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ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 1
OUTLINE
• Questions?
• News?
• Recommendation
• New homework
• IRR
• Depreciation
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 2
Recommendation
• National Gallery , Sainsbury wing– at Trafalgar Square
– Free
– Great impressionists (although there is a lot else there
such as medieval and renaissance paintings)
– Slide show at break
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 3
Practice Problem
You are considering
four types of
engineering designs.
The project lasts 10
years with the
following estimated
cash flows. The
interest rate (MARR) is
15%. Which of the four
is more attractive?
Project
A B C D
Initial cost $150 $220 $300 $340
Revenues/Year
$115 $125 $160 $185
Expenses/Year
$70 $65 $60 $80
IRR (%) 27.32 24.13 31.11 28.33
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 4
Items CMS Option FMS Option
Annual O&M costs:
Annual labor cost $1,169,600 $707,200
Annual material cost 832,320 598,400
Annual overhead cost
3,150,000 1,950,000
Annual tooling cost 470,000 300,000
Annual inventory cost
141,000 31,500
Annual income taxes 1,650,000 1,917,000
Total annual costs $7,412,920 $5,504,100
Investment $4,500,000 $12,500,000
Net salvage value $500,000 $1,000,000
Example 7.13 Incremental Analysis for Cost-Only Projects
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 5
n CMS Option FMS Option
Incremental
(FMS-CMS)
0 -$4,500,000 -$12,500,000 -$8,000,000
1 -7,412,920 -5,504,100 1,908,820
2 -7,412,920 -5,504,100 1,908,820
3 -7,412,920 -5,504,100 1,908,820
4 -7,412,920 -5,504,100 1,908,820
5 -7,412,920 -5,504,100 1,908,820
6 -7,412,920 -5,504,100
$2,408,820Salvage + $500,000 + $1,000,000
Incremental Cash Flow (FMS – CMS)
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 6
Solution:
PW i
P A i
P F i
IRR
FMS CMS
FMS CMS
( ) $8, ,
$1,908, ( / , , )
$2, , ( / , , )
.43%
000 000
820 5
408 820 6
0
12 15%,
select CMS.
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 7
Summary Rate of return (ROR) is the interest rate earned on
unrecovered project balances such that an investment’s cash receipts make the terminal project balance equal to zero.
Rate of return is an intuitively familiar and understandable measure of project profitability that many managers prefer to NPW or other equivalence measures.
Mathematically we can determine the rate of return for a given project cash flow series by locating an interest rate that equates the net present worth of its cash flows to zero. This break-even interest rate is denoted by the symbol i*.
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 8
Internal rate of return (IRR) is another term for ROR that stresses the fact that we are concerned with the interest earned on the portion of the project that is internally invested, not those portions that are released by (borrowed from) the project.
To apply rate of return analysis correctly, we need to classify an investment into either a simple or a nonsimple investment.
A simple investment is defined as one in which the initial cash flows are negative and only one sign change occurs in the net cash flow, whereas a nonsimple investment is one for which more than one sign change occurs in the net cash flow series.
Multiple i*s occur only in nonsimple investments. However, not all nonsimple investments will have multiple i*s either.
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 9
• For a pure investment, the solving rate of return (i*) is the rate of return internal to the project; so the decision rule is:
If IRR > MARR, accept the project.
If IRR = MARR, remain indifferent.
If IRR < MARR, reject the project.
IRR analysis yields results consistent with NPW and other equivalence methods.
• For a mixed investment, we need to calculate the true IRR, or known as the “return on invested capital.” However, if your objective is simply to make an accept or reject decision, it is recommended that either the NPW or AE analysis be used to make an accept/reject decision.
• To compare mutually exclusive alternatives by the IRR analysis, the incremental analysis must be adopted.
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 10
• Despite a strong academic preference for NPV, surveys
indicate that executives prefer IRR over NPV.[6] Apparently,
managers find it easier to compare investments of different
sizes in terms of percentage rates of return than by dollars of
NPV. However, NPV remains the "more accurate" reflection
of value to the business. IRR, as a measure of investment
efficiency may give better insights in capital constrained
situations. However, when comparing mutually exclusive
projects, NPV is the appropriate measure.
ENGINEERING ECONOMICS ISE460SESSION 16
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• Under Italian law, anyone who works at the same job for
more than 36 months is considered an employee, entitled by
law to four weeks of annual vacation, 11 paid holidays,
maternity leave, state pension contributions and severance
pay if dismissed.
