GATE CS 2012 Q.44 (CS Q.44): OSI/TCP-IP Layering, Physical Layer and Switching

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 $D$ ?

1005 ms

1010 ms

3000 ms

3003 ms

Official answer and solution

Official answer: A
From: Official GATE 2012 answer key

Solution

Calculate transmission delay, propagation delay, and store-and-forward pipelining delay across 3 links and 2 routers.

System parameters:

File size = $10^6$ bits, divided into 1000 packets of 1000 bits each.
Link bandwidth $B = 1\text{ Mbps} = 10^6\text{ bps}$ .
Link length $d = 100\text{ km} = 10^5\text{ m}$ .
Signal propagation speed $v = 10^8\text{ m/s}$ .
Number of links = 3 ( $S \to R_1$ , $R_1 \to R_2$ , $R_2 \to D$ ).

Transmission delay for one packet:

t_t = \frac{\text{Packet size}}{\text{Bandwidth}} = \frac{1000\text{ bits}}{10^6\text{ bps}} = 1\text{ ms}

Propagation delay per link:

t_p = \frac{d}{v} = \frac{10^5\text{ m}}{10^8\text{ m/s}} = 1\text{ ms}

Total propagation delay for any packet across 3 links = $3 \times 1\text{ ms} = 3\text{ ms}$ .

Pipelining analysis:

First packet leaves source $S$ at $t = 0$ .
Arrives at $R_1$ at $t_t + t_p = 1 + 1 = 2\text{ ms}$ .
Forwarded by $R_1$ and arrives at $R_2$ at $2 + 1 + 1 = 4\text{ ms}$ .
Forwarded by $R_2$ and arrives at destination $D$ at $4 + 1 + 1 = 6\text{ ms}$ .
Due to pipelining, the remaining 999 packets arrive at destination $D$ consecutively one packet every $1\text{ ms}$ .

Total delay:

\text{Total time} = 6\text{ ms} + (1000 - 1) \times 1\text{ ms} = 6 + 999 = 1005\text{ ms}

This matches option (A).

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2012 Computer Science and Information Technology (CS)
Question: 44 of 65, printed as CS Q.44
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source