Econ 101E (Module 3) Hand-out
I. Time-Value of Money(Interest) and Equivalence
Capital refers to wealth in the form of money or property that can be used to produce more wealth.
The majority of engineering economy studies involve commitment of capital for extended periods of
time, so the effect of time must be considered.
It is recognized that a dollar today is worth more than a dollar one or more years from now because
of the interest (or profit) it can earn. Therefore, money has a time value. It has been said that often the
riskiest thing a person can do with money is nothing! Money has value, and if money remains
uninvested (like in a piggy bank or hidden under the bed), value is lost.
Interest – the amount paid for the use of capital.
Two types of computation:
1. Simple interest
2. Compound interest
1. Simple Interest
– type of interest computation where only the principal or amount borrowed bears or earns
interest.
Formula :
I = (P)(i)(n)
where: I = total interest
P = principal amount lent or borrowed
i = interest rate per interest period
n = number of interest period (e.g., years)
Let F = future amount to be paid
F=P+I
F = P + Pin
F = P(1 + in)
Cash flow representation:
i
●
│
P
I
n
●
│
F
Type of simple interest computation:
1. Exact simple interest: n = 365 days in a year (use when specific times or dates are considered)
2. Ordinary simple interest : n = 360 days in a year (known as bankers discount)
Example 1) Php1,000.00 were loaned for three years at a simple interest rate of 10% per year. How
much interest will be paid at the end of the third year?
Solution: I = (P)(i)(n)
I = (1,000)(0.10)(3)
I = 300
Example 2) Determine the exact simple interest on Php100,000.00 for 9 months and 10 days if the rate
of interest is 12%.
Solution: I = P(i)(n)
I = 100,000(0.12)(9x30+10)/365
I = 9,205.48
Example 3) Bob borrowed Php12,000.00 from a lending company for 18 months. If the lending
company charges 18% simple interest per year, find the total amount that Bob will have to pay
back.
Solution: F = P + I
F = 12,000 + 12,000(0.18)(18/12)
F = 15, 240
Example 4) Mr. David has Php10,000.00 and invests Php3,000.00 for one year at 8% simple interest.
At what rate of simple interest should the remainder be invested if the estimated total income from
the Php10,000.00 is Php660.00?
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Solution: IT = I1 + I2
660 = 3,000(0.08)(1) + (10,000 – 3,000)(i)(1)
i = (660 – 240)/(7,000)
i = 0.06 of 6%
Practice Problems:
1. John deposited Php10,000.00 in a 26-week certificate of deposit that paid 14.9% ordinary simple
interest. How much interest will be receive at the end of 26 weeks?
a)Php1,506.56 b) Php753.28
c) Php10,753.28 d) Php11,506.56
2. You borrowed Php100,000.00 from a lending institution that charges a 20% simple interest rate.
The interest was deducted from the loan at the time the money was borrowed. If at the end of one
year, you have to pay the full amount of Php100,000.00 what is the actual rate of interest?
a) 25.0%
b) 27.5%
c) 30.0%
d) 18.8%
3. A man invested part of Php20,000.00 at 18% simple interest and the rest at 16%. The annual income
from the 16% investment was Php 620.00 less than three times the annual income of the 18%
investment. How much was invested at 18%?
a)Php5,546.45 b) Php 4,547.43 c) Php5,475.13 d) Php5,457.14
4. A time deposit of Php110,000.00 for 31 days earns Php890.39 on maturity date after deducting the
20% withholding tax on interest income. Find the interest rate per annum.
a)12.5%
b) 12.25%
c) 11.75%
d) 11.95%
5. Mr. Reyes borrowed from the bank. He received Php1,842.00 and promise to repay Php2,000 at
the end of ten months. Determine the simple interest rate.
a) 19.45%
b) 15.70%
c) 16.10%
d) 10.29%
2. Compound Interest
– type of interest computation where the principal or amount borrowed and the interest earned will bear
or earn another interest.
