eng. economy

8/10/2019 Eng. Economy

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One-half of the machines of the kind will be replaced after 8 years and the rest will be sold after

12 years. Compute the total annual straight-line depreciation charges by (a) the group

depreciation method, and (b) the composite depreciation method.

SOLUTION:

a.)

Group Depreciation Method

⁄ ⁄

DEPRECIATION FOR EACH YEAR IS SHOWN IN THE TABLE BELOW.

YEAR NUMBER IN SERVICE ANNUAL DEPRECIATION ACCOMULATED

1 2 3 4 DEPRECIATION1 – 8 8 6 4 4 P49, 600 P396, 400

9 – 12 4 3 2 2 P24, 000.05 P198, 400.40

P595, 200.05

DEPRECIATION FOR YEARS 9 – 12 = P2254.55(4 + 3 + 2 + 2) = P24, 800.05

b.) Composite Depreciation Method:

The annual depreciation amounts for each machine are:

⁄

⁄

⁄

⁄

Composite Depreciation amounts:

Years 1 – 8:

P18, 600.67 + P14, 400 + P5, 600 + P9, 000 = P47, 666.67

Accumulated Depreciation after 8 years = 8(47, 666.67) = P381, 333.36

Years 9 – 10:

18, 666.67 + P14, 400 + P5,600 = P38, 666.67

Accumulated Depreciation after 10 years = P381, 333.36 + 2(38, 666.67) = P458, 666.70



Years 11 – 12: (Machine 1 only)

Accumulated Depreciation after 12 years = P450, 666.70 + 2(P18, 666.67) = P496, 000.00

The total accumulated depreciation by either method after 12 years = P496, 000.00

4. A firm plans to market a new minicomputer that will sell for $10,000. This financing scheme is

being considered by the company: Payments of $872 each year for 20 years would be made by

the purchaser after an initial down payment is made. If their interest is charge is 6%

compounded monthly, what down deposit should the company request?

GIVEN:

Cost = $10, 000

A =$872

n = 20 yearsi = 6% compounded monthly

REQ’D:

DOWN DEPOSIT = ?

SOLUTION:

* +

⁄

5. A man deposit $2,000 an a savings account when his son was born. The nominal interest rate

was 8% per year, compounded continuously. On the son’s birthday, the accumulated sum is

withdrawn from the account. How much would this accumulated amount be?

GIVEN:

P = $2000

Nominal rate of interest = 8% per year compound continuously

N = 18

REQ’D:

F = ?

SOLUTION:

F = $8, 441



6. Maintenance costs for a new bridge with an expected 50-year life are estimated to be $1,000

each year for the first 5 years, followed by a $10,000 expenditure in the 15th year and a $10,000

expenditure in year 30. If I = 10% per year, what is the equivalent uniform annual cost over the

entire 50-year period?

GIVEN:

i = 10%

n = 50 years

A1 = $ 1,000 maintenance cost each year for the first 5 years

* + money worth after 5 years

C2 = $ 10,000 expenditure during the 15th year

C3 = $ 10,000 expenditure during the 30th year

Required:

P = ?& A = ?

SOLUTION:

Since,

Therefore;

7. Uncle Wilbur’s trout ranch is now for sale for $40,000. Annual property taxes, maintenance,

supplies, and so on, are estimated to continue to be $3,000 per year. revenues from the ranch

are expected to be $10,000 next year and then decline by $500 per year thereafter through the

tenth year. if you bought the ranch, you would plan to keep it only 5 years and at the time sell it

for the value of the land, which is $15,000. If your desired annual rate of return is 12% should

you become a trout rancher?

SOLUTION:

TOTAL EXPENSES = 15, 000+ 40, 000 = $55, 000 for 5 years

An expenses every year = 11, 000

TOTAL REVENGE = 45, 000 + 15, 000 = $60, 000 for 5 years

An expenses every year = 12, 000

FPROFIT EVERY YEAR = 12, 000 – 11, 000 = 1, 000

8. The heat loss through the exterior walls of a certain poultry processing plant is estimated to cost

the owner $3,00 next year. a salesman from super fier Insulation, Inc. has told you, the plant

engineer, that he can reduce the heat loss by 80% with the insallation of $15,000 worth of Superfiber now. If the cost of heat loss rises by $200 per year (gradient) after next year and the



ownder plans to keep the present building for 10 more years, what would yoy recommended if

the money is 12% per year?

