Short Answer

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
$\bar{x} (t) = [\begin{matrix} w (t) \\ m (t] \end{matrix}]$

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" $\bar{x} (t + + 1) = A \vec{x} (t)$ . 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 $A = [\begin{matrix} 0.8 & 0.1 \\ 0.2 & 0.9 \end{matrix}]$
$\begin{array}{l} w (t) = 400 . 1^{t} + 800 . {(- 0.7)}^{t} = 400 + 800 . {(- 0.7)}^{t} \\ m (t) = 400 . 1^{t} - 800 . {(0.7)^{t} = 800 - 800 . (- 0.7)}^{t} \end{array}$
The wipfs will never have to close.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Positive Transition Matrix

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.

Step 2: Finding out the matrix and no. of families that will shop at each store after two weeks :

(A) Clearly we see that,

$w (t + 1) = 0.8 w (t) + 0.1 m (t), m (t + 1) = 0.2 w (t) + 0.9 m (t) A = [\begin{matrix} 0.8 & 0.1 \\ 0.2 & 0.9 \end{matrix}]$

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

$\begin{aligned} v_{1} = [\begin{matrix} 0.1 \\ 0.2 \end{matrix}] \\ v_{2} = [\begin{array}{c} 1 \\ - 1 \end{array}] \\ λ_{1} = 1, λ_{2} = - 0.7 \\ x_{0} = [\begin{array}{c} 1200 \\ 0 \end{array}] = 4000 [\begin{array}{c} 0.1 \\ 0.2 \end{array}] + 800 [\begin{array}{c} 1 \\ - 1 \end{array}] \\ w (t) = {400.1}^{t} + 800 \cdot (- 0.7)^{t} = 400 + 800 \cdot (- 0.7)^{t} \\ m (t) = 400 \cdot 1^{t} - 800 \cdot (- 0.7)^{t} = 800 - 800 \cdot (- 0.7)^{t} \\ w (t) = 250 \\ 400 + 800 \cdot (- 0.7)^{t} = 250 \\ t = \log_{- 0.7} (\frac{- 3}{16}) \end{aligned}$

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

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

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