Linear Algebra Exercises-n-Answers.pdf

154 Linear Algebra, by Hefferon
0.54442
0.33091
0.29076
This is close to a steady state.
4 (a) This is the relevant system of equations.
(1 − 2p) · s U (n) + p · t A (n) + p · t B (n) = s U (n + 1)
p · s U (n) + (1 − 2p) · t A (n) = t A (n + 1)
p · s U (n) + (1 − 2p) · t B (n) = t B (n + 1)
p · t A (n) + s A (n) = s A (n + 1)
p · t B (n) + s B (n) = s B (n + 1)
Thus we have this. ⎛
⎜
1 − 2p p p 0 0 ⎞ ⎛ ⎞ ⎛ ⎞
s U (n) s U (n + 1)
⎜⎜⎜⎝ p 1 − 2p 0 0 0
t A (n)
p 0 1 − 2p 0 0
⎟ ⎜t B (n)
⎟
0 p 0 1 0⎠
⎝s A (n) ⎠ = t A (n + 1)
⎜t B (n + 1)
⎟
⎝s A (n + 1) ⎠
0 0 p 0 1 s B (n) s B (n + 1)
(b) This is the Octave code, with the output removed.
octave:1> T=[.5,.25,.25,0,0;
> .25,.5,0,0,0;
> .25,0,.5,0,0;
> 0,.25,0,1,0;
> 0,0,.25,0,1]
T =
0.50000 0.25000 0.25000 0.00000 0.00000
0.25000 0.50000 0.00000 0.00000 0.00000
0.25000 0.00000 0.50000 0.00000 0.00000
0.00000 0.25000 0.00000 1.00000 0.00000
0.00000 0.00000 0.25000 0.00000 1.00000
octave:2> p0=[1;0;0;0;0]
octave:3> p1=T*p0
octave:4> p2=T*p1
octave:5> p3=T*p2
octave:6> p4=T*p3
octave:7> p5=T*p4
Here is the output. The probability of ending at s A is about 0.23.
⃗p 0 ⃗p 1 ⃗p 2 ⃗p 3 ⃗p 4 ⃗p 5
s U 1
t A 0
t B 0
s A 0
s B 0
0.50000
0.25000
0.25000
0.00000
0.00000
0.375000
0.250000
0.250000
0.062500
0.062500
0.31250
0.21875
0.21875
0.12500
0.12500
0.26562
0.18750
0.18750
0.17969
0.17969
(c) With this file as learn.m
# Octave script file for learning model.
function w = learn(p)
T = [1-2*p,p, p, 0, 0;
p, 1-2*p,0, 0, 0;
p, 0, 1-2*p,0, 0;
0, p, 0, 1, 0;
0, 0, p, 0, 1];
T5 = T**5;
p5 = T5*[1;0;0;0;0];
w = p5(4);
endfunction
issuing the command octave:1> learn(.20) yields ans = 0.17664.
(d) This Octave session
octave:1> x=(.01:.01:.50)’;
octave:2> y=(.01:.01:.50)’;
octave:3> for i=.01:.01:.50
> y(100*i)=learn(i);
0.22656
0.16016
0.16016
0.22656
0.22656