Suppose the automobile rental company described in Example 5 has a fleet of 600 cars. Initially an equal number of cars is based at each location, so that p(0) = 200, q(0) = 200, and r(0) = 200. As in Example 5, let the week-by-week distribution of cars be governed by xk+1 = Axk , k = 0,1,...,

Question

Suppose the automobile rental company described in Example 5 has a fleet of 600 cars. Initially an equal number of cars is based at each location, so that p(0) = 200, q(0) = 200, and r(0) = 200. As in Example 5, let the week-by-week distribution of cars be governed by [latex]x _{k+1}=A x _{k}, k=0,1[/latex],..., where

[latex]x _{k}=\left[\begin{array}{l}p(k) \q(k) \r(k)\end{array} ight], \quad A=\left[\begin{array}{ccc}.6 & .1 & .1 \.1 & .8 & .2 \.3 & .1 & .7\end{array} ight], \quad ext { and } \quad x _{0}=\left[\begin{array}{l}200 \200 \200\end{array} ight][/latex].

Find [latex]\lim _{k ightarrow \infty} x _{k}[/latex]. Determine the number of cars at each location in the kth week, for k = 1, 5, and 10.

Accepted Answer

If A is not defective, we can use Eq. (6) [latex]x _{k}=A^{k} x _{0} =A^{k}\left(a_{1} u _{1}+a_{2} u _{2}+\cdots+a_{n} u _{n} ight) =a_{1} A^{k} u _{1}+a_{2} A^{k} u _{2}+\cdots+a_{n} A^{k} u _{n} =a_{1}\left(\lambda_{1} ight)^{k} u _{1}+a_{2}\left(\lambda_{2} ight)^{k} u _{2}+\cdots+a_{n}\left(\lambda_{n} ight)^{k} u _{n}.[/latex] to express [latex]x _{k}[/latex] as

[latex]x _{k}=a_{1}\left(\lambda_{1} ight)^{k} u _{1}+a_{2}\left(\lambda_{2} ight)^{k} u _{2}+a_{3}\left(\lambda_{3} ight)^{k} u _{3}[/latex],

where [latex]\left\{ u _{1}, u _{2}, u _{3} ight\}[/latex] is a basis for [latex]R^{3}[/latex], consisting of eigenvectors of A.
It can be shown that A has eigenvalues [latex]\lambda_{1}=1, \lambda_{2}=.6, ext { and } \lambda_{3}=.5[/latex]. Thus A has three linearly independent eigenvectors:

[latex]\lambda_{1}=1, \quad u _{1}=\left[\begin{array}{l}4 \9 \7\end{array} ight] ; \quad \lambda_{2}=.6 \quad u _{2}=\left[\begin{array}{r}0 \1 \-1\end{array} ight][/latex];
[latex]\lambda_{3}=.5, \quad u _{3}=\left[\begin{array}{r}-1 \-1 \2\end{array} ight][/latex].

The initial vector, [latex]x _{0}=[200,200,200]^{T}[/latex], can be written as

[latex]x _{0}=30 u _{1}-150 u _{2}-80 u _{3}[/latex].

Thus the vector [latex]x _{k}=[p(k), q(k), r(k)]^{T}[/latex] is given by

[latex] \begin{matrix} x _{k}=A^{k} x _{0} \ =A^{k}\left(30 u _{1}-150 u _{2}-80 u _{3} ight) \ =30\left(\lambda_{1} ight)^{k} u _{1}-150\left(\lambda_{2} ight)^{k} u _{2}-80\left(\lambda_{3} ight)^{k} u _{3} \ =30 u _{1}-150(.6)^{k} u _{2}-80(.5)^{k} u _{3}. \end{matrix} [/latex] (9)

From the expression above, we see that

[latex]\underset{k ightarrow \infty }{\lim } x _{k}=30 u _{1}=\left[\begin{array}{l}120 \270 \210\end{array} ight][/latex].

Therefore, as the weeks proceed, the rental fleet will tend to an equilibrium state with 120 cars at P, 270 cars at Q, and 210 cars at R. To the extent that the model is valid, location Q will require the largest facility for maintenance, parking, and the like.
Finally, using Eq. (9), we can calculate the state of the fleet for the kth week:

[latex]x _{1}=\left[\begin{array}{l}160 \220 \220\end{array} ight], \quad x _{5}=\left[\begin{array}{c}122.500 \260.836 \216.664\end{array} ight], \quad ext { and } \quad x _{10}=\left[\begin{array}{c}120.078 \269.171 \210.751\end{array} ight][/latex].

Question 4.8.6: Suppose the automobile rental company described in Example 5...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions