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DSS 5202 Sustainable Systems Analysis

Environmental Life Cycle Costing

POH, Kim Leng, PhD
Associate Professor
Department of Industrial Systems Engineering & Management
National University of Singapore



1 What is Life Cycle Costing (LCC)?............................................................................................ 2
2 Conventional Life Cycle Costing................................................................................................ 3
2.1 What is Conventional Life Cycle Costing .............................................................................. 3
2.2 Key Components of Life Cycle Costing ................................................................................. 4
2.3 Time Value of Money ............................................................................................................. 5
2.4 Cash Flows Analysis Diagrams .............................................................................................. 8
2.5 Equivalent Values of Cash Flows Considering Time Value of Money .................................. 9
2.6 Excel Financial Functions ..................................................................................................... 16
2.7 Python Financial Functions................................................................................................... 17
2.8 Economic Evaluation of Investment Projects ....................................................................... 18
3 Environment Life Cycle Costing (eLCC) ................................................................................. 32
3.1 Extending the System Boundaries in LCC ........................................................................... 32
3.2 Economic Cost vs. Environmental Impacts Trade-off Analysis ........................................... 40
4 Economic Analysis of Clean Energy Investment ..................................................................... 45
4.1 Levelized Cost of Energy...................................................................................................... 45
4.2 Case Study: Investment in Solar PV ..................................................................................... 45






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1 What is Life Cycle Costing (LCC)?

• Life Cycle Costing (LCC), also known as whole-life costing, is a method used to assess the total
economic cost of ownership of a product, asset, or project over its entire lifespan.

• LCC can complement LCA in providing the economic pillar in a full life cycle sustainability
assessment comprising the environmental, economic, and social dimensions.

• LCC can be applied in different ways for a range of purposes to support decision-making in
sustainable development.

• LCC can be conducted as a stand-alone analysis, e.g., for a financial feasibility study or for
budgeting purposes, as well as for comparative analysis, e.g., to compare alternative product or
service systems.

• There are three variants of LCC:

1. Conventional LCC, also known as financial and economic LCC, is the original method
developed for capital investment and budgeting analysis. It assesses costs over the life cycle
of a product or a system, i.e., the Total Cost of Ownership.

2. Environmental LCC is aligned with LCA in terms of system boundaries, functional units,
and methodological steps. It includes cost of environmental impacts and externalities that
are otherwise not included in a conventional LCC.

3. Societal LCC includes monetarization of other externalities, including both environmental
impacts and social impacts.


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2 Conventional Life Cycle Costing
2.1 What is Conventional Life Cycle Costing
• Conventional Life Cycle Costing (LCC), also known as economic LCC or whole-life costing, is
a method used to assess the total economic cost of ownership of a product, asset, or project over
its entire lifespan.

• This approach includes not only the initial acquisition costs but also the costs associated with
operation, maintenance, replacement, and ultimate disposal of a product, asset or project.

• The goal of LCC is to provide a comprehensive financial analysis that helps in making more
informed decisions by considering all costs involved in the life cycle of an asset.

• LCC can be used for the evaluation of design alternatives that satisfy a required level of
performance requirements, but that may have different initial investment costs; different
maintenance & operating (O&M) costs (including energy and water usage); and possibly
different lives.

• Examples (Built Environment)

 LCC can analyze if a proposed design that has a higher initial capital cost but with lower life-
time operating and maintenance costs is economically better or worse off than an alternative
design that has a lower initial capital cost but with higher life-time operating and maintenance
costs?

 LCC can also analyze if an increase in the initial capital cost of a new building or the retrofit
costs in an existing building is justified by reduced energy costs and other cost implications
over the project life or the investor's time horizon?

• Hence, LCC provides a significantly better assessment of the long-term cost effectiveness of a
project than alternative economic methods that focus only on first costs or on operating-related
costs in the short run.

• LCC can also be used in Capital Budgeting to prioritize the allocation of funding to a number of
independent capital investment projects within an organization when insufficient funding is
available to support all of them.

• In general, it can be used for ranking the profitability, liquidity or cost-effectiveness of project
proposals using different criteria.











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2.2 Key Components of Life Cycle Costing
• Cost components or categories that are frequently accounted for in LCC are:

1. Acquisition or Initial Investment Costs

1. These include the purchase price, installation costs, and any other costs incurred to bring the
assets into service. They are also known as capital expenditure costs and are usually one-off
in nature.

2. Operations Costs

These encompass the day-to-day expenses needed to operate the asset, such as energy
consumption, labor, and supplies. These are usually recurring in nature.

3. Maintenance Costs

These are the costs for routine maintenance, repairs, and any unforeseen issues that might
arise during the assets’ life span. These are usually recurring nature.

4. Disposal Costs

At the end of the asset's useful life, there may be costs associated with decommissioning,
disposal, or recycling.

5. Financing Costs

If the asset is financed through loans or other financial instruments, the interest and other
financing charges are included.

6. Replacement Costs

If the asset or parts of it needs to be replaced during its life span, these costs are also
considered.

7. Market Value

This is the value the asset is worth in the market if it is sold or acquired by another
organization.

8. Revenues

These may be actual cash flows or the estimated value of benefits that a project brings during
the assets’ lifetime. These are considered negative costs.


9. Depreciation or Capital Allowance

This is the deduction allowed by tax regulations to reduce the taxable income.

10. Book Value

This is the value of the asset in the accounting book and financial statements after deducting
depreciation according to tax regulations.
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2.3 Time Value of Money
• Consider the following scenarios:

Scenario 1

• You have won in a recent lucky draw a cash prize of $10,000. The prize will be forfeited if it is
not claimed by January 31, next year. Which of the following options would you choose?

1. Claim the $10,000 cash prize now.

2. Claim the $10,000 cash prize one year from now.

• Your Choice: ________ Why? _______________________________________________


Scenario 2

• You have won in a recent lucky draw a cash prize of $10,000. The terms and conditions provided
the following options for you:

1. You will receive $10,000 cash if you claim the prize now.

2. You will receive $10,500 cash if you claim the prize one year from now.

• Your Choice: ________ Why? _______________________________________________


Summary

• The value of money changes with time. It is always better to receive a dollar now than to receive
it later (Scenario 1).

• How the value of money changes with time depends on individuals - it will vary from person to
person, or from company to company.


Representing Time Value of Money

• The value of money over time may be represented by an “interest rate”:



• There are two types of interests:
1. Simple interests
2. Compound interests

Investment
Future
Amount
F
After N time periods
Interest rate per period i
Present
Amount
P
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Simple Interests
• The amount of interest earned is directly proportional to the principal, the number of periods for
which the principal is committed, and the interest rate per period.

• Let P = Principal amount
N = number of periods
i = simple interest rate per period

• Then total interest I = P N i

• Future amount, F = P + I
= P + P N i
= P (1 + N i )


Example

• Suppose you borrow $1,000 for 5 years at a simple interest rate of 10% per year. How much do
you have to return at the end of 5 years?

• Given P = $1,000, i = 0.1, N = 5 years.

