#OK to post homework 
#Natalya Ter-Saakov, Jan 30, Assignment 4 
read "C4.txt":
read "hw2NatalyaTer-Saakov.txt":

G1:= RandDisGame(30,30):
PureNashEqui(G1);
# The equilibrium is {[10, 18]}

FP(BR12(G1));FP(BR21(G1));
#The fixed point of B12 is {10} and the fixed point of B21 is {18}

G2:= RandDisGame(30,30):
PureNashEqui(G2);
# The equilibrium is {[21, 2]}

FP(BR12(G2));FP(BR21(G2));
#The fixed point of B12 is {21} and the fixed point of B21 is {2}

G3:= RandDisGame(30,30):
PureNashEqui(G3);
# The equilibrium is {[27, 9]}

FP(BR12(G3));FP(BR21(G3));
#The fixed point of B12 is {27} and the fixed point of B21 is {9}


# Problem 5 
# In a monopoly, each player makes A/4. 
# The total profit is TotalProfit = (A/4)(A-A/2) + (A/4)(A-A/2) = A^2/4. 

# In the Stackelberg model, if player 2's profits are P2(q1) = q2(A-q1-q2) which are optimized by making q2 = (A-q1)/2 worth of product. 
# Then q1's profits are P1 = q1(A/2-q1/2) which is optimized by q1 = A/2. 
# Then q2= A/4 and the total profit is TotalProfit = A/2(A-3A/4) + A/4 (A-3A/4) = 3A^2/16. 

# In the Cournot model, we have that q1* = (A-q2*)/2 and q2* = (A-q1*)/2, so q1*=q2*= A/3 and the total profit is TotalProfit = 2A/3(A-2A/3) = 2A^2/9.


# Problem 6 
# In a monopoly, the total profit is TP = q*(A-q)^d which is maximized by taking the derivative and setting it equal to 0:
# (A-q)^d-d*q*(A-q)^(d-1) = 0. 
# So q/2=q1=q2=A/(2d+2) and the total profit is TP = A/(d+1) * (d*A/(d+1)^d. 

# In the Stackelberg model, player 2's profits are P2 = q2*(A-q1-q2)^d. 
# Taking the derivative and setting it equal to 0:
# (A-q1-q2)^d - d*q2*(A-q1-q2)^(d-1)=0 <=> A-q1-q2 = d*q2, so this is optimized by q2(q1) = (A-q1)/(d+1).
# Then, player 1's profits are P1 = q1*(A-q1-(A-q1)/(d+1))^d = q1*(d*A-d*q1)^d. 
# This is optimized by q1 = A/(d+1) and so q2 = d*A/(d+1)^2. 

# The Nash Equilibrium, we have q1 = (A-q2)/(d+1) and q2 = (A-q1)/(d+1).
# So q1 = (A-(A-q1)/(d+1))/(d+1) = (d*A+q1)/(d+1)^2 which means that the optimal strategies are q1=q2=A/(d+2).