Transcript MIE 754 Manufacturing & Engineering Economics

MIE 754 - Class #7
Manufacturing & Engineering
Economics
• No Class Meeting this Thurs (10/4/01)
• Concerns and Questions
• Quick Recap of Previous Class
• Today’s Focus:
– Chapter 6: Depreciation (continued) and
Taxes
• Assignment 1 - Graded Due in 2 Weeks
– Refer to course web site for instructions and
materials
• Homework Problems Due in 1 Week (next
pg)
Concerns and Questions?
 No
Class Meeting this Thurs
(10/4)
 Homework
Due in 1 Weeks:
• Chap 3 - 5, 9, 14, 19, 21, 23
• Chap 4 - 6(a), 12, 16, 26(IRR only)
• Chap 6 - 6(a, c, d), 17(c), 20, 25, 30
 Homework
and Assignment Tracking
Quick Recap of Previous Class
 Internal
Rate of Return
• Comparing Alternatives
• Basic Rule and Steps
 Depreciation
• What is it? Why bother?
• Straight Line
• Declining Balance
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
200% Declining Balance
BVk
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.
Remember: MACRS is spread 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
After-Tax Cash Flow Analysis (Fig 6-4)
EOY
(A)
BTCF
(B)
dk
(C) = (A) - (B)
TI or NIBT
0
-I
---
---
1
2
3
:
R1-E1
R2-E2
R3-E3
:
N
RN-EN
N
MVN
---
MVN - BVN
After-Tax Cash Flow Analysis (Fig 6-4)
(D) = -t(C)
Tk
(E) = (A) + (D)
ATCF
EOY
---
-I
0
1
2
3
:
N
-t(MVN - BVN)
MV-t(MVN - BVN)
N
An after-tax evaluation of a project's aftertax cash flows requires an after-tax MARR
After - tax MARR
Before-tax MARR = (1 - tax rate )
Equation (6-18)
Example:If before-tax MARR = 20% and t = 40%.
What is the approximate after-tax MARR?
MARR AT
MARRBT = 1 - t
MARR AT
0.20 = 1 - 0.40
MARRAT = 0.12
When an asset is disposed of for more (less) than
its book value, the resulting gain (loss) is taxed
 Depreciation
recapture and capital gains
(losses) are taxed as ordinary income.
 Capital
 If
Gain (Loss) = MV - BV
an asset is sold for less than its current
book value (MV < BV), it is termed a
capital loss, and taxes on the loss represent
a tax credit.
Example
Investment
Net Annual Receipts
Study Period
Market Value at EOY 4
After-tax MARR
Effective income tax rate
MACRS recovery period
$10,000
$4,000/yr
4 years
$5,000
15%
40%
5 years
Is this a worthwhile investment after taxes?
Side Question: What is the approximate
before-tax MARR?
MARR AT
0.15
MARRBT = 1 - t = 1 - 0.4
= 25%
Develop ATCF for Example
EOY BTCFk
dk
k
0
- $10,000
-1
4,000 $2,000
2
4,000 3,200
3
4,000 1,920
4a
4,000
576
4b
5,000
--
TIk
-$2,000
800
2,080
3,424
2,696
Income
Tax
(t = 0.4)
-- 800
- 320
- 832
- 1,369.6
- 1,078.4
ATCF
-$10,000.0
3,200.0
3,680.0
3,168.0
2,630.4
3,921.6
Is Investment Worthwhile?
PW(15%) = - $10,000 + $3,200(P/F,15%,1)
+ $3,680(P/F,15%,2)
+ $3,168(P/F,15%,3)
+ ($2,630.4 + $3,921.6) (P/F,15%,4)
= $1,394
PW(MARRAT) > 0
investment is attractive
Looking Closer at Year 4
d4 = (0.1152) ($10,000) (0.5) = $576
Capital Related Gain in Year 4 = $5,000 - $2,304 = $2,696
BV4 = B - d4* = $10,000 ($2,000 + $3,200 + $1,920 + $576)
= $2,304
NOTE:
Only in years 1 - 4a is TI = BTCF - d
TI4b (sale of asset) = MV4 - BV4 = $5,000 - $2,304
= $2,696
Lease vs. Purchase Example

Determine the more economic means of acquiring a copier in
your business if you may either (a) purchase the copier for
$5,000 with a probable resale value of $1,000 at the end of 5
years or (b) rent the copier for an annual fee of $900 per year
for 5 years with an initial deposit of $500 refundable upon
returning the copier in good condition. If you own the copier,
you will depreciate it by using the MACRS method (class life
of 5 years). All rental fees are deductible for income tax
purposes. As the owner or lessee, you will pay all expenses
associated with the operation of the copier. A deposit does
not affect taxes when paid out or received back.

Compare these alternatives by using the equivalent uniform
annual cost method. The after-tax MARR is 10% per year,
and the effective income tax rate is 40%.
Option A - Purchase Copier
Investment = $5,000
MV5 = $1,000
MACRS recovery period = 5 years
EOY
0
1
2
3
4
5
5
BTCF
- $5,000
0
0
0
0
0
1,000
d
-$1,000
1,600
960
576
288
--
TI
-- $1,000
-1,600
-960
-576
-288
424
Tax
-$400.0
640.0
384.0
230.4
115.2
-169.6
ATCF
- $5,000.0
400.0
640.0
384.0
230.4
115.2
830.4
d5 = (0.1152)($5,000)(0.5) = $288
BV5 = $5000 - ($1,000 + $1,600 + $960 + $576 + $288) = $576
Capital Related Gain at EOY 5 = $1,000 - $576 = $424
Option A - Purchase Copier
PW(10%) = - $5,000 + $400(P/F,10,1) + $640(P/F,10,2) + $384(P/F,10,3)
+ $230.4(P/F,10,4) + ($115.2 + $830.4)(P/F,10,5) = - $3,074
AW(10%) = - $3,074(A/P,10,5) = - $811
Option B - Rent Copier
Rent = $900 / year
Deposit = $500
(returned at EOY 5)
EOY
BTCF
d
TI
Tax
ATCF
0
1-5
5
- $500
- 900
500
----
-- $900
--
-$360
--
- $500
- 540
500
AW(10%) = - $500(A/P,10,5) - $540 + $500(A/F,10,5) = - $590
Based on the annual equivalent of the after-tax cash flows, leasing the copier is
the minimum cost alternative.
Note: In general,
After-tax cost of lease = Leasing Cost (1 - t)
= $900 (1 - 0.4) = $540
Over what range of before-tax leasing costs would you choose the purchase
option based on an after-tax analysis?
Let
X = before-tax annual leasing cost of computer
After-tax cost of lease = X (1 - 0.4) = 0.6X
AW(10%)purchase = - $811
AW(10%)lease = - $500(A/P,10,5) - 0.6X + $500(A/F,10,5)
AWpurchase = AWlease
- $811 = - $50 - 0.6X
X = $1268.33 / year
If X = $1,268.33/yr, the annual equivalent costs for the alternatives are equal.
For X > $1,268.33/yr, the purchase option minimizes costs.