Equivalence for Repeated Cash Flows
Autor: deepika surapaneni • July 12, 2016 • Study Guide • 1,599 Words (7 Pages) • 985 Views
Name: Piyush Akkineni
EKU ID : 901589527
Chapter 4: Equivalence for Repeated Cash Flows
- [5 points] If i = 10%, for what value of B is the present value = 0?
[pic 1]
Solution
P = F(1+i)-n
1st Year
Present Value of B1 = B × (1+0.1)-1
= 0.909B
2nd Year
Present Value of B2 = -1.25B × (1+0.1)-2
= -1.033B
Present Value of 900 = 900 × (1+0.1)-2
= 743.8
3rd Year
Present Value of B3 = -B × (1+0.1)-3
= -0.75B
Present Value of 900 = 900 × (1+0.1)-3
= 676.18
4th Year
Present Value of B4 = -1.75B × (1+0.1)-4
= -1.19B
Present Value of 900 = 900 × (1+0.1)-4
= 614.7
On solving for the value of B using Excel Spreadsheet
B = 982.6387.
[pic 2]
- [5 points] These cash flow transactions are said to be equivalent in terms of economic desirability at an interest rate of 15% compounded annually. Determine the unknown value of A?
[pic 3]
Solution
[pic 4][pic 5]
[pic 6][pic 7]
P = + [pic 8][pic 9]
= + [pic 10][pic 11]
= 1155 + 670.4
= 1825.4 … (1)
[pic 12]
= 2.855A … (2)[pic 13]
On equating equations (1) and (2),
2.855A = 1825.4
A = [pic 14]
A = 639.369
Chapter 5: Present Worth Analysis
- [10 points] An industrial engineer has been studying a lean process line to determine if the company should switch from a labor-intensive line to a more automated system. If the company wants to earn at least 12% over a 10-year planning horizon, given the following cash flow estimates and using a Present Worth Analysis, which alternative is preferred?
Labor Intensive Line | Automated Line | |
Initial Cost | $ 0 | $ 100,000 |
Installation Cost | ×$ 0 | $ 100,000 |
1st-Year Maintenance Costs | $ 2,000 | $ 15,000 |
Annual Increase in Maintenance Costs | $ 250 | $ 500 |
1st-Year Labor Costs | $ 95,000 | $ 50,000 |
Annual Increase in Labor Costs | 3% | 3% |
Salvage Value (End of Year 10) | $ 10,000 | $ 20,000 |
...