1

ENGINEERING ECONOMY AND ACCOUNTING FOR

ENGINEERS

SME 4833

EXERCISE CHAPTER 2

Prepared by:

KHAIRUL RAZMIN B. ABDURAKMAN BACHELOR OF MECHANICAL ENGINEERING

SX105558MMD03 SPACE PENANG

2

PROBLEM 1

SOLUTION

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (year)

= + + + +

= 10,000 = ( , %,) = 2,000( , 12%, 15 ) = 2,000(6.8109) = 13,621.8 = ( , %,) = 1,000( , 12%, 15 ) = 1,000(6.8109) = 6,810.9 = ( , %,) = 2,000( , 12%, 15 ) = 2,000(0.1827) = 365.4 = ( , %,)( , %,) = 60( , 12%, 11 )( , 12%, 4 ) = 60(23.1288)(0.6355) = 886.48

= 10,000 13,621.8 6,810.9 365.4 886.48 = 31,684.58

, = ,. #

P = 10,000 F = 2,000

Am = 1,000

Ao = 2,000 G = 60

i = 12%

3

PROBLEM 2

SOLUTION

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 (year)

= ( , %,) = + + + +

= 20,000 = ( , %,) = 80,000( , 12%, 14 ) = 80,000(6.6282) = 530,256.00 = ( , %,) = 3,000( , 12%, 14 ) = 3,000(6.6282) = 19,884.60 = ( , %,) = 8,000( , 12%, 14 ) = 8,000(31.3624) = 250,899.20 = ( , %,)( , %,) = 500( , 12%, 13 )( , 12%, 1 ) = 500(28.7024)(0.8929) = 12,814.19

= 20,000 + 530,256.00 19,884.60 + 250,899.20 12,814.19 = 728,456.41 = 728,456.41( , 12%, 14) = 728,456.41(0.15087) = 109,902.22

, = ,. #

P = 20,000

A2 = 3,000

A1 = 80,000

G 2= 500

G 1= 8,000

i = 12%

4

PROBLEM 3

SOLUTION

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 (year)

a)

= + + + + + +

= 5,000 = ( , %,)( , %,) = 20,000( , 12%, 19 )( , 12%, 1) = 20,000(7.3658)(0.8929) = 131,538.46 = ( , %,) = 5,000( , 12%, 8 ) = 5,000(4.9676) = 24,838.00 = ( , %,) = 15,000( , 12%, 7 ) = 15,000(0.4523) = 6,784.50 = ( , %,)( , %,) = 8,000( , 12%, 11 )( , 12%, 8) = 8,000(5.9377)(0.4039) = 19,185.90 = ( , %,)( , %,) = 1,000( , 12%, 11 )( , 12%, 8 ) = 1,000(23.1288)(0.4039) = 9,341.72 = ( , %,) = 8,000( , 12%, 20 ) = 5,000(0.1037) = 518.50

= 5,000.00 + 131,538.46 24,838.00 6,784.50 19,185.90 9,341.72 518.50 = 65,869.84

, = ,. #

P = 5,000 A2 = 5,000

A1 = 20,000

G = 1,000 15,000

5,000

A3 = 8,000

i = 12%

5

b)

= ( , %,) = 65,869.84( , 12%, 20) = 65,869.84(0.13388) = 8,818.65

, = ,. #

c)

= ( , %,) = 65,869.84( , 12%, 12) = 65,869.84(3.8960) = 256,628.90

, = ,. #

d) According to the calculation from (c), the expected value of the project for year 12 is RM 256,628.90. If some company offered to buy the project at that year at the price of RM 25,000.00, the offer should be rejected because the price offered was below from the expected value.

6

PROBLEM 4

SOLUTION

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 (year)

i)

= + + + + + +

= 100,000.00 = ( , %,) = 10,000( , 12%, 20 ) = 10,000(7.4694) = 74,694.00 = ( , %,)( , %,) = 2,000( , 12%, 19 )( , 12%, 1) = 2,000(7.3658)(0.8929) = 13,153.85 = ( , %,) = 2,000( , 12%, 20 ) = 2,000(44.9676) = 89,935.20 = ( , %,)( , %,) = 1,000( , 12%, 19 )( , 12%, 1 ) = 1,000(42.9979)(0.8929) = 38,392.82 = ( , %,) = 40,000( , 12%, 10 ) = 40,000(0.3220) = 12,880.00 = ( , %,) = 8,000( , 12%, 20 ) = 20,000(0.1037) = 2,074.00

