Profit maximization and allocative efficiency result

Question

I completely stuck to the last part of this exercise. I cannot understand how can I determine an allocative efficiency result.

Any help will be appreciated. Thank you.

Answer 1

With no tolls, how many people will use the free way during rush hours?

Since during rush hours hundreds of cars travel, so the number of commuters $n$ using the freeway satisfy the following condition :

$10 + n = 60$ i.e. $n= 50$.

Why is it inefficient?

It is inefficient because average commuting time per person in equilibrium is 60 minutes which is the worst possibility. Reducing the number of cars using the freeway during rush hours will necessarily reduce the average commuting time to below 60.

Find the optimal toll charged by a profit maximizing entity that owns the highway.

Let $t$ denote the toll. Average cost of using the highway if $x$ people use it is :

$$\text{AC} = \begin{cases} 3 + t, & \text{if } x\leq 20 \\ 1 + 0.1x + t, & \text{if } x > 20\end{cases} $$

Profit maximization problem of the entity for peak time is :

\begin{eqnarray*} \max_{t, x} & \ tx \\ \text{s.t} & \ \text{AC} \leq 6 \end{eqnarray*}

Solving this problem we get $x^*_p = 25$ and $t^*_p =2.5$.

Profit maximization problem of the entity for non peak time is :

\begin{eqnarray*} \max_{t, x} & \ tx \\ \text{s.t} & \ \text{AC} \leq 6 & \ \\ & x \leq \overline{x}\end{eqnarray*} where $\overline{x} < 20$ is exogenously given demand during non peak time.

Solving this problem we get $x^*_n = \overline{x}$ and $t^*_n =3$.

Is the outcome efficient?

Yes, it is efficient. In fact, it minimizes the average commute time. If $x$ people use highway during the peak time, then the remaining $(50 - x)$ will be using side streets. So, the average commute time of these 50 people (who were using highway in case of no toll) is :

\begin{eqnarray*} \text{AC of 50 commuters} = \begin{cases} \frac{30x + 60(50-x)}{50} & \text{if } x \leq 20 \\ \frac{(10+x)x + 60(50-x)}{50} & \text{if } x > 20 \end{cases} \end{eqnarray*}

Minimizing it with respect to $x$, we get $x^* = 25$.