• F = $1,000 (1 + 5 × 0.10) = $1,500

• The year-by-year computations are shown below:

(1) (2) = P × 10% (3) = (1)+(2)
Period
(year)
Amount owed at
beginning of year
Interest amount for
the year
Amount owed at
end of year
1 $1,000 $100 $1,100
2 $1,100 $100 $1,200
3 $1,200 $100 $1,300
4 $1,300 $100 $1,400
5 $1,400 $100 $1,500


• Hence the amount to be repaid at the end of 5 years = $1,500.

• Note that the interest payable is constant each year. It is always equal to 10% of the $1,000
principal.

Compound Interests

• In the case of compound interests, the interest for any period is based on the remaining principal
amount plus any accumulated interest up to the beginning of that period.

• An example to illustrate compound interest is given below. We will derive the general formula later.


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Example

• Suppose you borrow $1,000 for 5 years at an interest rate of 10% compounded annually. How
much do you have to return at the end of 5 years?

• The interests accumulated and the amounts owned at the end of each year are given below:

(1) (2) = 0.10×(1) (3)=(1)+(2)
Period
(Year)
Amount owned at
beginning of year
Interest amount for
the year
Amount owned at
end of the year
1 1,000.00 100.00 1,100.00
2 1,100.00 110.00 1,210.00
3 1,210.00 121.00 1,331.00
4 1,331.00 133.10 1,464.10
5 1,464.10 146.41 1,610.51

• The amount to be repaid at the end of 5 years = $1,610.51

• Note that the interest amount for each year is not constant but increases over each year. This is
because it is equal to 10% of the total accumulated amount owned at the beginning of the year.


Comparison of Simple and Compound Interests

• Comparing the value of $1,000 with simple and compound interests at 10% per year for 30 years:






Observations:
1. Simple interest:
The amount grows linearly





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2.4 Cash Flows Analysis Diagrams

Cash Flow Diagrams
• In cash flow analysis, a useful tool for developing and visualizing cash flow models is the cash
flow diagram.

• A Cash Flow Diagram is a graphical representation of all the cash flows that occur at different
time on a time axis.

• An upward arrow in the cash flow diagram normally indicates a cash inflow at a specific point
in time, while a downward arrow indicates a cash outflow.

Example

• A person needs $8,000 now and borrows the amount from a bank at an interest rate of 10% per
year. (All interests are assumed to be compound interests unless otherwise stated).

• Suppose he pays only the interests at the end of each year, and the principal at the end of four
years, then the cash flows over four years may be computed as follows:

(a) (b) = 0.1× (a) (c) (d)=(a)+(b)–(c)
Year Amount owned at beginning of year Interests for the year
Repayment at the
end of year
Amount owned at the
end of year
1 8,000.00 800.00 800.00 8,000.00
2 8,000.00 800.00 800.00 8,000.00
3 8,000.00 800.00 800.00 8,000.00
4 8,000.00 800.00 8,800.00 0


• The cash flow from the borrower’s point of view is as follows:






$8,000
0
1
$800
Year
2
$800
3
$800
4
$8,800
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2.5 Equivalent Values of Cash Flows Considering Time Value of Money
2.5.1 Present Equivalent Value of a Single Future Cash Flow





• Let

F = cash flow at end of period N.
i = constant interest or discount rate per period for N periods”.


• The Present Equivalent Value at time zero of the single cash flow F is

Ni
FP
)1( +
=

• The factor 





+ Ni)1(
1
is called the Single Payment Present Worth Factor



Example

• An investment is to be worth $10,000 in six years.

• If the return on investment 8% per year compounded yearly, how much should be invested today?

• Given F = $10,000, i = 8%, and N = 6 years.

70.301,6$
)630170.0(000,10
)08.01(
1000,10 6
=
=






+
=P


F
0 N-2 N-1 N 1 2 3
P?
dss5202 10

2.5.2 Future Equivalent Value of a Single Future Cash Flow




• Let

P = Cash flow at present time 0.
i = Constant interest or discount rate per period for N periods”.

• The Future Equivalent Value at time N of cash flow P is

F = P (1 + i) N


• The factor (1 + i)N is called the Single Payment Compound Amount Factor.


Example

• Suppose you invest $8,000 in a saving account that earns 10% compound interest per year.

• What is the amount in the account at the end of 4 years?

• Given P = $8,000, i = 10% and N = 4 years.

• The amount accumulated at the end of 4 years is:

F = 8,000 (1 + 0.10)4
= 8,000 (1.4641)
= $ 11,712.80


P
0 N-2 N-1 N 1 2 3
F?
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2.5.3 Present Equivalent Value of a Uniform (End-of-Period) Series Cash Flows





• Let

A = Uniform cash flows at starting at end of period 1 and ending at end of period N.
i = Constant interest or discount rate per period for N periods”.


• The Present Equivalent Value at time 0 of the Uniform Series Cash Flows is







+
−+
= N
N
ii
iAP
)1(
1)1(


• The factor 





+
−+
N
N
ii
i
)1(
1)1( is called the Uniform Series Present Value Worth Factor


Example

• What is the present equivalent value of a series of end-of-year equal incomes valued at $20,000
each for 5 years if the interest rate is 15% per year?



• Given A = $20,000, i = 15%, and N = 5 years.

$67,043.10 =
)15.01(15.0
1)15.01( 5
5






+
−+
= AP


• By substitutions or inverting the above three formulas, we can derive other relationships below.

A
0 N-2 N-1 N 1 2 3
A A A A A
P?
20,000
0 1 2 3 4 5
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2.5.4 Future Equivalent Value of a Uniform Series Cash Flows




• Let

A = Uniform cash flows at starting at end of period 1 and ending at end of period N.
i = Constant interest or discount rate per period for N periods”.


• The Future Equivalent Value at time N of a Uniform Series Cash Flows A is






 −+
=
i
iAF
N 1)1(


• The factor 




 −+
i
i N 1)1( is called the Uniform Series Compound Amount Factor.


Example
• 15 equal deposits of $1,000 each will be made into a bank account paying 5% compound interest
per year, the first deposit being one year from now. What is the balance exactly 15 years from
now?


• Given A = $1,000, i=5%, and N = 15 years.

56.578,21$
05.0
1)05.01( 000,1
15
=





 −+
=F




A
0 N-2 N-1 N 1 2 3
A A A A A
F?
1,000
0 13 14 15 1 2 3
1,000
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2.5.5 Equivalent Uniform End-of-Period Cash Flows of a Present Amount


• Let P = Cash flow at present time 0.
i = Constant interest or discount rate per period for N periods”.

• The Equivalent Uniform End-of-Period Cash Flows of a Present Amount P is:







−+
+
=
1)1(
)1( N
N
i
iiPA

• The factor 





−+
+
1)1(
)1(
N
N
i
ii is called the Capital Recovery Factor.


Example
• Consider a loan of $8,000 to be paid back with 4 equal end-of-year payments?
• What is the yearly repayment amount if the interest rate is 10%?
• Given P = $8,000, i = 10%, and N = 4 years.

