Homework Answers
| Alternative J | |||||||
| Cash Outflows | 41000000 | ||||||
| Depreciation | 6833333 | ||||||
| Year 0 | 41000000 | 0 | -41000000 | 1 | -41000000 | ||
| Savings Due to Dep = A | Expenses = B | Net Cash Flow ( A- B) | PV Factor @ 13% | Net Present Value | |||
| Year 1 | 6833333 | 1100000 | -5733333 | 0.884955752 | -5073746 | ||
| Year 2 | 6833333 | 1100000 | -5733333 | 0.783146683 | -4490041 | ||
| Year 3 | 6833333 | 1100000 | -5733333 | 0.693050162 | -3973488 | ||
| Year 4 | 6833333 | 1100000 | -5733333 | 0.613318728 | -3516361 | ||
| Year 5 | 6833333 | 1100000 | -5733333 | 0.542759936 | -3111824 | ||
| Year 6 | 6833333 | 1100000 | -5733333 | 0.480318527 | -2753826 | ||
| Net PV of Cash Outflows | -63919285 | ||||||
| Alternative K | |||||||
| Cash Outflows | 66000000 | ||||||
| Depreciation | 6600000 | ||||||
| Year 0 | 66000000 | 0 | -66000000 | 1 | -66000000 | ||
| Savings Due to Dep = A | Expenses = B | Net Cash Flow ( A- B) | PV Factor @ 13% | Net Present Value | |||
| Year 1 | 6600000 | 1800000 | -4800000 | 0.884955752 | -4247788 | ||
| Year 2 | 6600000 | 1800000 | -4800000 | 0.783146683 | -3759104 | ||
| Year 3 | 6600000 | 1800000 | -4800000 | 0.693050162 | -3326641 | ||
| Year 4 | 6600000 | 1800000 | -4800000 | 0.613318728 | -2943930 | ||
| Year 5 | 6600000 | 1800000 | -4800000 | 0.542759936 | -2605248 | ||
| Year 6 | 6600000 | 1800000 | -4800000 | 0.480318527 | -2305529 | ||
| Year 7 | 6600000 | 1800000 | -4800000 | 0.425060644 | -2040291 | ||
| Year 8 | 6600000 | 1800000 | -4800000 | 0.376159862 | -1805567 | ||
| Year 9 | 6600000 | 1800000 | -4800000 | 0.332884833 | -1597847 | ||
| Year 10 | 6600000 | 1800000 | -4800000 | 0.294588348 | -1414024 | ||
| Net PV of Cash Outflows | -92045969 | ||||||
| Differenc of Cash Dflow between Alternatvie | Alternatie K - Alternatvie J | ||||||
| 92045969 - 63919285 = 28 Millions | Lower Cash Fow on Choosing Alternatuve J | ||||||
| Here from apparnet look the Outflow is heigher in Case of Alternative K , as the Period is higher both Plant will do the Same work | |||||||
| then why you Spend more cash outflow for Alternative K as compared to Alternative J. | |||||||
| It is Advisable to Select the Alternatvie J as that Alternative has les Cash Outflow. |
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Use the imputed market value technique to determine the better alternative below. The MARR is 13%...
Use the imputed market value technique to determine the better alternative below. The MARR is 10%...
Use the imputed market value technique to determine the better alternative below. The MARR is 10% per year and the study period is six years. 0 Alternative Alternative K Capital Investment, millions 47 62 Annual Expenses, millions 11 15 Useful Life, years 6 9 Market Value (End of useful life) Click the icon to view the interest and annuity table for discrete compounding when the MARR is 10% per year. The present worth of Alternative J over six years is...
Problem 6-2 (book/static) Question Help Years SS.OCC $6,000 58.000 511000 s14000 517000 Click the icon to...
Problem 6-2 (book/static) Question Help Years SS.OCC $6,000 58.000 511000 s14000 517000 Click the icon to view the interest factors for discrete compounding when i = 13% per year. i More Info The equivalent annual worth is $ . (Round to the nearest doller.) Single Payment Equal Payment Series Gradient Series N Compound Amount Factor (F/P, I, N) 1.1300 1.2769 1.4429 1.6305 1.8424 Present Worth Factor (P/F, I, NJ 0.8850 0.7831 0.6931 0.6133 0.5428 Compound Amount Factor (F/A,,N) 1.0000 2.1300...
1 year= $5,000 2 year=$5,000 3 year= $8,000 4 year= $11,000 5 year= $14,000 6 year=...
1 year= $5,000
2 year=$5,000
3 year= $8,000
4 year= $11,000
5 year= $14,000
6 year= 17,000
Problem 6-2 (book/static) Question Help Years SS.OCC $6,000 58.000 511000 s14000 517000 Click the icon to view the interest factors for discrete compounding when i = 13% per year. i More Info The equivalent annual worth is $ . (Round to the nearest doller.) Single Payment Equal Payment Series Gradient Series N Compound Amount Factor (F/P, I, N) 1.1300 1.2769 1.4429 1.6305 1.8424...
A firm has decided to manufacture biodegradable golf tees. These are two production processes av...
A firm has decided to manufacture
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MARRequals=1313 %
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Use the imputed market value technique to determine the better alternative below. The MARR is 12%...
Use the imputed market value technique to determine the better alternative below. The MARR is 12% per year and the study period is four years. Capital Investment, millions Annual Expenses, millions Useful Life, years Market Value (End of useful life) Alternative J 41 10 4 0 Alternative K 56 20 9 o Click the icon to view the interest and annuity table for discrete compounding when the MARR is 12% per year. The present worth of Alternative J over four...
