Computer Sciences > GATE 2026 SET-1 > Data Link Layer
Consider the implementation of sliding window protocol over a lossless link, with a window size of π‘Š frames, where each frame is of size 1000 bits (including header). The bandwidth of the link is 100 kbps (1 k = 103),and the one-way propagation delay is 100 milliseconds. Assume that processing times at the sender and receiver are zero and the transmission time of acknowledgements is also zero. Which one of the following options gives the minimum size of π‘Š (in number of frames) required to achieve 100% link utilization?
A
21
B
10
C
11
D
20

Correct : a

The correct answer is Option A: 21 frames.
To achieve 100% link utilization, the sender must keep transmitting without ever going idle while waiting for an acknowledgement.
Transmission time of one frame:
Tt = Frame size / Bandwidth = 1000 bits / 100,000 bps = 10 ms
Round Trip Time (RTT):
RTT = 2 × Propagation delay = 2 × 100 ms = 200 ms
(ACK transmission time = 0, processing time = 0)
Minimum window size for 100% utilization:
W = (Tt + RTT) / Tt = (10 + 200) / 10 = 210 / 10 = 21 frames
With W = 21, the sender continuously transmits 21 frames in the time it takes for the first ACK to arrive back, keeping the link fully busy at all times.

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Related Topics

GATE CS 2026 GATE CS 2026 Set-1 Q24 Sliding Window Protocol Minimum Window Size 100% Link Utilization Frame Size 1000 bits Bandwidth 100 kbps Propagation Delay 100ms Data Link Layer GATE Window Size 21 GATE CS 2026 Solved Bandwidth Delay Product

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