Day 14 in math modeling

  1. Homework:

  2. CINSAM poster session, part II: abstract due date, April 6th.

  3. Markov Chains, cont.

    1. Example, cont.: Markovian squirrels in Scotland (Project 3.4, p. 141).
      • Interpretation of eigenvector, and solution 
            (0.579274246838739    0.3843173554990237)
    (0.07885798724711165  0.05231803983031641)
    (0.8103825941753516   0.5376453333383054)
    (0.038766168941149134 0.025719271332354497)

    ( 0.384317      0.384317      0.384317      0.384317    )
    (  5.231804E-2   5.231804E-2   5.231804E-2   5.231804E-2)
    ( 0.537645      0.537645      0.537645      0.537645    )
    (  2.571927E-2   2.571927E-2   2.571927E-2   2.571927E-2)
    2. Regular matrices and the search for steady state solutions (p. 124). Example illustrating that steady state solutions need not exist: 
          [0   1]
A=  [     ]
    [1   0]
    3. Absorbing Markov Chains
      1. Figure 3.14 and T
      2. Matrix T (p. 125) and eigenvalues = 1
      3. Project 3.5: Hermit crabs (here's another relevant page) 
        T^500
(7.24E-104  0.00      0.00      0.00      0.00      0.00      0.00    )
( 4.99E-29  3.57E-29  2.03E-29  1.84E-29  9.16E-30  0.00      0.00    )
( 8.24E-28  5.90E-28  3.35E-28  3.04E-28  1.51E-28  0.00      0.00    )
( 1.74E-27  1.25E-27  7.09E-28  6.43E-28  3.20E-28  0.00      0.00    )
( 3.59E-27  2.57E-27  1.46E-27  1.32E-27  6.60E-28  0.00      0.00    )
( 0.12      0.13      0.13      0.14       7.29E-2   1.0      0.00    )
( 0.88      0.87      0.87      0.86      0.93      0.00       1.0    )
F
(  1.10526      0.000000      0.000000      0.000000      0.000000    )
( 0.428715       1.10252       4.591603E-2   7.348154E-3   9.929937E-4)
(  1.03850      0.831930       1.49227      0.238815       3.227230E-2)
( 0.682290      0.878382      0.415254       1.53877      0.207942    )
( 0.560319      0.627212      0.528814      0.748123       1.69569    )
Website maintained by Andy Long. Comments appreciated.