please show work step by step Question 5 300000 points Save

please show work step by step

Question 5 3.00000 points Save Answer A local food canning plant is considering to purchase a tomato-peeling machine. The purchasing manager prepared the following proformas. If the the manager decides to use MARR of 12%, which machine is the best alternative based on present worth analysis? Hint: consider the least common multiple as the study period. Cash flow Initial cost Maintenance & operating costs $15,000 Annual benefits Salvage value Useful life, in years Machine A Machine B Machine C $52,000 $63,000 $67,000 $9,000 $12,000 $38,000 $31,000 $37,000 $13,000 $19,000 $22,000 4 6 12

Solution

As the life of all the machines are unequal, use the common multiple method and convert the unequal life into equal life. The common multiple for 4, 6 and 12 years is 12. So the study period is 12 years. Machine A is to be repeated 3 times, machine B is to be repeated 2 times and machine C do not need any repetitions.

Machine A

Machine B

Machine C

Initial Cost =52,000

M&O Cost = 15,000

Annual Benefits = 38,000

Salvage Value = 13,000

MARR = 12%

NPW = -52,000 – 52,000 (P/F, 12%, 4) – 52,000 (P/F, 12%, 8) +23,000 (P/A, 12%, 12) + 13,000 (P/F, 12%, 4) + 13,000 (P/F, 12%, 8) + 13,000 (P/F, 12%, 12)

NPW = -52,000 – 52,000 (.6355) – 52,000 (.4039) +23,000 (6.194) + 13,000 (.6355) + 13,000 (.4039) + 13,000 (.2567)

= 53,263

Initial Cost =63,000

M&O Cost = 9,000

Annual Benefits = 31,000

Salvage Value = 19,000

MARR = 12%

NPW = -63,000 - 63,000 (P/F, 12%, 6) + 22,000 (P/A, 12%, 12) + 19,000 (P/F, 12%, 6) + 19,000 (P/F, 12%, 12)

NPW = -63,000 - 63,000 (.5066) + 22,000 (6.194) + 19,000 (.5066) + 19,000 (.2567)

= 55,855

Initial Cost =67,000

M&O Cost = 12,000

Annual Benefits = 37,000

Salvage Value = 22,000

MARR = 12%

NPW = -67,000 + 25,000 (P/A, 12%, 12) + 22,000 (P/F, 12%, 12)

NPW = -67,000 + 25,000 (6.194) + 22,000 (.2567)

= 93,497

Machine C should be selected as the Machine C has more NPV than Machine A and B.

Machine A	Machine B	Machine C
Initial Cost =52,000 M&O Cost = 15,000 Annual Benefits = 38,000 Salvage Value = 13,000 MARR = 12% NPW = -52,000 – 52,000 (P/F, 12%, 4) – 52,000 (P/F, 12%, 8) +23,000 (P/A, 12%, 12) + 13,000 (P/F, 12%, 4) + 13,000 (P/F, 12%, 8) + 13,000 (P/F, 12%, 12) NPW = -52,000 – 52,000 (.6355) – 52,000 (.4039) +23,000 (6.194) + 13,000 (.6355) + 13,000 (.4039) + 13,000 (.2567) = 53,263	Initial Cost =63,000 M&O Cost = 9,000 Annual Benefits = 31,000 Salvage Value = 19,000 MARR = 12% NPW = -63,000 - 63,000 (P/F, 12%, 6) + 22,000 (P/A, 12%, 12) + 19,000 (P/F, 12%, 6) + 19,000 (P/F, 12%, 12) NPW = -63,000 - 63,000 (.5066) + 22,000 (6.194) + 19,000 (.5066) + 19,000 (.2567) = 55,855	Initial Cost =67,000 M&O Cost = 12,000 Annual Benefits = 37,000 Salvage Value = 22,000 MARR = 12% NPW = -67,000 + 25,000 (P/A, 12%, 12) + 22,000 (P/F, 12%, 12) NPW = -67,000 + 25,000 (6.194) + 22,000 (.2567) = 93,497