TM%20661%20Engineering%20Economics

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TM 661 Engineering Economics

• Replacement Analysis

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Replacement / Challenge

• Example Car grows older and needs repairs
• at engine overhaul time should we fix or
  replace?

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Replacement / Challenge

• Example Car grows older and needs repairs
• at engine overhaul time should we fix or
  replace?
• Sunk costs are unrecoverable.
• Example Just put 800 in car, engine needs
  overhaul, should we repair or replace?
• The 800 just invested is not part of analysis.

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Example Replacement
Chemical Plant owns filter press purchased 3
years ago. Operating expense started at 4,000
per year 2 years ago and has increased by 1,000
per year. The press could last 5 more years with
an estimated salvage of 2,000 at that time.
Current market value of the press is 9,000. A
new press can be purchased for 36,000 with an
estimated life of 10 years. Annual operating
costs are 0 in year 1 growing by 1,000 per year.
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Cash Flow Approach
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Replacement (Cash Flow)
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Replacement (Cash Flow)
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Replacement (Cash Flow)

• NPW -7,000 (P/A, 15,5)
• - 1,000 (P/G, 15, 5)
• 2,000 (P/F, 15, 5)
• (28,246)

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Replacement (Cash Flow)
NPW 9,000 - 36,000 -1,000
(P/G, 15,5) 12,000 (P/F, 15, 5)
(26,809)
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Replacement (Cash Flow)
NPWK (28,246) NPWR (26,809)
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Replacement (Cash Flow)
NPWK (28,246) NPWR (26,809)
Choose Replace
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Replacement (Cash Flow)
NPWK (28,246) NPWR (26,809)
Note NPWR - NPWK 1,437
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Outsider viewpoint

• The outsider viewpoint approach considers the
  salvage value of the existing asset to be its
  investment cost if it is retained in service.
• The outsider viewpoint and the cash flow approach
  provide the same decision.

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Replacement (Outsider View)
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Replacement (Outsider View)

• NPW - 9,000
• -7,000 (P/A, 15,5)
• - 1,000 (P/G, 15, 5)
• 2,000 (P/F, 15, 5)
• (37,246)

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Replacement (Outsider View)
NPW - 36,000 -1,000 (P/G, 15,5)
12,000 (P/F, 15, 5) (35,809)
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Replacement (Outsider View)
NPWK (37,246) NPWR (35,809)
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Replacement (Outsider View)
NPWK (37,246) NPWR (35,809)
Choose Replace
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Replacement (Outsider View)
NPWK (37,246) NPWR (35,809)
Note NPWR - NPWK 1,437
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With 10 year Horizon
Suppose we now consider a 10 year
planning horizon. We estimate that the old press
will still have a salvage value of 2,000 5
years from now but that the new press will only
cost 31,000 5 years from now. Further,
estimated salvage 5 years hence is
15,000. Then
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With 10 Year Planning Horizon
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Replacement (Cash Flow)
NPW -7,000(P/A, 15,5) - 1,000(P/G,15,5)
-29,000(P/F,15,5) -1,000(P/A,15,5)(P/
F,15,5) 12,000(P/F,15,10) (42,821)
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Replacement (Cash Flow)
NPW -27,000 - 1,000(P/G,15,10)
3,000(P/F,15,10) (43,237)
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10-Year Horizon
NPWK (42,821) NPWR (43,237)
Choose Keep, trade in 5 years
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Multiple Alternatives
Suppose Dealer offers a 10,000 trade-in. In
addition, we identify 2 new alternatives 3.
New press for 40,000 with salvage after 5
years of 13,000. Trade-in on this machine
is 12,000. 4. Lease a press for 7,500
per year during the 5 year horizon.
Existing press will be sold on the open
market.
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Trade - In / Lease Options
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Outsider Viewpoint Approach
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Optimal Replacement
Suppose we have a compressor which costs 2,000
and has annual maintenance costs of 500
increasing by 100 per year. MARR20. Then
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Optimal Replacement
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Optimal Replacement
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Class Problem
The new president of Angstrom Technologies feels
the company must use the newest and finest
equipment in its labs. He has recommended that a
2-year-old piece of precision measurement
equipment be replaced immediately. Besides, he
feels it can be shown that his proposed equipment
is economically advantageous at a 15-per-year
return and a planning horizon of 5 years.
Perform the replacement analysis for a 5-year
period.
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Class Problem

• Current Proposed
• Original purchase price 30,000 40,000
• Current market value 15,000 ...
• Estimated useful life, years 5 15
• Estimated value, 5 years 7,000 10,000
• Salvage after 15 years ... 5,000
• Annual operating cost 5,000 3,000

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Solution
Keep
Replace
EUAW -25,000(A/P,15,5) -3,000
10,000(A/F,15,5) (8,975)
EUAW -5,000 7,000(A/F,15,5)
(3,962)
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Example (88, Page 249)

• Fluid Dynamics Company owns a pump that it is
  contemplating replacing. The old pump has annual
  operating and maintenance costs of 8000/year it
  can be kept for 4 years more and will have a
  salvage value at that time.
• The old pump can be traded in on a new pump. The
  trade-in value is 4000. The new pump will cost
  18,000 and have a value of 9000 in 4 years and
  will have annual operating and maintenance costs
  of 4500/year.
• Using a MARR of 10, evaluate the investment
  alternative based upon the present worth method
  and a planning horizon of 4500/year.
• Use the cash flow approach.
• Use the outsider approach.

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Solution

• Cash Flow Approach
• EOY NCF(Keep) NCF(Replace)
• 0 0 -14,000
• 1 -8,000 -4,500
  
• 2 -8,000 -4,500
• 3 -8,000 -4,500
• 4 -8,000 -4,500
• PW -25,358.92 -22,117.27 Replace
  

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Solution

• Outsider viewpoint
• EOY NCF(Keep) NCF(Replace)
• 0 -4,000 -18,000
• 1 -8,000 -4,500
• 2 -8,000 -4,500
• 3 -8,000 -4,500
  
• 4 -8,000 -4,500
• PW -29,358.92 -26,117.27
  Replace