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GE2250 Understand Global Project for Business and Engineering Professionals
Instructor: Jiayu Chen Ph.D.
Project Cost Estimation and Budgeting
© Jiayu Chen, Ph.D. 2
COURSE STRUCTURE
L4. Project Cost Estimation
and Budgeting
L5. Project Planning,
Control and Crashing
L6. Risk Management for Global Project
Project Management Triangle Strategic Management
L2. Cross-culture Collaboration
L3. Organizational Structures
Culture
Organization
Global Team
New Challenges
L8. Supply Chain Management
L10. Lean Project Management
L11. Mergers and Acquisitions
L9 Leadership and Team Building
L12 Virtual Environment and
Global Team
© Jiayu Chen, Ph.D. 3
Buy a lot of food !
Hide in your pocket!Save in the bank !
Lend it to your friend !
© Jiayu Chen, Ph.D. 4
INVESTMENT
Any currency have both purchasing power and earning power!
Investment Activity
The act of committing money or capital to an endeavor with the expectation of
obtaining an additional income or profit.
Simple Problems:
Can generally be worked in one's head without extensive analysis…
• Should I buy a subway pass or pay daily rates?
• Should I buy the textbook for 3129 or an iPod?
© Jiayu Chen, Ph.D. 5
Why Bother Investing?
INVESTMENT
• The rule of 7-10 and 10-7
• Inflation
• Future security
• Large purchase
© Jiayu Chen, Ph.D. 6
• Must be organized and analyzed.
• Sufficiently important to justify serious thought and action.
• The economic aspects are significant.
• Single criteria decision making is generally adequate.
However there are some decisions much difficult to make!Because investment for projects are much complex.
INVESTMENT
© Jiayu Chen, Ph.D. 7
INVESTMENT
Situation (given) Criterion
Fixed input Maximize output
Fixed output Minimize input
Neither fixed(Variable input and output)
Maximize difference(output –input) or (benefits –costs)
• I have $500 & want a ski
vacation at winter break!
• The aqueduct must provide
10,000 m3/d!
• The citizens of Hong Kong
need a more efficient way to
get to Macau!
Most Common
© Jiayu Chen, Ph.D. 8
ENGINEERING ECONOMICS
Engineering Economics is defined as the analysis of costs, benefits
and revenues occurring over time (for problems of intermediate
complexity).
• Examples at work– Should we rent a dredging machine to pump sand from
the riverbed for this 2 yr project or should we buy the sand
from a local source?
– Should we train and use in-house resources to develop
our 3D and 4D CAD models for the next fiscal year or
outsource it to a company abroad?
• Examples at home– How much can I afford to borrow in student loans based
on my expected future earnings if I plan to be debt free in 5
years?
© Jiayu Chen, Ph.D. 9
Money is a valuable commodity !
– Can be leased or rented (borrowed)
You’re probably renting some money to or from a bank right now!
– Amount borrowed/loaned is the Principal, P
– Amount paid for its use is called Interest, I
– Interest for 1 time period is called interest rate, i
TIME VALUE
If I borrow $100 (=P) for 1 year, and I pay
back a total of $109 (=P+I),
then the interest paid for 1 year (=I) = $9
and the interest rate (i) = $9 $100 = 9%
© Jiayu Chen, Ph.D. 11
• Interest that is computed only on the original sum or principal
• Total interest earned
𝐈 = 𝐏 × 𝐢 × 𝐧
SIMPLE INTEREST
I = $100 x 9% per period x 2 periods = $18
Where,
P = present sum of money
i = interest rate
n= number of periods
© Jiayu Chen, Ph.D. 12
SIMPLE INTEREST
• Amount of money due at the end of a loan
• Would you want a loan with simple interest terms?
• Would a bank want to give you one?
