GATE CSE 2024 | Set 1 | Question: 26

Question

Consider a network path $\mathrm{P}-\mathrm{Q}-\mathrm{R}$ between nodes $\mathrm{P}$ and $\mathrm{R}$ via router $\mathrm{Q}$. Node $\mathrm{P}$ sends a file of size $10^6$ bytes to $\text{R}$ via this path by splitting the file into chunks of $10^3$ bytes each. Node $\text{P}$ sends these chunks one after the other without any wait time between the successive chunk transmissions. Assume that the size of extra headers added to these chunks is negligible, and that the chunk size is less than the $\text{MTU}$.

Each of the links $\mathrm{P}-\mathrm{Q}$ and $\mathrm{Q}-\mathrm{R}$ has a bandwidth of $10^6 \mathrm{bits} / \mathrm{sec}$, and negligible propagation latency. Router $\text{Q}$ immediately transmits every packet it receives from $\text{P}$ to $\text{R}$, with negligible processing and queueing delays. Router $\text{Q}$ can simultaneously receive on $\operatorname{link} \mathrm{P}-\mathrm{Q}$ and transmit on $\operatorname{link} \mathrm{Q}-\mathrm{R}$.

Assume $\text{P}$ starts transmitting the chunks at time $t=0$.

Which one of the following options gives the time (in seconds, rounded off to $3$ decimal places) at which $\mathrm{R}$ receives all the chunks of the file?

• A. $8.000$
• B. $8.008$
• C. $15.992$
• D. $16.000$

Answer 1

Correct answer Option - B

The network has three nodes P,Q,R connected as P-Q-R. P sends 1000 packets each of size 1000 bytes. 

The transmission time of one packet between two nodes is packet size / Bandwidth. 

we are given bandwitdth is 1 mega bits per second. 

hence we have 8000 /  10⁶ = 8 milliseconds. 

The total transmission time = ( Number of packets + n ) * T_t  | here n is the number of nodes on the path in between source and destination. 

Hence total time taken = (1000+1)* 8ms = 8008ms = 8.008s 

Comments

Why are we adding the no. of nodes with the no. of pkts? Here the router isn't doing any thing specifically right?

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An intermediate router can only send a packet only after it has completely recieved a packet, because if it tries to start a new transmission with a packet that it is still recieving, any delay and corruption in the previous link can propogate further. So at intermediate links atleast for the first packet we have to add another transmission time so that we can allow time to completely recieve a packet and thent try to forward.

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phaniphani

correct it by 8008 ms = 8008 * 10^-3 s = 8.008 sec

Answer 2

This may help.