Chapter 9

Needs["DiscreteMath`RSolve`"]

Needs["Graphics`MultipleListPlot`"]

Needs["Graphics`Graphics3D`"]

Needs["Graphics`PlotField`"]

Question 1

Clear[p, Q, q1, q2, pi1, pi2]

Q = q1 + q2
p = A - B Q
pi1 = p q1 - a q1
pi2 = p q2 - a q2

Solve[D[pi1, q1] == 0, q2]

Solve[D[pi2, q2] == 0, q2]

solq1 = 
      Simplify[Solve[-((a - A + 2 B q1)/B) == -((a - A + 
                    B q1)/(2 B)), q1]]

solq2 = 
      Simplify[-((a - A + 2 B q1)/B)]  /. solq1

Simplify[Expand[((-a + A - (2/3) (-a + A))/B)]]

Hence, equilibrium quantities are identical.

Question 2

This question is best done with paper and pencil. Summations over j not equal to i are not conveniently dealt with in software packages.

Question 3

(i)

Clear[p, Q, q1, q2, pi1, pi2]

Q = q1 + q2
p = A - B Q
pi1 = p q1 - a1 q1
pi2 = p q2 - a2 q2

Solve[D[pi1, q1] == 0, q2]

Solve[D[pi2, q2] == 0, q2]

solq1 = 
      Simplify[Solve[-((-A + a1 + 
                    2 B q1)/B) == -((-A + a2 + 
                    B q1)/(2 B)), q1]]

solq2 = 
      Simplify[-((-A + a1 + 2 B q1)/B)] /. solq1

Simplify[Expand[((A - a1 - (2/3) (A - 2 a1 + a2))/B)]]

(ii)

We already have the following as the reaction functions in the static model:
	q1 = ((A - (Subscript[a, 1]))/(2  B)) - ((Subscript[q, 2])/2) for firm 1

	q2 = ((A - (Subscript[a, 2]))/(2  B)) - ((Subscript[q, 1])/2) for firm 2

So the dynamic reaction functions are:
	(Subscript[q, 1])(
          t) = ((A - (Subscript[a, 1]))/(2  B)) - (((Subscript[q, 2])(t))/2)

	(Subscript[q, 2])(
          t) = ((A - (Subscript[a, 2]))/(2  B)) - (((Subscript[q, 1])(t))/2)

RSolve[{q1[t] == ((A - (Subscript[a, 1]))/(2  B)) - q2[t - 1]/2, 
        q2[t] == ((A - (Subscript[a, 2]))/(2  B)) - q1[t - 1]/2, q1[0] == q10, 
        q2[0] == q20}, {q1[t], q2[t]}, t]

This could be simplified but we do not do this here. What matters is that the stability is governed purely by the coefficients of q2(t-1) and q1(t-1) in the reaction functions of firms 1 and 2 respectively. Since these are less than unity, then the system is asymptotically stable. In fact, the coefficients must each take the value of -1/2. Consequently the coefficients of q10 and q20 in the results involve only ((((1/2)))^t) or (((-(1/2)))^t) and so tend to zero in the limit regardless of the values of q10 and q20. (Note that the term B cancells.)

Question 4

Clear[p, Q, q1, q2, pi1, pi2]

(i)

p = 9 - Q
Q = q1 + q2
TC1 = a1 q1
TC2 = a2 q2
pi1 = p q1 - TC1
pi2 = p q2 - TC2

Solve[D[pi1, q1] == 0, q2]

Solve[D[pi2, q2] == 0, q2]

solq1 = 
      Simplify[Solve[9 - a1 - 2 q1 == (1/2) (9 - a2 - q1), 
          q1]]

solq2 = 9 - a1 - 2 q1  /. solq1

Simplify[Expand[9 - a1 - (2/3) (9 - 2 a1 + a2)]]

Simplify[
      (1/3) (9 - 2 a1 + a2) - (1/3) (9 + a1 - 2 a2)]

Hence, equilibrium (Subscript[q, 1]) is greater than equilibrium (Subscript[q, 2]) if (Subscript[a, 1]) < (Subscript[a, 2]).

(ii) (Subscript[a, 1]) = 3 and (Subscript[a, 2]) = 5

The model is then
	p = 9 - Q
Q = (Subscript[q, 1]) + (Subscript[q, 2])
(Subscript[TC, 1]) = 3 (Subscript[q, 1])
(Subscript[TC, 2]) = 5 (Subscript[q, 2])

pi1  /. a1 -> 3

pi2  /.  a2 -> 5

Solve[D[-3 q1 + q1 (9 - q1 - q2), q1] == 0, 
      q1]

Solve[D[-5 q2 + (9 - q1 - q2) q2, q2] == 0, 
      q2]

Our reaction functions are therefore
(Subscript[q, 1])(t) = 
        3 - (1/2) (Subscript[q, 2])(t - 1)

(Subscript[q, 2])(t) = 2 - (1/2) (Subscript[q, 1])(t - 1)

The monopoly points are (3, 0) for firm 1 and (0, 2) for firm 2.

