Please help!!! Engineering Economics class homework!!
When MARR = 9%
NPW of the first system = -100000 + (45000-22000)*(P/A,9%,10) - (100000 - 20000) * (P/F, 9%,5) + 20000 * (P/F, 9%,10)
= -100000 + 23000*6.417658 - 80000 * 0.649931 + 20000 * 0.422411
= 4059.86
NPW of system 2 = -150000 + (60000-40000)*(P/A,9%,10)
= -150000 + 20000* 6.417658
= -21646.85
First option should be selected as NPW is positive
When MARR = 18%
NPW of the first system = -100000 + (45000-22000)*(P/A,18%,10) - (100000 - 20000) * (P/F, 18%,5) + 20000 * (P/F, 18%,10)
= -100000 + 23000*4.494086 - 80000 * 0.437109 + 20000 * 0.191064
= -27783.44
NPW of system 2 = -150000 + (60000-40000)*(P/A,18%,10)
= -150000 + 20000* 4.494086
= -60118.27
First option should be selected as present cost is less
Please help!!! Engineering Economics class homework!! attached below are the tables needed!! first picture is alterna...
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