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