Engineering Economics

(TEMPLET TUGASAN - VERSI BAHASA MALAYSIA) (MUKA SURAT HADAPAN)

IJAZAH SARJANA MUDA DALAM PENGURUSAN TEKNOLOGI

SEPTEMBER 2013

EBMC 3103

ENGINEERING ECONOMICS

NO. MATRIKULASI : 810128016402002

NO. KAD PENGNEALAN : 810128-01-6402

NO. TELEFON : 07-7716368

E-MEL : [email protected]

PUSAT PEMBELAJARAN : JOHOR LEARNING CENTRE

EBMC 3103 ENGINEERING ECONOMICS

QUESTION 1

(a) The idea that money available at the present time is worth more than the same amount in

the future due to its potential earning capacity. This core principle of finance holds that provided money can earn interest, any amount of money is worth more the sooner it is received.Also referred to as “ Present Discounted Value”. Everyone knows that money deposited in a savings account will earn interest. Because of this universal fact, we would prefer to receive money today rather than the same amount in the future.For example, assuming a 5% interest rate, $100 invested today will be worth $105 in one year ($100 multiplied by 1.05). Conversely, $100 received one year from now is only worth $95.24 today ($100divided by 1.05), assuming s 5% interest rate.

(b)

P = 400n= 5 YEARSi= 10%F = ?

i= 10%

RM400 RM400 RM400 RM400 RM400

0 1 2 3 4 5

F = ?

1


(c)

A Nominal Interest Rate, r, can be defined as an interest Rate that does not include any consideration of compounding Nominal means, “in name only”, not the real rate in this case. Nominal interest rate is rate which includes inflation.

• Mathematically, we have the following definition:

r = (interest rate per period) x (No. of Periods)

An examples of Nominal Interest Rate are as follow:-

• 1.5% per month for 24 months– Same as: (1.5%) (24) = 36% per 24 months

• 1.5% per month for 12 months– Same as (1.5%) (12 months) = 18%/year

• 1.5% per month for 6 months– Same as: (1.5%) (6 months) = 9%/6 months or semi-annual period

• 1% per week for 1 year– Same as: (1%)(52 weeks) = 52% per year

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QUESTION 2

(a)

A measure of the decrease in value of an asset over a specific period of time in the certain asset life. This usually pertains to property such as asset, real estate, vehicle and etc. that can lose value due to indirect causes such as a decline in the quality or other external factors. It could be no market value after a certain life period. Or Our text book explain: It is the anticipated period of time in which the investment depreciates from the present cost to a salvage value or resale value.

(b) (i) i = 10%

RM22000 RM22000 RM2200 RM22000 RM22000

0 1 2 3 4 5

RM1516.31

(b) (ii)

Calculating the initial investment

Calculating the present value

Justified amount to purchase equipment $83,397.31

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QUESTION 3

(a)

Calculate the effective rate per semi-annual period.

Calculating effective rate:

r = (1+i/n)^n - 1 = (1 + 0.24/2)^2 - 1= 25.44%

(b)

Semi Annual compundingr = (1+i/n)^n - 1= (1 + 0.20/2)^2 - 1= 21.00%

Annual Compounding

r = (1+i/n)^n - 1= (1+0.20/1) ^1-1= 20.00%

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QUESTION 4

(i)

(ii)

G = - ((105-47500)/7) = - 7500 m.u.

A' = 105

n=8 & i=10%F=F'+F*= A'(F/A, 10, 8) + G(F/G, 10, 8)

=105((1.1

8-1)/0.1) – 7500/0.1(((1.1

8-1)/0.1)-8)

= 885897.2

1 2 3 4

0

5

i= 10 %

6 7

years

8

F = ??

100 000 m.u.

47 500 m.u.

5


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Engineering Economics

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