Illustration:

1 2 3 4 5 6 7 8 9 10

Without Super Fiber Insulation

1 2 3 4 5 6 7 8 9 10

With Super Fiber Insulation

SOLUTION:

When i = 12%

Without Super Fiber Insulation

* +

* +

With Super Fiber Insulation

* +

* +

Therefore:

The owner must not accept the offer of the Salesman.

9.

Solve for the value of F below so that the left-hand cash flow diagram is equivalent to the one

on the right. Let I = 8% per year.

P = ?

3000

F = ?

3200

36003400

3800 40004200

44004600

P = ?

F = ?

600800

12001000

14001600

18002000

22002400

15000

4800



10. A corporations floats P200,000 worth of ten-year callable bonds is P1,000 denominations. The

bond rate is 7% compounded annually. Prepare an amortization.

GIVEN:

F = P200, 000 n = 10 i = 7%

SOLUTION:

A * + *

+

28475.50

14000 = P14, 475.50

YEAR PRINCIPALINTEREST AT

7%

NO. OF BONDS

RETIRED

AMOUNT OF

PRINCIPAL

REPAID

YEAR END

PAYMENT

1 P200, 000 P14, 000 14 P14, 000 P28, 000

2 186, 000 13, 020 15 15, 000 28, 020

3 171, 000 11, 970 17 17, 000 28, 970

4 154, 000 10, 780 18 18, 000 28, 780

5 136, 000 9, 520 19 19, 000 28, 520

6 117, 000 8, 190 20 20, 000 28, 190

7 97, 000 6, 790 22 22, 000 28, 790

8 75, 000 5, 250 23 23, 000 28, 250

9 52, 000 3, 640 25 25, 000 28, 640

10 27, 000 1, 890 27 27, 000 28, 890

TOTALS P1, 215, 000 P85, 120 200 P200, 000 P285, 120

11. A man borrowed P150,000 from a bank for home improvement, to be repaid by month-end

payment for 60 months. The current rate of the interest charge by banks is 19% compounded

monthly. Based on this rate, prepare an amortization schedule.

GIVEN:

P = 150, 000.00

n = 60 months (period)

I = 19% compounded monthly =

SOLUTON:

