UNIVERSITY OF SAINT LOUIS

Tuguegarao City

SCHOOL OF ENGINEERING, ARCHITECTURE and

INFORMATION TECHNOLOGY EDUCATION
Summer
A.Y. 2020-2021

CORRESPONDENCE LEARNING MODULE

ENGE 1013 – ENGINEERING ECONOMICS

Prepared by:

ENGR. DARREL JAY M. BALBIN

ENGR. FRANCES LORANE T. CALAPINI, MSCE
ENGR. AIRA JANE C. BASSIG
Instructors

Reviewed by:

ENGR. WILLIAM M. VEGA, ME-CE

CE Program Chair

ENGR. ARIEL M. LORENZO, ME-ECE

ECE Program Chair

Recommended by:

ENGR. VICTOR C. VILLALUZ, MEE

Academic Dean

Approved by:

DR. EMMANUEL JAMES P. PATTAGUAN

VP for Academics

ENGE 1013 – ENGINEERING ECONOMICS | 1

Midterms Week 2

1. Weekly Study and Assessment Guide

CORRESPONDENCE LEARNING MODULE

ENGE 1013 – Engineering Economics
A.Y. 2020-2021

Lesson 2

Topic: A. Simple Interest

a. Ordinary Simple Interest
b. Exact Simple Interest
B. Compound Interest
a. Nominal Rate of Interest
b. Effective Rate of Interest
C. Discrete and Continuous Compounding
D. Equivalence
E. Discount
F. Inflation

Learning Outcomes: At the end of this module, you are expected to:

1. Solve problems involving interest and the time value of money

2. Expound the concept of equivalence cash flows

LEARNING CONTENT

ENGE 1013 – ENGINEERING ECONOMICS | 2

INTEREST

- amount of money earned by a given capital

A. SIMPLE INTEREST
- interest directly proportional to the length of time and the amount of principal borrowed

I = Pni
F = P + I = P + Pni
F = P (1 + ni)

where:
I = interest
P = principal amount or present worth
n = number of interest period (in years)
i = simple interest rate (per year)
F = accumulated amount or future worth

a. ORDINARY SIMPLE INTEREST

- computed on the basis of one banker’s year
1 banker’s year = 12 months (30 days each)
= 360 days

ENGE 1013 – ENGINEERING ECONOMICS | 3

 b. EXACT SIMPLE INTEREST
- computed based on exact number of days
1 year = 365 days (ordinary year)
1 year = 366 days (leap year)

EXAMPLES:

1. If a man borrowed money from his boyfriend with simple interest rate of 12%, determine the present
worth of P75,000 which is due at the end of seven months.
SOLUTION:
F = P (1 + ni)
75,000 = P [1 + (7/12)(0.12)]
P = 70,093.46 (answer)

2. Ivana has invested P1000, part at 5% and the remainder at 10% simple interest. How much is invested
at a higher rate if the total annual interest from this investment is P95?
SOLUTION:
P5 = principal of 5% simple interest
P10 = principal of 10% simple interest
P5 + P10 = 1000  equation 1

I5 + I10 = 95
I = Pni
P5(1)(.05) + P10(1)(.10) = 95  equation 2

Solving the 2 equations:

P5 = P100
P10 = P900 (answer)

3. Determine the accumulated amount using exact simple interest on P1000 for the period from January
20, 1990 to November 28 of the same year at 15% interest rate.
SOLUTION:
January 20-31 = 11 (excluding January 20)
February = 28
March = 31
April = 30
May = 31
June = 30
July = 31
August = 31
September = 30
October = 31
ENGE 1013 – ENGINEERING ECONOMICS | 4
 November = 28
312 days
F = P (1 + ni)
F = 1,000 [1 + (312/365)(0.15)]
F = P1,128.22 (answer)

CASH FLOW DIAGRAMS

- horizontal line with vertical upward downward arrows located at pits in time shown either as cash inflow
or outflow

receipt (positive cash flow or cash inflow)

disbursement (negative cash flow or cash outflow)

B. COMPOUND INTEREST
- interest is computed every end of each interest period and the interest earned for that period is added
to the principal

ENGE 1013 – ENGINEERING ECONOMICS | 5

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

Single-payment-compound-amount factor: (F/P, i, n) = (1 + i)n

Single-payment-present-worth factor: (P/F, i, n) = (1 + i)-n

where:
F = accumulated amount or future worth
P = principal amount or present worth
i = r/m
= interest rate per interest period
n = mt
= total number of interest period for n-years
r = nominal interest rate
m = number of interest period per year
t = number of years of investment

ENGE 1013 – ENGINEERING ECONOMICS | 6

RATE OF INTEREST

a. NOMINAL RATE OF INTEREST (r)

- specifies the rate of interest and the number of interest periods per year
EXAMPLE:

If the nominal rate of interest is 10% compounded quarterly find the rate of interest per interest period.

