<p>You need to think about what the total costs are and what the marginal costs are. The social optimum is where marginal costs are equal to the outside option which is riding the bus.</p>
<p>The story goes like this: Driving cars results in congestion because individual drivers do not take into account the cost they put on other drivers in terms of the driving time being an increasing function of the number of cars on the road.</p>
<p>If there are <span class="math-container">$n$</span> cars on the road then to total costs are</p>
<p><span class="math-container">$$TC(n) = n \left[\frac{15 + 0.25(n+1)}{60}  \cdot 15 + 5\right],$$</span></p>
<p>and obviously by definition <span class="math-container">$TC(n) = n \cdot AC(n)$</span> where <span class="math-container">$AC(n)$</span> are equal to the average costs. The costs you have found and set equal to the cost of riding the bus to solve for equilibrium are the average costs and that approach is quite correct.</p>
<p>To find the social optimum you find the marginal cost</p>
<p><span class="math-container">$$\frac{\partial TC(n)}{\partial n} = AC(n) + n \frac{\partial AC(n)}{\partial n},$$</span>
the first term is what the individual driver takes into account but the second term <span class="math-container">$$n \frac{\partial AC(n)}{\partial n}$$</span> is what has to be added to the price of the individual trip to attain social optimum (it is the Pigout tax).</p>
<p>In this case, you get</p>
<p><span class="math-container">$$MC(n) := \frac{\partial TC(n)}{\partial n}  =  \left[\frac{15 + 0.25(n+1)}{60}  \cdot 15 + 5\right] + \frac{15 \cdot 0.25}{60}n,$$</span></p>
<p>which you set equal to the bus price and solve for <span class="math-container">$n$</span>.</p>

CC BY-SA 4.0
2021-12-08T22:15:52.120Z
2021-12-08T22:15:52.120Z
Jesper Hybel
<p>I am struggling with the following question. Could someone please explain how to do it?</p>
<blockquote>
<p>People are trying to go to the city centre. A bus takes 1 hour and will always take 1 hour,
unaffected by traffic. If there is just one car on the road, it will take 20 minutes. Each additional car
adds 15 seconds to all car journeys. A bus ticket costs £2.50 and petrol for a car costs £5.00
(independent of the time). The opportunity cost of time is £15 an hour.</p>
</blockquote>
<blockquote>
<p>a) What is the equilibrium number of cars?</p>
</blockquote>
<blockquote>
<p>b) What is the socially optimum number of cars?</p>
</blockquote>
<p>Here is my attempt at part a:</p>
<ul>
<li>I found the cost of taking the bus to be £17.50 (£15 in time and £2.50 for the ticket).</li>
<li>I also found the cost of using a car to be £15*(time in hours) + 5 (for fuel). I found the time in hours as (20 + 0.25(n+1))/60 so the costs of using a car to be 5 + (20 + 0.25(n+1))/4.</li>
<li>I found the equilibrium number of cars by setting these equal to each other and solving for n to get 119. Is this correct?</li>
</ul>
<p>I have not attempted part b yet as I have no clue what to do!</p>
<p>Any help will be appreciated!</p>
<p>Thanks in advance</p>

<optimization><general-equilibrium>
How do I find the socially optimum and equilibrium value?
Arnav Sharma
# How do I find the socially optimum and equilibrium value?

(Asked by: username_0 on 2021-12-08)

I am struggling with the following question. Could someone please explain how to do it?

> People are trying to go to the city centre. A bus takes 1 hour and will always take 1 hour,
> unaffected by traffic. If there is just one car on the road, it will take 20 minutes. Each additional car
> adds 15 seconds to all car journeys. A bus ticket costs £2.50 and petrol for a car costs £5.00
> (independent of the time). The opportunity cost of time is £15 an hour.

> a) What is the equilibrium number of cars?

> b) What is the socially optimum number of cars?

Here is my attempt at part a:

* I found the cost of taking the bus to be £17.50 (£15 in time and £2.50 for the ticket).
* I also found the cost of using a car to be £15\*(time in hours) + 5 (for fuel). I found the time in hours as (20 + 0.25(n+1))/60 so the costs of using a car to be 5 + (20 + 0.25(n+1))/4.
* I found the equilibrium number of cars by setting these equal to each other and solving for n to get 119. Is this correct?

I have not attempted part b yet as I have no clue what to do!

Any help will be appreciated!

Thanks in advance

**Tags:** optimization, general-equilibrium

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## 1 answers

### Answer by username_1 on 2021-12-08. Score: 5 | Accepted answer

You need to think about what the total costs are and what the marginal costs are. The social optimum is where marginal costs are equal to the outside option which is riding the bus.

The story goes like this: Driving cars results in congestion because individual drivers do not take into account the cost they put on other drivers in terms of the driving time being an increasing function of the number of cars on the road.

If there are $n$ cars on the road then to total costs are

$$TC(n) = n \left[\frac{15 + 0.25(n+1)}{60} \cdot 15 + 5\right],$$

and obviously by definition $TC(n) = n \cdot AC(n)$ where $AC(n)$ are equal to the average costs. The costs you have found and set equal to the cost of riding the bus to solve for equilibrium are the average costs and that approach is quite correct.

To find the social optimum you find the marginal cost

$$\frac{\partial TC(n)}{\partial n} = AC(n) + n \frac{\partial AC(n)}{\partial n},$$
the first term is what the individual driver takes into account but the second term $$n \frac{\partial AC(n)}{\partial n}$$ is what has to be added to the price of the individual trip to attain social optimum (it is the Pigout tax).

In this case, you get

$$MC(n) := \frac{\partial TC(n)}{\partial n} = \left[\frac{15 + 0.25(n+1)}{60} \cdot 15 + 5\right] + \frac{15 \cdot 0.25}{60}n,$$

which you set equal to the bus price and solve for $n$.


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