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```					Cost and Benefit

Avinash Kishore
(avinash.kishore@gmail.com)

Based on notes from Andrew Foss

Economics 1661 / API-135

February 11, 2011
Review Section
Agenda

 Basic Econometrics
– Bi-variate regression
– Multivariate linear regression
– Special cases

 Elasticity
 Social Welfare Cost
 Benefits
– Aggregation of Demand Curves
– Travelling Cost Method

Note: A good general reference on costs and
benefits is EPA, Guidelines for Preparing Economic
Analyses, September 2000 (link), Chapters 7 and 8    1
Measures of Central Tendency and Dispersion

 Measures of Central Tendency: Mean
– Mean = ∑Xi /n
– Mean takes all data points into account
– Mean is sensitive to outliers. Outliers have a lot of weight on
mean

 Measures of Dispersion
– Variance = (Xi – mean)2/N
– Standard Deviation =[(Xi – mu)2/N]1/2 = (Variance)1/2

 More dispersed data have higher variance
– Like mean, standard deviation is also sensitive to outliers

2
It is possible that two datasets have the same mean
but different standard deviations

From Wikipedia
3
OLS regression (2 Variables)       http://www.ats.ucla.edu/stat/Stata/examples/rwg/rwgstata2/rwgstata2.htm

Y (wateruse) = 1201.124 + 47.54*income
4
Bi-variate Regression Model

 Yi = β0 + β1Xi + εi
– i = each observation; Y = Dependent Variable (water use); X =
Independent Variable (income); εi = Error term
– β0 = intercept. It tells us the predicted value of Y when X = 0.
– β1 = The coefficient that tells us how Y changes for unit
change in X.
   water81       Coef.   Std. Err.     t     P>t     [95% Conf. Interval]

   income    47.54869    4.652286    10.22   0.000   38.40798   56.6894
   _cons     1201.124    123.3245    9.74    0.000   958.8191   1443.43

– N = 496; R-square = 0.1745

 3 Sources of Error Term:
– Omitted variables, Measurement error and Chance or
Randomness                                                            5
Multiple Regression

 More than one independent variables
– Yi = β0 + β1Xi1 + β2Xi2 + β3Xi3 + εi

 Now, β1 is the change in value of Y for a unit change
in X1 while holding constant (or controlling for) X2 and
X3 (the marginal interpretation)
 Example:. reg water81 income educat peop81
       water81   Coef.         Std. Err.           t   P>t     [95% Conf.   Interval]

       income    32.05796    4.285087       7.48       0.000   23.63864     40.47729
       educat    -42.61099    17.23512      -2.47      0.014   -76.4745     -8.747484
       peop81    480.5194    31.73071      15.14       0.000   418.175      542.8639
       _cons     678.8889     245.195       2.77       0.006   197.1305     1160.647

N = 496, R-squared = 0.38

6
The Coefficient and the Confidence Interval                                                                         (precision)

Student-t Distribution
0.1                         μ = 32.06, σ = 4.28
0.09
Probability Density

0.08

0.07

0.06

0.05

0.04

0.03

0.02

0.01

0
14.88   18.304   21.728    25.152   28.576   32   35.424   38.848   42.272   45.696   49.12

7
Some Special Cases: binary independent variables

   Normally, continuous dependent and independent variables in OLS.
   But we can also have binary independents. Also Dummy Variables.
– Βeta coefficient of a dummy variable is not interpreted as its slope.
   Let there be a dummy variable Female s.t. Female = 1 for female
employees; 0 for male employees.
   Regression equation: Earnings = β0 + β1*Female + εi
– Earnings = 16.99 – 3.45*Female + εi
   Interpretation: Constant, β0 = Mean earning of men = 16.99
β1 = Difference in mean earnings of men and women. Here, β1 = -3.45. So,
mean earning of women = 16.99 – 3.45 = 13.44.
   Multivariate Example:
–   Earnings = β0 + β1 Years of Education + β2Female + εi

– Earnings = 15.55 + 3.00* Years of Education –3.55*Female

 Show graphically                                                             8
Special Cases: Interaction Terms

