34525895 engineering economics notes

of 55

By Prof. Amrit Nakarmi I/II MSREE 09 July 2008

Introduction to Economics of Energy Projects (EG854ES)

What Is Economics?

Scarcity

All economic questions arise from a single and inescapable fact: you can't always get what you want. We live

in a world of scarcity.

Scarcity means that wants always exceed resources available to satisfy them.

People get involved in Economic Activity to cope with Scarcity.

Economics is the study of how people use their limited resources to try to satisfy unlimited wants.

Faced with scarcity, we have to make choices because we can't have all what we want. Balancing the

wants and the resources available is called economizing or optimizing.

What Is Energy Economics?

What is competition?

Competition is the contest for command over scarce resources.

For human life and the production processes, a sufficiently available of energy is the highest priority.

Human beings can live without other things, but not without energy resources. Energy resources are also

scarce and hence, needs its optimization and it is dealt by energy economics.

What Is Economics of Energy Projects?

Economics of Energy Projects deals with how economically an energy project can be established and

operated. Cost structure, financing, capital budgeting (project evaluation), and financial performance analysis

of the energy projects or firms are looked into.

Types of Firms

Three basic types of firm

Sole proprietorship

Partnership

Corporation or limited company

Sole proprietorship

It is the oldest form of business organization. A single person owns the business, holds title to all its

assets, and is responsible for all of its liabilities.

Advantages

Simplicity

Quicker decision‐making

Easy to establish

of 55

Disadvantages

Good for small firms and not good for big firms such as energy Cos.

Responsible for all liabilities

Difficult to raise capital

Cost of capital is high

Partnership

A partnership is similar to a proprietorship in all aspects except that there is more than one owner.

Advantages

Decision made through consensus/agreement , hence low risk

Can raise higher capital

Easy to establish but more complex than single proprietorship

Disadvantages

Responsible for all liabilities

Slower decision‐making process than single proprietorship

Difficult to raise capital

Corporation or Limited Company

A company is an impersonal entity created by law, which can own assets and liabilities. The main

feature of this form is that the Co. is separate from its owners. A owner’s liability is limited to his/her

shareholding only.

Advantages

Limited liability

Can raise higher capital (a kind for energy Cos.)

Lower cost of capital

Decision‐making through consensus

Disadvantages

Slow decision‐making

Difficult to set up

RAISING

Equity CEquity cap

Rights of

R

th

re

R

ev

G FINANCES

Capital pital represen

Authorized

memorand

Issued cap

Subscribed

Paid‐up ca

Par value –

Book valu

outstandin

Market Va

market.

Initial Pub

Subsequen

Rights Issu

existing sh

f equity shar

ight to incom

he debt obliga

etained earni

ight to contr

very resolutio

S

nts ownershi

d capital –

dum of assoc

pital – the am

d capital – Th

apital – The a

– It is the valu

e – It is the s

ng shares.

alue – It is th

blic Offering (

nt offering is

ue –It is the

hareholders.

reholders

me – The equi

ation and pre

ngs or paid o

ol – Equity sh

on placed bef

p capital as e

The amoun

iation.

mount offered

he part of the

ctual amount

ue stated in t

um of the pa

he value of t

(IPO) – The in

called Second

e selling of th

ity sharehold

eferred share

ut as dividen

hareholders a

fore the Co.

quity (comm

t of capital

by the Co. to

issued capita

t paid up by t

he memoran

aid‐up capital

the share at

nitial public is

dary public of

he security i

ers have resi

dividends. Th

d.

are the actual

on) sharehold

that a Co.

o the shareho

al which has b

the investors.

dum and the

and retained

which it is tr

ssue of the sh

ffering (SPO).

n the primar

dual claim to

he residual in

l owners of th

ders collectiv

can potent

olders.

been subscrib

share certific

d earnings div

raded in the

hares to the m

ry market by

o the income

ncome can be

he Co. and ha

Pa

ely own the c

ially issue, a

bed to the inv

cate.

vided by the

stock exchan

members of t

y issuing sha

of the firm af

e withheld by

ave the right

age 3 of 55

company.

as per its

vestors.

number of

nge or the

the public.

res to the

fter paying

the Co. as

to vote on

of 55

Pre‐emptive rights – It enables the existing shareholders to maintain their proportional ownership of

the shares if the Co. issued additional shares in the market.

Sources of Finance

Preferential Capital (Preferred shares)

Preferred shares are hybrid forms of capital. They have the characteristics of both the equity (common

shares) and the debt such as debentures.

Main features are :

1. preferred share dividend is payable after net income,

2. it is cumulative (dividend if not paid in year, will be accumulated next year), and

3. it is taxable and has no voting rights.

Internal accumulation (retained earnings)

The internal accruals consist of depreciation and retained earnings. Retained earnings are

much more expensive than bank loans, because they are retained without paying out the dividend and cost

of capital (interest rate) of equity is higher than that of the loan.

Term Loans/debentures

Terms are given by financial institutions such as banks and have term of less than 10 years.

Debentures (bonds) are loans raised from the public and the interest (called here as coupon) is paid every six

months. It can be secured and unsecured. Debentures can be convertible into common shares.

Working capital advances (loans)

Under a cash credit or overdraft arrangement, a company can borrow required amount if it is within its limit

in the agreement with the financial institution or the bank.

Weighted Average Cost of Capital (WACC)

Cost of Debt

A firm with a 40% tax rate issues $1,000 bonds at a face value with coupon rate of 16%. Ignoring

underwriting and issuing expenses,

Market yield (market rate of return) = rd =160/1000 = 16%

Cost of debt (to the company)=Rd=160*(1‐0.4)/1000 =9.6%

If people invest in bonds for long‐term, then

Pb =Sum(I/(1+rb)t + F/(1+rb)

n

of 55

Cost of debt (to the Co)

NPb =Sum(I*(1‐Tax)/(1+kb)t + F/(1+kb)

n

Cost of Preferred Share A corporation issues new $100 preferred shares that provide $12 in annual dividends. The firm has identical

preferred shares outstanding that also trade at $100/share. Issuing and underwriting expenses are 5% of the

issue price and assumed to be tax deductible. The firm’s tax rate is 40%.

Net Proceeds of preferred share (to the Co.) =NPp=100‐(1‐0.4)*5 =$97

rp= 12/100 = 12%

kp=12/97 = 12.37% (cost of preferred share to the Co.)

Cost of Preferred Share (for long‐term investment)

Pp=Dp*Sum(1/(1+ rp)t)

Cost of Equity A corporation issues new $100 common shares that provide $16 in annual dividends. The firm has identical

common shares outstanding that also trade at $100/share. Issuing and underwriting expenses are 5% of the

issue price and assumed to be tax deductible. The firm’s tax rate is 40%.

Net Proceeds of common share (to the Co.) = NPe= 100‐(1‐0.4)*5 = $97

re= 16/100 = 16%

ke = 16/97 = 16.49% (cost of equity to the Co.)

Cost of Equity (long‐term)

Market capitalization rate

Pe= SUM(Dt/(1+ re)t)

Cost of new shares

NPe = SUM(Dt/(1+ ke)t)

Cost of Equity

Usually cost of equity is not known, then we have to use Capital Asset Pricing Model (CAPM) to find

out cost of equity.

Cost of

Cost of

In practice

•

•

•

eq

WeightIf ‘I’ is the

I =

where

B

P

Then,

W

Example o

A firm pla

that rough

Lo

P

N

Equity (CA

Equity

Cost

wher

E

e,

Short term g

Historical ris

Betas are es

quity is riskie

ed Average total investm

=B+P+E

is borrowing

is preferred s

WACC =Rb(1‐T

on WACC

ans on financ

hly correspon

ong –term de

referred shar

ew common

APM)

t of Equity (k

re,

Rf = Risk fre

E(R m) = Expec

government s

k premiums a

stimated by r

r than the ma

ge Cost of Cment, then

g (loans and b

shares, and E

Tax)*B/I +kp*P

cing major ne

nds to its curr

ebt $30

res $10

shares $40

ke) = Rf + Equ

ee rate

cted Return o

ecurity rates

are used for t

regressing sto

arket)

Capital (W

onds),

E is equity.