= 100,000.00 + 74,694.00 13,153.85 + 89,935.20 38,392.82 12,880.00 + 2,074.00 = 2,276.53

, = ,. #

P = 100,000

A1 = 10,000

G2 = 1,000

40,000 A2 = 2,000

F = 20,000

G1 = 2,000

i = 12%

7

ii)

= ( , %,) = 2,276.53( , 12%, 20) = 2,276.53(0.13388) = 304.78

, = . # iii)

= ( , %,) = 2,276.53 10,000.00( , 12%, 15) = 2,276.53 10,000(0.1827) = 449.53

= . ,. . iv)

i1-10= 12%, i11-20 = 15%

= + + + + + +

= 100,000.00 = ( , %,) + ( , %,)( , %,) = 10,000( , 12%, 10 ) + 10,000( , 15%, 10 )( , 15%, 10) = 10,000(5.6502) + 10,000(5.0188)(0.2472) = 68,908.47 = ( , %,)( , %,) ( , %,)( , %,) = 2,000( , 12%, 9 )( , 12%, 1) 2,000( , 15%, 10 )( , 15%, 10) = 2,000(5.3282)(0.8929) 2,000(5.0188)(0.2472) = 11,996.39 = ( , %,) + ( , %,)( , %,) = 2,000( , 12%, 10 ) + 2,000( , 15%, 10 )( , 15%, 10) = 2,000(20.2541) + 2,000(16.9795)(0.2472) = 48,902.86 = ( , %,)( , %,)( , %,)( , %,) = 1,000( , 12%, 9 )( , 12%, 1 ) 1,000( , 15%, 10 )( , 15%, 10 ) = 1,000(17.3563)(0.8929) 1,000(16.9795)(0.2472) = 19,694.77 = ( , %,) = 40,000( , 12%, 10 ) = 40,000(0.3220) = 12,880.00 = ( , %,) = 20,000( , 15%, 20 ) = 20,000(0.0611) = 1,222.00

= 100,000.00 + 68,908.47 11,996.39 + 48,902.86 19,694.77 12,880.00 + 1,222.00 = ,. The performance of the machine become worsen after the growth rate increasing to 15% after year 10 because the present value given a negative sign means the machine is unprofitable.

8

PROBLEM 5

SOLUTION

0 1 2 3 4 5 6 7 8 9 10 (year)

0 1 2 3 4 5 6 7 8 9 10 (year)

P = 50,000,000

A2 = 2,000,000

A1 = 40,000,000

G 2= 120,000

G 1= 1,000,000 F = 5,000,000

8,000,000

P = 50,000,000

A2 = 2,000,000

A1 = 40,000,000

G 2= 120,000

G 1= 1,000,000

F = 5,000,000

8,000,000

i = 12%

i = 12%

9

)i

+ + + + + + =

00.000,000,5 = 02.199,103,091 = )9298.0()2823.5(000,000,04 = )1 ,%21 , () 9 ,%21 , (000,000,04 = ),% , (),% , ( = 00.004,003,11 = )2056.5(000,000,2 = ) 01 ,%21 , (000,000,2 = ),% , ( = 72.044,794,51 = )9298.0()3653.71(000,000,1 = )1 ,% , () 9 ,%21 , (000,000,1 = ),% , (),% , ( = 00.294,034,2 = )1452.02(000,021 = ) 01 ,%21 , (000,021 = ),% , ( = 00.002,935,4 = )4765.0(000,000,8 = ) 5 ,%21 , (000,000,8 = ),% , ( = 00.000,016,1 = )0223.0(000,000,5 = ) 01 ,%21 , (000,000,5 = ),% , ( =

09.854,441,351 = 00.000,016,1 + 00.002,935,4 00.294,034,2 72.044,794,51 00.004,003,11 02.199,103,091 + 00.000,000,5 =

.,, = ,

)ii

03.904,401,291 = )4452.1(09.854,441,351 = )2 ,%21 , (04.895,859,841 = ),% , ( =

#.,, = ,

)iii

= 000,000,2 = )000,000,1()2( = )()1 3( = )( = 000,042,2 = 000,000,2 + )000,021()2( = + )()1 3( = )(

