Determine the FW of the following engineering project when the MARR is 15% per year. Is...
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.
Homework Answers
PW(15%) = 10,000 + ( 8,000 - 4000)(P/A,15%,5) - 1,000 (P/F,15%,5)
= - 10,000 + 4000(3.3522) - 1,000 (0.4972) = 2911.6
FW(15%) = - 10,000 (F/P,15%,5) + ( 8,000 - 4000) (F/A,l5%,5) - 1,000
= - 10,000 (2.0114) + 4000(6.7424) - 1,000 = 5855.6
AW(15%) = 8,000 - 4000 - [ 10,000(A/P,15%,5) - (- 1000)(A/F,15%,5)]
= 4000 - [ 10,000(0.2983) + 1,000(0.1483)] = 868.70
Since the IRR> MARR, the project is acceptable.
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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.
Determine the FW of the following engineering project when the MARR is 17% per year. Is...
Determine the FW of the following engineering project when the MARR is 17% per year. Is the project acceptable? a 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 project is $. (Round to the nearest dollar.)
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!
(:
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...
Determine whether either of the alternatives below should be selected. Use an MARR of 15% per...
Determine whether either of the alternatives below should be
selected. Use an MARR of 15% per year. Incremental Rate of Return
Analysis must be used. (PLEASE DO NOT USE EXCEL).
First Cost Annual Operating Cost Annual Repair Cost Annual Increase in Repair Cost Salvage Value Life (years) Project A -$60,000 -$15,000 -$5,000 -$1,000 $8,000 15 Project B -$90,000 -$8,000 -$2,000 -$1,500 $12,000 15
please show work and box answers 5. (15 points) A project is estimated in the face of a MARR = 15% as follows: Opti...
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
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NOT IN EXCEL
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engineering economics A public utility district (PUD) has retained your services to provide a perpetual water...
engineering economics
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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...
Determine the FW of the following engineering project when the MARR is 17% per year. Is the project acceptable? a 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 project is $. (Round to the nearest dollar.)
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...
Determine whether either of the alternatives below should be
selected. Use an MARR of 15% per year. Incremental Rate of Return
Analysis must be used. (PLEASE DO NOT USE EXCEL).
First Cost Annual Operating Cost Annual Repair Cost Annual Increase in Repair Cost Salvage Value Life (years) Project A -$60,000 -$15,000 -$5,000 -$1,000 $8,000 15 Project B -$90,000 -$8,000 -$2,000 -$1,500 $12,000 15
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
NOT IN EXCEL
7.For the following table assume a MARR of 6% per year, and a useful life for each alternative of six years. Which equals the study period. The rank order of alternatives from least capital to greatest capital investment is A, C, B. Complete the PW-Incremental Analysis by selecting the preferred alternative. (5ptos) V A Capital Investment A Annual Revenues A Annual Cost A Market Value APW A $15,000 4,000 1,000 6,000 $3,982 A(C-A) A(B-C) $2,000 $3,000 900...
engineering economics
A public utility district (PUD) has retained your services to provide a perpetual water service to a small resort community: Option 1 Option 2 $150,000 $235,000 45 Pipe First Cost Expected Life, years Salvage Value (S) Pump First Cost Expected Life, years 15,000 23,500 100,000 125,000 15 10,000 1,000 Pump House First Cost Expected Life, years 15,000 20,000 50 50 0 Energy (Annual) Costs First Year Gradient 50,000 1,000 increasing 20,000 1,000 decreasing 15,000 600 5,000 200 System...
a machine that have the
equivalent A manufacturing equipment has the following costs. The MARR is 10% compounded sen annualy. It is an acceptable alternative? Compu ethe annual worth First cost: $70,000 Semiannual operating cost: 8,000 Semiannual income: 18,000 Semiannual income gradient: 100 Salvage value: 10,000 Life in years: 4 years a. EAW-$54 b. EAW $741 G. EAW $541 d.EAW $44
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