### **Step 1: Introduction to the Concept** In computer networks, the total **one-way delay** experienced by a packet is the sum of several types of delays: - **Transmission Delay ($D_{trans}$):** Time to push all bits of the packet onto the link. - **Propagation Delay ($D_{prop}$):** Time for a bit to propagate through the medium. - **Processing Delay ($D_{proc}$):** Time to process the packet header. - **Queuing Delay ($D_q$):** Time the packet waits in queue before being transmitted. The **one-way delay** across a path is the sum of these delays for each segment (link + switch) in the path. ### **Given Data** - **Packet size = 1500 bytes = 1500 × 8 = 12,000 bits** - **Propagation speed = $3 × 10^8$ m/s** - **All distances, link speeds, and delays are given in the diagram.** - **You will add up all delays for each link and at each switch.** --- ## **Step 2: (a) One-way delay from A to B** ### **Path: A → S1 → S2 → B** **Link 1: A to S1** - Distance: 3000 km = 3,000,000 m - Speed: 2 Mbps - $D_{trans,1} = \frac{12,000}{2×10^6} = 0.006$ s = 6 ms - $D_{prop,1} = \frac{3,000,000}{3×10^8} = 0.01$ s = 10 ms **Switch S1** - $D_{proc, S1} = 2$ ms, $D_{q, S1} = 5$ ms **Link 2: S1 to S2** - Distance: 4500 km = 4,500,000 m - Speed: 3 Mbps - $D_{trans,2} = \frac{12,000}{3×10^6} = 0.004$ s = 4 ms - $D_{prop,2} = \frac{4,500,000}{3×10^8} = 0.015$ s = 15 ms **Switch S2** - $D_{proc, S2} = 1$ ms, $D_{q, S2} = 20$ ms **Link 3: S2 to B** - Distance: 1500 km = 1,500,000 m - Speed: 3.5 Mbps - $D_{trans,3} = \frac{12,000}{3.5×10^6} ≈ 0.00343$ s ≈ 3.43 ms - $D_{prop,3} = \frac{1,500,000}{3×10^8} = 0.005$ s = 5 ms **Total Delay (A to B):** \[ \begin{align*} \text{A to S1:} & \quad 6 + 10 = 16\ \text{ms} \\ \text{S1:} & \quad 2 + 5 = 7\ \text{ms} \\ \text{S1 to S2:} & \quad 4 + 15 = 19\ \text{ms} \\ \text{S2:} & \quad 1 + 20 = 21\ \text{ms} \\ \text{S2 to B:} & \quad 3.43 + 5 ≈ 8.43\ \text{ms} \\ \hline \text{Total:} & \quad 16 + 7 + 19 + 21 + 8.43 = \boxed{71.43\ \text{ms}} \end{align*} \] --- ## **Step 3: (b) One-way delay from A to D** ### **Path: A → S1 → S2 → S3 → D** **Links and switches:** - A to S1: as above (16 ms) - S1: as above (7 ms) - S1 to S2: as above (19 ms) - S2: as above (21 ms) - S2 to S3: - Distance: 6000 km = 6,000,000 m - Speed: 5 Mbps - $D_{trans} = \frac{12,000}{5×10^6} = 0.0024$ s = 2.4 ms - $D_{prop} = \frac{6,000,000}{3×10^8} = 0.02$ s = 20 ms - Total: 2.4 + 20 = 22.4 ms - S3: - $D_{proc, S3} = 5$ ms, $D_{q, S3} = 8$ ms ⇒ 13 ms - S3 to D: - Distance: 2500 km = 2,500,000 m - Speed: 2 Mbps - $D_{trans} = \frac{12,000}{2×10^6} = 0.006$ s = 6 ms - $D_{prop} = \frac{2,500,000}{3×10^8} = 0.00833$ s ≈ 8.33 ms - Total: 6 + 8.33 = 14.33 ms **Total Delay (A to D):** \[ \begin{align*} 16 + 7 + 19 + 21 + 22.4 + 13 + 14.33 = \boxed{112.73\ \text{ms}} \end{align*} \] --- ## **Step 4: (c) One-way delay from C to B** ### **Path: C → S3 → S2 → B** **C to S3:** - Distance: 3500 km = 3,500,000 m - Speed: 4 Mbps - $D_{trans} = \frac{12,000}{4×10^6} = 0.003$ s = 3 ms - $D_{prop} = \frac{3,500,000}{3×10^8} ≈ 0.01167$ s ≈ 11.67 ms - Total: 3 + 11.67 = 14.67 ms **S3:** 5 + 8 = 13 ms **S3 to S2:** as above (22.4 ms) **S2:** as above (21 ms) **S2 to B:** as above (8.43 ms) **Total Delay (C to B):** \[ 14.67 + 13 + 22.4 + 21 + 8.43 = \boxed{79.5\ \text{ms}} \] --- ## **Step 5: (d) One-way delay from D to B** ### **Path: D → S3 → S2 → B** **D to S3:** - Distance: 2500 km = 2,500,000 m - Speed: 2 Mbps - $D_{trans} = \frac{12,000}{2×10^6} = 0.006$ s = 6 ms - $D_{prop} = \frac{2,500,000}{3×10^8} ≈ 0.00833$ s ≈ 8.33 ms - Total: 6 + 8.33 = 14.33 ms **S3:** 5 + 8 = 13 ms **S3 to S2:** as above (22.4 ms) **S2:** as above (21 ms) **S2 to B:** as above (8.43 ms) **Total Delay (D to B):** \[ 14.33 + 13 + 22.4 + 21 + 8.43 = \boxed{79.16\ \text{ms}} \] --- ## **Step 6: Final Answers (Summary)** - (a) **A to B:** **71.43 ms** - (b) **A to D:** **112.73 ms** - (c) **C to B:** **79.5 ms** - (d) **D to B:** **79.16 ms** --- ### **Step 7: What We Did in Each Step** 1. Introduced the delay components and described the problem. 2. Calculated total delay for A to B by summing all delays for each segment. 3. Repeated the process for A to D, adding delays for each link and switch. 4. Did the same for C to B. 5. Did the same for D to B. 6. Summarized all answers concisely. 7. Explained our overall step-by-step approach.