+ + + + =

10

= ( , %,) = 40,000,000( , 12%, 7 ) = 40,000,000(4.5638) = 182,552,000.00 = ( , %,) = 2,000,000( , 12%, 7 ) = 2,000,000(4.5638) = 9,127,600.00 = ( , %,) = 2,240,000( , 12%, 7 ) = 2,240,000(4.5638) = 10,222,912.00 = ( , %,) = 1,000,000( , 12%, 7 ) = 1,000,000(11.6443) = 11,644,300.00 = ( , %,) = 120,000( , 12%, 7 ) = 120,000(11.6443) = 1,397,316.00 = ( , %,) = 8,000,000( , 12%, 3 ) = 8,000,000(0.7118) = 5,694,400.00 = ( , %,) = 5,000,000( , 12%, 7 ) = 5,000,000(0.4523) = 2,261,500.00

= 182,552,000.00 9,127,600.00 10,222,912.00 11,644,300.00 1,397,316.00 5,694,400.00 + 2,261,500.00 = 146,726,972.00

, = ,,.#

iv)

= ( , %,) = 148,958,598.40( , 12%, 7) = 153,144,458.90(2.2107) = 338,556,455.30 = ,,.

From the calculation above, found that the value of project at year 7 of project life is RM 338,556,455.30. If some company offered to buy the project with price of RM 150 million, the offer should be rejected because the offered price was below from the expected value.

11

PROBLEM 6

SOLUTION

A Alternative:

0 1 2 3 4 5 6 7 8 9 10 (year)

() = 12.90%2 = 6.45% = 0.0645 () = 6.45% = 0.0645

For every 6 month payment, n = 2 X 10 = 20

= ( , %,) = 50,000( , 6.45%, 20)

i% A/P, n = 20 6.00 0.08718 6.45 x 7.00 0.09439

0.09439 0.09439 0.08718 = 7.00 6.457.00 6.00 0.09439 0.00721 = 0.551.00 = 0.09439 0.55(0.00721) = 0.0904245

= ( , %,) = 50,000(0.0904245) = 4,521.23 = ,.

P = 50,000

AA

12

B Alternative:

0 1 2 3 4 5 6 7 8 9 10

() = 12.85%4 = 3.2125% = 0.032125 () = 3.2125% 2 = 0.06425 () = 1 + 0.064252 1 = 0.065282 = 6.5282%

For every 6 month payment, n = 2 X 10 = 20

= ( , %,) = 50,000( , 6.45%, 20)

i% A/P, n = 20 6.00 0.08718

6.5282 x 7.00 0.09439

0.09439 0.09439 0.08718 = 7.00 6.52827.00 6.00 0.09439 0.00721 = 0.47181.00 = 0.09439 0.4718(0.00721) = 0.0909883

= ( , %,) = 50,000(0.0909883) = 4,549.42 = ,.

From the calculation above, found that the A alternative is more economic compare to the B alternative because A alternative given the lowest installment for every 6 month payment method.

P = 50,000

AB

13

PROBLEM 7

SOLUTION

0 1 2 3 4 5 6 7 8 9 10 11 12 (month)

() = 12%12 = 1% = 0.01 () = 1 + 0.112 1 = 0.104713 = 1.04713% () = 1.04713% 12 = 12.56556%

For 1st year outstanding debt, n = 1

= ( , %,) = 1,500( , 12.56556%, 1)

i% F/P, n = 1 12.00 1.1200

12.56556 x 14.00 1.1400

1.1400 1.1400 1.1200 = 14.00 12.5655614.00 12.00 1.1400 0.02 = 0.71722 = 1.1400 0.71722(0.02) = 1.1256556

= ( , %,) = 1,500(1.1256556) = 1,688.48

, = ,.#

P = 1,500

F

14

PROBLEM 8

SOLUTION

0 1 2 3 4 5 6 7 8 9 10 11 12 (month)

() = 12%12 = 1% = 0.01 () = 1 + 0.112 1 = 0.104713 = 1.04713% () = 1.04713% 12 = 12.56556%

For 1 year outstanding debt, n = 1

= ( , %,) = 1,500( , 12.56556%, 1)

i% F/P, n = 1 12.00 1.1200

12.56556 x 14.00 1.1400

1.1400 1.1400 1.1200 = 14.00 12.5655614.00 12.00 1.1400 0.02 = 0.71722 = 1.1400 0.71722(0.02) = 1.1256556

= ( , %,) = 1,500(1.1256556) = 1,688.48

, = ,.#

P = 1,500

F

15

PROBLEM 9

SOLUTION

A Alternative:

0 1 2 3 4 5 6 7 8 9 10 (year)

() = 12.90%2 = 6.45% = 0.0645 () = 6.45% = 0.0645

For every 6 month payment, n = 2 X 10 = 20

= ( , %,) = 50,000( , 6.45%, 20)

i% A/P, n = 20 6.00 0.08718 6.45 x 7.00 0.09439

0.09439 0.09439 0.08718 = 7.00 6.457.00 6.00 0.09439 0.00721 = 0.551.00 = 0.09439 0.55(0.00721) = 0.0904245

= ( , %,) = 50,000(0.0904245) = 4,521.23 = ,.

