1. **Problem Statement:** Calculate the probability $P_{is}(t)$ for Service (i) and Service (ii) paths using the given formula and determine which server will contain the 4th connection and how many connections each server will have by the 5th connection. 2. **Given:** - Euclidean distance = 22 units (not directly used in probability formula here) - Speed = 2.5 units/s (not directly used in probability formula here) - Service (i) has 2 connections - Service (ii) has 1 connection - Formula for probability: $$P_{is}(t) = \frac{(t_s + q_{is}(t))^{\alpha}}{(t_s + q_{is}(t))^{\alpha} + (t_s + q_u(t))^{\alpha}}$$ - Standardized weight formula: $$sW = sW \times \frac{P(t)}{\beta} \quad \text{where} \quad \beta = |sW - P(t)|$$ 3. **Step 1: Understand variables** - $t_s$ is the service time or base time (not given explicitly, assume constant or 0 for simplification) - $q_{is}(t)$ is the queue length or load on server $i$ at time $t$ - $q_u(t)$ is the queue length or load on the other server at time $t$ - $\alpha$ is a parameter (not given, assume $\alpha=1$ for simplicity) 4. **Step 2: Assign initial weights and loads** - Virtual Server A weight $sW_A = 5.79$ - Virtual Server B weight $sW_B = 2.25$ - Initial connections: Service (i) has 2 connections, Service (ii) has 1 connection 5. **Step 3: Calculate probabilities for Service (i) path** Assuming $t_s=0$ and $\alpha=1$: - For Server A: $$P_A = \frac{q_A}{q_A + q_B}$$ - For Server B: $$P_B = \frac{q_B}{q_A + q_B}$$ Assuming $q_A = 2$ (connections on A), $q_B = 1$ (connections on B): $$P_A = \frac{2}{2+1} = \frac{2}{3} \approx 0.6667$$ $$P_B = \frac{1}{3} \approx 0.3333$$ 6. **Step 4: Calculate standardized weights for each server** For Server A: $$\beta_A = |5.79 - 0.6667| = 5.1233$$ $$sW_A = 5.79 \times \frac{0.6667}{5.1233} \approx 0.753$$ For Server B: $$\beta_B = |2.25 - 0.3333| = 1.9167$$ $$sW_B = 2.25 \times \frac{0.3333}{1.9167} \approx 0.391$$ 7. **Step 5: Determine which server gets the 4th connection** The server with higher standardized weight $sW$ gets the next connection. - $sW_A = 0.753$ - $sW_B = 0.391$ Therefore, Server A gets the 4th connection. 8. **Step 6: Calculate connections by the 5th connection** After 4th connection: - Server A connections = 3 - Server B connections = 1 Recalculate probabilities for 5th connection: $$P_A = \frac{3}{3+1} = 0.75$$ $$P_B = \frac{1}{4} = 0.25$$ Recalculate standardized weights: $$\beta_A = |5.79 - 0.75| = 5.04$$ $$sW_A = 5.79 \times \frac{0.75}{5.04} \approx 0.861$$ $$\beta_B = |2.25 - 0.25| = 2.0$$ $$sW_B = 2.25 \times \frac{0.25}{2.0} = 0.281$$ Server A again has higher $sW$, so 5th connection goes to Server A. Final connections: - Server A: 4 connections - Server B: 1 connection **Answer:** - i. The 4th connection will be assigned to Server A. - ii. By the 5th connection, Server A will have 4 connections and Server B will have 1 connection.