7.01_econ_12 - Replacement Analysis Fundamentals Chapter 14...

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1   ©M Pore  -  7.01  -  econ 12.ppt Replacement Analysis  Replacement Analysis  Fundamentals  Fundamentals  Chapter 14 Contemporary Engineering Economics
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2   ©M Pore  -  7.01  -  econ 12.ppt • Sunk cost : any past  cost unaffected by  any future decisions • Trade-in allowance value offered by the  vendor to reduce the  price of a new  equipment • Defender : an old  machine  • Challenger : a new  machine • Current market  value : selling price  of the defender in the  market place Replacement Terminology
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3   ©M Pore  -  7.01  -  econ 12.ppt Sunk Cost associated with an Asset’s Disposal     $0 $5000 $10,000 $15,000 $20,000 $25,000 $30,000 Original investment $10,000 $5000 Market value $10,000 Lost investment (economic depreciation) Repair cost $20,000 Sunk costs = $15,000
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4   ©M Pore  -  7.01  -  econ 12.ppt Replacement Decisions  Cash Flow Approach Treat the proceeds from  sale of the old machine  as down payment   toward purchasing the  new machine. This approach is  meaningful when both  the defender and  challenger have the  same service life . • Opportunity Cost  Approach – Treat the proceeds  from sale of the old  machine as the  investment required   to keep the old  machine. – This approach is more  commonly practiced  in replacement  analysis.
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5   ©M Pore  -  7.01  -  econ 12.ppt Example • Defender – Market price:  $10,000 – Remaining useful  life: 3 years – Salvage value:  $2,500 – O&M cost: $8,000 • Challenger – Cost: $15,000 – Useful life: 3  years – Salvage value:  $5,500 – O&M cost: $6,000
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6   ©M Pore  -  7.01  -  econ 12.ppt 0 1 2 3 0 1 2 3 $8000 $2500 $15,000 $6000 $5500 (a) Defender (b) Challenger $10,000 Sales proceeds from defender Replacement Analysis Cash Flow Approach
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7   ©M Pore  -  7.01  -  econ 12.ppt Defender: PW(12%) D = $8,000 ( P / A , 12%, 3) -$2,500 ( P / F , 12%, 3) = $17,434.90 AEC(12%) D = PW(12%) D ( A / P , 12%, 3) = $7,259.10 Challenger : PW(12%) C = $5,000 + $6,000 ( P / A , 12%, 3) - $5,500 ( P / F , 12%, 3) = $15,495.90 AEC(12%) C = PW(12%) C ( A / P , 12%, 3) = $6,451.79 Replace the defender now! Annual Equivalent Cost Cash Flow Approach
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8   ©M Pore  -  7.01  -  econ 12.ppt Comparison of Defender and Challenger Based on Opportunity Cost Approach
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9   ©M Pore  -  7.01  -  econ 12.ppt Defender: PW(12%) D = -$10,000 - $8,000( P / A , 12%, 3) + $2,500( P / F , 12%, 3) = -$27,434.90 AEC(12%) D = -PW(12%) D ( A / P , 12%, 3) = $11,422.64 Challenger : PW(12%) C = -$15,000 - $6,000( P / A , 12%, 3) + $5,500( P / F , 12%, 3) = -$25,495.90 AEC(12%) C = -PW(12%) C ( A / P , 12%, 3) = $10,615.33 Replace the defender now! Annual Equivalent Cost Opportunity Cost Approach
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10   ©M Pore  -  7.01  -  econ 12.ppt Economic Service Life  Economic Service Life 
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11   ©M Pore  -  7.01  -  econ 12.ppt Economic Service Life Definition Economic  service life  is the  remaining useful life of an  asset that results in the  minimum annual  equivalent cost.  Why do we need it? : We  should use the respective  economic service lives of  the defender and the  challenger when  conducting a replacement  Ownership (Capital) cost Operating cost + Minimize Annual Equivalent Cost
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12   ©M Pore  -  7.01  -  econ 12.ppt
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7.01_econ_12 - Replacement Analysis Fundamentals Chapter 14...

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