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Two alternative machines are being considered for a cost reduction project. Mach

ID: 2662860 • Letter: T

Question

Two alternative machines are being considered for a cost reduction project.
Machine A has a first cost of $60,000 and a salvage value of $22,000 at the end of it 6 year service life. Annual operating costs of this machine and the associated probabilities are estimated as follows:
Annual O&M Costs         Probability
$5,000                           0.20
8,000                            0.30
10,000                            0.30
12,000                            0.20
Machine B has a first cost of $35,000 and its salvage value at the end of it 4-year service life is negligible. Annual operating costs of this machine and the associated probabilities are estimated as follows:
Annual O&M Costs           Probability
$8,000                           0.10
10,000                           0.30
12,000                            0.40
14,000                            0.20
The MARR on this project is 10%. The required service period is estimated to be 12 years. Assuming independence, calculate the mean and variance for the equivalent annual cost of operating each machine.

Explanation / Answer

Solution:

    (a)

        • Machine A:

                  CR(10%) of A =         ($60,000 - $22,000) (A/P, 10%,6)

                                         + (0.10) ($22,000)

                                     =   $10,924

           E[NAW(10%)] of A =            ($5,000) (0.20) + ($8,000) (0.30)

                                         + ($10, 000) (0.30) + ($12,000) (0.20)

                                         + $10,924

                                         = $19,725

        Var[NAW(10%)] of A =             (15,924 - 19,725)2 (0.20)

                                         + (18,924 - 19, 725)2 (0.30)

                                  + (20,924 - 19, 725)2 (0.30)

                                  + (22,924 - 19, 725)2 (0.20)

                              =   5,560,000

      • Machine B:

          CR(10%) of B =     $35,000 (A/P, 10%, 4)

                         =   $11,042

     E[NAW(10%)] of B=       ($8,000) (0.10) + ($10,000) (0.30)

                             + ($12,000) (0.40) + ($14,000) (0.20)

                             + $11,042

                         =   $22,442

Var[NAW(10%)] of B =        (19,042 - 22,442)2 (0.10)

                             + (21,042 - 22,442)2 (0.30)

                             + (23,042 - 22,442)2 (0.40)

                             + (25,042 - 22,442)2 (0.20)

                         =   3,240,000

(b) Prob[NAW(10%) of A> NAW(10%) of B]:

                      Joint event                               Joint probability