what is the marginal cost divided by the average cost

Question

Marginal cost and average cost are both a u-curve. They cross in the competitive equilibrium. If we divide one by the other, is the ratio the degree of competition? Ratio is 1 in perfect competition. What is the name of this quantity, MC/AC?

Answer 1

$MC/AC$ is not a quantity, it is a function. More precisely it would be written as $MC(y)/AC(y)$.

It is true that in a competitive setting in the long-run equilibrium $MC(y^*) = AC(y^*)$.

However, this can also be true in a monopoly setting. For example in the special case when $C(y) = c y$ where $c \in \mathbb{R}_+$, you have $$ \forall y: \ MC(y) = c = AC(y). $$ Also the monopolist's profit-maximizing $y^*$ is independent of any fixed costs $F$, but $AC(y^*)$ is not. For example if the inverse demand function is $p(y) = 1 - y$ and the cost function is $C(y) = y^2 + F$ then \begin{align*} y^* & = \arg\max_y (1 - y)y - y^2 - F = 1/4 \\ MC(y^*) & = 2y^* = 1/2 \\ AC(y^*) & = y^* + F/y^* = 1/4 + 4F. \end{align*} In this case $$ \begin{array}{lcl} MC(y^*)/AC(y^*) < 1 & \mbox{ if } & F > 1/8 \\ MC(y^*)/AC(y^*) = 1 & \mbox{ if } & F = 1/8 \\ MC(y^*)/AC(y^*) > 1 & \mbox{ if } & F < 1/8. \end{array} $$ Yet there is exactly one firm in the market, so there is no competition in the classical sense. Nor is it the case that the firm is a natural monopoly at $F = 1/8$. This could mean that this is the 'efficient level of competition', but it is not the case, because when $F = 1/8$ we have \begin{align*} y^* & = 1/4 \\ p(y^*) & = 3/4 \\ AC(y^*) & = 1/2 \\ \text{Profits} & = 1/16 > 0. \end{align*}