#OK to post homework
#Julian Herman, 9/27/21, Assignment 6

1.
Survival:
n_1(t) = 0.95*n_0(t-1)
n_2(t) = 0.97*n_1(t-1)
n_3(t) = 0.90*n_2(t-1)

Fertility:
n_0(t) = 0.1*n_0(t-1)+1.2*n_1(t-1)+0.9*n_2(t-1)+0.1*n_3(t-1)

n_1(t-1) = 0.95*n_0(t-2)
#found by plugging t=t-1 into n_1(t)

n_2(t-1) = 0.97*n_1(t-2) = 0.97*0.95*n_0(t-3) = 0.9215*n_0(t-3)
#found by plugging t=t-1 into n_2(t) and then t=t-2 into n_1(t)

n_3(t-1) = 0.90*n_2(t-2) = 0.90*0.97*n_1(t-3) = 0.873*0.95*n_0(t-4) = 0.82935*n_0(t-4)
#found by plugging t=t-1 into n_3(t) and then t=t-2 into n_2(t) and then
t=t-3 into n_1(t)

Substituting these equations back into the above Fertility equations:
n_0(t) = 0.1*n_0(t-1)+1.2*0.95*n_0(t-2)+0.9*0.9215*n_0(t-3)+0.1*0.82935*n_0(t-4)

Final recurrence:
n_0(t) = 0.1*n_0(t-1)+1.14*n_0(t-2)+0.82935*n_0(t-3)+0.082935*n_0(t-4)

REC:=[0.1, 1.14, 0.82935, 0.082935]

GrowthCe([0.1, 1.14, 0.82935, 0.082935]);
                          1.385732629



2.
L := Matrix([[0.1, 1.2, 0.9, 0.1], [0.95, 0, 0, 0], [0, 0.97, 0, 0], [0, 0, 0.9, 0]]);
                       [0.1   1.2   0.9  0.1]
                       [                    ]
                       [0.95   0     0    0 ]
                  L := [                    ]
                       [ 0    0.97   0    0 ]
                       [                    ]
                       [ 0     0    0.9   0 ]

with(LinearAlgebra);
E := Eigenvalues(L);
     [            1.38573262885364 + 0. ⅈ             ]
     [                                                           ]
     [&uminus0;0.583351516086360 + 0.403125877485025 ⅈ]
E := [                                                           ]
     [&uminus0;0.583351516086360 - 0.403125877485025 ⅈ]
     [                                                           ]
     [       &uminus0;0.119029596680917 + 0. ⅈ        ]

abs(E[1]);
                        1.38573262885364



3.
P := Matrix([[0.5, 1/6, 1/6, 1/6], [1/5, 0.4, 1/5, 1/5], [7/30, 7/30, 0.3, 7/30], [4/15, 4/15, 4/15, 0.2]]);
                        [      1    1    1 ]
                        [0.5   -    -    - ]
                        [      6    6    6 ]
                        [                  ]
                        [ 1         1    1 ]
                        [ -   0.4   -    - ]
                        [ 5         5    5 ]
                   P := [                  ]
                        [ 7    7         7 ]
                        [--   --   0.3  -- ]
                        [30   30        30 ]
                        [                  ]
                        [ 4    4    4      ]
                        [--   --   --   0.2]
                        [15   15   15      ]

with(MTM);
sum(P, 2);
                         [1.000000000]
                         [           ]
                         [1.000000000]
                         [           ]
                         [1.000000000]
                         [           ]
                         [1.000000000]


S := P^1000;
S := [[0.315196998123827, 0.262664165103189, 0.225140712945591, 0.196998123827392],
      [0.315196998123827, 0.262664165103190, 0.225140712945591, 0.196998123827392],
      [0.315196998123827, 0.262664165103189, 0.225140712945591, 0.196998123827392],
      [0.315196998123827, 0.262664165103189, 0.225140712945591, 0.196998123827392]]

for i in S[1] do
    convert(i, fraction);
end do;
                              168
                              ---
                              533

                              140
                              ---
                              533

                              120
                              ---
                              533

                              105
                              ---
                              533

#0.315196998123827 of the surfers stay in website 1 which is 168/533
#0.262664165103189 of the surfers stay in website 2 which is 140/533
#0.225140712945591 of the surfers stay in website 3 which is 120/533
#0.196998123827392 of the surfers stay in website 4 which is 105/533

#In this mini-website, here is the rankings from most popular to least popular:
#         site1 > site2 > site3 > site4
#site1 is the most popular and site4 is the least popular.