ce 231 engineering economy problem set 6 problem...

CE 231

ENGINEERING ECONOMY

PROBLEM SET 6

PROBLEM 1

An asset has a first cost of 13.000 TL, an estimated life of 15 years, and an estimated

salvage value of 1.000 TL. Using the straight-line method, find:

a) The annual depreciation charge,

b) The annual depreciation rate expressed as a percentage of first cost, and

c) The book value at the end of 9 years.

SOLUTION 1

P = 13.000 TL

n = 15 yrs

F = 1.000 TL

a) Annual depreciation charge = 15

000.1000.13 = 800 TL

b) 000.13

800= 0.0615 = 6,15%

c) 13.000 – (9 x 800) = 5.800 TL

PROBLEM 2

An asset has a first cost of 22.000 TL, an estimated life of 30 years, and an estimated

salvage value of 2.000 TL. Using the double declining balance method, find:

a) The depreciation charge in the first year,

b) The depreciation charge in the 6th year, and

c) The book value at the end of 6th.

SOLUTION 2

P = 22.000 TL

n = 30 yrs r = n

2 =

30

2 = 0,07

F = 2.000 TL

End of Year Depreciation Book Value

0 - 22.000

1 22.000 x 0,07 = 1540 20.460

2 20.460 x 0,07 = 1432 19.027,80

3 19.027,8 x 0,07 = 1331,95 17.695,85

4 17.695,85 x 0,07 = 1238,71 16.457,14

5 16.457,14 x 0,07 = 1152 15.305,14

6 15.305,14 x 0,07 = 1071,36 14.233,78

PROBLEM 3

An asset has a first cost of 9.000 TL, an estimated life of 12 years, and an estimated salvage

value of 1.200 TL. It is to be depreciated by the sum-of-the-years-digit method. What will

be the depreciated charge;

a) in the first year and;

b) in the 7th year?

c) What will be the book value at the end of 6 years?

SOLUTION 3

P = 9.000 TL

n = 12 yrs

F = 1.200 TL

s = 2

)1( nn=

2

1312x= 78

a) 1d = 2(P-F)

nn

in2

1= 2 (9000-1200)

1212

11122

= 1200 TL

b) 7d = 2 (9000-1200)

1212

71122

= 600 TL

c) 6BV = 9000 – (9000–1200) 1212

6661222

2

xx= 3300 TL

OR,


0 - 9.000

1 (9000-1200) x 12/78 = 1200 7.800

2 7.800 x 11/78 = 1100 6.700

3 7.800 x 10/78 = 1000 5.700

4 7.800 x 9/78 = 900 4.800

5 7.800 x 8/78 = 800 4.000

6 7.800 x 7/78 = 700 3.300

7 7.800 x 6/78 = 600 2.700

PROBLEM 4

A company has purchased a numerically controlled machine for 150.000 TL. It is estimated

that it will have a salvage value of 50.000 TL four years from now. What rate must be used

with the declining-balance method of depreciation so that the book value of the machine

will be equal to its salvage value at the end of its life?

a) using the rate just calculated with declining-balance depreciation find the

depreciation and book value for each year of the machine’s life,

b) compare those figures with similar figures for straight-line and sum-of-the

years-digits depreciation.

SOLUTION 4

P = 150.000 TL

n = 4 yrs

F = 50.000 TL

a) R = 1 - n PF /

R = 1 - 4 000,150/000,50 = 1-0,7598 = 0,2402


0 - 150.000

1 150.000 x 0,24 = 36.000 114.000

2 114.000 x 0,24 = 27.360 86.640

3 86.640 x 0,24 = 20.793,6 65.846

4 65.846 x 0,24 = 15.803 50.042

b) Straight line depreciation = (150.000-50.000)/4 = 25.000


0 - 150.000

1 25.000 125.000

2 25.000 100.000

3 25.000 75.000

4 25.000 50.000

Sum-of-the-years digits: s = 2

54x= 10


0 - 150.000

1 (150.000-50.000) x 4/10 = 40.000 110.000

2 100,000 x 3/10 = 30.000 80.000

3 100,000 x 2/10 = 20.000 60.000

4 100,000 x 1/10 = 10.000 50.000

PROBLEM 5

A piece of equipment that cost 5.000 TL was found to have a trade-in value of 4.000 TL at

