Adobas ALOC Cadelina CRUZ Garcia Assignment 2
Electronics and Communication Engineering (BS ECE 1)
Polytechnic University of the Philippines
College of Engineering and Architecture
Department of Computer Engineering
Engineering Economics
ENSC 20093
In Partial Fulfilment
Of the Requirement for
Bachelor of Science in Computer Engineering
Submitted to:
Engr. Roland C. Viray
Submitted by:
Adobas, John Loyd C.
Aloc, Jhon Robert M.
Cadeliña, Allyza Ruth P.
Cruz, Francesca L.
Garcia, Chelsea L.
November 25, 2021 (Thursday)
0
0
1. Find the accumulated amount of the ordinary annuity paying an amortization of 1000P per
month at a rate of 12% compounded monthly for 5 years.
Given:
Required:
j = 12% = 0.12
n = 1 year = 12 month
n1 = 5 yrs.
F=?
Solution:
Get i:
i=j/n
= 12% / 12
= 1%
Then get F:
� =� (
(1+i) �−1
)
i
� = 1000 (
(1+0.01)5−1
)
0.01
� = 81669.66
81670
2. What present sum is equivalent to a series of 1000P annual end-of-year payments, if a total of
20 payments are made and interest is 12%?
Given:
Required: P = ?
A = 1000P
i = 12% annually = 0.25
n = 20 year
Solution:
P=A{
+−
+−
+−
+−
1−(1+ )^−
�
P = 1000 {
}
1−(1+0.12)^−20
0.12
}
P = 7469.44 P or 7470P
0
0
3. A man made ten annual-end-of year purchases of 1000P common stock. At the end of 10th
year he sold all the stock for 12000P. What interest rate did he obtain on his investment? 4%
Given:
F = 12,000
CFD :
F = 12000
A = 1000
n = 10
10
Required:
Interest Rate (i)=??
A = 1,000
Solution:
(1 + i)� − 1
�=� (
)
i
(1 + i)10 − 1
)
i
12000 =
1000 (
1200
(1 + i)10 − 1
= (
)
0
i
100
0
12i = (1+i)10 -1
i = 0.03989 or 4%
4.
A piece of property is purchased for 10,000P and yields a 1000P yearly profit. If the
property is sold after 5 years, what is the maximum price to break-even if the interest is 6% per
annum?
Given:
Required:
A = 10000P
n = 5yrs
PT = ?
i = 6% = 0.06
Solution:
PT = P( 1+i )n – F
F = �(
(+
+
+
1+ ) �−1
�
= 10000P ( 1+0.06)5 – 5637.09296P
)
F = 1000P �(
PT
(1+0.06) −1
0.0
5 )
PT = 7745.1628P or 7745P
F = 5637.09296P
0
0
5. A condominium unit can be bought at a down payment of 150000P and a monthly payment of
10000P for 10 years starting at the end of 5th year from the date of purchase. If money is worth
12% compounded monthly, what is the cash price of the condominium unit?
Given:
Required: P = ?
Down Payment = 150,000P
A = 10,000P
j = 12% compounded monthly = 0.12
n1 = monthly = 12
i = j/n1 = 0.12/12 = 0.01
t = 10 years
t1 = 5th year = 5
n = n1(t)+1 = 12(10)+1 = 121
m = n1(t1)–1 = 12(5)–1 = 59
Solution: draw CFD
P=?
m=5
n = 10
12345
0
0
12345678910
A
P = A{
A
A
A
A
A
1−(1 + �)
�
− } (1 + �)−−− + 150,000
1−(1 +
−121
P = 10,000
0.01)
{
0.01
}(1 + �) + 150,000
−59
P = 389,170.752 + 150,000
P = 539,170.752P or 539,171P
0
0
A
A
A
A = 10,000P
6. The owner of the quarry signs a contract to sell his stone on the following basis. The purchaser
is to remove the stone from the certain portion of the pit according to a fixed schedule of volume,
price and time. The contract is to run 18 years as follows. Eight years excavating a total of
20,000 m per year at 10P per meter, the remaining ten years, excavating a total of 50,000 m per
year at 15P per meter. On the basis of equal year end payments during each period by the
purchaser, what is the present worth of the pit to the owner on the basis of 15% interest?
Given:
Required:
8 years: 20,000m per year at 10P
10 years: 50,000m per year at 15P
P1 = ?
P2 = ?
i = 15% or 0.15
n1 = 8 years
n2 = 10 years
PT = ?
Solution:
Get each contract first: 8 and 10 years.