𝐅 = 𝐏 + 𝐏 × 𝐢 × 𝐧 or 𝐅 = 𝐏[𝟏 + (𝐢 × 𝒏)]
Where,
F = future value
F = $100 [1 + (0.09 x 2)] = $118
© Jiayu Chen, Ph.D. 13
COMPOUND INTEREST
• Interest that is computed on the original unpaid debt AND the unpaid interest
• Unless otherwise specified, in this course (and in real life) assume you are
dealing with Compound Interest
𝐈𝐧 = 𝐏 × 𝟏 + 𝐢𝐧 − 𝐏
Where,
P = present sum of money
I = interest rate
n = number of periods (e.g., years)
𝐈𝟐 = $𝟏𝟎𝟎 × 𝟏 + 𝟎. 𝟎𝟗 × (𝟏 + 𝟎. 𝟎𝟗) − $𝟏𝟎𝟎 = $𝟏𝟖. 𝟖𝟏
First year
Second year
© Jiayu Chen, Ph.D. 14
Period Simple Compound
0 $100.00 $100.00
1 $109.00 $109.00
2 $118.00 $118.81
3 $127.00 $129.50
4 $136.00 $141.16
5 $145.00 $153.86
6 $154.00 $167.71
7 $163.00 $182.80
8 $172.00 $199.26
9 $181.00 $217.19
10 $190.00 $236.74
15 $235.00 $364.25
20 $280.00 $560.44
25 $325.00 $862.31
COMPOUND INTEREST vs. SIMPLE INTEREST
Principal = $100.00 Interest rate= 9.00%
Over a short period the difference
between simple and compound interest is
negligible.
Over a long period of time the
difference could be considerable.
© Jiayu Chen, Ph.D. 15
PAYMENT PLANS
Borrowing $5,000 at 8% for 5 years
• Which plan would you choose as a borrower?
• Which plan would you choose as a lender?
• Why?
End of year Plan 1 Plan 2 Plan 3
1 $400 $1,252 $0
2 $400 $1,252 $0
3 $400 $1,252 $0
4 $400 $1,252 $0
5 $5,400 $1,252 $7,347
Total $7,000 $6,260 $7,347
© Jiayu Chen, Ph.D. 16
Debt of $5,000 and i = 8%
You pay almost twice the interest in Plan 3
vs. Plan 2… how can these be equivalent?
© Jiayu Chen, Ph.D. 17
Debt of $5,000 and i = 9%
None of these plans are equivalent to the original $5,000, nor are
they equivalent to each other (note the balances).
© Jiayu Chen, Ph.D. 18
EVALUATION – Cash Flows
Present Value is the value of a sum of money in today’s
dollars–Future sums are discounted when moved backward in time(i.e., to the present)
Future Value is the value of a sum of money at a specified
time and interest rate–Present sums are compounded when moved forward in time(i.e., to the future)
Cash flows are equivalent if they are of equal value at a particular point in time
and a particular interest rate–All 3 plans are equivalent to $5,000 now at 8%
© Jiayu Chen, Ph.D. 19
INTEREST FORMULAS
Notation:
i = Interest rate per interest period
n = Number of interest periods
P = Present sum of money (Present worth)
F = Future sum of money (Future worth)
PeriodBeginning Balance
Interest for Period
Ending Balance
0 P
1 P iP P(1 + i)
2 P(1 + i) iP(1 + i) P 1 + i 2
3 P 1 + i 2 iP 1 + i 2 P 1 + i 3
n P 1 + i n−1 iP 1 + i n−1 P 1 + i n
© Jiayu Chen, Ph.D. 20
INTEREST FORMULAS
Period Interest this period Value
0 $100.00
1 $8.00 $108.00
2 $8.64 $116.64
3 $9.33 $125.97
4 $10.08 $136.05
5 $10.88 $146.93
6 $11.75 $158.69
7 $12.69 $171.38
A future sum of $171.38 at a time 7 periods from now is equivalent to a
present value of $100, at 8% interest.
