Lind on Measuring Benefits
© Allen C. Goodman, 2009
Suppose
• We have 2 types of land, half
polluted (p), half not (u).
• Pp = 100; Pu = 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?
Diagram
Pp
Pu
200
150
100
100
100
200
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.
Analysis
aij = maximum rent individual i would pay for parcel j.
Alternatively:
aij = Net earnings of firm i at parcel j exclusive of land cost.
n activities, n parcels
pj = rental value of parcel j.
Activity locates at parcel j ONLY IF:
(2.3) aij  pj AND
aij
(2.4) aij - pj  aik - pk
pj
pj,
aij
Analysis
Define consumer surplus (or profit) as sij.
sij aij – pj.
SO, the total benefit from land is
(2.5) aij = sij + pj
Land is ENHANCED if there is some aij > aij for some
parcels, and aij = ij for the rest.
Let:
A = ajj
sij
(2.7) A a jp(j)
(2.8) A - a jp(j)- a j j
pj,
aij
Analysis
(2.5) aij = sij + pj
Let:
A = ajj
(2.7) A a jp(j)
(2.8) A - a jp(j)- a j j
From (2.5)
(2.9) A - s  jp(j) - s j j ) + p  j - pj)
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.
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.
Suppose we have 3 unimproved parcels
Din productivity = [a12 – a11] + [a23 – a22] + [a31 – a33]
Since there is no improvement,
Din productivity = [a12 – a11] + [a23 – a22] + [a31 – a33]
From the equilibrium conditions:
(2.3) aij  pj AND
(2.4) aij - pj  aik - pk
[ajj+1 – ajj]  p  j+1 – p  j
New rents
AND
[ag1 – p 1]  [agg – p g]
Suppose we have 3 unimproved parcels
[ajj+1 – ajj]  p  j+1 – p  j
AND
[ag1 – p 1]  [agg – p g]
ALSO
[ajj+1 – ajj]  p j+1 – p j
AND
[ag1 – p 1]  [agg – p g]
New rents
Old rents
Suppose we have 3 unimproved parcels
a12 – a11  p2 – p1
a23 – a22  p3 – p2
a31 – a33  p1 – p’3
Sum

0
New rents
a12 – a11  p2 – p1
a23 – a22  p3 – p2
a31 – a33  p1 – p3
Sum

Old rents
0
Benefits must equal 0!
Let’s improve a parcel (#1)
D = Benefits. Assume activity 1 moves to parcel 2, activity 2
to parcel 3, etc.
g -1
D   (a' jj 1 -a jj )  (a' g1 -agg )
1
With no improvement on parcels 2 through g:
g -1
 (a'
1
pj,
aij
g -1
jj 1
-a jj )   (a jj 1 - a jj )
1
“Winning” aij changes;
Eq’m rent pj changes.
From eq’m for
th
j
firm
p ' j 1 - p ' j  a jj 1 - a jj  p j 1 - p j
Why they’re here now
Why they weren’t before
So, for g-1 firms:
g -1

g -1
g -1
1
1
p' j 1 - p' j   a jj1 - a jj   p j 1 - p j
1
g -1
We know:
 p'
j 1
- p ' j  p ' g - p '1
1
g -1
p
1
j 1
- p j  p g - p1
From eq’m for
th
j
firm
p ' g - p '1  D - (a ' g1 -a gg )  p g - p1
(a ' g1 -a gg )  p ' g - p '1  D  p g - p1  (a ' g1 -a gg )
(a ' g1 -a gg ) - ( p '1 - p ' g )  D  (a' g1 -a gg ) - ( p1 - p g )
Increased profit at new prices
Increased profit at old prices
If we’re operating on the margin, we get equalities above:
p' j 1 - p' j  a jj 1 - a jj  p j 1 - p j
(a' g1 -a gg ) - ( p'1 - p' g )  D  (a' g1 -a gg ) - ( p1 - p g )
D  ( p '1 - p1 )
Earlier Problem
Activity is on polluted land p rather than unpolluted land u.
aip – pp aiu – pu
pu – pp aiu – aip
If pollution is eliminated, all land is same and net benefits are:
 (for half of parcels) (aiu – aip)
From above,
 (for half of parcels) (aiu – aip)  (n/2)(p2 – p1)
Here, with 200 parcels, 100 polluted, 100 not
 (for half of parcels) (aiu – aip)  100 (200 – 100) = 10000
Another example
You should
graph this
and do the
calculations
Two types of land
Demand for unpolluted land: P = 250 – 0.5Q
Demand for polluted land: P = 110 –0.1Q
100 acres of each. Pu = 200; Pp = 100. Su = 2500; Sp = 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  jp(j) - s j j ) + p  j - pj)
A -  7000 + 0
From above,  (for half of parcels) (aiu – aip)  10000
Polinsky and Shavell
© Allen C. Goodman 2009
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))
where
k = distance, y = income, T = transportation
costs, a(k) is an amenity.
We have V1 > 0, V2 < 0, V3 > 0. Why?
What can we do with this?
V = V (y - T(k), p(k), a(k))
Within a city dV = 0.
dV/dk = -V1T´ + V2p´ +V3 a´ = 0.
Leads to p´ = (V1/V2)T´- (V3/V2)a´.
(V1/V2) = [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.
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.
Regression analysis w/ PS
Cobb-Douglas Example
U = Axqba(k)d;  + b = 1.
 x(k) =  (y – T(k))
q(k) = b (y – T(k))/p(k)
Putting x and q into U 
V(k) = C[y-T(k)]p(k)-ba(k)d C is a constant
Solve for p(k) as:
log p(k) = (1/b) log (C/V*) + (1/b) log [Y – T(k)] + (d/b) log a(k)
= b0 + b1 log [Y – T(k)] + b2 log a(k)
Regression analysis w/ PS
log p(k) = (1/b) log (C/V*) + (1/b) log [Y – T(k)] + (d/b) log a(k)
= b0 + b1 log [Y – T(k)] + b2 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.