Engineering Economics LESSON 3b
September 1, 2022 | Author: Anonymous | Category: N/A
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LESSON 3b – Economic Study Methods I. Other Method The Rate of Return Method – Method – is is the measure of the effectiveness of an investment of capital. Formula:
= The Annual Worth (AW) Method – Method – the the excess of annual cash inflows over annual cash outflow should be not less than zero on the proposed investment to considered justified – justified – is is valid. The Present Worth (PW) Method – Method – the the present worth of the net cash flows should be equal to, or greater than zero to considered the project justified economically. The Future Worth (FW) Method – Method – exactly exactly comparable to the PW method except that all cash inflows and outflows are compounded forward to a reference point in time called the future. The Payback (Payout) Period Method – Method – is is the length of time required to recover the first cost of an investment from the net cash flow produced by that investment for an interest rate of zero. Formula:
() () = ℎ Sample Problem: 1. An investment of ₱270,000 can be made in a project that will produce a uniform annual revenue of ₱185,400 for 5 years and then have a salvage value of 10% of the investment. Out-of-pocket cost for operation and maintenance will be ₱81,000 per year. Taxes and insurance will be 4% of the first cost per year. The company expects capital to earn not less than 25% before the income taxes. Is this a desirable investment? What is the payback period of the investment? Solution: By the rate of return method: Annual revenue Annual costs: ₱185,400 ₱ , − ₱ , ₱ , = . = ₱ 29,609 Depreciation = ⁄,%,
Operation and maintenance = ₱ 81,000 Taxes and insurance = ₱270,000(4%) = ₱ 10,800 ₱121,409 Total annual cost: ₱ 63,991 Net annual profit:
63,991 = . = ₱₱270,000 .. % Since the rate of return is less than 25%, the investment is not justified. justified.
By the annual worth method: Annual revenue Annual costs:
₱185,400
Depreciation
− ₱, ₱, = ₱, ⁄ ,%, = . = ₱ 29,609
Operation and maintenance = ₱ 81,000 Taxes and insurance = ₱270,000(4%) = ₱ 10,800 = ₱ 67,500 Interest on capital = ₱270,000(25%) Total annual cost: ₱188,909 Since the excess of annual cash inflows over annual cash outflows is less than zero ( – ( –₱3,509), the investment is not justified. justified.
By the present worth method:
₱185,400 ,25%,5+ ₱27,000 ,25%,5 = ₱185,400(2.6893) + ₱27,000(0.3277) = ₱506,370
PW of cash inflows =
Annual costs (excluding depreciation) d epreciation) = ₱81,000+ ₱270,000 270,000((4% 4%)) = ₱91,800
PW of cash outflows = ₱270,000+ ₱91,800 ,25%,5 = ₱516,880
justified. Since the PW of the net cash flows is less than zero ( – –₱10,510), the investment is not justified.
By the future worth method:
FW of cash inflows = ₱27,000+ ₱185,400 ,25%,5 = ₱27,000+ ₱185,400(8.2070) = ₱1,548,580
₱91,800 ,25%,5+ ₱270,000 ,25%,5 = ₱91,800(8.2070)+ ₱270,000 270,000((3.0518 3.0518)) = ₱1,577,390
FW of cash outflows =
justified. Since the FW of the net cash flows is less than zero ( – –₱28,810), the investment is not justified.