SOLUTION:

i = r/m = 10/4

i = 2.5% (answer)

b. EFFECTIVE RATE OF INTEREST (ER)

- the actual rate of interest on the principal for one year
ER = interest earned in one year/principal at the beginning of the year

ER = (1 + r/m)m – 1

Note: If compounded annually,

i = r = ER

EXAMPLES:

1. If the nominal rate of interest is 10% compounded quarterly find the actual rate of interest after one
year.
SOLUTION:
ER = (1 + r/m)m – 1
ER = (1 + 0.10/4)4 – 1
ER = 0.1038 or 10.38% (answer)

2. Find the nominal rate which if converted quarterly could be used instead of 12% compounded monthly.
What is the corresponding effective rate?
SOLUTION:
r% compounded quarterly = 12% compounded monthly

Note: For two or more nominal rates to be equivalent, the corresponding effective rates must be equal.
ERQ = ERM
(1 + r/4)4 – 1 = (1 + 0.12/12)12 – 1
r = 0.1212 or 12.12% compounded quarterly (answer)

ER = (1 + 0.12/12)12 – 1
ER = 0.1268 or 12.68% (answer)

EXAMPLES ON COMPOUND INTEREST:

ENGE 1013 – ENGINEERING ECONOMICS | 7
1. Aling Marie deposited a sum of P12,000 in an account earning interest rate of 9% compounded
quarterly.
a. What will it become after 1 year?
b. What is the effective rate?
c. What is then the equivalent nominal interest rate if compounded monthly?
SOLUTION:

a. F = 12,000 (1 + 0.09/4)4(1)
F = P13,117 (answer)

b. ER = interest earned in one year/principal at the beginning of the year

ER = (13,117 – 12,000)/12,000
ER = 0.0931 or 9.31% (answer)
or
ER = (1 + 0.09/4)4 – 1
ER = 0.0931 or 9.31% (answer)

c. ERM = ERQ
(1 + r/12)12 – 1 = 0.0931
r = 0.0893 or 8.93% (answer)

2. What payment X ten years from now is equivalent to a payment of P1000 six years from now, if interest
is 15% compounded monthly?
SOLUTION:

6
0 10
P1,000

=
10

0 6

X
F10 = P6 (1 + i)n
F10 = 1,000 (1 + 0.15/12)12(4)
F10 = P1,815.35 (answer)

3. 20k
Withdrawals
5k

0 1 3 4

2 5

ENGE 1013 – ENGINEERING ECONOMICS | 8

10k
13k 15k
 Deposits

Based on the cash flow diagram above, compute the amount saved in the bank at the end of 5 years if interest
is 12% compounded yearly.

SOLUTION:

F = P (1 + i)n

Deposits:

F10k = 10,000(1 + 0.12)4 = 15,735.19

F13k = 13,000(1 + 0.12)2 = 16,307.20

F15k = 15,000(1 + 0.12)1 = 16,800.00

Total Deposits: 48,842.39

Withdrawals:

F5k = 5,000(1 + 0.12)3 = 7,024.64

20,000.00

Total Withdrawals: 27,024.64

Savings = Total Deposits – Total Withdrawals = 48,842.39 - 27,024.64

Savings = P21,817.75 (answer)

C. DISCRETE COMPOUNDING
- The interest is compounded at the end of each finite – length period such as a month, a quarter or a
year.

ENGE 1013 – ENGINEERING ECONOMICS | 9

 CONTINUOUS COMPOUNDING
- cash payments are assumed to occur once per year but the compounding is continuous throughout the
year

F = P (1 + i)n
F = P (1 + r/m)mt

Let m/r = k, then m = rk, as m increases so must k

(1 + r/m)mt = (1 + 1/k)rkt = [(1 + 1/k)k]rt
1 k
k→∞
( )
lim 1+
k
=e

[(1 + 1/k)k]rt = ert

Thus,
F = Pert
P = Fe-rt

ER = (1 + r/m)m – 1
Let m/r = k, then m = rk, as m increases so must k
(1 + r/m)m = (1 + 1/k)rk = [(1 + 1/k)k]r
1 k
k→∞
( )
lim 1+
k
=e

[(1 + 1/k)k]r = er

Thus,

ER = er – 1

EXAMPLE:

How many years are required for P1000 to increase to P2000 if invested at 9% per year compounded
continuously?

