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Skip to main content x Sign In. This text covers the basic techniques and applications of engineering economy for all disciplines in the engineering profession. The third edition concentrates on fundamental techniques and their applications, the efficient use of spreadsheets, and a rich coverage of personal financial situations in which engineering economy techniques can be applied easily and rapidly.
The text presents the topics in condensed formats when compared to the larger text Engineering Economy. Lowest Price. Rent now. Listing page for eBook Subscription Program. Your monthly rental cart is already full. Click continue to view and update your selected titles. Use the challenger for all three years. Therefore, keep the defender 1 year and then replace it with the challenger b The PW values are placed in the year cell prior to when the year starts for challenger.
Spreadsheet screen shot shows the AW of the current system defender D for its retention period with close-down cost in last year, followed by annual contract cost challenger C for years in effect.
Do not replace. Four options are present, but they have the same conclusion. Answer is a The defender should be kept for 2 more years and then replaced with another used machine just like the one presently owned. The ESL is 13 years. Year 13 is predicted to require the 4th rebuild; the pump will not be used beyond 13 years anyway.
Compare the results in 2 and 3 with that in 1 and comment on them. No bundle with X and Y are listed; 12 remain. Considering the selection restrictions, the 9 viable bundles are: DN 4 34 1 13 3 23 Not acceptable bundles: 2, 12, 14, 24, , , Select bundle 4.
Select projects A, B and C. Alternatively, use a spreadsheet. Develop a scatter chart. Largest PW is for project Q. Answer is b The breakeven point falls substantially to , when the lower FC is in effect. Note: To guarantee that the cell computations in column C correctly track when the breakeven point falls below ,, the same IF statement is used in all cells.
With this feature, sensitivity analysis on the , estimate may also be performed. It has the smaller slope of 5 versus 25 for the buy option. Painting and blasting is not done at end of year Set the AW relations equal. See note above. A spreadsheet solution for all three parts follows. Payback between 6 and 7 months. Solve for np in PW relation. Payback just over 7 months. Resolve for N. Therefore, the cost difference would increase. If hardness removal were discontinued, the extra cost for its removal column 4 in Table 1 would be zero for all alternatives.
Therefore, the best alternative continues to be number 3. Therefore, in order for the dollars to have the same value i. Subtract the future value of each payment from the bond face value 5 years from now. Both amounts take purchasing power into account. Do the answers seem reasonable? Stocks: Possibly, if the economy and selected corporate stocks do very well.
Bonds: Probably not, the discount required is far more than given when a bond is purchased. This is why, in part, the fixed-income investments are losers when inflation is a sincere factor. If a single allocation basis had been used throughout, the rate would have had to increase for each basis.
Make has annual indirect costs. Cost driver is the number of bags handled, some of which are lost or damaged. Significant change took place, especially at MEX.
Another basis could be guest-dollars, that is, total amount of money a guest spends. Price for standard should go up. Premium model makes a good profit at current price under ABC Overall — Not a correct impression when costs are examined. Market value is the amount that could be received if the asset is offered for sale on the open market. Tax depreciation is used to determine the annual tax-deductible amount. They are not necessarily the same amount. The adjusted basis means some depreciation has been charged.
These are commonly different, usually shorter, than the anticipated useful life of an asset used in the economic evaluation. It is an allowance for the wear and tear, deterioration, or obsolescence of the property. The recovery period and method of depreciation are the primary differences. Even if not using the property, it is in service when it is ready and available for its specific use. Depreciating stops when property is retired from service, even if its cost is not fully recovered.
The taxpayer can elect the Section deduction instead of recovering the cost through depreciation deductions. Develop the scatter chart to plot book values. The round-off errors are noticeable. MACRS is accelerated compared to SL depreciation because more of the first cost is written off in the early years of the recovery period. Book values are close for both ways. The relations used to determine the values row 50 are indicated first row 3.
By Equation [16A. S indicates the selected method and amount. Marginal rate: The portion of each taxable dollar of TI that is paid in taxes on the last dollar of income, e. Indexing: Updating of the TI limits not the rates each year to account for inflation and other factors.
Use Eq. The CFAT expression removes the capital investment or adds salvage but does not consider depreciation, since it is a non-cash flow. Spreadsheet solution follows. Taxes are best when low country 1. This increases the CFAT considerably in the last year. This will reduce taxes on the gains. If there are no gains, carry-forward and carry-back allowances may apply.
For a capital gain, it is the difference between the sales price and the unadjusted basis first cost of the asset. If it is a capital loss, the netting of losses against gains can affect AW. The tax savings will be applied to the challenger, since the savings is realized only if the challenger is bought. Thus, a capital loss will render the challenger more attractive.
DR is involved on the defender trade in. Defender trade depreciation recapture must be included. Succession options Option Defender Challenger 1 2 years 1 year 2 1 2 3 0 3 2. Find AW for defender and challenger for 1, 2 and 3 years of retention. Calculate CFAT for 1, 2, and 3 years of ownership.
There is DR each year. Engineering VP has the better economic strategy. The only option is the defender for 5 years and the challenger for 5 years. Spreadsheet is below with Solver parameters. Engineering managers like CFAT because it indicates actual cash flow of the project. The AW values are the same. CFAT shows a big cost in year 0 and positive cash flows thereafter.
Spreadsheet solution is easier. The CFAT series is shown, for information only. Use Equation [ As mentioned in the chapter, the largest D-E financing option will always offer the largest return on the invested equity capital. But, too high of D-E mixes are risky. Subtract 2 different equity CFAT totals. Conclusion: The added business in Victoria should turn positive the third year and remain a contributor to the business after that, as indicated by the EVA values.
Otherwise, keep the current one. R and n. Variations in P and R have about the same effect on PW in opposite directions, and more effect than variation in n.
AW is most sensitive to variations in revenue R and least sensitive to variations in life n. Top node: 0. Conclusion: Company should implement the test program option and delay the full-scale decision for 1 year. Spreadsheet analysis used for changes in MARR. Sensitivity to changes in life is performed by hand. Not very sensitive; plan A has the best PW for all life estimates.
Treat output as discrete or continuous variable. If discrete, center points on cells, e. P L is triangular like the distribution in Figure with the mode at 5. A larger sample is needed to observe all values of X. Some samples are: RN X p As an illustration, assume the probabilities that are assigned by a student are: 0. Assume you want 25 values. Repeat for sample of 25 grades. Substitute Y for DY. Substitute M for DM. The limit to the series N 0.
CFAT given for years 0 through 6. Uniform for both i and CFAT values. Set up a spreadsheet. Plot the PW values for as large a sample as desired.
For risk, the result depends on the preponderance of positive PW values from the simulation, and the distribution of PW obtained in step 6.
The RNG screen image is shown below. The spreadsheet above is the same as that in Problem The decision to accept the plan uses the same logic as that described in Problem The 7-step procedure from Section For each sample cash flow series, calculate the AW value for each alternative.
To obtain a final answer of which alternative is the best, it is recommended that the number of positive and negative AW values be counted as they are generated. Then the alternative with the most positive AW values indicates which one to accept.
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