#OK to post Homework
#Jeton Hida, September 6, 2021, Assignment 1
#Number 1
#Only 1 year old, 2 year old, and 3 year old females can have children 
#The prob that 1 y.o. female has children is p1 
#The prob that 2 y.o. female has children is p2 
#The prob that 3 y.o. female has children is p3 
#Assume only females of this species are born 
#R(n) is the number of females born at year n 
#R(0)=c0, R(1)=c1, R(2)=c2 
#R(n,p1,p2,p3,c0,c1,c2) is the expected number of females born at year n
#R(n)=p1*R(n-1)+p2*R(n-2)+p3*R(n-3) RECURRENCE RELATION
#R(3)=p1*R(2)+p2*R(1)+p3*R(0) =p1*c2+p2*c1+p3*c0 
#R(4)=p1*R(3)+p2*R(2)+p3*R(0) =p1*(p1*c2+p2*c1+p3*c0)+p2*c2+p3*c1 

#Number 2
F := proc(n, p1, p2, p3, c0, c1, c2) option remember;
 if n = 0 then
 c0;
 elif n = 1 then
 c1;
 elif n = 2 then
 c2;
 else p1*F(n - 1, p1, p2, p3, c0, c1, c2) + p2*F(n - 2, p1, p2, p3, c0, c1, c2) + p3*F(n - 3, p1, p2, p3, c0, c1, c2); 
end if;
end proc;
seq(F(n, p1, p2, p3, c0, c1, c2), n = 4);
                  p1 (c0 p3 + c1 p2 + c2 p1) + p2 c2 + p3 c1

#Number 3 #c0=c1=c2=1 n=1000 
#i. Extinction
F(1000, 0.12, 0.12, 0.01, 1, 1, 1);
                                            -354
                              2.266843675 10    
F(1000, 0.25, 0.25, 0.25, 1, 1, 1);
                                            -61
                              1.095771968 10   
#Extinction is reached when p1+p2+p3 < 1 

#ii. Stable Population
F(1000, 0.25, 0.25, 0.5, 1, 1, 1);
                                 1.000000000
F(1000, 0.01, 0.55, 0.44, 1, 1, 1);
                                 1.000000000
#A stable population is reached when p1+p2+p3=1 

#iii. Population Explosion
F(1000, 0.55, 0.45, 0.6, 1, 1, 1);
                                            104
                              6.719690806 10   
F(1000, 0.50, 0.50, 0.01, 1, 1, 1);
                                 707.3273879
F(1000, 0.5, 0.5, 0.10, 1, 1, 1);
                                             25
                               2.775908316 10  
#A population explosion is achieved when p1+p2+p3 > 1