Cash flow representation:
●
│
P
1
2
3 . .
●
│
F1
●
│
F2
●
●
│
│
F3 . . . Fn
n
F1 = P(1 + in) ; n = 1
F1 = P(1 + i)
F2 = F1(1 + in); n =1
F2 = F1 (1 + i)
F3 = F2(1 + in); n =1
F3 = F2 (1 + i)
Therefore: F3 = P(1 + i) (1 + i) (1 + i)
Fn = P(1 + i)n
The quantity (1 + i)n is commonly called the single payment compound amount factor. The
functional symbol of (1 + i)n is (F/P, i%, n).
Therefore the equation can be expressed as F = P (F/P, i%, n) where the factor in parentheses
is read “find F given P at i% interest per period for n interest periods.”
Example 2.1) Find the compound amount and the compound interest on Php2,500.00 invested for 10
years at 16% compounded annually.
Solution: F10 = P(1 + i)10
I=F-P
F10 = 2,500(1 + 0.16)10
I = 11,028.59 – 2,500
F10 = 11,028.59
I = 8,528.59
Example 2.2) If Php1,000.00 were invested at 6% compounded annually, how long will it take for the
investment to double itself?
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Fn = P(1 + i)n
2,000 = 1,000(1.06)n
n = [ ln(2,000/1,000)]/ln1.06
n = 11.896 years
Types of Interest Compounding:
1. effective interest (ieff) – compounding is one year
F = P(1 + ieff)n
where: F = future sum or amount
P = principal amount lent or borrowed
ieff = interest rate compounded per year
n = number of years
2. nominal interest (r) – compounding is less than one year
F = P(1 + r/m)mn
where: r = nominal interest rate
m = number of compounding in a year
ex. For semi-annual compounding: m = 2
For quarterly compounding: m = 4
For monthly compounding: m = 12
Solution:
Relationship between effective and nominal interest rate:
effective
nominal
n
F = P(1 + ieff)
F = P(1 + r/m)mn
For n = 1:
(1 + ieff) = (1 + r/m)m
ieff = (1 + r/m)m – 1
3. continuous compounding – daily compounding
From nominal interest,
F = P( 1 + r/m)mn
For n = 1, let: r/m = 1/p
pr = m
substituting: F = P( 1 + 1/p)pr
The limit of ( 1 + 1/p)p as p∞ is e = 2.71828 . . . . (the Naperian base of natural logarithm)
Therefore: F = Pern
Relationship between effective and continuous compounding rate:
effective
continuous
F =
For n = 1:
P(1 + ieff)n
F = Pern
(1 + ieff) = er
ieff = er – 1
Example 2.3) A company loaned Php500,000.00 from a lending institution to expand its manufacturing
facilities. What is the effective interest rate and the amount to be paid after five years, if interest
rate is:
a) 15% compounded monthly;
b) 15% compounded continuously.
Solution:
a) ieff = (1 + r/m)m – 1
b) ieff = er – 1
12
ieff = (1 + 0.15/12) – 1
ieff = e0.15 – 1
ieff = 0.160754517
ieff = 0.161834242
F = P(1 + ieff)n
F = 500,000(1 + 0.160754517)5
F = 1,053,590.67
or
F = P(1 + r/m)mn
F = 500,000(1 + 0.15/12)12(5)
F = 1,053,590.67
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F = P(1 + ieff)n
F = 500,000(1 + 0.161834242)5
F = 1,058,500.01
or
F = Pern
F = 500,000e0.15(5)
F = 1,058,500.01
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II. The Concept of Equivalence
"Economic equivalence is established, in general, when we are indifferent between a future
payment, or series of future payments, and a present sum of money." Economic equivalence is a
combination of interest rate and time value of money to determine the different amounts of
money at different points in time that are equal in economic value.
As an illustration, if the interest rate is 6% per year, $100 today (present time) is equivalent to $106
one year from today.
