Question: Data table begin{tabular}{|c|c|c|c|c|c|c|} hline multicolumn{7}{|c|}{ Discrete Compounding; i=12%} hline & multicolumn{2}{|c|}{ Single Payment } & multicolumn{4}{|c|}{ Uniform Series } hline & begin{tabular}{l} Compound




Data table \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{7}{|c|}{ Discrete Compounding; i=12%} \\ \hline & \multicolumn{2}{|c|}{ Single Payment } & \multicolumn{4}{|c|}{ Uniform Series } \\ \hline & \begin{tabular}{l} Compound \\ Amount \\ Factor \end{tabular} & \begin{tabular}{c} Present \\ Worth Factor \end{tabular} & \begin{tabular}{l} Compound \\ Amount \\ Factor \end{tabular} & \begin{tabular}{c} Present \\ Worth Factor \end{tabular} & \begin{tabular}{l} Sinking \\ Fund \\ Factor \end{tabular} & \begin{tabular}{c} Capital \\ Recovery \\ Factor \end{tabular} \\ \hline N & \begin{tabular}{c} To Find F \\ Given P \\ F/P \end{tabular} & \begin{tabular}{c} To Find P \\ Given F \\ P/F \end{tabular} & \begin{tabular}{c} To Find F \\ Given A \\ F/A \end{tabular} & \begin{tabular}{c} To Find P \\ Given A \\ P/A \end{tabular} & \begin{tabular}{l} To Find A \\ Given F \\ A/F \end{tabular} & \begin{tabular}{c} To Find A \\ Given P \\ A/P \end{tabular} \\ \hline 1 & 1.1200 & 1.1200 & 1.0000 & 0.8929 & 1.0000 & 1.1200 \\ \hline 2 & 1.2544 & 0.7972 & 2.1200 & 1.6901 & 0.4717 & 0.5917 \\ \hline 3 & 1.4049 & 0.7118 & 3.3744 & 2.4018 & 0.2963 & 0.4163 \\ \hline 4 & 1.5735 & 0.6355 & 4.7793 & 3.0373 & 0.2092 & 0.3292 \\ \hline 5 & 1.7623 & 0.5674 & 6.3528 & 3.6048 & 0.1574 & 0.2774 \\ \hline 6 & 1.9738 & 0.5066 & 8.1152 & 4.1114 & 0.1232 & 0.2432 \\ \hline 7 & 2.2107 & 0.4523 & 10.0890 & 4.5638 & 0.0991 & 0.2191 \\ \hline 8 & 2.4760 & 0.4039 & 12.2997 & 4.9676 & 0.0813 & 0.2013 \\ \hline 9 & 2.7731 & 0.3606 & 14.7757 & 5.3282 & 0.0677 & 0.1877 \\ \hline 10 & 3.1058 & 0.3220 & 17.5487 & 5.6502 & 0.0570 & 0.1770 \\ \hline \end{tabular} Potable water is in short supply in many countries. To address this need, two mutually exclusive water purification systems are being considered for implementation in China. Doing nothing is not an option. Assume the repeatability of cash flows for alternative 1. a. Use the PW method to determine which system should be selected when MARR =9% per year. b. Which system should be selected when MARR =12% per year? Click the icon to view the alternatives description. Click the icon to view the interest and annuity table for discrete compounding when i=9% per year. Click the icon to view the interest and annuity table for discrete compounding when i=12% per year. \begin{tabular}{|c|c|c|c|c|c|c|} \hline \multicolumn{7}{|c|}{ Discrete Compounding; i=9%} \\ \hline & \multicolumn{2}{|c|}{ Single Payment } & \multicolumn{4}{|c|}{ Uniform Series } \\ \hline & \begin{tabular}{c} Compound \\ Amount \\ Factor \end{tabular} & \begin{tabular}{c} Present \\ Worth Factor \end{tabular} & \begin{tabular}{l} Compound \\ Amount \\ Factor \end{tabular} & \begin{tabular}{c} Present \\ Worth Factor \end{tabular} & \begin{tabular}{l} Sinking \\ Fund \\ Factor \end{tabular} & \begin{tabular}{c} Capital \\ Recovery \\ Factor \end{tabular} \\ \hline N & \begin{tabular}{c} To Find F \\ Given P \\ F/P \end{tabular} & \begin{tabular}{c} To Find P \\ Given F \\ P/F \end{tabular} & \begin{tabular}{c} To Find F \\ Given A \\ F/A \end{tabular} & \begin{tabular}{c} To Find P \\ Given A \\ P/A \end{tabular} & \begin{tabular}{c} To Find A \\ Given F \\ A/F \end{tabular} & \begin{tabular}{c} To Find A \\ Given P \\ A/P \end{tabular} \\ \hline 1 & 1.0900 & 0.9174 & 1.0000 & 0.9174 & 1.0000 & 1.0900 \\ \hline 2 & 1.1881 & 0.8417 & 2.0900 & 1.7591 & 0.4785 & 0.5685 \\ \hline 3 & 1.2950 & 0.7722 & 3.2781 & 2.5313 & 0.3051 & 0.3951 \\ \hline 4 & 1.4116 & 0.7084 & 4.5731 & 3.2397 & 0.2187 & 0.3087 \\ \hline 5 & 1.5386 & 0.6499 & 5.9847 & 3.8897 & 0.1671 & 0.2571 \\ \hline 6 & 1.6771 & 0.5963 & 7.5233 & 4.4859 & 0.1329 & 0.2229 \\ \hline 7 & 1.8280 & 0.5470 & 9.2004 & 5.0330 & 0.1087 & 0.1987 \\ \hline 8 & 1.9926 & 0.5019 & 11.0285 & 5.5348 & 0.0907 & 0.1807 \\ \hline 9 & 2.1719 & 0.4604 & 13.0210 & 5.9952 & 0.0768 & 0.1668 \\ \hline 10 & 2.3674 & 0.4224 & 15.1929 & 6.4177 & 0.0658 & 0.1558 \\ \hline \end{tabular}
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