financing corporation charged him?

4. P1,000,000.00 was deposited in a bank and left for 8 years, at which time the principal
is withdrawn. The interest has accrued and is left for another 8 years. If the effective
annual interest is 5%, what will be the withdrawal amount at the end of the 16th year?
5. Mr. Gomez plans to deposit P525,000.00 in the bank now and another P1,050,000.00
for the next 2 years. If Mr. Gomez plans to withdraw P1,750,000.00 3 years after his last
deposit for the purpose of buying a house and lot, what will be the amount of money
left in the bank after one year of his withdrawal? Effective annual interest rate is 10%.
Lesson Proper

THE TIME VALUE OF MONEY

The dependence of the economic value of a loan or investment on time is a

consequence of interest. Interest may be thought of as money paid for the use of borrowed
money or the return obtained by the productive investment of capital. Interest is defined as:

interest = total amount accumulated - original amount (2-1)

The interest rate, i, is by definition

𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑎𝑐𝑐𝑟𝑢𝑒𝑑 𝑑𝑢𝑟𝑖𝑛𝑔 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑

𝑖= (2-2)
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡

The interest rate is usually expressed as a percent, that is, 100% times i; however, the fraction
𝑖 as defined by Eq. (2-2) is the quantity to be used in all formulas that are presented. The unit
time period or interest period is typically one year, but shorter interest periods such as one
month are also used. The total number of interest periods is denoted by n, and the number of
interest compounding periods per year is denoted by m.

If more than one interest period is involved, the type of interest, either simple or
compound, must be considered. Simple interest ignores the time value of the interest
accrued in preceding interest periods; therefore, it is almost never used in engineering
economy.

Simple Interest

Simple interest is calculated using the principal only, ignoring any interest accrued in
preceding interest periods. The total simple interest over several periods is computed as:

𝑖 = 𝑃𝑛𝑖 (2-3)

𝐹 = 𝑃 + 𝐼 = 𝑃 + 𝑃𝑛𝑖 (2-4)
𝐹 = 𝑃(1 + 𝑛𝑖) (2-5)
 where: 𝐼 = 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡

𝑃 = 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑜𝑟 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑤𝑜𝑟𝑡ℎ

𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑝𝑒𝑟𝑖𝑜𝑑𝑠
𝑖 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑖𝑚𝑡𝑒𝑟𝑒𝑠𝑡 𝑝𝑒𝑟 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑝𝑒𝑟𝑖𝑜𝑑
𝐹 = 𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑟 𝑓𝑢𝑡𝑢𝑟𝑒 𝑤𝑜𝑟𝑡ℎ

Ordinary and Exact Simple Interest

(a) Ordinary Simple Interest is computed on the basis of one banker’s year, which is

1 𝑏𝑎𝑛𝑘𝑒𝑟 ′ 𝑠𝑦𝑒𝑎𝑟 = 12 𝑚𝑜𝑛𝑡ℎ𝑠, 𝑒𝑎𝑐ℎ 𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑖𝑛𝑔 𝑜𝑓 30 𝑑𝑎𝑦𝑠 = 360 𝑑𝑎𝑦𝑠

(b) Exact Simple Interest is based on the exact number of days, 365 for an ordinary year
and 366 days for a leap year. The leap years are those which are exactly divisible by
4, but excluding the century years such as the years 1900, 2000 etc.

If d is the number of days in the interest period, then for ordinary simple interest,

𝐼 = 𝑃𝑖𝑛

𝑑
where: 𝑛= (2-6)
360

For exact simple interest:

𝐼 = 𝑃𝑖𝑛

𝑑
where : 𝑛= (2-7)
366
Examples:

2-01 A bank charges 12% simple interest on a P3,000.00 loan. How much will be repaid if the
loan is paid back in one lump sum after three years?

Solution:

𝐹 = 𝑃(1 + 𝑛𝑖)

= 3000[1 + ((3)(0.12)]
Thus ; 𝐹 = 𝑃 4,080.00

2-01Determine the (a) ordinary simple interest, (b) exact simple interest on P5,000.00 for
the period from January 18 to October 26 2013, if the rate of interest is 24%.

