Compilation Test 2

COEB 442: ENGINEERING ECONOMICS

SEMESTER 2, 2014/2015 TEST 2

ANSWER ALL QUESTIONS (TIME: 1.5 HOURS)

WRITE YOUR ANSWERS ON THE SPACE PROVIDED

Section A: True / False Questions

1. An initial amount of money in transactions involving debt or investments is called the

principal. True

2. Simple interest is interest earned on only the principal amount during each interest period.

True

3. Economic equivalence exists when two cash flows could be traded for one another because

there have the same amount. False

4. The process of finding future value from a present value is called discounting process. False

5. An annuity is a series of uniform amounts over a period of time. True

6. The symbol for present worth is P and it always represents a cash inflow. False

7. The annual percentage yield (APY) does not explain precisely the amount of interest that

will accumulate in a year. False

8. When compounding takes place once annually, the effective interest is equal to the

nominal interest. True

9. If loan is to be repaid in equal periodic amount (e.g.monthly) it is said to be an amortized

loan. True

10. In a typical amortized loan, the amount of interest owed for a specified period is calculated

on the basis of the remaining balance on the loan at the beginning of the period. True

ANSWERS-SECTION A

1. 2. 3. 4. 5.

6. 7. 8. 9. 10.

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Section B: Problem Solving Questions [30 marks]

QUESTION 1 [4 marks] Chapter 3

Find the equivalent value of the savings at year 5 (F) with interest at 5% per year.

F=?

-5 -4 -3 -2 -1 0 1 2 3 4 5

RM600 RM300 RM400

ANSWERS

F=600(F/P,5%,10) + 300 (F/P,5%,8) + 400(F/P,5%,5) [3 marks]

F=600(1.6289) + 300(1.4775) + 400(1.2763) = RM1931.11 [1 mark]


Keith manufacturing, a glue producing company borrowed RM120, 000 at a simple interest of 10% per year for six years to expand its facilities.

How much simple interest would be repaid as a lump - sum amount at the end of year 6 when half of the loan principal is repaid at the end of year 3 and the other half will be repaid in one lump –sum amount at the end of year 6? (Hint: Re calculate the interest at year 4 to year 6)

ANSWERS

YEAR AMOUNT OWED AT BEGINNING OF YEAR-RM

INTEREST ACCRUED FOR YEAR- RM

TOTAL AMOUNT OWED AT END OF YEAR-RM

PRINCIPAL PAYMENT-RM

1 120,000 12,000 132,000 02 120,000 12,000 132,000 03 120,000 12,000 132,000 60,0004 60,000 6,000 66,000 05 60,000 6,000 66,000 06 60,000 6,000 66,000 60,000

Total simple interest : (RM12,000*3)[2 marks] + (RM6,000*3) [2 marks] = RM54,000[1 mark]

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A future amount of RM200,000 is to be accumulated through annual payments , A, over 20years so that a cash payment can be made to settle for a housing loan. If the interest rate is 9%, what is the value of A?

ANSWERS

A = RM200,000 (A/F, 9%,20) [2 marks]= 200,000(0.0195) = RM3900 [1 mark]


Determine the current amount of money that must be invested now at 12% interest to provide an annuity of RM10,000 (per year) f or six years, starting 12 years from now.

ANSWERSP11 = RM10,000 (P/A,12%, 6) = 10,000(4.1114)= 41114 [3 marks]

P0 = 41114(P/F,12%,11) = 41114(0.2875) = RM11820.275 [3 marks]


Peter borrows RM50,000 to purchase a machine for producing mincemeat. He must repay the loan in 36 equal monthly payments. Interest is 0.5% per month. Calculate:

(a) the nominal annual interest rate(b) the annual percentage yield (effective annual interest rate)(c) the amount of the monthly payment(d) the lump sum amount to be paid after the 20th payment has been made if Peter decided to pay

off the loan

ANSWERS(a) r = i x m = 0.5% x 12 = 6% [2 marks](b) (1 + 0.005)12 -1 = 6.168% [3 marks](c) $50,000(A/P, 0.5%,36) = 50,000(0.0304) = RM1520 [3 marks](d) 1520(P/A,0.5%,16) = 1520(15.3399) = RM23316.648 [4 marks]

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COEB 442: ENGINEERING ECONOMICS

SEMESTER 1, 2014/2015, TEST 2

ANSWER ALL QUESTIONS (TIME: 1.5 HOURS)

WRITE YOUR ANSWERS ON THE SPACE PROVIDED

Section A: True / False Questions [1 mark each]