PV
FV
© Jiayu Chen, Ph.D. 21
• Formula:
𝑭 = 𝑷 𝟏 + 𝒊 𝒏 is the single payment compound amount factor
• Functional notation:
𝑭 = 𝑷 (𝑭/𝑷, 𝒊, 𝒏)𝑭 = $𝟏𝟎𝟎 (𝑭/𝑷, 𝟖%, 𝟕) = $𝟏𝟕𝟏
• Read (F/P,i,n) as “the future value of a present value at interest rate i
for n periods.”
INTEREST FORMULAS
Calculating a Future Value
© Jiayu Chen, Ph.D. 22
INTEREST FORMULAS
Calculating a Present Value
• Formula:
𝑷 = 𝑭 𝟏 + 𝒊 𝒏 𝟏 + 𝒊 𝒏 is the single payment present worth factor
• Functional notation:
𝑷 = 𝑭 (𝑷/𝑭, 𝒊, 𝒏)𝑭 = $𝟏𝟕𝟏 (𝑷/𝑭, 𝟖%, 𝟕) = $𝟏𝟎𝟎
• Read (P/F,i,n) as “the present value of a future value at interest rate i
for n periods.”
© Jiayu Chen, Ph.D. 24
• An understanding of “P” and “F” calculations is fundamental tounderstanding cash flows.
• However, in reality cash flows are often more complex:
– Cash flows can be repeated over many periods
e.g., Car payments
– Cash flows can increase or decrease over many periods
e.g., Maintenance costs
– Interest rates can be identified on an annualized basis but
calculated on shorter intervals
e.g., Credit cards
– Interest rates can be compounded continuously (vs. annually)
e.g., Rarely used in practice… useful in financial modeling as
theoretical maximum
CASH FLOWS
© Jiayu Chen, Ph.D. 25
CASH FLOWS
Uniform Series
A uniform series of payments (A) that occur at the end of each period
P F
0 1 2 3 4 0 1 2 3 4
© Jiayu Chen, Ph.D. 26
CASH FLOWS
• Find the future value of an investment based on periodic, constant
payments and a constant interest rate
• Solve with Tables:
𝐅 = 𝐀 (𝐅/𝐀, 𝐢, 𝐧)– (F/A, i, n) converts a series of uniform amounts A into an equivalent
future amount F
– Called the Uniform Series Compound Amount Factor
– “Find F given A” in Interest Tables
• Solve with Equation:
𝑭 = 𝑨𝟏 + 𝒊 𝒏 − 𝟏
𝒊
© Jiayu Chen, Ph.D. 27
CASH FLOWS
Find F given A = 5000, I =5%, n=5
YearBegin
BalanceAdd Interest
End Balance
1 $0 $5000 $0 $5000
2 $5000 $5000 $250 $10250
3 $10250 $5000 $513 $15763
4 $15763 $5000 $788 $21551
5 $21551 $5000 $1078 $27628
F = A (F/A, i, n) = $5000 (F/A,5%,5)
F = $5000 (5.526)
F = $27,630
© Jiayu Chen, Ph.D. 28
• A = F (A/F, i, n)
– Called the Uniform Series Sinking Fund Factor
– “Find A given F” in Interest Tables
• A = P (A/P, i, n)
– Called the Uniform Payment Series Capital
Recovery Factor
– “Find A given P” in Interest Tables
• P = A (P/A, i, n)
– Called the Uniform Payment Series Present Worth
Factor
– “Find P given A” in Interest Tables
CASH FLOWS
𝐴 = 𝐹𝑖
1 + 𝑖 𝑛 − 1
𝐴 = 𝑃𝑖 1 + 𝑖 𝑛
1 + 𝑖 𝑛 − 1
𝑃 = 𝐴1 + 𝑖 𝑛 − 1
𝑖 1 + 𝑖 𝑛
© Jiayu Chen, Ph.D. 29
CASH FLOWS
Arithmetic Gradient
• An amount that increase uniformly over
time.
