Discrete Network Telemetry (Computer & Systems Engineering) Solution

Part 1: Calculate average data transfer rates from the table

• For the interval $[10, 14]$:
  $$\text{Average Rate} = \frac{D(14) - D(10)}{14 - 10} = \frac{274.00 - 150.00}{4} = \frac{124.00}{4} = 31\text{ GB/s}$$
• For the interval $[12, 14]$:
  $$\text{Average Rate} = \frac{D(14) - D(12)}{14 - 12} = \frac{274.00 - 206.00}{2} = \frac{68.00}{2} = 34\text{ GB/s}$$

Part 2: Shrinking intervals using continuous model $D(t) = 1.5t^2 - 5t + 50$

The baseline value at the fixed starting point $t = 14$ is:

• Interval $[14, 14.1]$:
  • Evaluate at $t = 14.1$:
    $$D(14.1) = 1.5(14.1)^2 - 5(14.1) + 50 = 1.5(198.81) - 70.5 + 50 = 277.715\text{ GB}$$
  • Calculate average rate:
    $$\text{Average Rate} = \frac{277.715 - 274.00}{14.1 - 14} = \frac{3.715}{0.1} = 37.15\text{ GB/s}$$
• Interval $[14, 14.01]$:
  • Evaluate at $t = 14.01$:
    $$D(14.01) = 1.5(14.01)^2 - 5(14.01) + 50 = 1.5(196.2801) - 70.05 + 50 = 274.37015\text{ GB}$$
  • Calculate average rate:
    $$\text{Average Rate} = \frac{274.37015 - 274.00}{14.01 - 14} = \frac{0.37015}{0.01} = 37.015\text{ GB/s}$$
• Interval $[14, 14.001]$:
  • Evaluate at $t = 14.001$:
    $$D(14.001) = 1.5(14.001^2) - 5(14.001) + 50 = 1.5(196.028001) - 70.005 + 50 = 274.0370015\text{ GB}$$
  • Calculate average rate:
    $$\text{Average Rate} = \frac{274.0370015 - 274.00}{14.001 - 14} = \frac{0.0370015}{0.001} = 37.0015\text{ GB/s}$$

Conclusion (Limit Inference): As the interval length $\Delta t$ approaches $0$, the average data transfer rates ($37.15 \to 37.015 \to 37.0015$) converge to exactly $37\text{ GB/s}$. This represents the estimated instantaneous data transfer rate (throughput) at $t = 14$ seconds.