GATE CS 2005 Q73 — Packet, Circuit and Virtual Circuit Switching

To find the optimum packet size, we analyze the total transmission time required to send the message from the source to the destination.

1. Identify the Parameters

• Message size ($M$): $24$ bytes
• Header size ($H$): $3$ bytes
• Number of intermediate nodes: $2$
• Number of hops ($k$): $3$ (Source $\to$ Node 1 $\to$ Node 2 $\to$ Destination)

Let:

• $x$ be the packet size (in bytes).
• $p = x - H = x - 3$ be the payload size per packet (in bytes).
• $N = \lceil M / p \rceil = \lceil 24 / p \rceil$ be the number of packets.
• $R$ be the transmission rate of each link (in bytes/second).

2. Total Transmission Time Formula

Using the pipelining concept in packet switching, the total time $T$ to transmit $N$ packets over $k$ hops is given by:

Substituting $k = 3$:

We want to find the packet size $x$ that minimizes the normalized transmission time $T \cdot R$.

3. Evaluate the Options

• Option A: $x = 4$ bytes
  • Payload $p = 4 - 3 = 1$ byte
  • Number of packets $N = 24 / 1 = 24$
  • $T \cdot R = (24 + 2) \times 4 = 104$
• Option B: $x = 6$ bytes
  • Payload $p = 6 - 3 = 3$ bytes
  • Number of packets $N = 24 / 3 = 8$
  • $T \cdot R = (8 + 2) \times 6 = 60$
• Option C: $x = 7$ bytes
  • Payload $p = 7 - 3 = 4$ bytes
  • Number of packets $N = 24 / 4 = 6$
  • $T \cdot R = (6 + 2) \times 7 = 56$
• Option D: $x = 9$ bytes
  • Payload $p = 9 - 3 = 6$ bytes
  • Number of packets $N = 24 / 6 = 4$
  • $T \cdot R = (4 + 2) \times 9 = 54$

Comparing the normalized times, the minimum time of $54$ is achieved when the packet size $x = 9$ bytes.