(a) firm 1 monopolist

RSolve[{q1[t] == 3 - (1/2) q2[t - 1], 
        q2[t] == 2 - (1/2) q1[t - 1], q1[0] == 3, 
        q2[0] == 0}, {q1[t], q2[t]}, t]

(b) firm 2 monopolist

RSolve[{q1[t] == 3 - (1/2) q2[t - 1], 
        q2[t] == 2 - (1/2) q1[t - 1], q1[0] == 0, 
        q2[0] == 2}, {q1[t], q2[t]}, t]

q1firm1 = 
        Table[(1/3) (2^(-1 - t)) (3 - (((-1))^t) + 
                (2^(4 + t))), {t, 0, 10}];

q1firm2 = 
        Table[(1/3) (2^(1 - t)) (-3 - (((-1))^t) + 
                (2^(2 + t))), {t, 0, 10}];

MultipleListPlot[q1firm1, q1firm2, PlotJoined -> True, 
        PlotRange -> All, AxesOrigin -> {0, 0}];

The equilibrium is reached sooner when firm 1 is the monopolist. This should not be surprising, since firm 1 has the lower unit costs.

Question 5

Clear[p, Q, q1, q2, q3, pi1, pi2, pi3]

(i)

pi1 = (9 - q1 - q2 - q3) q1 - 5  q1
pi2 = (9 - q1 - q2 - q3) q2 - 5  q2
pi3 = (9 - q1 - q2 - q3) q3 - 5  q3

D[pi1, q1]

D[pi2, q2]

D[pi3, q3]

Solve[{4 - 2 q1 - q2 - q3 == 0, 4 - q1 - 2 q2 - q3 == 0, 
        4 - q1 - q2 - 2 q3 == 0}, {q1, q2, q3}]

Solve[4 - 2 q1 - q2 - q3 == 0, q1]

Solve[4 - q1 - 2 q2 - q3 == 0, q2]

Solve[4 - q1 - q2 - 2 q3 == 0, q3]

Hence, the reaction curves are:
	(Subscript[q, 1])(t) = 
        2 - (1/2) (Subscript[q, 2])(t - 1) - (1/2) (Subscript[q, 3])(
              t - 1)

(Subscript[q, 2])(t) = 
        2 - (1/2) (Subscript[q, 1])(t - 1) - (1/2) (Subscript[q, 3])(
              t - 1)

(Subscript[q, 3])(t) = 
        2 - (1/2) (Subscript[q, 1])(t - 1) - (1/2) (Subscript[q, 2])(t - 1)

(iii)

RSolve[{q1[t] == 2 - (1/2) q2[t - 1] - (1/2) q3[t - 1], 
        q2[t] == 2 - (1/2) q1[t - 1] - (1/2) q3[t - 1], 
        q3[t] == 2 - (1/2) q1[t - 1] - (1/2) q2[t - 1], 
        q1[0] == q10, q2[0] == q20, q3[0] == q30}, {q1[t], 
        q2[t], q3[t]}, t]

Because of the occurrence of the terms (((-1))^t) then the system also oscillates, eventually oscillating with constant amplitude.

Question 6

Clear[p, Q, q1, q2, q3, pi1, pi2, pi3]

(i)

pi1 = (9 - q1 - q2 - q3) q1 - 3  q1
pi2 = (9 - q1 - q2 - q3) q2 - 2  q2
pi3 = (9 - q1 - q2 - q3) q3 - q3

D[pi1, q1]

D[pi2, q2]

D[pi3, q3]

Solve[{6 - 2 q1 - q2 - q3 == 0, 7 - q1 - 2 q2 - q3 == 0, 
        8 - q1 - q2 - 2 q3 == 0}, {q1, q2, q3}]

(ii)

Solve[6 - 2 q1 - q2 - q3 == 0, q1]

Solve[7 - q1 - 2 q2 - q3 == 0, q2]

Solve[8 - q1 - q2 - 2 q3 == 0, q3]

Hence, the reaction curves are:
	(Subscript[q, 1])(t) = 
        3 - (1/2) (Subscript[q, 2])(t - 1) - (1/2) (Subscript[q, 3])(
              t - 1)

(Subscript[q, 2])(t) = 
        (7/2) - (1/2) (Subscript[q, 1])(t - 1) - (1/2) (Subscript[q, 3])(
              t - 1)

(Subscript[q, 3])(t) = 
        4 - (1/2) (Subscript[q, 1])(t - 1) - (1/2) (Subscript[q, 2])(t - 1)

Monopoly points are (3, 0, 0) for firm 1, (0, 7/2, 0) for firm 2 and (0, 0, 4) for firm 3.

