#OK to post homework
#Blair Seidler, 2/6/22, Assignment 4

Help:=proc(): print(`No code for this assignment!`): end:

#2. Also used BR1d and BR2d to confirm.
(*
> G := RandDisGame(30, 30);
> BR12(G);
[2, 4, 19, 11, 11, 8, 26, 10, 4, 20, 13, 11, 4, 11, 4, 30, 8, 21, 11, 21, 21, 11, 23, 8, 4, 18, 4, 18, 20, 17]

Fixed points at 21 and 23

> BR21(G);
[22, 12, 9, 25, 5, 8, 29, 28, 9, 8, 19, 4, 18, 13, 1, 19, 11, 23, 9, 8, 28, 2, 28, 17, 9, 18, 23, 18, 12, 2]

Fixed points at 5 and 9

> PureNashEqui(G);
                       {[21, 9], [23, 5]}
> BR1d(G, 9), BR1d(G, 5);
                             21, 23
> BR2d(G, 21), BR2d(G, 23);
                              9, 5


> G := RandDisGame(30, 30);
> BR12(G);
[12, 2, 2, 26, 9, 14, 11, 9, 3, 18, 2, 13, 9, 12, 11, 9, 13, 21, 2, 30, 18, 9, 7, 30, 27, 27, 2, 6, 3, 12]

Fixed point at 2

> BR21(G);
[5, 14, 6, 1, 14, 1, 12, 6, 25, 13, 5, 14, 3, 14, 5, 3, 30, 3, 11, 16, 22, 21, 22, 5, 14, 1, 14, 12, 22, 2]

Fixed point at 14

> PureNashEqui(G);
                           {[2, 14]}
> BR1d(G, 14);
                               2
> BR2d(G, 2);
                               14



> G := RandDisGame(30, 30);
> BR12(G);
[2, 8, 12, 8, 27, 5, 21, 16, 2, 30, 7, 12, 5, 15, 13, 29, 2, 21, 2, 8, 27, 16, 27, 11, 9, 12, 2, 7, 27, 16]

Fixed point at 12

> BR21(G);
[21, 8, 5, 2, 21, 23, 14, 25, 26, 1, 19, 30, 9, 16, 21, 21, 25, 2, 2, 30, 2, 23, 28, 25, 16, 14, 14, 25, 9, 30]

Fixed point at 30

> PureNashEqui(G);
                           {[12, 30]}
> BR1d(G, 30);
                               12
> BR2d(G, 12);
                               30

*)

#3 & 4 Done

#5.
(*
(i) Total profit is (A/4+A/4)*(A-A/4-A/4)=A^2/4

(ii) The Stackelberg model is solved in the book, with q1=A/2 and q2=A/4
      Total profit is (A/2+A/4)*(A-A/2-A/4)=3*A^2/16

(iii) The Cournot model is solved in the book, with q1=q2=A/3 as the solution
      Total profit is (A/3+A/3)*(A-A/3-A/3)=2*A^2/9

The largest total profit comes from (i), the Monopoly model. Stackelberg is the lowest total.
*)

#6.
(*
(i) In a Monopoly, the profit for the (single) company is
P(q)=q*(A-q)^d
P'(q)=(A-q)^d-d*q*(A-q)^(d-1)
Setting this derivative equal to 0 yields
(A-q)^d=d*q*(A-q)^(d-1)
(A-q)/d=q

When d=1, this equation becomes A-q=q, so q=A/2. *check*

The profit at both endpoints (q=0, q=A) is zero
P((A-q)/d)=(A-q)/d*(A-(A-q)/d)^d=(A-q)*(((d-1)*A+q))^d/(d^(d+1))

(ii) In the Stackelberg model (treating q1 as a constant for the moment)
P2(q2)=q2*(A-q1-q2)^d
P2'(q2)=(A-q1-q2)^d-d*q2*(A-q1-q2)^(d-1)
Setting this derivative equal to 0 yields
(A-q1-q2)^d=d*q2*(A-q1-q2)^(d-1)
which has two solutions, (A-q1)/(d+1) and A-q1. We reject the second solution because if q2=A-q1,
the profit for both companies would be zero.
So we will use q2=(A-q1)/(d+1)

Now, we have P1(q1)=q1*(A-q1-(A-q1)/(d+1))^d
P1'(q1)=(A-(d+1)*q1)*(d*(A-q1)/(d+1))^d/(A-q1)
Setting this derivative to zero yields q1=A/(d+1)

The solution is now q1=A/(d+1), q2=dA/(d+1)^2.
When d=1, this gives us q1=A/2, q2=A/4 as expected. *check*

The total profit is then d^2*A^2*(2*d+1)/(d+1)^4. (skipping the algebra)

(iii) In the Cournot model (NE), we can reuse the computation of P2' from (ii) and its analog for P1'.
Solving both of these will give us the system:
q1=(A-q2)/(d+1)
q2=(A-q1)/(d+1)

Solving this system yields q1=q2=A/(d+2).
When d=1, this gives us q1=q2=A/3 as expected. *check*

The total profit is 2*d*A^2/(d + 2)^2