P/I + ke*E/I

ew expansion

rent capital st

0 mil

0 mil

0 mil

uity Beta * (

on the Market

are used as r

the risk premi

ock returns a

WACC)

n programs by

tructure:

(E(Rm ) ‐ Rf)

t Index (Diver

risk free rates

ium

against mark

y drawing on

rsified Portfo

s

et returns (it

n funds in the

Pa

lio)

t shows how

e following p

age 6 of 55

much the

roportions

of 55

Issuing and underwriting expenses can be ignored. Debt can be issued at a coupon rate of 12%, and the

dividend yield on preferred shares would be 9%. Common Shares currently trade at $45 per share. The

current dividend yield on preferred shares would be $2.25 per share. Management feels that, over long run,

growth in dividend match inflation rate, which is anticipated to be 10% per year. The corporate tax is 40%.

What is the firm’s weighted average cost of capital (WACC) ?

Example on WACC

The current interest on government debt is 10%, and the return on the market is expected to exceed this

rate by 7%. What value of beta do we have to assume for the firm if the cost of equity as derived from the

CAPM is to match the Ke =15% calculated according to the dividend growth model under above example?

kb = (1‐T) rb =0.6*12% =7.2%

kp = rp =9%

Ke =D1 / pe+g =2.25/45+0.1 = 0.15 or 15%

Source Proportion Cost in % Weighted Cost

Debt 30/80 =0.375 7.2% 2.7%

Preferred 10/80 =0.125 9% 1.13%

Common 40/80 =0.50 15% 7.5%

WACC = 2.7% +1.13% +7.5% =11.33%

Example on WACC

A firm plans on financing major new expansion programs by drawing on funds in the following proportions

that roughly corresponds to its current capital structure:

Long –term debt $30 mil

Preferred shares $10 mil

New common shares $40 mil

Issuing and underwriting expenses can be ignored. Debt can be issued at a coupon rate of 12%, and the

dividend yield on preferred shares would be 9%. Common Shares currently trade at $45 per share. The

current dividend yield on preferred shares would be $2.25 per share. Management feels that, over long run,

growth in dividend match inflation rate, which is anticipated to be 10% per year. The corporate tax is 40%.

What is the firm’s weighted average cost of capital (WACC) ?

Example on WACC

The current interest on government debt is 10%, and the return on the market is expected to exceed

this rate by 7%. What value of beta do we have to assume for the firm if the cost of equity as derived from

the CAPM is to match the Ke =15% calculated according to the dividend growth model under above example?

of 55

Financial Statements These are statements of financial information to the managers and the shareholders.

Income Statement (Profit and Loss Statement)

Balance Sheet

Cash flow statement / Source and Use of Funds Statement (Statement of changes in financial

positions)

Income Statement It shows the record of financial events between two points in time. It has revenue from sales and expenses

incurred during the period.

Net Worth

The value of total assets minus total liabilities or the value of the owners' claim on the assets.

Income Statements (P/L statement)

Expenditures are all cash outflows

Expenses are only those expenditures that affect net worth of the shareholders and appear in the Income

Statement.

Receipts are all cash inflows.

Revenues are only those receipts that affect net worth and thus appear in the income statement.

Balance Sheet

It gives snapshot summary of the firm's financial position at a single point in time.

The balance sheet shows the net worth of shareholders at a point in time, whereas income

statement measures changes in net worth.

Liabilities indicate what money has been made available to the firm.

Assets show how the firm has used the money made available to it.

Current Liabilities are the short ‐term debt obligations of a firm, with maturities of less than one

year.

Fixed liabilities are firm's long‐term finance such as long‐term debts from banks and the public.

Shareholders' equity is the money invested by the shareholders and the retained earnings.

Fixed Assets are acquired for long‐term uses in the firm such as plant, building, land, and

equipment.

Current Assets are cash, accounts receivables, and inventories of finished goods and raw materials.

Depreciation is the allocation of cost of an asset to different time periods.

Working Capital is composed of firm's current assets.

Net Working Capital is the difference between current assets and current liabilities.

of 55

Some Items in the Financial Statements

Profit is the excess of revenues over expenses during a given period of time.

Cash flow is the actual cash flowing in and out of a firm over a particular time period.

Operating cash flow is the flow of cash arising from the operation of a firm and consists of net

profit (profit after tax) plus non‐cash charges such as depreciation.

Cash Flow Statement

Source‐ and ‐use‐of‐funds Statement (cash flow statement) is a summary of the flow of the

financial activity of the firm. It shows where the firm obtains cash and how it uses it.

Sources of funds

Increase in liabilities

Increase net worth through retained earnings or capital contribution by the shareholders

Reduction in assets through sales of assets

Uses of funds

reduction in liabilities

reduction in net worth through payment of dividends or losses

increase in assets

Cash flow from operating activities

Net profit

Depreciation

Decrease in account receivables

Increase in accounts payables

Cash flow from investing activities

Sales of fixed assets

Investment in new fixed assets

Cash flow from financial activities

Increase in debt (cash inflow)

Issuance of new shares (cash inflow)

Dividend payment (cash outflow)

Balanc

Income S

R

C

G

Ex

N

ce Sheet

Statement

evenue

osts of Good

Gross Profit

xpenses

Net Income

ds Sold (COGGS)

Pag

ge 10 of 55

of 55

INTEREST AND INTEREST RATE

• Interest is a rental amount charged by financial institutions for the use of money.

• Interest rate, or the rate of capital growth, is the rate of gain received from an investment.

• Usually this rate of gain is stated on a per‐year basis, and it represents the percentage gain realized on the

money committed to the undertaking. Thus, an 11% interest rate indicates that for every dollar of money

used, an additional $0.11 must be returned as payment for the use of that money.

THE TIME VALUE OF MONEY

Because money can earn at a certain interest rate through its investment for a period of time, a rupee

received at some future date is not worth as much as a rupee in hand at present. This relationship between

interest and time leads to the concept of the time value of money.

• A rupee in hand now is worth more than a rupee received n years from now. Why?

Because having the rupee now provides the opportunity for investing that rupee for n years more than the

rupee to be received n years hence. Since money has earning power, this opportunity will earn a return, so

that after n years the original rupee plus its interest will be a larger amount than the rupee received at that

time. Thus, the fact that money has a time value means that equal rupee amounts at different points in time

have different value as long as the interest rate that can be earned exceeds zero.

It is also true that money has time value because the purchasing power of a rupee changes through time.

During periods of inflation the amount of goods that can be bought for a particular amount of money

decreases as the time value of money it is important to recognize both the earning power of money and the

purchasing power of money.

THE PURCHASING POWER OF MONEY

Inflation (price rises) and deflation (depreciation, devaluation) are terms that describe changes in price levels

in an economy.

SIMPLE AND COMPUND INTEREST

Simple Interest:

Under simple interest, the interest owed (payable, billed) upon repayment of a loan is proportional to the

length of time the principal sum has been borrowed. The interest earned may be found in the following

of 55

manner. Let I represent the interest earned, P the principal amount, n the interest period, and i the interest

rate. Then,

I = P* n* i

Suppose that NRS1, 000 is borrowed at a simple interest rate of 12% per annum. At the end of the year, the

interest owed would be,

I = NRS 1,000 *1*0.12 = NRS 120

The principal plus interest would be NRS 1,120 and would be due at the end of the year. Interest and

principal become due (payable) only at the end of the time period.