$2,523.77
)1.01(1
1.0)000,8(
)1(1

1)1(
)1(
4
=






+−
=






+−
=






−+
+
=
−
−N
N
N
i
iP
i
iiPA

• To verify that the 4 equal annual amounts of $2,533.77 each do indeed repay the loan after 4
years, we can compute the year-by-year balance as follows:

(a) (b) = 0.1× (a) (c) (d)=(a)+(b)–(c)
Year Amount owned at beginning of year
Interests for the
year
Repayment at
the end of year
Amount owned at
the end of year
1 8,000.00 800.00 2523.77 6,276.23
2 6,276.23 627.62 2523.77 4,380.09
3 4,380.09 438.01 2523.77 2,294.33
4 2,294.33 229.43 2523.77 0

A?
0 N-2 N-1 N 1 2 3

P
A?
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2.5.6 Equivalent Uniform End-of-Period Cash Flows of a Future Amount




• Let

F = cash flow at end of period N.
i = constant interest or discount rate per period for N periods”.

• Equivalent Uniform End-of-Period Cash Flows of a Future Amount F is:







−+
=
1)1(
Ni
iFA

• The factor 





−+ 1)1( Ni
i is called the Sinking Fund Factor.


Example
• If you need a lump sum of $1 million at your retirement 45 years from now, how much must you
save per year if the interest rate is 7% per year?

• Given F = $1,000,000, i = 7% and N = 45 years.

57.499,3$
1)07.01(
07.0000,000,1
1)1(
45
=






−+
=






−+
= Ni
iFA






A?
0 N-2 N-1 N 1 2 3

F
A?
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2.5.7 Summary of formulas for discrete cash flows with discrete compounding
• The six formulas and their symbolic notations (in engineering economics) are given below:


Factor Find Given Symbol Formula
Single Payment Compound Amount F P [F/P, i%, N] (1+i)N
Single Payment Present Worth P F [P/F, i%, N] Ni)1(
1
+

Sinking Fund A F [A/F, i%, N] 1)1( −+ Ni
i
Uniform Series Compound Amount F A [F/A, i%, N]
i
i N 1)1( −+
Capital Recovery A P [A/P, i%, N]
1)1(
)1(
−+
+
N
N
i
ii
Uniform Series Present Worth P A [P/A, i%, N] N
N
ii
i
)1(
1)1(
+
−+


Relations between Interest Formulas

• The six formulas involving P, A and F are not independent. Only two independent ones are
needed, and the other four may be derived from these two using the following relations:

1. Reciprocal Rules:
•
]%,,/[
1]%,,/[
NiFP
NiPF =
•
]%,,/[
1]%,,/[
NiAF
NiFA =
•
]%,,/[
1]%,,/[
NiAP
NiPA =

2. Chain Rules:

• [F/A, i%, N] = [F/P, i%, N] × [P/A, i%, N]
• [F/P, i%, N] = [F/A, i%, N] × [A/P, i%, N]

• [P/A, i%, N] = [P/F, i%, N] × [F/A, i%, N]
• [P/F, i%, N] = [P/A, i%, N] × [A/F, i%, N]

• [A/F, i%, N] = [A/P, i%, N] × [P/F, i%, N]
• [A/P, i%, N] = [A/F, i%, N] × [F/P, i%, N]



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2.6 Excel Financial Functions

Excel Functions pv(), pmt(), fv(), rate(), nper()

• The five Excel functions pv(), pmt(), fv(), rate(), and nper() compute the value of one variable,
given the values of the other four variables so that the equivalent value of the cash flows is
zero.



Inputs (up to any four):
rate = interest rate per period
nper = total number of periods
pv = a single cash flow at time zero
pmt = uniform cash flow per period
fv = a single cash flow at end of period nper periods

Parameters:



=
flowscash period of beginning uniform1
(default) flowscash periods of end uniform0
type

Outputs for uniform end-of-period cash flows:

• PV (rate, nper, pmt, fv [,0] )

returns – ( pmt [P/A, rate, nper] + fv [P/F, rate, nper] )

• PMT (rate, nper, pv, fv [,0] )

returns – ( pv [A/P, rate, nper] + fv [A/F, rate, nper] )

• FV(rate, nper, pmt, pv [,0] )

returns – ( pv [F/P, rate, nper] + pmt [F/A, rate, nper] )


• RATE(nper, pmt, pv, fv , 0 , guess)

returns r such that pv + pmt [P/A, r, nper] + fv [P/F, r, nper] = 0

• NPER(rate, pmt, pv, fv)

returns N such that pv + pmt [P/A, rate, N] + fv [P/F, rate, N] = 0


• Take note of the negative sign in front of the quantity returned by PV, PMT and FV.
pv
0 1
fv
2 3 4 nper-2 nper-1 nper
pmt pmt
type=0 pv
0 1
fv
2 3 4 nper-2 nper-1 nper
pmt pmt
type=1
dss5202 17



2.7 Python Financial Functions
Functions pv(), pmt(), fv(), rate() and nper()

• The five numpy_financial functions pv(), pmt(), fv(), rate() and nper() are similar to their
corresponding Excel functions with the same names. They also compute the value of one
variable, given the values of the other 4 variables so that the overall equivalent value of the cash
flows is zero.



Inputs (up to any four of the following):
rate = interest rate per period
nper = total number of periods
pv = a single cash flow at time zero
pmt = uniform cash flow per period
fv = a single cash flow at end of nper periods

Parameter:



=
flowscash period of beginning uniform1
(default) flowscash periods of end uniform0
when

Uniform end-of-period cash flows

import numpy_finanical as npf

• npf.pv (rate, nper, pmt, fv)

returns – ( pmt [P/A, rate, nper] + fv [P/F, rate, nper] )

• npf.pmt (rate, nper, pv, fv)

returns – ( pv [A/P, rate, nper] + fv [A/F, rate, nper] )

• npf.fv (rate, nper, pmt, pv)

returns – ( pv [F/P, rate, nper] + pmt [F/A, rate, nper] )

• npf.rate (nper, pmt, pv, fv, guess, tolerance)

returns r such that pv + pmt [P/A, r, nper] + fv [P/F, r, nper] = 0

• npf.nper (rate, pmt, pv, fv)

returns N such that pv + pmt [P/A, rate, N] + fv [P/F, rate, N] = 0

pv
0 1
fv
2 3 4 nper-2 nper-1 nper
pmt pmt
when=0 pv
0 1
fv
2 3 4 nper-2 nper-1 nper
pmt pmt
when=1
dss5202 18

• Take note of the negative signs in front of the quantity returned by PV, PMT and FV.

2.8 Economic Evaluation of Investment Projects
2.8.1 Methods for Economic Evaluation of Project
Methods for Economic Analysis:

• Discounted Cash Flow methods
o Net present value (NPV) method

• Rate of Return methods
o Internal rate of return (IRR)
o Modified rate of return (MIRR)

• Discounted payback period method

• Benefit-cost ratio methods



Interest or Discounting Rate project evaluation

• Minimum Attractive Rate of Return (MARR)

 MARR is the minimum rate return required by the investors or stakeholders for the project to
be acceptable.


