Transcript MIE 754 Manufacturing & Engineering Economics

MIE 754 - Class #6
Manufacturing & Engineering
Economics
• Concerns and Questions
• Quick Recap of Previous Class
• Today’s Focus:
– Chap 4 Finish Rate of Return Methods
– Chap 4, Appendix 4B - Payback Period
Method and Liquidity
– Chapter 6 - Depreciation Methods
Concerns and Questions?
Quick Recap of Previous Class
 Useful
Life versus Study Period
 Internal
Rate of Return
• Single Alternative
• Comparing Alternatives
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
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
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
Example Problem Cont.
Step 1. DN -> 2 -> 3 -> 1
Step 2. Compare DN -> 2
 cash flows
 Investment
-16,000 - 0 = -16,000
 Annual Receipts
3,300 - 0 = 3,300
 Salvage Value
0-0=
0
Compute  IRRDN->2
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 Receipts 4,800 - 3,300 =
1,500
 Salvage Value
500 - 0 =
500
Computing  IRR2->3
PW(i') = 0
0= -7,500 + 1,500(P|A, i'%, 10) +
500(P|F, i'%, 10)
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 Receipts 5,500 - 4,800 =
700
 Salvage Value
1,500 - 500 =
1,000
Compute  IRR3->1
PW(i') = 0
0= -4,500 + 700(P|A, i'%, 10) +
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.
Graphical Interpretation of Example
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).
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 @ EOY 5:
10,000
• Study Period:
5 years
• MARR
20%
 Find
Simple Payback Period
 Find Discounted Payback Period
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
Chapter 6 - Consideration of
Depreciation and Taxes
Why consider taxes in economic
analysis?
BTCF versus ATCF?
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).
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)

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.
Depreciation Methods
Depreciable Life
If useful life *Straight line (includes SVN)
not given, use
*Declining balance (no SVN
class life
considered)
(Table 6-2)
use Property *MACRS (General Depreciation
Class (Table 6System)
2 or 6-3)
(no SVN considered)
*Skip ADS
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
BVk = B - dk*
(6-2)
(6-3)
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* = B[1 - (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
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
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
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* = 22,000{1-(1-0.33)6} = $20,069
Note: BV6 = B - d6* = 22,000 - 20,069 = $1,931
SL and DB Comparison
Straight Line
EOY
dk
BVk
200% Declining Balance
dk
BVk
0
0 22,000
0
22,000
1
3,000 19,000
7,333
14,667
2
3,000 16,000
4,889
9,778
3
3,000 13,000
3,259
6518
4
3,000 10,000
2,173
4,345
5
3,000
7,000
1,448
2,897
6
3,000
4,000
966
1,931
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 asset’s cost basis by the appropriate
recovery rate, rk.
MACRS over N + 1 years due to half-year
convention
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:
Previous Example with MARCS
EOY
rate, rk
dk
0
BVk
$22,000
1
0.2000
$4,400
17,600
2
0.3200
7,040
10,560
3
0.1920
4,224
6,336
4
0.1152
2,534
3,802
5
0.1152
2,534
1,268
6
0.0576
1,268
0
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 straightline 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
 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.
Example Straight-Line Method

Assume SV9 = $5,000
dk = (95,000-5,000)/9= $10,000/yrfor 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
Example 200% Declining Balance Method
R = 2/9
1.
2 
d1 = $95,000 9  = $21,111
3
1 - 2  2 
d4 = $95,000  9  9  = $9,933
2. d4* = $95,000
 
2 4 
1 - 1 
9  
 

= $60,235
4
1 - 2 


9
3. BV4 = $95,000
= $95,000 - $60,235 = $34,765
Example MACRS Method
Note: A class life = 9 years corresponds to an MACRS recovery
period = 5 years.
1. d1= (0.2) ($95,000) = $19,000
d4= (0.1152) ($95,000) = $10,944
2. d4*= $95,000 (0.2 + 0.32 + 0.192 + 0.1152) = $78,584
3. BV4 = $95,000 - $78,584 = $16,416
What happens if the asset is sold before the end of the recovery
period?
Can only take half of the normal MACRS deduction
If the bus is sold in yr 4: d4 = $95,000 (0.1152) (0.5) = $5,472