Consider a source computer (S) transmitting a file of size $10^6$ bits to a destination computer (D) over a network of two routers ($R_1$ and $R_2$) and three links ($L_1$, $L_2$, and $L_3$). $L_1$ connects S to $R_1$; $L_2$ connects $R_1$ to $R_2$; and $L_3$ connects $R_2$ to D. Let each link be of length 100 km. Assume signals travel over each link at a speed of $10^8$ meters per second. Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D?

To find the total time taken to transmit the entire file from source $S$ to destination $D$ through two store-and-forward routers ($R_1, R_2$) over three links ($L_1, L_2, L_3$):

1. Given Parameters:

   • Packet size, $L = 1000\text{ bits}$

   • Number of packets, $N = 1000$

   • Bandwidth of each link, $B = 1\text{ Mbps} = 10^6\text{ bps}$

   • Distance of each link, $d = 100\text{ km} = 10^5\text{ m}$

   • Propagation speed, $v = 10^8\text{ m/s}$

   • Number of links, $k = 3$

2. Delays per Packet per Link:

   • Transmission delay for one packet:

     $$T_t = \frac{L}{B} = \frac{1000}{10^6} = 10^{-3}\text{ s} = 1\text{ ms}$$

   • Propagation delay on one link:

     $$T_p = \frac{d}{v} = \frac{10^5}{10^8} = 10^{-3}\text{ s} = 1\text{ ms}$$

3. Total Time using Pipelined Store-and-Forward Transmission:

   • The first packet takes $3$ hops to reach destination $D$:

     $$T_{\text{first packet}} = 3 \times T_t + 3 \times T_p = 3(1\text{ ms}) + 3(1\text{ ms}) = 6\text{ ms}$$

   • Due to pipelining, each of the remaining $N - 1 = 999$ packets arrives at the destination $1\text{ ms}$ ($T_t$) after the previous one:

     $$T_{\text{remaining}} = (N - 1) \times T_t = 999 \times 1\text{ ms} = 999\text{ ms}$$

   • Total delay:

     $$T_{\text{total}} = T_{\text{first packet}} + T_{\text{remaining}} = 6\text{ ms} + 999\text{ ms} = 1005\text{ ms}$$

Therefore, the correct option is (A).