• Weighted Average Cost of Capital (WACC)

 Capital used in a project may be a combination of various sources including both equity
capitals or debt capitals in different proportions.

 WACC is computed based on weighted average of the cost of these capitals. The weights are
usually based on the proportion of capital contributed by the sources.



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2.8.2 Discounted Cash Flow Methods
• Consider a proposed investment project with the following cash flows:



where

Fk = Cash flow at end of period k, for k =0, 1, 2, .., N.
N = Project life.



• The Net Present Value (NPV) of a Project at MARR = i is

∑
= +
=
+
++
+
+
+
+=
N
k
k
k
N
N
i
F
i
F
i
F
i
FFiNPV
0
2
21
0
)1(
)1()1()1(
%)( 



Investment Decision Rule for NPV:

• Given a project with the cash flows:


where N = Project life and Fk = Cash flow at end of period k, for k =0, 1, 2, .., N.


If NPV(MARR) ≥ 0 then

The project is financially feasible at the given MARR

Else

The project is not financially feasible at the given MARR.



F1
0 1
F0 F2
2
FN
N …
F3
3
FN-1
N-1
F1
0 1
F0 F2
2
FN
N …
F3
3
FN-1
N-1
dss5202 20

Example (ABC Company Problem)

• ABC Company is considering investing in an air pollution control system that will reduce its
carbon footprint. The proposed equipment costs $25,000 and has a salvage value of $5,000 at the
end of its 5-year useful life. Annual net benefits from this investment is expected to be $8,000
for 5 years. Determine if the investment is financially feasible if the MARR is 20%. Assume a
study period of 5 years.
• The data may be summarized as follows:

 Study period 5 years
 Initial investment cost $25,000
 Annual benefits $ 8,000
 Salvage value at end of 5 years $ 5,000
 MARR 20%


• Cash flow diagram:




• 5432 )2.01(
000,5000,8
)2.01(
000,8
)2.01(
000,8
)2.01(
000,8
)2.01(
000,8000,25%)20(
+
+
+
+
+
+
+
+
+
+
+−=NPV
= $ 934.28 > 0


• The proposed project is financially feasible at MARR = 20%.

Using Excel Functions:
1. PV
2. NPV

Using Python numpy_finanical Functions:
1. npf.pv
2. npf.npv


0
1
25,000
8,000
2 3 4 5
5,000
8,000 8,000 8,000
8,000
+
dss5202 21

Example (XYZ Company Problem)

• XYZ Company is considering whether to invest in a machine to manufacture a new product.
The machine has an initial cost of $12,000. Revenue from sales of the product is expected to be
$5,310 per year. Annual operating and maintenance expense for the machine is expected to be
$3,000. The product life is expected to be 5 years, at which time the machine can be sold for
$2,000 (salvage value). Is the project financially feasible if the company’s MARR is 10%?

• Summary of data:
 Study period 5 years
 Initial investment cost $12,000
 Annual revenue or income $ 5,310
 Annual expenses $ 3,000
 Annual profit $ 5,310 – $3,000 = $2,310
 Salvage value at end of 5 years $ 2,000
 MARR = 10%

 Cash flow diagram:




• 5432 )1.01(
000,2310,2
)1.01(
310,2
)1.01(
310,2
)1.01(
310,2
)1.01(
310,2000,12%)10(
+
+
+
+
+
+
+
+
+
+
+−=NPV
= – $ 2,001.44 < 0


• The project is not financially feasible at MARR = 10%.



Using Excel Functions:
• PV
• NPV

Using Python numpy_finanical Functions:
• npf.pv
• npf.npv

0
1
12,000
2,310
2 3 4 5
2,000
dss5202 22

The Capitalized Worth Formula

• Consider an infinite series of uniform cash flows starting from end of period 1 through infinity:



• What is the Present Equivalent value of this infinite series of cash flows at reference point time
zero if the MARR = i? Is it infinity?

• The present equivalent value at time zero is

i
A
i
A
ii
iA
i A,[PAiPV
Ni
N
N
N
N
=







 −
=






+
−+
=
∞
+
∞→
∞→
)1(
11
lim
)1(
1)1(lim
] %,/ = %)(


• The present equivalent value at time=0 of an infinite uniform series of cash flows A starting at
end of period 1 with time value of money = i% is known as the Capitalized Worth (CW(i%))
where
i
AiCW =%)(

Example (Condo rental)

• John is considering purchasing in cash, a free-hold property costing $1 million. If he intends to
rent it out perpetually, what minimum equal monthly rental income is required to justify the
investment, if the MARR is 6% per year compounded monthly? Ignore property taxes and other
unstated costs.

• i = 6% per year compounded monthly = 6%/12 or 0.5% per month.

• Assume that the property has perpetual life.

• Let A = monthly rental income

• Using the CW formula:

005.0
000,000,1 A=

⇒ A = $ 5,000 per month.



A
0 1 ∞
…
A
2
A
3
A A A A A
dss5202 23

2.8.3 The Internal Rate of Return Method (IRR)
• Consider an investment project with the following cash flows:



where
Fk = cash flow at end of period k, for k =0, 1, 2, .., N.
N = life of the project.

• The Internal Rate of Return (IRR) of a project is defined as the interest rate at which the NPV
of the project is zero.

• Mathematically, IRR is the value of i such that 0
)1(0
=
+∑=
N
k
k
k
i
F




Investment Decision Rule for IRR

• Given a project with the cash flows:


where N = Project life and Fk = Cash flow at end of period k, for k =0, 1, 2, .., N.


• Let the Internal Rate of Return of the cash flows = IRR.


If IRR ≥ MARR then

The project is financially feasible at this MARR
Else
The project is not financially feasible at this MARR




F1
0 1
F0 F2
2
FN
N …
F3
3
FN-1
N-1
F1
0 1
F0 F2
2
FN
N …
F3
3
FN-1
N-1
dss5202 24

Example (ABC Company Problem)

• Summary of data:

 Study period 5 years
 Initial investment cost $25,000
 Annual benefits $8,000
 Salvage value at end of 5 years $5,000
 MARR 20%


• Cash flow diagram:


• To find the IRR, we solve the equation:

0
)r1(
000,5000,8
)r1(
000,8
)r1(
000,8
)r1(
000,8
)r1(
000,8000,25)r(NPV 5432 =+
+
+
+
+
+
+
+
+
+
+−=


• An equation solver like Excel Goal Seek or Python Root function can be used to find the root of
the equation.


Solution: IRR = 21.578% > 20% = MARR


• Hence the project is financially infeasible at MARR=20%.

Using Excel Functions:
• RATE
• IRR
• GoalSeek


Using Python numpy_finanical Functions:
• npf.rate
• npf.irr

0
1
25,000
8,000
2 3 4 5
5,000
8,000 8,000 8,000
8,000
+
dss5202 25

Example (XYZ Company Problem)

• Summary of data:
 Study period 5 years
 Initial investment cost $12,000
 Annual profit $ 2,310
 Salvage value at end of 5 years $ 2,000
 MARR = 10%

• Cash flow diagram



• To find the IRR, we solve the equation:

0
)1(
000,2310,2
)1(
310,2
)1(
310,2
)1(
310,2
)1(
310,2000,12)( 5432 =+
+
+
+
+
+
+
+
+
+
+−=
rrrrr
rNPV


• An equation solver like Excel Goal Seek or Python Root function can be used to find the root of
the equation.