 It is possible that both intercepts and slopes (returns per year of
education in this case) are different for two groups.
 e.g. Earnning = β0 + β1Years of Education + β2Female + β3
Female*Years of Education + εi
 So, for females: Earnings = β0 + β1Years of Education + β2*1 +
β3*1*Years of Education+ εi
 = (β0 + β2) + (β1+ β3)*Years of Education+ εi
 for males: Earnings = β0 + β1Years of Education + β2*0 +
β3*0* Years of Education + εi
                      = β0 + β1Years of Education + εi
 So, for females, Slope = (β1+ β3) and Intercept = β0 + β2
 For males, Slope = β1 and intercept = β0
9
We can also capture non-linear relations through
linear regression

 Quadratic: If the dependent variable is a parabolic
function of the independent, we can still capture the
relationship by adding a squared term, e.g.
Environmental Kuznets Curve hypothesis:
– Pollution = β0 + β1GDPPC + β2 (GDPPC)2 + εi

 Categorical Dependent Variable (as in RUM)
– We get estimates of probability
– OLS is not appropriate in such cases
– Logit or probit models

10
Cost and Benefits
Costs Estimation Methods and Elasticity

 Direct Compliance Cost Method
 Partial Equilibrium Analysis (behavioral response)
 General Equilibrium Analysis
 Want to know how consumers and firms will react to
changes in prices for the good/service being regulated
 Depends on price elasticity of supply and demand
 Elasticity = % ΔQ/% ΔP = (ΔQ/Q)/ (ΔP/P) =
(ΔQ/ΔP)*(P/Q) = (1/Slope)*(P/Q)
 Higher the elasticity, greater the behavioral response
to regulation and greater the social welfare cost.
12
If demand is very responsive to change in price,
good is price elastic. If demand does not respond
much to price changes, demand is price inelastic

Elasticity =   ∞               Elasticity = 0

13
Social Welfare Costs

 Illustration of social welfare costs
– Losses in consumer and producer surplus from increased
marginal cost across regulated industry

Welfare Effects of Industry-Wide Increase in Marginal Cost
Price                                Price

D                                    D           MC1 = S1
CS1
CS0             MC0 = S0                               MC0 = S0
P1
P0                                   P0    PS1
PS0

Q0     Quantity                      Q1 Q0    Quantity   14
Benefits: Aggregation of Demand Curves (Private Goods)
Let’s say Q is a private good. Person 1 demand for Q is: Q1 = 100 – P
Person 2 demand for Q is : Q2 = 100 – P
So, what is the aggregate market demand for Q?
120                                                              120

100                                                              100

80                                                              80

60                                                              60

40                                                              40

20                                                              20

0                                                                0
0      20     40      60       80        100         120         0   20    40     60      80      100    120
120
Algebraically, it is
100
QT = 200 – 2P
80

60                                                           Note: The algebra and
the graph won’t be this
40
simple if demand
20
functions are different

0
0        50         100          150      200    250                          15
Benefits: Aggregation of Demand Curves (Public Goods)
Now assume Q is a public good (non-rivalrous and non-excludable)

Does that change how we aggregate demand curves?
120                                              120

100                                              100

80                                               80

60                                               60

40                                               40

20                                               20

0                                                0
0    20   40   60         80   100   120         0        20     40    60     80     100     120
250

200
Algebraically, it is
150
P = 200 – 2Q
100
You work with inverse
50                                                         demand curve

0
0        20     40   60        80        100    120                           16
One more example: public park provision

 There are 500 people in a locality
 Each has demand function: Q = 100 – P
– P = \$ price per unit area of park people are willing to pay for Q
sq. yards of the park preserved
– MC = \$ 10,000/sq. yard

 How many sq yard of park area should be preserved?
– P = 100 – Q
– Aggregate demand: P =500[100 – Q]
– or P = 50,000 – 500Q

 50,000 – 500Q = 10,000, Q* = 80 sq. yards
 What is the total benefit @ Q*? Net benefit = ?
17
Travel Cost Method:
Example Problem

 Ruritania is a country with three cities and a beautiful
park at its center
 Environmental economists in Ruritania have collected
the following data on park visitors from the three cities
Origin        Population   Visitors                  Travel Cost
1             100,000      15,000                    \$12.50
2             750,000      75,000                    \$25.00
3             1,000,000    50,000                    \$37.50

 The environmental economists want to estimate the
recreational value (i.e., non-market use value) of the
park to the people of Ruritania
18
Travel Cost Method:
Example Problem

