ENM 530 Review for Quiz 2 Fall 2008

ENM 530 Review for Quiz 2 Fall 2008

1. Choose the better of mutually exclusive alternatives at a MARR of 5%.

Year 0 1 2 3IRR (%)

A-2000 800 800 8009.70

B-28001100110011008.69

B-A-800 300 300 3006.13

Since B – A exceeds 5%, B is the preferred alternative for MARR  6.13%.

(IRR '(-800 300 300 300))6.128562

(cubic -800 300 300 300)1.061286=> 6.13% return on B - A

2. Invest $3810 now, spend $500 at n = 1,250 at n = 2 to find the RoR in perpetuity.

[3810(1 + R) -500](1 + R) - 250 = 3810

3810(1 + R)2 -500 (1 + R) = 4060

(quadratic 381 -50 -406)1.099987 -0.968754) => 10%

500

200 perpetuity

$3810

3.n 0 1 2 3
cf-100360-428168

Find the MIRR using 6% for borrowing and 12% for investing.

3 sign changes => 3 possible RoRs

(cubic -100 360 -428 168)(1.4 1.2 1.0) => 40%, 20%, 0%.

(MIRR '(-100 360 -428 168) 6 12)  8.8%

4. Find the IRR for the cash flow -50 20 20. (quadratic -50 20 20)

(IRR '(-50 20 20)) -13.67%

5. Consider the following set of mutually exclusive alternatives;

n A B C

0-$2000-$1000 -$3000

1 1500 800 1500

2 1000 500 2000

3 800 500 1000

IRR (%) 34.7 40.76 24.81

What project would you select based on rate of return on the incremental

investment with the MARR unknown.

ans. A - B ~ 27.6% C - A ~ 8.8% A C – B ~ 17.2%

IRRA - B= (IRR '(-1000 700 500 300))  27.6%

IRRC - A= (IRR '(-1000 0 1000 200))  8.8%

IRRC - B= (IRR '(-2000 700 1500 500))  17.2%

0  MARR  8.8%: Choose C;

8.8%  MARR  27.60%: Choose A;

27.6%  MARR  40.76%: Choose B;

MARR > 40.76%: Choose Do Nothing.

6. Benefit/Cost 9-33 with 6-year life and 10% MARR

A B C

First Cost$560$340$129

UAB 140 100 40

Salvage Value 40 0 0

Must use incremental approach. AEA = 16.60; AEB = 78.06; AEC = 29.61

CRA = (P – S) (A/P, 10%, 6) + Si = $123.40; CRB = $78.07; CRC = $29.62

B/CA = 140/123.40 = 1.134; B/CB = 100/78.07 = 1.281; B/CC = 40/29.62 = 1.35.

Notice that B is best yet C's B/C ratio is higher.

BCC- = 40/29.62 = 1.35 => C is better than Do Nothing or 1

BCB-C = 60/48.45 = 1.24 => B is better than C.

BCA-B = 40/45.33 = 0.88 => B is best.

7. n 0 1 2 3

cf-500200-5001200

(cum+ '(-500 200 -500 1200)) (-500 -300 -800 400) => a unique cash flow.

(IRR ‘(-500 200 -500 1200)) 21.09%

(cubic -5 2 -5 12)  1.211 or 21.1%

8. Given that P = $900, N = 5, and S = $70; compute the following for straight line depreciation:

a. Annual depreciation charge ______.

b. Total depreciation ______.

c. Book Value for end of 3rd year ______.

9. Given that P = $900, N = 5, and S = $70; compute the following for SOYD.

a. Annual depreciation charge ______.

b. Total depreciation ______.

c. Book Value for end of 3rd year ______.

10. Given that P = $900, N = 5, and S = $70; compute the following for double declining balance.

a. Annual depreciation charge ______.

b. Total depreciation ______.

c. Book Value for end of 3rd year ______.

Note that in any year n,DDB depreciation = (2P/N)(1 – 2/N)n – 1.

d. The Total DDB depreciation for the first 4 years ______.

Total DDB depreciation is given by P[1 – (2/N)n]

11. A truck costing $3000 with a 5-year life and $750 salvage saves $800 a year. Using straight line depreciation, compute PW and AE of both the before and after tax cash flows.

Use i = 10% for an ATCF interest rate and use a tax rate of 34%. As a review exercise, compute the after tax rate of return as well.

12. The following 5 depreciation schedules are based on the same initial cost, useful life, and salvage value. Identify which are the following:

a) Straight Lineb) SOYDc) 150% DBd) DDB

e) DDB with conversion to St Linef) MACRS

Year / A / B / C / D / E / F
1 / $58.00 / 58.00 / 27 / $29.00 / $45.00 / $43.50
2 / 34.80 / 34.80 / 27 / 46.40 / 36.00 / 30.45
3 / 20.88 / 20.88 / 27 / 27.84 / 27.00 / 21.32
4 / 12.53 / 12.53 / 27 / 16.70 / 18.00 / 14.92
5 / 8.79 / 7.52 / 27 / 16.70 / 9.00 / 10.44
6 / 8.36

135 133.73 135 145 135 120.63

ED A FB C

Ans. COST = 145; S = 10

P – S = 135; P*.0576 = 8.36 => P = $145.00

13. A company buys land for $220,000 and builds a $900,000 factory and installed $650,000 worth of chemical and packaging equipment. The plant was completed and operations began on April 1. The gross income for the year was $450,000. Supplies and operating expenses were $100,000. Using MACRS depreciation a) the 1styear depreciation charge is

b) The 1st year taxable income is

c) The federal taxes for the first year is .

