Apiece of equipment has a first cost of $170,000, a maximum useful life of 7 years, and a market (salvage) value described by the relation s = 120,000 – 11,000k, where k is the number of years since it was purchased. the salvage value cannot go below zero. the aoc series is estimated using aoc = 60,000 + 14,000k. the interest rate is 12% per year. determine the economic service life and the respective aw.

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

PV. 133,363.7

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

113,914.8

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

the answer is true.  i do believe this is the answer because of the amount of wrecks people get into each year.  if you double this with avoiding those so said wrecks, that would make you a "skilled" driver.

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