 First calculate the visitation rate for each origin
Origin         Population   Visitors     Visit Rate    Travel Cost
1              100,000      15,000       0.15          \$12.50
2              750,000      75,000       0.10          \$25.00
3              1,000,000    50,000       0.05          \$37.50

19
Travel Cost Method:
Example Problem

 Next plot the relationship between visitation rate and
travel cost and express it in an equation

\$50
Origin 3
\$40
Travel Cost (TC)

TC = - (50 / 0.20) * R + 50
Origin 2
\$30
= -250 * R + 50
\$20
Origin 1     or
\$10
R = - (0.20 / 50) * TC + 0.2
\$0
= -0.004 * TC + 0.2
0.00     0.05          0.10           0.15         0.20
Visitation Rate (R)
20
Travel Cost Method:
Example Problem

 Suppose a fee were charged to enter the park
 Calculate the relationship between Origin 1’s visitation
rate and the hypothetical fee

\$40
Hypothetical Fee

\$30                                                  R = -0.004 * (TC + Fee) + 0.2

\$20                                                  TC1 = 12.5

\$10                                                  R1 = -0.004 * (12.5 + Fee) + 0.2

\$0                                                       = -0.004 * Fee + 0.15
0.00   0.05          0.10           0.15   0.20
or
Visitation Rate (R)
Fee = -250 * R1 + 37.5
Origin 1                                                        21
Travel Cost Method:
Example Problem

 Calculate the relationship between Origin 2’s visitation
rate and the hypothetical fee

\$40
Hypothetical Fee

\$30                                                    R = -0.004 * (TC + Fee) + 0.2

\$20                                                    TC2 = 25

\$10                                                    R2 = -0.004 * (25 + Fee) + 0.2

\$0                                                         = -0.004 * Fee + 0.10
0.00     0.05          0.10           0.15   0.20
or
Visitation Rate (R)
Fee = -250 * R2 + 25
Origin 2                                                         22
Travel Cost Method:
Example Problem

 Calculate the relationship between Origin 3’s visitation
rate and the hypothetical fee

\$40
Hypothetical Fee

\$30                                                    R = -0.004 * (TC + Fee) + 0.2

\$20                                                    TC3 = 37.5

\$10                                                    R3 = -0.004 * (37.5 + Fee) + 0.2

\$0                                                         = -0.004 * Fee + 0.05
0.00     0.05          0.10           0.15   0.20
or
Visitation Rate (R)
Fee = -250 * R3 + 12.5
Origin 3                                                        23
Travel Cost Method:
Example Problem

 Calculate the relationship between all three origins’
visitation rate and the hypothetical fee

\$40
Hypothetical Fee

\$30                                                      R = -0.004 * (TC + Fee) + 0.2

\$20                                                      R1 = -0.004 * Fee + 0.15

\$10                                                      R2 = -0.004 * Fee + 0.10

\$0                                                      R3 = -0.004 * Fee + 0.05
0.00       0.05          0.10           0.15   0.20
Visitation Rate (R)

Origin 1       Origin 2      Origin 3                                          24
Travel Cost Method:
Example Problem

 Calculate the relationship between the number of
visitors from Origin 1 and the hypothetical fee

\$40
Per capita demand function:
Hypothetical Fee

\$30                                                      R1 = -0.004 * Fee + 0.15

\$20                                                      Population = 100,000

\$10                                                      So, demand function is:
= 100,000*(-0.004 * Fee + 0.15)
\$0
0     30,000   60,000   90,000   120,000 150,000   = -400 * Fee + 15,000
Visitors (Q)

Origin 1

25
Travel Cost Method:
Example Problem

 Calculate the relationship between the number of visitors
from Origin 2 and the hypothetical fee

Per capita demand function:
\$40                                                    R1 = -0.004 * Fee + 0.10
Hypothetical Fee

\$30                                                    Population = 75,000
\$20
So, total demand function is:
\$10
= 75,000*(-0.004 * Fee + 0.10)

\$0                                                    Q2 = -3,000 * Fee + 75,000
0   30,000   60,000   90,000   120,000 150,000
Visitors (Q)

Origin 2                                                       26
Travel Cost Method:
Example Problem

 Calculate the relationship between the number of
visitors from Origin 3 and the hypothetical fee

