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\begin{LARGE}
\author{Maria Arbatskaya, Kaushik Mukhopadhaya, and Eric Rasmusen}
\title{The Parking Lot Problem}
\date{12 November 2007}
\maketitle
\begin{abstract}
\noindent If property rights are not assigned over individual goods such
as
parking spaces, competition for them can eat up the entire surplus. We
find
that when drivers are homogeneous, there is a discontinuity in social
welfare
between \textquotedblleft enough\textquotedblright\ and
\textquotedblleft not
enough.\textquotedblright\ Building slightly too small a parking lot is
worse
than building much too small a parking lot, since both have zero net
benefit
and larger lots cost more to build. More generally, the welfare losses
from
undercapacity and overcapacity are asymmetric, and parking lots should
be
\textquotedblleft overbuilt.\textquotedblright\ That is, the optimal
parking
lot size can be well in excess of mean demand. Uncertainty over the
number of
drivers, which is detrimental in the first-best, actually increases
social
welfare if the parking lot size is too small.
\vspace{0.1in}
\noindent\emph{Keywords:} Property rights; commons; planning; rent-
seeking;
all-pay auction; timing game; capacity; queue.
\noindent\emph{JEL classification numbers: }R4; L91; D72; C7.
\vspace{0.1in}
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{\small Maria Arbatskaya, Department of Economics, Emory University,
Atlanta,
GA 30322-2240. Phone: (404) 727 2770. Fax: (404) 727 4639. Email:
marbats@emory.edu.\smallskip}
{\small Kaushik Mukhopadhaya, Department of Economics, Emory University,
Atlanta, GA 30322-2240. Phone: (404) 712 8253. Fax: (404) 727 4639.
Email:
kmukhop@emory.edu\smallskip.}
{\small Eric Rasmusen, Dan R. and Catherine M. Dalton Professor,
Department of
Business Economics and Public Policy, Kelley School of Business, Indiana
University, BU 456, 1309 E 10th Street, Bloomington, Indiana, 47405-
1701.
Phone: (812) 855 9219. Fax: (812) 855 3354. Erasmuse@indiana.edu.
http://www.rasmusen.org.}
\end{abstract}
\newpage
\noindent
{\bf The Model}
A set of players -- the drivers -- is indexed by $I\equiv\left\{
1,...,N\right\} $.
The drivers are workers who must show up for work no
later
than time $t=T$.
Each driver demands one space in the parking lot, and
his
value for it is $v>0$.
Let $K>0$ be the size of the parking lot, and let $cK$, however, the drivers compete for spaces, and they may wish to
arrive
early to secure spots.
A driver who arrives earlier than
time $T$
incurs a cost of $w>0$ per unit of time, so his cost of arrival at time
$t$ is
$L(t)=(T-t)w$. We will model time as discrete, with the interesting case
being
what happens as the time interval $\Delta>0$ shrinks to zero. Thus, a
time
grid includes times $t=k\Delta\in\lbrack0,T]$ where
$k\in\{0,...,T/\Delta\}$.
\newpage
We compare two alternative assumptions on whether a
driver
knows the number of spaces still open in the parking lot at time $t$.
When a
player making his arrival decision at time $t$ observes all arrivals
prior to
$t$, this is called \textit{full observability}; when he must decide
without
knowing if the parking lot is full, this is called \textit{
unobservability}.
For any time $t$, define $N_{t}$ as the number of unarrived drivers and
$K_{t}$ as the number of unoccupied parking spaces; $N_{t}=N-K+K_{t}$.
At time
$t$, the remaining drivers simultaneously decide whether to arrive at
the
parking lot at that instant. Under full observability, the decision can
be
conditioned on the number of parking spaces still available, $K_{t}$.
If player $i$ arrives at $t$ he obtains a spot with
probability
$p_{i,t}=\min\left\{ K_{t}/n_{t},1\right\} $, where $n_{t}\leq N_{t}$
denotes the number of players arriving at exactly time $t$.