(a) firm 1 monopolist

Clear[q1, q2, q3]

RSolve[{q1[t] == 3 - (1/2) q2[t - 1] - (1/2) q3[t - 1], 
        q2[t] == (7/2) - (1/2) q1[t - 1] - (1/2) q3[t - 1], 
        q3[t] == 4 - (1/2) q1[t - 1] - (1/2) q2[t - 1], 
        q1[0] == 3, q2[0] == 0, q3[0] == 0}, {q1[t], q2[t], 
        q3[t]}, t]

q1[t_] := -3 (-(1/4) - (2^(-t)) + 
            (1/4) (E^(I π t)))

q2[t_] := 
      (7/4) - (2^(-t)) - 
        (3/4) (E^(I π t))

q3[t_] := (11/4) - ((3 (((-1))^t))/4) - (2^(1 - t))

points6a = Table[{q1[t], q2[t], q3[t]}, {t, 0, 10}];

ScatterPlot3D[points6a, PlotJoined -> True, 
        AxesLabel -> {"q1", "q2", "q3"}];

(b) firm 2 as monopolist

Clear[q1, q2, q3]

RSolve[{q1[t] == 3 - (1/2) q2[t - 1] - (1/2) q3[t - 1], 
        q2[t] == (7/2) - (1/2) q1[t - 1] - (1/2) q3[t - 1], 
        q3[t] == 4 - (1/2) q1[t - 1] - (1/2) q2[t - 1], 
        q1[0] == 0, q2[0] == 7/2, q3[0] == 0}, {q1[t], 
        q2[t], q3[t]}, t]

q1[t_] := (1/12) (9 - 7 (((-1))^t) - (2^(1 - t)))

q2[t_] := -(7/12) (-3 - 
            (2^(2 - t)) + (E^(I π t)))

q3[t_] := 
      (1/3) (2^(-2 - t)) (-26 - 7 (((-2))^t) + 
            33 (2^t))

points6b = Table[{q1[t], q2[t], q3[t]}, {t, 0, 10}];

ScatterPlot3D[points6b, PlotJoined -> True, 
        AxesLabel -> {"q1", "q2", "q3"}];

(c) firm 3 monopolist

Clear[q1, q2, q3]

RSolve[{q1[t] == 3 - (1/2) q2[t - 1] - (1/2) q3[t - 1], 
        q2[t] == (7/2) - (1/2) q1[t - 1] - (1/2) q3[t - 1], 
        q3[t] == 4 - (1/2) q1[t - 1] - (1/2) q2[t - 1], 
        q1[0] == 0, q2[0] == 0, q3[0] == 4}, {q1[t], q2[t], 
        q3[t]}, t]

q1[t_] := 
      (3/4) - ((2^(-t))/3) - 
        (5/12) (E^(I π t))

q2[t_] := 
      (1/12) (21 - 5 (((-1))^t) - (2^(4 - t)))

q3[t_] := 
      (11/4) + ((5 (2^(-t)))/3) - 
        (5/12) (E^(I π t))

points6c = Table[{q1[t], q2[t], q3[t]}, {t, 0, 10}];

ScatterPlot3D[points6c, PlotJoined -> True, 
        AxesLabel -> {"q1", "q2", "q3"}];

(iii)

Clear[q1, q2, q3]

RSolve[{q1[t] == 3 - (1/2) q2[t - 1] - (1/2) q3[t - 1], 
        q2[t] == (7/2) - (1/2) q1[t - 1] - (1/2) q3[t - 1], 
        q3[t] == 4 - (1/2) q1[t - 1] - (1/2) q2[t - 1], 
        q1[0] == q10, q2[0] == q20, q3[0] == q30}, {q1[t], 
        q2[t], q3[t]}, t]

Question 7

Clear[p, Q, q1, q2, q3, pi1, pi2, pi3]

(i)

pi1 = (15 - 2  q1 - 2  q2 - 2  q3) q1 - 
        5  q1
pi2 = (15 - 2  q1 - 2  q2 - 2  q3) q2 - 
        3  q2
pi3 = (15 - 2  q1 - 2  q2 - 2  q3) q3 - 2  q3

D[pi1, q1]

D[pi2, q2]

D[pi3, q3]

Solve[{10 - 4 q1 - 2 q2 - 2 q3 == 0, 
        12 - 2 q1 - 4 q2 - 2 q3 == 0, 
        13 - 2 q1 - 2 q2 - 4 q3 == 0}, {q1, q2, q3}]

Solve[10 - 4 q1 - 2 q2 - 2 q3 == 0, q1]

Solve[12 - 2 q1 - 4 q2 - 2 q3 == 0, q2]

