Math Challenges

Submissions for Problem #17

Problem #17

A car rental company has a fleet of 300 cars and two locations, an east-side and a west-side office. Each weekday cars are returned to either location.

The given matrix represent percentages of cars rented from (the columns) and returned to (the rows) the two locations (East and West).

i.e. 95% of cars rented from the East location are returned there, and 90% of cars rented from the West location are returned to the West location.

On Monday 200 cars are rented from the East location, and 100 cars are rented from the West location. On Wednesday how many cars do you expect to be returned to each location?

Now instead of the previous distribution, suppose that on Monday 100 cars are rented from the East location and 200 cars are rented from the West location. How many cars do you see at each location on Wednesday?

\[ M=\begin{pmatrix}.95 & .10\\ .05 & .90\end{pmatrix} \]

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zapwai

Solution: \( \displaylines{\text{Given }\mathbf{x}=\begin{pmatrix}200\\ 100\end{pmatrix}\text{, we compute }M^2\mathbf{x}.\\ \begin{pmatrix}.95 & .10\\ .05 & .90\end{pmatrix}\begin{pmatrix}.95 & .10\\ .05 & .90\end{pmatrix}\begin{pmatrix}200\\ 100\end{pmatrix}=\begin{pmatrix}200\\ 100\end{pmatrix}\\ \text{Given }\mathbf{x}=\begin{pmatrix}100\\ 200\end{pmatrix}\text{ instead, we get }M^2\mathbf{x}\text{ as }\\ \begin{pmatrix}.95 & .10\\ .05 & .90\end{pmatrix}\begin{pmatrix}.95 & .10\\ .05 & .90\end{pmatrix}\begin{pmatrix}100\\ 200\end{pmatrix}=\begin{pmatrix}127.75\\ 172.25\end{pmatrix}} \)

Explanation:
We simply multiply M (twice, for two days passed) by the given column vector x (cars rented from East and West locations). Interestingly the first choice (200, 100) happens to be an eigenvector for the given matrix, and there is no change, it's actually a fixed point. The second point indicates a slow shift toward the East company.