Compound Interest

• When loan is made for several interest periods, interest is calculated and payable at the end of each

interest period. There are number of loan repayment plans.

• These range from paying the interest when it is due to accumulating the interest until the loan is due.

• If the borrower does not pay the interest earned at the end of each period and is charged interest on the

total amount owed (principal plus interest), the interest is said to be compounded.

• The interest owed in the previous year becomes part of the total amount owed for this year. This year’s

interest charge includes interest that has been earned on previous interest charges.

CALCULATION OF COMPOUND INTEREST WHEN INTEREST IS PAID ANNUALLY

CALCULATION OF COMPOUND INTEREST WHEN INTEREST IS PERMITTED TO COMPOUND

of 55

INTEREST FORMULAS

Let,

I = the annual interest rate

N = the number of annual interest periods;

P = a present principal amount;

A = a single payment, in a series of n equal payments, made at the end of each annual payments.

F = a future amount in n annual interest periods.

Assumptions:

1. End of one year is the beginning of the next year

2. P is at the beginning of a year from a time regarded as being the present

3. F is at the end of the nth year from a time regarded as being the present.

4. An A occurs at the end of each year of the period under consideration. When P and A are

involved, the first A of the series occurs one year after P. when F and A are involved, the last A of

the series occurs simultaneously with F.

Single‐Payment Compound‐Amount Factor (F/P, i, n)

If an amount P is invested now and earns at the rate i per year, how much principal and interest are

accumulated after n years?

Derivation of Single‐payment Compound‐amount Factor

• The resulting factor, (1 + i)n, is known as the single payment compound‐amount factor and is denoted by

(F/P, i, n).

• The future amount, F, of a present principal amount, P is given by: F = P(1 + i)n

of 55

Example:

If $1,000 is invested at 16% interest compounded annually at the beginning of year one, the compound

amount at the end of the fourth year will be: F = $1,000(1 + 0.16)4 = $1,000(1.811) = $1,811

Single‐Payment Present‐Worth Factor (P/F, i, n)

Find the present worth P of a future amount, F received at the end of n periods if the interest rate is i.

•The present worth in this case is given by: P = F [1/ (1 + i)n]

•The resulting factor, 1/ (1 + i) n, is known as the single payment present‐worth factor and is denoted by:

(P/F, i, n).

Example:

How much must be invested now at 16% compounded annually so that $1,811, can be received 4 years?

P= $1,811 [1/(1 + 0.16)4] = $1,811(0.5523) = $1,000

Equal‐Payment‐Series Compound‐Amount Factor (F/A, i, n)

To find the single future value that would accumulate from a series of equal payments occurring at the end

of succeeding interest periods.

If A represents a series of n equal payments

F = A + A (1 +i) + ….. + A (1 + i)n‐2 + A(1 + i)n‐1

F = A [(1+i)n – 1]/i

The resulting factor, [(1 + i)n ‐1] /i, is known as the equal payment‐ series compound‐amount factor and is

designated as (F/A,i,n).

Example:

The Compound Amount of a Series of Year‐End Payments of $100 of Annual Interest Rate Of 12%

Equal‐Payment‐Series Sinking Fund Factor (A/F, i, n)

of 55

The equal‐payment‐series compound‐amount relationship may be solved for A as follows:

The resulting factor, i/ [(1 +i)n – 1], is known as the equal‐payment series sinking‐fund factor

Example:

It is desired to accumulate $635 by making a series of five equal annual payments at 12% interest

compounded annually, the required amount of each payment will be

Equal‐Payment‐Series Capital Recovery Factor (A/P, i, n)

A deposit of amount P is made now at an annual interest rate i. The depositor wishes to withdraw the

principal, plus earned interest, in a series of equal year‐end amounts over the next n years. When the last

withdrawal is made, there should be no funds left on deposit. The value of equal year‐end amount (annuity)

is given by:

The resulting factor i(1+i)n / [(1 + i) n 1] is known as the equal payment series capital‐recovery factor and is

designated as (A/P, i, n).

Example:

$1,000 invested at 15% interest compounded annually will provide for eight equal year‐end payments of

Equal‐Payment‐Series Present‐Worth Factor, (P/A, i, n)

• To find what single amount must be deposited now so that equal end‐of‐period payments can be made, P

must be found in terms of A

of 55

The resulting factor, [(1 + i)n‐1]/i (1 +i)n, is known as the equal‐payment series present‐worth factor and is

designated as (P/A, i, n).

Example:

The present worth of a series of eight equal annual payments of $223 at an interest rate of 15% compounded

annually will be

Uniform‐Gradient‐Series Factor, (A/G, i, n)

In some cases, periodic payments do not occur in an equal series.They may increase or decrease by a

constant amount.

Let

G = annual change or gradient

n = the number of years;

A = the equal annual payment

The resulting factor is called the uniform gradient series factor and is designated as (A/G, i, N)

Geometric Gradient‐Series Factor, (A/G, i, n)

In some situations, annual payments increase or decrease, not by a constant amount, but by a constant

percentage. If g is used to designate the percentage change in the magnitude of the tth payment is related to

payment A1 as

of 55

Two types of linear gradient series as composites of a uniform series of N payments of A1 and the gradient

series of increments of constant amount G

Project Cash Flows

Typically a capital project initially requires investment outlay and produces annual net cash inflows.

Cash Outflows:

Purchase of new equipment

Working capital

Manufacturing, operating, and maintenance costs

Leasing expenses

Interest and repayment of borrowed funds

Income taxes and tax credits.

Cash Inflows:

• Borrowed funds

• Operating revenues

• Cost savings (or cost reductions)

• Salvage value

• Working capital release (cost recovery by liquidation)

Elements of Cash Outflows

Working Capital:

It is an investment in non‐depreciable assets.

of 55

Some projects require such investment so that more funds are available in order to increase firm’s revenues.

In accounting, working capital is the amount carried in cash, accounts receivable, and inventory (account,

stock) that is available to meet day‐to‐day operating needs.

For example, if a company is going to a market a new product, inventories of the product and larger

inventories of raw materials will be needed. Accounts receivables from sales will increase, and management

might also decide to carry more cash because of higher volume of activities.

These are working capital but these have no tax effect and the cash flows always sum to zero over the life of

a project but the inflows and outflows are shifted in time so, they do affect the net present worth.

Overhead:

The cost that occurs irrespective to the specific units of output of a firm is called as overhead cost.

In manufacturing, all costs other than direct material and labors are treated as overhead costs.

For example, costs to indirect material and labor, maintenance and repairs on production equipment; heat

and light, property taxes, royalty, depreciation and insurance, overtime premiums.

Depreciation of factory buildings is unaffected by the amount of production during any particular period. If

however, a new building is required to meet any increased production, manufacturing overhead will certainly

increase.

Indirect materials like solder used to make electrical connections in a computer circuit board and the glue

used to bind the books, these are treated as a part of manufacturing overhead.

Typical examples of indirect labor include the wages of janitors, supervisors, material handlers and night

security guards.

Gains Tax:

It is a tax applicable to gains (or losses) occurred on the earning or loss as a result of difference in the book

value and salvage value of an asset. If the salvage value is higher than the book value then the tax needs to

be paid to incremental earning. Similarly, if the salvage value is less than the book value then tax needs to be

adjusted to the loss.

of 55

Income Statement Cash Flow Approach

Once the cash flow elements (both inflows and outflows) are determined, these may be grouped into three

categories:

Operating activities

• Cash flow from operations (e.g., current sales revenues, the cost of goods sold, operating expenses (also

includes Interest payments) and income taxes)

Cash flow from operation = Net income + Depreciation

Investing activities

In general, three investment flows are associated while buying equipment they are: original investment,

salvage value and working capital investment or recovery.

Financing activities

• It includes:

• The amount of borrowing,

• The repayment of principal.

