Aerotron Electronics is considering the purchase of a water filtration system to assist in circuit board manufacturing. The system costs $70,000. It has an expected life of 7 years at which time its salvage value will be $7,500. Operating and maintenance expenses are estimated to be $7,000 per year. If the filtration system is not purchased, Aerotron Electronics will have to pay Bay City $44,000 per year for water purification. If the system is purchased, no water purification from Bay City will be needed. Aerotron Electronics must borrow 1/2 of the purchase price, but they cannot start repaying the loan for 2 years. The bank has agreed to 3 equal annual payments, with the 1st payment due at the end of year 2. The loan interest rate is 8% compounded annually. Aerotron Electronics, MARR is 10% compounded annually. Click here to access the TVM Factor Table Calculator What is the annual worth of this investment?$ Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is ±10. What is the decision rule for judging the attractiveness of investments based on annual worth? Should Aerotron Electronics buy the water filtration system?

Respuesta :

Answer:

Yes It will be accepted.

As the present worth (cost) of the water filtration

is far less than paying to Bay City for the purification

It decrease to $96,890.53‬ from  $214,210.43  of the Bay Ciity Option

Explanation:

Present worth if the system is not purchased:

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

C $ 44,000.00

time 7 years

rate 0.1

[tex]44000 \times \frac{1-(1+0.1)^{-7} }{0.1} = PV\\[/tex]

PV $214,210.4280

We will now compare the value of the water filtration system:

F0 investment: 70,000 / 2 = 35,000

loan payment:

principal after two-years grace period:

rate 0.08000

[tex]35000 \: (1+ 0.08)^{2} = Amount[/tex]

Amount 40,824.00

Then there is 3 payment annuity-due as it begins at the beginning right away after the grace period:

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

PV 40,824

time: 3 years

rate 0.08

[tex]40824 \div \frac{1-(1+0.08)^{-3} }{0.08}(1+.08) = C\\[/tex]

C  $ 14,667.667

Present value of the annuity discounted at Minimim accepted rate of return of 10%:

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

C 14,667.67

time years 3

rate 0.1

[tex]14667.6668309512 \times \frac{1-(1+0.1)^{-3} }{0.1}(1+0.10) = PV\\[/tex]

PV $40,123.9481

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

Maturity  40,123.95

time  2.00

rate  0.10000

[tex]\frac{40123.9481078087}{(1 + 0.1)^{2} } = PV[/tex]  

PV   33,160.29

present value of the maintenance expenses:

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

C $ 7,000.00

time 7 years

rate 0.1

[tex]7000 \times \frac{1-(1+0.1)^{-7} }{0.1} = PV\\[/tex]

PV $34,078.9317

Present value of the salvage value:

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

Salvage: $7,500.00

time  7 years

rate  0.10000

[tex]\frac{7500}{(1 + 0.1)^{7} } = PV[/tex]  

PV   3,848.69

Present worth of the water filtration system:

33,500 + +33,160.29 + 34,078.93 - 3,848.69 = 96,890.53‬