Solve[13 - 2 q1 - 2 q2 - 4 q3 == 0, q3]

Hence, the reaction functions are:

	(Subscript[q, 1])(t) = 
        (5/2) - (1/2) (Subscript[q, 2])(t - 1) - (1/2) (Subscript[q, 3])(
              t - 1)

(Subscript[q, 2])(t) = 
        3 - (1/2) (Subscript[q, 1])(t - 1) - (1/2) (Subscript[q, 3])(
              t - 1)

(Subscript[q, 3])(t) = 
        (13/4) - (1/2) (Subscript[q, 1])(t - 1) - (1/2) (Subscript[q, 2])(
              t - 1)

Monopoly points are ((5/2), 0, 0) for firm 1, (0, 3, 0) for firm 2 and (0, 0, (13/4)) for firm 3.

(a) firm 1 monopolist

RSolve[{q1[
            t] == (5/2) - (1/2) q2[t - 1] - (1/2) 
              q3[t - 1], 
        q2[t] == 3 - (1/2) q1[t - 1] - (1/2) q3[t - 1], 
        q3[t] == (13/4) - (1/2) q1[t - 1] - (1/2) 
              q2[t - 1], q1[0] == 5/2, q2[0] == 0, 
        q3[0] == 0}, {q1[t], q2[t], q3[t]}, t]

q1[t_] := -5 (2^(-3 - t)) (-4 - (2^t) + 
            (2^t) (E^(I π t)))

q2[t_] := (13/8) - ((5 (((-1))^t))/8) - (2^(-t))

q3[t_] := 
      (2^(-3 - t)) (-12 - 5 (((-2))^t) + 
            17 (2^t))

points7a = Table[{q1[t], q2[t], q3[t]}, {t, 0, 10}];

ScatterPlot3D[points7a, PlotJoined -> True, 
        AxesLabel -> {"q1", "q2", "q3"}];

(b) firm 2 as monopolist

Clear[q1, q2, q3]

RSolve[{q1[
            t] == (5/2) - (1/2) q2[t - 1] - (1/2) 
              q3[t - 1], 
        q2[t] == 3 - (1/2) q1[t - 1] - (1/2) q3[t - 1], 
        q3[t] == (13/4) - (1/2) q1[t - 1] - (1/2) 
              q2[t - 1], q1[0] == 0, q2[0] == 3, 
        q3[0] == 0}, {q1[t], q2[t], q3[t]}, t]

q1[t_] := 
      (1/3) (2^(-3 - t)) (-4 - 11 (((-2))^t) + 
            15 (2^t))

q2[t_] := 
      (1/3) (2^(-3 - t)) (44 - 11 (((-2))^t) + 
            39 (2^t))

q3[t_] := 
      (1/24) (51 - 11 (((-1))^t) - 5 (2^(3 - t)))

points7b = Table[{q1[t], q2[t], q3[t]}, {t, 0, 10}];

ScatterPlot3D[points7b, PlotJoined -> True, 
        AxesLabel -> {"q1", "q2", "q3"}];

(c) firm 3 monopolist

Clear[q1, q2, q3]

RSolve[{q1[
            t] == (5/2) - (1/2) q2[t - 1] - (1/2) 
              q3[t - 1], 
        q2[t] == 3 - (1/2) q1[t - 1] - (1/2) q3[t - 1], 
        q3[t] == (13/4) - (1/2) q1[t - 1] - (1/2) 
              q2[t - 1], q1[0] == 0, q2[0] == 0, 
        q3[0] == 13/4}, {q1[t], q2[t], q3[t]}, t]

q1[t_] := 
      (2^(-3 - t)) (-2 - 3 (((-2))^t) + 
            5 (2^t))

q2[t_] := 
      (2^(-3 - t)) (-10 - 3 (((-2))^t) + 
            13 (2^t))

q3[t_] := 
      (1/8) (17 + 3 (2^(2 - t)) - 
            3 (E^(I π t)))

points7c = Table[{q1[t], q2[t], q3[t]}, {t, 0, 10}];

ScatterPlot3D[points7c, PlotJoined -> True, 
        AxesLabel -> {"q1", "q2", "q3"}];

(iii)

Clear[q1, q2, q3]

RSolve[{q1[
            t] == (5/2) - (1/2) q2[t - 1] - (1/2) 
              q3[t - 1], 
        q2[t] == 3 - (1/2) q1[t - 1] - (1/2) q3[t - 1], 
        q3[t] == (13/4) - (1/2) q1[t - 1] - (1/2) 
              q2[t - 1], q1[0] == q10, q2[0] == q20, 
        q3[0] == q30}, {q1[t], q2[t], q3[t]}, t]

No. Because of the presence of (((-1))^t), the system will eventually oscillate.