Solution:

January 18 – 31 = 13 (excluding January 18)

February = 28
March = 31

April = 30
May = 31
June = 30
July = 31

August = 31

September = 30
October = 26 (including October 26)

Total = 292 days

 (a) Ordinary simple interest

𝐼 = 𝑃𝑛𝑖
292
𝐼 = 𝑃5,000 ( ) (0.24)
360
𝐼 = 𝑃973.33

(b) Exact simple interest

𝐼 = 𝑃𝑛𝑖
292
𝐼 = 𝑃5,000 ( ) (0.24)
365
𝐼 = 𝑃960.00

2-01Marian buys a television set from a merchant who asks P 15,000 at the end of 60 days.
Marian wishes to pay immediately and the merchant offers to compute the cash price on the
assumption that money is worth 8% simple interest. What is the cash price today?
Solution:

𝐹 =𝑃+𝐼
𝐹 = 𝑃 + 𝑃𝑟𝑡
𝑃(0.08)(60)
15,000 = 𝑃 +
360
𝑃 = 14,851.49

2-01Paul borrows P20,000 from KASAPI Lending Corporation. The interest is to be deducted
from the loan at the time the money is borrowed. What is the actual rate of interest if Paul
has to pay back P20,000 at the end of one year if the rate of simple interest is 15%?

Solution:
𝑃 = 𝐹 − 𝐼 ; 𝐹 = 𝑃20,000; 𝑛 = 1

𝑃 = 𝑃20,000 − 0.15(𝑃20,000) = 𝑃 17,000.00

𝑎𝑛𝑑 𝐹 = 𝑃(1 + 𝑛𝑖)

𝑃20,000 = 𝑃17,000(1 + 𝑖)

𝑖 = 0.1765 = 17.65%
Compound Interest

Compound interest is the practice of charging an interest rate to an initial sum and to
any previously accumulated interest that has not been withdrawn. It is calculated using the
principal plus the total amount of interest accumulated in previous periods. Thus, compound
interest means “interest on top of interest”.

Cash flow Diagrams

A cash flow diagram is simply a graphical representation of cash flows drawn on a time scale.
Cash flow diagram for economic analysis problems is analogous to that of free body diagrams
for mechanics problems.

receipt (positive cash flow or cash inflow)

Disbursement (negative cash flow or cash outflow)

A loan of P1000 at a simple interest of 10% will become P1500 after 5 years

Fig. 2.1 Cash flow diagram on the viewpoint of the lender

Fig. 2.2 Cash flow diagram on the viewpoint of the borrower

Fig. 2.3 Compound interest (borrower’s viewpoint)

Interest Principal at Interest Compound amount at the

period beginning of period earned end of period
during
period

1 P Pi P+Pi =P(1+i)

2 P(1+i) P(1+i)i P(1+i)+P(1+i)i =P(1+i)2

3 P(1+i)2 P(1+i)2i P(1+i)2+P(1+i)2i

=P(1+i)3

---- --------- -------- ---------

n P(1+i)n-1 P(1+i)n-1i P(1+i)n-1+P(1+i)n-

1i=P(1+i)n=F

Table 2.1 Derivation of Formula

𝐹 = 𝑃(1 + 𝑖)𝑛 (2-8)

 The quantity (1+I)n is commonly called the “single payment compound factor” and is
designated by the functional symbol F/P,i%,n. Thus,
𝐹 = P(F/P,i%,n) (2-9)

The symbol F/P, i%,n is read as “F given P at I per cent in n interest periods”. From
Equation (2-8),

𝑃 = 𝐹(1 + 𝑖)−𝑛 (2-10)

The quantity (1+i)-n is called the “single payment present worth factor” and is
designated by the functional symbol P/F,i%,n. Thus,
P=F(P/F,i%,n) (2-11)

The symbol P/F,i%,n is read as “P given F at I per cent in n interest periods”.