A = * +*

+

PERIOD

PRINCIPAL AT

THE BEGINNING

OF EACH 6

M0NTHS

INTEREST AT 4%

PER PERIOD

PAYMENT AT

END OF EACH

PERIOD

PERIODIC

PAYMENT TO

PRINCIPAL

1 150000 2374.999995 3891.08 1516.080005



2148483.92 2350.995395 3891.08 1540.084605

3146943.8354 2326.610722 3891.08 1564.469278

4 145379.3661 2301.839959 3891.08 1589.240041

5143790.1261 2276.676991 3891.08 1614.403009

6142175.7231 2251.11561 3891.08 1639.96439

7140535.7587 2225.149508 3891.08 1665.930492

8138869.8282 2198.772275 3891.08 1692.307725

9137177.5205 2171.977403 3891.08 1719.102597

10135458.4179 2144.758278 3891.08 1746.321722

11133712.0961 2117.108184 3891.08 1773.971816

12131938.1243 2089.020297 3891.08 1802.059703

13130136.0646 2060.487685 3891.08 1830.592315

14128305.4723 2031.503307 3891.08 1859.576693

15

126445.8956 2002.06001 3891.08 1889.01999

16124556.8756 1972.150526 3891.08 1918.929474

17122637.9461 1941.767477 3891.08 1949.312523

18120688.6336 1910.903362 3891.08 1980.176638

19118708.457 1879.550565 3891.08 2011.529435

20116696.9275 1847.701349 3891.08 2043.378651

21 114653.5489 1815.347854 3891.08 2075.732146

22112577.8168 1782.482095 3891.08 2108.597905

23110469.2188 1749.095961 3891.08 2141.984039

24108327.2348 1715.181214 3891.08 2175.898786

25106151.336 1680.729483 3891.08 2210.350517

26103940.9855 1645.732267 3891.08 2245.347733

27101695.6378 1610.180928 3891.08 2280.899072

2899414.7387 1574.066693 3891.08 2317.013307

2997097.72539 1537.380649 3891.08 2353.699351

3094744.02604 1500.113743 3891.08 2390.966257

3192353.05979 1462.256777 3891.08 2428.823223

32 89924.23656 1423.800409 3891.08 2467.279591



3387456.95697 1384.735149 3891.08 2506.344851

3484950.61212 1345.051356 3891.08 2546.028644

3582404.58348 1304.739236 3891.08 2586.340764

3679818.24271 1263.78884 3891.08 2627.29116

3777190.95155 1222.190064 3891.08 2668.889936

3874522.06162 1179.93264 3891.08 2711.14736

3971810.91426 1137.00614 3891.08 2754.07386

4069056.8404 1093.399971 3891.08 2797.680029

4166259.16037 1049.10337 3891.08 2841.97663

4263417.18374 1004.105407 3891.08 2886.974593

4360530.20914 958.3949761 3891.08 2932.685024

4457597.52412 911.9607966 3891.08 2979.119203

4554618.40492 864.7914094 3891.08 3026.288591

4651592.11633 816.8751734 3891.08 3074.204827

4748517.9115 768.2002638 3891.08 3122.879736

4845395.03176 718.7546681 3891.08 3172.325332

4942222.70643 668.5261837 3891.08 3222.553816

5039000.15261 617.5024151 3891.08 3273.577585

5135726.57503 565.6707701 3891.08 3325.40923

52

32401.1658 513.0184574 3891.08 3378.061543

5329023.10426 459.5324831 3891.08 3431.547517

5425591.55674 405.1996475 3891.08 3485.880352

5618564.60293 293.9395458 3891.08 3597.140454

5714967.46248 236.984822 3891.08 3654.095178

5811313.3673 179.1283152 3891.08 3711.951685

59 7601.415613 120.3557469 3891.08 3770.724253

603830.69136 60.65261307 3891.08 3830.427387

TOTAL5271477.736 83465.06397 233464.8 150000

12. (M.E. Board, November 1983) On January 1, 1978 the purchasing manager of a cement company

bought a new machine costing P140,000. Depreciation has been computed by the straight-line

method, based on an estimated useful life of 5 years and residual scrap value)12,800.



SOLUTION:

* +

By trial and error:

effective rate compounded annually

To get for the equivalent effective rate compounded quarterly:

( )

√

THEREFORE:

15. A debt of P10,000 with interest at the rate law of 8% payable semi-annually is to be amortized

by equal payments at the end of each 6 months for 4 years. Find the semi-annual payment and

contract and amortization schedule.

GIVEN:

P = 10, 000 n = 4(2) = 8 quarters

SOLUTION:

A = P ( ⁄ * +

PERIODPRINCIPAL AT THE

BEGINNING OF

EACH 6 M0NTHS

INTEREST AT 4%

PER PERIOD

PAYMENT AT END

OF EACH PERIOD

PERIODICPAYMENT TO

PRINCIPAL

1 P10, 000.00 P400.00 P1,485.28 P1, 085.28

2 8, 917.72 356.59 P1,485.28 1,228.69

3 8, 786.03 311.44 P1,485.28 1, 173.84

4 6, 612.19 264.49 P1,485.28 1, 220.79

5 5, 391.70 451.66 P1,485.28 1, 269.62

6 4, 121.78 164.87 P1,485.28 1, 320.41



7 2, 801.37 112.05 P1,485.28 1, 373.23

8 1, 428.14 57. 13 P1,485.28 1, 428.15

TOTALS P47, 055.63 P1882.23 P11, 882.24 P10, 000.00

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