the end of the first year, 3.200 TL at the end of the second year, 2.560 TL at the end of the

third year, 2.048 TL at the end of the fourth year. Determine the depreciation that occurred

during each year.

SOLUTION 5

P = 5.000

1P = 4.000 1d = 5.000 – 4.000 = 1.000 TL

2P = 3.200 2d = 4.000 – 3.200 = 800 TL

3P = 2.560 3d = 3.200 – 2.560 = 640 TL

4P = 2.048 4d = 2.560 – 2.048 = 512 TL

PROBLEM 6

A firm purchases a computer for 2.000.000 TL. It has a life of 9 years and a salvage value

of 200.000 TL at that time. Determine the depreciation charge for year 6 and the book value

at the beginning of year 6, using:

a) Straight line depreciation

b) Declining balance depreciation

c) Sum of the years digits depreciation

SOLUTION 6

P = 2.000.000 TL

n = 9 yrs

F = 200.000 TL

a) d = 9

000.200000.000.2 = 200.000

6D = 200.000

5BV = 2.000.000 – 5 x 200.000 = 1.000.000

b) R = 1 - n PF / = 1 - 9 000.000.2/000.200 = 0,2257


0 - 2.000.000

1 2 x 610 x 0,2257 = 0,4519 x 610 1,5485 x 610

2 1,5485 x 610 x 0,2257 = 0,3495 x 610 1,1990 x 610

3 0,2706 x 610 0,9285 x 610

4 0,2096 x 610 0,7189 x 610

5 0,1623 x 610 0,5566 x 610

6 0,1256 x 610

c) s = 2

)1( nn=

2

109x= 45


0 - 2.000.000

1 1,8 x 610 x 9/45 = 0,36 x 610 1.640.000

2 1,8 x 610 x 8/45 = 0,32 x 610 1.320.000

3 1,8 x 610 x 7/45 = 0,28 x 610 1.040.000

4 1,8 x 610 x 6/45 = 0,24 x 610 800.000

5 1,8 x 610 x 5/45 = 0,20 x 610 600.000

6 1,8 x 610 x 4/45 = 0,16 x 610

PROBLEM 7

A new 250.000 TL automobile will depreciate over the next 5 years approximately

according to the sum of the years digit method with the first year depreciation being 50.000

TL.

a) Determine the salvage value at the end of 5-year period.

b) Determine the year-end book value for each year.

SOLUTION 7

P = 250.000 TL

n = 5 yrs

s = 2

)1( nn=

2

65x= 15

a) (250.000 – F) 5/15 = 50.000

F = 100.000 TL

b) End of Year Depreciation Book Value

0 - 250.000

1 50.000 200.000

2 150.000 x 4/15 = 40.000 160.000

3 150.000 x 3/15 = 30.000 130.000

4 150.000 x 2/15 = 20.000 110.000

5 150.000 x 1/15 = 10.000 100.000

PROBLEM 8

A bulldozer has a first cost of 350.000 TL with an estimated life of 5 years. The salvage

value at the end of 5 years is estimated to be 50.000 TL. What is the book value of this

bulldozer at the end of 3rd year. Use:

a) Straight line method

b) Double-declining-balance method

c) Declining balance method

d) Sum-of-the-years-digit method

SOLUTION 8

P = 350.000 TL

n = 5 yrs

F = 50.000 TL

a) d = 5

000.50000.350 = 60.000

3BV = 350.000 – (3 x 60.000)

