Based on the cash flow shown over 10 years for three projects:Which should be selected using the concept of IRR and Delta IRR?Explain.
Based on the cash flow shown over 10 years for three projects, which should be selected using the concept of IRR and Delta IRR?
MARR 14%
Year Project P Project Q Project R
0 $ (29,200.00) $(44,400.00) $(67,800.00)
1 $ 6,200.00 $ 8,825.00 $ 13,825.00
2 $ 8,200.00 $ 10,825.00 $ 16,825.00
3 $ 10,200.00 $ 12,825.00 $ 18,825.00
4 $ 12,200.00 $ 14,825.00 $ 20,825.00
5 $ (31,120.00) $ 11,825.00 $(24,815.00)
6 $ 12,504.00 $ (1,244.44) $ 14,825.00
7 $ 12,504.00 $ 6,618.88 $ 16,825.00
8 $ 12,504.00 $ 7,030.77 $ 18,825.00
9 $ 12,504.00 $ 7,483.84 $ 20,825.00
10 $ 12,504.00 $ 7,982.23 $ 22,825.00
Group of answer choices
First project
Second project
Third project
None. All should be rejected.
Flag question: Question 2
Question 210 pts
The net cash flows for 4 machines are as shown. Match the Delta IRR with the correct values.
Year Project A Project B Project C Project D
0 $ (45,500) $ (52,900) $ (70,500) $ (90,000)
1 $ 7,200 $ 14,325 $ 8,000 $ 14,200
2 $ 9,200 $ 12,325 $ 10,000 $ 24,200
3 $ 11,200 $ 15,325 $ 12,000 $ 24,200
4 $ 13,200 $ 18,325 $ 16,325 $ 24,200
5 $ 15,200 $ 20,325 $ 18,325 $ 24,200
6 $ 17,200 $ 22,325 $ 18,325 $ 24,200
7 $ 13,325 $ 8,119 $ 21,198 $ 24,200
8 $ 15,325 $ 8,531 $ 25,301 $ 24,200
9 $ 17,325 $ 8,984 $ 29,404 $ 18,325
10 $ 19,325 $ 9,482 $ 33,507 $ 20,325
Group of answer choices
The Delta IRR between machines A and B is:
The Delta IRR between machines B and C is:
The Delta IRR between machines A and C is:
The Delta IRR between machines C and D is:
Flag question: Question 3
Question 310 pts
For machines U and V, details are provided. What will be the Delta IRR and which machine should be selected?
Machine U Machine V
Initial cost $320,000 $410,000
Life in years 5 6
Inflation (for costs and benefits) 2% p. y.
MARR 7% p. y. c. y.
Project life 12 years
Salvage value of machine today $44,800 $57,400
Machine market value with 2 years of use $153,600
First year estimated costs $39,500 $53,800
First year estimated benefits $113,900 $128,435
Group of answer choices
3.83%, go with Machine U
8.42%, go with Machine U
9.69%, go with either Machine U or V
11.61%, go with Machine V
Flag question: Question 4
Question 410 pts
For Machines X and Y shown, what is the difference between the EUAW for the HICP and the LICP?
Machine X Machine Y
Initial cost $571,000 $933,000
Life 4 6
Inflation 3.5% p. y.
MARR 8% p. y. c. y.
Project life 14 years
First year estimated costs $199,850 $363,870
First year estimated benefits $411,120 $606,450
Salvage value of machine today $143,500 $172,400
Market value of machine today with 2 years of use $256,950 $419,850
Group of answer choices
$13,144
$26,830
$39,682
$10,219
Flag question: Question 5
Question 510 pts
The City of Omniville has two options for a viaduct. One is a permanent cement canal and the other a cast iron (CI) pipe. If details for the options are as shown, at what rate of interest should the city be indifferent to either choice?
Cement Canal CI PIPE
Initial cost ($3,000,000) ($1,000,000)
Life Perpetual 15 years
Annual maintenance cost $250,000 $350,000
Salvage $0 $50,000
Initial Rate 5.0% p. y.
Group of answer choices
6.9%
10.66%
16.31%
35.04%
Flag question: Question 6
Question 610 pts
Data for two machines P and Q are as shown below. At what MARR will both machines be equally attractive for installation?
Machine P Machine Q
Initial cost $125,000 $275,000
Life in years 4 6
Inflation per year 2.00%
Benefit increase per year 4.50%
MARR per year compounded yearly 12.00%
Project life in years 12
First year estimated costs $33,750 $63,250
First year estimated benefits $65,000 $112,750
Salvage value of machine 12.00% 8.00%
Group of answer choices
11.44%
12.17%
13.19%
17.98%
Flag question: Question 7
Question 710 pts
Data for two machines are shown. Also, four graphs of NPW for the two machines is shown. Which graph represents the data shown below? Quiz 7-7(1).pptx
Machine P Machine Q
Initial cost $105,000 $155,000
Life in years 4 6
Inflation per year 2.00%
Benefit increase per year 4.50%
MARR per year compounded yearly 12.00%
Project life in years 12
First year estimated costs $28,350 $35,650
First year estimated benefits $54,600 $63,550
Salvage value of machine 12.00% 8.00%
Group of answer choices
Graph 1
Graph 2
Graph 3
Graph 4
Flag question: Question 8
Question 810 pts
Data for 3 machines are given. For the given MARR, which machine will have the highest EUAW and hence be selected for purchase?
MARR 8.0%
Year Project J Project K Project L
0 $ (101,500) $ (117,435) $ (153,612)
1 $ 6,950 $ 35,850 $ 64,950
2 $ 6,950 $ 35,850 $ 42,950
3 $ 6,950 $ 32,450 $ 52,950
4 $ 6,950 $ 31,950 $ 42,950
5 $ 32,950 $ 31,950 $ 6,950
6 $ 51,950 $ 24,950 $ (13,900)
7 $ 51,950 $ (49,900)
8 $ 51,950 $ 16,350
9 $ 51,950
10 $(103,900.00)
Group of answer choices
Machine L with an EUAW of $1,755
Machine J with an EUAW of $819
Machine K with an EUAW of $1,029
Any will do as all 3 machines have the same EUAW of $1,926