The Net Cash Flow for a given year is simply the sum of the net cash flows from operating, investing and

financing activities.

of 55

Generalized Cash Flow Approach

• It is an approach in which net cash flow is obtained by aggregating individual items.

E.g. A computerized machining center has been proposed for a small tool manufacturing company. If the new

system, which costs $125,000 is installed, it will generate annual revenues of $100,000 and will require

$20,000 in annual labor, $12,000 in annual material expenses, and another $8,000 in annual overhead

(power and utility) expenses. The automation facility would be depreciated at the rate of 20%.

• The company expects to phase out the facility at the end of five years, at which time it will be sold for

$10,000. Find the year‐by‐year after‐tax net cash flow for the project at a 30% marginal tax rate based on the

net income and determine the after‐tax net present worth of the project at the company’s MARR of 15%.

• Solve it using both Income Statement Cash Flow and Generalized Cash Flow Approach.

of 55

Example with Financing (Borrowing)

In the previous example, it is assumed that $62,500 of the $125,000 paid for the investment is obtained

through debt financing (debt ratio =0.5). The loan is to be repaid in equal annual installments at 10% interest

over five years. The remaining $62,500 will be provided by equity (e.g., from retained earnings). Find the net

after‐tax cash flows.

Solution

Computing the annual loan repayment installments:

$62,500 (A/P, 10%,5) = $16,487.

of 55

Now, computing the component of interest and loan repayment installments for 5 years as below:

For, n = 1,

The interest due at n = 1 is $6,250 (10% of 62,500)

Which leaves 16,487 ‐ 6,250 = 10,237 as left over for principal payment with loan balance as: 52,263.

For, n = 2 is $5,226 (10% of 52,263)

Which gives 16,487 – 5,226 = 11,261 as left over for principal payment and with loan balance as: 41,002 …

of 55

Depreciation Definition

• It is a loss in value over the time the property is being used.

• Depreciable property includes buildings, machinery, equipment and vehicles. Exceptions are land (no

definite life) and collectible items/inventories (primary for sale)

Economic Depreciation (ED)

Example: A car’s reliability and appearance usually decline with age. The vehicle is worth less with each

passing year.

ED accumulated = Purchase price – market value

Accounting Depreciation (AD)

is based on the idea of ED

The systematic allocation of the initial cost of an asset in parts over a time, known as its depreciable

life and the process is AD.

Sometimes refer to it as “asset depreciation”

AD concept is popular in engineering economic analysis because AD provides a basis for determining

the income taxes associated with any project undertaken.

of 55

What constitutes a depreciable property?

• It must be used in business or held for production of income.

• It must have definite service life and that life must be longer than 1 year.

• It must be something that wears out, decays, gets used up, becomes obsolete, or loses value from

natural causes.

Claiming Depreciation

• The tax payer must be the owner of the property for the claim

• In the case of leased property, lessee is not entitled to depreciate that property (e.g., leased

automobiles)

Accounting for the Depreciation of Capital Assets

There are two aspects of depreciation:

1. The actual lessening in value of an asset with use and the passage of time and

2. The accounting for this lessening in value

Depreciation views the cost of an asset as a prepaid expense that is to be charged against profits over some

reasonable period of time.

Rather than charging the entire cost as an expense at the time the asset is purchased, depreciation is a

systematic way to spread the anticipated loss in value over the life of the asset.

This is the concept of amortizing the cost of an asset so that the profit ad loss statement is a more accurate

reflection of capital consumption which is basic to financial reporting and income tax calculation.

of 55

Value time function and Book value

• The value of an asset decreases yearly in accordance with one of several mathematical functions.

• Selection of particular function involves decisions as to the life of the asset, its salvage value and the form

of the mathematical function. A general value‐time function is shown below:

Book value is the acquisition cost of an asset less its accumulated depreciation charges. A function similar

to above can represent book value.

Book Value Calculation

The book value at the end of any year (Bt) is equal to the book value at the beginning of the year (initial cost,

P) less the depreciation expenses charged during the year.

Mathematically,

of 55

Basic Depreciation Methods Three important methods will be discussed:

1. Straight‐Line Methods

2. Accelerated Methods

3. Units‐of Production Method

Straight‐Line (SL) Method

This model assumes that the value of an asset decreases at the constant rate.

The expression for depreciation charge in any year is:

And the book value at end of year t is:

Where,

P = initial cost of the asset,

F = Salvage value and

N = useful life

Example 2: SL Depreciation

Cost basis of the asset, P = $ 12,000

Useful Life, N = 5 years,

Salvage value, F =20% of the initial cost = $2400

of 55

Accelerated Depreciation Methods

• The mechanical efficiency of an asset tends to decline with age, because maintenance costs tend to

increase with age, or because of the increasing likelihood that better/efficient equipment will become

available and make the original asset obsolete.

• This reasoning leads to a method named “accelerated depreciation method “ that charges a larger fraction

of the cost as an expense of the early years than that of the later years

• The two most widely used accelerated methods are :

a. Declining Balance

b. Sum‐of‐the‐year‐digits

Declining Balance Method (DB)

• Calculation by DB depreciation method allocates a fixed fraction of the beginning book balance each year.

The fraction, α, is obtained as follows:

α = (1/N) . (Multiplier)

• The most commonly used multipliers are 1.5 ( i.e., 150% DB) and 2.0 (called 200% or double declining

balance , DDB)

• When “n” increases , α decreases. This results in a situation in which depreciation is highest in the first year

and decreases over the asset’s depreciable life.

• For a depreciation rate α ,the depreciation charge in any year for declining –balance can be derived as

follows:

• Total DB (TDB) depreciation at the end of t years is computed as follows:

• The book value, Bt, at the end of year t is the cost of the asset, minus the depreciation at the end of t years

of 55

Example 3 : DB Method for example 2

Here, α = (1/N) . (multiplier) = (1/5) . 200% = 40% (Double DB)

Cases of DB Methods

When final book value ≠ es mated salvage value, we must re‐adjust our analysis because either tax law does

not permit us to depreciate assets below their salvage value or one have not taken full advantage of

depreciation’s tax‐deferring benefits. Hence, two cases will be discussed:

Case 1 : When the final book value is less than the estimated salvage value.

Case 2: When the final book value is greater than the estimated salvage value.

Case 1: When Bt < F

Example 4: The final book value $ 933 is less than the estimated salvage value of $2,400.

To avoid deducting depreciation charges that would drop the book value (BV) below the salvage value F,

adjust depreciation amounts such that Bt = F in the period when the implied BV is lower than F.

of 55

Note that B4 would be less than F = $ 2,400, if the full deduction ( $1,037) had been taken. We adjusted D4

to $ 192 , making B4 = $ 2,400.

Case 2: When Bt > F

Example 5: The final book value $ 933 is greater than the estimated salvage value of $ 0 (assume)

To reduce the book value (BV) of an asset to its salvage value as quickly as possible, it can be done by

switching from DB to SL whenever SL depreciation results in larger depreciation charges i.e., switch from DB

to SL if depreciation by DB in any year is less than ( or equal to ) it would be by SL.

Note that , switching takes place at year 4 as DB depreciation is less than SL depreciation .The resulting

depreciation schedule is :

of 55

Sum‐of‐the‐year‐digits (SOYD) Method of Depreciation

• As compared to SL‐depreciation, SOYD also results in larger depreciation charges during the early years of

an asset’s life and smaller charges as the asset reaches the end of its estimated useful life.

• Here, if N is the estimated years of useful life, the numbers 1, 2, 3,…,N are summed:

SOYD = 1 + 2 + 3 +…….+ N = N (N+1) / 2

• Depreciation rate Dn is computed here as a fraction in which the denominator is the SOYD and the

numerator is, for the first year, N; for the second year, N‐1 and so on. Each year the depreciation

charge is computed by dividing the remaining useful life by the SOYD and by multiplying this ratio by the total

amount to be depreciated (P‐F). Mathematically expressing:

of 55

Example 6 : SOYD Depreciation

Units‐of‐Production Method

• SL method can be applicable only if the machine/asset be used for exactly the same amount of time each

year. What happens when a machine is run 2000 hours one year and 800 hours the next?

