Two 180-horsepower water pumps are being considered for installation in a farm. Financial data for these pumps are in the table below. If power cost is a flat 6 cents per kWh over the study period, determine the minimum number of hours of full-load operation per year that would justify the purchase of the more expensive pump at an interest rate of 10% (1 HP=746 watts = 0.746 kilowatts). Click the icon to view the additional information. Click the icon to view the interest factors for discrete compounding when i= 10% per year. The minimum number of hours of full-load operation required per year to justify the purchase of the more expensive motor is hours. (Round to the nearest whole number.) More Info Item Initial cost Efficiency Useful life Annual operating cost Salvage value Print Pump I $5,600 80% 15 years $475 $0 Done Pump II $3,300 75% 15 years $460 $0 D X More Info N 1 2 3 4 5 6 7 8 9 10 11 12 Compound Amount Factor (F/P, i, N) 1.1000 1.2100 1.3310 1.4641 1.6105 1.7716 1.9487 2.1436 2.3579 2.5937 2.8531 3.1384 Present Worth Factor (P/F, I, N) 0.9091 0.8264 0.7513 0.6830 0.6209 0.5645 0.5132 0.4665 0.4241 0.3855 0.3505 0.3186 Compound Amount Factor (F/A, i, N) 1.0000 2.1000 3.3100 4.6410 6.1051 7.7156 9.4872 11.4359 13.5795 15.9374 18.5312 21.3843 magasra megvww.m Sinking Present Fund Worth Factor Factor (P/A, I, N) (A/F, i, N) 1.0000 0.9091 0.4762 0.3021 0.2155 0.1638 0.1296 0.1054 0.0874 0.0736 0.0627 0.0540 0.0468 1.7355 2.4869 3.1699 3.7908 4.3553 4.8684 5.3349 5.7590 6.1446 6.4951 6.8137 Capital Recovery Factor (A/P, I, N) 1.1000 0.5762 0.4021 0.3155 0.2638 0.2296 0.2054 0.1874 0.1736 0.1627 0.1540 0.1468 - X
Two 180-horsepower water pumps are being considered for installation in a farm. Financial data for these pumps are in the table below. If power cost is a flat 6 cents per kWh over the study period, determine the minimum number of hours of full-load operation per year that would justify the purchase of the more expensive pump at an interest rate of 10% (1 HP=746 watts = 0.746 kilowatts). Click the icon to view the additional information. Click the icon to view the interest factors for discrete compounding when i= 10% per year. The minimum number of hours of full-load operation required per year to justify the purchase of the more expensive motor is hours. (Round to the nearest whole number.) More Info Item Initial cost Efficiency Useful life Annual operating cost Salvage value Print Pump I $5,600 80% 15 years $475 $0 Done Pump II $3,300 75% 15 years $460 $0 D X More Info N 1 2 3 4 5 6 7 8 9 10 11 12 Compound Amount Factor (F/P, i, N) 1.1000 1.2100 1.3310 1.4641 1.6105 1.7716 1.9487 2.1436 2.3579 2.5937 2.8531 3.1384 Present Worth Factor (P/F, I, N) 0.9091 0.8264 0.7513 0.6830 0.6209 0.5645 0.5132 0.4665 0.4241 0.3855 0.3505 0.3186 Compound Amount Factor (F/A, i, N) 1.0000 2.1000 3.3100 4.6410 6.1051 7.7156 9.4872 11.4359 13.5795 15.9374 18.5312 21.3843 magasra megvww.m Sinking Present Fund Worth Factor Factor (P/A, I, N) (A/F, i, N) 1.0000 0.9091 0.4762 0.3021 0.2155 0.1638 0.1296 0.1054 0.0874 0.0736 0.0627 0.0540 0.0468 1.7355 2.4869 3.1699 3.7908 4.3553 4.8684 5.3349 5.7590 6.1446 6.4951 6.8137 Capital Recovery Factor (A/P, I, N) 1.1000 0.5762 0.4021 0.3155 0.2638 0.2296 0.2054 0.1874 0.1736 0.1627 0.1540 0.1468 - X
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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Transcribed Image Text:Two 180-horsepower water pumps are being considered for installation in a farm. Financial data for these pumps are in the table below. If power cost is a flat 6 cents per kWh over the study period,
determine the minimum number of hours of full-load operation per year that would justify the purchase of the more expensive pump at an interest rate of 10% (1 HP = 746 watts = 0.746 kilowatts).
Click the icon to view the additional information.
Click the icon to view the interest factors for discrete compounding when i= 10% per year.
The minimum number of hours of full-load operation required per year to justify the purchase of the more expensive motor
is hours. (Round to the nearest whole number.)
More Info
Item
Initial cost
Efficiency
Useful life
Annual operating cost
Salvage value
Get more help.
Print
Pump I
$5,600
80%
15 years
$475
$0
Done
Pump II
$3,300
75%
15 years
$460
$0
- X
D
More Info
N
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
wag sjunum,
Compound
Amount
Factor
(F/P, i, N)
1.1000
1.2100
1.3310
1.4641
1.6105
1.7716
1.9487
2.1436
2.3579
2.5937
2.8531
3.1384
3.4523
3.7975
4.1772
Present
Worth
Factor
(P/F, I, N)
0.9091
0.8264
0.7513
0.6830
0.6209
0.5645
0.5132
0.4665
0.4241
0.3855
0.3505
0.3186
0.2897
0.2633
0.2394
Compound
Amount
Factor
(F/A, I, N)
1.0000
2.1000
3.3100
4.6410
6.1051
7.7156
9.4872
11.4359
13.5795
15.9374
18.5312
21.3843
24.5227
27.9750
31.7725
www.megaSIMIES MALIUM
Sinking Present
Fund
Worth
Factor Factor
(A/F, I, N)
(P/A, i, N)
1.0000
0.9091
0.4762
1.7355
0.3021
2.4869
0.2155
3.1699
0.1638
3.7908
0.1296
0.1054
0.0874
0.0736
0.0627
0.0540
0.0468
0.0408
0.0357
0.0315
4.3553
4.8684
5.3349
5.7590
6.1446
6.4951
6.8137
7.1034
7.3667
7.6061
Capital
Recovery
Factor
(A/P, I, N)
1.1000
0.5762
0.4021
0.3155
0.2638
0.2296
0.2054
0.1874
0.1736
0.1627
0.1540
0.1468
0.1408
0.1357
0.1315
- X
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