= 170.000 TL

b) r = 2 x 1/5 = 0,40


0 - 350.000

1 350.000 x 0,40 = 140.000 210.000

2 210.000 x 0,40 = 84.000 126.000

3 126.640 x 0,40 = 50.400 75.600

4

5

c) R = 1 - n PF /

R = 1 - 5 000.350/000.50 = 0.32

0.32 = 1 - 3 000.350/F => 3 000.350/F = 1 – 0,32

F = 350.000 3)32,01(

F = 350.000 x 3)68,0( = 110.051,20 TL

d) s = 2

)1( nn=

2

65x= 15

(P-F) = 350.000 – 50.000 = 300.000


0 - 350.000

1 300.000 x 5/15 = 100.000 250.000

2 300.000 x 4/15 = 80.000 170.000

3 300.000 x 3/15 = 60.000 110.000

4

5

PROBLEM 9

A machine has a first cost of 100.000 TL with an estimated life of 10 years. The salvage

value is estimated to be 10.000 TL. Determine the yearly book values of this machine by

using:

a) Straight line method

b) Double-declining-balance method

c) Declining balance method

d) Sum-of-the-years-digit method

SOLUTION 9

P = 100.000 TL

n = 10 yrs

F = 10.000 TL

a) yearly depreciation = 10

000.10000.100 = 9.000 TL/yr


0 - 100.000

1 9.000 91.000

2 9.000 82.000

3 9.000 73.000

4 9.000 64.000

5 9.000 55.000

6 9.000 46.000

7 9.000 37.000

8 9.000 28.000

9 9.000 19.000

10 9.000 10.000

b) r = 2 x 1/10 = 0,20


0 - 100.000

1 100.000 x 0,20 = 20.000 80.000

2 80.000 x 0,20 = 16.000 64.000

3 64.000 x 0,20 = 12.800 51.200

4 51.200 x 0,20 = 10.240 40.960

5 40.960 x 0,20 = 8.192 32.768

6 32.768 x 0,20 = 6.553,6 26.214,4

7 26.214,4 x 0,20 = 5.242,88 20.971,52

8 20.971,52 x 0,20 = 4.194,30 16.777,22

9 16.777,22 x 0,20 = 3.355,44 13.421,78

10 13.421,78 x 0,20 = 2.684,36 10.737,42

10* 2.684,36 – (10.000 – 10.737,42) = 3.421,78 10.000

c) R = 1 - n PF /

R = 1 - 10 000.100/000.10 = 0,2057


0 - 100.000

1 100.000 x 0,2057 = 20.570 79.430

2 79.430 x 0,2057 = 16.338,75 63.091,25

3 63.091,25 x 0,2057 = 12.977,87 50.113,38

4 50.113,38 x 0,2057 = 10.308,32 39.805,06

5 39.805,06 x 0,2057 = 8.187,90 31.617,16

6 31.617,16 x 0,2057 = 6.503,65 25.113,51

7 25.113,51 x 0,2057 = 5.165,85 19.947,66

8 19.947,66 x 0,2057 = 4.103,23 15.844,43

9 15.844,43 x 0,2057 = 3.259,20 12.585,23

10 12.585,23 x 0,2057 = 2.588,78 ~10.000

d) s = 2

)1( nn=

2

1110x= 55

(P–F) = 100.000 – 10.000 = 90.000


0 - 100.000

1 90.000 x 10/55 = 16.363,64 83.636,36

2 90.000 x 9/55 = 14.727,27 68.909,09

3 90.000 x 8/55 = 13.090,91 55.818,18

4 90.000 x 7/55 = 11.454,55 44.363,63

5 90.000 x 6/55 = 9.818,18 34.545,45

6 90.000 x 5/55 = 8.181,82 26.363,63

7 90.000 x 4/55 = 6.545,46 19.818,17

8 90.000 x 3/55 = 4.909,09 14.909,08

9 90.000 x 2/55 = 3.272,73 11.636,35

10 90.000 x 1/55 = 1.636,36 10.000

0

20000

40000

60000

80000

100000

0 1 2 3 4 5 6 7 8 9 10

Straight Line

Double Declining Balance

Declining Balance

Sum of the Years Digit

PROBLEM 10

A machine has a first cost of 800 TL with an estimated life of 8 years. Determine the

