Question

Consider the implementation of sliding window protocol over a lossless link, with a
window size of 𝑊 frames, where each frame is of size 1000 bits (including header).
The bandwidth of the link is 100 kbps (1k = 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 𝑊 (in number of frames) required to
achieve 100% link utilization?

• A. 10
• B. 21
• C. 20
• D. 11

Hint:

Compute transmission time \(T_t = \frac{1000}{100000} = 10\text{ ms}\) and round trip time \(2T_p = 200\text{ ms}\). For 100% utilization, \(W \geq \frac{T_t + 2T_p}{T_t}\).

Answer 1

The Correct Option is B

An alternative way to think about this problem is in terms of how many frames the sender can 'fit' into one full round trip before the first acknowledgement returns.

Setting up the timeline: After the sender transmits the first bit of frame 1, it takes \(T_t = \frac{1000}{100 \times 10^3} = 0.01\text{ s} = 10\text{ ms}\) to push the entire frame onto the link. The last bit then needs \(T_p = 100\text{ ms}\) to reach the receiver, and since the ACK is instantaneous (zero transmission time) and processing is zero, another \(T_p = 100\text{ ms}\) is needed for the acknowledgement to travel back.

So the total time from the start of transmitting frame 1 until its ACK is received back at the sender is:

\[T_{cycle} = T_t + 2T_p = 10 + 200 = 210\text{ ms}\]

Counting frames sent in this cycle: During this 210 ms window, the sender keeps transmitting new frames back-to-back (since it has not yet received any ACK). The number of frames it can push out in 210 ms, each taking 10 ms to transmit, is:

\[N = \frac{T_{cycle}}{T_t} = \frac{210}{10} = 21\]

For the link to never sit idle (i.e., 100% utilization), the window must be large enough to allow exactly this many frames to be in flight (unacknowledged) at once. If the window were smaller, say 20, the sender would exhaust its window and stall for a short time before the first ACK arrives, causing the link to go idle briefly.

Hence the minimum window size needed is:

\[W_{min} = 21 \text{ frames}\]

This matches option (b): 21.