20,000m per year at 10P = 20,000 x 10P = 200,000P
50,000m per year at 15P = 50,000 x 15P = 750,000P
1−(1+0.15)−8
�1 = 200,000 (
�1 =
200,000 (
0.15
0.673098
0.15
)
)
�1 = 897,46
4
4
4
�2 = 750,000 (
1−(1+0.15)
�2 = 750,000 (
0.15
0.752815
0.15
−10)(1 + 0.15)−8
)(0.326902)
�2 = 1,230,483.6
5
5
5
P = P1 + P2
P = 897,46
4
4
4 + 1,230,483.6
5
5
5
P = 2,127,948P
0
0
7. A wealthy man donated a certain amount of money to provide scholarship grants to deserving
students. The fund will grant 10,000P per year for the first 10 years and 20,000P per year on the
years thereafter. The scholarship grants started one year after the money was donated. How much
was donated by the man if the fund earns 12% interest.
Given:
Required: P = ?
P1 A = 10000P per year for the first 10 years
P2 A = 20000P per year on the years thereafter
i = 12% annually = 0.12
n = 10 year
Solution:
CFD Diagram
1 2 3 4 5 6 7 8 9 10
∞
,
10000P
20000P
P1 = A {
((1+�^�
) ) −1
+^
+^
+^
+^
(1+ )^
P1 = 10000 {
}
((1+0.12)^10) −1
(1+0.12)^10
} = 56502.2303P
P2 = � = 20000 = 166666.6667P
�
0.12
�2
P3 =+^
+^
+^
+^
(1+ ^
)
P3 =
166666.6667
= 53662.2061P
(1+0.12)^10
P = P1 + P3
P = 56502.2303 + 53662.2061
P = 110,164.4364P or 110,165P
0
0
8. What amount of money deposited 40 years ago at 12% interest would now provide a perpetual
payment of 10,000P per annum? 896P
Given:
CFD :
A = 10,000
P10000
p=
0.12
i = 12% or 0.12
A = 10000
40
1
n = 40
Required:
2
3 4
�(1.12 40)
Amount of money deposited (P) = ??
Solution:
10000
= �(1.1240)
0.12
83333.33333 = P(1.12)40
�
=
83333.33333
93.05097044
P = 895.5665152 or 896P
0
0
∞
9. A company rent a building for 50,000P per month for a period of 10 years. Find the
accumulated amount of the rentals if the rental for each month is being paid at the start of each
month and money is worth 12% compounded monthly.
Given:
Required: F = ?
j = 12% = 0.12
m = monthly =
12 t = 10 years
n = m(t) = 12(10) = 120
i = j/m = 0.12/12 = 0.01
A = 50,000P
Solution: draw CFD
P
0
1
2
3
4
5
A
A
A
A
A
A
6
A
7
8
9
10
A
A
A
A = 50,000P
F=?
F = � [(1 + �)
�
− 1](1 + �)
F = 50,000 [(1 + 0.01)120 − 1](1 + 0.01)
0.01
F = 11,616,953.82P or 11,616,954P
0
0
10. The amount of the perspective investor pay for a bond if he desires an 8% return on his
investment and the bond will return 1000P per year for 20 years and 20,000P after 20 years is?
Given:
I = 1000P/yr
i = 8% = 0.08
n = 20yrs
C = 20,000
P = required
Solution:
P=I{
1−(1+◻)−◻
} + C (1+i)
◻
-n
P = 1000 {1−(1+0.08)−20} + 20000(1 + 0.08)−20
0. 0
P = 14109.1156P or 14109P
11. A machine costs 50,000P. Find the capitalized cost if the annual maintenance and operational
cost is 5000P and money worth 15% per annum.
Given:
Required:
FC = 50,000
MC = 5,000P
i = 15%
CR = 0
RC = 0
CC = ?
Solution:
��
= �� +
��
��
�
�� = 50,000 +
+
(1+
+
+ ) � −1
5,000
��
+
+
0.15
(1+
+
+) � −1
0
(1+◻)◻−1
+
0
(1+◻)◻−1
�� = 83,333.33333 �
0
0
12. A machine cost 50,000P. Find the capitalized cost if the annual maintenance cost is 5000P
and cost of repair is 4000P every 4 years and money worth 12% per annum.
Given:
Required: CC= ?
FC = 50,000P MC
= 5,000P CR =
4,000P
k=4
i = 12% annually = 0.12
L=0
SV = 0
RC = 0
Solution:
��
��
= �� +
��
+
+
(1
+
++) � −1
�
CC = 50,000 + 5000 +
0.12
4000
��
(1
+
++) � −1
+0
((1+0.12)^4) − 1
CC = 98,641.14788 or 98,641P
0
0
13. A building cost 10 million and the salvage value is 150,000P after 25 years. The annual
maintenance cost is 60,000P costs of repair is 200,000P every 5 years. Find the capitalized cost if
money worth 15% per annum.
Given:
Required:
FC= 10,000,000
Capitalized Cost (CC) = ?
SV= 150,000
L= 25 years
MC= 60,000
CR= 200,000
i = 15% or 0.15 per year
Solution:
RC = FC – SV – CR = 10,000,000 – 150,000 – 200,000 = 9,650,000
�� = 10,000,000
+
60,000
200,000
9,650,000
0.15
+
(1.15)5
(1.15)25 − 1
+
CC = 10900082.29 or 10.9M
0
0
14.