SOLUTION:

F = Pert

2,000 = 1,000e0.09t

t = 7.70 years (answer)

ENGE 1013 – ENGINEERING ECONOMICS | 10

 D. EQUIVALENCE
- is a condition that exists when the value of a cost at one time is equivalent to the value of the related
benefit received at a different time.

1. MATHEMATICAL EQUIVALENCE
- Equivalence is a consequence of the mathematical relationship between time and money. This is the
form of equivalence used in F = P (1 + i)n
2. DECISIONAL EQUIVALENCE
- Equivalence is a consequence of indifference on the part of a decision maker among available choices.

3. MARKET EQUIVALENCE
- Equivalence is a consequence of the ability to exchange one cash flow for another at zero cost.

EXAMPLES:

1. Sarawat has had $800 stashed under his mattress for 30 years. How much money has he lost by not
putting it in a bank account at 8 percent annual compound interest all these years?
SOLUTION:

Since Sarawat has kept the $800 under his mattress, he has not earned any interest over the 30 years. Had he
put the money into an interest-bearing account, he would have far more today. We can think of the $800 as a
present amount and the amount in 30 years as the future amount.

F = P (1 + i)n
F = 800 (1 + 0.08)30
ENGE 1013 – ENGINEERING ECONOMICS | 11
 F = $8,050.13

Sarawat would have $8050.13 in the bank account today if he had deposited his $800 at 8 percent annual
compound interest. Instead, he has only $800. He suffered an opportunity cost of $8050.13 – $800 = $7250.13
(answer) by not investing the money.

2. You want to buy a new computer, but you are $1000 short of the amount you need. Your aunt has
agreed to lend you the $1000 you need now, provided you pay her $1200 two years from now. She
compounds interest monthly. Another place from which you can borrow $1000 is the bank. There is,
however, a loan processing fee of $20, which will be included in the loan amount. The bank is
expecting to receive $1220 two years from now based on monthly compounding of interest.
(a) What monthly rate is your aunt charging you for the loan? What is the bank charging?
(b) What effective annual rate is your aunt charging? What is the bank charging? (c) Would you prefer
to borrow from your aunt or from the bank?

SOLUTION:

a. Your aunt
F = P (1 + i)n
1,200 = 1,000 (1 + i)12(2)
i = 0.007626 or 0.7626%
Your aunt is charging interest at a rate of approximately 0.76 percent per month.

The bank
F = P (1 + i)n
1,220 = 1,020 (1 + i)12(2)
i = 0.007488 or 0.7488%
The bank is charging interest at a rate of approximately 0.75 percent per month.

b. Your aunt
ER = (1 + r/m)m – 1
ER = (1 + 0.007626)12 – 1
ER = 0.09544 or 9.54%
The effective annual rate your aunt is charging is approximately 9.54 percent.

The bank
ER = (1 + r/m)m – 1
ER = (1 + 0.007488)12 – 1
ER = 0.09365 or 9.37%
The effective annual rate for the bank is approximately 9.37 percent.

c. The bank appears to be charging a lower interest rate than that offered by your aunt. This can be
concluded by comparing the two monthly rates or the effective annual rates. If you were to base
ENGE 1013 – ENGINEERING ECONOMICS | 12
 your decision only on who charged the lower interest rate, you would pick the bank, despite the fact
that it has a fee. However, although you are borrowing $1020 from the bank, you are getting only
$1000, since the bank immediately gets its $20 fee. The cost of money for you from the bank is
better calculated as
F = P (1 + i)n
1,220 = 1,000 (1 + i)12(2)
i = 0.008320 or 0.8320%
From this point of view, the bank is charging interest at a rate of approximately 0.83 percent per
month and you would be better off borrowing from your aunt.

E. DISCOUNT
- Interest paid in advance

SIMPLE DISCOUNT RATE

- Is an interest transaction where the price of the corresponding loan is set down by subtracting the so
called discount from the amount due
- The corresponding interest is credited at the beginning of the discount period (interest-in-advance),
while in the simple interest model the interest is credited in arrears at the end of the interest period.

