By Donald G. Newnan
Now in a 9th version, Engineering monetary research bargains accomplished assurance of economic and fiscal decision-making for engineering tasks, with an emphasis on challenge fixing, lifestyles cycle charges, and the time price of cash. The spreadsheet fabric from the former variation has been extended, permitting scholars to create and learn extra life like cash-flow types. The authors' concise, obtainable writing type and sensible emphasis make this article perfect for undergraduate engineering financial system classes. gains of the 9th version New pedagogical positive factors comprise: --Chapter-opening vignettes --Chapter targets the internal layout will enhance clarity, generate scholar curiosity, and facilitate comprehension of the cloth. a brand new bankruptcy has been further: bankruptcy 18, Accounting and Engineering financial system. bankruptcy 10, chance and Uncertainty, has been thoroughly rewritten to stress easy methods to make stable offerings via contemplating the uncertainty that's a part of each engineering economic climate program. bankruptcy thirteen, alternative research, has been rewritten to explain the comparability of latest resources with more recent choices. Appendix 7A, problems fixing for an rate of interest, has been completely revised to exploit the ability of spreadsheets to unravel difficulties. End-of-chapter difficulties are reorganized and up-to-date all through. A significant other site is out there: http://www.oup.com/us/engineeringeconomy. more advantageous pupil aid Packaged with each replica of the textual content: research consultant, via Edward Wheeler CD that includes interactive ExcelRG spreadsheets and tutorials more suitable teacher help CD containing examination records through Meenakshi Sundaram, ExcelRG spreadsheets, and PowerPointRG slides ideas handbook
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Additional resources for Engineering Economic Analysis, Solutions
Sample text
17 From compound interest tables, i = 10%. 45 (b) A = $140 i=? 95 From the 10% interest table, n = 16. 5% interest table, n = 35. 037 = $494 4-19 F=? 50%, 180) Since the ½% interest table does not contain n = 180, the problem must be split into workable components. On way would be: A = $20 n = 90 n = 90 F¶ F F = $20 (F/A, ½%, 90) + $20 (F/A, ½%, 90)(F/P, ½%, 90) = $5,817 Alternate Solution Perform linear interpolation between n = 120 and n = 240: F = $20 ((F/A, ½%, 120) ± (F/A, ½%, 240))/2 = $6,259 Note the inaccuracy of this solution.
5% interest table we see that n is between 15 and 16. This indicates that there will be 15 payments of $10 plus a last payment of a sum less than $10. 20 The final payment is the present worth of the three unpaid payments. 35 4-37 A=? 61 4-38 (a) A=? 35 (b) The quantities in Table 4-38 below are computed as follows: Column 1 shows the number of interest periods. n = 15 years Column 2 shows the equal annual amount as computed in part (a) above. 55 is the total payment which includes the principal and interest portions for each of the 15 years.
A=? 03313) = $977 4-40 Amortization schedule for a $4,500 loan at 6% Paid monthly for 24 months P = $4,500 Pmt. # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 i = 6%/12 mo = 1/2% per month Amt. 64 Int. 60 Monthly Pmt. 005 Column D = Column B + Column C (principal + interest) Column E = Column F - Column C (payment - interest owed) Column F = Uniform Monthly Payment (from formula for A/P) 4-41 Amortization schedule for a $4,500 loan at 6% Paid monthly for 24 months P = $4,500 Pmt. # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 i = 6%/12 mo = 1/2% per month Amt.