Chemical equipment is probably in 7-year class => $650,000 * 14.29% = $92,885.

Building is in the 39-year real property class on April 1 => $900,000 * 2.249% = $20,241.

Land is a non-depreciable asset.

Total 1st year MACRS depreciation is $113,126.

b) Taxable income = $450,000 – 100,000 – 113,126 = $236,874.

c) Federal Income tax = $22,250 + 39%(236,874 – 100,000) = $75,631.

14. The continuous uniform density function for RV X is given by f(x) = cx2 on [0, 1]. Then

a) c = .b) P(X < 1/3) = .c) E(X) = .

d) V(X) = .e) Given P(X > x) = ½, x = .

15. Given discrete distribution density shown below, calculate the variance.

X123456

P(x)c/4c/4c2c/4c/2c/4

P(X < 3.5) = ______.

16. Which alternative should be selected using B/C analysis?

AlternativeFirst CostB/C RatioJKLM

J201.10 --

K250.960.40 --

L331.221.42 2.14 --

M450.890720.800.08--

17. Benefits = $60K/yr, Disbenefits = $16K/yr and costs = $35K/yr. B/C ratio is closest to

a) 0.92b) 0.96c) 1.257d) 2.00

18.Two mutually exclusive alternatives B/C ratios are 1.5 for the lower first cost and 1.4 for the higher first cost. The B/C ratio for the increment is

a)< 1.4. b) > 1.4 c) between 1.4 and 1.5, d) the lower cost is best.

(1.4X – 1.5Y)/(X – Y) = 1.4 – 0.1Y/(X – Y) = 1.4 – positive quantity => < 1.4

19. Fill in the blanks using incremental B/C analysis to complete unfinished analysis.

AlternativeFirst CostB/C RatioPQRS

P101.10 --2.83

Q402.402.83---

R501.40 --

S801.50 --

20.AlternativePQRS
Benefits119670120 A $10K bond with coupon rate 8% /year payable quarterly was bought for $9200 and kept

until due for a total of 7 years. Find rate of return per 3 months and per year.

(UIRR 9200 200 28 10E3) 2.395% => 9.58% APR => 9.93% effective rate of return.

21. Complete the depreciation schedules below for an initial cost of $5000 and salvage of $1000 for an asset with a 4-year life.

Year DDB SOYD SLN

1$ 2,500.00 $1600 $1000.00

2 1,250.00 1200 1000.00

3 250.00 800 1000.00

4 400 1000.00

22. Which is the best of the following projects with the same service life and only investment in year 0 with the MARR at 29%, given:

A B CB – CB – AA - C

First cost-1250-1000-1200

IRR(%) 43 56 67 146 144 155

a) Ab) Bc) Cd) Do Nothing

23. Determine the acceptability of the following project’s cash flow at a MARR of 8%, using 8% for reinvesting if needed.

n 0 1 2

CF$-100,000310,000-220,000

The two IRRs are 100% and 10%. Determine the acceptability of the project.

New cash flow -100,000106296.300 => 6.29% not suitable.

NPW = -$1577.50

24. A machine is bought for $50K and kept for 5 years before it salvaged for $0. It produces $20K of revenue at an annual cost of $5K per yearb. If the machine is depreciated straight line and the effective tax rate is 30%, the present value of the after tax cash flow using a MARR of 10% is

a) $6800b) -$31,000c) $1200d) $0

30%

nBTCFCostDepTITax RateATCF

0-50K-50K
1-5 20K 5K 10K 5K -1.5K 13.5K

-50K + 13.5K(P/A, 10%, 5) = $1175.62

25. A $100K piece of equipment has no salvage value. It nets $15K per year and the interest rate is 10%. The breakeven time for the investment is closest to

a) 14 yearsb) 12 yearsc) 10 yearsd) 8 years

nBTCFDepTITax RateATCF

0-100K-100K

1-? 15K15K-1.5K 13.5K

100K = 13.5K(P/A, 10%, n) => (P/A, 10%, n) = 7.4074 => n = 14 years

26. For a 10-year life use incremental analysis pick the best of the lot. MARR at 10%

ABCDE
First cost $5K$6K$8K$9K$12K

Annual cost170016001200 1000$700

B - A (UIRR 1000 100 10) 0%

C - A(UIRR 3E3 500 10)10.56%

D – C(UIRR 1E3 200 10)15.10% D is best

E-D(UIRR 3E3 300 10) 0

27. For cash flow -500 200 -400 1100 in consecutive years, find the RoR.

(cum+ '(-500 200 -400 1100))(-500 -300 -700 400)=> unique positive RoR.

(IRR '(-500 200 -400 1100)) 22.08%

28. 10-year Life; MARR is 9%, use RoR analysis to selectbest alternative.

A B C D E

First Cost $4,000 $5,000 $2,000 $3,000 $6,000

UAB $797 $885 $259 $447 $1,063

A is best

29.A poor analysis technique for ranking alternatives is ______.

a) Benefit-cost analysis b) Incremental rate of return

c) Payback period d) Sensitivity analysis

30. Should we use asphalt or concrete for a road? Concretecosts $15,000/mile and lasts for 20 years. Annual maintenance cost for either is $500/mile. Asphalt lasts for 10 years. What is the maximum that can be spent on asphalt? MARR = 12%.

a) $30,000 b) $7,500 c) $12,923 d) $11,347

X (A/P, 12%, 10) = 15,000(A/P, 12%, 20)

X (0.1770) = 15,000 (0.1339) => X= $11,347