F−P
i= [ ]
P 1 year

F −P
d= [ ]
F 1 year

d = 1 – (1 + i)-1

i
d=
1+i

where:

P = proceeds of the loan or principal

F = amount to be paid or future value

d = simple discount rate

i = simple interest rate

EXAMPLE:

1. Tine was granted a loan of P10,000 but the bank deducted an interest of P1,000 thereby giving him a
net loan amount of P9,000 payable after one year. What is the rate of a.) interest? b.) discount?
SOLUTION:

ENGE 1013 – ENGINEERING ECONOMICS | 13

 F−P
i= [ P ] 1 year

10,000 −9,000
i= [ 9,000 ]
i = 0.1111 or 11.11%

F −P
d= [ F ] 1 year

10,000 − 9,000
d= [ 10,000 ]
d = 0.10 or 10

or

0.1111
d=
1+0.1111

d = 0.10 or 10%

F. INFLATION
- Increase in the amount of money needed to purchase same amount of good and services
- Results in a decrease in purchasing power

a. FC = PC (1 + f)n

where:

PC = present cost of a commodity/good/service

FC = future cost of the same commodity/good/service

f = annual inflation rate

n = number of years

EXAMPLE:

A pocket wifi presently costs P1000. If inflation is at the rate of 8% year, what will be the cost of the item in two
year?

ENGE 1013 – ENGINEERING ECONOMICS | 14

SOLUTION:

FC = PC (1 + f)n = 1000 (1 + 0.08)2

FC = P1,166.40

b. In an inflationary economy, the buying power of money decreases as costs increase. Thus,

F = P (1 + f)-n

where:

F = future worth in today’s pesos

P = present amount

EXAMPLE:

The Philippine economy is experiencing inflation at an annual rate of 8%. If this continues, what will P1,000 be
worth two years from now in terms of today’s pesos?

SOLUTION:

F = P (1 + f)-n = 1000 (1 + 0.08)-2

F = P857.34

c. If interest is being compounded at the same time that inflation is occurring, the future worth will be

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

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

EXAMPLE:

Loonyo invested P10,000 at an interest rate of 10% compounded annually. What will be the final amount of his
investment, in terms of today’s pesos, after five years, if inflation remains the same at the rate of 8% per year?

SOLUTION:

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

1+0.10 5
F=10000( )
1+0.08

ENGE 1013 – ENGINEERING ECONOMICS | 15

F = P10,960.86

d. ic = i + f + if

where:

i = real interest rate

f = inflation rate

= rate of change in the value of currency

ic = market or inflation-adjusted rate

= rate that takes inflation into account

= combined interest-inflation rate

EXAMPLES:

1. A P2,000 in 2 years has an average inflation rate of 6% and the real interest rate of money is 10%.
Determine the inflation-adjusted interest rate.
SOLUTION:

ic = i + f + if

ic = 0.10 + 0.06 + 0.10(0.06)

ic = 0.166 or 16.6%

2. A company invests P10,000 today to be repaid in 5 years in one lump sum at 12% compounded
annually. If the rate of inflation is 3% compounded annually, how much profit is relieved over 5 years?
SOLUTION:
I=F-P
F = P (1 + i)n

ic = i + f + if

0.12 = i + 0.03 + i(0.03)

i = 0.08738

F = 10000 (1 + 0.08738)5

F = 15,202.21

I = 15202.21 – 10000

I = 5,202.21

*** END of LESSON 2***

ENGE 1013 – ENGINEERING ECONOMICS | 16
SEATWORK

PROBLEM-SOLVING. Solve each of the problems below and show your complete, systematic, and neat
solution on a clean sheet of paper.
1. Determine the ordinary simple interest on P10,000 for 9 months and 10 days if the rate of interest is 12%
2. Determine the ordinary and exact simple interest on P5,000 for the period from January 15, to June 20,
1993 if the rate of simple interest is 14%.
3. A man borrows P6,400 from a loan association. In repaying this debt, he has to pay P400 at the end of
every 3 months on the principal and a simple interest of 16% on the principal outstanding at that time.
Determine the total amount he has paid after paying all his debt.
4. If the sum of P12,000 is deposited in an account earning interest at the rate of 9% compounded quarterly,
what will it become at the end of 8 years?
5. A man possesses a promissory note, due 3 years hence, whose maturity value is P6,700.48. If the rate of
interest is 10% compounded semi-annually, what is the value of this note now?
6. What payment X 10 years from now is equivalent to a payment of 10,000 six years from now, if interest is
15% compounded (a) annually, and (b) monthly
7. If P1,000 becomes P1,811.36 after 5 years when invested at an unknown rate of interest compounded
bimonthly (every 2 months), determine the unknown nominal rate and the corresponding effective rate.
8. Calculate the effective rate corresponding to 9% compounded
a. semi-annually
b. quarterly
c. bi-monthly
d. monthly
e. continuously
ENGE 1013 – ENGINEERING ECONOMICS | 17
9. Compare the accumulated values at the end of 10 years if P100 is invested at the rate of 12% per year
compounded annually, semi-annually, quarterly, monthly, daily and continuously.
10. How many years are required for P1,000 to increase to P2,000 if invested at 9% per year compounded
annually, semi-annually, quarterly, monthly, daily and continuously.
11. An advertisement of an investment firm states that if you invest P500 in their firm today you will get P1,000
at the end of 4½ years. What nominal rate is implied if interest is compounded (a) quarterly? (b) monthly?
Determine also the effective rate of interest in each case.
12. A man wishes to bequeath to his daughter P20,000 10 years from now. What amount should he invest now
if it will earn interest of 8% compounded annually during the first 5 years and 12% compounded quarterly
during the next 5 years?
13. A debt of P15,000 was paid for as follows: P4,000 at the end of 3 months, P5,000 at the end of 12 months,
P3,000 at the end of 15 months and a final payment F at the end of 21 months. If the rate of interest was
18% compounded quarterly, find F.

