ENM 530 Final Review

ENM 530 Final Review

1. Given first cost is $25K and MARR is 12%, find the economic life of machine with the following cash flow :

n12345

Salvage$22K21K20K16K10K

O&M$6K8K10K12K14K

n EUAC

0 $0.00

1 $6.00 ***

2 $6.94

3 $7.85

4 $8.72

5 $9.55

EUAC$60004887448248835361

2. First cost of a machine is $1050K with salvage value $225K when sold. Yearly maintenance and operating costs are $235K with a gradient of $75K. MARR is 10%. Find the economic life.

P = 1050KS = 225KAOC = 235K with 75K gradient MARR = 10%.

EUAC = -1050K(A/P, 10%, n) + 225K(A/F, 10%, n) - 235K - {75K(A/G, 10%, n)}

EUAC1 = 1165K; EUAC2 = 769K; EUAC3 = 659,487; EUAC4 = 621,351

EUAC5 = 610,892;EUAC6 = 613,693

3. Find the B/C ratio of a $1K investment, which returns $500 at the end of every year for 5 years with the MARR at 10%

500/1000(A/P, 10%, 5) = 500/263.80 = 1.9

4. nDefenderMARR = 15% Challenger

Op Cost Book Value Op costBook value

0 $12K $35K

1$3,400 7K0.2K 30K

2 3,900 4K1K 27K

3 4,600 2.5K1.2K 24K

4 5,600 1.0K1.5K 20K

52K 17K

62.6K 15K

Is it worth replacing the defender if required only for the next 4 years?

Defender

YearMV lossInterestOp-costMarginalEUAC

1 5K1800 3400 10,20010,200

2 3K1050 3900 7,950 9153

3 1.5K 600 4600 6,700 8447

4 1.5K 375 5600 7,475 8252

Not Decreasing

Challenger

YearMV lossInterestOp-costMarginalEUAC

15K525020010,45010450

23K45001K 8,500 9543

33K40501200 8,250 9171

44K36001500 9,100 9158

8985

8774

Do not replace Defender

5. 5. What amount should you deposit into a savings that pays 8% with f = 10% to purchase after 10 years an article currently costing $10,000.

10K(1.1)10(1.08)-10 = $12,014 or 10K(1 - 0.01818)-10 = 12.014 with ir = -1.818

6. You want a true rate of 10% a year after allowing for inflation at 4% per year. What after tax return must you earn? 14.4%. 14.4%

7. Over 8 years prices have increase by a total of 65%. Find the annual inflation rate.

1.65 = (1 + f)8 => = 6.46%

8. Given RV X has density f(x) = cx2 on [0, 1],

a) c = 3,b) P( ½ < X < ¾) = ¾3 – ½3

c) E(X) = 3/4,d) V(X) = ______.

9. Budget is $600,000. Find the best allocation for the following 5 independent projects:

ProjectFirst CostNPW

A$200K$50K

B 250K 60K

C 300K 70K

D 150K 80K

There are 4 + 6 + 4 + 1 = 15 combinations: The 4 individually as shown in table.

There is the Do Nothing as well.

CombinationAB AC AD BC BDCD ABC ABDBCD ACD ABCD

Required $450 500 350 550 400450 750 600700 650 950

Total Net PW110 120 130 130 140150 180 190210 200 260

10. A 100-room hotel is bought for $5M. A 25-year loan is available for 10%. Study data shows

OccupancyProbability

65% full0.4

70%0.3

75%0.2

80%0.1

The operating hotel costs are: Taxes and insurance$70K annually

Maintenance$60K annually

Operating$260K annually

The life of the hotel is 25 years when operating 365 days per year with salvage value $500K. Neglect tax credit and income taxes. Determine the average rate to charge per room per night to return 20% of the initial cost per year.

AE = 71.5% occupancy 70K + 60K + 260K + 1M – 500K(A/F,10%, 25)

= 1,334,915.96

= 100 * 0.70 * 365 * X = 1,334,915.96 => X = $52.21.

11. Retain or Replace given the following data:

CHALLENGER DEFENDER

nAW Replace $/YearAW Retain $/Year

1-25.5K-27K

2-25.5K-26.5K

3-26.9K-25K

4-27K-25.9K

The defender should be replaced:

a) After 4 more yearsb) after 3 more yearsc) after 1 more yeard) Now

12. To make, see AOC table given the first cost is $2M with $50K salvage value after 10 years with MARR at 10%. To buy is $1.5M per year.

Department / Basis / Rate/hour / Allocated / Direct Material / Direct Labor
A / Labor / $10 / 25,000 / $200,000 / $200,000
B / Machine / 5 / 25,000 / 50,000 / 200,000
C / Labor / 15 / 10,000 / 50,000 / 100,000
$300,000 / $500,000

Indirect cost

Dept A25,000 * 10 = $250,000

Dept B25,000 * 5 = 125,000

Dept C 10,000 * 15 = 150,000

$525,000

AOC = direct labor + direct material + indirect costs

= $500,000 + $300,000 + $525,000

= $1,325,000

To make: AC = (2M – 50K)(A/P, 10%, 10) + 50K * 0.10 + 525K

= $1,647,353.52

Conclude it is still cheaper to buy at $1.5M per year.

