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One of the two methods should be used to produce expansion stabilizers. Method A initially costs $80,000 and will have a salvage value of $15,000 after 3 years. The cost to operate this way would be $30,000 a year.


Question: One of the two methods should be used to produce expansion stabilizers. Method A initially costs $80,000 and will have a salvage value of $15,000 after 3 years. The cost to operate this way would be $30,000 a year. The method will have an initial cost of $420,000, an operating cost of $8,000 per year, and a salvage value of $40,000 after its 3-year useful life. At an interest rate of 12% annually, specify the following: a. PW for method A $> b. PW from method B, $ C. Select method + → A Go to another question that will save this answer. Question 5 One of the two methods that should be used to produce the expansion anchor method would be $30,000 per year. Method B will have an initial cost of 12% per annum, determine the following: a. PW of method A, - 152,000 b. PW of method B, -162677-141377 C. Select which one complied -340472-110742 A None B 312000 C-162677.

Method A -

Calculate the present value for method A -
  • PW = - Initial cost - Annual operating cost (P / A, i, n) + Salvage value (P / F, i, n)
    PW = -80000 - 30,000 (P / A, 12%, 3) + 15000 (P / F, 12%, 3)
    PW = -80000 - [30000 * 2.4018] + [15000 * 0.7118]
    PW = -80000-72054 + 10677
    PW = -141377
    And so,
    PW for method A is- $141,377

    Method B.

    Calculate the present value of method B -
    PW = - Initial cost - Annual operating cost (P / A, i, n) + Salvage value (P / F, i, n)
    PW = -120000-8000 (P / A, 12%, 3) + 40000 (P / F, 12%, 3)
    PW = -120000 - [8000 * 2.4018] + [40000 * 0.7118]
    PW = -120000 - 19,214.4 + 28472
    PW = -110742.4
    And so,
    The PW for method B is $ -110,742
    (C)  The PW of method B is numerically greater.
    And therefore,
    Method b.
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