Solution: IRR = 3.8040% < 10% = MARR


• Hence the project is not financially infeasible.



Using Excel Functions:
• RATE
• IRR
• GoalSeek


Using Python numpy_finanical package:
• npf.rate
• npf.irr


0
1
12,000
2,310
2 3 4 5
2,000
dss5202 26

Limitation of the IRR Method

1. Reinvestment and Financing Rates Assumption

• Discounted cash flows methods such as the NPV method assume that the cash flows are either
reinvested or financed at the MARR during the study period.

• IRR method on the other hand assumes that cash flows are either reinvested or financed at a rate
equal to the project’s IRR and not at the MARR.

• The assumption that reinvestment rate = IRR is hard to realize in practice especially for a project
with a very high IRR as it may not be possible to reinvest cash flows at other projects with
similarly high IRR.



2. Valid only when the projects NPV(i) is a decreasing function of i.

• When NPV(i) is a decreasing function, IRR agrees with NPV in accepting or rejecting a project.

• When NPV(i) is an increasing function, IRR disagrees with NPV in accepting or rejecting a
project.


(a) When NPV is a decreasing function (b) When NPV is an increasing function



NPV(MARR) > 0 ⇔ IRR > MARR NPV(MARR) < 0 but IRR > MARR



3. Possible Multiple IRR Solutions

• The solution to the equation 0
)'1(0
=
+∑=
N
k
k
k
i
F may be non-unique, i.e., there may be multiple
solutions.

• When this happens, use the NPV method to resolve.


i IRR
NPV(i)
0
NPV(MARR)
MARR
i IRR
NPV(i)
0
NPV(MARR)
MARR
dss5202 27

Example: Multiple IRR solutions

• Consider a project with the following cash flows over 5 years.

k 0 1 2 3 4 5
Fk 500 -1,000 0 250 250 250


• We compute NPV(i) for 0% ≤ i ≤ 100%:


i'% NPV(i'%)
0 $250.00
5 $165.14
10 $104.72
15 $62.05
20 $32.38
25 $12.32
30 -$0.58
35 -$8.11
40 -$11.62
45 -$12.09
50 -$10.29
55 -$6.77
60 -$1.98
65 $3.76
70 $10.19
75 $17.11
80 $24.36
85 $31.82
90 $39.41
95 $47.05
100 $54.69



• Two values of IRR are obtained at ≈ 30% and 62%.


• In practice, to find all possible roots using Excel or Python, use a number of initial guesses at
different start points and let the solver search for a solution from there.


-$50
$0
$50
$100
$150
$200
$250
$300
0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%
i%
PW
(i%
)
dss5202 28

Using Python root function to find multiple roots

# -*- coding: utf-8 -*-
# multiple_IRR_example.py
import numpy as np
import numpy_financial as npf
from scipy.optimize import root
import matplotlib.pyplot as plt

CF = [500, -1000, 0, 250, 250, 250]

# Define the function whose root we wish to find
NPV = lambda r: npf.npv(r, CF)

# Plot NPV(r) vs r
fig, ax = plt.subplots()
r = np.linspace(0, 1, 101)
ax.plot(r, [NPV(y) for y in r], 'r', lw=2)
ax.set_xlabel("Rate")
ax.set_xticks(np.linspace(0,1,11))
ax.set_ylabel("NPV")
ax.set_xlim(0,1)
ax.grid()
plt.show()

guess=0.2
s1 = root(NPV, x0=guess, options={'xtol': 1E-10})
print(f"guess = {guess}, IRR = {s1.x[0]:.6f}")

guess=0.7
s2 = root(NPV, x0=guess, options={'xtol': 1E-10})
print(f"guess = {guess}, IRR = {s2.x[0]:.6f}")

dss5202 29

2.8.4 Modified Internal Rate of Return Method (MIRR)

• The IRR method assumes reinvestment of cash flows at IRR which is highly unrealistic,
especially when IRR is high or when IRR >> MARR.

• Usually, the reinvestment rate is just around MARR.

• The Modified Internal Rate of Return (MIRR) fixes the problem by explicitly stating two more
interest rates:

1. Financing rate if.

 This is the cost of capital for the fund needed for the project.
 This applies to all negative annual net cash flows.
 The weighted average cost of capital (WACC) can be used for this rate.
 We also expect MARR ≥ WACC.


2. Reinvestment rate ir.

 This is the rate at which cash inflows are reinvested in other projects.
 This applies to all positive annual net cash flows.
 We also expect ir ≥ MARR.


Procedure for Computing MIRR

• Let N = project life or study period
if = financing rate
ir = reinvestment rate


1. For each year k = 0 to N, determine the net cash flows Fk :
• If Fk ≥ 0, it is a net cash inflow in year k.
• If Fk < 0 , it is a net cash outflow in year k.


2. Compute the absolute value of Present Value (PV) of all the -ve net cash flows at the financing
rate if:
∑
<
= +
=
N
0F
0k
k
f
k
f
k
)i1(
|F|)iCF,net ve-(PV


3. Compute the Future Value (FV) of all the +ve net cash flows at the reinvestment rate ii:
kN
r
N
0F
0k
kr )i1(F )iCF,net ve(FV
k
−
>
=
∑ +=+

dss5202 30



4. The Modified Internal Rate of Return (MIRR) is defined as the interest rate that establishes
equivalence between |PV(-ve net CF, if)| at time zero and FV(+ve net CF, ir) at time N.

kNr
N
0F
0k
kr )i1(F )iCF,net ve(FV
k
−
>
=
∑ +=+

∑
<
= +
=
N
0F
0k
k
f
k
f
k
)i1(
|F|)iCF,net ve-(PV



Hence )MIRR1( )iCF,net ve-(PV )iCF,net ve(FV Nfr +=+

Therefore 1
)iCF,net ve(PV
)iCF,net ve(FVMIRR N
f
r −
−
+
=



Investment Decision Rule for MIRR

• Given a project with the cash flows:


where N = Project life and Fk = Cash flow at end of period k, for k =0, 1, 2, .., N.

• Let MIRR = Modified internal rate of return at given financing rate and reinvestment rate.

If MIRR ≥ MARR then
The project is financially feasible.
Else
The project is not financially feasible.


0
MIRR=?
N
4
2
3
F1
0 1
F0 F2
2
FN
N …
F3
3
FN-1
N-1
dss5202 31

Example: MIRR Analysis of the ABC Company Problem

• Summary of data:
 Study period 5 years
 Initial investment cost $25,000
 Annual benefits $ 8,000
 Salvage value at end of 5 years $ 5,000
 MARR 20%

• Assume
 Financing rate if = 15%
 Reinvestment rate ir =20%

• Cash flow diagram:


• |PV(-ve net cash outflows, if )| = 25,000.00

• FV(+ve net cash flows, ir) = 8,000 [F/A, 20%,5] + 5,000
= 8,000 (7.4416) + 5,000
= 64,532.80
• 2088.01
00.000,25
80.532,64
5 =−=MIRR

• MIRR = 20.88% > MARR = 20% ⇒ Project is financially feasible.