• G = the amount by which it increase
from one period to the next.
Period Cash Flow
1 $100
2 $110
3 $120
4 $130
5 $140
6 $150
7 $160
8 $170
P
0 1 2 3 4
© Jiayu Chen, Ph.D. 30
CASH FLOWS
P
0 1 2 3 4
A
A+GA+2G
A+3G
P1
A A A A
P2
0G
2G3G
= +
• We can split this into “A”s and an
“xG” series
• Converting series of “A”s to P1 is
straightforward
• Note that the first amount in the gradient series is always 0 (at period 1)
t = 0 --
t = 1 0
t = 2 G
t = 3 2G
t = 4 3G
etc.
© Jiayu Chen, Ph.D. 31
http://www.investopedia.com/calculator/
http://www.moneychimp.com/calculator/present_value_calculator.htm
• Find the present value of a series of uniformly increasing payments at a
constant interest rate
P = G (P/G, i, n)– Called the Arithmetic Gradient Present Worth Factor
– “Find P given G” in Interest Tables
• Find the value of a series of uniform payments from a series of uniformly
increasing payments at a constant interest rate
A = G (A/G, i, n)– Called the Arithmetic Gradient Uniform Series Factor
– “Find A given G” in Interest Tables
CASH FLOWS
© Jiayu Chen, Ph.D. 32
CASH FLOWS
Find A given i = 6% and the following CFD:
𝐀 = 𝟏𝟎𝟎𝟎 + 𝟏𝟎𝟎𝟎 × (𝐀/𝐆, 𝟔%, 𝟒) = 𝟏𝟎𝟎𝟎 + 𝟏𝟎𝟎𝟎 × (𝟏. 𝟒𝟐𝟕) = $2427
© Jiayu Chen, Ph.D. 33
CASH FLOWS
The estimated maintenance expense for
a new machine is as follows:
Year Maintenance Cost
1 $1000
2 $2000
3 $3000
4 $3000
What is the equivalent uniform annual maintenance cost if the interest rate
is 6%?
Example 1
© Jiayu Chen, Ph.D. 34
ANALYSIS PERIOD
• Sometimes called the planning horizon
• What is the useful life of each alternative?
– Depends on technology and business
• Three possibilities:
1.Useful life of each alternative equals the analysis period.
2.Alternatives have useful lives different from the analysis period.
3.Analysis period is infinite (n = ∞)
© Jiayu Chen, Ph.D. 35
ANALYSIS PERIOD
Useful Lives = Analysis Period
Let’s face it, commuting on the subway stinks. Why not board in? You’re
considering two purchase options… a Classic board for $200 and an
ÜberCool board for $300. Apple is Thinking Different and decided to have
iTunes sponsor you if you put their logo on your board. They’ll pay $45/yr for
Classic (cool) and $60/yr for ÜberCool (cooler). At the end of 6 years you plan
to sell the board (assume a salvage value of $10 for Classic and $70
ÜberCool). Interest rate is 8%… which do you choose?
Example 2
© Jiayu Chen, Ph.D. 36
Assume an Analysis Period that is least common multiple
when it is feasible.
Assume Alternative 1 has a useful life of 5 years, and Alternative 2
has a useful life of 10 years.
ANALYSIS PERIOD
© Jiayu Chen, Ph.D. 37
ANALYSIS PERIOD
Useful Lives ≠Analysis Period
Crane A costs $150,000, lasts 5 yrs, and has a salvage value of $20,000.
Crane B costs $160,000, lasts 10 yrs, and has a salvage value of $32,500. At
7% interest, which crane do you recommend?
Example 2
© Jiayu Chen, Ph.D. 38
Assume an Analysis Period that is least common multiple when
it is feasible.
Assume Alternative 1 has a useful life of 7 years, and Alternative 2 has a
useful life of 13 years. The least common multiple is 91 years… not very
feasible!
