GATE CS 2014 Set 2 Q26 — Packet, Circuit and Virtual Circuit Switching

The network consists of 3 hops ($A \to R1 \to R2 \to B$), so the number of links is $k = 3$. The bandwidth of each link is $B = 10^6$ bytes/sec. The total transmission time for sending $n$ packets over $k$ hops is given by:

where $t$ is the transmission delay of a single packet on one link, $t = \frac{\text{Packet Size}}{B}$.

• Case 1 ($T1$): Single packet containing the complete file.
  • Data size = $1000$ bytes, Header = $100$ bytes $\implies$ Packet size = $1100$ bytes.
  • Number of packets, $n = 1$.
  • $T1 = (1 + 3 - 1) \times \frac{1100}{10^6} = 3 \times 1.1\text{ ms} = 3.3\text{ ms}$.
• Case 2 ($T2$): File split into $10$ equal parts.
  • Data size per packet = $100$ bytes, Header = $100$ bytes $\implies$ Packet size = $200$ bytes.
  • Number of packets, $n = 10$.
  • $T2 = (10 + 3 - 1) \times \frac{200}{10^6} = 12 \times 0.2\text{ ms} = 2.4\text{ ms}$.
• Case 3 ($T3$): File split into $20$ equal parts.
  • Data size per packet = $50$ bytes, Header = $100$ bytes $\implies$ Packet size = $150$ bytes.
  • Number of packets, $n = 20$.
  • $T3 = (20 + 3 - 1) \times \frac{150}{10^6} = 22 \times 0.15\text{ ms} = 3.3\text{ ms}$.

Comparing the transmission times, we get $T1 = T3$ and $T3 > T2$.