Question (1) Use FW method, MARR 15% Project Investment Cost 10,000 Expected Life 5 years Market...
Question (1) Use FW method, MARR 15%
|
Project |
|
|
Investment Cost |
10,000 |
|
Expected Life |
5 years |
|
Market (Salvage) Value* |
-1,000 |
|
Annual Receipts |
8,000 |
|
Annual Expenses |
4,000 |
Answer:
FW(15%) = - 10,000 (2.0114) + 4000(6.7424) - 1,000 = 5,855.60
Is this project acceptable? Yes, but I would like to see more profit.
Homework Answers
let, investment cost (I) = -10000 , expected life (n) = 5 years , annual receipts (r) = 8000 annual expenses (c) = 4000, salvage value (s) = -1000 , MARR (m) = 15% or 0.15
then, future worth (fw) = -10000*(1+0.15)5 + (8000-4000)*(1+0.15)4 + (8000-4000)*(1+0.15)3 + (8000-4000)*(1+0.15)2 + (8000- 4000)*(1+0.15)1 +(8000-4000)*(1+0.15)0 - 1000
= -20113.57188 + 26969.525 - 1000
= 5855.95312 (approximately)
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Question (1) Use FW method, MARR 15%
Project
Investment Cost
10,000
Expected Life
5 years
Market...
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Determine the FW of the following engineering project when the MARR is 15% per year. Is the project acceptable? Project Investment Cost $10,000 Expected life 5 years Market (Salvage) Value* -$1,000 Annual Receipts $8,000 Annual Expenses $4,000 * A negative market value means that there is a cost to dispose of an asset.
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Determine the FW of the following engineering project when the MARR is 17% per year. Is the project acceptable? Investment cost Expected life Market (salvage) value Annual receipts Annual expenses Proposal A $9,500 6 years -$1,100 $7,000 $4,000 A negative market value means that there is a net cost to dispose of an asset. Click the icon to view the interest and annuity table for discrete compounding when the MARR is 17% per year. The FW of the following engineering...
Please answer both questions, I would appreciate it greatly! (: If you cannot see the picture, I...
Please answer both questions, I would appreciate it greatly!
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If you cannot see the picture, I have transcribed below:
1.
Consider the following project and its cash flow:
Investment cost $10,000
Expected life 5 years
Market (salvage) value* -$1,000
Annual receipts $8,000
Annual expense -$4,000
* A negative market value means that there is a net cost to
dispose of an asset.
a. Determine its PW and FW with MARR 15% per year. Is the
project acceptable?
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Determine the FW of the following engineering project when the MARR is 17% per year. Is the project acceptable? Investment cost Expected life Market (salvage) value Annual receipts Annual expenses Proposal A $9,500 6 years -$1,100 $7,000 $4,000 A negative market value means that there is a net cost to dispose of an asset. Click the icon to view the interest and annuity table for discrete compounding when the MARR is 17% per year. The FW of the following engineering...
Please answer both questions, I would appreciate it greatly!
(:
If you cannot see the picture, I have transcribed below:
1.
Consider the following project and its cash flow:
Investment cost $10,000
Expected life 5 years
Market (salvage) value* -$1,000
Annual receipts $8,000
Annual expense -$4,000
* A negative market value means that there is a net cost to
dispose of an asset.
a. Determine its PW and FW with MARR 15% per year. Is the
project acceptable?
b. What...
please show work and box answers
5. (15 points) A project is estimated in the face of a MARR = 15% as follows: Optimistic Most likely Pessimistic Investment $45,000 $50,000 $55,000 Service life 6 years 5 years 4 years Salvage value $15.000 $12,000 $10,000 Net annual return $16,000 $13,000 $10,000 Find the annual equivalent amount for each possible outcome
Most likely estimates for a project are as follows. MARR Useful life Initial investment Receipts - Expenses (R-E 12% per year 6 years $7,000 $1,300/year Determine whether the statement "If the profit (R-E) is decreased by 6%, this project is not profitable." is true or false, Click the icon to view the relationship between the PW and the percent change in parameter. Click the icon to view the interest and annuity table for discrete compounding when the MARR is 12%...
Nules. Enter the information exactly as below. Factor Annuity (05"/1.ES/"E71/((1+5)*E7-1) =3*G3 30,000.00 Investment Save: 300000 0.10 MARR Market value $120,000.00 (A/P =300000*0.1 0.15 $8,000.00 AFI, E5/((1+5)*87-1) 1666 Benefit *H4/H3-H6) (do not forget to write the = sign) To format columns with numbers: right click the mouse in the cell you whic Cancel S 1. When the MARR is 15% and the life of the heating system asset is 6 years is the alternative acceptable? a. True b. False 2. When...
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