Using Excel Functions:
• MIRR


Using Python numpy_finanical Functions:
• npf.mirr

0
1
25,000
8,000
2 3 4 5
5,000
8,000 8,000 8,000
8,000
+
dss5202 32

3 Environment Life Cycle Costing (eLCC)
3.1 Extending the System Boundaries in LCC
• The main difference between environmental life cycle costing and conventional (economic) life
cycle costing lies in the scope and inclusion of environmental impacts and externalities.

• Conventional (Economic) LCC considers only the direct internal costs of a product borne by
the producer, consumer, and other stakeholders. These include costs that an actor (i.e., a
manufacturer, transporter, consumer, or other directly involved stakeholder) is paying for the
production, use, or end-of-life expenses.
• More specifically, conventional LCC focuses on the direct financial costs throughout the life
cycle of an asset.

1. Acquisition costs
2. Installation costs
3. Operating costs
4. Maintenance costs
5. Replacement costs
6. Disposal costs

• Environmental LCC extends conventional (economic) LCC with the monetary valuation of
environmental impacts across a product’s entire life cycle, i.e. from raw material extraction to
disposal or recycling.

• More specifically, environmental LCC expands the scope by including environmental
externalities which are the costs of environmental and social impacts (e.g., greenhouse gas
emissions, air pollution, water contamination, biodiversity loss) expressed in monetary terms.

• These costs are beyond the economic consideration of those involved in the system and are
normally not directly borne by any actor in the product life cycle.

• Environmental costs are often calculated using environmental impact assessment results (from
LCA) and damage cost factors or carbon prices.


• Hence an environmental LCC can be conducted as an extension to a conventional LCC by
incorporating environmental costs, which can include, for example:

1. Costs of environmental emissions
2. Costs of resource depletion (e.g., water, minerals)
3. Costs of pollution control and mitigation
4. Costs associated with environmental regulations and compliance
5. Costs of waste management and recycling



dss5202 33

Steps in Performing an Environmental Life Cycle Costing for a Product








Define Goal, Scope,
System Boundary and
impact categories
Collect data
Perform Life Cycle
Inventory (LCI)
Analysis
Perform Life Cycle
Costing (LCC)
Analysis
Perform Life Cycle
Impact Analysis
(LCIA)
Monetize
environmental
impacts
Identify areas of
environmental
concern (hot spots)
Combine economic
and environmental
costs
Interpret results and
perform sensitivity
analysis
Identify both economic
and environmental costs
to be included
Start
End
dss5202 34


Key Differences: Economic LCC vs. Environmental LCC

Aspect Economic LCC Environmental LCC
Scope of costs Direct financial costs (capital, O&M, replacement, disposal, etc.)
Direct financial costs plus
monetized environmental costs
(externalities)
Data source Project budgets, accounting records, market prices
Combination of economic data and
environmental life cycle
inventory/impact assessment data
Purpose Minimize total ownership cost Minimize total ownership + environmental damage cost
Stakeholder
perspective
Usually single stakeholder (owner,
operator)
Broader perspective including
environmental impacts
Output Net present (NPC) or equivalent uniform annual cost
NPC + Net present environmental
cost







dss5202 35


Case Study (Washing Machine)

Goal & Scope

• To conduct environmental life cycle costing for a typical washing machine.


Functional Unit

• A side-loading washing machine for a family of three providing annually 120 wash cycles for
10 years.

System boundary

• Cradle to Grave


Discounting interest rate

• 3% per year.


Study period

• 10 years


Mid point Impact Categories

1. Global Warming Potential (GWP)
2. Acidification Potential (AP)
3. Eutrophication Potential (EP)
4. Photochemical Ozone Creation Potential (POCP)
5. Water Usage


Economic Life Cycle Cost Analysis

Initial Cost

• Purchase price = $1,000
• Present equivalent cost = $1,000


Electricity Consumption

• Energy consumption = 2.3 kWh per cycle
• Cost of electricity = $0.30 per kWh
• Annual number of wash cycles = 120

dss5202 36


• Annual electricity consumption = 2.3 × 120 = 276 kWh
• Annual electricity cost = 0.3 × 276 = $82.8
• Present equivalent cost = 82.8 [P/A, 3%, 10] = $ 706.30


Water Consumption

• Water consumption per cycle = 70 liters
• Cost of water = $ 0.003 per liter
• Annual water consumption = 120 ×70 = 8,400 liters
• Annual water consumption cost = 0.003 × 8,400 = $ 25.20
• Present equivalent cost = 25.20 [P/A, 3%, 10] = $ 214.96

Maintenance

• Annual maintenance cost = $ 50
• Present equivalent cost = 50 [P/A, 3%, 10] = $ 426.51

End-of-life

• Disposal cost = $ 30
• Present equivalent cost = 30 [P/F, 3%, 10] = $ 22.32


Present equivalent life cycle cost

= 1,000 + (82.8 + 25.2 + 50) [P/A, 3%, 10] + 30 [P/F, 5%, 10]
= $ 2,370.09

Equivalent uniform annual cost (EUAC)

= $ 2,370.09 [A/P, 3%, 10]
= $ 277.85



Summary of Costs














Phase of Life Cycle Cost ($) Present equivalent Cost ($)
1 Initial investment 1,000 1,000
2 Annual electricity consumption 82.8 706.30
3 Annual water consumption 25.2 214.96
4 Annual maintenance 50 426.51
5 End of life disposal 30 22.32
Total 2,370.09
dss5202 37






• We note that the capital expenditure cost is the highest, followed by the electricity consumption
cost.

• The end-of-life disposal cost is not very significant.



Life Cycle Impact Analysis for Global Warming

• LCIA results for Global Warming are given below:




















Life cycle phase GWP impact
1 Manufacturing 400 kg CO2 eq
2 Transportation 804 kg CO2 eq
3 Use Phase (Electricity) 300 kg CO2 eq
4 Use Phase (Water) 15 kg CO2 eq
5 End-of-Life 4.6 kg CO2 eq
Total life cycle 1,523.6 CO2 eq
dss5202 38

• Comparing the GWP with the Economic Cost for each phase of the life cycle
















• We note that although present equivalent cost of transportation cost, it contributes the most to
the GWP.



Life cycle phase GWP impact Present equivalent Cost ($)
1 Manufacturing 400 kg CO2 eq 1,000
2 Transportation 804 kg CO2 eq 706.30
3 Use Phase (Electricity) 300 kg CO2 eq 214.96
4 Use Phase (Water) 15 kg CO2 eq 426.51
5 End-of-Life 4.6 kg CO2 eq 22.32
Total life cycle 1,523.6 kg CO2 eq 2,370.09
dss5202 39


Life Cycle Impact Analysis for other mid-point impact categories

Midpoint impact category Impact score Unit
1 Global Warming Potential (GWP) 1,523.6 kg CO2 eq
2 Acidification Potential (AP) 30 kg SO2 eq
3 Eutrophication Potential (EP) 6 kg PO₄³⁻eq
4 photochemical Ozone Creation Potential (POCP) 3 kg C2H4 eq

Monetization of environmental impact
• The eco-costs based on marginal prevention costs, meaning they represent the cost of avoiding
the damage in the first place using best-available technologies, are given in the table below.







