Question

The cash flows of two alternatives for an electronic machine are given below. Which one should be selected on the basis of AW-Based rate of return analysis? MARR is 20% per year. A First Cost ($) AOC ($/year) Salvage Value ($) Life (years) -270,000 -135,000 75,000 3 -245,000 - 139,000 35,000 Both O None OB

Answer 1

Solution: For Alternative A, 757000 no 135,000 135,000 135,000 270,000 & a 20% cash How diagram. pwa pw of cash in flow -pw of cash out flow = 75000 (P/F, 20%,3)- 135000/P/A, 20%,3) - -270,000 =175000*0.3787) - (135000x2.106] - 270,000 From compound interest Teable . = -510got.5 .. AW = pw (A/P, &, n) =-510907.5 (A/P, 20%,3) =-57ogor.5x0.4747 - - 242527.79 $

For Alternative B. n=0 1 2 3 4 5 3500o 139,000 245,000 r = 200. Chash cash flow diagsam PW=35000 (P/F, 20%,6)_139,000 (P/A, 20%,6) - 245,000 =135000x 0.3349) - (139,000 x 3.326) - 245000 = -695592.5 ₂ AW = pW (Alp, sin) =t695592.3-80.3007) -209164.66$ It Awto the project is economically justified. Hese both the cases we got awco so, None of the project should selected If you have to choose one then you should choose B to minimazine the loss.