Chapter 12, Problem 12.6P

a)

To determine

To write: the expression for A’s Profit if B produces zero as a function of $q_A$ ,labeled ${\pi }_M$ .

Answer to Problem 12.6P

A’s profit as a function of $q_A$ ,when B produces nothing is ${\pi }_M\text{ =}{q}_A\left(120-q_A\right)$

Explanation of Solution

When the firm A moving first and firm B then market demand: $Q=120-P$ and production is costless.

Firm B’s best-response function: $q_B\text{ =}\frac{120-q_A}{2}$ .

Substitute the above function into A’s profit equation express a $q_A$ labeled ${\pi }_A$ .Substitute the above function into B’s profit equation express a $q_A$ labeled ${\pi }_B$ .

The market demand is

  $Q=120-P\text{ or P=}120-Q\mathrm{......}\left(1\right)$

Where $Q=q_A+q_B$ , $q_A$ is the output produced by firm A, and $q_B$ is the output produced by firm B.

Firm A’s total revenue is the product of quantity produced by firm A and market price.

  $T{R}_A\text{ =P}{q}_A$

  $=\left(120-q_A-q_B\right)q_A$

Since production is costless, a firm’s total revenue equals its profit.

Thus, firm A’s profit function: ${\pi }_A\text{ =}{q}_A\left(120-q_A-q_B\right)\mathrm{.......}\left(2\right)$

Also, firm B’s total revenue is the product of the quantity produced by firm B and the market price.

  $T{R}_B\text{ =P}{q}_B$

  $=\left(120-q_A-q_B\right)q_B$

Since production is costless, a firm’s total revenue equals its profit.

Thus, firm B’s profit function: ${\pi }_B\text{ =}{q}_B\left(120-q_A-q_B\right)\mathrm{.......}\left(3\right)$

Firm B’s best-response function is

  $q_B\text{ =}\frac{120-q_A}{2}\mathrm{........}\left(4\right)$

Substitute $\left(4\right)\text{ in }\left(2\right)$

  ${\pi }_A\text{ =}{q}_A\left[120-q_A-\left(\frac{120-q_A}{2}\right)\right]$

  $\text{=}{q}_A\left(\frac{240-2q_A-120+q_A}{2}\right)$

  $\text{=}{q}_A\left(\frac{120-q_A}{2}\right)$

  $\text{=60}{q}_A-\frac{1}{2}{q}_A{}^2$

Thus, A’s profit as a function of $q_A$ is ${\pi }_A\text{ =60}{q}_A-\frac{1}{2}{q}_A{}^2$

Substitute $\left(4\right)\text{ in }\left(3\right)$

  ${\pi }_B\text{ =}\left(\frac{120-q_A}{2}\right)\left[120-q_A-\left(\frac{120-q_A}{2}\right)\right]$

  $\text{=}\left(\frac{120-q_A}{2}\right)\left(\frac{240-2q_A-120+q_A}{2}\right)$

  $\text{=}\left(\frac{120-q_A}{2}\right)\left(\frac{120-q_A}{2}\right)$

  $\text{=}{\left(\frac{120-q_A}{2}\right)}^2$

Thus, A’s profit as a function of $q_A$ is ${\pi }_B\text{ =}{\left(\frac{120-q_A}{2}\right)}^2$

Set $q_B=0\text{ in}\left(2\right)$

  ${\pi }_M=q_A\left(120-q_A-0\right)$

  $=q_A\left(120-q_A\right)$

Economics Concept Introduction

Introduction: In game theory, one of the strategic game is Stackelberg model. In that case, there are two players’ leader and follower. First move by the leader firm and the follower firm moves sequentially. Both they are compete with quantities.

b)

To determine

the table of entries using the formula from subpart (a).

Answer to Problem 12.6P

$q_A$	${\pi }_A$	${\pi }_B$	${\pi }_M$
0	0	0	0
20	1000	2500	2000
40	1600	1600	1600
60	1800	900	1200
80	1600	400	800
100	1000	100	400
120	0	0	0

Explanation of Solution

The following table shows that the profits of firm A and Firm B of the monopoly:

$q_A$	${\pi }_A$	${\pi }_B$	${\pi }_M$
0	0	0	0
20	$\text{60}\left(20\right)-\frac{1}{2}{\left(20\right)}^2$   $=1200-200$   $=1,000$	${\left(\frac{120-20}{2}\right)}^2$   $={\left(50\right)}^2$   $=2,500$	$20\left(120-20\right)$   $=2,000$
40	$\text{60}\left(40\right)-\frac{1}{2}{\left(40\right)}^2$   $=2,400-800$   $=1,600$	${\left(\frac{120-40}{2}\right)}^2$   $={\left(40\right)}^2$   $=1,600$	$20\left(120-40\right)$   $=1,600$
60	$\text{60}\left(60\right)-\frac{1}{2}{\left(60\right)}^2$   $=3,600-1,800$   $=1,800$	${\left(\frac{120-60}{2}\right)}^2$   $={\left(30\right)}^2$   $=900$	$20\left(120-60\right)$   $=1,200$
80	$\text{60}\left(80\right)-\frac{1}{2}{\left(80\right)}^2$   $=4,800-3,200$   $=1,600$	${\left(\frac{120-80}{2}\right)}^2$   $={\left(20\right)}^2$   $=400$	$20\left(120-80\right)$   $=800$
100	$\text{60}\left(100\right)-\frac{1}{2}{\left(100\right)}^2$   $=6,000-5,000$   $=1,000$	${\left(\frac{120-100}{2}\right)}^2$   $={\left(10\right)}^2$   $=100$	$20\left(120-100\right)$   $=400$
120	$\text{60}\left(120\right)-\frac{1}{2}{\left(120\right)}^2$   $=7,200-7,200$   $=0$	${\left(\frac{120-120}{2}\right)}^2$   $=0$	$20\left(120-120\right)$   $=0$

Economics Concept Introduction

Introduction:

Best-response function refers to the best strategies of the game theory. It is defined that making the most wanted outcome of the player who take the other’s strategies is known as best response.

c)

To determine

the following:

1. Suppose Firm A would choose $q_A\text{ =60}$ in the stackelberg game.
2. If B had a fixed cost of entry equal to the bit more than 400, how much Firm A produce to deter B’s entry.
3. If B had the fixed cost of the entry a bit more than 100.
4. The firm A $q_A\text{ =60}$ to the worthwhile when the B’s entry in these above cases.

Answer to Problem 12.6P

1. Yes, because Firm A earns the highest profit at $q_A\text{ =60}$
2. If firm B’s fixed cost is a bit more than 400,Firm A should produce $q_A\text{ =80}$
3. If firm B’s fixed cost is a bit more than 100,Firm A should produce $q_A\text{ =100}$
4. It might be worthwhile for firm A, as it can occupy the entire market and earn positive profits.

Explanation of Solution

If firm B’s fixed cost is a bit more than 400, Firm A should produce $q_A\text{ =80}$ to deter B’s entry. This is because, the outcome of the Firm B would be negative profit.

If firm B’s fixed cost is a bit more than 100, Firm A should produce $q_A\text{ =100}$ to deter B’s entry. This is because, the outcome of the Firm B would be negative profit.

Economics Concept Introduction

Introduction: One of the contribution of the game theory is best-response function. In Nash equilibrium point of view, every player of the game selected the best response to the opponent player’s strategies.