If player $i$ arrives at time $t$, given that $K_{t}\in\{0,...,K\}$
parking
spaces are unoccupied at $t$ and $n_{t}\in\{0,...,N_{t}\}$ players
arrive at
$t$, his payoff is
\begin{equation}
u_{i,t}=p_{i,t}v-L(t). \tag{1}%
\end{equation}
\newpage
Let us define the \textit{indifference arrival time}, $t^{\ast}$, as the
time
at which the arrival cost, $L(t^{\ast})$, equals the prize value, $v$.
It
follows that
\begin{equation}
t^{\ast}\equiv T-v/w. \tag{2}
\end{equation}
A player who arrives at $t^{\ast}$ and finds a parking space receives a
payoff
of zero since the disutility of the early arrival equals the value of
the
parking space. When looking for equilibria, we can restrict our
attention to
$t\geq t^{\ast}$ since arriving before $t^{\ast}$ yields a negative
payoff and
is strictly dominated by arriving at $T$.
Define $t_{p}^{\ast}$ as the time at which a player receives
a zero
payoff from participating in a lottery with probability $p$ of winning a
parking space; $t_{p}^{\ast}$ is determined from equation
$pv-L(t_{p}^{\ast
})=0$. It follows that
\begin{equation}
t_{p}^{\ast}\equiv T-pv/w. \tag{3}%
\end{equation}
Finally, let $\underline{t}$ and $\overline{t}$ denote the earliest and
the
latest time any player arrives with a positive probability along the
equilibrium path.
\newpage
\noindent\textbf{Definition 1.}\textit{\ A time grid is fine if }
\begin{equation}
\Delta<\frac{v}{Nw}. \tag{4}%
\end{equation}
\medskip
When a time grid is fine, there are more than $N\geq K+1$\ periods
between
$t^{\ast}$\ and $T$ since $T-t^{\ast}=v/w>N\Delta$. No more than $K$
players
can profitably enter at time $t=t^{\ast}+\Delta$ when the time grid is
fine.
Otherwise, even at the highest odds of obtaining a space, $K/(K+1)$, the
payoff of a player is negative at $t=t^{\ast}+\Delta$ for a fine time
grid.
When a player is not guaranteed to obtain a parking spot, the player
would
choose to arrive one period earlier to secure a parking space, because
for a
fine time grid the cost of one-period earlier arrival, $w\Delta$, is
justified
by the gain associated with the higher probability of obtaining
parking.
In the first-best
outcome, all
$N$ players arrive at $T$, and $K$ of them get to park in the parking
lot.
This cannot be
part
of an equilibrium strategy profile when the time grid is fine, since
each
player would prefer to deviate by arriving just before $T$ and increase
his
probability of finding a spot to 100\%.
\newpage
\noindent\textbf{The Simplest Case: Two Drivers}
\noindent\textbf{Claim 1.} \textit{Consider the parking game on a fine
time
grid under full observability when there are two drivers and one parking
spot
(}$N=2$\textit{ and }$K=1$\textit{). The following arrival strategies
constitute a subgame-perfect pure-strategy equilibrium:\ contingent on
the
parking lot not being full at time }$t$\textit{, (i) at }$t=t^{\ast}%
+k\Delta\in\lbrack t^{\ast},t_{1/2}^{\ast})$,\textit{ player} $1$\textit
{
arrives if} $\ k=0,2,4,...$\textit{ and player }$2$\textit{ arrives if
}$k=1,3,5,...$;\textit{ (ii) at }$t\in\lbrack
t_{1/2}^{\ast},T)$\textit{, both
players arrive. }
When Claim 1's strategies are played out, player 1 arrives at
$t^{\ast}$,
player 2 arrives at $T$, and both players obtain zero payoffs. Thus, the
parking lot yields zero social benefit; and if it is costly to construct
or
the land has other uses, the lot has a negative social value.
