Question

A new machine will cost $10,000. Net savings will be
$1000 per year for 4 years, and then it will climb by
$200 per year until the machine wears out at the end
of year 12. If i = 8%, what is the machine’s EAW?

Answer 1

As per the information provided in the question

Initial cost of machine (I) = $10000

Net savings (A) =$ 1000

Useful life (N) = 12 years

Interest rate (i) = 8%

The net savings increases over $1000 from 4^th year, by $200 per year till 12^th year

So A1 = $1000

G = 200

N = 9 years

A = A1 + G(A/G,8%, 9)

A = 1000 +200(3.4910) = 1000+698.2 = $1698.2

Present worth of savings from 4^th year to 12^th year = A(P/A,8%,9)(P/F,8%,3)

Present worth of savings from 4^th year to 12^th year = $1698.2 (6.2469)(0.7938) =$8421.02

Present worth of savings from 1^st year to 3^rd year =$1000(P/A,8%,3) = $1000(2.5771) =$2577.1

Net present worth = -Initial cost of machine + Present worth of savings from 1^st year to 3^rd year + Present worth of savings from 4^th year to 12^th year

Net present worth = -10000 + $8421.02 + $2577.1 = 998.12

Equivalent annual worth (EAW) = Net present worth(A/P,8%,12)

Equivalent annual worth (EAW) = 998.12 (0.13270) = $132.45

The machine’s Equivalent annual worth (EAW) =$132.45