Bài giảng nhập môn quản lý xây dựng

– All costs and benefits occur within a brief period of time within 1 year – No investment of capital is required, or differences are only in annual costs/revenues – Long-term costs

PUBLIC WORKS ECONOMICS

Department of Construction Management - Faculty of

Economic and Management

A bottling company installed a conveyor system 10

years ago at the cost of $20,000 The company is

considering replacing the system because maintenance costs have been increasing and new technology is

available that will be less costly to

operate and provide greater

performance

You have been assigned to

analyze both systems and

recommend the most

economical decision – retain

the old system or replace it

with the new system

A petroleum company needs to paint the vessels and

pipes in its refinery periodically to prevent rust

“

• Tuff-Coat, a durable paint, can be purchased for $8.05 a gallon

• Quick-Cover, a less durable paint, costs $3.25 a gallon

Both paints are equally easy to apply and will cover the same area per gallon Quick-Cover is expected to last 5 years

How long must Coat last to justify

Tuff-its use?

You are the city engineer for a small but

growing community and are applying for a

grant to help fund an upgrade to the

wastewater treatment plant The new design is based on population growth over the next 20 years

The grant requires

You and your spouse have just had a beautiful baby boy

You want to start saving for his college

college, how much should you save each month to ensure you will have enough to fund his education in 18 years?

You have found your dream home, but you

need to decide how to finance the purchase Should you obtain a

• 30-year fixed rate mortgage?

• 15-year fixed rate mortgage?

• A creative financing option that your uncle is

offering that requires you pay no interest for the first

5 years of the loan?

You have saved $20,000 in your

401K Should you withdraw that

money and use it as a down

payment or keep it invested?

Course Objectives

• Be able to prepare economic justifications

for engineering proposals

• Be able to manage personal finances and

make wise investment decisions

Course Objectives (cont.)

• Be able to evaluate economic justifications

performed by others

• Be comfortable using financial language

• Be aware of tradeoffs involving expenses

and capital costs under tax and inflation

conditions

• Be familiar with advanced techniques of

modeling decision problems under risk and uncertainty conditions

What is Engineering Economic

Analysis?

• Definition: applying economic analysis

techniques to compare alternative engineering projects

• Other names

– Engineering economy

– Economic decision analysis

– Capital investment analysis

• Cash flow analysis

- $

Engineering Economic Analysis

• Engineers must

– understand how money works in order to make

informed design decisions

– make decisions over the entire product life cycle

(cradle to)

– consider the economics of the design in order to

“sell” it to management / public

• Determine if a project is “economically

viable” grave

• Choose the “best” project between multiple

alternatives

What makes Engineering Economic Decisions hard?

• Uncertainty/risk

• Multiple, conflicting objectives

– Maximize quality while minimizing cost

– Environmental impacts

– Safety

– How do you quantify these things?

Why is time an important factor

in economic analysis?

• Many investment decisions have a life cycle

of several (or many) years

• Would you rather choose a project that has a

large up-front cost and low annual cost, or a project with low up-front costs and higher

annual costs?

– Answer: It depends

• It is useful to compare alternative projects at

a common point in time, therefore we must consider the time value of money

Time Value of Money

• Would you rather receive $1000 today or

$1000 a year from today?

• Would you rather receive $1000 today or

$1050 a year from today?

• Would you rather receive $1000 today or

$1500 a year from today?

• Would you rather receive $1000 today or

$2000 a year from today?

Time Value of Money

• The value of a sum of money is dependent

not only on the amount, but also on the time

at which the money is received

• This is because money has “earning power,”

or opportunity cost

Time Value of Money is represented with interest rates

Opportunity Cost

• Receiving cash now, rather than in the

future, allows you to earn interest on that money until that point in the future

– Alternatively, you need to be compensated for

waiting by receiving a larger sum (with interest)

Time Value of Money

• When would the time value of money not be

important to your decision?

– All costs and benefits occur within a brief period

of time (within 1 year)

– No investment of capital is required, or

differences are only in annual costs/revenues

– Long-term costs of the alternatives are the same

Economic Justification Seven Questions to Answer

1 What investment alternatives are available?

2 What is the length of time over which the

decision is to be made?

3 What TVOM will be used to move monies

forward and/or backward in time?

4 What are the best estimates of the cash

flows for each alternative?

5 Which investment alternative seems best,

based on the economic criterion chosen?

6 How sensitive is the decision to changes or

errors in the estimates used in the analysis?