A salesman earns 1000P on the 1st month, 1500P on the 2nd month, 2000P on the 3rd
month and so on. Find the accumulated amount of his income at the 10th month if money worth
12% compounded monthly.
Given:
Required:
A = 1000P
G = 500P
F=?
n = 10
j = 12% = 0.12
i = 0.12/12 = 0.1
Solution:
draw the CFD
0K
2K
1K
3K
use, F = FG
5K
4.5K
�
�
F+
(1
++)=
−1
+ FA
G
500
(1+0.01
F
= )10−1
G
3.5K
2.5K
1.5K
4K
0.01
�
A =
{
�
−�
F
− 10
0.01
{
= 23110.6271P
(1+0.01
)10−1
F = 1000
A
(
0.01
X
)
= 10462.21254P
F = 23110.6271P + 10462.21254P
= 33572.8396P or 33573P
0
0
�(
(1
+
++) � −1
)
�
15. A man wishes to accumulate a total of 500,000P at the age of 30. On his 20th birthday, he
deposited a certain amount of money at a rate of 12% per annum. If he increases his deposit by
10% each year until the 30th birthday, how much should his initial deposit be?
Given:
Required: initial deposit = x = ?
P = 500,000P
R = 12% = 0.12
i = 10% = 0.10
n = 30th – 20th = 10yrs
i = 12% = 0.12
Solution: draw CFD
500,000P
20
0
22
X=?
24
2
4
2
X(1.1)
26
28
30
6
8
10
4
X(1.1)
X(1.1)
6
X(1.1)
8
10
X(1.1)
+
+
w = 1+
= 1.12 = 1.0182
1+
+
+
1. 1
x + PGG = 500,000(1.1)-10
x 1+.12)
�(
1.1
1−(1.0182)1
} = 500,000(1.1)-10
0
{
1−1.0182
12.0566x = 192771.6447
x =15,988.9156P or 15,989P
0
0
16. If 2000P is deposited in a savings account at the beginning of each of 15 years and the
account draws interest at 7% per year, compounded annually. Find the value of the account at the
end of 15 years.
Given:
Required:
A = 2000P
n = 15
i = 7%
i=7/1
i = 7% or 0.07
F=?
Solution:
(1+0.07)16− 1
�=
2000
(
− 1)
0.07
� = 53,776�
17. A man deposits 1000P every year for 10 years in a bank. He makes no deposit during the
subsequent 5 years. If the bank pays 8% interest, find the amount of the account at the end of 15
years.
Given:
Required: F = ?
A = 1000P per year for the last 10 years
i = 8% annually = 0.08
n = 10 years for the last 10 payments, and 5 years for the no deposit
Solution:
F=A{
+^+
+^
+^
+^
((1
+^ )^ )−1
�
F = 1000 {
}
((1+0.08)^10)−1
0.08
}
F = 14,486.56247P
F = 14,486.56247*((1 + 0.08)^5)
F = 21,285.5130 or
21 286P
0
0
18. Twenty-five thousand pesos is deposited in a savings account that pays 5% interest,
compounded semi-annually. Equal annual withdrawals are to be made from the account,
beginning one year from now and continuing forever. Find the maximum amount of the equal
annual withdrawal. 1265P/yr
Given:
P = 25000
i = 5/2% = 2.5% or 0.025 semi-annually
n=2
Required:
Maximum amount of equal annual withdrawal (F) = ??
Solution:
�=
�
�
A = Pi
A = 25000*0.025 A =
625
�
� = � (1 + i) − 1
)
(
i
(1 + 0.025)2 − 1
)
� = 625
0.025
(
(1.025)2 − 1
)
0.025
� = 625
(
F = 1265.625 or 1265P/yr
0
0
19.
What amount of money deposited 50 years ago at 8% interest would now provide a
perpetual payment of 10000P per year?
Given:
Required:
P1 = ?
P=?
A = 10000P/yr
i = 8% = 0.08
t = 50yrs
Solution:
draw the CFD
50
1
2
P(1.08)50
∑ =∑
P1 = 10000 / 0.08
= 12500P
P = P1 / (1.08)50
= 2665.15P or 2665P
0
0
3
20. A man buys a motor cycle. There will be no maintenance cost the first year as the motor
cycle is sold with one year free maintenance. The 2nd year the maintenance is estimated at
2000P. In subsequent years the maintenance cost will increase by 2000P per year. How much
would need to be set aside now at 5% interest to pay the maintenance costs of the motor cycle for
the first 6 years of ownership?
Given:
Required: P = PG = ?
i = 5% = 0.05
n=6
G = 2,000P
Solution: draw CFD
0
1
2
3
4
5
6
0
2,00
0
4,000
6,000
8,000
10,000
�
+
+
1−(1+ )
−
−} − �(1 + �)−
P=P = {
G
�
2,000
P = 0.05 {
�
1−(1.05)−6
}
0.05
− 6(1.05)−6
P = 23,935.9875P or 23,936P
0
0