13. Total mechanical and electrical unit costs for a library are listed at $34 per ft2 while total project cost is estimated at $114 per ft2. Based on these values, the percentage of total costs represented by mechanical and electrical costs is closest to: c

a) 21%b) 29%c) 34%d) 38%e) none of the above

14. Given that the cost to make the first auto chassis is $5000 and the cost to make the 13th one is $1300.90, find the slope of the learning cost and the learning rate.

T13 = T1 * nb

1300.9 = 5000 * 13b

b = -0.5251852

LC% = (log LC% 2) = -0.5251852 => 69.5% = 2-0.5251852

15. Develop an income statement given the account balances shown below with taxes at 34% combine federal and state.

Payroll and administrative expenses $16,000

Subcontract expenses 20,000

Development expenses 18,000

Interest payment 400

Sales revenue 65,000

Selling expenses 9000

Revenue

Sales revenue$65,000

Total 65,000

Expenses

Payroll and administrative expenses $16,000

Subcontract expenses 20,000

Development expenses 18,000

Selling expenses 9000

Total$63,000

Total operating income = $65,000 - $63,000 = $2000

Non-operating revenues and expenditures

Interest payments $400

Taxable income2000 – 400 = $1600

Income taxes@34% of $1600 = $544

Total = 400 + 544 = $944

Net profit after taxes = Total operating profit = $1056

16. Determine the economic service life at i = 10% per year for a machine with first costs of $10K given the following data:

nO&MSalvage value

1-10007000

2-12995000

3-15004200

4-20003000

5-30002000

17. Plan 1: Receive $100K now

Plan 2: Receive $15K per year for 8 years beginning 1 year from now.

Plan 3: Receive $45K now, another $45K 4 years from now, and $45K 8 years from now.

The real rate is 6% and inflation is at 4% per year.

Im = 10.24%

Which plan would you choose based on present worth?

P1~PW = $100K

P2~PW = 15K(P/A, 10.24, 8) = $79,329.48

P3~PW = 45K + 45K(P/F, 10.42, 8) + 45K(P/F, 10.42, 8)

= $96,098.80

18.A machine costing $250,000 is bought with a down payment of $160,000 and 10% compound annual interest for 6 years for the balance. Annual revenue for each of the 6 years is $40,000. The machine is in the 5-year MACRS class. Salvage value is $25,000.

Write the 3rd year BTCF to ATCF line. Use 34% combined federal & state tax rate.MACRS %s = (20.0 32.0 19.2 11.52 11.52 5.76)

Loan Amount = $ 90000.00 at 10.00 % for N = 6 pay periods

Pay Num Payment Principal Interest Balance Total Interest

1 20664.66 11664.66 9000.00 78335.34 9000.00

2 20664.66 12831.13 7833.53 65504.21 16833.53

3 20664.66 14114.24 6550.42 51389.97 23383.95

4 20664.66 15525.66 5139.00 35864.31 28522.95

5 20664.66 17078.23 3586.43 18786.08 32109.38

6 20664.66 18786.05 1878.61 0.03 33987.99

nBTCFDepInterest TI 34%Tax RateATCF

0-250K-$250K

1 40K50K 9K-19K 6460 25,795.34

2 40K80K 7833.53-47.833.53 16263.40 35598.74

3 40K48K 6550.42 -14550.42 4947.14 24282.48

4 40K28.8K 5139.00 6061 -2060.74 17274.60

5 40K28.8K 3586.43 7613.57 -2588.61 16746.73

6 40K14.4K 1878.6123721.39 -8065.27 11270.07

6 25K 25K

19. You buy a $5000 par-value bond at 7% annual coupon rate with inflationat 4%. Compute your real rate of return if you sell the bond for $4850 after 2 years.

Cash flow –5K 350 5200  5.54% actual => real rate of return = 1.48%.

(IRR ‘(-5000 350 5200)  5.54%

20. Plan 1: $50K now

Plan 2: Receive $8K for 8 years

Plan 3: Receive $23K now, another $23K 4 years from now and another $23K 8 years from now.

The real rate of interest is 7% and inflation is 3%.

a) Which plan has the highest PW?

b) Determine FW of each in inflated dollars 8 years from now.

c)Determine future value of each in terms of today's purchasing power.