7 Which investment is recommended?

SEAT

Systematic Economic Analysis Technique

A Seven-Step Procedure

In this class, we will most often be comparing

mutually exclusive alternatives

Chapter 1 Basic Concepts

•INTEREST

•Interest is a fee that is charged for the use of

someone else's money The size of the fee will

depend upon the total amount of money borrowed and the length of time over which it is borrowed

•Example 1.1 An engineer wishes to borrow $20

000 in order to start his own business A bank will lend him the money provided he agrees to repay

$920 per month for two years How much interest

is he being charged?

The total amount of money that will be paid to the bank is 24 x $920 =

$22 080 Since the original loan is only $20 000, the amount of interest is

$22 080 - $20 000 = $2080

•Whenever money is borrowed or invested, one

party acts as the lender and another party as the

borrower The lender is the owner of the money, and the borrower pays interest to the lender for the use of the lender's money

•For example, when money is deposited in a

savings account, the depositor is the lender and

the bank is the borrower The bank therefore pays interest for the use of the depositor's money

•(The bank will then assume the role of the lender,

by loaning this money to another borrower, at a

higher interest rate.)

INTEREST RATE

If a given amount of money is borrowed for a

specified period of time (typically, one year), a certain percentage of the money is charged as

interest This percentage is called the interest

rate

Example 1.2 (a) A student deposits $1000 in a

savings account that pays interest at the rate of 6% per year How much money will the student have after one year?

(b) An investor makes a loan of $5000, to be repaid

in one lump sum at the end of one year What

annual interest rate corresponds to a lump-sum payment of $5425?

The student will have his original $1000, plus an

interest payment of 0.06 x $1000 = $60 Thus, the student will have accumulated a total of $1060

after one year (Notice that the interest rate is

expressed as a decimal when carrying out the

calculation.)

The total amount of interest paid is $5425 - $5000 =

$425 Hence the annual interest rate is

(425/5000)%=8.5%

SIMPLE INTEREST

Simple interest is defined as a fixed percentage of the principal (the

amount of money borrowed), multiplied by the life of the loan Thus,

I = niP (1.1)

where

I = total amount of simple interest

n = life of the loan

i = interest rate (expressed as a decimal)

Example A student borrows $3000 from his uncle in order to finish school His uncle agrees to charge him simple interest

at the rate of 5.5% per year Suppose the student waits two years and then repays the entire loan How much will he have

to repay?

By (1.2) F = $3000[1 + (2)(0.055)] = $3330

COMPOUND INTEREST

When interest is compounded, the total time period is subdivided into several

interest periods (e.g., one year, three months, one month) Interest is credited

at the end of each interest period, and is allowed to accumulate from one

interest period to the next

F = P(1 + i) n (1.3)

Year 0: F0 = P0Year 1: F1 = P0 + I1 = P0 + P0.i = P0(1+i) Year 2: F2 = F1+F1.i = F1(1+i) = P0(1+i) 2

During a given interest period, the current interest is determined as

a percentage of the total amount owed (i.e., the principal plus the previously accumulated interest) Thus, for the first interest period,

the interest is determined as I 1 = iP

and the total amount accumulated is F1 = P + I1 = P + iP = P(l + i)

For the second interest period, the interest is determined as

I2=iF1= i (1+i)P

and the total amount accumulated is

F 2 = P + I1 + I 2 = P(1+i) 2

For the third interest period, I 3 =i(l + i) 2 P F3=P(1 + i)3

and so on In general, if there are n interest periods, we have

(dropping the subscript): F = P(l + i)n (1.3)

which is the so-called law of compound interest Notice that F, the

total amount of money accumulated, increases exponentially with

h, the time measured in interest periods

Example A student deposits $1000 in a savings account that

pays interest at the rate of 6% per year, compounded

annually If all of the money is allowed to accumulate, how much will the student have after 12 years? Compare this with the amount that would have accumulated if simple

interest had been paid

F = $1000(1 + 0.06)12 = $2012.20

Thus, the student's original investment will have more than doubled over the 12-year period

If simple interest had been paid, the total amount that would

have accumulated is determined by (1.2) as

F = $1000[1 + (12)(0.06)] = $1720.00

THE TIME VALUE OF MONEY

Example 1.5 A student who will inherit $5000 in three years has a savings account that pays 5.5% per year, compounded annually What is the present worth of the student's inheritance?