ENGINEERING ECONOMICS ISE460SESSION 16
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ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
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DEPRECIATION
• FIXED ASSETS LOOSE THEIR VALUE - BEST EXEMPLIFIED
BY PERSONAL AUTOMOBILES
• THIS LOSS IS CALLED DEPRECIATION
• ECONOMIC DEPRECIATION - GRADUAL LOSS OF UTILITY
WITH TIME = PURCHASE PRICE - MARKET VALUE
• ACCOUNTING (OR ASSET) DEPRECIATION -- ALLOCATING
THE COST OF THE ASSET OVER ITS DEPRECIABLE LIFE.
TRIES TO MATCH ECONOMIC DEPRECIATION
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WHAT IS DEPRECIABLE?
ITEMS THAT ARE:
1. USED IN BUSINESS AND HELD FOR THE PRODUCTION
OF INCOME
2. HAVE A DEFINITE SERVICE LIFE LONGER THAN A
YEAR
3. MUST WEAR OUT, DECAY, BE USED UP, BECOME
OBSOLETE
4. MUST COST MORE THAN $1000
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WHO CAN DEPRECIATE ASSETS?
• ANYONE WHO USES THE ASSET FOR BUSINESS
PURPOSES AND MEETS THE RULES ON THE PREVIOUS
SLIDE
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COST BASIS
• COST BASIS IS:
• THE TOTAL COST OF THE ASSET OVER ITS LIFE
– INITIAL INVESTMENT
– FREIGHT
– SITE PREPARATION
– INSTALLATION
– TRADE-IN ALLOWANCE OF THE PREVIOUS EQUIPMENT
• IT DOES NOT INCLUDE OPERATING EXPENSES
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OTHER DEFINITIONS
• THE BOOK VALUE EQUALS THE COST BASIS LESS THE ACCUMULATED TOTAL DEPRECIATION
• SALVAGE VALUE - ESTIMATED VALUE AT THE END OF AN ASSET’S LIFE
• DEPRECIABLE LIFE - NUMBER OF YEARS THAT THE ASSET CAN BE USED
• ASSET DEPRECIATION RANGE (ADR) - PRESCRIBED GUIDELINES BY THE IRS FOR DEPRECIABLE LIFE
• IRS - INTERNAL REVENUE SERVICE -- ENFORCES TAX RULES CREATED BY CONGRESS AND COLLECTS THE TAXES
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DEPRECIATION METHODS• DEPRECIATION IS CALCULATED TWO WAYS, EACH WITH ITS OWN
PURPOSES
• BOOK DEPRECIATION
– FINANCIAL REPORTS
– INCOME STATEMENTS
– REFLECTS ACTUAL LOSS IN VALUE
– STATE TAXES AND FEDERAL TAXES BEFORE 1981
• TAX DEPRECIATION
– TO CALCULATE TAXES FOR THE IRS
– TAKES ADVANTAGE OF TAX RULES
– USUALLY LEADS TO BETTER CASH POSITION IN EARLIER YEARS
ENGINEERING ECONOMICS ISE460SESSION 16
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BOOK DEPRECIATION METHODS
• STRAIGHT LINE (SL)
• ACCELERATED
– DECLINING BALANCE (DB)
– DECLINING BALANCE WITH CONVERSION TO SL
• UNIT OF PRODUCTION
• DEPLETION
).......(
)(
21 nn
N
DDDIBN
SID
ENGINEERING ECONOMICS ISE460SESSION 16
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DECLINING BALANCE METHOD
• USES A FIXED FRACTION (alpha) OF THE BEGINNING BOOK
BALANCE EACH YEAR:
• MULTIPLIERS:
– 1.5 OR 150% DB
– 2.0 OR 200% OR DOUBLE DECLINING METHOD (DDB)
)(/1 multiplierN
nn
n
NN
IB
DDDTotalDB
ID
)1(
....
)1(
21
1
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
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DECLINING BALANCE WITH CONVERSION TO STRAIGHT LINE
• WHAT DO WE DO IF THE DB METHOD DOES NOT RESULT
IN THE ESTIMATED SALVAGE VALUE?