Question 8

Clear[p, Q, q1, q2, pi1, pi2]

(i)

(a)

Q = q1 + q2
p = 20 - 3  Q
pi1 = p q1 - 4  q1
pi2 = p q2 - 4  q2

D[pi1, q1]

D[pi2, q2]

Solve[{16 - 3 q1 - 3 (q1 + q2) == 0, 
        16 - 3 q2 - 3 (q1 + q2) == 0}, {q1, q2}]

(b)

Clear[p, Q, q1, q2, q3, pi1, pi2, pi3]

Q = q1 + q2 + q3
p = 20 - 3  Q
pi1 = p q1 - 4  q1
pi2 = p q2 - 4  q2
pi3 = p q3 - 4  q3

D[pi1, q1]

D[pi2, q2]

D[pi3, q3]

Solve[{16 - 3 q1 - 3 (q1 + q2 + q3) == 0, 
        16 - 3 q2 - 3 (q1 + q2 + q3) == 0, 
        16 - 3 q3 - 3 (q1 + q2 + q3) == 0}, {q1, q2, 
        q3}]

(c)

Clear[p, Q, q1, q2, pi1, pi2]

Q = q1 + q2
p = 20 - 3  Q
pi1 = p q1 - 4  (q1^2)
pi2 = p q2 - 4  (q2^2)

D[pi1, q1]

D[pi2, q2]

Solve[{20 - 11 q1 - 3 (q1 + q2) == 0, 
        20 - 11 q2 - 3 (q1 + q2) == 0}, {q1, q2}]

(d)

Clear[p, Q, q1, q2, q3, pi1, pi2, pi3]

Q = q1 + q2 + q3
p = 20 - 3  Q
pi1 = p q1 - 4  (q1^2)
pi2 = p q2 - 4  (q2^2)
pi3 = p q3 - 4  (q3^2)

D[pi1, q1]

D[pi2, q2]

D[pi3, q3]

Solve[{20 - 11 q1 - 3 (q1 + q2 + q3) == 0, 
        20 - 11 q2 - 3 (q1 + q2 + q3) == 0, 
        20 - 11 q3 - 3 (q1 + q2 + q3) == 0}, {q1, q2, 
        q3}]

(ii)

Here we are comparing models (a) and (b).

Clear[p, Q, q1, q2, pi1, pi2]

Solve[16 - 3 q1 - 3 (q1 + q2) == 0, q1]

Solve[16 - 3 q2 - 3 (q1 + q2) == 0, q2]

The two reaction functions are therefore:

	(Subscript[q, 1])(t) = 
        (8/3) - (1/2) (Subscript[q, 2])(t - 1)

(Subscript[q, 2])(t) = 
        (8/3) - (1/2) (Subscript[q, 1])(t - 1)

If firm 1 is the monopolist then the initial point is (8/3, 0).

RSolve[{q1[t] == (8/3) - (1/2) q2[t - 1], 
        q2[t] == (8/3) - (1/2) q1[t - 1], q1[0] == 8/3, 
        q2[0] == 0}, {q1[t], q2[t]}, t]

q1[t_] := 
      (1/9) (2^(2 - t)) (3 - (((-1))^t) + (2^(2 + t)))

q2[t_] := 
      (1/9) (2^(2 - t)) (-3 - (((-1))^t) + 
            (2^(2 + t)))

points8a = Table[{q1[t], q2[t]}, {t, 0, 10}];

ListPlot[points8a, PlotJoined -> True, 
        AxesOrigin -> {1.8, 0}];

Clear[p, Q, q1, q2, q3, pi1, pi2, pi3]

Solve[16 - 3 q1 - 3 (q1 + q2 + q3) == 0, q1]

Solve[16 - 3 q2 - 3 (q1 + q2 + q3) == 0, q2]

Solve[16 - 3 q3 - 3 (q1 + q2 + q3) == 0, q3]

The three reaction functions are therefore:

	(Subscript[q, 1])(t) = 
        (8/3) - (1/2) (Subscript[q, 2])(t - 1) - (1/2) (Subscript[q, 3])(
              t - 1)

(Subscript[q, 2])(t) = 
        (8/3) - (1/2) (Subscript[q, 1])(t - 1) - (1/2) (Subscript[q, 3])(
              t - 1)

(Subscript[q, 3])(t) = 
        (8/3) - (1/2) (Subscript[q, 1])(t - 1) - (1/2) (Subscript[q, 2])(
              t - 1)

If firm 1 is the monopolist then the initial point is (8/3, 0, 0).