P = 50,000

AA

16

B Alternative:

0 1 2 3 4 5 6 7 8 9 10

() = 12.85%4 = 3.2125% = 0.032125 () = 3.2125% 2 = 0.06425 () = 1 + 0.064252 1 = 0.065282 = 6.5282%

For every 6 month payment, n = 2 X 10 = 20

= ( , %,) = 50,000( , 6.45%, 20)

i% A/P, n = 20 6.00 0.08718

6.5282 x 7.00 0.09439

0.09439 0.09439 0.08718 = 7.00 6.52827.00 6.00 0.09439 0.00721 = 0.47181.00 = 0.09439 0.4718(0.00721) = 0.0909883

= ( , %,) = 50,000(0.0909883) = 4,549.42 = ,.

From the calculation above, found that the A alternative is more economic compare to the B alternative because A alternative given the lowest installment for every 6 month payment method.

P = 50,000

AB

17

PROBLEM 10

SOLUTION

0 1 2 3 4 (year)

() = 12%4 = 4% = 0.04 () = 4% 1 = 4% () = 1 + 0.041 1 = 0.04 = 4%

For monthly installment, n = 4 X 12 = 48

= ( , %,) = 200( , 4%, 48)

n = 20 P/A 45 20.7200 48 x 50 21.4822

21.4822 21.4822 20.7200 = 50 4850 45 21.4822 0.7622 = 0.4 = 21.4822 0.4(0.7622) = 21.17732

= ( , %,) = 200(21.17732) = 4,235.46

, = ,.#

P

A = 200

18

0 1 2 3 4 (year)

() = 12%2 = 6% = 0.06 () = 6% 1 = 6% () = 1 + 0.061 1 = 0.06 = 6%

For monthly installment, n = 4 X 12 = 48

= ( , %,) = 4,235.46( , 6%, 48)

n P/A 45 0.06470 48 x 50 0.06344

0.06344 0.06344 0.06470 = 50 4850 45 0.06344 0.00126 = 0.4 = 0.06344 0.4(0.00126) = 0.063944

= ( , %,) = 4,235.46(0.063944) = 270.83

, = .#

P = 4,235.46

A

19

PROBLEM 11

SOLUTION

1st Alternative:

0 1 2 3 4 5 6 7 8 9 10 11 12 (month)

() = 12%12 = 1% = 0.01 () = 1 + 0.0112 1 = 0.010046 = 1.0046% () = 1.0046% 12 = 12.0552%

For 1 year payment, n = 1

= ( , %,) = 100,000( , 12.0552%, 1)

i% F/P, n = 1 12.00 1.1200

12.0552 x 14.00 1.1400

1.1400 1.1400 1.1200 = 14.00 12.055214.00 12.00 1.1400 0.02 = 0.9724 = 1.1400 0.9724(0.02) = 1.120552

= ( , %,) = 100,000(1.120552) = 112,055.20 = ,.

P = 100,000

F

20

2nd Alternative:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 (month)

() = 12%12 = 1% = 0.01 () = 1 + 0.0124 1 = 0.010048 = 1.0048% () = 1.0048% 24 = 24.1152%

For 2 year payment, n = 2

= ( , %,) = 100,000( , 24.1152%, 2)

i% F/P, n = 2 24.00 1.5376

24.1152 x 25.00 1.5625

1.5625 1.5625 1.5376 = 25.00 24.115225.00 24.00 1.5625 0.0249 = 0.8848 = 1.5625 0.8848(0.0249) = 1.54046848

= ( , %,) = 100,000(1.54046848) = ,.

P = 100,000

F

21

Investment:

0 1 2 (year)

= ( , %,) = 100,000( , 15%, 2) = 100,000(8.1371) = 813,710.00 = ,.

= = ,. ,. = ,.

From the calculation above, the best solution for the company is 2nd alternative because its more profitable compare to the 1st alternative. Even the total payment for 2nd alternative is much higher, the company still can earn a profit about RM 659,663.15 from the investment after loan payment.

P = 100,000

F

i = 15%