ENGG 209 Lecture Notes - Lecture 20: Islamic Iran Participation Front, Mansfield, Sunk Costs

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Published on 11 Dec 2016
Department
Course
Replacement Analysis
Terms and Methodology
Defender: An existing piece of equipment
Challenger: The best available replacement piece of equipment
Always assume asset class continues in replacement analysis
Always assume the replaced asset is sold at the beginning of year N+1
Sunk costs should be ignored
EXAMPLE: A pump (class 8-20% CCA) bought three years ago cost $5000. A major overhaul was just
completed at a cost of $2000. The annual operating cost is $1000, and the current market value is
$2500. If t=40%, what is the relevant opportunity cost of not replacing the pump now?
$2000 before tax.
Basic Replacement Decision (same useful life)
Calculate PE and AE of cash flows over the useful life of the defender and challenger
Choose the lowest cost alternative
EXAMPLE: Suppose the pump from the previous example is expected to last three more years after
which it is salvaged for $1000. A new pump (also expected to last three years) could be purchased
for $4000. This new pump would reduce operating costs by 50% and could be salvaged for $2000
after three years. MARR = 10%, t = 40%, CCA = 20%, CTFII = CTF1/2 = 0.7455. Should the existing pump
be replaced now?
Defender:
PE operating costs = 0.6x1000x(P/A, 10, 3) = 1492
PE opportunity of not replacing pump = 2500xCTFII = 1864
PE salvage = 1000xCTFIIx(P/F, 10, 3) = 560
PE defender = 1492+1864-560 = 2796
Challenger:
PE capital costs = 4000xCTF1/2 = 2982
PE operating costs = 1492/2 = 746
PE salvage = 2000xCTFiix(P/F, 10, 3)=1120
PE Challenger = 2982+746-1120 = 2608
2608<2796, so the defender should be replaced immediately
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