RSolve[{q1[
            t] == (8/3) - (1/2) q2[t - 1] - (1/2) 
              q3[t - 1], 
        q2[t] == (8/3) - (1/2) q1[t - 1] - (1/2) q3[t - 1], 
        q3[t] == (8/3) - (1/2) q1[t - 1] - (1/2) q2[t - 1], 
        q1[0] == 8/3, q2[0] == 0, q3[0] == 0}, {q1[t], 
        q2[t], q3[t]}, t]

q1[t_] := -(1/9) (2^(2 - t)) (-4 + (((-2))^t) - 
            3 (2^t))

q2[t_] := 
      (4/9) (3 - (2^(-t)) (2 + (((-2))^t)))

q3[t_] := 
      (4/9) (3 - (2^(-t)) (2 + (((-2))^t)))

points8b = Table[{q1[t], q2[t], q3[t]}, {t, 0, 10}];

ScatterPlot3D[points8b, PlotJoined -> True, 
        AxesLabel -> {"q1", "q2", "q3"}];

points8b1 = Table[{t, q1[t]}, {t, 0, 10}];

ListPlot[points8b1, PlotJoined -> True, 
        AxesLabel -> {"t", "q1"}];

When n = 2 the system was stable. However, for n = 3 the system soon begins to oscillate. Here we show it only for q1 but it applies equally to q2 and q3, which is apparent from the presence of (((-1))^t) in the solutions.

(iii)

Part (ii) already compares (a) and (b) and so here we consider just (c) and (d).

Clear[p, Q, q1, q2, pi1, pi2]

Solve[20 - 11 q1 - 3 (q1 + q2) == 0, q1]

Solve[20 - 11 q2 - 3 (q1 + q2) == 0, q2]

Hence the reaction curves are:

	(Subscript[q, 1])(t) = 
        (10/7) - (3/14) (Subscript[q, 2])(t - 1)

(Subscript[q, 2])(t) = 
        (10/7) - (3/14) (Subscript[q, 1])(t - 1)

If firm 1 is a monopoly then the initial point is (10/7, 0).

RSolve[{q1[t] == (10/7) - (3/14) q2[t - 1], 
        q2[t] == (10/7) - (3/14) q1[t - 1], 
        q1[0] == 10/7, q2[0] == 0}, {q1[t], q2[t]}, t]

q1[t_] := 
      (5/17) (2^(-t)) (7^(-1 - t)) (17 (3^t) + 
            (2^(2 + t)) (7^(1 + t)) - 
            11 (3^t) (E^(I π t)))

q2[t_] := 
      (5/119) (28 - 11 (((-(3/14)))^t) - 
            17 ((((3/14)))^t))

points8c = Table[{q1[t], q2[t]}, {t, 0, 10}];

ListPlot[points8c, PlotJoined -> True, 
        AxesLabel -> {"q1", "q2"}];

Clear[p, Q, q1, q2, q3, pi1, pi2, pi3]

Solve[20 - 11 q1 - 3 (q1 + q2 + q3) == 0, q1]

Solve[20 - 11 q2 - 3 (q1 + q2 + q3) == 0, q2]

Solve[20 - 11 q3 - 3 (q1 + q2 + q3) == 0, q3]

Hence the reaction curves are:

	(Subscript[q, 1])(t) = 
        (10/7) - (3/14) (Subscript[q, 2])(t - 1) - (3/14) (Subscript[q, 3])(
              t - 1)

(Subscript[q, 2])(t) = 
        (10/7) - (3/14) (Subscript[q, 21])(t - 1) - (3/14) (Subscript[q, 3])(
              t - 1)

(Subscript[q, 3])(t) = 
        (10/7) - (3/14) (Subscript[q, 1])(t - 1) - (3/14) (Subscript[q, 2])(
              t - 1)

With firm 1 the monopolist, then the initial point is (10/7, 0, 0).

RSolve[{q1[
            t] == (10/7) - (3/14) q2[t - 1] - (3/14) 
              q3[t - 1], 
        q2[t] == (10/7) - (3/14) q1[t - 1] - (3/14) 
              q3[t - 1], 
        q3[t] == (10/7) - (3/14) q1[t - 1] - (3/14) 
              q2[t - 1], q1[0] == 10/7, q2[0] == 0, 
        q3[0] == 0}, {q1[t], q2[t], q3[t]}, t]

q1[t_] := -(1/3) (2^(-t)) (7^(-1 - t)) (11 (((-6))^t) - 20 (3^t) - 3 (2^t) (7^(1 + t)))

q2[t_] := -(1/3) (2^(-t)) (7^(-1 - t)) (11 (((-6))^t) + 10 (3^t) - 3 (2^t) (7^(1 + t)))

q3[t_] := -(1/3) (2^(-t)) (7^(-1 - t)) (11 (((-6))^t) + 10 (3^t) - 3 (2^t) (7^(1 + t)))

points8d = Table[{q1[t], q2[t], q3[t]}, {t, 0, 10}];

ScatterPlot3D[points8d, PlotJoined -> True, 
        AxesLabel -> {"q1", "q2", "q3"}];

points8d1 = Table[{t, q1[t]}, {t, 0, 10}];

ListPlot[points8d1, PlotJoined -> True, 
        AxesLabel -> {"t", "q1"}];

When marginal costs are not constant then the system is more stable and equilibrium is reached sooner. The oscillatory pattern observed for n = 3 under constant marginal costs now disappears and the system is stable.

