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$[\mathbf{M}]$ In Detroit, Hertz Rent $A$ Car has a fleet of about 2000 cars. The pattern of rental and return locations is given by the fractions in the table below. On a typical day, about how many cars will be rented or ready to rent from the downtown location?$$\left[\begin{array}{ccc}{.90} & {.01} & {.09} \\ {.01} & {.90} & {.01} \\ {.09} & {.09} & {.90}\end{array}\right]$$

on a typical day, about $( .090909 ) ( 2000 ) = 182$ cars will be rented or available from the downtown location.

Calculus 3

Chapter 4

Vector Spaces

Section 9

Applications to Markov Chains

Vectors

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in this example were considering a stochastic Matrix P that will describe the migration for rental cars that come from the airport downtown and a metro airport area. So, for example, if we consider this entry, that means the car came from the Metro Air area and there is a 1% chance it ends up in the downtown area. So what we want to do next is to determine the steady state vector Q of P. And to do that, we have to first solve the matrix equation that p Times X equals X. If we subtract X Trible sites and factor it out, were then solving the equation p minus the identity matrix Times X equals the zero vector. So let's write down what the augmented matrix will be for this system of equations. I first noticed that when we subtract the identity matrix frumpy will just be subtracting one from each one of these entries, and that will give us a negative 10.1 in each one of those positions. So are augmented. Matrix is going to now be of the form negative 0.1, then copy 0.1 point 09 Calling to is 0.1 Then, from this subtraction, we have a negative 0.1 again and copy 0.9 0.9 0.1 And then we have a negative 0.1 once again. And once again these entries here came from P minus I. Now we're augmenting with a zero vector. It's all placed, though, that here and now we need to reduce this system. This is a very tedious system to do by hand, and so it's heavily encouraged to use a calculator or a computer algebra system in a calculator. I found the road reduce echelon form is 10 negative. 0.919 zero Row two is 01 then negative point 192 zero and then the third row contains Onley zeros. This tells us then that x one is equal to positive. 0.919 x three x two Next is positive point 192 x three and x three is a free variable. Saw right x three equals x three. This tells us then that the solution X to this matrix equation is the following. I'm going to pick a one in place of x three, then that will produce a 0.192 and a point 919 for the entries of X. The next thing we need to do to find the steady state vector is take the entries of X 0.919 and add them together. So at 0.192 to the first century plus one, and this is 10.2111 altogether now, with this value 0.2111 we can assert that the steady state vector Q is now equal to 1/2 0.111 scaling that Vector X. So multiply each entry of X by the reciprocal of 2.111 And after rounding to three decimal places, we obtained point for 35 in the first entry, 0.91 and then point 474 for their remaining entries. Next, let's say what this means in particular and will focus on the second entry here, which corresponds to the downtown area. So we'll say here that if there are, let's say, 2000 cars in this rental car company fleet, then in the long run, there will be the 2000 cars multiplied by the 0.91 that we have highlighted here. This funny is equal to 182 cars, but that tells us that the 102 cars ends up in the downtown area, so this will be in downtown. So this will allow the rental car company to plan on how to redistribute cars in an equal way, looking at the steady state vector and where the cars will eventually end up.

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