Rates of Interest

(a) Nominal rate of interest

The nominal rate of interest specifies the rate of interest and a number of interest periods in
one year.
𝑟
𝑖= (2-12)
𝑚

where: 𝑖 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑝𝑒𝑟 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑝𝑒𝑟𝑖𝑜𝑑

𝑟 = 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒

𝑚 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑖𝑛𝑔 𝑝𝑒𝑟𝑖𝑜𝑑𝑠 𝑝𝑒𝑟 𝑦𝑒𝑎𝑟
If the nominal rate of interest is 15% compounded quarterly, then i=15%/4=3.75%, the rate
of interest per interest period.
(b) Effective rate of interest
The effective rate of interest is the actual rate of interest on the principal for one year. It is
equal to the nominal rate if the interest is compounded annually, but greater than the nominal
rate if the number of interest periods per year exceeds one, such as for interest compounded
semi-annually, quarterly or monthly. If P1.00 is invested at a nominal rate of 15%
compounded quarterly, after one year this will become,
0.15 4
𝑃1(1 + ) = P1.1586
4
The actual interest earned is P0.1586, therefore, the rate of interest after one year is 15.86%.
Hence,
Effective rate = F1 – 1 =(1+i)m – 1 (2-13)

where: F1 = the amount P1.00 will be after one year

F1

0 1 2 m

I year

P1.00

Fig. 2.4 Effective Rate of Interest

Examples

2-05 Find the nominal rate which if converted quarterly could be used instead of 15%
compounded semi-annually. What is the corresponding effective rate?

Solution

Let r = the unknown nominal rate

For two or more nominal rates to be equivalent, their corresponding effective rates must be
equal.
Nominal rate Effective rate

r% compounded quarterly (1 + 𝑟/4)4 -1

0.15 2
15% compounded semi-annually (1 + ) -1
2

𝑟 0.15 2
(1 + )4 -1 = (1 + ) -1
4 2
𝑟
1+ = (1.075)1/2
4
𝑟 = 0.1473 𝑜𝑟 14.73% 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑𝑒𝑑 𝑠𝑒𝑚𝑖 − 𝑎𝑛𝑛𝑢𝑎𝑙𝑙𝑦
2-06 Mr. and Mrs. Galang decided that for every child that will be born to them they will place
a deposit in the bank so that on the child’s 18th birthday, the child will receive the amount of
P600,000.00. If the bank will pay an interest of 18% compounded yearly, how much deposit
will Mr. and Mrs. Galang have to make on the birth of a child born to them?
Solution

0 18

𝐹 = 𝑃(1 + 𝑖)𝑛

600,000 = 𝑃(1.18)18

𝑃 = 𝑃30,498.26

2-07 How long will it take P10,000.00 to be four times its value if invested at the rate of 7%
compounded semi-annually?

Solution

0 1 2 3 n

𝐹 = (1 + 𝑖)𝑛 =4P
0.07 P
𝑖= = 0.035
2
𝑃 = 𝑃10,000

𝐹 = 4𝑃 = 4(10,000) = 40,000
𝐹 = 𝑃(1 + 𝑖)𝑛

40,000 = 10,000(1 + 0.035)𝑛

(1.035)𝑛 = 4

𝑛𝑙𝑜𝑔(1.035) = 𝑙𝑜𝑔4
𝑙𝑜𝑔4
𝑛=
log 1.035
𝑛 = 40.29 𝑦𝑒𝑎𝑟𝑠 ≈ 41 𝑦𝑒𝑎𝑟𝑠
Effective rate = (1 + 𝑖)𝑚 -1

Effective rate = (1.035)2 − 1

Effective rate = 0.0712 or 7.12%

2-08 Find the amount at the end of 2 years and 3 months if P10,000 is invested at 10%
compounded quarterly using simple interest for anytime less than a year interest period.