\newpage
\noindent\textbf{The General Case: $N$ Drivers}
Denote by $\widehat{n}_{t}$ the critical number
of
players for whom entry yields a nonnegative payoff at time $t$, given
that
$K_{t}$ spaces remain unclaimed by time $t$.
\noindent\textbf{Claim 2.} \textit{Consider the parking game on a fine
time
grid under full observability and with more drivers than parking spaces
(}$N>K$\textit{).}
\noindent\textbf{A.}\textit{\ In any pure-strategy equilibrium to the
game,
the parking lot is full after }$t^{\ast}+\Delta$\textit{. The
equilibrium
outcome is for }$K_{0}\in\{0,...,K\}$\textit{\ players to arrive at }
$t^{\ast}$;
$K-K_{0}$ \textit{players to arrive at} $t^{\ast}+\Delta$\textit{, and }
$N-K$
\textit{players to arrive at }$T$.\textit{\smallskip}
\noindent\textbf{B.}\ \textit{All the pure-strategy equilibria
have
the following arrival schedule:}
(i) at $t=t^{\ast}$, $K_{0}
\in\{0,...,K_{t}\}$\textit{\ players arrive;}
(ii) at $t=t^{\ast}+\Delta
$\textit{, }$K_{t}$ \textit{players arrive; if player }$i\in I$\textit{
deviated by not arriving at }$t^{\ast}$,\ \textit{the set of arriving
players
excludes player }$i$;
(iii) at $t\in\lbrack t^{\ast}+2\Delta
,t_{1/2}^{\ast})$\textit{, if player }$i\in I$\textit{ deviated by not
arriving at }$t-\Delta$\textit{, one other player arrives;}
(iv) at
$t\in\lbrack t_{1/2}^{\ast},T)$\textit{, if player }$i\in I$\textit{
deviated
by not arriving at }$t-\Delta$, \textit{then }$\min\left\{ \widehat{n}%
_{t},N_{t}\right\} \geq2$\textit{\ players arrive at }$t$
(v)
at
$t=T $\textit{\ all players arrive who have not yet arrived.}$\medskip
\medskip$
\newpage
There are many sizes of this initial shrinkage that support
equilibria.
1. For example, there is an
equilibrium
outcome in which one group of $K$ players arrive at time $t^{\ast}$ and
they
park in the lot, while a second group of $N-K$ players arrive at $T$ and
park
elsewhere.
Both groups receive the same payoff of zero.
2. Another polar
equilibrium outcome is for no drivers to arrive at $t^{\ast}$ and for
$K$
players to arrive at $t^{\ast}+\Delta$. In all pure-strategy equilibria,
the
parking lot fills up no later than $t^{\ast}+\Delta$ and no players
arrive
between $t^{\ast}+\Delta$ and $T$.
\bigskip
\noindent\textbf{Corollary to Claim 2.} \textit{As the time grid becomes
infinitely
fine,
drivers dissipate all the rents from parking in any pure-strategy
equilibrium
of the parking game under full observability when there are more drivers
than
parking spaces.}
\medskip$\medskip$
\newpage
\noindent
{\bf Nonexistence of Pure-Strategy Equilibria under
Unobservability}
Claim 2 said that under full observability some players
arrive at $t\leq t^{\ast}+\Delta$ and others arrive at $T$, both having
nearly
zero expected payoffs.
Under unobservability, Claim 2 does not apply
because
one of the players who is supposed to arrive at $t\leq t^{\ast}+\Delta$
could
deviate and arrive at $T-\Delta$ instead, reducing his early-arrival
cost.
The
other players would not observe that he had failed to arrive as
scheduled at
$t$, so they would be unable to respond by taking \textquotedblleft
his\textquotedblright\ spot before $T-\Delta$.
