A company is trying to decide between two machines for its manufacturing line. T

Question

A company is trying to decide between two machines for its manufacturing line. The Process1000 has an initial cost of $56776 and operating costs of $14 per hour. It will allow the company to produce a higher quality product, providing an annual benefit of $70474. The company will use the Process1000 10204 hours per year. The Process1000 has a useful life 4 years and an estimated salvage value of $16802 at the end of its useful life.

The Super X has an initial cost of $77901. It is a bigger machine and can process more at a time so the company will ony need to use it for 8600 hours per year. The Super X has an estimated benefit of $92417 per year. The Super X will have a useful life of 8 years and a salvage value of $29183 at the end of its useful life.

Using a MARR of 7%, what is the maximum operating cost per hour at which the Super X could run to make it equally desirable to the Process1000?

*note- please use compound interest factors if possible and use annual worth instead of present worth. I have tried this problem a couple of times with present worth and keep getting it wrong and would like to see another way of doing it.

Explanation / Answer

Process 1000 Super X Initial Outflow                                  (56,776.00)     (77,901.00) Annual Benefit                                    70,474.00       92,417.00 Salvage Value                                    16,802.00       29,183.00 Life (Years)                                                     4                        8 Years Capital(Outflow/Inflow Annual Benefit Net Inflow Dis. Factor PV 0                                  (56,776.00)    -56,776.00 1    -56,776.00 1       70,474.00      70,474.00 0.934579      65,863.55 2       70,474.00      70,474.00 0.873439      61,554.72 3       70,474.00      70,474.00 0.816298      57,527.78 4                                  (39,974.00)       70,474.00      30,500.00 0.762895      23,268.30 5       70,474.00      70,474.00 0.712986      50,246.99 6       70,474.00      70,474.00 0.666342      46,959.80 7       70,474.00      70,474.00 0.62275      43,887.67 8                                    16,802.00       70,474.00      87,276.00 0.582009      50,795.43 Assume no Taxes Total 5.971299 3,43,328.23 EUAC      57,496.41 Assume After four years same machine is available at same rate Years Capital(Outflow/Inflow Annual Benefit Net Inflow Dis. Factor PV 0                                  (77,901.00)    -77,901.00 1    -77,901.00 1       92,417.00      92,417.00 0.934579      86,371.03 2       92,417.00      92,417.00 0.873439      80,720.59 3       92,417.00      92,417.00 0.816298      75,439.80 4                                                   -         92,417.00      92,417.00 0.762895      70,504.49 5       92,417.00      92,417.00 0.712986      65,892.04 6       92,417.00      92,417.00 0.666342      61,581.35 7       92,417.00      92,417.00 0.62275      57,552.66 8                                    29,183.00       92,417.00 1,21,600.00 0.582009      70,772.31 Assume no Taxes Total 5.971299 4,90,933.27 EUAC      82,215.50 EUAC of Super X       82,215.50 EUAC of Process 1000       57,496.41 Addional Annual Benefit in Super X       24,719.08 No fo Hrs Process 1000 work       10,204.00 Operating Cost Per hour                14.00 Total annual operating cost in Process 1000    1,42,856.00 Total Operating cost of Super X(24719.08+142856)    1,67,575.08 No of Hrs Machine Super X works          8,600.00 Operating Cost Per hour in case of Super X                19.49

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.