Consider the sliding window flow-control protocol operating between a sender and a receiver over a full-duplex error-free link. Assume the following: • The time taken for processing the data frame by the receiver is negligible. • The time taken for processing the acknowledgement frame by the sender is negligible. • The sender has infinite number of frames available for transmission. • The size of the data frame is 2,000 bits and the size of the acknowledgement frame is 10 bits. • The link data rate in each direction is 1 Mbps ( $= 10^6$ bits per second). • One way propagation delay of the link is 100 milliseconds. The minimum value of the sender’s window size in terms of the number of frames, (rounded to the nearest integer) needed to achieve a link utilization of 50% is ________.

Your answerNumerical answer

Enter a value to check

Solution

Step-by-step verified solution

Official answer: 51 (±1) (±1)

Data Link Layer

Given parameters:

Data frame size, $L_f = 2000\text{ bits}$
ACK frame size, $L_a = 10\text{ bits}$
Bandwidth, $B = 1\text{ Mbps} = 10^6\text{ bps}$
Propagation delay, $T_p = 100\text{ ms} = 0.1\text{ s}$
Desired link utilization, $\eta = 50\% = 0.5$

Calculate transmission times:

T_{t(\text{frame})} = \frac{L_f}{B} = \frac{2000}{10^6}\text{ s} = 2\text{ ms}

T_{t(\text{ack})} = \frac{L_a}{B} = \frac{10}{10^6}\text{ s} = 0.01\text{ ms}

Total cycle time to send a frame and receive its acknowledgement:

T_{\text{total}} = T_{t(\text{frame})} + T_p + T_{t(\text{ack})} + T_p = 2 + 100 + 0.01 + 100 = 202.01\text{ ms}

Link efficiency/utilization for window size $W$ is:

\eta = \frac{W \times T_{t(\text{frame})}}{T_{\text{total}}}

Setting $\eta = 0.50$ :

0.50 = \frac{W \times 2}{202.01} \implies W = \frac{0.50 \times 202.01}{2} = 50.5025

Rounding to the nearest integer gives $W = 51$ (acceptable range is 50 to 52).

Try the question first, then check your work.

Think the official key or this write-up is wrong?

Report error or challenge

Examiner trap

The misconception this question tests

This Data Link Layer previous year question is Q49 of the GATE CSE 2021 Set 1 paper. Find more GATE CSE PYQs on Data Link Layer in Computer Networks, or browse all GATE CSE PYQs.

Free diagnostic

Where do you stand for GATE?

Take the 20-minute diagnostic: the 3 topics that cost you the most marks, and a rank range. Or sit 5 related Data Link Layer PYQs now with no account.

One GATE CSE PYQ every morning

An official question by email at 7 AM IST, with a link to its verified solution. Unsubscribe in one click.

Data Link Layer in GATE CSE

Data Link Layer has 27 GATE CSE PYQs across 17 papers from 2013 to 2026. 12 of them are 2-mark NAT questions like this one, and GATE CSE 2021 Set 1 has 3.

How GATE tests Data Link Layer

More from GATE CSE 2021 Set 1

Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.