A low-cost noncontact temperature measuring tool may be able to identify railroad car wheels that are in need of repair long before a costly structural failure occurs. If the BNF railroad saves $100,000 in year 1, $110,000 in year 2, and amounts increasÂing by $10,000 each year for 5 years, what is the future worth of the savings in year 5 at an interest rate of 10% per year?
What will be an ideal response?
A = 100,000 + 10,000(A/G,10%,5)
= 100,000 + 10,000(1.8101)
= $118,101
F = 118,101(F/A,10%,5)
= 118,101(6.1051)
= $721,018
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