• The units‐of‐production method takes care of such situation where depreciation charge for a period is

related to the number of service units consumed in that period i.e., depreciation varies with production

volume.

• This method gives more accurate picture of machine usage.

• By this method, the depreciation in any year is given by:

• A disadvantage of this method is that the collecting of data on machine use and the accounting methods

are somewhat tedious.

Example 6: Units‐of‐Production Method

A truck for hauling coal has an estimated cost of $55,000 and is expected to give service for 250,000 miles,

resulting is a $5,000 salvage value. Compute the allowed depreciation amount for truck usage of 30,000

miles.

Solution :

of 55

Tax Depreciation

• It is a depreciation method by which a company depreciates a fixed asset for tax purposes. This is one of

the many tax incentive options that increases cash flows for reinvestment.

• The depreciation methods are country specific and these are formulated by Tax Authority for estimating

depreciation for Tax purpose.

• Usually in depreciation, the asset is expected to be fully depreciated at approximately the end of its useful

life. However, for tax purpose, it is not necessary.

• Simple methods are established which defines several classes of assets with allocated rate of depreciation.

Book Depreciation vs Tax Depreciation

Some of the Tax Depreciation Methods

Modified Accelerated Cost Recovery System (MACRS) Method:

• MACRS is a method for tax depreciation practiced in developed countries e.g. US.

• MACRS scheme formulated simpler guidelines with eight classes of assets/properties, each with a more or

less arbitrary life called a recovery period. The recovery period do not necessarily bear any relationship to

expected useful lives. The salvage value of property is always treated as zero.

of 55

MACRS Property Classifications

Note:

• ADR = Asset Depreciation range; guidelines published by the IRS.

• Automobiles have a midpoint life of 3 years in the ADR guidelines, but are classified into a 5 yr property

class.

MACRS Depreciation Schedules :

MACRS recovery percentage as shown is according to Half Year‐Convention (HYC),

Declining Balance Method

• HYC assumes that all assets are placed in service at midyear and they will have zero salvage value.

• Half of one year’s depreciation in the first year, full year’s depreciation in each remaining years and

remaining half‐year’s depreciation in the year following the end of recovery period is taken.

(See table in book)

MACRS Depreciation: An Example

# A taxpayer wants to place in service a $10,000 asset that is assigned to the 5‐yr class. Compute the MACRS

% and the depreciation amounts for the asset.

Solution:

Given: 5‐yr asset, ½ yr convention

We compute, α = 40% (200%DB), and S = 0

Since,

Straight‐line rate = 1/5 = 0.20

200% DB rate = 2(0.20) = 40%

Under MACRS, salvage value (S) = 0

Find: MACRS depreciation % Dn for $10,000 asset.

of 55

Then, beginning with the first taxable year and ending with the sixth year, MACRS deduction % is computed

as follows:

Depreciation Amounts (Dn):

Yr 1: 20% x $10,000 = $ 2,000; Yr 4: 11.52% x $10,000 = $ 1,152

Yr 2: 32% x $10,000 = $ 3,200. Yr 5: 11.52% x $10,000 = $ 1,152

Yr 3: 19.20% x $10,000 = $ 1,920; Yr 6: 5.76% x $10,000 = $ 576

Other Tax Depreciation Methods

Maximum Tax Depreciation rate classification in Thailand

• Based on acquisition price (SL method?)

– The purchaser will depreciate the asset as if it is acquired new.

– In the case of a second hand asset, depreciation of the asset, while with the previous owner, would not

affect the minimum depreciation period for the asset to the new owner. It will be treated as if it is a new

asset (using the asset price paid by the new owner).

Corporate Taxes

Corporate rate is How are the taxes reflected in Cash Flow?

Tax applied to the taxable income of a corporation.

of 55

Tax Rates in Thailand

The maximum corporate income tax rate in Thailand is 30% on net profit.

However, the rates vary depending on types of tax payers.

Cash Flow vs Net Income

Depreciation = Expense or Income ?

A company buys a numerically controlled NC machine for

Baht 3 million (year 0) and uses it for 5 years, after which it is scrapped.

of 55

The allowed depreciation rate is 20% in the acquisition price. Suppose the company estimates the following

revenues and expenses, including depreciation, for the first operating year:

Gross income = Baht 4 million

Cost of goods sold = Baht 2 million

Operating Expenses = Baht 1 million

a) If the company pays taxes at the rate of Thai Tax rates, what is its net income from the project?

b) Assume that

1) all sales are cash sales and

2) operating expenses will drop to 80% of that in Year 1 in subsequent years. Show the net cash flow.

Cost and Time value of Money

Classifications of Cost

A number of cost classifications have come into use to serve as a basis for economic analysis.

First Cost

• First cost is the initial cost of capitalized property, including transportation, installation, and other related

initial expenditures.

• First cost is usually made up of a number of cost elements that do not recur after an activity is initiated. For

purchased equipment, these include engineering design and development cost, test and evaluation cost, and

construction or production cost as well as shipping, installation, and training costs.

of 55

• Many activities that otherwise may be profitable cannot be undertaken because their associated first cost

represents too high a level of investment. Many engineering proposals that are otherwise sound are not

initiated because the first cost involved is beyond the reach of the controlling organization.

Operation and Maintenance Cost

• Whereas first cost occurs only once in getting an activity started,

• Operation and maintenance cost is that group of costs experienced continually over the useful life of the

activity.

Fixed Cost

• Fixed cost is that group of costs involved in an on going activity whose total will remain relatively constant

throughout the range of operational activity.

• Fixed costs are made up of such cost items as depreciation, maintenance, taxes, insurance, lease rentals,

interest on invested capital, sales programs, certain administrative expense, and research.

Variable Cost

• Variable Cost is that group of costs that vary in some relationship to the level of operational activity.

• In general, all costs such as direct labor, direct material, direct power, and the like, which can readily be

allocated to each unit produced, are considered to constitute variable costs, and the balance of the costs of

the enterprise are regarded as fixed.

Incremental and Marginal Cost

• The terms incremental cost and marginal cost refer to essentially the same concept. The word increment

means increase.

• Incremental cost is the additional cost that will be incurred as the result of increasing output by one more

unit.

Cost Behaviors

Fixed cost and variable costs are the two most common cost behavior patterns.

of 55

Fixed Costs

The costs of providing a company’s basic operating capacity are known as fixed cost or capacity cost. For a

cost item to be classified as fixed, it must have a relatively wide span of output where costs are expected to

remain constant. This span is called the relevant range. In other words, fixed costs do not change within a

given time period although volume may change. For our automobile example, the annual insurance

premium, property tax and license fee are fixed costs since they are independent of the number of miles

driven per year. Some typical examples would be building rents, depreciation buildings, machinery and

equipment, and salaries of administrative and production personnel.

Variable Costs

In contrast to fixed operating costs, variable operating costs have a close relationship to the level of volume.

If, for example, volume increases 10%, a total variable cost will also increase by approximately 10%. Gasoline

is a good example of a variable automobile cost, as fuel consumption is directly related to miles driven.

Similarly, the tire replacement cost will also increase as a vehicle is driven more. In a typical manufacturing

environment, direct labor and material costs are major variable costs.

Mixed Costs

Some costs do not fall precisely into either the fixed or the variable category, but contain elements of both.

