MIE 754 Manufacturing

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MIE 754 - Class 6 Manufacturing Engineering
Economics

• Concerns and Questions
• Quick Recap of Previous Class
• Todays Focus
• Chap 4 Finish Rate of Return Methods
• Chap 4, Appendix 4B - Payback Period Method and
  Liquidity
• Chapter 6 - Depreciation Methods

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Concerns and Questions?
3
Quick Recap of Previous Class

• Useful Life versus Study Period
• Internal Rate of Return
• Single Alternative
• Comparing Alternatives

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Comparing Mutually Exclusive Alternatives (MEAs)
with RR Methods

• Fundamental Purpose of Capital Investment
• Obtain at least the MARR for every dollar
  invested.
• Basic Rule
• Spend the least amount of capital possible unless
  the extra capital can be justified by the extra
  savings or benefits.
• (i.e., any increment of capital spent above the
  minimum must be able to pay its own way)

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Rate of Return Methods for Comparing Alternatives
MUST use an Incremental Approach!

• Step 1. Rank order alternatives from least to
  greatest initial investment.
• Step 2. Compare current feasible alternative with
  next challenger in the list
• Step 3. Compute RR (IRR or ERR) and compare with
  MARR. If RR lt Marr choose the least initial
  investment alternative. If RR ? MARR choose the
  greater initial investment alternative
• Step 4. Remove rejected alternative from list.
  Continue with next comparison

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Example Problem Given three MEAs and MARR 15

• 1 2 3
• Investment (FC) -28,000 -16,000 -23,500
• Net Cash Flow/yr 5,500 3,300 4,800
• Salvage Value 1,500 0 500
• Useful Life 10 yrs 10 yrs 10 yrs
• Study Period 10 yrs 10 yrs 10 yrs
• Use the IRR procedure to choose the best
  alternative.

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Example Problem Cont.

• Step 1. DN -gt 2 -gt 3 -gt 1
• Step 2. Compare DN -gt 2
• ? cash flows
• ? Investment -16,000 - 0 -16,000
• ? Annual Receipts 3,300 - 0 3,300
• ? Salvage Value 0 - 0 0
• Compute ? IRRDN-gt2
• PW(?i') 0 -16,000 3,300(PA, ?i', 10)
• ?i'DN-gt2 ? 15.9

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• Step 3. Since ?i' gt MARR, keep alt. 2 (higher FC)
  as current best alternative. Drop DN from
  further consideration.
• Step 4. Next comparison 2 -gt 3
• ? Investment -23,500 - (-16,000) -7,500
• ? Annual Receipts 4,800 - 3,300 1,500
• ? Salvage Value 500 - 0 500
• Computing ? IRR2-gt3 PW(?i') 0
• 0 -7,500 1,500(PA, ?i', 10) 500(PF, ?i',
  10)

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• ?i'2-gt3 ? 15.5
• Since ?i' gt MARR, keep Alt. 3 (higher FC) as
  current best alternative. Drop Alt. 2 from
  further consideration.
• Next comparison 3 -gt 1 ? cash flows
• ? Investment -28,000 - (-23,500) -4,500
• ? Annual Receipts 5,500 - 4,800 700
• ? Salvage Value 1,500 - 500 1,000
• Compute ? IRR3-gt1 PW(?i') 0
• 0 -4,500 700(PA, ?i', 10) 1,000(PF, ?i',
  10)
• ?i'3-gt1 ? 10.9

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• Since ?i' lt MARR, keep alt. 3 (lower FC) as
  current best alternative. Drop alt. 1 from
  further consideration.
• Step 5. All alternatives have been considered.
• Recommend alternative 3 for investment.

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Graphical Interpretation of Example
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Measures of Liquidity

• Simple Payback Period (?) - how many years it
  takes to recover the investment (ignoring the
  time value of money).
• Discounted Payback Period (?') - how many years
  it takes to recover the investment (including the
  time value of money).

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Measures of Liquidity Example

• Given the following
• Cost/Revenue Estimates
• Initial Investment 50,000
• Annual Revenues 20,000
• Annual Operating Costs 2,500
• Salvage Value _at_ EOY 5 10,000
• Study Period 5 years
• MARR 20
• Find Simple Payback Period
• Find Discounted Payback Period

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Example

• Simple Payback Discounted Payback
• (Cumulative PW) (Cumulative PW)
• EOY (i 0) (i MARR 20)
• 0 -50,000 -50,000
• 1 -32,500 -35,417
• 2 -15,000 -23,264
• 3 2,500 -13,137
• 4 20,000 -4,697
• 5 47,500 6,354.50
• ? 3 years ?' 5 years

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Chapter 6 - Consideration of Depreciation and
Taxes

• Why consider taxes in economic analysis?
• BTCF versus ATCF?

