Answer:The answer is economic service life is 8 Years, The Annual worth is $27,225.7
Explanation:
Given the equation
S = 120,000 - 11,000K,and AOC is 60,000 - 14,000K, where K is the number of years since the machine was purchased
Substitute the value of K into the equation
S= 120,000 - 11,000 (7)
S = 120,000 - 11,000 ×7
S= 120,000 - 77,000
S= 43,000
AOC = 60,000 - 14,000 (7)
AOC = 60,000 - 14,000 × 7
AOC = 60,000 - 98,000
A
To calculate the economic useful life,we use the equivalent annual cost( EAC)
Yr particular. CF DF. PV
0. Purchase cost 170,000. 1. ( 170,000)
1-7. Running cost. (38,000) 0.4523. 17,187.4
1-7. Scrap proceed. 43,000. 0.4523. 19,448.9
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PV. 133,363.7
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Discount Factor is calculated thus
(1 + r)-∧n , since r = 12% ,n = 7 (1+0.12)∧-7 = 0.4523
8years replacement
Yr. Particular. CF. DF. PV
0. Purchase cost. 170,000 1. (170,000)
1-7. Running cost. (38,000) 0.4523. 17,187.4
1-7. Scrap proceed 43,000. 0.4523. 19,448.9
8 Scrap proceed 43,000. 0.4523. 19,448.9
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113,914.8
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Finding the 7years replacement at the cost of capital using (1+r)∧-n/r
= (1+0.12)∧-7 -1/0.12
= (1.12)∧-7 -1/0.12
= 0.4523-1/0.12
=0.5477/0.12
= 4.5642
Therefore Equivalent Annual Cost
= 133,363.7/4.5642
= $29,219.51
8years replacementat the cost of capital using
(1+r)∧n -1/r
(1+0.12)∧-8 -1/0.12
=(1.12)∧-8 -1/0.12
= 0.4038 -1/0.12
=0.5962/0.12
=4.9683
EAC = 113,914.8/4.9683
= $22,928.33
Since they are both cost we prefer the cheapest, that is 8 years replacement
Therefore the economic life of the machine is 8years
To calculate the AW, we use the formula
AW = -P ( P-S)(A/P r n ) - Sr
AW= -(170,000 - 43,000) ( A/P 12% 7) - 43,000 (0.12)
To calculate (A/P r n ),we use the formula
A= P i (1+ r)∧n/ (1 + r )∧n - 1
0.12 (1+0.12)∧7/(1+0.12)∧7-1
= 0.12 ( 1.12)∧7/(1.12)∧7 -1
0.12 (2.210681407)/1.21061407
= 0.265281768/1.210681407
= 0.2191
AW = -127,000,000 (0.2191) --38,000 - 43,000 (0.12)
= 27,825.7 - 5,000 (0.12)
= 27,825.7 - 600
= $27,225.7
Therefore the economic life is 8years, Annual worth is $27,225.7