salvage value of this machine by using:

a) Straight line method if the book value of the machine is 500 TL at the end of

4th year

b) Double-declining-balance method if the corrected depreciation is 10 TL

c) Declining balance method if the book value of the machine is 500 TL at the end

of 4th year

d) Sum-of-the-years-digit method if the book value of the machine is 500 TL at

the end of 4th year

SOLUTION 10

P = 800

n = 8 yrs

F = ?

a) yearly depreciation = 8

800 F = 100 – F/8


0 - 800

1 100 – F/8 800 – (100 – F/8)

2 100 – F/8 800 – 2*(100 – F/8)

3 100 – F/8 800 – 3*(100 – F/8)

4 100 – F/8 800 – 4*(100 – F/8)

5 100 – F/8 800 – 5*(100 – F/8)

6 100 – F/8 800 – 6*(100 – F/8)

7 100 – F/8 800 – 7*(100 – F/8)

8 100 – F/8 800 – 8*(100 – F/8)

9 100 – F/8 800 – 9*(100 – F/8)

10 100 – F/8 800 – 10*(100 – F/8)

4BV = 800 – 4 x (100 – 8

F)

500 = 800 – 4 x (100 – 8

F)

4 x (100 – 8

F) = 300

F = 200 TL

b) r = 2 x 1/8 = 0,25


0 - 800

1 800 x 0,25 = 200 600

2 600 x 0,25 = 150 450

3 450 x 0,25 = 112,5 337,50

4 337,50 x 0,25 = 84,38 253,12

5 253,12 x 0,25 = 63,28 189,84

6 189,84 x 0,25 = 47,46 142,38

7 142,38 x 0,25 = 35,60 106,78

8 106,78 x 0,25 = 26,70 80,08

9 80,08 x 0,25 = 20,02 60,06

10 60,06 x 0,25 = 15,02 45,04

10* 10 60,06 – 10 = 50,06

F = 50,06 TL

c) R = 1 - n PF /

R = 1 - 8 800/F

End

of

Year

Depreciation Book Value

0 - 800

1 800R 800 – 800R= 800(1–R)1

2 800R – 800R2 800 (R2 – 2R + 1)= 800(1–R)2

3 800 (R3 – 2R2 + R) 800 (– R3 + 3R2 –3R + 1)= 800(1–R)3

4 800 (– R4 + 3R3 – 3R2 + R) 800 (+ R4 – 4R3 + 6R2 – 4R

+1)= 800(1–R)4

4BV = 800 (1–R)4

500 = 800 (1–R)4

R = 0,11086

1 - 8 800/F = 0,11086

F = 312,5 TL

d) s = 2

)1( nn=

2

98x= 36

(P–F) = 800 – F


0 - 800

1 (800 – F) x 8/36 800 – (800 – F) x 8/36

2 (800 – F) x 7/36 800 – (800 – F) x (8/36+7/36)

3 (800 – F) x 6/36 800 – (800 – F) x (8/36+7/36+6/36)

4 (800 – F) x 5/36 800 – (800 – F) x (8/36+7/36+6/36+5/36)

4BV = 500

800 – (800 – F) x (26/36) = 500

300 = (800 – F) x (26/36)

F = 384,6 TL

PROBLEM 11

A construction equipment has an estimated life of 4 years and a salvage value of 360.000

TL at the end of these 4 years. It is assumed that the equipment will depreciate according

to the Double-Declining Balance Method. In this method, correction is necessary only in

the calculation of the last year’s (4th year’s) depreciation and this corrected depreciation

value is 180.000 TL.

a) What is the initial value of equipment?

b) Calculate and tabulate the yearly depreciations and book values of equipment.

c) The income in the first three years is 2.500.000 TL/yr, and the income in the last

year is 1.250.000 TL. The tax which will be paid is 60% of the net income, ie., 60%

of (income-depreciation). Calculate the present worth of these tax payments

assuming a MARR of 15%.

