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The cash flows in the table below represent the potential annual savings associated with two different...

The cash flows in the table below represent the potential annual savings associated with two different types of production​ processes, each of which requires an investment of ​$34,000. Assume an interest rate of 6​%.

                                                                                                                                                                                                                                                                    

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(a) Determine the equivalent annual savings for each process.

The equivalent annual savings for process A are ​$_________. ​(Round to the nearest​ dollar.)

The equivalent annual savings for process B are ​$_________. ​(Round to the nearest​ dollar.)

(b) Determine the hourly savings for each process if will be in operation of 5,000 hours per year.

The hourly savings for process A are $______. (Round to the nearest cent.)

The hourly savings for process B are $______. (Round to the nearest cent.)

(c) Which process should be selected? Choose the correct answer below.

Yes or No

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Answer #1

(a)

First, Present Worth (PW) of savings is computed as follows.

PW, Process A ($) = -34,000 + 16,910 x P/F(6%, 1) + 15,330 x P/F(6%, 2) + 13,750 x P/F(6%, 3) + 12,170 x P/F(6%, 4)

= -34,000 + 16,910 x 0.9434** + 15,330 x 0.8900** + 13,750 x 0.8396** + 12,170 x 0.7921**

= -34,000 + 15,953 + 13,644 + 11,545 + 9,640

= 16,782

PW, Process B ($) = -34,000 + 12,000 x P/A(6%, 4) = -34,000 + 12,000 x 3.4651** = -34,000 + 41,581 = 7,581

Therefore,

Equivalent Annual savings, Process A ($) = PW of Process A / P/A(6%, 4) = 16,782 / 3.4651** = 4,843

Equivalent Annual savings, Process B ($) = PW of Process B / P/A(6%, 4) = 7,581 / 3.4651** = 2,188

(b)

Hourly saving, Process A ($) = Equivalent Annual savings / Number of hours per year = 4,843 / 5,000 = 0.97

Hourly saving, Process B ($) = Equivalent Annual savings / Number of hours per year = 2,188 / 5,000 = 0.44

(c)

Since Process A has higher hourly saving, Process A should be selected.

**From P/A and P/F Factor tables

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