Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is $10^6$ bytes / sec. A user on host A sends a file of size $10^3$ bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these packets are sent from A to B. Each packet contains 100 bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let T1, T2 and T3 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT?

The network consists of $3$ links: $\text{A} \to \text{R1} \to \text{R2} \to \text{B}$. Bandwidth $B = 10^6\text{ bytes/sec} = 1\text{ byte/}\mu\text{s}$.

For a store-and-forward packet network with $k$ packets over $m$ links, the total transmission time is:

Here, the number of links $m = 3$.

1. Case 1: Single packet ($k = 1$):

   • Packet size = $1000\text{ bytes (data)} + 100\text{ bytes (header)} = 1100\text{ bytes}$.

   • Transmission time per packet = $\frac{1100}{10^6}\text{ s} = 1100\,\mu\text{s}$.

   • Total time $T_1 = (3 - 1 + 1) \times 1100 = 3 \times 1100 = 3300\,\mu\text{s}$.

2. Case 2: File split into $10$ equal parts ($k = 10$):

   • Data per packet = $\frac{1000}{10} = 100\text{ bytes}$.

   • Packet size = $100 + 100 = 200\text{ bytes}$.

   • Transmission time per packet = $\frac{200}{10^6}\text{ s} = 200\,\mu\text{s}$.

   • Total time $T_2 = (3 - 1 + 10) \times 200 = 12 \times 200 = 2400\,\mu\text{s}$.

3. Case 3: File split into $20$ equal parts ($k = 20$):

   • Data per packet = $\frac{1000}{20} = 50\text{ bytes}$.

   • Packet size = $50 + 100 = 150\text{ bytes}$.

   • Transmission time per packet = $\frac{150}{10^6}\text{ s} = 150\,\mu\text{s}$.

   • Total time $T_3 = (3 - 1 + 20) \times 150 = 22 \times 150 = 3300\,\mu\text{s}$.

Comparing the times:

Hence, $T_1 = T_3$ and $T_3 > T_2$, which corresponds to option (D).