We refer to these as mixed costs (or semi‐variable costs). In our automobile example, depreciation (loss of

value) is a mixed cost. Some depreciation occurs simply from passage of time, regardless of how many miles

a car is driven, and this represents the fixed portion of depreciation. On the other hand, the more miles an

automobile is driven a year, the faster it losses its market value, and this represents the variable portion of

depreciation. A typical example of a mixed cost in manufacturing is the cost of electric power. Some

components of power consumption, such as lighting are independent of operating volume (e.g. number of

machine‐hours operated).

Sunk Cost

A sunk cost is a past cost that cannot be altered by future action and is therefore irrelevant.

Life‐cycle Cost

Life‐cycle cost is defined as all costs, both nonrecurring and recurring, that occur over the life cycle.

Many systems and products are planned, designed, produced and operated with very little concern of their

life‐cycle cost. In generation, the elements of life‐cycle cost fall into categories that are based on

organizational activity over the life cycle. These are described in the following list:

o Research and development cost: initial planning; market analysis; feasibility studies; product research;

requirements analysis; engineering design; design data and documentation; software; test and evaluation of

engineering models; and associated management functions.

o Production and construction cost: manufacturing; facility construction; process development; production

operations; quality control

of 55

o Operation and support cost: consumer or user operations of the system or product in the field; product

distribution; and sustaining maintenance and logistics support throughout the system or product life cycle

o Retirement and disposal cost: disposal of non repairable items throughout the life cycle; system/product

retirement; material recycling; and applicable logistic support requirements.

• Experience has indicated that a large portion of the total cost for many systems is the direct result of

activities associated with their operation and support.

• Throughout the system/product life cycle, there are many actions required, both technical and non‐

technical.

• The majority of the actions, particularly those at the earlier stages, has life‐cycle implications and definitely

affects life‐cycle cost.

Financial and Economic Evaluation of Projects Financial vs. Economic Analysis

• Both financial and economic analyzes appraise the profit/return of an investment, but from differing

perspectives:

– Financial analysis estimates the profit accruing to the project from the points of view of the

operating entity (the utility and the firm)

– Economic analysis measures economic benefits and returns from the national economic

point of view and assesses the effect of the project on the overall economy of the country.

• Both types of analysis are conducted in monetary terms, but have different definitions of costs and

benefits.

• For a project to be economically viable, it must be financially sustainable, as well as economically

efficient. If a project is not financially sustainable, economic benefits will not be realized. Financial

analysis and economic analysis are therefore two sides of the same coin and complementary.

Financial Evaluation of Projects

Financial evaluation of projects is necessary to:

• evaluate the commercial viability and financial sustainability of a project from the viewpoint of the

project operating entity;

• assess the degree to which a project will generate sufficient revenues to meet its financial

obligations;

• assess incentives for producers;

• take into account all expenditures and revenues (costs and benefits) resulting from the project

Financial evaluation of costs and benefits of large projects

It is carried out on three levels:

• Owner’s evaluation

• Banker’s evaluation

of 55

• Economic evaluation

Owner’s Evaluation:

The owner is concerned only with the return to investment (equity) and thus would be more interested to

know net benefits and their net present value in comparison with the value of the investment.

Banker’s Evaluation:

The banker evaluates the return on the total investment (equity + loans) of the project applied for the loan

and considers its profitability i.e., the banker will consider the NPV of the whole investment and not just the

investors’ equity.

Economic Evaluation:

This includes all the economic costs and benefits i.e., social and environmental/external costs that can be

evaluated .Such evaluation is carried out specifically by development banks and similar institutions and also

by the concerned planning departments in the government.

Project Financial Cost The three main types are:

1. Investment costs:

– Initial Cost (construction and commissioning including land, civil works, equipment and

installations)

– Replacement cost

– Residual values (value of the investment items at the end of the project’s useful life)

2. Operating costs

– Fixed costs: ‐ remuneration and benefits, administrative costs, insurance payments, part of

the maintenance cost, etc.

– Variable cost (depends upon the level of production):‐ fuel and energy, water, lubricants and

part of the maintenance cost, raw materials (for industrial projects)

3. Working capital:

– Capital required for the stock (spare parts ,fuel, raw materials) to ensure continuous

production

4. Financial Evaluation of Projects:

Some Methods

5. Financial Evaluation of Projects:

Some Methods

6. Least Cost Solution

– Present Value method

– Annual Cost Method

7. Measuring worth of the investment

– Internal Rate of Return

– Net Present Value

– Benefit/Cost Ratio

– Payback Period

8. Least Cost Solution:

of 55

– The least cost solution aims at evaluating all realistic alternatives (candidates) financially and

economically before deciding the alternative that can achieve the project benefits at the least cost

i.e., least discounted overall cost over the useful life span of the project or over the specified

planning horizon.

– In the case of energy supply industry, when there are many alternatives for meeting the need for

increased electricity demand, the least cost solution aims at finding out the alternative technical

arrangement that meets the requirement of electrical energy with the least cost to the utility, its site

and timing.

– In choosing the least cost solution we compare the present value of the cost of the alternatives and

or discounted energy cost (US¢ /kWh) depending upon the costs and technical characteristics of the

alternatives.

9. The most popular methods for financial evaluation and comparing alternatives:

Present Value method

– Here all costs and benefits of the project or net benefits (Cash flows) are discounted to the base year.

The net present value (NPV) is nothing but the difference between the discounted benefits and the

discounted costs.

Here, benefits represent the income or the revenue produced by the project (financial flows only)

Example 1:

A company is considering buying an efficient boiler for its manufacturing process. It would cost $ 40,000

immediately but would save $7500 each year for the next 10 years. However at then end of the fifth year a

one time maintenance expense of $ 5000 would be incurred .If the owner’s opportunity cost of capital is

8%, should the company buy the boiler?

Solution:

The machine is a good investment; it produces a discounted savings of $ 6922.2 .

Year Benfits,$ Cost,$ Benfits,$ Cost,$0 0 40000 0.0 40000.01 7500 0 6944.4 0.02 7500 0 6430.0 0.03 7500 5953.7 0.04 7500 0 5512.7 0.05 7500 5,000 5104.4 3402.96 7500 0 4726.3 0.07 7500 0 4376.2 0.08 7500 0 4052.0 0.09 7500 0 3751.9 0.010 7500 0 3474.0 0.0

Total 50325.61 43402.92NPV, $ 6922.7

Discounted Values

0 0(1 ) (1 )

n nt t

t tt t

C B

r r

of 55

Example 3:

Financial evaluation of energy efficient motor over a 20 year life time and at 10% discount factor

Good Motor (GM)

Cost: $700

Input Power : 7.9 kW

Running hours: 1600 h/year

Electricity cost : $ 0.095 / kWh

Energy Efficient Motor (EEM)

Cost: $846

Input Power : 7.5 kW

Running hours: 1600 h/year

Electricity cost : $ 0.095 / kWh

Solution :

The annual electricity cost (AEC) for the two motors is:

GM : 7.9kW x 1600h/yr x $ 0.095/kWh = $1201/yr

EEM: 7.5kW x 1600h/yr x $ 0.095/kWh = $1140/yr

NPV = Discounted benefits – Discounted costs

Here,

discounted benefits = present value of future fuel/energy savings

= ΔAEC x PVF(@10%,20yrs) = (1201‐1140)/yr x 8.5/yr = $518.5

and discounted cost = incremental cost of better product (initial cost)

= ($846‐$700)= $146

Therefore, NPVEEM = $518.5 ‐ $ 146 = $372.5 :> The EEM is the better investment !

Annual Cost method

This is a useful and quick method that gives rapid results and allows the attention of the evaluator to focus

on a few alternatives supposing that certain assumptions and approximations are possible.