THE TIME VALUE OF MONEY

Equation (1.3) may be solved for P, given the value of F:

P = $ 4258 07

) 055

0 1 (

5000

$ )

• Interest Rate (i)

– Rate charged for use of funds (assume to be

annual rate unless otherwise specified)

• Life of Loan/Time Period (n)

– It is assumed that i and n have same units of time

• Number of compounding periods /year (m)

• Annual Amount (A)

• Uniform Gradient Amount (G)

Terms

Types of Cash Flows

• First cost (capital cost, initial investment) = expense to

build or to buy and install

• Operations and maintenance (O&M) = annual

expense, such as electricity, labor and minor repairs

• Salvage value = receipt at project termination for sale or

transfer of the equipment (can be a salvage cost, e.g., disposal cost)

• Revenues = annual receipts due from sale of products

or services

• Overhaul = major capital expenditure that occurs during

the asset’s life

Cash Flow Analysis

• Cash Flow Tables

– Typically, benefits/revenues are positive and costs/expenses

are negative

– Best handled in spreadsheets

• Cash Flow Diagrams

End of Year

Capital Cost

O&M Costs Revenue

Net Cash Flow

Cash Flow Analysis

• Cash Flow Diagrams

– Horizontal line divided into period time units

– Vertical arrows represent cash flows; up for benefits/revenues

and down for costs/expenses

– Similar to a force diagram for a beam

• Rules

– Cash flows cannot be added/subtracted unless they occur at

the same point(s) in time

– Cash flows occurring during a period are counted at the end of

the period

– Only capital cost occur in the beginning of period

0 1 2 3 4 5

Example

On December 1, Al Smith purchased a car for $18,500 He paid

$5000 immediately and agreed to pay three additional payments of

$6000 each at the end of 1, 2 and 3 years Maintenance for the car

is projected at $1000 at the end of the first year and $2000 at the end of each subsequent year Al expects to sell the car at the end

of the fourth year (after paying for the maintenance work) for

$7000 Using these facts, prepare a table of cash flows and a cash flow diagram

Example

Consider a mechanical device that will cost $20,000 when

purchased Maintenance will cost $1000 each year The device will generate revenues of $5000 each year for five years, after which the salvage value is expected to be $7000 Draw the cash flow diagram

• Graphic description of cash flow

– Horizontal (time) axis is marked off in equal

increments, one per period, up to the duration of the project

– Shows receipts (income) – disbursements (costs)

• Receipts (income) represented by upward pointed

arrows

• Expenditures represented by downward pointed arrows

• Arrow length proportional to the magnitude of cash flow

– Two or more transfers in the same period are

combined (Simplified Cash Flow Diagram)

Cash Flow Diagrams

Simplified Cash Flow Diagram

Example

• A mechanical device costs $20,000

when purchased Maintenance is

$1000/year The device will generate

revenues of $5000/yr for five years,

after which the salvage value is

expected to be $7000

Cash Flow

• Three types of analysis

– Single payment cash flow – Uniform series cash flow

– Gradient series cash flow

Cash Flow

• Single Payment cash flow

– Can occur at:

• Beginning of time line (t=0)

• End of time line (t=n)

• Any time in between

i = 8%

n = 1 yr

F = $1,080

Cash Flow

• Uniform series cash flow

– Series of equal transactions starting at t=1

and ends at t=n

– The symbol A (an annual amount) is given

to the magnitude of each individual cash flow

t=0

t=1

t=n

A each

Cash Flow

• Geometric series cash flow

– Series of increasing or decreasing

transactions

– The symbol G is given to the uniform

gradient amount

Arithmetic Gradient combined

with a Uniform Series

• Decompose the cash flows into a uniform

series and a pure gradient Then add or

subtract the Present Value of the gradient to the Present Value of the Uniform series

Example of Cash Flow

Scenarios

• If $1,000 is borrowed for a four year period (n=4) at

8% the borrower and lender might select one of the following four alternative plans for repayment

1 Pays interest only on the $1,000 loan at the end of each

year At the end of four years the loan is repaid with the interest owed for that year

$1,080

$80 $80 $80

Cash Flow

• If $1,000 is borrowed for a four year period (n=4) at

8% the borrower and lender might select one of the following four alternative plans for repayment

2 Pays equal fractions of principal each year (one-fourth or $250 in

this case) plus interest on the amount of principal owed during the year

$270

$330 $310 $290

Cash Flow

• If $1,000 is borrowed for a four year period (n=4) at

8% the borrower and lender might select one of the following four alternative plans for repayment

3 Pays equal annual amounts as called for in most standard

installment loan contracts

$301.9

$301.9 $301.9 $301.9 P=$1,000

Cash Flow