ie 343 midterm exam 2 - weeklyjoys · pdf filename (printed): _,_ (last) (first)...

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Purdue University Spring 2012 – IE 343 – Section 1 Instructor: Tian Ni Midterm Exam #2, April 6, 2012

Version A 1

IE 343 Midterm Exam 2

April 6, 2012

Version A

Closed book, closed notes.

50 minutes

Write your printed name in the spaces provided above on every page.

Show all of your work in the spaces provided.

Interest rate tables are provided for you to use in questions that require numerical answers.

Exam 2 has three Parts totally 5 Problems with 105 Points:

Part I – 3 Old Problems, totally 76 Points. The old problems are selected from

Homework, Quizzes, Lecture Notes Examples and Textbook Examples with numbers

changed.

Part II – 1 New Problem, 24 Points.

Part III – 1 Bonus Problem, 5 Points.

You are suggested to do Part I first, and then part II. Part III is optional.



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Part I – 3 Old Problems, totally 76 Points: old problems selected from Homework, Quizzes,

Lecture Notes Examples or Textbook Examples

Question 1 (35 Points)

We consider four design alternatives D1, D2, D3, D4 each with a useful life of 5 years. The

MARR is 20%.

D1 D2 D3 D4

Initial Investment: $100 $300 $150 $200

Annual Revenues: $50 $90 $70 $65

Annual Expenses: $20 $15 $15 $20

Salvage Value: $30 $100 $50 $50

Useful Life: 5 Year 5 Year 5 Year 5 Year

(a) Are the four design alternatives D1, D2, D3, D4 investment alternatives or cost

alternatives? Please explain why? (3 Points)

Solution:

(a) Investment alternative, because there are annual revenues.

(b) We claim that the design alternative D4 can be directly eliminated from the analysis.

Why? Please explain the reason. (4 Points)

Solution:

(b) D4 has higher initial investment than D3 ($200 > $150), but D4 has lower net annual

revenues less expenses than D3 ($45 < $55). D3 and D4 have the same Salvage Value

$50, So D3 is a strictly better alternative than D4. So D4 can be directly eliminated.



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(c) You are calculating the IRR of D1. You get PW(15%) = 15.5, and PW(25%) = -9.5 by

trial and error. What is the approximated IRR using linear interpolation? Based on this

approximated IRR, is D1 profitable? Explain the reason why? (8 Points)

Solution:

(c) By linear interpolation:

D1 is profitable.

Circle exactly one solution for part (d) to (h) (20 Points)

After eliminating D4, we can perform the incremental analysis procedure using IRR. The

table we get is as follows,

Increment Considered △(D1 – DN) ____(d)____ ____(f)____

△Capital Investment $100 $50 $150

△Annual Revenues Less Expenses $30 $25 $20

△Salvage Value $30 $20 $50

Useful life(Years) 5 5 5

IRR△ 21.2% 45.01% 0.00%

Is increment justified by IRR△? Yes Yes No

Which alternative will you select? D1 ____(e)____ ____(g)____

(d) △(D1 – D3) / △(D2 – D3) / △(D3 – DN) / △(D3 – D1)

Solution:

(d) △(D3 – D1)

(e) D1 / D2 / D3 / DN

Solution:

(e) D3



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(f) △(D2 – D3) / △(D3 – D2) / △(D3 – DN) / △(D3 – D1)

Solution:

(f) △(D2 – D3)

(g) D1 / D2 / D3 / DN

Solution:

(g) D3

(h) Which design alternative is the best?

D1 / D2 / D3 / D4

Solution:

(h) D3


Two mutually exclusive alternatives are being considered for the environmental protection

equipment at a petroleum refinery. One of these alternatives must be selected. The estimated case

flows for each alternative are follows:

Alternative A Alternative B

Capital investment $6,000 $5,000

Annual expenses $2,000 $2,500

Salvage value at end of useful life $1,000 $500

Useful life 4 years 8 years



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(a) Which environmental protection equipment alternative should be selected? The firm’s

MARR is 10% per year. Assume the equipment will be needed indefinitely. (10 Points)

Solution:

(a) Since the equipment will be needed indefinitely, the repeatability assumption holds.

AW(A) = -6000(A/P,0.1,4)-2000+1000(A/F,0.1,4) = -6000*0.3155-2000+1000*0.2155 =

-3677.5

AW(B) = -5000(A/P,0.1,8)-2500+500(A/F,0.1,8) = -5000*0.1874-2500+500*0.0874 =

-3393.3

Select Alternative B, since AW(B) = -3393.3 > AW(A) = -3677.5

(b) Assume the study period is shortened to four years. The firm’s MARR is 10% per year.

The market value of Alternative B at the end of year four is estimated to be $3,000. Use

the imputed market value technique to determine which alternative you would

recommend. (10 Points)

Solution:

(b) AW(A) = -3677.5

AW(B) = -5000(A/P,0.1,4)-2500+3000(A/F,0.1,4) = -5000*0.3155-2500+3000*0.2155 =

-3431

Select Alternative B, since AW(B) = -3431 > AW(A) = -3677.5



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Suppose that an asset with a cost basis of $60,000 is expected to produce net revenues of

$30,000 per year during the four year period, and its terminal market value is negligible. The

effective income tax rate is 40% and the after-tax MARR is 10% per year. The asset is being

depreciated under the Straight Line method over a period of three years.

