Question

Solve for A and B, Engineering Economy

Question Help %) Problem 6-52 (algorithmic) Compare alternatives A and B with the present worth method if the MARR is 15% per year. Which one would you recommend? Assume repeatability and a study period of 20 years. $40,000 $7,000 at end of year 1 and increasing by $700 per year thereafter $7,000 every 5 years $15,000 $14,000 at end of year 1 and increasing by $1,400 per year thereafter Capital Investment Operating Costs Overhaul Costs None Life 10 years 20 years $14,000 if just overhauled Salvage Value negligible Click the icon to view the interest and annuity table for discrete compounding when the MARR is 15% per year. The PW of Alternative A is s.(Round to the nearest dollar.)

please solve it right!

Answer 1

The answer for above problem is explained below.

(Pw) Capital perahing oveshaul Sal voge Net Cash flouw Cost nvestmen Value C$) 4 o, ood o 40,000 6086.95 ,000 (115 1100 58. 30 T100 5523.13 8,400 8 400 84400 [。15)

50 2.95 , IOO 4 a, oo (1-15) 6,9 0O Boo 5 835.56 6,800 5 Ísl5リ 4539.4i3 0,500 1o, 500 4210. 4 , 200 20O 1I, 200 1.15)

,900 3810. 13 11,900 11,900 358 1 . To 9 12, 600 12,600 5017. 344 2o, 300 10 3 007,00 3009. 205 141000 14,000 2747. 535 14,100 4To0 12 2502.930 5, 400 13 15,400 2275. 39 14 6, 100 に、100 294338 3, 800 1 5

7, 500 870. 133 7, 500 16 61, 251 18, 200 18, 2o0 1孑 18, 900 18 18,90 377. 200 19, 600 19,600 19 a 523.44\ 20,300 | 1,40,300 | 14,000 | 00ס Q0 1146 76.679 尸 Pん 丰 The Pw of Altesahve A s 114, 676. 6