Adobas ALOC Cadelina CRUZ Garcia Assignment 2 PDF

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Polytechnic University of the PhilippinesCollege of Engineering and ArchitectureDepartment of Computer EngineeringEngineering EconomicsENSC 20093In Partial FulfilmentOf the Requirement forBachelor of Science in Computer EngineeringSubmitted to:Engr. Roland C. ViraySubmitted by:Adobas, John Loyd C.Al...

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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)

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

}

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

F = 5637.09296P

5 )

PT = 7745.1628P or 7745P

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

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

A

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

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

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 n = 40 Required: 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

40 1

2

3 4

�(1.12 40)

∞

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

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

�� = 83,333.33333 �

(1+ + +) � −1 0

(1+◻)◻−1

+

0 (1+◻)◻−1

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

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

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

0.01

G

�

{

− 10 0.01

{

= 23110.6271P (1+0.01 )10−1 F = 1000 A

4.5K

� � F+ (1 ++)= −1

+ FA

500 (1+0.01 F = )10−1 G

3.5K

2.5K

1.5K

4K

(

0.01

)

= 10462.21254P F = 23110.6271P + 10462.21254P

= 33572.8396P or 33573P

5K X

A =

�

−�

F

�(

(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

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

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

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 P(1.08)50

∑ =∑ P1 = 10000 / 0.08 = 12500P P = P1 / (1.08)50

= 2665.15P or 2665P

2

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...