Problem 2: (30 points) alternatives are given below. Use an interest rate of 9%

Question

Problem 2: (30 points) alternatives are given below. Use an interest rate of 9% and The cost data for two equipment answer the following questions. Initial Cost Net Annual Benefits Salvage Value Life Machine A $200,000 $30,000 $45,000 20 vears Machine B $180,000 $24,000 $20,000 15 years a) What is the Equivalent Uniform Annual Worth (EUA W) of Machine A? 26 4.624 504 0. . 213303+45900-200000 113903 "-H A L b) What is the Equivalent Uniform Annual Worth (EUAW) of Machine B? 424-13,34 19132 1345 t0 120,ooo 1440 c) What option should the company select, why? opton A is the bettei choce ption

Explanation / Answer

EUAW of Machine-A = -200,000/{(1-(1+0.09)-20)/0.09} + 30,000 + 45,000/{((1+0.09)20-1)/0.09}

                               = -200000/9.1285 + 30,000 + 45,000/51.1601

                               = -21,909.40 + 30,000 + 879.59

                               = 8,790.18

EUAW of Machine-B = -180,000/{(1-(1+0.09)-15)/0.09} + 24,000 + 20,000/{((1+0.09)15-1)/0.09}

                               = -180,000/8.0607 + 24,000 + 20,000/29.3609

                               = -22,330.57 + 24,000 + 681.18

                               = 2,350.61

Since, EUAW of Machine-A > EUAW of Machine-B,

Machine-A will add more value per year and hence it should be selected one.

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