1. IF BN > S
– WE HAVE NOT DEPRECIATED THE ASSET FULLY
– SWITCH TO STRAIGHT LINE THE FIRST YEAR THAT THE
STRAIGHT LINE DEPRECIATION IS GREATER THAN THE
DB METHOD
2. BN < S
– STOP DEPRECIATING WHEN YOU GET TO S
EXAMPLES 9.3 AND 9.4
ENGINEERING ECONOMICS ISE460SESSION 16
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Example 9.3
• Cost basis of the asset I = $10,000
• Useful life N = 5 years
• Estimated salvage value S = $2,000
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Example 9.4 –Declining Balance
• Cost basis of the asset I = $10,000
• Useful life N = 5 years
• Estimated salvage value, S = $778
• D1 =
ENGINEERING ECONOMICS ISE460SESSION 16
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Changing to straight line
WE DID NOT DEPRECIATE THE ASSET FULLY AND MUST
SWITCH TO STRAIGHT LINE
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UNITS - OF - PRODUCTION METHOD
• ALLOCATES THE DEPRECIATION IN PROPORTION TO THE
UNITS PRODUCED AS A FRACTION OF THE TOTAL UNITS
EXAMPLE 9.7
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HISTORY OF DEPRECIATION METHODS
• UNTIL 1954 - STRAIGHT LINE ONLY
• 1954 - 1981 --ADDED ACCELERATED METHODS:
– DECLINING BALANCE (DB)
– DOUBLE DECLINING METHOD (DDB)
– SUM OF THE YEARS’ DIGITS (SOYD) (WE DID NOT COVER)
• 1981
– REPLACED BY ACCELERATED COST RECOVERY SYSTEM (ACRS)
– 1986 MODIFIED ACRS = MACRS
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MODIFIED ACCELERATED RECOVERY SYSTEM (MACRS)
• SALVAGE VALUE ALWAYS ZERO
• RECOVERY PERIOD -- ARBITRARY LIFE FOR THE
INVESTMENT, NOT NECESSARILY EQUAL TO ACTUAL LIFE
– EIGHT CATEGORIES: 3, 5, 7, 10, 15, 20, 27.5, 39 YEARS
• 3, 5, 7 AND 10 YEARS - USE 200%DDB, SWITCH TO SL
• 15 AND 20 YEARS -- USE 150%DDB AND SWITCH TO SL
• 27.5 YEARS RESIDENTIAL RENTAL -- SL
• 39 YEARS COMMERCIAL BUILDINGS -- SL
• USE TABLE 9.3 ON PAGE 448
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 28
MACRS TABLE
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HALF YEAR CONVENTION
• FOR EQUIPMENT PLACED IN SERVICE USING MACRS:
• ALL ASSETS ARE ASSUMED TO BE PLACED IN SERVICE IN
THE MIDDLE OF THE YEAR
• ALL SALVAGE VALUE IS ZERO
• ONLY A HALF YEARS’ DEPRECIATION IS ALLOWED THE
FIRST YEAR
• THE REMAINING HALF YEAR IS ALLOWED FOLLOWING THE
END OF THE RECOVERY PERIOD
• FOR REAL PROPERTY, USE MID - MONTH INSTEAD OF
HALF YEAR
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 30
EXAMPLE 9.8
• ASSET = $10,000 MACRS CLASS 5 YEARS
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DEPLETION
• APPLIES TO NATURAL RESOURCES
• OBJECTIVE IS THE SAME AS DEPRECIATION
• COST DEPLETION WORKS JUST LIKE UNITS OF PRODUCTION - EQUATION 9.9:
COST DEPLETION =(ADJUSTED BASIS OF MINERAL PROPERTY)x(NUMBER OF UNITS SOLD) / (TOTAL NUMBER OF RECOVERABLE UNITS)
EXAMPLE 9.10
• PERCENTAGE DEPLETION - PORTION OF GROSS INCOME, LIMITED BY 50% OF THE TAXABLE INCOME WITHOUT THE DEPLETION
• ALLOWABLE PERCENTAGES ARE GIVEN IN TABLE 9.5. THESE RANGE FROM 5% TO 22%
EXAMPLE 9.11
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Cost depletion vs. percentage depletion
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REPAIRS OR IMPROVEMENTS
• For tax depreciation purposes, repairs or improvements made to
any property are treated as separate property items.
• The recovery period for a repair or improvement to the initial
property normally begins on the date the repaired or improved
property is placed in service.
• The recovery class of the repair or improvement is the recovery
class that would apply to the property if it were placed in service
at the same time as the repair or improvement.
• Example 9.12 (modified)
ENGINEERING ECONOMICS ISE460SESSION 16
June 23, 2015
Geza P. Bottlik Page 34
EXAMPLE 9.12
• In January 2004, Kendall Manufacturing Company purchased a
new numerical control machine at a cost of $60,000. The machine
had an expected life of 10 years at the time of purchase and a
zero expected salvage value at the end of the 10 years.
• For book depreciation purposes, no major overhauls had been
planned over the 10-year period, and the machine was being
depreciated toward a zero salvage value, or $6,000 per year, with
the straight-line method.
• For tax purposes, the machine was classified as a 7-year MACRS
property.
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Geza P. Bottlik Page 35
EXAMPLE 9.12 CONTINUED
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EXAMPLE 9.12 CONTINUED
• In December 2006, however, the machine was thoroughly
overhauled and rebuilt at a cost of $15,000. It was estimated that
the overhaul would extend the machine’s useful life by 5 years.
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Geza P. Bottlik Page 37
EXAMPLE 9.12 CONTINUED