Landing Gear
The landing gear on an aircraft requires a $2,000 part upgrade with an annual maintenance cost of $25. This component can also be salvaged after 5 years for $400. If the owner finances the purchase at 2%, what is the equivalent uniform annual cost?
Expand Hint
The problem is asking for the uniform amount per interest period,
$$A$$
, given the future worth,
$$F$$
.
Hint 2
The problem is also asking for the uniform amount per interest period,
$$A$$
, given the present worth,
$$P$$
.
The problem is asking for the uniform amount per interest period,
$$A$$
, given the future worth,
$$F$$
. Uniform Series Sinking Fund:
$$$A=F\cdot \frac{i}{(1+i)^{n}-1}$$$
where
$$i$$
is the interest rate per interest period, and
$$n$$
is the number of compounding periods.
$$$A=\$400\cdot \frac{0.02}{(1+0.02)^5-1}$$$
$$$=\$400\cdot \frac{0.02}{1.10408-1}=\frac{\$8}{0.10408}=\$76.86$$$
The problem is also asking for the uniform amount per interest period,
$$A$$
, given the present worth,
$$P$$
. Capital Recovery:
$$$A=P\cdot\frac{i(1+i)^n}{(1+i)^n-1}$$$
where
$$i$$
is the interest rate per interest period, and
$$n$$
is the number of compounding periods.
$$$A=\$2,000\cdot\frac{(0.02)(1+0.02)^5}{(1+0.02)^5-1}$$$
$$$=\frac{(\$40)(1.10408)}{(1.10408)-1}=\frac{\$44.16}{0.10408}=\$424.32$$$
The salvage value helps offset costs, while the $25 maintenance adds to it. Thus, the equivalent uniform annual cost is:
$$$\$424.32+\$25-\$76.86=\$372.46$$$
Alternatively, interest tables (commonly found in the FE Handbook) could have been used to solve this problem:
$$$\$2,000(A/P,\:2\%,\:5)+\$25-\$400(A/F,\:2\%,\:5)$$$
Referring to the 2% Factor Table:
where
$$n$$
is the number of compounding periods,
$$A$$
is the uniform amount per interest period,
$$P$$
is the present worth, and
$$F$$
is the future worth. Substituting the intersecting values of A/P and A/F with the 5 year row:
$$$\$2,000(0.2122)+\$25-\$400(0.1922)$$$
$$$=\$424.40+\$25-\$76.88=\$372.52\approx \$372.46$$$
$372.46
Time Analysis
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