Example 4: Comparing alternative projects by evaluating cost of production and prices:

Discount factor ( r )

Cost

Expected life ( n )

Fixed and running cost

Full load operating hrs

Calculation:

Annuity factor (r%, n)

Cost of production =

Combined Cycle Steam Turbine

10%

$ 1000/kW

30 yrs

0.02 $/kWh

8000

9.427

($1000/9.427)/8000+ $0.02 = 3.33

¢/kWh

Combined Cycle Gas Turbine

10%

$ 500/ kW

20 yrs

0.024 $/kWh

7000

8.514

($500/8.514)/7000+ $0.024 = 3.24 ¢/kWh

of 55

Example 5:

Comparison of cost of energy production from Wind Turbine, Photovoltaic ,and small Gas Turbine system

Photovoltaic Wind Turbine Gas Turbine

Owner discount rate

Installation Cost (P)

Capacity (kW)

Capacity factor

Heat rate

Fuel Cost

Variable O&M Cost

Annual O&M

Annual loan payment @

Annual operating hours

The cost of electricity

10%

$ 10,000

3.0

0.25

‐

‐

‐

‐

(6% interest, 20 yr loan)

P x CRF = $ 872/yr

8760 x 0.25 =2190 h

=872/(2190 x 3)

= 13.3 ¢/kWh

10%

$ 2500

0.90

0.385

‐

‐

‐

$100/yr

(7% interest,15 yr loan)

P x CRF = $ 275/yr

8760 x 0.385 = 3373h

(275+100) / (0.9 x 3373)

=12.4 ¢/kWh

10%

$850/kW

0.70

12500 Btu/kWh

$ 7 /106 Btu

$ 0.002/kWh

‐

(10% interest, 20 yr loan)

P x CRF =$99.8/yr

8760 x 0.7 = 6132 h

Fuel + O&M + Fixed Cost

=(12500* 7 /106) +

0.002/kWh+($99.8/6132)

=10.6 ¢/kWh

Measuring worth of the investment for financial decision making :

• The important criterion are :

– Internal rate of return

– NPV of the project

– Benefit/Cost ratio

– Other criteria (pay back, profit/investment ratio ,commercial return on equity capital)

Note:

All the above criteria except for the last, involve discounting

ExamWhat wis this

As a firs

Spreadshvalues)

Since th

RepeatiOCC of cost of

mple 6 onwould be the project desi

The c

st estimate, le

heet could be u

he sum is positi

ing this iteratif 11%, the projborrowing. Th

n IRRIRR ,if the o

rable ?

condition requi

et IRR = 10%,

used to compute

ive, let IRR = 1

ive process yiject will generhis is an attrac

opportunity c

red is :

e IRR more eas

2%, this would

elds an IRR orate more thanctive investme

cost of capita

sily using the sy

d give :

f 17.5%.Sincen enough incoent.

al (OCC) is 11

yntax: IRR(cash

e this exceedsome to cover t

Pag

1%t,

h-flow

the he

ge 45 of 55

of 55

Capital Cost

O&M

Costa

Volume of crude

oilPrice of crude oil

Cost of

crude oilb

Product sales /

revenuecNet

BenefitsYear $ million $ million b/d $/bl $ million $ million $ million

0 6 0 0 0 0 0 -6.01 45 0 0 0 0 0 -45.02 55 0 0 0 0 0 -55.03 52 0 0 0 0 0 -52.04 0 18 500 65.0 10.7 27 -1.95 0 35 2500 66.6 55.0 139 49.56 0 35 2000 68.3 45.1 116 35.97 0 35 2500 70.0 57.7 151 58.18 0 35 2800 71.7 66.3 141 39.29 0 2800 73.5 68.0 146 78.210 0 3000 75.4 74.6 163 88.211 0 3000 77.3 76.5 169 92.912 0 3000 79.2 78.4 176 97.713 0 3000 81.2 80.4 183 102.814 0 3000 83.2 82.4 191 108.115 0 3000 85.3 84.4 198 113.7

PV@10% $119.48 $88.07 $12,341 $302.02 $704.21 $166.08

Notes: IRR 23%

b/d: barrels per daya O&M include the cost of fuel used by the refineryb based on 330 days per yearcbased on the yield of various products (gasoline,jet fuel,gasoil,fuel oil and butane) from the refinery and the market prices of theses products

Financial Analysis of a Petroleum Refining Project

Example 7 (IRR)

Benefit-Cost RatioThis method compares the discounted total benefits of the project to its discounted costs:

0

0

(1 )/

(1 )

nt

ttn

t

tt

Br

B CC

r

Only projects of B/C > 1.0 are adopted. The criteria is useful in capital constraint situation e.g., utility has a lot of feasible projects but limited investment budget. In this case, projects are ranked in accordance with their B/C ratio and are adopted accordingly until their combined costs equal the capital investment budget.

Example 8:

B/C of example-1 = 1.15

of 55

Payback period

It is the time required for a project's total benefits to exceed its total cost. At that time project can be said to

have “paid back” its initial cost.

The most common applications is in the analysis of energy conservation programs.

Example 9:

Energy efficiency retrofit of large building reduces the annual electricity demand for heating and cooling

from 2.3 GWh to 0.8 GWh and the peak demand for power by 150 kW. Electricity costs $ 0.06/kWh and

demand charges are $7/kW‐month.If the project costs $ 500,00,what would be the payback period of the

investment ?

Energy Savings (A)

= (2.3‐0.8) x 106 kWh/yr x $ 0.06/kWh = $ 90,000/yr

Demand Savings (B)

= 150 kW x $ 7/kW‐month x 12 months/yr = $ 12,600/yr

Total Annual Savings: (A ) + (B) = $ 102,600/yr

Simple payback period = Initial Investment / Annual Savings

= $500,000/$102,600/yr

= 4.87 yr.

An Overview of Renewable Energy Project Financing through CDM

Brief Background

Formulation of the United Nations Framework Convention on Climate Change (UNFCCC) in 1992.

UNFCC sets an “ultimate objective” of stabilizing atmospheric concentrations of greenhouse gases at

safe levels.

Categorization of countries into two groups: Annex I Parties i.e. the industrialized countries who have

historically contributed the most to climate change, and non‐Annex I Parties, i.e. developing

countries.

Conference of Parties (COP)

United Nations Framework Convention on Climate Change‐ As the Precursor of Clean Development

Mechanism

Kyoto Protocol

The Protocol was adopted in December 1997.

It creates legally binding obligations on Annex I countries to reduce their emissions of GHGs.

A

in

Th

Kyoto pro

below to t

C

C

de

What ar

En

fl

A

de

In

H

de

What do

Dual Obje

• lo

• w

s of 27 Febru

nstruments of

he protocol e

otocol requir

the level of 1

DM is one of

DM allows

eveloping cou

re the aspe

nables Annex

exible and co

ssist develop

evelopment o

nvestors bene

ost Countries

evelopment.

oes CDM ai

ectives:

ower the over

while also sup

uary 2006, 16

f ratifications

entered into f

es developed

990 by 2012

the Kyoto me

emission re

untries to gen

cts of CDM

x‐1 countries

ost‐effective m

ping countrie

objectives.

efit by obtaini

s benefit in t

im to achie

rall cost of red

porting susta

62 states and

s, accessions,

force on 16 Fe

d countries t

(the first com

echanisms to

duction proj

nerate “certif

M ?

(developed c

manner.

es (non‐Ann

ing Certificat

he form of in

ve?

ducing GHG e

inable develo

d regional eco

approvals or

ebruary 2005

o reduce the

mmitment per

o achieve the

jects that a

fied emission

countries) to

ex I or the

es of Emissio

nvestment, ac

emissions

opment initia

onomic integr

acceptances

5.

eir GHG emis

riod :2008‐20

objective of

ssist in crea

reductions (C

meet their em

host count

ons Reduction

ccess to bette

tives within d

ration organi

.

ssions ( CO2 e

012).

reducing GHG

ating sustain

CER)” for use

mission reduc

ries) in mee

ns (CER).

er technology

developing co

Pag

zations have

equivalent) a

G emissions.

nable develo

by the invest

ction commit

eting their s

y, and local s

ountries.

ge 48 of 55

deposited

t least 5%

opment in

tor.

ments in a

sustainable

sustainable

of 55

Benefits of CDM for developing countries

• Attract foreign investment to countries engaged in the trading of CERs

• Increase the profitability of cleaner and more efficient technology in energy, industry, and transport

sectors.