\medskip\medskip
\newpage
\noindent\textbf{Claim 3.}\textit{ There does not exist a pure-strategy
Nash
equilibrium under unobservability when the time grid is fine and there
are
more drivers than parking spaces}.\medskip\medskip
\noindent\textbf{Proof. }Denote by $t^{\prime}$ the time when
the
parking lot becomes full in a pure-strategy equilibrium. If at
$t^{\prime}$
not all arriving drivers obtain parking, then each of them would find it
profitable to deviate by arriving a period earlier.
If parking is
guaranteed
at $t^{^{\prime}}$, then there are drivers who arrive at $T$, because
arriving
between $t^{\prime}$ and $T$ yields a negative payoff. Each of these
drivers
benefits from arriving shortly before $t^{\prime}$ unless
$t^{\prime}\leq
t^{\ast}+\Delta$.
A fine time grid implies that only $K$ drivers would
arrive
at $t^{\prime}=t^{\ast}$ or $t^{\prime}=t^{\ast}+\Delta$. Then, each of
them
can arrive later and still obtain the parking space. Hence, no pure-
strategy
equilibrium exists in this case either. Q.E.D.\medskip$\medskip$
\newpage
This is an all-pay auction with multiple prizes. The arrival
times
are like bids, and the empty parking spaces are like prizes. The reason
for
the nonexistence of a pure-strategy Nash equilibrium under
unobservability
when the time grid is fine is essentially the same as in a multi-prize
all-pay
auction with continuous strategy space (see Barut and Kovenock, 1998,
Clark
and Riis, 1998, or the general characterization by Baye et al. [1996]).
Bulow-Klemperer (1999, AER) is different. It is about a war of attrition by N players with K prizes. They show that K+1 players will mix and the rest give up immediately. But in a pre-emption game like this one,
``doing nothing" means staying in, and not incurring costs.
Mixed-strategy equilibria under unobservability are complicated to
describe
fully. In the two-driver game there is, for example, an alternating
equilibrium in which for $k\in\{0,1,2...\}$ and $t\leq T$ player $1$
arrives
at $t=t_{0}+2k\Delta$ with probability $(v/(2w\Delta)-k)^{-1}$ and
player $2$
arrives at $t=t_{0}+(2k+1)\Delta$ with the same probability.
In the case
of
$w=1$, $v=10$, $T=10$, and $\Delta=1$, the probabilities of player $1$
arriving
at $t=1,3,5,7,9$ and of player $2$ arriving at $t=2,4,6,8,10$ are 1/5, 1/4,
1/3,
1/2, and $1$. After $t=T-\Delta=9$ the parking lot is full with
probability
one.
\newpage
\section{Full Rent Dissipation}
We first establish that in any equilibrium, players arriving at
$\underline
{t}$\textit{\ }and\textit{\ } $\overline{t}$ (the earliest and the
latest
times any player arrives with a positive probability along the
equilibrium
path) obtain nearly zero payoffs whenever there are more drivers than
parking
spaces. Then, we will show that all the players in between must almost
fully
dissipate the value from parking. Our propositions will apply under
either
full observability or unobservability.
\medskip$\medskip$
\noindent\textbf{Claim 4. }\textit{Consider the parking game on a fine
time
grid with more drivers than parking spaces. In any subgame-perfect
equilibrium, the probability that the parking lot is full at time }
$\overline{t}\leq T$\textit{ is one under full observability and
approaches
one as the time grid becomes infinitely fine under unobservability. The
equilibrium payoffs of players who arrive at }$\overline{t}$ \textit{
with a
positive probability are zero under full observability and tend to zero
under
unobservability as the time grid becomes infinitely fine. Any player who
arrives at }$\underline{t}$ \textit{has a payoff that tends to zero as
the
time grid becomes infinitely fine.}$\medskip\medskip$
\noindent\textbf{Corollary.} \textit{The parking lot is almost surely
full by
time} $T$\textit{.}
\newpage
\noindent\textbf{Proposition 1. }\textit{If there are more drivers than
parking spots (}$N>K$\textit{), then under either full observability or
unobservability, players fully dissipate rents from the parking lot in
any
equilibrium as the time grid becomes infinitely fine. }
\medskip
\noindent\textbf{Proof of Proposition 1.} The proof is by contradiction.
