Question

One of two methods must be used to produce expansion anchors. Method A costs $70,000 initially and will have a $18,000 salvage value after 3 years. The operating cost with this method will be $22,000 per year. Method B will have a first cost of $105,000, an operating cost of $18,000 per year, and a $30,000 salvage value after its 3-year life. The interest rate for both the methods is 10%. Which method should be used on the basis of a present worth analysis? The present worth of method A is $______ and that of method B is $______ . Method __________ is selected by the company.

Answer 1

Ans: The present worth of method A is $111,188.4 and that of method B is $127,225.2. Method A is selected by the company.

Explanation:

Method A :

Initial costs = $70,000

Annual operating cost = $22,000

Salvage value (F ) = $18,000

n = 3 years

i = 10%

Present worth of method A = 70000 + 22000 ( P / A , i , n ) - 18000 ( P / F , i , n )

  = 70000 + 22000 ( P / A , 10% , 3 ) - 18000 ( P / F , 10% , 3)

= 70000 + 22000 (2.4869 ) - 18000 ( 0.7513)

= 70000 + 54711.8 - 13523.4

= 124711.8 - 13523.4

= $111,188.4

Method B:

Initial costs = $105,000

Annual operating cost = $18,000

Salvage value (F ) = $30,000

n = 3 years

i = 10%

Present worth of method B = 105000 + 18000 ( P / A , i , n ) - 30000 ( P / F , i , n )

  = 105000 + 18000 ( P / A , 10% , 3 ) - 30000 ( P / F , 10% , 3)

= 105000+ 18000 (2.4869 ) - 30000 ( 0.7513)

= 105000 + 44764.2 - 22539

= 149764.2 - -22539

= $127,225.2

The total present worth cost of method A is less than that of method B. So method A should be selected.