Download presentation

Presentation is loading. Please wait.

Published byLyndsey Dobbs Modified about 1 year ago

1
Lind on Measuring Benefits © Allen C. Goodman, 2009

2
Suppose We have 2 types of land, half polluted (p), half not (u). P p = 100; P u = 200 If a program of air pollution control equalizes productivity everywhere, rent may end up at 150. If so total property values remain same. Does this mean there were no benefits?

3
Diagram 200 100 PpPp PuPu 200 150

4
Analysis Assume class of programs w/ output specific to particular locations. How do we measure benefits? Two major findings –Benefits of a project can be approximating by measuring net increase in profits of activities located on land directly affected by project. –W/ zero profits, total change in productivity equals change in land values.

5
Analysis a ij = maximum rent individual i would pay for parcel j. Alternatively: a ij = Net earnings of firm i at parcel j exclusive of land cost. n activities, n parcels p j = rental value of parcel j. Activity locates at parcel j ONLY IF: (2.3) a ij p j AND (2.4) a ij - p j a ik - p k p j, a ij pjpj

6
Analysis Define consumer surplus (or profit) as s ij. s ij a ij – p j. SO, the total benefit from land is (2.5) a ij = s ij + p j Land is ENHANCED if there is some a ij > a ij for some parcels, and a ij = ij for the rest. Let: A = a jj (2.7) A a j (j) (2.8) A a j (j) - a j j p j, a ij s ij

7
Analysis (2.5) a ij = s ij + p j Let: A = a jj (2.7) A a j (j) (2.8) A a j (j) - a j j From (2.5) (2.9) A s j (j) - s j j ) + p j - p j ) What does this mean? Change in total land values = change in productivity ONLY if surplus terms = 0, or surplus terms don’t change! Essentially an open city argument.

8
Benefit Measurement Lind shows that any cycle of relocation NOT involving a parcel of land affected by the project will leave net productivity unchanged. We need to consider ONLY parcels that are directly affected, where land is improved. Benefits can be measured by considering in surplus of those activities alone that move onto improved land.

9
Suppose we have 3 unimproved parcels in productivity = [a 12 – a 11 ] + [a 23 – a 22 ] + [a 31 – a 33 ] Since there is no improvement, in productivity = [a 12 – a 11 ] + [a 23 – a 22 ] + [a 31 – a 33 ] From the equilibrium conditions: (2.3) a ij p j AND (2.4) a ij - p j a ik - p k [a jj+1 – a jj ] p j+1 – p j AND [a g1 – p 1 ] [a gg – p g ] New rents

10
Suppose we have 3 unimproved parcels [a jj+1 – a jj ] p j+1 – p j AND [a g1 – p 1 ] [a gg – p g ] ALSO [a jj+1 – a jj ] p j+1 – p j AND [a g1 – p 1 ] [a gg – p g ] New rents Old rents

11
Suppose we have 3 unimproved parcels a 12 – a 11 p 2 – p 1 a 23 – a 22 p 3 – p 2 a 31 – a 33 p 1 – p’ 3 Sum 0 New rents Old rents a 12 – a 11 p 2 – p 1 a 23 – a 22 p 3 – p 2 a 31 – a 33 p 1 – p 3 Sum 0 Benefits must equal 0!

12
Let’s improve a parcel (#1) D = Benefits. Assume activity 1 moves to parcel 2, activity 2 to parcel 3, etc. With no improvement on parcels 2 through g: p j, a ij “Winning” a ij changes; Eq’m rent p j changes.

13
From eq’m for j th firm Why they’re here nowWhy they weren’t before So, for g-1 firms: We know:

14
From eq’m for j th firm Increased profit at new pricesIncreased profit at old prices If we’re operating on the margin, we get equalities above:

15
Earlier Problem Activity is on polluted land p rather than unpolluted land u. a ip – p p a iu – p u p u – p p a iu – a ip If pollution is eliminated, all land is same and net benefits are: (for half of parcels) (a iu – a ip ) From above, (for half of parcels) (a iu – a ip ) (n/2)(p 2 – p 1 ) Here, with 200 parcels, 100 polluted, 100 not (for half of parcels) (a iu – a ip ) 100 (200 – 100) = 10000

16
Another example Two types of land Demand for unpolluted land: P = 250 – 0.5Q Demand for polluted land: P = 110 –0.1Q 100 acres of each. P u = 200; P p = 100. S u = 2500; S p = 500. Total surplus = 3000. If all land is now unpolluted, bidders will use first demand curve Q = 200; P = 150. Property values unchanged. New surplus = 10000 From (2.5) (2.9) A s j (j) - s j j ) + p j - p j ) A + From above, (for half of parcels) (a iu – a ip ) 10000 You should graph this and do the calculations

17
Polinsky and Shavell © Allen C. Goodman 2009

18
Polinsky and Shavell Examine the distinction between closed and open cities when looking at the measurement of benefits. Take a little different tack in the modeling but with similar results. They use an indirect utility function: V = V (y - T(k), p(k), a(k)) wherek = distance, y = income, T = transportation costs, a(k) is an amenity. We have V 1 > 0, V 2 0. Why?

19
What can we do with this? V = V (y - T(k), p(k), a(k)) Within a city dV = 0. dV/dk = -V 1 T´ + V 2 p´ +V 3 a´ = 0. Leads to p´ = (V 1 /V 2 )T´- (V 3 /V 2 )a´. (V 1 /V 2 ) = [Utility/$]/[Utility/(acres/$)] (1/Land). We have negative price-distance function. With an open city V is fixed at V*, so suppose there is an increase in a(k). For V to stay equal to V*, p(k) must rise.

20
What can we do with this? Suppose we have a closed city. V** = V (y - T(k), p(k), a(k)) V starts at V**. Suppose amenities increase everywhere but k. V** must rise. Since amenities haven’t improved at k, p(k) must fall relative to elsewhere. This is an indirect effect. Now increase a(k). p(k) must rise to maintain V**. This is the direct effect.

21
Regression analysis w/ PS Cobb-Douglas Example U = Ax q a(k) ; + = 1. x(k) = (y – T(k)) q(k) = (y – T(k))/p(k) Putting x and q into U V(k) = C[y-T(k)]p(k) - a(k) C is a constant Solve for p(k) as: log p(k) = (1/ ) log (C/V*) + (1/ ) log [Y – T(k)] + ( ) log a(k) = b 0 + b 1 log [Y – T(k)] + b 2 log a(k)

22
Regression analysis w/ PS log p(k) = (1/ ) log (C/V*) + (1/ ) log [Y – T(k)] + ( ) log a(k) = b 0 + b 1 log [Y – T(k)] + b 2 log a(k) In an open city, since V* is fixed, a change in a(k) will predict change in log p(k). In closed city V* V**. Must know what happens to a(k) all over city. Gen’l eq’m model is necessary. SO: Changes in aggregate land values correspond to WTP only with an open city model. Eq’m rent schedule will give enough information to identify demand for a(k), all else equal; in “closed city” all else may not be equal.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google