Suppose player $i$ is earning a positive payoff bounded away from zero.
Let
$t\in(\underline{t},\overline{t})$ be the earliest moment the player
arrives
with a positive probability; $t>\underline{t}$. Consider a player (
player $j$)
who arrives at $\overline{t}$ (with pure strategy) and earns an almost
zero
payoff (by Claim 4). If player $j$ follows his equilibrium strategy
until
$t-\Delta$ and arrives with probability one at $t-\Delta$, he would
obtain a
positive payoff bounded away from zero. The costs of arriving at $t$ and
$t-\Delta$ differ by $\Delta w$, a difference that becomes negligible as
$\Delta\rightarrow0$. It suffices to prove that player $j$'s odds of
obtaining
parking at $t-\Delta$ are no worse than player $i$'s odds of obtaining
parking
at $t$. Note that by the definition of time $t$, player $i$ has not
attempted
to arrive before $t$. Player $j$ may have been mixing before $t$.
Therefore,
player $j$ has the expected payoff from arriving at $t-\Delta$ that is
no less
than that of player $i$. Since there is a profitable deviation, there
cannot
exist an equilibrium with a player obtaining a positive payoff at a time
$t$
between $\underline{t}$ and $\overline{t}$. Hence, all players earn
almost
zero payoffs. Q.E.D.
\newpage
\noindent
{\bf Welfare and Parking Lot Size}
The
drivers
have a value $v>0$ for each of the parking spaces in use. Let us denote
the
cost of providing $K$ parking spaces by an increasing function $C(K)$.
The
welfare from a parking lot of size $K$ is
\begin{equation}
W(K)=\left\{
\begin{tabular}
[c]{lll}%
$vN-E_{\sigma(N,K)}\left( \sum_{i=1}^{N}L(t_{i})\right) -C(K)$ & if &
$K\geq
N$\\
$vK-E_{\sigma(N,K)}\left( \sum_{i=1}^{N}L(t_{i})\right) -C(K)$ & if &
$KK$ the benefit is
dissipated
by rent-seeking so it does not matter whether there are $K$ or $K-1$
spaces.
Since the $K$th space matters only if $N=K$, the change in the expected
welfare is the probability that there are exactly $K$ drivers multiplied
by
the benefit from eliminating rent-seeking behavior, $vK$, net of the
change in
construction costs.
On
the one hand, at larger parking lot sizes, it is more important to have
a
sufficiently big parking lot because there are more people who could get
benefit from it. On the other hand, it could be less likely that larger
parking lots are filled up. When the first effect dominates and $Kf(K)$
increases in $K$, the marginal benefit increases with $K$ too. For a
constant-marginal-cost technology, $C(K)=cK$, this implies a corner
solution
to the problem: the parking lot should accommodate all potential
drivers, even
if that is much greater than the expectation of the number of drivers,
or not
be built at all.
\newpage
\noindent
{\bf Example}
Consider a discrete uniform distribution for the number
of
drivers on the support $\{0,...,\overline{N}\}$. For the uniform
distribution,
$Kf(K)$ does increase in $K$, so we have a corner solution. Under the
uniform
distribution, it is equally likely at any capacity that the parking
demand
will be barely met. At a larger capacity benefits accrue to more people,
so at
larger capacity levels, the benefit from building an additional space is
higher, and planners should design the parking lot for the
\textquotedblleft
peak demand,\textquotedblright\ accommodating all potential
drivers.
Designing for peak demand is not always the optimal
choice.
