Question

A pipeline contractor can purchase a needed truck for $42,000. Its estimated life is 6 years, and it has no salvage value. Maintenance is estimated to be $3,100 per year. Operating expense is $60 per day. The contractor can hire a similar unit for $190 per day. MARR is 7%.

a. How many days per year must the truck’s services be needed such that the two alternatives are equally costly?

b. If the truck is needed for 180 days/year, should the contractor buy the truck or hire the similar unit? Determine the dollar amount of savings generated by using the preferred alternative rather than nonpreferred.

Answer 1

Solution :-

Initial Cost = $42,000

Annual Maintenance Cost = $3,100

Useful Life = 6 Years

Daily Operating Exps = $60

let truck service in a year be X days in a year

Annual Operating Cost = 60 X

MARR = 7%

Now Annual Expenditure = $42,000 * A/P( 7% , 6 ) + 60 X + $3,100

= ( $42,000 * 0.20979 ) + 60 X + $3,100

= $11,911.42 + 60 X

Rent of truck per day = $190 X

Now take these two equal

= $11,911.42 + $60 X = $190 X

X = 91.63 Days

Two alternatives are equal costly if truck service are required approx 92 Days .

(b)

If truck is needed for 180 days per year

Total Annual Expenditure if the truck is bought = $11,911.42 + 60 X

= $11,911.42 + ( 60 * 180 )

= $22,711.42

Total Annual Expenditure if the truck is Rented = ( $190 * 180 ) = $34,200

Dollar amount of saving = $34,200 - $22,711.42 = $11,488.58

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