Question 9

Clear[p, Q, q1, q2, pi1, pi2]

(i)

The Cournot solution is the same as for model 8(i)(a), namely ((16/9), (16/9)).

(ii)

In question 8(ii) we derived the reaction functions, which in their continuous dynamic form are:

(Subscript[x, 1])(t) = 
        (8/3) - (1/2) (Subscript[q, 2])(t)

(Subscript[x, 2])(t) = (8/3) - (1/2) (Subscript[q, 1])(t)

Hence

	\(TraditionalForm\`\(q\_1\%'\)(t) = 
        0.2 \((8\/3 - \(1\/2\) \(\(q\_2\)(t)\) - \(q\_1\)(
                t))\)
\)
\(TraditionalForm\`\(q\_2\%'\)(t) = 
        0.2 \((8\/3 - \(1\/2\) \(\(q\_1\)(t)\) - \(q\_2\)(t))\)\)

Simplify[0.2 ((8/3) - (1/2) q2[t] - q1[t])]

Simplify[0.2 ((8/3) - (1/2) q1[t] - q2[t])]

The matrix of the system is therefore

mA = {{-0.2, -0.1}, {-0.1, -0.2}}

Tr[mA]

Det[mA]

Tr[mA]^2 - 4 Det[mA]

Hence the equilibrium is dynamically stable.

(iii)

The initial points are (8/3, 0) for firm 1 a monopolist and (0, 8/3) for firm 2 being a monopolist and (0, 0) indicating both firms are to enter the industry.

arrows9 = 
        PlotVectorField[{0.2 ((8/3) - (1/2) q2 - q1), 
            0.2 ((8/3) - (1/2) q1 - q2)}, {q1, 0, 5}, {q2, 0, 5}, 
          Axes -> True, AxesLabel -> {"q1", "q2"}, 
          ScaleFunction -> (1 &), 
          AspectRatio -> 0.8];

path91 = 
      NDSolve[{q1'[t] == 
            0.2 ((8/3) - (1/2) q2[t] - q1[t]), q2'[
              t] == 0.2 ((8/3) - (1/2) q1[t] - q2[t]), 
          q1[0] == 8/3, q2[0] == 0}, {q1, q2}, {t, 0, 
          20}]

path92 = 
      NDSolve[{q1'[t] == 
            0.2 ((8/3) - (1/2) q2[t] - q1[t]), q2'[
              t] == 0.2 ((8/3) - (1/2) q1[t] - q2[t]), 
          q1[0] == 0, q2[0] == 8/3}, {q1, q2}, {t, 0, 
          20}]

path93 = 
      NDSolve[{q1'[t] == 
            0.2 ((8/3) - (1/2) q2[t] - q1[t]), q2'[
              t] == 0.2 ((8/3) - (1/2) q1[t] - q2[t]), 
          q1[0] == 0, q2[0] == 0}, {q1, q2}, {t, 0, 
          20}]

traj91 = 
        ParametricPlot[Evaluate[{q1[t], q2[t]}  /. path91], {t, 0, 20}, 
          DisplayFunction -> Identity, 
          PlotStyle -> {{RGBColor[1, 0, 
                  0]}, {Thickness[ .02]}}];

traj92 = 
        ParametricPlot[Evaluate[{q1[t], q2[t]}  /. path92], {t, 0, 20}, 
          DisplayFunction -> Identity, 
          PlotStyle -> {{RGBColor[1, 0, 
                  0]}, {Thickness[ .02]}}];

traj93 = 
        ParametricPlot[Evaluate[{q1[t], q2[t]}  /. path93], {t, 0, 20}, 
          DisplayFunction -> Identity, 
          PlotStyle -> {{RGBColor[1, 0, 
                  0]}, {Thickness[ .02]}}];

lines9 = 
        Plot[{16/3 - 2  q1, 8/3 - (1/2) q1}, {q1, 0, 5}, 
          PlotRange -> {0, 5}, 
          PlotStyle -> {{RGBColor[0, 0, 1]}, {RGBColor[0, 0, 
                  1]}, {Thickness[ .02]}}, 
          DisplayFunction -> Identity];

Show[arrows9, traj91, traj92, traj93, lines9, 
        PlotRange -> {{0, 5}, {0, 5}}];

The diagram reveals the stability of the Cournot solution.

Question 10

Clear[p, Q, q1, q2, pi1, pi2, mA]

(i)

The Cournot solution is the same as for model 8(i)(c), namely ((20/17), (20/17)).