ANALYSIS PERIOD
© Jiayu Chen, Ph.D. 40
ANNUAL WORTH ANALYSIS / PRESENT VALUE ANALYSIS
• Similar process to Present Worth Analysis
• Convert cash flows to equivalent
– EUAB = Equivalent Uniform Annual Benefit
– EUAC = Equivalent Uniform Annual Cost
– EUAW = Equivalent Uniform Annual Worth EUAB-EUAC
• The Analysis Period or Planning Horizon is an important consideration
• However, unlike Present Worth Analysis, it is not necessary to compute
each alternative over the same period in most cases.
© Jiayu Chen, Ph.D. 41
ANNUAL WORTH ANALYSIS
Situation Criterion
Fixed Input
Fixed amount of capital
to investMaximize Benefits
(EUAB)
Fixed Output
Fixed task or benefit to
achieveMinimize Costs
(EUAC)
Neither Fixed
Neither capital to
invest, nor the final end
result is fixed
Maximize Benefits –
Costs
(EUAW = EUAB –
EUAC)
© Jiayu Chen, Ph.D. 42
ANNUAL WORTH ANALYSIS
Net Present Value (NPV) of a time series of cash flows, both incoming and
outgoing, is defined as the sum of the present values of the individual cash flows
of the same entity.
NPV is a central tool in discounted cash flow analysis and is a standard method
for using the time value of money to appraise long-term projects.
𝐍𝐏𝐕 𝐢, 𝐍 =
𝐧=𝟎
𝐍𝐂𝐅𝐧𝟏 + 𝐢 𝐧
© Jiayu Chen, Ph.D. 43
Compared alternatives using:PW -Present Worth of cash flowsEUAW –Equivalent Uniform Annual Worth of cash flows
Analysis Periods and useful lives of alternatives are important
Interest rate is assumed
ANNUAL WORTH ANALYSIS
So far…
Which is a Good Deal?
1. The Net Present Value of the project is $50,000.
2. The Equivalent Uniform Annual Benefit of the project is
$5,000.
3. The project will produce a 25% of Return.
© Jiayu Chen, Ph.D. 44
ANNUAL WORTH ANALYSIS
1. Rate of return, ROR, is the interest rate earned by an
investment; the i at which you recover your money
2. Internal rate of return, IRR, is the i at which the PW of
the revenue equals the PW of the costs
3. Minimum attractive rate of return, MARR, is the
lowest i investors will accept
4. Incremental rate of return, DROR, is the i earned on
the extra cost of a higher-cost alternative
© Jiayu Chen, Ph.D. 45
ANNUAL WORTH ANALYSIS
1. Rate of return, ROR, is the interest rate earned by an
investment; the i at which you recover your money
2. Internal rate of return, IRR, is the i at which the PW of
the revenue equals the PW of the costs
3. Minimum attractive rate of return, MARR, is the
lowest i investors will accept
4. Incremental rate of return, DROR, is the i earned on
the extra cost of a higher-cost alternative
© Jiayu Chen, Ph.D. 46
INTERNAL RATE OF RETURN
IRR
The interest rate paid on the unpaid balance of a loan such that the
payment schedule makes the unpaid loan balance equal to zero when
the final payment is made.
Repayment Plan
Year A B C D
0 ($5,000.00) ($5,000.00) ($5,000.00) ($5,000.00)
1 $1,400.00 $400.00 $1,252.28 $0.00
2 $1,320.00 $400.00 $1,252.28 $0.00
3 $1,240.00 $400.00 $1,252.28 $0.00
4 $1,160.00 $400.00 $1,252.28 $0.00
5 $1,080.00 $5400.00 $1,252.28 $7,346.64
IRR 8.00% 8.00% 8.00% 8.00%
© Jiayu Chen, Ph.D. 47
INTERNAL RATE OF RETURN
The interest rate earned on the unrecovered investment such that the payment
schedule makes the unrecovered investment equal to zero at the end of the life
of the investment.