Computation of Total Life Cycle Environmental Costs (undiscounted)
 GWP: 1,523.6 × 0.163 = $ 248.35

 AP: 30 × 8.42 = $ 252.60

 EP: 6 ×5.92 = $ 35.52

 POCP 3 × 6.73 = $ 20.19
Total Environmental Life Cycle Cost = $ 556.66

• Note that we should not add directly add the undiscounted life cycle environmental costs to the
present equivalent economic costs

• Undiscounted total economic costs = 1,000 + (82.8 + 25.2 + 50) × 10 + 30
= $ 2,610.00
• Total (undiscounted) Economic + Environmental cost = 2,610.00 + 556.66
= $ 3,166.66
Midpoint Impact Category Eco Cost
1 Global Warming Potential (GWP100) $ 0.163 per kg CO₂ eq
2 Acidification Potential (AP) $ 8.42 per kg SO₂ eq
3 Eutrophication Potential (EP) $ 5.92 per kg PO₄³⁻ eq
4 photochemical Ozone Creation Potential (POCP) $ 6.73 per kg C₂H₄ eq
dss5202 40

3.2 Economic Cost vs. Environmental Impacts Trade-off Analysis
• In the previous example, we assume that it is possible to monetize the environmental impacts or
damages as costs and add the dollar values to the economic costs.

• When the environmental impacts or damages cannot be monetized reliably or it is not realistic to
do so, we can do economic cost vs environmental impact trade-off analysis and gain some insight
into the problem.

Two-Attribute Dominance Analysis
• Consider a problem with 2 attributes x1 and x2 whose values we would like to minimize.

• For example, x1 could be life cycle economic cost in dollars and x2 could be a specific
environmental impact measured on some midpoint or endpoint scale.

• Suppose it is not possible to monetize the value of x2 and add it to x1.

• Consider two alternative product systems A and B with values for x1 and x2 are (a1, a2) and
(b1, b2) respectively.


a1 < b1 and a2 < b2 a1 = b1 and a2 < b2 a1 < b1 and a2 = b2
A strongly dominates B A weakly dominates B A weakly dominates B



a1 > b1 and a2 < b2
No dominance




x1
x 2

0
A (a1, a2)
B (b1, b2)
x1
x 2

0
A (a1, a2)
B (b1, b2)
x1
x 2

0
A (a1, a2)
B (b1, b2)
x1
x 2

0
A (a1, a2)
B (b1, b2)
dss5202 41

Definition: Multiple Attribute Dominance (n-attribute)
• Consider a n-attribute problem with outcome vector x = (x1, x2, …, xn).

• Without loss of generality, we assume that we prefer less to more in all the n attributes.

• Let alternatives A and B be two alternatives with attribute vector x = (a1, a2, …, an) and (b1, b2, …,
bn), respectively. Then

A Dominates B if and only if

ai ≤ bi ∀ i ∈ {1, 2, …, n} and
ai < bi ∃ i ∈ {1, 2, …, n}

• That is, A dominates B if and only if A is not worse off than B in all attributes, and is strictly
better in at least one attribute.

• The strict inequality ensures that the attribute vectors of A and B are not identical.


Strong Dominance
• If A dominates B and all the attribute values of A are strictly better than those of B, we say that
A strongly dominates B.

Weak Dominance
• If A dominates B and at least one attribute value of A is equal to that of B, we say that A weakly
dominates B.


Efficient, Non-Dominated or Pareto-Optimal Solutions
• Let S ={A1, A2, … Am} be a set of feasible alternatives.

• An alternative Aj ∈ S is Efficient, Non-dominated or Pareto-optimal with respect to S and the
n attributes if and only if there does not exist another alternative Ak ∈ S that dominates Aj.

Non-Efficient or Dominated Alternatives can be Eliminated from further consideration
• Given a set of alternatives, a decision maker should not choose a dominated one from the set as
he/she can always be better off by choosing another alternative from the set that performs better
in at least in one other attribute without degrading any other attributes.





dss5202 42

Efficient Frontier Analysis
• Consider a 2-attribute problem with alternatives A to F, and we prefer less to more for all
attributes.

• Let the 2-attribute values be denoted by x1 and x2 and are plotted as follows:


• Observations:

 D is dominated by C and B
 E is dominated by A and B
 F is weakly dominated by A

• Removing D, E and F, the efficient or Pareto-optimal set = {A, B, C}.

• Alternatives D, E, and F can be eliminated from further consideration.

• If we connect up all the efficient solutions in the 2-attribute space, we obtain the Efficient
Frontier:

The Efficient Frontier


• All alternatives that lie in the interior of the efficient frontier are dominated by at least one other
alternative on the efficient frontier.

• Multiple attribute dominance analysis allows us to eliminate inefficient alternatives from further
consideration.


x1
x2
A
B
0
C
D
E
F
x1
x2
A
B
0
C
D
E
F
dss5202 43

Example (3 Washing Machines)

Goal and Scope

• Compare the economic life cycle cost of owning three different types of washing machines with
their respective life cycle environmental impacts.

Functional unit

• 1840 washing cycles for a household of 3 over 10 years.

Product Alternatives

1. Inexpensive machine, average washing behavior.
2. Average price machine, optimized washing behavior.
3. Expensive machine, highly optimized washing behavior.

Impact Category

• CML-IA, GWP100 (kg CO2 eq)


Life Cycle Inventory (LCI) Analysis


Product GWP100 (kg CO2 eq)
1 Inexpensive machine 2,100
2 Average machine 1,800
3 Expensive machine 1,500


Life Cycle Costing Analysis

Product Normalized Net Present Cost
1 Inexpensive machine 1,000
2 Average machine 1,200
3 Expensive machine 1,800




dss5202 44

Efficient Frontier Analysis





• We see a clear trade-off between economic cost and environmental impact (GWP) by the
ownership of the three types of washing machines.

• The solid line segments joining the three points on the plot is the efficient frontier.




dss5202 45

4 Economic Analysis of Clean Energy Investment
4.1 Levelized Cost of Energy
• The Levelized Cost of Energy (LCOE) is a popular measure for assessing the economic
feasibility as well as the overall competitiveness of investing in renewable energy systems and
technologies, such as for examples solar PV, wind farms, hydroelectric stations, geothermal
energy, etc.

• LCOE is defined as the minimum revenue per unit of electricity generated that would be required
to recover the costs of investing, building, and operating an energy-generating plant during an
assumed system useful life or study period.

• Intuitively, LCOE is the minimum equivalent constant price at which electricity must be sold in
the energy market for the investment to financially break even over the lifetime of the project.