(ii)

In question 8(iii) we derived the reaction functions, which in their continuous dynamic form are:

	(Subscript[x, 1])(t) = 
        (10/7) - (3/14) (Subscript[q, 2])(t)

(Subscript[x, 2])(t) = 
        (10/7) - (3/14) (Subscript[q, 1])(t)

Hence

	\(TraditionalForm\`\(q\_1\%'\)(
          t) = \(\(0.2\) \((10\/7 - \(3\/14\) \(\(q\_2\)(t)\) - \(q\_1\)(
                t))\)\(
\)\)\)
\(TraditionalForm\`\(q\_2\%'\)(t) = 
        0.2 \((10\/7 - \(3\/14\) \(\(q\_1\)(t)\) - \(q\_2\)(t))\)\)

Simplify[0.2 ((10/7) - (3/14) q2[t] - q1[t])]

Simplify[0.2 ((10/7) - (3/14) q1[t] - q2[t])]

The matrix of the system is therefore

mA = {{-0.2, -0.0428571}, {-0.0428571, -0.2}}

Tr[mA]

Det[mA]

Tr[mA]^2 - 4 Det[mA]

Hence the equilibrium is dynamically stable.

(iii)

The initial points are ((10/7), 0) for firm 1 a monopolist and (0, (10/7)) for firm 2 being a monopolist and (0, 0) indicating both firms are to enter the industry.

arrows10 = 
        PlotVectorField[{0.2 ((10/7) - (3/14) q2 - q1), 
            0.2 ((10/7) - (3/14) q1 - q2)}, {q1, 0, 5}, {q2, 0, 
            5}, Axes -> True, AxesLabel -> {"q1", "q2"}, 
          ScaleFunction -> (1 &), 
          AspectRatio -> 0.8];

path101 = 
      NDSolve[{q1'[t] == 
            0.2 ((10/7) - (3/14) q2[t] - q1[t]), q2'[
              t] == 0.2 ((10/7) - (3/14) q1[t] - q2[t]), 
          q1[0] == 10/7, q2[0] == 0}, {q1, q2}, {t, 0, 
          20}]

path102 = 
      NDSolve[{q1'[t] == 
            0.2 ((10/7) - (3/14) q2[t] - q1[t]), q2'[
              t] == 0.2 ((10/7) - (3/14) q1[t] - q2[t]), 
          q1[0] == 0, q2[0] == 10/7}, {q1, q2}, {t, 0, 
          20}]

path103 = 
      NDSolve[{q1'[t] == 
            0.2 ((10/7) - (3/14) q2[t] - q1[t]), q2'[
              t] == 0.2 ((10/7) - (3/14) q1[t] - q2[t]), 
          q1[0] == 0, q2[0] == 0}, {q1, q2}, {t, 0, 
          20}]

traj101 = 
        ParametricPlot[Evaluate[{q1[t], q2[t]}  /. path101], {t, 0, 20}, 
          DisplayFunction -> Identity, 
          PlotStyle -> {{RGBColor[1, 0, 
                  0]}, {Thickness[ .02]}}];

traj102 = 
        ParametricPlot[Evaluate[{q1[t], q2[t]}  /. path102], {t, 0, 20}, 
          DisplayFunction -> Identity, 
          PlotStyle -> {{RGBColor[1, 0, 
                  0]}, {Thickness[ .02]}}];

traj103 = 
        ParametricPlot[Evaluate[{q1[t], q2[t]}  /. path103], {t, 0, 20}, 
          DisplayFunction -> Identity, 
          PlotStyle -> {{RGBColor[1, 0, 
                  0]}, {Thickness[ .02]}}];

lines10 = 
        Plot[{20/3 - (14/3) q1, 10/7 - (3/14) q1}, {q1, 0, 5}, 
          PlotRange -> {0, 5}, 
          PlotStyle -> {{RGBColor[0, 0, 1]}, {RGBColor[0, 0, 
                  1]}, {Thickness[ .02]}}, 
          DisplayFunction -> Identity];

Show[arrows10, traj101, traj102, traj103, lines10, 
        PlotRange -> {{0, 5}, {0, 5}}];

The diagram reveals the stability of the Cournot solution for the chosen initial points.

DSolve[{q1'[t] == 
          0.2 ((10/7) - (3/14) q2[t] - q1[t]), q2'[
            t] == 0.2 ((10/7) - (3/14) q1[t] - q2[t]), 
        q1[0] == q10, q2[0] == q20}, {q1[t], q2[t]}, t]

Since all coefficients of q10 and q20 involve (e^(-0.242857 t)), (e^(0.242857  t)), or (e^(0.0857143  t)) then the system converges on the Cournot equilibrium regardless of the initial values.

Questions 11-15

These questions are more easily done with a spreadsheet and are therefore not undertaken here with Mathematica.