Future amount = 100 + 100(0.06) = 100(1 + 0.06) = $106
If someone offered you a gift of $100 today or $106 one year from today, it would make no
difference which offer you accepted from an economic perspective. In either case you have $106 one
year from today. However, the two sums of money are equivalent to each other only when the interest
rate is 6% per year.
At a higher or lower interest rate, $100 today is not equivalent to $106 one year from today.
Case 1: i = 5%
F = 100(1 + 0.05) = 105 < 106
Case 2: i = 7%
F = 100(1 + 0.07) = 107 > 106
In addition to future equivalence, we can apply the same logic to determine equivalence for previous
years.
Present amount now, P = Amount one year ago, P-1 ( 1 + i) ; if P = 100 and i = 0.06
100 = P-1 ( 1 + 0.06)
P-1 = 100/(1.06) = 94.34
.From these illustrations, we can state the following: $94.34 last year, $100 now, and $106 one year
from now are equivalent at an interest rate of 6% per year. The fact that these sums are equivalent can
be verified by computing the two interest rates for 1-year interest periods.
Example: Manufacturers make backup batteries for computer systems available to Batteries+ dealers
through privately owned distributorships. In general, batteries are stored throughout the year, and a 5%
cost increase is added each year to cover the inventory carrying charge for the distributorship owner.
Assume you own the City Center Batteries_ outlet. Make the calculations necessary to show which of
the following statements are true and which are false about battery costs.
(a) The amount of $98 now is equivalent to a cost of $105.60 one year from now.
(b) A truck battery cost of $200 one year ago is equivalent to $205 now.
(c) A $38 cost now is equivalent to $39.90 one year from now.
(d) A $3000 cost now is equivalent to $2887.14 one year earlier.
(e) The carrying charge accumulated in 1 year on an investment of $20,000 worth of batteries is $1000.
Solution:
(a) Total amount accrued _= 98(1.05) = $102.90 ≠ $105.60; therefore, it is false.
Another way to solve this is as follows: Required original cost is 105.60/1.05 = $100.57 ≠ $98.
(b) Equivalent cost 1 year ago is 205.00/1.05 = $195.24 ≠ $200; therefore, it is false.
(c) The cost 1 year from now is $38(1.05) = $39.90; true.
(d) Cost now is 2887.14(1.05) = $3031.50 ≠ $3000; false.
(e) The charge is 5% per year interest, or $20,000(0.05) = $1000; true.
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Practice Problems:
1. A loan of Php50,000.00 taken now, is to be paid at the end of 3 years. If interest rate is 10%
compounded quarterly, how much is the final payment?
a) Php66,550.00
b) Php67,244.44 c)Php156,921.42
d)Php67,492.94
2. A 40-year old engineer wants to set-up a retirement fund to be used at the age of 65. Php20,000.00 is
invested now which earns 12% compounded annually? How much is his retirement fund at his 65th
birthday?
a) Php395,769.33 b) Php384,372.64
c)Php389,992.06 d)Php340,001.29
3. A firm desires to have P1,000,000 five years from now to buy a certain equipment. What amount should
be invested now to provide for this amount if an investment that earns 15% interest compounded semiannually is being offered in the money market?
a) Php476,742.69 b) Php478,892.34
c)Php485,193.93 d)Php497,176.74
4. At a certain interest rate compounded quarterly, Php10,000.00 will amount to Php45,000.00 in 15
years. What is the corresponding amount at the end of 10 years?
a) Php27,257
b) Php25,272
c) Php27,522
d) Php25,722
5. Find the nominal rate compounded monthly which is equivalent to 12% compounded quarterly.
a) 8.112%
b) 8.821%
c) 11.882%
d) 18.82%
References:
1. Blank, L., & Tarquin, A. (2018). Engineering Economy (8th ed.). McGraw-Hill Education.
2. Sullivan, W. et al, (2015). Engineering Economy (16th ed.), Pearson Education-South Asia Pte. Ltd
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