Harry, owner of an automobile battery distributorship has decided to place surge
ID: 2794494 • Letter: H
Question
Harry, owner of an automobile battery distributorship has decided to place surge protectors in-line for all his major pieces of testing equipment. Two different manufacturers’ protectors need to be compared using a MARR of 15%.
Scenario 1: PowrUp and Lloyd’s have submitted the following economic information (See table below)
PowrUp
Lloyd's
Cost and installation, $
-26,000
-36,000
Annual maintenance cost,
$ per year
-800
-300
Salvage value, $
2,000
3,000
Annual repair savings,
$ per year
25,000
35,000
Useful life, years
6
10
Q 1a) Calculate the annual worth of the “PowrUp” protector’s (10 points)
Q 1b) What is the capital recovery amount for the PowrUp protectors? (3 points)
Q 1c) Calculate the annual worth of the “Lloyd” protector’s (10 points)
Q 1d) What is the capital recovery amount for the Lloyd’s protectors? (3 points)
Q 1e) Which manufacturers’ protectors should be selected by Harry using annual worth analysis (4 points)
Scenario 2: Lloyd’s have recently revised their economic estimates which are shown in the table below.
Lloyd's (new estimates)
Initial investment
Upgrade cost
Salvage value
Annual maint.
Repair savings
Year
0
-$36,000
1
-$300
$35,000
2
-$300
$32,000
3
-$300
$28,000
4
-$1,200
$26,000
5
-$1,320
$24,000
6
-$9,000
-$1,452
$22,000
7
-$1,597
$20,000
8
-$1,757
$18,000
9
-$1,933
$16,000
10
$1,000
-$2,126
$14,000
Using these new estimates, answer the following questions:
Q 2a) What is the recalculated AW for the Lloyd’s protectors? (36 points)
Q 2b) What is the recalculated capital recovery amount for the Lloyd’s protectors? (10 points)
Q 2c) Which manufacturers’ protectors should now be selected by Harry using annual worth analysis (4 points)
Hints
Q2a) You need to convert all ten yearly lines first to Present Worth and then back to Annual Worth. Following is a small hint:
For the initial investment of 36,000 convert it into AW using the (A/P,15%,10).
For the salvage value of 1,000 convert it into AW using the (A/F,15%,10).
For the first line convert (35,000 - 300) into PW using (P/F,15%,1) and then into AW using the (A/P,15%,10).
For the second line convert (32,000 - 300) into PW using (P/F,15%,2) and then into AW using the (A/P,15%,10).
And so on for each of the remaining 8 lines.
Q2b) For the CR only consider the following:
For the initial investment of 36,000 convert it into AW using the (A/P,15%,10).
For the salvage value of 1,000 convert it into AW using the (A/F,15%,10).
For the upgrade cost of 9,000 convert it into PW using (P/F,15%,6) and then into AW using the (A/P,15%,10).
PowrUp
Lloyd's
Cost and installation, $
-26,000
-36,000
Annual maintenance cost,
$ per year
-800
-300
Salvage value, $
2,000
3,000
Annual repair savings,
$ per year
25,000
35,000
Useful life, years
6
10
Explanation / Answer
Question 1 1a) Calculate the annual worth of the “PowrUp” protector’s (10 points) AWp=-26,000(A/P,15%,6)+(25,000-800)+2,000(A/F,15%,6)=-26,000(0.26424)+24,200+2000(0.11424)=$17,558 Q 1b) What is the capital recovery amount for the PowrUp protectors? (3 points) CRp=-26,000(A/P,15%,6)+2,000(A/F,15%,6)=-26,000(0.26424)+2,000(0.11424)=$-6,642 Q 1c) Calculate the annual worth of the “Lloyd” protector’s (10 points) AW=-36,000(A/P,15%,10)+(35,000-300)+3,000(A/F,15%,10)=-36,000(0.19925)+34,700+3,000(0.04925)=$27,675 Q 1d) What is the capital recovery amount for the Lloyd’s protectors? CRp=-36,000(A/P,15%,10)+3,000(A/F,15%,10)=-36,000(0.19925)+3,000(0.04925)=$- 7025
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