ie 2324 chap 11 homework with answers

7/24/2019 IE 2324 Chap 11 Homework With Answers

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IE 3301 ENGINEERING ECONOMIC ANALYSIS

Ch11 Problems - Solutions

11.1A milling machine has a first cost of $50,000. Use a) Straight line (SL) b) sum-of-years

digits (SOYD) depreciation accounting method with a $5,000 salvage value and a 10-

year depreciation life, to determine the depreciation schedule and the milling machine’s

book value at the end of year eight.

Solution:

SL Method

B = $50,000, N = 10, S = $5000

Dt = (B-S)/N = (50,000 – 5000) / 10 = 4500

d-B=BV

8

1t

t

=50,000 – 8*4500 = 14000

SOYD Method

B = $50,000. N = 10. S = $5,000.

dt = [ {N-t+1}/ {N × (N + 1) / 2 }](B - S)

Year dt cumulative dt BV

1 8181.818 8181.818182 41,818

2 7363.636 15545.45455 34,455

3 6545.455 22090.90909 27,909

4 5727.273 27818.18182 22,182

5 4909.091 32727.27273 17,2736 4090.909 36818.18182 13,182

7 3272.727 40090.90909 9,909

8 2454.545 42545.45455 7,455

9 1636.364 44181.81818 5,818

10 818.1818 45000 5,000

Book Value at the end of year 8 is $7,455

Another Method to solve

dt = (B - S) / {N × (N + 1) / 2 } = (50,000 - 5,000) / {10 × (10 + 1) / 2} =

$818.18.

AD8 = [{n × (n + 1) / 2} - { (n - 8) × (n - 8 + 1) / 2 }] × dt =

[{10 × (10 + 1) / 2} - {(10 - 8) × (10 - 8 + 1) / 2}] × 818.18 = $42,545.45.

BV j = B - AD j. BV8 = 50,000 - 42,545.45 = $7,454.55.



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11.2An account manager in XYZ Company is using DDB method for a Special Handling

Device (SHD) that cost $12,000 and has a $600 salvage value at the end of 4 years.

Calculate the depreciation schedule for the SHD.

Solution:

Double Declining Balance

DDB = ),(2 1

1

t

j

d B N

11.3Kutz Petroleum Corporation bought drilling equipment used for petroleum extraction 4

years ago. Due to recent development in drilling technology, KPC decided to buy a new

equipment selling the existing one which was bought for $200,000. It is recently sold for

$45000. Assuming the equipment sold is MACRS GDS property class of 5 years

a) Determine the MACRS depreciation schedule for 4 years of ownership.

b) Determine recaptured depreciation or loss on the sale of drilling equipment

Solution:

a)First cost = $200,000. Depreciation schedule for 4 years as follows:

Year MACRS % MACRS% x FirstCost

Depreciation

1 20 20%*200,000 40,000

2 32 32%*200,000 64,000

3 19.2 19.2%*200,000 38,400

4 11.52/2* 5.76%*200,000 11,520

*we divide it by 2 due to half a year convention that halves the percentage for the4th year.

b)Total Depreciation = 40,000+64,000+38400+11,520 =$153920Book Value at the end of year 4 = 200,000 – 153920 = $46080

Book Value at the end of year 4 is greater than sold amount therefore, loss of the sale ofdrilling equipment.

Loss = $(46080 – 45000) = $1,080.



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11.4Robert, a real estate businessman, bought a $15 million new commercial building and put

it in business on March 1, 2004. He sold it for $13.5 million on September 1, 2009.

Calculate the modified accelerated cost recovery system depreciation for the time he

retained the commercial building.

Solution:Use Table 11-4 for MACRS Depreciation for Real Property (real estate).

Depreciation for 2004 = 2.033% * $15 million = $304,950Depreciation for 2005 through 2008 = 2.564% * $15 million = $384,600Depreciation for 2009 = 1.819% * $15 million = $272850

11.5Greg’s Coal Company bought equipment worth $115,000 for coal extraction. It is

reported that there are 100,000 tons of raw coal left in the coal mine. The company plans

to extract all the coal in next 5 years as followsYear Coal (tons)

1 35000

2 25000

3 15000

4 15000

5 10000

The salvage value of the equipment used would be zero at the end of year 5. Thecompany considers Unit of Production depreciation method. What would be the PresentValue of Unit of Production depreciation schedule? Assume i=6%.

Solution:UOP depreciation in any year = {Production for year / Total lifetime production forassets } (B-S)

Coal (tons)

UOP

Depreciation($)

Present Worth = UOP

Deprc. *(P/F, 6%, n)

1 35000 35000*(115000-0)/100000 40250

40250*0.9434

=37971.852 25000 25000*(115000-0)/100000 28750 28750*0.89=25587.5

3 15000 15000*(115000-0)/100000 17250 17250*0.8396=14483.1

4 15000 15000*(115000-0)/100000 17250 17250*0.7921=13663.73

5 10000 10000*(115000-0)/100000 11500 1150*0.7473=8593.95

Total 100000 Present Worth 100,300.10



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11.6Buyers Choice Pulp Factory has entered into a contract to harvest a timber for $950,000.

The total estimated available timber harvest is 250,000,000 board feet. What would be

the depletion allowances given for years 1 to 3, if 50, 60, 70 million board-feet are

harvested by Buyers Choice Pulp Factory in those years.

Solution:Depletion rate = 950,000/250,000,000 = $3800 per million board feet.

Depletion allowance year 1 = 3800 * 50 = $190000Depletion allowance year 2 = 3800*60 = $228000Depletion allowance year 3 = 3800*70 =$266000

Partial Answers 11.1 a) On your own b) BV8 = 7,455

11.2

On your own11.3 Loss, because BV4 is greater than selling price of an equipment. Loss =$1,080.

11.4 On your own

11.5 PW of depreciation = $100,300.1

11.6 On your own

ie 2324 chap 11 homework with answers

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