SOLUTION 11

a)

Double – Declining – Balance-rate of depreciation = 2 n

1 = 2

4

1 = 0,5

Initial cost = P

Salvage value = F = 360.000 TL

n= 4 yrs

Corrected depreciation amount at the th4 year : 4D = 180.000 TL

End of Year Yearly Depreciation Book Value

0 0 P

1 0,5 P P – 0,5 P = 0,5 P

2 0,5 0,5 P = 0,25 P 0,5 P – 0,25 P = 0,25 P

3 0,5 0,25 P = 0,125 P 0,25 P – 0,125 P = 0,125 P

4 0,5 0,125 P = 0,0625 P 0,125 P – 0,0625 P = 0,0625 P *4 180.000 360.000

Corrected depreciation = 3BV - Salvage Value

180.000 = 0,125 P – 360.000

P = 4.320.000 TL

b)

End of Year Yearly Depreciation Book Value

0 0 4.320.000

1 0,5 4.320.000= 2.160.000 4.320 000 – 2.160.000 = 2.160.000

2 0,5 2.160.000 = 1.080.000 2.160 000 – 1.080.000 = 1.080.000

3 0,5 1.080.000= 540.000 1.080 000 – 540 000 = 540.000

4 0,5 540.000 = 270.000 540.000 – 270.000 = 270.000 *4 180.000 360.000

c)

End of Year Yearly income Yearly Depreciation Taxable Profit Tax (60 %)

1 2.500.000 2.160.000 340.000 204.000

2 2.500.000 1.080.000 1.420.000 852.000

3 2.500.000 540.000 1.960.000 1.176.000

4 1.250.000 180.000 1.070.000 642.000

Present worth of taxes:

PW (15%) = -204.000 ( P/F, 15%, 1 ) – 852.000 ( P/F, 15%, 2 ) – 1.176.000 ( P/F, 15%,

3) – 642.000 ( P/F, 15%, 4)

= - 204.000 0,8696 – 852.000 0,7561 – 1.176.000 0,6575 – 642.000 0,5718

= - 177.398,4 – 644.197,2 – 773.220 – 367.095,6

= - 1.961.911,2 TL

1

2

3

852.000

4

204.000

0

Salvage Value Corrected Depreciation

642.000

1.176.000

(1) (2) (1 – 2) (3 0.60)

PROBLEM 12

A contractor has an equipment of having a first cost of 1.000.000 TL and a useful life of 7

years. The yearly income for this equipment will be 400.000 TL/yr and the yearly operating

cost will be 100.000 TL/yr in the first two years, 150.000 TL in the third year and 225.000

TL/yr in the remaining 4 years. The salvage value of the equipment at the end of 7 years is

150.000 TL and MARR=6%.

a) Assuming that the equipment will depreciate according to the Declining Balance

Method, calculate the depreciation and book values.

b) If you were the contractor, at the end of which year you should replace the

equipment? (Hint: Use the book values as salvage values)

c) At the end of its economic life it is possible to replace this equipment by new

equipment whose purchase value is 1.500.000 TL, its estimated life is 6 years and

its salvage value at the end of those 6 years is 300.000 TL. The annual operating

cost and annual incomes are 75.000 TL/yr and 600.000 TL/yr respectively. Should

you keep the old equipment or should you replace it with the new equipment. Solve

the problem by using the Receipts and Disbursements Method. Use Present

Worth Approach in your calculations.

