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Q28E

Expert-verifiedFound in: Page 337

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**28 :** **Consider the isolated Swiss town of Andelfingen, inhabited by 1,200 families. Each family takes a weekly shopping trip to the only grocery store in town, run by Mr. and Mrs. Wipf, until the day when a new, fancier (and cheaper) chain store, Migros, opens its doors. It is not expected that everybody will immediately run to the new store, but we do anticipate that 20% of those shopping at Wipf’s each week switch to Migros the following week. Some people who do switch miss the personal service (and the gossip) and switch back: We expect that 10% of those shopping at Migros each week go to Wipf’s the following week. The state of this town (as far as grocery shopping is concerned) can be represented by the vector**

$\overline{\mathbf{x}}{\mathbf{\left(}}{\mathit{t}}{\mathbf{\right)}}{\mathbf{=}}{\left[\begin{array}{c}w\left(t\right)\\ m\left(t\right]\end{array}\right]}$

** **

**where w(t) and m(t) are the numbers of families shopping at Wipf’s and at Migros, respectively, t weeks after Migros opens. Suppose w(0) = 1,200 and m(0) = 0.**

**a. Find a 2 × 2 matrix A such that role="math" localid="1659586084144" $\overline{\mathbf{x}}{\mathbf{(}}{\mathit{t}}{\mathbf{+}}{\mathbf{+}}{\mathbf{1}}{\mathbf{)}}{\mathbf{=}}{\mathit{A}}\overrightarrow{\mathbf{x}}{\mathbf{\left(}}{\mathit{t}}{\mathbf{\right)}}$****. Verify that A is a positive transition matrix. See Exercise 25.**

**b. How many families will shop at each store after t weeks? Give closed formulas. c. The Wipfs expect that they must close down when they have less than 250 customers a week. When does that happen?**

- 2 × 2 matrix $A=\left[\begin{array}{cc}0.8& 0.1\\ 0.2& 0.9\end{array}\right]$
- $\begin{array}{l}w\left(t\right)=400.{1}^{t}+800.{(-0.7)}^{t}=400+800.{(-0.7)}^{t}\\ m\left(t\right)=400.{1}^{t}-800.{\left(0.7{)}^{t}=800-800.(-0.7\right)}^{t}\end{array}$

Transition matrix may refer to: The matrix associated with a change of basis for a vector space. Stochastic matrix, a square matrix used to describe the transitions of a Markov chain. State-transition matrix, a matrix whose product with the state vector. at an initial time.

(A) Clearly we see that,

$\mathrm{w}\left(\mathrm{t}+1\right)=0.8\mathrm{w}\left(\mathrm{t}\right)+0.1\mathrm{m}\left(\mathrm{t}\right),\phantom{\rule{0ex}{0ex}}\mathrm{m}\left(\mathrm{t}+1\right)=0.2\mathrm{w}\left(\mathrm{t}\right)+0.9\mathrm{m}\left(\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}A=\left[\begin{array}{cc}0.8& 0.1\\ 0.2& 0.9\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

(B) The matrix A is positive transition matrix, we know its eigenvectors are

$\begin{array}{r}{v}_{1}=\left[\begin{array}{c}0.1\\ 0.2\end{array}\right]\\ {v}_{2}=\left[\begin{array}{c}1\\ -1\end{array}\right]\\ {\lambda}_{1}=1,{\lambda}_{2}=-0.7\\ {x}_{0}=\left[\begin{array}{c}1200\\ 0\end{array}\right]=4000\left[\begin{array}{c}0.1\\ 0.2\end{array}\right]+800\left[\begin{array}{c}1\\ -1\end{array}\right]\\ w\left(t\right)={400.1}^{\mathrm{t}}+800\cdot (-0.7{)}^{\mathrm{t}}=400+800\cdot (-0.7{)}^{\mathrm{t}}\\ \mathrm{m}\left(\mathrm{t}\right)=400\cdot {1}^{\mathrm{t}}-800\cdot (-0.7{)}^{\mathrm{t}}=800-800\cdot (-0.7{)}^{\mathrm{t}}\\ w\left(\mathrm{t}\right)=250\\ 400+800\cdot (-0.7{)}^{\mathrm{t}}=250\\ \mathrm{t}={\mathrm{log}}_{-0.7}\left(\frac{-3}{16}\right)\end{array}$

Hence, the final answer is : (a) $A=\left[\begin{array}{cc}0.8& 0.1\\ 0.2& 0.9\end{array}\right]$

(b) $\begin{array}{l}w\left(t\right)=400.{1}^{t}+800.{(-0.7)}^{t}=400+800.{(-0.7)}^{t}\\ m\left(t\right)=400.{1}^{t}-800.{\left(0.7{)}^{t}=800-800.(-0.7\right)}^{t}\end{array}$

(c) The wipfs will never have to close.

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