Introduction To Transportation Economic Analysis

Discounting

Importance of Highway Investment Decisions

Resources for public projects are scare What else?

Development Around the west system interchange in West DSM

West DSM 1930s

WEST DSM 1990sWest DSM 2005

Why is it important to make efficient transportation investment choices

Public sector investments leverage much greater private sector investments– Transportation costs (travel time)– Locational decisions

Public Sector investment impact the cost of development

Lincoln, Nebraska - Population 250K

Des Moines, IA, City Population 190,000, Urban Area Population 500,000

Equivalence

For Comparison Purposes– Cash flows through time are made equivalent

through the use of interest rates.

Interest rates and inflation are not the same. Interest is the time value of money

Definition

Constant dollars - real dollars– Constant purchasing power over time

Nominal dollars– Fluctuate in purchasing power overtime

Real discount rates– The time value of money with no inflation premium

Nominal discount rate– The time value of money including an inflation premium

Minimum time increment is one year

Definitions

Interest rate used to borrow money include Time Value of Money Inflation Premium Risk Premium Profit for lender

Inflation – change in purchasing power of money with time

Social discount rate – the societal time value of money

Opportunity Cost – the benefits forgone by using the resource on the next most efficient use

Cash Flow Comparisons – which on is preferable

$10,000

$20,000

Ten Years

$10,000

$1,0

00

$1,0

00

$1,0

00

$1,0

00

$1,0

00

$1,0

00

$1,0

00

$1,0

00

$1,0

00 $11,000

Present Worth

If i = time value Suppose that the value of $100 over 1 year is $4 (no

risk)– Then $100 is equivalent to $104 in 1 year or $100 in one year

is equivalent to $96 today– 4% = I

PW = 100/(1-i)= 96 one year in the future PW = 100/(1-i)2= 92.16

Work problem 1

At 4%, which on is the best deal?

What happens when the interest rate is increased or decreased?

Year Cash Flow 1 Present worth Cash Flow 1 Present worth0 (10,000.00)$ (10,000.00)$ (10,000.00)$ (10,000.00)$ Rate 0.041 -$ -$ 1,000.00$ 960.00$ 2 -$ -$ 1,000.00$ 921.60$ 3 -$ -$ 1,000.00$ 884.74$ 4 -$ -$ 1,000.00$ 849.35$ 5 -$ -$ 1,000.00$ 815.37$ 6 -$ -$ 1,000.00$ 782.76$ 7 -$ -$ 1,000.00$ 751.45$ 8 -$ -$ 1,000.00$ 721.39$ 9 -$ -$ 1,000.00$ 692.53$

10 20,000.00$ 13,296.65$ 11,000.00$ 7,313.16$ 3,296.65$ 4,692.34$

First Problem (assume 4% discount rate)

Present Worth

Equivalence

Interest rates create comparable cash flow Equivalence depends on the interest rate used

Interest Formulas

Definition Simple interest - interest is accumulated on

the principal but not on the interest.

Suppose $100 is borrowed at 10% per year simple interest for two years – the loan is repaid in two years with 100 + 10 + 10 = $120

Definitions

Compound interest – interest is paid on the interest.

Suppose you were loaned $100 at 10% per year compounded annually – at the end of two year, you would be owed

100 + 10 + 10*0.1 + 10 = $121

100*(1+i)^N = F

Definitions

Nominal versus effective interest– Nominal is the arithmetic sum of interest charged at

a rate less than one year – Nominal interest rate of 18% compounded monthly

is 1.5% per month.

$100 at a 18% nominal rate compounded month is

100*(1+0.015)^12 = $119.56– The effective rate is 19.56% per year

Formulas

nn iPFiF

i

)1( )1(/ P

period theof end

at theamount uniform a Represents A

money of sum future a Represents F

money of sumpresent a Represents P

years)(usually periods ofNumber n

periodper RateInterest

Formulas Continued

n

n

n

n

n

n

ii

iAP

ii

iAF

ii

iPA

i

iFA

)1(

1)1(

)1(

1)1(

1)1(

1)1(

Discounting problem

Example: an operator of a taxi cab company has the option of purchasing two types of vehicles. On type is a cheaper model and has a shorter expected life while the other is more expensive and has a longer expected life. The required information is listed below: (project level analysis)

Type 1 Type IIFirst cost = $600,000 First costs = $800,000O&M = $100,000/year O&M = $80,000Life = 3 years Life = 5 yearsSalvage = $40,000 Salvage = $50,000

Project with dissimilar lives

Present worth analysis – pick a minimum common multiple of lives

Present worth analysis – end at common year and calculate residual value

Estimate the annual uniform equivalent cost – and compare one year

Present worth Comparison

At an interest rate of 10%, which alternative is the most cost effective?First pick a the least common multiple of years for comparison – 15 year

$600,000 $600,000

$100,000 $100,000 $100,000 $100,000 $100,000

$100,000

$40,000 $40,000 $40,000 $40,000$40,000

$600,000 $600,000 $600,000

$100,000 $100,000 $100,000

$80,000 $80,000 $80,000

$800,000$800,000 $800,000

$80,000 $80,000

$50,000 $50,000 $50,000

Work Problem 2

Sensitivity Analysis Questions

At what interest rate are the present worths about the same

Why did you have to change the interest rate upwards or downward to make they equivalent?

