An existing asset that cost $16,000 twoyears ago has a market value of $12,000 today, anexpected salvage value of $2,000 at the end of itsremaining useful life of six more years, and annualoperating costs of $4,000. A new asset under consideration as a replacement has an initial cost of$10,000, an expected salvage value of $4,000 at theend of its economic life of three years, and annualoperating costs of $2,000. It is assumed that thisnew asset could be replaced by another one identical in every respect after three years at a salvagevalue of $4,000, if desired. Use a MARR of 11%,a six-year study period, and PW calculations todecide whether the existing asset should be replacedby the new one
An existing asset that cost $16,000 two
years ago has a market value of $12,000 today, an
expected salvage value of $2,000 at the end of itsremaining useful life of six more years, and annual
operating costs of $4,000. A new asset under consideration as a replacement has an initial cost of
$10,000, an expected salvage value of $4,000 at the
end of its economic life of three years, and annual
operating costs of $2,000. It is assumed that this
new asset could be replaced by another one identical in every respect after three years at a salvage
value of $4,000, if desired. Use a MARR of 11%,
a six-year study period, and PW calculations to
decide whether the existing asset should be replaced
by the new one
existing asset that cost $16,000 two years ago
operating cost = 4000
market value = 12000
slavage value = 2000
useful years = 6
replacement asset cost 10000
salvage value = 4000
operating cost = 2000
MARR of 11%, a six-year study period or present worth will decide by below calculation
pw(i) = d+a(p l A, i, N) - F(P l F i,N)
where
p = present value
f = future value
A = annual payments
D = initial value of asset
i = interest rate
N = time period
substuting the values in the given formula
PW(0.11) = [12000 +4000(Pl A, 0.11,6)- 2] - 2000(PlF, 0.11,6)
= 12000 + 4000(4.2305) - 2000(0.5346)
= 12000 + 16922 - 1069.2
PW(0.11) = 27852.80
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