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Depreciation Terms

• Depreciation an annual non-cash charge against
  income. It represents an estimate of the dollar
  cost of fixed assets used in the production of a
  good or service.
• Cost Basis (B) actual cash cost plus book value
  of trade-in (if any) plus costs of making asset
  servicable (e.g., installation).
• Book Value (BVk) value of asset as shown on the
  accounting records (EOY k).
• BVk cost basis - cumulative depreciation
  through year k

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Depreciation Terms

• SVN estimated salvage value in year N (used in
  depreciation calculations where applicable)
• MVN market (resale) value from the disposal of
  an asset
• dk depreciation in year k (claimed at EOY)
• dk cumulative depreciation through year k
• Recovery Period number of years over which the
  basis of a property is recovered through the
  accounting process. Depreciable life, based on
  useful life (SL DB), property class (GDS), or
  class life (ADS) recovery period. (see Table
  7-2)

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What is Depreciable?

• 1. Must be used in business or held to produce
  income.
• 2. Must have a determinable life greater than one
  year.
• 3. Must wear out or get used up over time.
• 4. Is not inventory, stock in trade, or
  investment property.

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Depreciation Methods
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Straight Line Method

• A constant amount is depreciated each year over
  the asset's life.
• dk (B - SVN) / N for k 1, 2, ..., N (6-1)
• dk k(dk) for 1 ? k ? N (6-2)
• BVk B - dk (6-3)

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Declining Balance Method

• Annual depreciation is a constant percentage of
  the asset's value at the BOY.
• d1 B(R) (6-4)
• dk B(1-R)k-1(R) BVk-1(R) (6-5)
• dk B1 - (1 - R)k (6-6)
• BVk B(1 - R)k (6-7)
• BVN B(1 - R)N (6-8)
• R 2/N 200 declining balance, or
• R 1.5/N 150 declining balance
• Uses the useful life (or class life) for N
• Does not consider SVN

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SL and DB Example

• A computer was purchased for 20,000 and 2,000
  was spent installing it. The computer has an
  estimated salvage value of 4,000 at the end of
  its class life. Compute the depreciation
  deduction in year 3 and the book value at the end
  of year 6 using
• a) straight-line method
• b) 200 declining balance method

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• Step 0. Compute the Cost Basis (B) B
  20,000 2,000 22,000
• Step 1. Determine the Class Life From Table
  6-2, N 6 years
• Straight Line Method
• BV6 B - dk 22,000 - (6(3,000)) 4,000

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200 Declining Balance

• R 2/6 1/3 0.33
• d3 B (1-R)k-1(R) 22,000(0.67)2(0.33) 3,259
• BV6 B (1-R)k 22,000(0.67)6 1,931
• d6 22,0001-(1-0.33)6 20,069
• Note BV6 B - d6 22,000 - 20,069 1,931

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SL and DB Comparison
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MACRS (GDS) METHOD

• Annual depreciation is a fixed percentage of the
  cost basis (percentage specified by the IRS).
  Mandatory for most assets.
• dk rkB
• Step 1. Determine the property class (recovery
  period) from Table 6-2 or Table 6-3
• Step 2. Use Table 6-4 to obtain GDS rates, rk
• Step 3. Compute depreciation deduction in year k
  by multiplying the assets cost basis by the
  appropriate recovery rate, rk.
• MACRS over N 1 years due to half-year convention

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Previous Example by MACRS Method

• Step 0. Compute the Cost Basis (B) for the
  Computer B 20,000 2,000 22,000
• Step 1. Determine the Property Class (Recovery
  Period) From Table 6-2 5 year Recovery Period
• Steps 2 and 3 shown in the following table

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Previous Example with MARCS
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Example

• The La Salle Bus Company has decided to purchase
  a new bus for 85,000, with a trade-in of their
  old bus. The old bus has a trade-in value of
  10,000. The new bus will be kept for 10 years
  before being sold. Its estimated salvage value
  at that time is expected to be 5,000.
• Compute the following quantities using (a) the
  straight-line method, (b) the 200 declining
  balance method, and (c) the MACRS method.
• depreciation deduction in the first year and the
  fourth year
• cumulative depreciation through year four
• book value at the end of the fourth year

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• First, calculate the cost basis.
• B 10,000 85,000 95,000
• Next, determine the depreciable life.
• From Tables 6-2 and 6-3the class life 9 years
  and the GDS recovery period 5 years for buses.

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Example Straight-Line Method

• Assume SV9 5,000
• dk (95,000-5,000)/9 10,000/yr for k 1 to 9
• 1. d1 d4 10,000
• 2. d4 4 (10,000) 40,000
• 3. BV4 B - d4 95,000 - 40,000 55,000

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Example 200 Declining Balance Method
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Example MACRS Method