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MIE 754 - Class #6 Manufacturing & Engineering Economics Concerns and Questions Concerns and Questions Quick Recap of Previous Class Quick Recap of Previous Class Today’s Focus: Today’s 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?

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Quick Recap of Previous Class b Useful Life versus Study Period b Internal Rate of Return Single AlternativeSingle Alternative Comparing AlternativesComparing 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.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.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 < 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 0500 Useful Life 10 yrs10 yrs 10 yrs Study Period 10 yrs10 yrs 10 yrs Use the IRR procedure to choose the best alternative.

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Example Problem Cont. Step 1. DN -> 2 -> 3 -> 1 Step 2. Compare DN -> 2 cash flows cash flows Investment-16,000 - 0 = -16,000 Annual Receipts 3,300 - 0 = 3,300 Salvage Value 0 - 0 = 0 Compute IRR DN->2 PW( i') = 0 = -16,000 + 3,300(P|A, i'%, 10) i' DN->2 15.9%

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Step 3. Since i' > MARR, keep alt. 2 (higher FC) as current best alternative. Drop DN from further consideration. Step 4. Next comparison: 2 -> 3 Investment-23,500 - (-16,000) = - 7,500 Annual Receipts4,800 - 3,300 = 1,500 Salvage Value 500 - 0 = 500 Computing IRR 2->3 PW( i') = 0 0= -7,500 + 1,500(P|A, i'%, 10) + 500(P|F, i'%, 10)

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i' 2->3 15.5% Since i' > MARR, keep Alt. 3 (higher FC) as current best alternative. Drop Alt. 2 from further consideration. Next comparison: 3 -> 1 cash flows Investment-28,000 - (-23,500) = - 4,500 Annual Receipts5,500 - 4,800 = 700 Salvage Value 1,500 - 500 = 1,000 Compute IRR 3->1 PW( i') = 0 0= -4,500 + 700(P|A, i'%, 10) + 1,000(P|F, i'%, 10) i' 3->1 10.9%

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Since i' < 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:$50,000Initial Investment:$50,000 Annual Revenues:20,000Annual Revenues:20,000 Annual Operating Costs:2,500Annual Operating Costs:2,500 Salvage Value @ EOY 5:10,000Salvage Value @ EOY 5:10,000 Study Period: 5 yearsStudy Period: 5 years MARR20%MARR20% Find Simple Payback Period Find Discounted Payback Period

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Example Simple PaybackDiscounted 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 (BV k ) = value of asset as shown on the accounting records (EOY k). BV k = cost basis - cumulative depreciation through year k

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Depreciation Terms SV N = estimated salvage value in year N (used in depreciation calculations where applicable) MV N = market (resale) value from the disposal of an asset d k = depreciation in year k (claimed at EOY) d k* = 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. d k = (B - SV N ) / N for k = 1, 2,..., N(6-1) d k * = k(d k ) for 1 k N(6-2) BV k = B - d k * (6-3)

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Declining Balance Method Annual depreciation is a constant percentage of the asset's value at the BOY. d 1 = B(R)(6-4) d k = B(1-R) k-1 (R) = BV k-1 (R)(6-5) d k * = B[1 - (1 - R) k ](6-6) BV k = B(1 - R) k (6-7) BV N = B(1 - R) N (6-8) R = 2/N200% declining balance, or R = 1.5/N150% declining balance Uses the useful life (or class life) for N Does not consider SV N

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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 BV 6 = B - d k * = 22,000 - (6(3,000)) = $4,000

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200% Declining Balance R = 2/6 = 1/3 = 0.33 d 3 = B (1-R) k-1 (R) = 22,000(0.67) 2 (0.33) = $3,259 BV 6 = B (1-R) k = 22,000(0.67) 6 = $1,931 d 6 * = 22,000{1-(1-0.33) 6 } = $20,069 Note: BV 6 = B - d 6 * = 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. d k = r k B 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, r k Step 3. Compute depreciation deduction in year k by multiplying the asset’s cost basis by the appropriate recovery rate, r k. 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 b 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. b 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 yeardepreciation deduction in the first year and the fourth year cumulative depreciation through year fourcumulative depreciation through year four book value at the end of the fourth yearbook 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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b Assume SV 9 = $5,000 d k = (95,000-5,000)/9= $10,000/yrfor k = 1 to 9 1. d 1 = d 4 = $10,000 2. d 4 * = 4 ($10,000) = $40,000 3.BV 4 = B - d 4 * = $95,000 - $40,000 = $55,000 Example Straight-Line Method

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Example 200% Declining Balance Method

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Example MACRS Method

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