For an interest rate of 13% compounded annually, determine the following (a) How much can be...
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Please Calculate the expected PW value for building and widening a two-lane bridge. Will rate highly....
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A bridge is to be constructed now as part of a new road. An analysis has shown that traffic density on the new road will justify a two-lane bridge at the present time. Because of uncertainty regarding future use of the road, the time at which an extra two lanes will be required is currently being studied. The estimated probabilities of having to widen...
Suppose that you have an old car that is a real gas guzzler. It is 10...
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A company is considering the purchase of a capital asset for $110,000. Installation charges needed to...
A
company is considering the purchase of a capital asset for
$110,000.
Installation
charges needed to make the asset serviceable will total
$25,000.
The
asset will be depreciated over six years using the
straight-line method and an estimated
salvage value
(SV6)
of
$24,000.
The
asset will be kept in service for six years, after which it
will be sold for
$34,000.
During
its useful life, it is estimated
that the asset will produce annual revenues of
$25,000.
Operating
and maintenance...
ABC Brand 1,600 XYZ Brand 5,600 Purchase Price $ Life in Years Salvage Value $ Annual...
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Use the imputed market value technique to determine the better alternative below. The MARR is 10% per year and the study period is six years. 0 Alternative Alternative K Capital Investment, millions 47 62 Annual Expenses, millions 11 15 Useful Life, years 6 9 Market Value (End of useful life) Click the icon to view the interest and annuity table for discrete compounding when the MARR is 10% per year. The present worth of Alternative J over six years is...
Problem 6-2 (book/static) Question Help Years SS.OCC $6,000 58.000 511000 s14000 517000 Click the icon to view the interest factors for discrete compounding when i = 13% per year. i More Info The equivalent annual worth is $ . (Round to the nearest doller.) Single Payment Equal Payment Series Gradient Series N Compound Amount Factor (F/P, I, N) 1.1300 1.2769 1.4429 1.6305 1.8424 Present Worth Factor (P/F, I, NJ 0.8850 0.7831 0.6931 0.6133 0.5428 Compound Amount Factor (F/A,,N) 1.0000 2.1300...
1 year= $5,000
2 year=$5,000
3 year= $8,000
4 year= $11,000
5 year= $14,000
6 year= 17,000
Problem 6-2 (book/static) Question Help Years SS.OCC $6,000 58.000 511000 s14000 517000 Click the icon to view the interest factors for discrete compounding when i = 13% per year. i More Info The equivalent annual worth is $ . (Round to the nearest doller.) Single Payment Equal Payment Series Gradient Series N Compound Amount Factor (F/P, I, N) 1.1300 1.2769 1.4429 1.6305 1.8424...
A firm has decided to manufacture
biodegradable golf tees. These are two production processes
available for consideration.
Assuming material and direct labor costs are the only variable
costs and the
MARRequals=1313 %
per year, over what range of annual production volume is Process
A preferred. Assume repeatability.
& Problem 6-33 (algorithmic) Question Help A firm has decided to manufacture biodegradable golf tees. These are two production processes available for consideration. Assuming mater al and direct labor costs are the only...
Use the imputed market value technique to determine the better alternative below. The MARR is 12% per year and the study period is four years. Capital Investment, millions Annual Expenses, millions Useful Life, years Market Value (End of useful life) Alternative J 41 10 4 0 Alternative K 56 20 9 o Click the icon to view the interest and annuity table for discrete compounding when the MARR is 12% per year. The present worth of Alternative J over four...
For an interest rate of 13% compounded annually, determine the following (a) How much can be lent now if $18,000 will be repaid at the end of four years? (b) How much will be required in six years to repay a $26,000 loan received now? Click the icon to view the interest factors for discrete compounding when ,-13% per year (a) The amount to be lent now is $ . (Round to the nearest dollar) More Info Compound Present Worth...
Please Calculate the expected PW value for building and widening
a two-lane bridge. Will rate highly.
A bridge is to be constructed now as part of a new road. An analysis has shown that traffic density on the new road will justify a two-lane bridge at the present time. Because of uncertainty regarding future use of the road, the time at which an extra two lanes will be required is currently being studied. The estimated probabilities of having to widen...
Suppose that you have an old car that is a real gas guzzler. It is 10 years old and could be sold to a local dealer for $500 cash. The annual maintenance costs will average $800 per year into the foreseeable future, and the car averages only 10 miles per gallon. Gasoline costs $3.40 per gallon, and you drive 17,000 miles per year. You now have an opportunity to replace the old car with a better one that costs $6,000....
A
company is considering the purchase of a capital asset for
$110,000.
Installation
charges needed to make the asset serviceable will total
$25,000.
The
asset will be depreciated over six years using the
straight-line method and an estimated
salvage value
(SV6)
of
$24,000.
The
asset will be kept in service for six years, after which it
will be sold for
$34,000.
During
its useful life, it is estimated
that the asset will produce annual revenues of
$25,000.
Operating
and maintenance...
ABC Brand 1,600 XYZ Brand 5,600 Purchase Price $ Life in Years Salvage Value $ Annual Maintenance $ Efficiency % None 190 None 350 Discrete Compounding; i = 11% Uniform Series Single Payment Capital Recovery Factor To Find A Given P А/Р Compound Amount Present Factor Worth Factor To Find F To Find P Given P Given F P F/ P /F 1.1100 0.9009 1.2321 0.8116 1.3676 0.7312 1.5181 0.6587 1.6851 0.5935 1.8704 0.5346 2.0762 0.4817 2.3045 0.4339 2.5580 0.3909...
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