21. AlternativeCost ($M)B/C vs. DNJKLM

J201.1-

K250.960.4-

L331.221.422.14-

M450.890.720.800.08-

The 4 mutually exclusive alternatives are compared using B/C analysis. The chosen alternative is a) J b) K c) L d) M

22. Find the interest on the 3rd payment of a $12K loan at 6% for 10 years.

(- (AGP 12e3 6 10) (PGF (AGP 12e3 6 10) 6 8))  $607.47

23. Given EVA = NPAT – cost of invested capital

= NPAT – (after tax interest rate)(book value in year n -1)

= TI(1 – Trate) – i * BVn-1

First cost$500KMARR = 12% tax rate 40%

Gross income – expenses = $170K/year

Life4 years

Salvage value = $0

Find EVA evaluation using straight line depreciation

nBTCFDepBVTITaxNPATCICEVAATCF

0-500K-500K

1 170K 125K 375K45K-18K 27K -$60K -$33K 152K

2250K 27K- 45K -18K

3125K- 30K -3K

4 0-15K 12K

Cost of Invested capital (CIC) = 12% of Book Value of previous year.

EVA (list-pgf '(-33e3 -18e3 -3e3 12e3) 12)  -$38,322.90

(AGP 38322.90 12 4)12617.217024

(+ -500e3 (PGA 152e3 12 4)) -$38,322.88

24. The AT RoR with a 50% tax rate and 3 year straight-line depreciation (zero salvage value)

is closest to

a) 5% b) 7%c) 9%d) 11%

n 0123

cf-21K9K9K9K

Repeat with inflation set at 4%.

25.AutoCurrent Price 3-year inflation %
X$27K4

Y 30K1.5

Z 25K8

26. Problem 14-55.

Year / A / B
0 / -420 / -300
1 / 200 / 150
2 / 200 / 150
3 / 200 / 150

No salvage with = 5% and 25% tax rate using straight line depreciation, choose the better after tax mutually exclusive alternative using MARR of 7%.

Cash flows are in Year 0-$.

nBTCF Year-nDepTI25% Tax RateATCFR$ n=0

0-420-420 -420

1 200 21014070 - 17.5192.5 183.33

2 200220.514080.5 - 20.13200.38 181.75

3 200213.5314091.53 - 22.88208.64 180.23

nBTCF Year-nDepTI25% Tax RateATCFR$ n=0

0-300-300 -300

1 150 157.5010057.5 - 14.38143.12 136.30

2 150165.3810065.38 - 16.34149.04 135.18

3 150173.6410073.64 - 18.41155.23 134.09

Find RoR for A - B

(IRR ‘(-120 47.03 46.57 46.14))  8.05% => A is better

27. Problem 15-3 Lease: Pay $267/month for 24 months or Buy: A = 9400(A/P, 1%, 24) = $442.49

Salvage value = 9400/2 = $4700.

Purchase: (442.29 – 267) = $175.74 saved per month plus salvage

175.74(F/A, i%, 24) = 4700 => (F/A, i%, 24) = 426.744700/175.74 = 26.744 => i = 0.93% or

11.16% APR => Prefer leasing about 11.2% MARR.

28.Problem 17-8

Project A: A1 Sell for $500K; A2 Lease to get annual $98,700 for 20 years and $750K salvage value

A2 – A1 (UIRR 500e3 98700 20 750e3)  20%

A3 – A1 Expected rental income (EV '(1 1.1 1.2 1.9) '(0.1 0.3 0.4 0.2))  $1.29M

-500K 0 0 1.29e6 from 3-20

(IRR '(-5e6 0 0 1.29e6 1.29e6 1.29e6 1.29e6 1.29e6 1.29e6 1.29e6 1.29e6 1.29e6 1.29e6 1.29e6 1.29e6 1.29e6 1.29e6 4.29e6))  17.67%

Note $4.5M to build - $500K to get -$5M

A3 – A2 (IRR '(-4.5e6 -98700 -98700 1191300 1191300 1191300 1191300 1191300

1191300 1191300 1191300 1191300 1191300 1191300 1191300 1191300 1191300 1191300 1191300 1191300 2250000)) 17.65%

Project B 90 days at 13¾% per year compounded continuously => e0.1375 – 1 = 14.74%

Project C (UIRR 2e6 500e3 10 2e6) 25%

Project D 16%

Project E 14.06 APR compounded monthly =>15%

Project F (IRR '(-2e6 1e6 1.6048e6) 18%

Rank order C25%

A220%

F18%

A3-A217.65%

D16%

E15%

B14.74%

Budget $4M or $4.5M with land, choose C, A2, and F

Budget = $9M + $500K land, choose C, F, A3, and D.

29. 14-10 An economist predicts 8% inflation for next 5 years and 6% for the following 5 years.

Find the average price change per year for the 10-year period/

F = 1(1 + 0.08)5(1 + 0.06)5 = 1.9663

(1.9663 – 1)/ 10 = 9.663 cents per year

30.14-11 In the last 5 years, prices have increased a total of 50%. What is the equivalent annual inflation rate?

1.5 = 1(1 + i)5 => I = 8.447%

31.15-2 10-year life and no salvage value, find range of MARR for choosing C.

ABCD
First cost$0$100$50$25

UAB 0 16.27 9.96 5.96
RoR0% 10% 15% 20%

C – D (UIRR 25 4 10) 9.6%

B – C (UIRR 50 6.31 10)  4.47%