Question

A car rental agency has three rental locations, denoted by 1, 2, and 3. A customer may rent a car from any of the three locations and return the car to any of the three locations. The manager finds that customers return the cars to the various locations according to the following probabilities [{0.8, 0.3, 0.2}, {0.1, 0.2, 0.6}, {0.1, 0.5, 0.2}] If a car is rented initially from location 2, then the initial state vector is x = {0, 1, 0}. Calculate x (1) , x (2) , and x (3) to three decimal places

Answer 1

The solution to the problem is to calculate the state vector of the transition matrix of the Markov chain.

Simulate this problem using the following matrix:

T = $\begin{bmatrix} 0.8 & 0.3&0.2 \\ 0.1&0.2&0.6\\ 0.1&0.5&0.2\\ \end{bmatrix}$

where initial state vector is X₀ = [0, 1, 0]

If X_n+1 and X_n are two consecutive state vectors of a Markov chain with transition matrix T, then X_n+1=T X_n

calculate X₁:

X₁ = X₀*T =[0.8*0+0.3*1+0.2*0, 0.1*0+0.2*1+0.6*0, 0.1*0+0.5*1+0.2*0] =[0.3,0.2,0.5]

X₁ = [0.3,0.2,0.5]

calculate X₂ to three decimal places:

X₂ =X₁*T = [0.8*0.3+0.3*0.2+0.2*0.2, 0.1*0.3+0.2*0.2+0.6*0.5, 0.1*0.3+0.5*0.2+0.2*0.5] =

[0.4 , 0.37, 0.23]

X₂ = [0.4 , 0.37, 0.23]

calculate X₃ to three decimal places:

X₃ =X₂*T = [0.8*0.4+0.3*0.37+0.2*0.23, 0.1*0.4+0.2*0.37+0.6*0.23 0.1*0.4+0.5*0.37+0.2*0.23] =

[0.477, 0.252, 0.271]

X₃ = [0.477, 0.252, 0.271]