Science:Math Exam Resources/Courses/MATH221/April 2010/Question 12 (b)

MATH221 April 2010

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Question 12 (b)

Suppose that the yearly movement of people between Alberta, BC, and Manitoba is given as follows: 5% of the people in Alberta move to BC, 5% move to Manitoba, and the rest remain 7% of the people in BC move to Alberta, 2% move to Manitoba, and the rest remain 6% of the people in Manitoba move to Alberta, 7% move to BC, and the rest remain. Suppose that there are 1 million people in BC. Find the populations of Manitoba and Alberta, given that the populations are in a steady state (meaning that they don’t change from year to year)

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Hint

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Solution

Whoops, no solution has been written yet. Be the first to submit a suggestion. Given that we are in steady state, and with the transition matrix $M={\begin{bmatrix}0.9&0.07&0.06\\0.05&0.91&0.07\\0.05&0.02&0.87\end{bmatrix}}$, we have that $Ms=s$. Reworking this a bit, we get: $Ms=sMs-s=0(M-1)s=0(M-I)s=0$ Solving for the null space of $(M-I)$, we get our steady state vector $s={\begin{bmatrix}0.6700\\0.6505\\0.3578\end{bmatrix}}$ Given that the population of BC is 1 million, we need to multiply by $1,000,000/0.6505$ to get our final population vector. This turns out to be: $s*1,000,000/0.6505={\begin{bmatrix}1,030,000\\1,000,000\\550,000\end{bmatrix}}$ So the population of Manitoba is 550,000, and the population of Alberta is 1,030,000.