By the payback period: Total annual cost = ₱81,000+ ₱270,000(4%) = ₱91,800 Net annual cash flows = ₱185,400 ₱91,800 = ₱93,600 Payback period =
− ₱, − ₱, = 2.6 years years = ₱,
2. An engineer is considering building a 25-unit apartment in a place near a progressive commercial center. He felt that because of the location of the apartment it will be occupied 90% at all time. He desires a rate of return of 20%. Other pertinent data are the following: Land investment Building investment Study period Cost of land after 20 years Cost of building after 20 years
₱5,000,000
7,000,000 20 years 20,000,000 2,000,000
Rent per unit per month Upkeep per unit per year Property taxes Insurance
6,000 500 1% 0.50%
Is this a good investment? Solution: Annual income: = Rental = (₱6,000)(12)(25)(0.90) ₱,, − ₱,, ₱,, ₱1,620,000 Land = = ₱ =
⁄,%,
₱,
Total annual income:
Annual costs: Depreciation
− ₱,, = ₱,, ⁄,%,
80,350 ₱1,700,350
= ₱ 26,780
Upkeep = ₱500(25) = ₱ 12,500 = ₱120,000 Taxes = ₱12,000,000(1%) = ₱ 35,000 Insurance = ₱7,000,000(0.5%) Total annual cost: ₱194,280 Net annual profit: ₱1,506,070
1,506,070 = . .% .% = ₱₱12,000,000 Since the rate of return is less than 20%, the engineer should not invest. invest. Other solution: Investment = ₱5,000,000 + ₱7,000,000 Amount of investment after af ter 20 years = ₱20,000,000 Annual income: Rental = (₱6,000)(12)(25)(0.90) Annual costs: Depreciation
− ₱,, = ₱,, ⁄,%,
= ₱12,000,000 + ₱2,000,000 = ₱22,000,000 = ₱1,620,000
= ₱ 53,570
Upkeep = ₱500(25) = ₱ = ₱ Taxes = ₱12,000,000(1%) = ₱ Insurance = ₱7,000,000(0.5%) Total annual cost: Net annual profit:
12,500 120,000 35,000 ₱113,930
₱1,506,070 ₱ 1,506,070 = ₱12,000,000 = . .% .% Since the rate of return is less than 20%, the engineer should not invest. invest.
3. A man is considering investing ₱500,000 to open a semi-automatic auto-washing business in a city of 400,000 populations. The equipment can wash, on the average, 12 cars per hour, using two men to operate it and to do small amount of hand work. The man plans to hire two men, in addition to himself, and operate the station on an 8-hour basis, 6 days per week, 50 weeks per year. He will pay his employees ₱25.00 per hour. He expects to charge ₱25.00 for a car wash. Out-of-pocket miscellaneous cost would be ₱8,500 per month. He would pay his employees for 2 weeks for vacations each year. Because of the length of his lease, he must write off his investment within 5 years. His capital now is earning 15% and he is employed at a steady job that pays ₱25,000 per month. He desires a rate of return of at least 20% on his investment. Would you recommend the investment? Solution:
By the rate of return method: Annual revenue = (12)( (12)(₱₱25.00)(8)(6)(50) Annual costs: ₱, ₱, Depreciation = ⁄,%, = .
Labor = (2)(48)(50)( (2)(48)(50)(₱₱25.00) Vacation pay = (2)(2)(48)( (2)(2)(48)(₱₱25.00) Miscellaneous = ₱8,500 (12) Owner’s salary = ₱25,000 (12) Total annual cost: Net annual profit:
₱720,000 = ₱ 74,160
= ₱120,000 = ₱ 4,800 = ₱102,000 = ₱102,000
₱600,960 ₱119,040
= ₱₱119,040 .% 500,000 = . .% Since the rate of return is greater than 20%, the man should invest invest..
By the annual worth method: Annual revenue = (12)( (12)(₱₱25.00)(8)(6)(50) Annual costs: ₱, ₱, Depreciation = ⁄,%, = .
Labor = (2)(48)(50)( (2)(48)(50)(₱₱25.00) Vacation pay = (2)(2)(48)( (2)(2)(48)(₱₱25.00) Miscellaneous = ₱25,000 (12) Interest on capital = ₱500,000 (0.20) Total annual cost: Net annual profit:
₱720,000 = ₱ 74,160 = ₱120,000 = ₱ 4,800 = ₱300,000 = ₱100,000
₱700,960 ₱ 19,040
Since the excess of annual revenue over annual cost is greater than zero, the investment is justified. The man should invest invest..
Assignment: 1. A young engineer is considering establishing his own small company. An investment of ₱400,000 will be required which will be recovered in 15 years. It is estimated that sales will be ₱800,000 per year and that operating expenses will be as follows: Materials Labor Overhead Selling expense
₱160,000 ₱280,000
per year per year ₱40,000 + 10% of sales per year ₱60,000 per year
The man will give up his regular job paying ₱216,000 per year and devote full time to the operation of the business; this will result in decreasing labor cost by ₱40,000 per year, material cost by ₱28,000 per year and overhead cost by ₱32,000 per year. If the man expects to earn at least 20% of his capital, should he invest?
2. A firm is considering purchasing equipment that will reduce cost by ₱40,000. The equipment costs ₱300,000 and has a salvage value of ₱50,000 and a life of 7 years. The annual maintenance cost is ₱6,000. While not in use by the firm, the equipment can be rented to others to generate an income of ₱10,000 per year. If money can be invested for an 8 per cent return, is the firm justified in buying the equipment?
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