EOY (A)

BTCF

(B)

Depreciation

(C)=(A)–(B)

Taxable Income

(D)= –t(C)

Income Taxes

(E)=(A)+(D)

ATCF

(a) What is the depreciation deduction d2 for year 2? (3 Points)

Solution:

(a) Since Straight Line method is applied, the depreciation deduction is the same every year,

d2 = d = $60,000 / 3 = $20,000

(b) Complete the table below with information given above. (12 Points)

EOY BTCF Depr TI Tax ATCF

0 -$60,000 -$60,000

1 $30,000 . . . .

2 $30,000 . . . .

3 $30,000 . . . .

4 $30,000 . . .



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Solution:

(b) .


0

-$60,000

-$60,000

1

$30,000

$20,000

$10,000

-$4,000

$26,000

2

$30,000

$20,000

$10,000

-$4,000

$26,000

3

$30,000

$20,000

$10,000

-$4,000

$26,000

4

$30,000

$30,000

-$12,000

$18,000

(c) Are you willing to invest this asset if the income tax is considered? (6 Points)

Solution:

(c) PW = -60,000 + 26,000 (P/A,10%, 3) + 18,000(P/F,10%,4) = -60,000 + 26,000*2.4869 +

18,000*0.6830 = 16,953 > 0

So I will invest this asset.



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Part II, 1 Problem, 24 Points: New Problem


XYZ Corp. is considering buying a computer workstation for $1,000. The equipment is in the

3-year MACRS (GDS) property class and is expected to reduce net operating expenses by $500

every year. The company’s effective income tax rate is 40%. Assume that the after-tax MARR is

10% per year and the study period is 4 years.

EOY (A)

BTCF

(B)

Depreciation

(C)=(A)–(B)

Taxable Income

(D)= –t(C)

Income Taxes

(E)=(A)+(D)

ATCF

(a) What are the depreciation deductions d2 and d3 for year 2 and 3? (4 Points)

EOY BTCF Depr

0 -$1,000

1 $500 d1 = $333.3

2 $500 d2 = .

3 $500 d3 = .

4 $500 d4 = $74.1

Solution:

(a) Under 3-year MACRS GDS r1 = 0.3333, r2 = 0.4445, r3 = 0.1481, r4 = 0.0741

EOY BTCF Depr

0 -$1,000

1 $500 d1 = $333.3

2 $500 d2 = $444.5 .

3 $500 d3 = $148.1 .

4 $500 d4 = $74.1



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(b) If you can sell the asset at end of year 2 for $500, what is your capital gain (loss) on this

disposal? What is the tax liability (or tax credit) on this disposal? (10 Points)

Solution:

(b) If the asset is sold at end of year 2, half-year convention will be applied to the depreciation

deduction in year 2. Now d2 = $444.5/2 = $222.25

BV2 = $1,000 - $333.3 - $222.25 = $444.45

Capital gain (loss) = MV2 – BV2 = $500 - $444.45 = $55.55

Tax liability = 0.4(Capital gain) = 0.4 ($55.55) = $22.22

(c) If you can sell the asset at the end of year 2 for $500, please complete the table below.

(10 Points)


0 -$1,000 -$1,000

1 $500 . . . .

2 $500 . . . .

2 $500 . . .

Solution:

(c) .


0 -$1,000 -$1,000

1 $500 $333.3 $166.7 -$66.68 $433.32

2 $500 $222.25 $277.75 -$110.1 $388.9

2 $500 $55.55 -$22.22 $477.78



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Part III, 1 Bonus Problem, 5 Points

Question 5 – Extension of Question 4 (5 Points)

XYZ Corp. is considering buying a computer workstation for $1,000. The equipment is in the

5-year MACRS (ADS) property class and is expected to reduce net operating expenses by $500

every year. The company’s effective income tax rate is 40%. Assume that the after-tax MARR is

10% per year and the study period is 4 years. The market value of the computer workstation at

the end of year 4 is considered to be negligible.

Please complete the table below.


0 -$1,000 -$1,000

1 $500 . . . .

2 $500 . . . .

3 $500 . . . .

4 $500 . . . .

4 $0 . . .

Solution:

For 5-year MACRS (ADS) property class, we need 6 years to fully depreciate the computer

workstation. Study period is 4 years and the market value of the computer workstation is

negligible after 4 years, which means there’s an early disposal in year 4. Then half year

convention should be applied to the 4th

year. For MACRS (ADS), straight line method with

half year convention to the 1st year is applied, so d1=1000/5/2=100, d2=d3=1000/5=200,

d4=1000/5/2=100

BV4 = 1000-d1-d2-d3-d4 = 1000-100-200-200-100 = 400

MV4 = 0

Capital gain (loss) = MV4 – BV4 = 0 – 400 = -400

Tax credit = -0.4 * (-400) = 160



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0 -$1,000 -$1,000

1 $500 $100 $400 -$160 $340

2 $500 $200 $300 -$120 $380

3 $500 $200 $300 -$120 $380

4 $500 $100 $400 -$160 $340

4 $0 -$400 $160 $160



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Formulas and Tables

Find F given P:

Find P given F:

Imputed Market Value:

Capital Investment, : Salvage Value, : Useful life, : Study Period

Depreciation using SL method:

;

Depreciation using DB method:

; ;

Depreciation of SL when using DB with switchover to SL:

GDS or ADS depreciation: k kd r B

Taxable Income (TI) = Gross income - all expenses (except capital investments) - depreciation

deductions.

[Gain(loss)ondisposal]N MVN BVN



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ie 343 midterm exam 2 - weeklyjoys · pdf filename (printed): _____,_____ (last) (first)...

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