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help pls begin{tabular}{|c|c|c|c|c|c|c|c|c|} hline multicolumn{9}{|c|}{ Discrete Compounding; i=7%} hline & multicolumn{2}{|c|}{ Single Payment } & multicolumn{4}{|c|}{ Uniform Series } & multicolumn{2}{|c|}{ Uniform Gradient }
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\begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline \multicolumn{9}{|c|}{ Discrete Compounding; i=7%} \\ \hline & \multicolumn{2}{|c|}{ Single Payment } & \multicolumn{4}{|c|}{ Uniform Series } & \multicolumn{2}{|c|}{ Uniform Gradient } \\ \hline & CompoundAmountFactor & PresentWorthFactor & CompoundAmountFactor & PresentWorthFactor & SinkingFundFactor & CapitalRecoveryFactor & GradientPresentWorthFactor & GradientUniformSeriesFactor \\ \hline N & ToFindFGivenPF/P & ToFindPGivenFP/F & ToFindFGivenAF/A & ToFindPGivenAP/A & ToFindAGivenFA/F & ToFindAGivenPA/P & ToFindPGivenGP/G & ToFindAGivenGA/G \\ \hline 1 & 1.0700 & 1.0700 & 1.0000 & 0.9346 & 1.0000 & 1.0700 & 0.0000 & 0.0000 \\ \hline 2 & 1.1449 & 0.7561 & 2.0700 & 1.8080 & 0.4831 & 0.5531 & 0.8734 & 0.4831 \\ \hline 3 & 1.2250 & 0.6575 & 3.2149 & 2.6243 & 0.3111 & 0.3811 & 2.5060 & 0.9549 \\ \hline 4 & 1.3108 & 0.5718 & 4.4399 & 3.3872 & 0.2252 & 0.2952 & 4.7947 & 1.4155 \\ \hline 5 & 1.4026 & 0.4972 & 5.7507 & 4.1002 & 0.1739 & 0.2439 & 7.6467 & 1.8650 \\ \hline 6 & 1.5007 & 0.4323 & 7.1533 & 4.7665 & 0.1398 & 0.2098 & 10.9784 & 2.3032 \\ \hline 7 & 1.6058 & 0.3759 & 8.6540 & 5.3893 & 0.1156 & 0.1856 & 14.7149 & 2.7304 \\ \hline 8 & 1.7182 & 0.3269 & 10.2598 & 5.9713 & 0.0975 & 0.1675 & 18.7889 & 3.1465 \\ \hline 9 & 1.8385 & 0.2843 & 11.9780 & 6.5152 & 0.0835 & 0.1535 & 23.1404 & 3.5517 \\ \hline 10 & 1.9672 & 0.2472 & 13.8164 & 7.0236 & 0.0724 & 0.1424 & 27.7156 & 3.9461 \\ \hline \end{tabular} You are the manager of a large crude-oil refinery. As part of the refining process, a certain heat exchanger (operated at high temperatures and with abrasive material flowing through it) must be replaced every year. The replacement and downtime cost in the first year is $185,000. This cost is expected to increase due to inflation at a rate of 7% per year for five years (i.e. until the EOY 6 ), at which time this particular heat exchanger will no longer be needed. If the company's cost of capital is 15% per year, how much could you afford to spend for a higher quality heat exchanger so that these annual replacement and downtime costs could be eliminated? Click the icon to view the interest and annuity table for discrete compounding when i=7% per year. Click the icon to view the interest and annuity table for discrete compounding when i=15% per year. You could afford to spend $ thousands for a higher quality heat exchanger. (Round to one decimal place.) \begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline \multicolumn{9}{|c|}{ Discrete Compounding; i=15%} \\ \hline & \multicolumn{2}{|c|}{ Single Payment } & \multicolumn{4}{|c|}{ Uniform Series } & \multicolumn{2}{|c|}{ Uniform Gradient } \\ \hline & CompoundAmountFactor & PresentWorthFactor & CompoundAmountFactor & PresentWorthFactor & SinkingFundFactor & CapitalRecoveryFactor & GradientPresentWorthFactor & GradientUniformSeriesFactor \\ \hline N & ToFindFGivenPF/P & ToFindPGivenFP/F & ToFindFGivenAF/A & ToFindPGivenAP/A & ToFindAGivenFA/F & ToFindAGivenPA/P & ToFindPGivenGP/G & ToFindAGivenGA/G \\ \hline 1 & 1.1500 & 0.8696 & 1.0000 & 0.8696 & 1.0000 & 1.1500 & 0.0000 & 0.0000 \\ \hline 2 & 1.3225 & 0.7561 & 2.1500 & 1.6257 & 0.4651 & 0.6151 & 0.7561 & 0.4651 \\ \hline 3 & 1.5209 & 0.6575 & 3.4725 & 2.2832 & 0.2880 & 0.4380 & 2.0712 & 0.9071 \\ \hline 4 & 1.7490 & 0.5718 & 4.9934 & 2.8550 & 0.2003 & 0.3503 & 3.7864 & 1.3263 \\ \hline 5 & 2.0114 & 0.4972 & 6.7424 & 3.3522 & 0.1483 & 0.2983 & 5.7751 & 1.7228 \\ \hline 6 & 2.3131 & 0.4323 & 8.7537 & 3.7845 & 0.1142 & 0.2642 & 7.9368 & 2.0972 \\ \hline 7 & 2.6600 & 0.3759 & 11.0668 & 4.1604 & 0.0904 & 0.2404 & 10.1924 & 2.4498 \\ \hline 8 & 3.0590 & 0.3269 & 13.7268 & 4.4873 & 0.0729 & 0.2229 & 12.4807 & 2.7813 \\ \hline 9 & 3.5179 & 0.2843 & 16.7858 & 4.7716 & 0.0596 & 0.2096 & 14.7548 & 3.0922 \\ \hline 10 & 4.0456 & 0.2472 & 20.3037 & 5.0188 & 0.0493 & 0.1993 & 16.9795 & 3.3832 \\ \hline \end{tabular}Step by Step Solution
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