An engineer made a development in the production process which will save his company 10,000$ per year. The following equation is used to calculate the savings in 8 years at i=10% per year. 10% 10% TABLE 15 Discrete Cash Flow Compound Interest Factors Arithmetic Gradients Single Payments Compound Present Amount Worth F/P P/F Sinking Fund Uniform Series Payments Compound Capital Amount Recovery F/A AP Present Worth P/A Gradient Present Worth P/G Gradient Uniform Series AG n A/F 1 2 3 4 5 1.1000 1.2100 1.3310 1.4641 1.6105 1.7716 1.9487 2.1436 2.3579 2.5937 0.9091 0.8264 0.7513 0.6830 0.6209 0.5645 0.5132 0.4665 0.4241 0.3855 1.00000 0.47619 0.30211 0.21547 0.16380 0.12961 0.10541 0.08744 0.07364 0.06275 1.0000 2.1000 3.3100 4.6410 6.1051 7.7156 9.4872 11.4359 13.5795 15.9371 1.10000 0.57619 0.40211 0.31547 0.26380 0.22961 0.20541 0.18744 0.17364 0.16275 0.9091 1.7355 2.4869 3.1699 3.7908 4.3553 4.8684 5.3349 5.7590 6.1446 0.8264 2.3291 4.3781 6.8618 9.6842 12.7631 16.0287 19.4215 22.8913 0.4762 0.9366 1.3812 1.8101 2.2236 2.6216 3.0045 3.3724 3.7255 6 7 8 9 10 Select one: O a. F=10,000(F/P.10.8) b. A=10,000(A/F, 10.8) O C. P=10.000(P/A,8,10) d. F=10,000(F/A, 10.8)