For example, in a case of the binomial distribution of $N$, if each of
$100$
drivers is in need of parking with probability $0.5$ and $c/v=1/5$, the
optimal parking size is $K^{\ast}=58$. Only about $8$ out of $58$ spaces
are
empty, on average. This corresponds to about $86\%$ utilization level.
\newpage
When the number
of
drivers is large, calculus can be used. Consider the
production
technology with a constant marginal cost, $C(K)=cK$ where $c0$, and $[\underline
{N},\overline{N}]\subset\lbrack0,\infty)$ are such that
$\int_{\underline{N}%
}^{\overline{N}}\alpha N^{-\beta}dN=1$. The first-order condition
implies the
optimal capacity $K^{\ast}=\left( \frac{1}{\alpha}\frac{c}{v}\right)
^{-1/(\beta-1)}$, and the second-order condition is satisfied if and
only if
$\beta>1$.
For $\beta\leq 1$, the parking lot should have $\overline{N}$
spaces, if it is built. The mean utilization of the parking lot of
optimal
size can be measured as a ratio of the mean number of parking spots
taken,
$E(X)$, to the optimal capacity size, $K^{\ast}$. If $N0$.
For example, in the case of linear demand, $v_{i}=a-bi$ for
some
constants $a$ and $b$.
Assume that there are also drivers who have zero
value
for parking and hence arrive at their preferred time, $T$.
As before,
the cost
of constructing a parking space is $c$, and the cost per unit time of
early
arrival is $w$.
In the first-best case, both the parking lot capacity and access to
parking
can be regulated. The first-best policy is to choose capacity to
accommodate
all players with values higher (or weakly higher) than $v_{i}=c$ and to
give
only them access to the parking lot.
\newpage
When access to parking cannot be restricted, planners should
design a
larger parking lot.
In the second-
best
world, the optimally designed parking lot provides parking even to
drivers who
value it less than the construction cost. In many cases, including that
of
linear demand for parking, the second-best policy is to choose $K=N$ and
guarantee parking to all players with positive values.
If $K0$.
\newpage
The welfare from a parking lot of size $K\geq N$ is equal to the total
value
minus construction costs. For any $K0$. In other
words,
allowing all drivers with positive values to park is not optimal when
$K\cdot
v^{\prime\prime}(K)/\left( -v^{\prime}(K)\right) >1$, that is, when
the
slope of the inverse demand is elastic.
\newpage
Another way to look at the problem is that if capacity is chosen at the
first-best optimum of $K=45$ there will be a discontinuous loss from
allowing
more than $N=45$ players access to the parking lot. Consider the problem
of
choosing the number of parking permits, $n$, for a lot of a given size,
$K$,
which arises when capacity is chosen in the long run while permits are
allocated in the short run. Assume that permits are rationed efficiently
to
the highest-value players. Clearly, the optimal number of permits cannot
be
less than capacity. For any $n>K$, the welfare is $K\cdot v_{K+1}$ lower
than
at $n=K$. Hence, the welfare loss from having too few or too many
parking
permits is not symmetric. Welfare jumps down by $K\cdot v_{K+1}$ when
one too
many parking permits are issued, but welfare falls gradually when too
few
drivers are served.
At the first-best capacity level $K=45$, welfare
drops due
to rent-seeking loss by $45\cdot5=380$ when $46$ permits (or more) are
allocated. In contrast, welfare drops by $1.2$ when $n=44$ and one
driver with
value $1.2$ is unserved. There is a discontinuity in welfare at the
optimal
number of permits, $n=K$; oversupply of permits is more dangerous than
undersupply.
\newpage
If there were no rentseeking, the expected welfare would be composed of the welfare from $N$ places being used if $N \leq K$ plus $K$ spaces being used if $N >K$:
\begin{equation}
EW(K)= \sum_{N=0}^{K}(vN)f(N)+ \sum_{N=K+1}^{\overline{N}}(vK)f(N)- C\left( K\right) .