QUIZ

PROBLEM-SOLVING. Solve each of the problems below and show your complete, systematic, and neat
solution on a clean sheet of paper.

1. A man borrows P10,000 from a loan firm. The rate of simple interest is 15%, but the interest is to be
deducted from the loan at the time the money is borrowed. At the end of one year, he has to pay back
P10,000. What is the actual rate of interest?
2. Based on news reports, it is predicted that there will be an average annual rate of inflation in prices of
commodities of 8.2% during the next ten years. Assuming this prediction to be accurate, a home currently
costing P51,000 now would have what price ten years hence?
3. An agent of real estate company stated that a house he sold in 1980 for P37,000 was sold by the buyer for
P90,000 in 1990. If the increase in price is due solely to inflation, determine the average rate of inflation
between 1980 and 1990.
4. On his son’s fifth birthday, a man decides to deposit a certain amount which will be equivalent to P28,00
with today’s purchasing power of the peso on his son’s eighteenth birthday when he will start his college
education. If the bank pays 5½% interest compounded annually but the rate of inflation is 8.7%
compounded annually, how much should the man deposit now?
5. Ten years ago, an item costs P2,500. The rate of inflation for the first 4 years was 4%, during the next 3
years, 6%; and for the last 3 years, 9%. Assuming that increases in price were due to inflation alone, what
would the item cost now? What is the average inflation rate during the 10 years?
6. Given: P = P1,000; F = P2,400; n = 6 years; f = 8%. Find (a) i c, the combined interest-inflation rate of
return; (b) i, the real after-inflation rate of return.
7. You have just won a lottery prize of P1,000,000 collectable in 10 yearly installments of P100,000 starting
today. Why is this prize not really P1,000,000? What is it really worth today if money can be invested at 10
percent annual interest, compounded monthly?
8. Suppose in Problem 7 that you have a large mortgage you want to pay off now. You propose an alternative
but equivalent payment scheme. You would like $300 000 today and the balance of the prize in five years
when you intend to purchase a large piece of waterfront property. How much will the payment be in five
years? Assume that annual interest is 10 percent, compounded monthly.
9. You are looking at purchasing a new computer for your four-year undergraduate program. Brand 1 costs
$4000 now, and you expect it will last throughout your program without any upgrades. Brand 2 costs $2500
now and will need an upgrade at the end of two years, which you expect to be $1700. With 8 percent
annual interest, compounded monthly, which is the less expensive alternative, if they provide the same
level of service and will both be worthless at the end of the four years?
10. Today, an investment you made three years ago has matured and is now worth $3000. It was a three-year
deposit that bore an interest rate of 10 percent per year, compounded monthly. You knew at the time that
you were taking a risk in making such an investment because interest rates vary over time and you “locked
in” at 10 percent for three years.
(a) How much was your initial deposit?

ENGE 1013 – ENGINEERING ECONOMICS | 18

 (b) Looking back over the past three years, interest rates for similar one-year investments did indeed vary.
The interest rates were 8 percent the first year, 10 percent the second, and 14 percent the third. Did you
lose out by having locked into the 10 percent investment? If so, by how much?

REFERENCES

Textbooks

Chan Park. (2018). Fundamentals of Engineering Economy. Pearson Prentice Hall.

Niall M. Fraser and and Elizabeth Jewkes. (2016). Engineering Economics Financial Decision Making for
Engineers. Pearson Canada.

R. Paneerselvam. (2014). Engineering Economics. PHI Learning Private Limited, New Delhi.

Leland Blank and Anthony Tarquin. (2013). Basics of Engineering Economy. McGraw-Hill Higher Education.

Hipolito Sta. Maria. (). Engineering Economy. National Bookstore.

Online Reference

ENGE 1013 – ENGINEERING ECONOMICS | 19