SOLUTION 12

a) r = 1- √FP⁄

𝑛=1-√150.000

1.000.000⁄7

= 1 – 0,7626

r = 0,2374


0 --- 1.000.000

1 237.400 762.600

2 181.041,24 581.558,76

3 138.062,05 443.496,71

4 105.286,12 338.210,59

5 80.291,19 257.919,40

6 61.230,07 196.689,33

7 46.694,05 149.995,28

400.000 TL/yr

225.000 TL

150.000 TL

100.000 TL 150.000 TL

0 1 2 3 4 5 6 7

1.000.000 TL

b) n = 1

AE(6%) = -1.000.000 (A P⁄ , 6%, 1) + 1.062.600

AE(6%) = -1.000.000 (1,06) + 1.062.600

AE(6%) = 2.600 TL/yr

n = 2

AE(6%) = -1.000.000 (A P⁄ , 6%, 2) + 300.000 + 581.558,76 (A F⁄ , 6%, 2)

AE(6%) = -1.000.000 (0,54544) + 300.000 + 581.558,76 (0,48544)

AE(6%) = 36.871,88 TL/yr

n = 3

AE(6%) = -1.000.000 (A P⁄ , 6%, 3) + 400.000 – 150.000 (A F⁄ , 6%, 3) – 100.000

(P A⁄ , 6%, 2) ( A P⁄ , 6%, 3) + 443.496,71 (A F⁄ , 6%, 3)

AE(6%) = -1.000.000 (0,37411) + 400.000 – 150.000 (0,31411) – 100.000 (1,8334)

(0,37411) + 443.496,71 (0,31411)

AE(6%) = 49.490,92 TL/yr

1x106

TL

400.000

TL

0 1

762.600

TL

100.000

TL

1x106

TL

0

1 2

581.558,76

100.000

400.000

1x106

TL

0

1 2

443.496,71

100.000

400.000

3

150.000

n=4

AE(6%) = -1.000.000(A P⁄ , 6%, 4) – 100.000 (P A⁄ , 6%, 2) ( A P⁄ , 6%, 4) – 150.000

(F P⁄ , 6%, 1)( A F⁄ , 6%, 4) – 225.000(A F⁄ , 6%, 4) + 400.000 + 338.210,59(A F⁄ , 6%, 4)

AE(6%) = -1.000.000 (0,28859) – 100.000 (1,8334) (0,28859) – 150.000 (1,06)

(0,22859) – 225.000 (0,22859) + 400.000 + 338.210,59 (0,22859)

AE(6%) = 48.032,91 TL/yr

Economic life is 3 years.

c)

1x106

TL

0

1 2

338.210,59

100.000

400.000

3

150.000

4

225.000

400.000 TL/yr

225.000 TL

150.000 TL

100.000 TL 150.000 TL

0 1 2 3 4 5 6 7

1.000.000 TL

Old Equipment

Economic Life

600.000 TL/yr

300.000 TL

75.000 TL

0 1 2 3 4 5 6

1.500.000 TL

443.496,71 TL

New Equipment

Common Multiple of Lives 12 years

AE(6%)OLD = -225.000 + 400.000 + 150.000 (A F⁄ , 6%, 4)

= -225.000 + 400.000 + 150.000 (0,22859) = 209.288,5 TL/yr

PW(6%)OLD = 209.288,5 (P A⁄ , 6%, 12) = 209.288,5 (8,3838) = 1.754.632,93 TL.

AE(6%)REP = -1.056.503,29 (A P⁄ , 6%, 6) + 600.000 – 75.000 + 300.000 (A F⁄ , 6%, 6)

= -1.056.503,29 (0,20336) + 600.000 – 75.000 + 300.000 (0,14336)

= -214.850,51 + 525.000 + 43.008 = 353.157,49 TL/yr.

PW(6%)REP = 353.157,49 (P A⁄ , 6%, 12) = 353.157,49 (8,3838) = 2.960.801,77 TL

Choose New Equipment.

ce 231 engineering economy problem set 6 problem...

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