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Ashley Foods, Inc. has determined that any one of five machines can be used in one phase of its chili canning operation. The costs of the machines are estimated below, and all machines are estimated to have a 4-year useful life. If the minimum attractive rate of return is 20% per year, determine which machine should be selected on the basis of a rate or return analysis.

Sagot :

Answer:

Machine 1 should be selected

Explanation:

Note: This question is not complete. The complete question is therefore presented before answering the question as follows:

Ashley Foods, Inc. has determined that any one of five machines can be used in one phase of its chili canning operation. The costs of the machines are estimated below, and all machines are estimated to have a 4-year useful life. If the minimum attractive rate of return is 20% per year, determine which machine should be selected on the basis of a rate or return analysis.

Machine             First Cost ($)        AOC ($)

      1                       -31000               -16000

     2                       -29000              -19300

     3                       -34500               -17000

     4                       -49000              -12200

     5                        -41000              -15500

Note: AOC means annual operating cost.

The explanation of the answer can now be given as follows:

The machine that should be selected can be determined by comparing the present worth (PW) of the 5 machines to one another which can be calculated using the following  formula :

PW = Co + (A * ((1 - (1 / (1 + r))^n) / r)) …………………………………. (1)

Where;

PW = Present worth of each machine

A = Annual operating cost of each machine

r = minimum attractive rate of return = 20%, or 0.20

n = useful life = 4

Using equation (1) and substituting the relevant values, we have:

PW of Machine 1 = -31000 + (-16000 * ((1 - (1 / (1 + 0.20))^4) / 0.20)) = -31000 + (-16000 * 2.58873456790123) = -31000 -  41,419.75 = -$72,419.75

PW of Machine 2 = -29000 + (-19300 * ((1 - (1 / (1 + 0.20))^4) / 0.20)) = -29000 + (-19300 * 2.58873456790123) = -29000 - 49,962.58 = -$78,962.58

PW of Machine 3 = -34500 + (-17000 * ((1 - (1 / (1 + 0.20))^4) / 0.20)) = -34500 + (-17000 * 2.58873456790123) = -34,500 - 44,008.49 = -$78,508.49

PW of Machine 4 = -49000 + (-12200 * ((1 - (1 / (1 + 0.20))^4) / 0.20)) = -49000 + (-12200 * 2.58873456790123) = -49,000 - 31,582.56 = -$80,582.56

PW of Machine 5 = -41000 + (-15500 * ((1 - (1 / (1 + 0.20))^4) / 0.20)) = -41000 + (-15500 * 2.58873456790123) = -49,000 - 40,125.39 = -$81,125.39

Since the PW of Machine 1 of -$72,419.75 is the least negative from the above calculations, Machine 1 should therefore be selected.

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