Partition the problem into three small problems and add the future worth
together.
First, find the future worth of the early annual deposits of $1100 from year 22 to
year 27, and move to year 65 with F/P factor.
At t =27, F1 = ($1100)(F/A, 10%, 6)
=$8487.16
At t = 65, F1 =$8487.16 (F/P, 10%, 38)
=$317,456.28
Second, find the future worth of the gradient deposits from year 28 to year 40,
and move to year 65 with the F/P factor.
At t =40,
F2 = ($2200)(F/A, 10%, 13) +$400 (P/G, 10%, 13)(F/P, 10%, 13)
=$100,041.18
At t = 65,
F2 =$100,041.18 (F/P, 10%,25)
F2 =$1,083,916.17
Third, find the future worth of the remaining equal deposit of $7400 from year
41 to year 60, and move to year 65 with the F/P factor.
At t= 60, F3 = ($7400)(F/A, 10%, 20)
=$423,835.00
At t = 65, F3 =$423,835.00 (F/P, 10%, 5)
=$682,586.27
Thus, the future of the fund at the age of 65 is:
F = F1 + F2 + F3
=$2,083,958.72
Longhorn Fabricators Inc. plans to expand its metals–forming facility over the next 5 years.
The company will add 20,000 square feet to its 100,000–square–foot plant as it adds robotic
welding units, additional laser technology, and automated loading facilities. Construction
at the plant is expected to start by the end of next year. The company expects to pay 5
equal payments of $250,000 every 12 months over the 5 year period. What is the future
value of the total improvement cost, if the interest rate is 18% per year, compounded every
12 months?
Effective interest rate =(1 +r/M)M– 1 =(1 +0.1800)1–1
=0.18 or 18.00 per 12 months
F =$250,000 (F/A, 18.00%, 5)
=$1,788,550.00