Example: What is the IRR if we buy a $5,000 tool with a 5 year useful life that
generates benefits of $1,252 per year?
© Jiayu Chen, Ph.D. 48
INTERNAL RATE OF RETURN
The interest rate earned on the unrecovered investment such that the payment
schedule makes the unrecovered investment equal to zero at the end of the life
of the investment.
Example: What is the IRR if we buy a $5,000 tool with a 5 year useful life that
generates benefits of $1,252 per year?
© Jiayu Chen, Ph.D. 49
INTERNAL RATE OF RETURN
The interest rate earned on the unrecovered investment such that the payment
schedule makes the unrecovered investment equal to zero at the end of the life
of the investment.
Example: What is the IRR if we buy a $5,000 tool with a 5 year useful life that
generates benefits of $1,252 per year?
© Jiayu Chen, Ph.D. 50
• MARR is the lowest i investors will accept
• IRR ≥MARR to proceed
• When two alternatives have equal first costs (or initial investments),
choose the one with higher IRR.
• If alternatives have IRR ≥MARR and different costs, then look at the cost
of each and decide whether the incremental investment is worthwhile
INTERNAL RATE OF RETURN
Minimum Attractive Rate of Return (MARR)
Make sure you understand a positive NPV and MARR
© Jiayu Chen, Ph.D. 51
INTERNAL RATE OF RETURN
• Where two mutually exclusive alternatives are being evaluated, performROR analysis using the incremental rate of return –DROR–on the
difference between the alternatives.
• The convention to calculate DROR is to subtract the lower initial cost
alternative from the higher initial cost alternative
© Jiayu Chen, Ph.D. 52
Option 1: Crane A costs $25,000 and will earn $7,460 per yr for 5 yrs.
Option 2: Crane B costs $60,000 and will earn $16,640 per yr for 5 yrs.
Neither has a salvage value & MARR is 8%, which do you choose?
INTERNAL RATE OF RETURN
© Jiayu Chen, Ph.D. 53
INTERNAL RATE OF RETURN
Is it worth it to invest the extra $35,000 in Crane B, or should we
just stick with Crane A?
Δ cost (A B) = $60,000 - $25,000 = $35,000
Δ benefit (A B) = $16,640 - $7,460 = $9,180 per year
© Jiayu Chen, Ph.D. 54
Benefit-Cost Graph Basics
• Plot (PWCost, PWBenefit) points
• Line where PWBenefit= PWCost represents IRR = MARR
• Area above/left of the NPW = 0 line is favorable
GRAPHICAL ANALYSIS
© Jiayu Chen, Ph.D. 55
Visual Analysis
• Slope of Origin-to-Alternative line = IRR of Alternative
• Slope of Increment line = IRR of Increment
Visually, is the incremental return worth the incremental investment of
purchasing B over A?
GRAPHICAL ANALYSIS
© Jiayu Chen, Ph.D. 61
RATIONING
• Until now all worthwhile projects were funded
– In reality companies have more projects than resources to complete them
• Until now options were interdependent (i.e., we selected one winner
from mutual exclusive alternatives)
– Now we will evaluate independent projects
• How do we ration capital among independent projects?
– Rate of Return or Present Worth methods
© Jiayu Chen, Ph.D. 65
RATIONING
Step 3 if ROR Method:
– Calculate ROR for each project being evaluated
– Sort by decreasing ROR
– Ration Capital to projects up to available budget
© Jiayu Chen, Ph.D. 66
You’re entrepreneurial efforts are raking in the dough. Now you must make
decisions about how you’re going to reinvest your earnings for next year’s
capital expenditures. You’ve arrived at the following lead alternatives for
each project.
If you have $200,000 in your budget, which projects will you select? What is
your Cutoff Rate of Return?
RATIONING Example 2
© Jiayu Chen, Ph.D. 67
You’re entrepreneurial efforts are raking in the dough. Now you must make
decisions about how you’re going to reinvest your earnings for next year’s
capital expenditures. You’ve arrived at the following lead alternatives for
each project.
