1. **Markov Chain with Products A and B** **Problem:** Given initial proportions and switching probabilities, find the transition matrix, proportions after two months, and long-run proportions. **Step 1: Define states and initial vector** States: A and B. Initial vector: $\mathbf{p_0} = \begin{bmatrix}0.4 \\ 0.6\end{bmatrix}$ **Step 2: Construct transition matrix $P$** - From A: 30% switch to B, 70% stay in A. - From B: 25% switch to A, 75% stay in B. $$P = \begin{bmatrix}0.7 & 0.25 \\ 0.3 & 0.75\end{bmatrix}$$ **Step 3: Find proportions after two months** Calculate $\mathbf{p_2} = P^2 \mathbf{p_0}$. First, compute $P^2$: $$P^2 = P \times P = \begin{bmatrix}0.7 & 0.25 \\ 0.3 & 0.75\end{bmatrix} \times \begin{bmatrix}0.7 & 0.25 \\ 0.3 & 0.75\end{bmatrix} = \begin{bmatrix}0.7\times0.7 + 0.25\times0.3 & 0.7\times0.25 + 0.25\times0.75 \\ 0.3\times0.7 + 0.75\times0.3 & 0.3\times0.25 + 0.75\times0.75\end{bmatrix} = \begin{bmatrix}0.55 & 0.325 \\ 0.3 & 0.6\end{bmatrix}$$ Then multiply by $\mathbf{p_0}$: $$\mathbf{p_2} = \begin{bmatrix}0.55 & 0.325 \\ 0.3 & 0.6\end{bmatrix} \begin{bmatrix}0.4 \\ 0.6\end{bmatrix} = \begin{bmatrix}0.55\times0.4 + 0.325\times0.6 \\ 0.3\times0.4 + 0.6\times0.6\end{bmatrix} = \begin{bmatrix}0.22 + 0.195 \\ 0.12 + 0.36\end{bmatrix} = \begin{bmatrix}0.415 \\ 0.48\end{bmatrix}$$ **Step 4: Find long-run proportions (steady state)** Solve $\pi P = \pi$ with $\pi = [\pi_A, \pi_B]$ and $\pi_A + \pi_B = 1$. Equations: $$\pi_A = 0.7\pi_A + 0.25\pi_B$$ $$\pi_B = 0.3\pi_A + 0.75\pi_B$$ From first: $$\pi_A - 0.7\pi_A = 0.25\pi_B \Rightarrow 0.3\pi_A = 0.25\pi_B \Rightarrow \pi_B = \frac{0.3}{0.25} \pi_A = 1.2 \pi_A$$ Using $\pi_A + \pi_B = 1$: $$\pi_A + 1.2\pi_A = 1 \Rightarrow 2.2\pi_A = 1 \Rightarrow \pi_A = \frac{1}{2.2} = 0.4545$$ $$\pi_B = 1 - 0.4545 = 0.5455$$ --- 2. **Customer Categories Markov Chain** **Step 1: Define states L, A, C and transition matrix $P$** $$P = \begin{bmatrix}0.8 & 0.15 & 0.05 \\ 0.3 & 0.6 & 0.1 \\ 0 & 0 & 1\end{bmatrix}$$ **Step 2: Probability At-Risk customer eventually churns** Churned (C) is absorbing. Use absorbing Markov chain theory. Partition $P$: Transient states: L, A Absorbing: C $$Q = \begin{bmatrix}0.8 & 0.15 \\ 0.3 & 0.6\end{bmatrix}, \quad R = \begin{bmatrix}0.05 \\ 0.1\end{bmatrix}$$ Fundamental matrix: $$N = (I - Q)^{-1} = \left(\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} - \begin{bmatrix}0.8 & 0.15 \\ 0.3 & 0.6\end{bmatrix}\right)^{-1} = \begin{bmatrix}0.2 & -0.15 \\ -0.3 & 0.4\end{bmatrix}^{-1}$$ Calculate determinant: $$det = 0.2 \times 0.4 - (-0.15) \times (-0.3) = 0.08 - 0.045 = 0.035$$ Inverse: $$N = \frac{1}{0.035} \begin{bmatrix}0.4 & 0.15 \\ 0.3 & 0.2\end{bmatrix} = \begin{bmatrix}11.43 & 4.29 \\ 8.57 & 5.71\end{bmatrix}$$ Probability of absorption: $$B = N R = \begin{bmatrix}11.43 & 4.29 \\ 8.57 & 5.71\end{bmatrix} \begin{bmatrix}0.05 \\ 0.1\end{bmatrix} = \begin{bmatrix}11.43\times0.05 + 4.29\times0.1 \\ 8.57\times0.05 + 5.71\times0.1\end{bmatrix} = \begin{bmatrix}0.5715 + 0.429 \\ 0.4285 + 0.571\end{bmatrix} = \begin{bmatrix}1.0 \\ 1.0\end{bmatrix}$$ So, an At-Risk customer will eventually churn with probability 1. **Step 3: Steady-state distribution** Solve $\pi P = \pi$ with $\pi_L + \pi_A + \pi_C = 1$. From $\pi P = \pi$: $$\pi_L = 0.8\pi_L + 0.3\pi_A$$ $$\pi_A = 0.15\pi_L + 0.6\pi_A$$ $$\pi_C = 0.05\pi_L + 0.1\pi_A + \pi_C$$ From last: $$\pi_C = 0.05\pi_L + 0.1\pi_A + \pi_C \Rightarrow 0 = 0.05\pi_L + 0.1\pi_A$$ Implies $\pi_L = \pi_A = 0$. Then $\pi_C = 1$. Long-run distribution is all customers churned. --- 3. **Economy States Markov Chain** **Step 1: Transition matrix $P$** States: B, N, R $$P = \begin{bmatrix}0.7 & 0.2 & 0.1 \\ 0.3 & 0.6 & 0.1 \\ 0.1 & 0.4 & 0.5\end{bmatrix}$$ **Step 2: Probability economy in Normal will be in Boom after two years** Calculate $P^2$ and look at element $(N,B)$. Calculate $P^2 = P \times P$: Row N: $$P^2_{N,B} = 0.3\times0.7 + 0.6\times0.3 + 0.1\times0.1 = 0.21 + 0.18 + 0.01 = 0.4$$ Answer: Probability is 0.4. --- 4. **Employee Levels Markov Chain** **Step 1: Transition matrix $P$** States: J, M, S $$P = \begin{bmatrix}0.7 & 0.2 & 0 \\ 0 & 0.5 & 0.4 \\ 0 & 0 & 0.8\end{bmatrix}$$ (Leaving and retiring are absorbing outside states, omitted here.) **Step 2: Probability Junior employee eventually reaches Senior** Transient states: J, M Absorbing: S Partition: $$Q = \begin{bmatrix}0.7 & 0.2 \\ 0 & 0.5\end{bmatrix}, \quad R = \begin{bmatrix}0 \\ 0.4\end{bmatrix}$$ Fundamental matrix: $$N = (I - Q)^{-1} = \begin{bmatrix}1-0.7 & -0.2 \\ 0 & 1-0.5\end{bmatrix}^{-1} = \begin{bmatrix}0.3 & -0.2 \\ 0 & 0.5\end{bmatrix}^{-1}$$ Determinant: $$0.3 \times 0.5 - 0 = 0.15$$ Inverse: $$N = \frac{1}{0.15} \begin{bmatrix}0.5 & 0.2 \\ 0 & 0.3\end{bmatrix} = \begin{bmatrix}3.33 & 1.33 \\ 0 & 2\end{bmatrix}$$ Probability of absorption: $$B = N R = \begin{bmatrix}3.33 & 1.33 \\ 0 & 2\end{bmatrix} \begin{bmatrix}0 \\ 0.4\end{bmatrix} = \begin{bmatrix}1.33 \times 0.4 \\ 2 \times 0.4\end{bmatrix} = \begin{bmatrix}0.533 \\ 0.8\end{bmatrix}$$ Starting from Junior, probability to reach Senior is approximately 0.533. --- 5. **Warehouse Inventory Levels Markov Chain** **Step 1: Transition matrix $P$** States: H, M, L $$P = \begin{bmatrix}0.6 & 0.3 & 0.1 \\ 0.2 & 0.5 & 0.3 \\ 0.1 & 0.5 & 0.4\end{bmatrix}$$ **Step 2: Probability starting at Medium reaches Low after 2 months** Calculate $P^2$ and look at element $(M,L)$. Calculate $P^2 = P \times P$: Row M: $$P^2_{M,L} = 0.2\times0.1 + 0.5\times0.3 + 0.3\times0.4 = 0.02 + 0.15 + 0.12 = 0.29$$ Answer: Probability is 0.29. --- 6. **Input-Output Table Economy** Given: $$Z = \begin{bmatrix}500 & 800 \\ 700 & 1000\end{bmatrix}, \quad d = \begin{bmatrix}350 \\ 200\end{bmatrix}$$ **Step 1: Technological coefficient matrix $A$** Calculate column sums: $$x_1 = 500 + 700 + 350 = 1550, \quad x_2 = 800 + 1000 + 200 = 2000$$ Calculate $A$ by dividing each element of $Z$ by total output of sector: $$A = \begin{bmatrix}\frac{500}{1550} & \frac{800}{2000} \\ \frac{700}{1550} & \frac{1000}{2000}\end{bmatrix} = \begin{bmatrix}0.3226 & 0.4 \\ 0.4516 & 0.5\end{bmatrix}$$ **Step 2: Leontief matrix $L = (I - A)^{-1}$** $$I - A = \begin{bmatrix}1-0.3226 & -0.4 \\ -0.4516 & 1-0.5\end{bmatrix} = \begin{bmatrix}0.6774 & -0.4 \\ -0.4516 & 0.5\end{bmatrix}$$ Determinant: $$det = 0.6774 \times 0.5 - (-0.4) \times (-0.4516) = 0.3387 - 0.1806 = 0.1581$$ Inverse: $$L = \frac{1}{0.1581} \begin{bmatrix}0.5 & 0.4 \\ 0.4516 & 0.6774\end{bmatrix} = \begin{bmatrix}3.16 & 2.53 \\ 2.86 & 4.29\end{bmatrix}$$ **Step 3: Gross output with increased final demand** New final demand: $$d' = \begin{bmatrix}600 \\ 200 \times 1.2 = 240\end{bmatrix}$$ Gross output: $$x = L d' = \begin{bmatrix}3.16 & 2.53 \\ 2.86 & 4.29\end{bmatrix} \begin{bmatrix}600 \\ 240\end{bmatrix} = \begin{bmatrix}3.16\times600 + 2.53\times240 \\ 2.86\times600 + 4.29\times240\end{bmatrix} = \begin{bmatrix}1896 + 607.2 \\ 1716 + 1029.6\end{bmatrix} = \begin{bmatrix}2503.2 \\ 2745.6\end{bmatrix}$$ --- Final answers summarized: - Task 1a: $P = \begin{bmatrix}0.7 & 0.25 \\ 0.3 & 0.75\end{bmatrix}$ - Task 1b: After 2 months $\approx \begin{bmatrix}0.415 \\ 0.48\end{bmatrix}$ - Task 1c: Long-run $\approx \begin{bmatrix}0.4545 \\ 0.5455\end{bmatrix}$ - Task 2a: $P = \begin{bmatrix}0.8 & 0.15 & 0.05 \\ 0.3 & 0.6 & 0.1 \\ 0 & 0 & 1\end{bmatrix}$ - Task 2b: Probability At-Risk eventually churns = 1 - Task 2c: Steady state all churned - Task 3i: $P = \begin{bmatrix}0.7 & 0.2 & 0.1 \\ 0.3 & 0.6 & 0.1 \\ 0.1 & 0.4 & 0.5\end{bmatrix}$ - Task 3ii: Probability Normal to Boom in 2 years = 0.4 - Task 4I: $P = \begin{bmatrix}0.7 & 0.2 & 0 \\ 0 & 0.5 & 0.4 \\ 0 & 0 & 0.8\end{bmatrix}$ - Task 4II: Probability Junior reaches Senior = 0.533 - Task 5I: $P = \begin{bmatrix}0.6 & 0.3 & 0.1 \\ 0.2 & 0.5 & 0.3 \\ 0.1 & 0.5 & 0.4\end{bmatrix}$ - Task 5II: Probability Medium to Low in 2 months = 0.29 - Task 6a: $A = \begin{bmatrix}0.3226 & 0.4 \\ 0.4516 & 0.5\end{bmatrix}$ - Task 6b: $L = \begin{bmatrix}3.16 & 2.53 \\ 2.86 & 4.29\end{bmatrix}$ - Task 6c: Gross output $\approx \begin{bmatrix}2503.2 \\ 2745.6\end{bmatrix}$