In: Economics
Problem #3: Sam is considering buying a new lawnmower. He has a choice between a Yard Machine mower and a Lawn Guy mower. Sam has a MARR of 10%. The salvage value of each mower at the end of its service life is zero.
Yard Machine mower |
Lawn Guy mower |
|
First Cost |
$450 |
$350 |
Life |
10 Years |
10 Years |
Annual Gas Cost |
$80 |
$100 |
Annual Maintenance Cost |
$30 |
$40 |
a) Ans: The Alternative - Yard Machine Mower is preferable.
Explanation:
Yard Machine Mower;
Initial cost ( P ) = $450
Annual cost ( A ) = $80 + $30 = $110
Interest rate ( i ) = 10%
Number of interest period ( n) = 10
PW = P + A ( P/A , i , n )
= $450 + $110 ( P/A , 10% , 10 )
= $450 + $110 (6.1446 ) =$450 + $675.91
=$1125.91
Lawn Guy Mower;
Initial cost ( P ) = $350
Annual cost ( A ) = $100 + $40 = $140
Interest rate ( i ) = 10%
Number of interest period ( n) = 10
PW = P + A ( P/A , i , n )
= $350 + $140 ( P/A , 10% , 10 )
= $350 + $140 (6.1446 ) =$350 + $860.24
=$1210.24
Here , The present worth cost of Yard Machine Mower is less than the Lawn Guy Mower . Therefore , Yard Machine Mower is preferable.
b) Ans: The Alternative - Yard Machine Mower is preferable.
Explanation:
Yard Machine Mower;
Initial cost ( P ) = $450
Annual cost ( A ) = $80 + $30 = $110
Interest rate ( i ) = 10%
Number of interest period ( n) = 10
AW = P ( A/P , i , n ) + A
= $450( A/P , 10% , 10 ) + $110
= $450 (0.1627 ) + $110 = $73.22 + $110
= $183.22
Lawn Guy Mower;
Initial cost ( P ) = $350
Annual cost ( A ) = $100 + $40 = $140
Interest rate ( i ) = 10%
Number of interest period ( n) = 10
AW = P ( A/P , i , n ) + A
= $350( A/P , 10% , 10 ) + $140
= $350 (0.1627 ) + $140 = $56.95 + $140
= $196.95
Here , The annual worth cost of the Yard Machine Mower is less than the Lawn Guy Mower . Therefore , Yard Machine Mower is preferable.