Question

Consider a source computer (S) transmitting a file of size ${10}^6$ bits to a destination computer (D) over a network of $\left(\text{two routers}{R}_{1}and{R}_2\text{and three links}\right)$
$(L_1,L_{2}and{L}_3).L_1\text{connects S to}$
$R_1;L_{2}connects{R}_{1}to{R}_2;and$
$L_{3}connects{R}_{2}toD.$
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 1 Mbps. 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?

• A. 1010 ms
• B. 3000 ms
• C. 3003 ms
• D. 1005 ms

Answer 1

The correct option is D 1005 ms



$\text{Propagation delay to travel from S to}{R}_1$
$=\frac{Distance}{LinkSpeed}=\frac{{10}^5}{{10}^8}=1ms$
Total propagation delay to travel from
$StoD=3\times 1ms=3ms$
Total transmission delay for 1 packet
$=3\times \frac{\left(Numberofbits\right)}{Bandwidth}$
$=3\times \frac{\left(1000\right)}{{10}^6}=3ms$
So the first packet will take 6 ms to reach D. While first packet was reaching D, other packets must have been processing in parallel. So D will receive remaining packets 1 packet per 1 ms form R₂.
So remaining 999 packets will take 999 ms and total time will be 999 + 6 = 1005 ms.