ce 231 engineering economy problem set 6 problem...
TRANSCRIPT
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,
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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
End of Year Depreciation Book Value
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.