4.2 Case Study: Investment in Solar PV
Problem Description
• A company is planning to invest in a solar PV system on the rooftop of a shopping mall that it is
developing.

• The system will require an initial cost of $288,000. It will have a useful life of 20 years and a
salvage value of $2,000 at the end of life.

• Annual operating and maintenance costs are estimated to be $1,500.


Energy output and system efficiency

• Based on Singapore's geographical location, orientation of the building, and weather conditions
throughout the year, the PV panels can produce 250,000 kWh of electricity in the first year of
operation.

• However, this output will degrade at a rate of 1% per year through its 20-year useful life. The
electricity output from the PV panel is DC and needs to be converted to AC via an inverter at an
efficiency of 95%.

• The electricity produced can be sold at $0.18 per kWh to either the shopping mall’s tenants or to
the power grid.


Project financing and cost of capital

• The initial investment cost of the project will be fully funded by the company’s equity capital.

• The company’s WACC is 10% for the purpose of evaluating the financial feasibility of this project.


Summary of Data
dss5202 46


Solar PV System Base Value
1 Useful life (years) 20 years
2 Initial cost $288000
3 Salvage value $2,000
4 Annual O&M cost $1,500
5 Energy output in year 1 (kwh) 250,000
6 System efficiency 95.00%
7 Annual degradation rate 1.00%
8 Electricity market price per kwh $0.180


Key Questions to be answered

1. Is the project financially feasible based on the base values of the data provided?

2. What is the levelized cost of energy produced for this project?

3. If the data are subject to uncertainty, which factors have the most significant impact on the
financial feasibility of the project?

4. What is the financial risk associated with this project?



Life Cycle Cash Flow Model
• Parameters:

I = Initial investment cost.
N = Study period.
Ek = Operations & maintenance cost in year k, for k = 1, 2, …, N.
i = WACC or MARR
rk = Annual PV panel degradation rate in year k, for k = 1, 2, …, N-1.
ηk = System efficiency in year k, for k = 1, 2, …, N.
pk = Price of electricity in the energy market in year k, for k = 1, 2, …, N.
SVN = Salvage value at end of year N.

• Annual electricity energy produced:

Let S1 = Total electricity produced by PV in year 1.

Due to annual degradation of the PV panels, total electricity produced in year k is

Sk = Sk-1 (1 – rk-1) for k = 2, …, N. (1)

dss5202 47

• Annual total electricity sold in year k is

Qk = Sk ηk for k = 1, 2, …, N. (2)

A more convenient recursive formula to compute Qk in Excel is




=η−
=η
=
−− NkrS
kS
Q
kkk
k ,...,2for )1(
1for
11
11 (3)

• Annual revenue from the sale of electricity in year k is
Rk = Qk pk for k = 1, 2, …, N. (4)


• NPV of the project is






+
+





+
−
+−=






+
+





+
−
+−=
∑
∑
=
=
N
N
N
k
k
kkk
N
N
N
k
k
kk
i
SV
i
EpQI
i
SV
i
ERIiNPV
)1()1(
)1()1(
%)(
1
1 (5)
Project Financial Feasibility
• Using the data given, we obtain
NPV(10%) = $ 40,914.20 > 0
IRR = 12.11% > 10%
• Hence, the project is financially feasible.

Levelized Cost of Energy
• Let pk = p for all year k.

• The LCOE is the value of p that makes project NPV = 0.
∑
∑
=
=
+
+
−





+
+
= N
k
k
k
N
N
N
k
k
k
i
Q
i
SV
i
EI
1
1
)1(
)1()1(
LCOE (6)
• Hence can be interpreted as:
sold (kwh)Energy of NPV
Project ofCost Cycle Life EquivalentPresent LCOE = (7)
• Using the data given:
LCOE = $ 0.15843 < $ 0.18

dss5202 48

Sensitivity Analysis
• Several factors in the project are subject to uncertainty.
• The company experts assessed their possible low and low values as follows:
Variables Low Value Base Value High Value
1 Initial cash flow -$347,000 -$288,000 -$229,000
2 Salvage value $1,000 $2,000 $3,000
3 Annual O&M cash flow -$2,000 -$1,500 -$1,000
4 Energy output in year 1 (kwh) 240,200 250,000 259,800
5 System efficiency 93.00% 95.00% 97.00%
6 Annual degradation rate 0.50% 1.00% 1.50%
7 Electricity selling price per kwh $0.161 $0.180 $0.200

 A one-way range sensitivity analysis is performed and the following tornado diagram for NPV
and IRR are generated:


• The two top most sensitive variables are
1. Initial investment cost
2. Electricity price

 The least sensitive variable is
Salvage value
-$347,000
$0.16100
240,200
1.50%
93.00%
-$2,000
$1,000
-$229,000
$0.20000
259,800
0.50%
97.00%
-$1,000
$3,000
-$30,000 -$20,000 -$10,000 $0 $10,000 $20,000 $30,000 $40,000 $50,000 $60,000 $70,000 $80,000 $90,000 $100,000 $110,000
Initial cash flow
Electricity selling price per kwh
Energy output in year 1 (kwh)
Annual degradation rate
System efficiency
Annual O&M cash flow
Salvage value
NPV
-$347,000
$0.16100
240,200
1.50%
93.00%
-$2,000
$1,000
-$229,000
$0.20000
259,800
0.50%
97.00%
-$1,000
$3,000
8.00% 9.00% 10.00% 11.00% 12.00% 13.00% 14.00% 15.00% 16.00% 17.00% 18.00%
Initial cash flow
Electricity selling price per kwh
Energy output in year 1 (kwh)
Annual degradation rate
System efficiency
Annual O&M cash flow
Salvage value
IRR
dss5202 49

Risk Analysis
 The company’s experts assessed the probability distributions for six variables as follows:
Variable Distribution Parameters
1 Initial cash flow normal mean= -288,000, sd= 30,000
2 Salvage value Fixed 2,000
3 Annual O&M cash flow triangular min= -2,000, mode= -1,500, max= -1,000
4 PV output in year 1 (kwh) normal mean= 250,000, sd=5,000
5 System efficiency truncated normal mean=95.0%,sd=1.0%,min=93.0%,max=97.0%
6 Annual degradation rate uniform min= 0.50%, max= 1.50%
7 Electricity price per kwh lognormal mean= 0.180, sd= 0.010

 Monte Carlo simulation is performed using the above probability distributions:
Risk profile for Project NPV


Risk Interpretations
• Downside risk:
 Prob (NPV ≤ 0) = 9.5%

 Upside potentials:
Prob (NPV ≥ $50,000) = 38.5%
Prob (NPV ≥ $75,000) = 13.8%
Prob (NPV ≥ $100,000) = 2.9%

 Value-at-Risk
 Net Present VaR (95%) = $10,571
dss5202 50

Risk profile for Project IRR



Risk Interpretations

 Downside risk:
 Prob (IRR ≤ 10%) = 9.5%

 Upside potentials:
 Prob (IRR ≥ 12%) = 52.6%
 Prob (IRR ≥ 14%) = 16.7%
 Prob (IRR ≥ 16%) = 3.4%










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