• Help in waste management operations.

• Contribute to sustainable development of the host country.

Project Sustainability Screening

Establishment of Sustainable Development criteria

Should reflect economic, social, and environmental sustainability dimensions.

The assessment of SD aspect of a project will involve a set of indicators.

A transfer of finances and contribution to sustainable development in the Host Country

Host Country

Flow of Finances

Flow of Credits

Annex I Country

of 55

The indicators should be :

Complete: adequate to indicate the degree to which the overall objective of sustainability has been

met.

Operational: can be used in a meaningful way in the analysis.

Decomposable: the decisions can be broken down into parts involving a smaller no. of indicators.

Non‐redundant: The indicators should be defined to avoid double counting of consequences.

Minimal: It is desirable to keep the set of indicators as small as possible. For instance it may be

possible to combine indicators to reduce the dimensionality of the decision problem.

Distribution of Registered CDM Projects by Host Countries

Distribution of Registered CDM Projects According to the Sectoral Scopes

Others, 16.31%Sri

Lanka, 2.13%Guatemala, 2.13%

Ecuador, 2.13%

Argentina, 2.13%

Republic of Moldova, 2.13%Panama, 2.1

3%

China, 4.26%

Chile, 4.96%

Honduras, 6.38%

Mexico, 9.22%

India, 19.86%

Brazil, 26.24%

Energy Demand, (2.96%)Fugitive emissions from fuels (solid, oil and gas), (1.18%)

Fugitive emissions from production and consumption of halocarbons and sulphur hexafluoride, (2.96%)Waste handling and disposal, (23.67%)

Chemical industries, (1.18%)

Manufacturing industries, (1.78%)

Agriculture, (10.65%)

Energy industries (renewable - / non-renewable sources), (55.62%)

of 55

According to the Sectoral Scopes

CDM Project Requirements

Baseline study for emissions

Emissions additionality & financial additionality

Host country government approval

Meets the sustainable development criteria

Demand for CERs ‐ price of CERs

Additionality and Baseline

Additionality

GHG reduced below what would have occurred without the project activity

Baseline scenario

Representing anthropogenic emissions by sources of GHGs that would occur without the proposed

project activity Additionality and Baseline

Energy in dustries24%

Energy d d9%

Energy distribution2%

Manufacturing industries12%

Chemical industries5%

Fugitive emissions from fuels 6%

Fugit ive emiss ions from halocarbons and sulphur

hexafluorid2%

Waste handling and 25%

Afforestat ion and reforestatio

3% Agriculture

8%

Transport2%

Mining2%

Identify B

Base line/

t0

E0

EC

C

= ton of e

Examples

Impact o

Baseline Em

/CER concept

0 = starting ye

0t =Total emis

CDMt =Total em

ER = E0t – ECDM

equivalent CO

s:

f CERs price o

mission Meth

ts

ear of CDM

ssion without

mission with C

Mt

O2 reduction

on the projec

hodology

CDM project

CDM

(traded in th

cts’ IRR

t

e internationnal market: e.g., 5 $ to 25$

Pag

$/tC )

ge 52 of 55

of 55

The substantial improvement in IRR observed in Biomass and Solid Waste management projects.

Case Study Presentation Financial Analysis of a Small Hydro Project with and without CDM Benefits

Characteristics of the proposed CDM Project

Type : Run‐of‐the river hydro project

Installed Capacity: 3.5 MW

Category : Renewable Energy project (<15 MW)

Load type : National grid

Developer : Private

Country’s Power Development Scenario

• The project is situated in a country where diesel and fuel oil based generation system supply 70% of

electricity and the remaining is supplied by hydro sources.

• In the last 5 years, all the new capacity added is based on diesel and fuel oil.

• LCGEP shows that future additions to power generation capacity will be based on fuel oil or diesel

though some exploitable hydro capacity is available.

• The baseline emission factor is the emission factor of the diesel based generation system of

appropriate capacity and load factor. (Method I)

Estimation of Base‐line CER‐Method I

Estimation of Base‐line CER‐Method II

The emission factor is the weighted average emissions (in kg CO2e/kWh) of all the generation units in the

system. The emission coefficient is calculated as sum of total emission from each of the generation units

divided by the sum of their generation in that year.

Country ProjectIRR without

Carbon finance (%)

IRR with Carbon finance

(%)

Change in IRR

(%)Costa Rica Wind power 9.7 10.6 0.9Jamaica Wind power 17 18 1.0Chile Hydro 9.2 10.4 1.2Costa Rice Hydro 7.1 9.7 2.6Gyana Bagasse 7.2 7.7 0.5Brazil Biomass 8.3 13.5 5.2India Solid Waste 13.8 18.7 4.9

Hydo Power Thermal power plant (Diesel)

Capacity 10 MW Input (1) 181,081 MWh/aOutput 67 GWh/a Emission factor (2) 0.202 tC/MWh

Emissions 0 t CO2/a Emissions (3)=(1)*(2) 36,506 tC/a

Efficiency 0.37Output 67 GWh/a

CER price (4) 10 US$/tC

CERs (3)*(4) 365,059 US$/a

CDM Paramters

of 55

Identification of Additionality of the Proposed CDM Project

• GHG emissions from this project activities is lower than that in the baseline

• Proposed CDM project activities is not a baseline scenario project.

• The project activity is not expected to get implemented in the absence of the CDM due to the

investment barrier and financial analysis is used to demonstrate this:

24

Detail Cash-flow analysiswithout CDM Credit

Detail Cash-flow analysis with CDM Credit

Compute FIRR and FNPV

Compute FIRR and FNPV

Is financiallyattractive ?

Not a CDM ProjectYES

NO

Is financiallyattractive ?

Consideration for CDM Project

YESNOInfeasible project

Financial Analysis

F u e l

N e t G e n e r a t io n

( G W h )

F u e l C o n s u m e d

( 1 0 3 to n n e s ) ( A )

N e t C a lo r i f ic V a lu e ( T J /1 0 3 to n n e s ) - ( B )

C a r b o n E m is s io n *

F a c to r ( I P C C ;

tC /T J ) - ( C )

E m is s io n tC O 2 ( D ) =

( A ) x ( B ) x ( C ) x 4 4 /1 2

C o a l 7 2 5 6 3 5 0 7 7 6 1 6 .2 2 2 5 .8 7 7 9 1 1 3 0 1 L ig n i t e 1 6 3 6 8 1 1 4 5 4 1 6 .2 2 2 7 .6 1 8 8 0 1 3 2 8 G a s 1 8 8 2 6 3 7 4 3 4 3 .3 3 1 5 .3 9 1 1 6 3 7 3 H y d r o 1 6 5 8 7 0 0 N u c le a r 4 1 2 2 0 0 T o ta l 1 2 8 4 6 6 1 0 5 8 2 9 0 0 2

( tC O 2 /M W h )o r tC /M W h

0 .8 2 40 .2 2 5

B a s e l in e e m is s io n f a c t o r

of 55

CER vs. FIRR

Example : Break‐even

0

4

8

12

16

20

24

0 (BaseCase)

3 5 10 15 25

CER, $/tC

FIR

R, %

Break-even CER

MARR

$0

$50,000

$100,000

$150,000

$200,000

$250,000

$300,000

$350,000

0 500 1000 1500 2000 2500

Annual Sales units

Pre

se

nt

Wo

rth

PV Benefits PV Costs

Loss

Profit

Break even- volume

34525895 engineering economics notes

Documents