1. An initial amount of money in transactions involving debt is called the principal. T

2. With simple interest, the interest earned during each interest period does not earn

additional interest in the remaining periods. T

3. Economic equivalence exists between cash flows that have the same economic effects but

could not be traded for one another. F

4. Finding the present worth of a future sum is known as the compounding process. F

5. Interest is a cost to the borrower. T

6. Money has a time value because it can earn more money over time. T

7. Wood is a direct raw material in furniture. T

8. Total costs will increase only if both fixed costs and variable costs increase. F

9. As volume increases, the fixed cost per unit will increase. F

10. Product costs are costs charged to expenses in the period in which they are incurred. F

ANSWERS FOR SECTION A

1 2 3 4 5

6 7 8 9 10

Section B: Problem Solving Questions


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StayHealthy Manufacturing produces jogging shoes. It has a fixed costs of RM600,000. In addition, it costs RM60 to produce each pair of shoes. The company has to produce 6000 pairs of shoes to breakeven.

(a) What is the minimum price per pair the company should charge? [3 marks]

600,000 + (6000)(60) = P(6000) ; P = RM160 per pair of shoes.

(b) Calculate the marginal contribution and interpret your answer. [3 marks]

160-60= 100, RM100 [2 m]; RM100 to absorb fixed costs and profits

(c) StayHealthy Sales Department has established a sales goal of 12000 pairs of shoes. If they achieve their sales goal, what is their total profit? [3 marks]

TR – TC = ProfitTC = 600,000 + 60(12,000) = RM1,320,000

TR = 160(12,000)= RM1,920,000

Profit = RM1,920,000 - RM1,320,000 = RM600,000

(d) Label the fixed cost , the breakeven quantity and the profit at 12,000 pairs of shoes on the cost- volume -profit graph below. [3 marks]

0.00

TR

1,320,000.00

Cost-Volume-Profit Breakeven Graph

QUANTITY

REVE

NUE

AND

EXPE

NSES

(RM

)

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Norbaini Abdul, 08/11/14,

0 6,000 12,0000.00

500,000.00

1,000,000.00

1,500,000.00

2,000,000.00

2,500,000.00

0.00

BE=960,000.00

1,920,000.00

FC=600,000.00

1,320,000.00

Cost-Volume-Profit Breakeven Graph

Sales Revenue Cost of Sales Fixed ExpensesBreakeven

Rev

enue

s an

d E

xpen

ses

(e) What is the effect on profit at 12,000 pairs of shoes if the fixed costs increases by RM170,000 and variable costs reduce by RM10? Calculate to answer. [3 marks]

FC = 600,000 + 170,000 = RM770,000; VC per unit = 60 – 10 = RM50 per unitProfit = TR – FC – VC 160(12,000) -770,000 - 50(12,000) = RM550,000 RM600,000 – RM550,000 = RM50,000Profit decline by RM50,000.


a) A small manufacturing company took RM20,000 loan from a bank at a simple interest rate of 12% per year for six years.

i. How much should be paid at the end of the sixth year? [3 marks] F = 20,000 + 20,000(0.12)(6) = 20,000 + 14,400 = RM34,400

ii. How much interest would be repaid as a lump sum amount at the end of the sixth year? [1 mark]

Interest = RM34,400 – RM20,000 = RM14,400

iii. Draw the cash flow diagram. [1 mark]

F=34,400

0 6

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P=20,000

b) A certain machine has an investment cost of RM150, 000, a life of 10 years and no resale value. Annual costs of materials, maintenance and electric power for the machine are expected to total RM20,000. A major servicing of the machine will occur once in year 5 at a cost of RM30,000. If the interest rate is 15% per year, what is the lump-sum cost of this machine at the present time? [5 marks]

150,000 + 20,000(P/A,15%,10) + 30,000(P/F,15%,5)

150,000 + 20,000(5.0188) + 30,000(0.4972) = RM265,292

c) You are considering whether to invest in an asset. The investment will cost you RM3,500 now. The expected annual rental income of the asset for 6 years is as in the table below. At an interest rate of 9%, should you invest in this asset? [5 marks]

YEAR ANNUAL RENTAL INCOME (RM)

1 800

2 700

3 600

4 750

5 750

6 750

800(P/A,9%,3) -100(P/G,9%,3)+750(P/A,9%,3)(P/F,9%,3)

800(2.5313) – 100(2.3860) + 750(2.5313)(0.7722)

2025.04 – 238.6 + 1466.00 = RM3,252.44 ]

The present equivalent of the rental income RM3,252.44 is LESS than the present equivalent investment of RM3,500, so the rental asset is NOT a good investment.

NOTE: 800(P/A,9%,3) -100(P/G,9%,3) can be replaced as:

800(P/F,9%,1)+ 700(P/F,9%,2) +600(P/F,9%,3)

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Compilation Test 2

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