Perform Sensitivity analysis on problem 2

Break Even Analysis

How high does the interest rate have to go before Type I is preferred option

Break even analysis

-$500,000.00

-$400,000.00

-$300,000.00

-$200,000.00

-$100,000.00

$0.00

$100,000.00

0% 10% 20% 30% 40% 50% 60% 70% 80%

Interest Rate

PW

of

Typ

e 1

- T

ype

II

Break even point

Why do high interest rates favor the low cost option?

Redo the analysis using a uniform annual cash flow

i

i

iPA

n 1)1(

Gradient Are Commonly Used in Transportation evaluation problems

Gradient can be expressed as a percentage– Truck traffic is expected to grow at a rate of 2

percent per year on I-80

Gradient can be expressed as a fixed growth rate– Truck traffic on I-80 is expected grow by 100 vehicle

per year

Percentage Gradient

If the widening of I-80 is expected to save the average truck traveling from boarder to boarder 15 minute and there are an average truck per day of 12,000, how many truck minutes will be saved over the next 20 years, assume an average annual increase in truck traffic of 2 percent per year.

Do problem 4

Economic Evaluation And Transportation System Development

Very seldom do we develop more than an increment change to the system– Since travel is possible between most any two

points, improvements simply reduce the cost of transportation

– Typical benefits Reduced travel time Reduced crash morbidity and mortality Reduced environmental degradation

Example – Six lane I-80

I will cost approximately $ 3 Billion to Six lane all of rural I-80.

Do the trucker travel time saving warrant the construction. – Assume a travel time cost of $30 per hour– Assume a 4 percent social discount rate

Do problem 5

Transportation Benefits

Direct benefits– Travel time savings– Safety saving– Trip reliability improvement– Trip quality improvement (e.g., smooth pavement,

quieter, more esthetically pleasing, etc.) Indirect benefits

– Economic development– Property value increases

Transportation benefits

Reduced externalities Externality are costs that no one pays

– Air pollution– Noise pollution – Reduced wild life

Transportation Benefits

Transfers– If a fast food restaurant locates at the new

interchange on I-35 have jobs just been created or are they transferred from somewhere else.

– Was economic development created because of the transportation improvement or just transferred from somewhere else.

Uncertainty and the future

Transportation Facility are typically long lived facilities– Great deal of uncertainty in traffic forecasts– Great deal of uncertainty in changes in technology– Great deal of uncertainty in sustainability of current

patterns

Ways of dealing with uncertainty

Typical discounting of future benefits and costs weights distant less heavily

Evaluate several likely scenarios– Under risk – where the likelihood of an outcome can

be assess (probability of an outcome), use the expected value

– Under uncertainty – where the likelihood of an outcome is unknown, us a decision rule like Max-Max of Mini-Max

Expected Value

Suppose you have to invest in one location and you two project alternatives to select between.

High Demand p = 0.5

Benefit1,900,000

Low Demand p= 0.5

-1,000,000

HighRisk Location

Low RiskLocation

High Demand P = 0.7

500,000

Low Demand p = 0.3

200,000

Treatment of Risk

Expected Value High Risk = 0.5(1,900,000) + 0.5(-1,000,000) = $450,000

Expected Value Low Risk = 0.7(500,000) + 0.3(200,000) = $410,000

However, people and entities are know to be risk adverse. Expected value assumes that risk is linear.

Risk Adversity

Certain Money Equivalent (subtracting out risk premium)

p

1

0

Expected Value

CME

$

Expected Monetaryvalue

Certain moneyequivalent

Decision Under Uncertainty

Maximin and Maximax Rules– Maximin rule – select an alternative on the basis of

comparing the lowest possible returns of each of the alternatives and choosing that alternative which has a minimum return larger than the minimums of the others.

Decision Making under Uncertainty

Maximax Rule – Select the alterative having the largest possible return.

Example: Suppose there are four alternatives with four possible outcomes (e.g., high, high medium, low medium, and low demand)

Returns from alternatives

Alternatives

Event A0 A1 A2 A3 A4

E1 $0 $5.84 $6.60 $6.50 $5.70

E2 $0 $7.15 $6.80 $6.65 $6.40

E3 $0 $7.40 $7.45 $7.80 $7.65

E4 $0 $9.00 $8.30 $8.75 $9.15

Decision making under Uncertainty

Maximin Decisions A0 A1 A2 A3 A4

$0 $5.84 $6.60 $6.50 $5.70

Maximax Decisions A0 A1 A2 A3 A4

$0 $9.00 $8.30 $8.75 $9.15