Solution

F2

F1

0 1 2 2 years 3 mnths

compound interest Simple

interest
P10,000
10%
For compound interest: 𝑖 = = 2.5% ; 𝑛 = 2(4)=8
4
3
For simple interest : 𝑖 = 10% ; 𝑛 =
12

F1=𝑃(1 + 𝑖)𝑛 = 𝑃10,000(1 + 0.025)8 = 12,184.03

3
F2=𝐹1 (1 + 𝑛𝑖) = 𝑃12,184.03 [1 + (0.10)] = 𝑃12,488.63
12

Equation of Value

An equation of value is obtained by setting the sum of the values on a certain comparison or
focal date of one set of obligations equal to the sum of the values on the same date of another
set of obligations.
Examples

2-09 Mr. Fajardo bought a house and lot worth P2,000,000.00 if paid in cash. On the
installment basis, he paid a down payment of P400,000.00; P600,000.00 at the end of one
year; P800,000.00 at the end of 3 years and a final payment at the end of 5 years. What was
the final payment if interest was 20%?
Solution

P2,000,000-P400,000=P1,600,000.00

0 1 2 3 4 5

P600,000

P800,000

X
Let x = the final payment

Using today as the focal date, the equation of value is

𝑃1,600,000 = 𝑃600,000(P/F,20%,1)+𝑃800,000(P/F,20%,3)+x(P/F,20%,5)

P1,600,000=P600,000(1.20)-1+P800,000(1.20)-3+x(1.20)-5
x=P1,585,152.00

Another Solution
Using after 5 years as the focal date, the equation of value is

P1,600,000(F/P,20%,5)=P600,000(F/P,20%,4)+P800,000(F/P,20%,2)+x
P1,600,000(1.20)5=P600,000(1.20)4+P800,000(1.20)2+x

x=P1,585,152.00

2-10 Oliver deposits P200,000 in the bank today at the rate of 4% per annum. After two years
he deposits another P400,000. In five years he will withdraw P600,000. How much money
does he have on the 6th year?

Solution

P600,000 x

0 2 5 6

P200,000 P400,000

Using after 6 years as the focal date, the equation of value is

P200,000(F/P,4%,6)+P400,000(F/P,4%,4)=P600,000(F/P,4%,1)+x

P200,000(1.04)6+P400,000(1.04)4=P600,000(1.04)1+x

x=P97,007.23
Continuous Compounding

If r is the nominal annual interest rate and m is the number of interest periods each year, then
the interest rate per interest period is i=r/m, and the number of interest periods in n years is
mn. Equation 2-8, the single payment compound amount factor, maybe written as
𝑟
𝐹 = 𝑃(1 + )𝑚𝑛 (2-14)
𝑚

Increasing m, the number of interest periods per year, without limit, it becomes very
large and approaches infinity and r/m approaches zero. This is the situation for continuous
compounding.
𝑟 𝑚𝑛
𝐹 = 𝑃 lim (1 + ) (2-15)
𝑚→∞ 𝑚

Set r/m =x; then m=(1/x)r and mn=(1/x)rn. As m approaches infinity, x approaches
zero. Equation (2-15) becomes

𝐹 = 𝑃 [lim (1 + 𝑥)1/𝑥 ]rn (2-16)

𝑥→0

From Calculus, we have the limit

lim(1 + x)1/x =2.71828……=e,

x→0

where e is the base of natural logarithms.

Thus, Equation (2-16) becomes

𝐹 = 𝑃𝑒 𝑟𝑛 (2-17)

which is the continuous compounding single payment compound amount formula.

Examples
2-11 Bessie wishes to have P400,000 in a certain fund at the end of 8 years. How much should
she invest in a fund that will pay 6% compounded continuously?