If you have $200,000 in your budget, which projects will you select? What is
your Cutoff Rate of Return?
RATIONING
© Jiayu Chen, Ph.D. 68
Step 3 if PV Method:
– At times the question the budget committee will ask how to rank the projects
(as opposed to just deciding which projects to fund)
– Divide the Net Present Value by the Present Value of the Costs at the
Minimum Attractive Rate of Return to arrive at a ratio as follows:
NPV PVCost
– Place the projects in order of decreasing NPV PVCost ratio values in order to
determine the projects’ rank order
(highest ratio is highest rank)
RATIONING
© Jiayu Chen, Ph.D. 69
One of your oil rigs just hit the largest oil deposit in the history of Hong Kong
oil exploration! You are considering an expanded set of projects and wish to
rank order them.
Assuming a MARR of 14.5% how would you rank these projects?
RATIONING
© Jiayu Chen, Ph.D. 70
RATIONING
One of your oil rigs just hit the largest oil deposit in the history of Hong Kong
oil exploration! You are considering an expanded set of projects and wish to
rank order them.
Assuming a MARR of 14.5% how would you rank these projects?
© Jiayu Chen, Ph.D. 71
RATIONING
One of your oil rigs just hit the largest oil deposit in the history of Hong Kong
oil exploration! You are considering an expanded set of projects and wish to
rank order them.
Assuming a MARR of 14.5% how would you rank these projects?
© Jiayu Chen, Ph.D. 72
RATIONING
One of your oil rigs just hit the largest oil deposit in the history of Hong Kong
oil exploration! You are considering an expanded set of projects and wish to
rank order them.
Assuming a MARR of 14.5% how would you rank these projects?
© Jiayu Chen, Ph.D. 73
RATIONING
One of your oil rigs just hit the largest oil deposit in the history of Hong Kong
oil exploration! You are considering an expanded set of projects and wish to
rank order them.
Assuming a MARR of 14.5% how would you rank these projects?
© Jiayu Chen, Ph.D. 74
RATIONING
One of your oil rigs just hit the largest oil deposit in the history of Hong Kong
oil exploration! You are considering an expanded set of projects and wish to
rank order them.
Assuming a MARR of 14.5% how would you rank these projects?
© Jiayu Chen, Ph.D. 75
It is a decline in value resulting from…
– Decline in market value of an asset
– Decline in value of an asset to its owner(due to
obsolescence or deterioration)
DEPRECIATION
Depreciation is a systematic allocation of the cost of an asset over its
depreciable life.
Depreciation vs. Expense !
© Jiayu Chen, Ph.D. 77
Total depreciation is spread evenly over the depreciable life.
Straight Line Depreciation
DEPRECIATION APPROACHES
𝐝𝐭 =𝐁 − 𝐒
𝐍
Where N is depreciable life
© Jiayu Chen, Ph.D. 78
DEPRECIATION APPROACHES
Assets lose majority of their value in the first several years of use.
Accelerated Depreciation
𝒅𝒕 = 𝑩 × 𝒓𝒕
Where 𝑟𝑡 is depreciation rate
© Jiayu Chen, Ph.D. 80
Unit of Production Depreciation
DEPRECIATION APPROACHES
Depreciation on some assets is more related to utilization than time.
Construction equipment is a typical example.
𝐝𝐭 =𝐏𝐫𝐨𝐝𝐮𝐜𝐭𝐢𝐨𝐧 𝐟𝐨𝐫 𝐲𝐞𝐚𝐫
𝐓𝐨𝐭𝐚𝐥 𝐥𝐢𝐟𝐞𝐭𝐢𝐦𝐞 𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐢𝐨𝐧 𝐟𝐨𝐫 𝐚𝐬𝐬𝐞𝐭× (𝐁 − 𝐒)