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A machine purchased three years ago for $317,000 has a current book value using straight-line depreciation of $188,000; its operating expenses are $37,000 per year. A replacement machine would cost $227,000, have a useful life of ten years, and would require $9,000 per year in operating expenses. It has an expected salvage value of $77,000 after ten years. The current disposal value of the old machine is $86,000; if it is kept 10 more years, its residual value would be $13,000.

Respuesta :

Answer:

At discount rate of 12% it is convinient to replace the machine as the net worth is lower.

Explanation:

We aren't given with any rate to work for we are going to assume a 12% rate of return

                      Current New machine

market value         86,000     227,000

expenses          37,000          9,000

useful life                  10                10

salvage                   13,000       86,000

F0                           -                     (141,000)*

PV expenses      (209,058)**  (50,852)***

PV salvage value     4,186****     27,690*****

Net worth       (204,873)          (164,162)

*227,000 cost less proceed from sale of the old machine

** annuity for 37,000 during 10 years discounted at 12%

[tex]C \times \frac{1-(1+r)^{-time} }{rate} = PV\\[/tex]

C 37,000.00

time 10

rate 0.12

[tex]37000 \times \frac{1-(1+0.12)^{-10} }{0.12} = PV\\[/tex]

*** annuity of 9,000 during 10 years discounted at 12%

[tex]C \times \frac{1-(1+r)^{-time} }{rate} = PV\\[/tex]

C 9,000.00

time 10

rate 0.12

[tex]9000 \times \frac{1-(1+0.12)^{-10} }{0.12} = PV\\[/tex]

**** present value of 13,000 value of the machine in 10 year discounted at 12%

[tex]\frac{Maturity}{(1 + rate)^{time} } = PV[/tex]  

Maturity  13,000.00

time  10.00

rate  0.12000

[tex]\frac{13000}{(1 + 0.12)^{10} } = PV[/tex]  

***** present value of 77,000 value of the machine in 10 year discounted at 12%

[tex]\frac{Maturity}{(1 + rate)^{time} } = PV[/tex]  

Maturity  77,000.00

time  10.00

rate  0.12000

[tex]\frac{77000}{(1 + 0.12)^{10} } = PV[/tex]