Exercise 12

Suppose that a city is starting a bicycle sharing program with bicycles at locations $B$ and $C$. Bicycles that are rented at one location may be returned to either location at the end of the day. Over time, the city finds that 80% of bicycles rented at location $B$ are returned to $B$ with the other 20% returned to $C$. Similarly, 50% of bicycles rented at location $C$ are returned to $B$ and 50% to $C$.

To keep track of the bicycles, we form a vector $$\xvec_k = \twovec{B_k}{C_k}$$ where $B_k$ is the number of bicycles at location $B$ at the beginning of day $k$ and $C_k$ is the number of bicycles at $C$. The information above tells us how to determine the distribution of bicycles the following day: $$\begin{alignedat}{3} B_{k+1} & {}={} 0.8B_k & {}+{} & 0.5 C_k \\ C_{k+1} & {}={} 0.2B_k & {}+{} & 0.5 C_k. \\ \end{alignedat}$$ Expressed in matrix-vector form, these expressions give $$\xvec_{k+1} = A\xvec_k$$ where $$A = \left[\begin{array}{rr} 0.8 & 0.5 \\ 0.2 & 0.5 \\ \end{array}\right]$$.

1. Let's check that this makes sense.

   1. Suppose that there are 1000 bicycles at location $B$ and none at $C$ on day 1. This means we have $\xvec_1 = \twovec{1000}{0}$. Find the number of bicycles at both locations on day 2 by evaluating $\xvec_2 = A\xvec_1$.

   2. Suppose that there are 1000 bicycles at location $C$ and none at $B$ on day 1. Form the vector $\xvec_1$ and determine the number of bicycles at the two locations the next day by finding $\xvec_2 = A\xvec_1$.

2. Suppose that one day there are 1050 bicycles at location $B$ and 450 at location $C$. How many bicycles were there at each location the previous day?

3. Suppose that there are 500 bicycles at location $B$ and 500 at location $C$ on Monday. How many bicycles are there at the two locations on Tuesday? on Wednesday? on Thursday?