\end{equation}
The size of the parking lot should be increased from $K-1$ to $K$ as long as
\begin{equation}
vKf(K) + \sum_{N=K+1}^{\overline{N}}(vK)f(N) - \sum_{N=K }^{\overline{N}}(v(K-1))f(N) - \left( C\left( K\right) -C\left( K-1\right) \right) \geq 0
\end{equation}
This is equivalent to:
\begin{equation}
vKf(K) + \sum_{N=K+1 }^{\overline{N}}v f(N) -f(K)(K-1)v - \left( C\left( K\right) -C\left( K-1\right) \right) \geq 0
\end{equation}
or
\begin{equation}
\sum_{N=K }^{\overline{N}}v f(N) - \left( C\left( K\right) -C\left( K-1\right) \right) \geq 0.
\end{equation}
In words, for the extra space to be worthwhile, the probability the extra parking space gets used times its use value, $v$, must be as great as its marginal cost. In the case of the uniform distribution example, this leads to a parking lot of size $K=80$, rather than the $K=100$ we found as the result of rentseeking.
\newpage
Whether the optimal parking lot size is greater with rentseeking than without is unclear to me now. I think it depends on the $f(N)$ function. The optimality condition for the no-rentseeking case gives a smaller marginal benefit from increasing $K$ than for the rentseeking case if, looking at the middle terms of the intermediate step in the equation above for the no-rentseeking case:
\begin{equation}
vKf(K) + \sum_{N=K+1 }^{\overline{N}}v f(N) -f(K)(K-1)v - \left( C\left( K\right) -C\left( K-1\right) \right) \geq 0
\end{equation}
versus for the rentseeking case:
\begin{equation}
vKf(K) - \left( C\left( K\right) -C\left( K-1\right) \right) \geq 0
\end{equation}
The difference is:
\begin{equation}
\sum_{N=K+1 }^{\overline{N}}v f(N) -f(K)(K-1)v < 0.
\end{equation}
If $f(N)$ is smaller for bigger $N$ beyond $K^*$, this condition will be true.
\newpage
Another case to consider is what happens if the drivers do not know the number of drivers at the beginning of the day, but they do know the distribution. That case is both better and worse than the case we just analyzed, in which the number of drivers is fixed after the parking lot size. It is better because drivers will not know that showing up early is necessary on a particular day, so they will moderate their rentseeking. It is worse because drivers will pursue the same policy every day, so they will pursue some rentseeking even when it is unnecessary.
\newpage
Drivers will pursue mixed strategies, since this is a version of the unobservable arrival game. Consider the payoff from showing up at exactly time $T$. With probability $\sum_{N=0}^{K} f(N)$ the parking lot does not fill up. Thus, the payoff to one driver must be $\sum_{N=0}^{K} f(N)$ if the support of the mixing distribution includes arrival at $T$. This must be multiplied by the expected number of players, which is $\sum_{N=0}^{\bar{N}} Nf(N)$, which yields
\begin{equation}
\begin{array}{ll}
EW(K) &= ( \sum_{N=0}^{\bar{N}} Nf(N) ) (\sum_{N=0}^{K} f(N)) - C\left( K\right) \\
& \\
&= (E(N)) (\sum_{N=0}^{K} f(N)) - C\left( K\right) \\
\end{array}
\end{equation}
The optimality condition is then
\begin{equation}
E(N) f(K) v - \left( C\left( K\right) -C\left( K-1\right) \right) \geq 0.
\end{equation}
Compare that with the optimality condition when drivers know $N$:
\begin{equation}
K f(K) v - \left( C\left( K\right) -C\left( K-1\right) \right) \geq 0.
\end{equation}
If $K^* > E(N)$, then optimal capacity is smaller when drivers do not know $N$.
%\includegraphics{parking2.pdf}
\end{LARGE}
\end{raggedright}
\end{document}