# what is the marginal cost divided by the average cost

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?

• I think it's related to the markup. MC=P and AC is the average cost. So markup is defined as price over cost. Markup is an indicator of market competition. Jan 1 '19 at 12:41

$$MC/AC$$ 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*}