\$40
Per capita demand function:
Hypothetical Fee

\$30                                                      R1 = -0.004 * Fee + 0.05
\$20
Population = 50,000
\$10
So, total demand is:
\$0                                                      = 50,000*(-0.004 * Fee + 0.05)
0     30,000   60,000   90,000   120,000 150,000   Q3 = -4,000 * Fee + 50,000
Visitors (Q)

Origin 3
27
Travel Cost Method:
Example Problem
– Aggregate curve (from horizontal summation) represents total
recreational demand as a function of hypothetic park fee
– 1st kink @ Fee = \$25, Qagg = 5000
– 2nd kink @ Fee = \$ 12.50, Qagg = 10,000 + 37,500 = 47,500
– X-intercept = 15,000 + 75,000 + 50,000 = 140,000

\$40
–
Hypothetical Fee

Q1 = -400 * Fee + 15,000
\$30
Q2 = -3,000 * Fee + 75,000
\$20
Q3 = -4,000 * Fee + 50,000
\$10

\$0
0       30,000   60,000   90,000     120,000 150,000
Visitors (Q)

Origin 1        Origin 2      Origin 3      Aggregate                                   28
Travel Cost Method:
Example Problem

 Calculate recreational value as area under the
aggregate curve
– The recreational value of the park is \$1,531,250

\$40
Hypothetical Fee

Area A
\$30                                                        Area A = ½ * (37.5 - 25) * 5,000
Area B                                              = 31,250
\$20
Area C                           Area B = ½ * (25 - 12.5) *
\$10                                                                 (5,000 + 47,500)
= 328,125
\$0
0      30,000    60,000   90,000   120,000 150,000   Area C = ½ * (12.5 - 0) *
(47,500 + 140,000)
Visitors (Q)                             = 1,171,875

Aggregate                         Total Area = 1,531,250         29
Travel Cost Method:
Example Problem

 Calculate recreational value as area under each of the
origin-specific demand curves (alternative method)
– The recreational value of the park is \$1,531,250

\$40
Hypothetical Fee

\$30                                                          Area 1 = ½ * 37.5 * 15,000
Area 1                                                = 281,250
\$20
Area 2
Area 2 = ½ * 25 * 75,000
Area 3
\$10                                                                 = 937,500

\$0                                                          Area 3 = ½ * 12.5 * 50,000
0      30,000     60,000   90,000    120,000 150,000          = 312,500

Visitors (Q)                       Total Area = 1,531,250

Origin 1      Origin 2       Origin 3                                        30
Travel Cost Method:
Example Problem

 Calculate the number of visitors and consumer surplus
if no fee were charged to enter the park
– Total park visitation is 15,000 + 75,000 + 50,000 = 140,000
– CS is equal to recreational value: \$1,531,250

31
Travel Cost Method:
Example Problem

 Calculate the number of visitors and consumer surplus
if a fee of \$20 were charged to enter the park
– Total park visitation is 7,000 (Or. 1) + 15,000 (Or. 2) = 22,000
– CS is area below demand curve and above price: \$98,750
\$40
Hypothetical Fee

Area A
\$30                                                       Area A = ½ * (37.5 - 25) * 5,000
Area D                                              = 31,250
Fee = \$20
\$20
Area D = ½ * (25 - 20) *
\$10                                                                (5,000 + 22,000)
= 67,500
\$0
0      30,000   60,000   90,000   120,000 150,000   Total Area = 98,750

Visitors (Q)

Aggregate                                                        32
Benefits: Mid-term 2005
   The City of Miami, Florida proposes to invest in a new water reservoir for its public water
system, and estimates its cost. To justify this substantial expenditure of public funds, the
Mayor explains that if the new reservoir is not constructed then the next most costly way to
increase Miami’s water supply will be to invest in a desalinization plant, which will be even
more expensive. Hence, the Mayor explains, the social benefits of building the new reservoir
clearly exceed its social costs. There are no environmental or other externalities involved with
either alternative. How would you assess this reasoning from an economic perspective?

– This is the “avoided cost” method of evaluating benefits.
It is incorrect because it ignores demand for the public good
(water), i.e., water’s real benefits to the society.

Avoided                          Willingness                        Willingness
Costs                             to Pay                            to Accept

33
To Be Continued…

 Next time: More on benefit estimation methods

34

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