Share
Explore BrainMass

# Replacement decision for a lathe machine

The Burnaby Machine Company makes small parts under contract for manufacturers in the Vancouver area. The company makes a group of metal parts on a turret lathe for a local ski manufacturer. The current lathe is now six years old. It has a planned further life of three years. The contract with the ski manufacturer has three more years to run as well. A new, improved lathe has become available. The challenger will have lower operating costs than the defender.

The defender can now be sold for \$1200 in the used-equipment market. The challenger will cost \$25,000 including installation. Its salvage value after installation, but before use, will be \$20,000. Further data for the defender and the challenger are shown in the tables in the 2 tables in the attached file.

Burnaby machine is not sure if the contract it has with the ski company will be renewed. Therefore, Burnaby wants to make the decision about replacing the defender with the challenger using a three-year study period. Burnaby Machine uses a 12% MARR for this type of investment.

a) What is the present worth of costs over the next three years for the defender?
b) What is the present worth of costs over the next three years for the challenger?
c) Now suppose that Burnaby did not have a good estimate of the salvage of the challenger at the end of three years. What minimum salvage value for the challenger at the end of three years would make the present worth of costs for the challenger equal to that of the defender?

#### Solution Preview

a) What is the present worth of costs over the next three years for the defender?

The relevant costs
Year
0 \$1,200 (Salvage value foregone if machine is not sold
1 \$20,000
2 \$20,500
3 \$21,012.50 less salvage \$150.00 = \$20,862.50

We fine the present value of these costs using a discount rate of 12% (MARR)

Year Cost Discount factor @ Discounted cost =
12%
0 \$1,200.00 1 1,200 =1200*1
1 \$20,000.00 0.8929 17,858.00 =20000*0.8929
2 \$20,500.00 0.7972 16,342.60 =20500*0.7972
3 \$20,862.50 0.7118 14,849.93 =20862.5*0.7118
Present worth= \$50,250.53

Answer : Present worth of costs over the next 3 years for the ...

#### Solution Summary

Evaluates a replacement decision for a lathe machine.

\$2.19