Solution
𝐹 = 𝑃𝑒 𝑟𝑛

400,000 = 𝑃𝑒 0.06(8)

𝑃 = 𝑃247,513.36
2−12 Compute the difference in the future amount of P500,000 compounded annually at
nominal rate of 5% and if it is compounded continuously for 5 years at the same rate.
Solution

For compounded annually:

𝐹 = 𝑃(1 + 𝑖)𝑛

𝐹 = 500,000(1.05)5

𝐹 = 𝑃638,140.78
For compounded continuously:

𝐹 = 𝑃𝑒 𝑟𝑛

𝐹 = 500,000𝑒 0.05(5)
𝐹 = 𝑃642,012.71

Difference in future worth:

= P642,012.71-P638,140.78

=P3,871.93

Discount

Discount on a negotiable paper is the difference between what it is worth in the future and
its present worth. It is the interest paid in advance. Thus,

𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡 = 𝐹𝑢𝑡𝑢𝑟𝑒 𝑣𝑎𝑙𝑢𝑒 − 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑉𝑎𝑙𝑢𝑒

𝐷 = 𝐹−𝑃 (2-18)

For example, if a negotiable paper such as bond can be sold for P1,000, six months from now,
but it is sold for P950 at present then the discount is P50.00.

The rate of discount is the discount on one unit of principal for one unit of time. If d is the rate
of discount, then

1
𝑑 = 1− = 1- (P/F,i%,1) (2-19)
1+𝑖

𝑖
𝑑= = (P/F,i%,1)I (2-20)
1=𝑖
For the equivalent rate of interest corresponding to a rate of interest I, we have from Equation
(2-20).
𝑑 𝑑
𝑖= = 𝑃 (2-21)
1−𝑑 (𝐹,𝑖%,𝑖)

Example
2-13 Allan borrowed from a bank under a promissory note that he signed in the amount of
P250,000.00 for a period of one year. He received only the amount of P219,150.00 after the
bank collected the advance interest and an additional amount of P850.00 for notary and
inspection fees. What was the rate of interest that the bank collected in advance?

Solution
𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 = 𝑃250,000 − (𝑃219,150 + 𝑃850)

𝐼𝑛𝑡𝑒𝑟𝑒𝑠𝑡 = 𝑃30,000
Amount he actually received

=250,000-30,000=P220,000
30,000
Rate of interest = (100)
220,000

Rate of interest = 13.64%

Another solution
𝑑
i=
1−𝑑
30,000
d= (100)
250,000

d= 12%
0.12
i=
1−0.12

i=0.1364
i=13.64%
Assessing Learning

Activity 1
Name: ________________________________________ Score: ______________

Course/Year/Section: _______________________ Date: _______________

1. Bianca needs P100,000 per year for four years to go to college. His father invested

P125,000 in an account earning an interest of 7% per annum for her education on the day

that she was born. If Bianca will withdraw P100,000 at the end of her 17 th, 18th, 19th and

20th birthday, how much money will be left in the account at the end of her 21st birthday?

2. Sebastian owes Paulo P75,000.00 due in one year and P225,000.00 due in 4 years. He

agrees to pay P150,000.00 today and the balance in two years. How much must he pay

at the end of two years if money is worth 5% compounded semi-annually?

3. The Hope Foundation is planning to put up its own building. There are two proposals that

are being considered:

Proposal 1: The construction of the building now, to cost P8,000,000.00.

Proposal 2: The construction of a smaller building now, to cost P6,000,000.00 and at the end

of 5 years, an extension to be added to cost P4,000,000.00.

If the interest rate is 20%, which proposal is more economical and by how much? Neglect

depreciation.

4. Mr. Co borrowed P400,000 from Mr. So on August 1, 2008 and P100,000 on August 1,

2010, agreeing that money is worth 5% compounded annually. Mr. Co paid P100,000 on

August 1, 2011, P80,000 on June 1, 2012 and P140,000 on August 1, 2013. What additional

sum should Mr. Co pay on August 1, 2016 to discharge all remaining liability?
5. Mr. and Mrs. Corpuz borrowed P600,000 from Mabuhay Lending Corp. for 6 years at

12%. At the end of 6 years, it renews the loan for the amount due plus P600,000 more

for 3 years at 12%. What is the lump sum due?