GATE CS 2024 Set 1 — Question 36

MCQ+2 / -0.67MediumStop-and-WaitData Link LayerComputer Networks

Computer Networks → Data Link Layer → Stop-and-Wait

Last updated 20 August 2026

Question

Consider a network path P—Q—R between nodes P and R via router Q. Node P sends a file of size

10^6

bytes to R via this path by splitting the file into chunks of

10^3

bytes each. Node 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 MTU.

Each of the links P—Q and Q—R has a bandwidth of

10^6

bits/sec, and negligible propagation latency. Router Q immediately transmits every packet it receives from P to R, with negligible processing and queueing delays. Router Q can simultaneously receive on link P—Q and transmit on link Q—R.

Assume 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 R receives all the chunks of the file?

8.000

8.008

15.992

16.000

Correct answer

(B) 8.008

Solution

Given:

File size $F = 10^6$ bytes
Chunk size $L = 10^3$ bytes
Bandwidth $B = 10^6$ bits/sec
Path: P $\to$ Q $\to$ R (2 hops)

First, calculate the number of packets (

N

N = \frac{10^6 \text{ bytes}}{10^3 \text{ bytes}} = 1000 \text{ packets}

Next, calculate the transmission time for one packet (

T_{trans}

):
Packet size in bits =

10^3 \times 8 = 8000

bits

T_{trans} = \frac{8000 \text{ bits}}{10^6 \text{ bits/sec}} = 8 \times 10^{-3} \text{ sec} = 8 \text{ ms}

In a packet-switched network with pipelining (store-and-forward at router Q), the total time to transfer

N

packets over

k

hops is given by:

T_{total} = (N + k - 1) \times T_{trans}

Here, the number of hops

k = 2

(P-Q and Q-R).

T_{total} = (1000 + 2 - 1) \times 8 \text{ ms}

T_{total} = 1001 \times 8 \text{ ms} = 8008 \text{ ms}

Converting to seconds:

T_{total} = 8.008 \text{ seconds}

Alternatively, we can trace the packets:

The first packet arrives at Q at $t = 8$ ms. Q immediately starts forwarding it to R. It arrives at R at $t = 8 + 8 = 16$ ms.
Due to pipelining, subsequent packets arrive at R every $T_{trans} = 8$ ms after the first packet.
There are $N-1 = 999$ remaining packets.
Total time = Time for 1st packet + Time for remaining 999 packets
Total time = $16 \text{ ms} + 999 \times 8 \text{ ms} = 16 + 7992 = 8008 \text{ ms} = 8.008 \text{ s}$ .

GATE CS 2024 Set 1 — Question 36

Question

Correct answer

Solution

More questions on Data Link Layer