Consider the implementation of sliding window protocol over a lossless link, with a
window size of π 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 π (in number of frames) required to
achieve 100% link utilization?
Consider the implementation of sliding window protocol over a lossless link, with a
window size of π 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 π (in number of frames) required to
achieve 100% link utilization?
Show Hint
- 10
- 21
- 20
- 11
The Correct Option is B
Solution and Explanation
This question tests the concept of sliding window protocol and the minimum window size required for 100% link utilization.
Step 1: Note down the given data
Frame size = 1000 bits
Bandwidth = 100 kbps = \(100 \times 10^3\) bits per second
One-way propagation delay \(T_p\) = 100 ms
Processing time and ACK transmission time = 0
Step 2: Compute the transmission time of a frame
\[ T_t = \frac{\text{Frame size}}{\text{Bandwidth}} = \frac{1000 \text{ bits}}{100 \times 10^3 \text{ bps}} = 0.01 \text{ s} = 10 \text{ ms} \]
Step 3: Compute the round trip time
Since ACK transmission time is zero and there is only propagation delay both ways:
\[ RTT = 2 T_p = 2 \times 100 = 200 \text{ ms} \]
Step 4: Apply the sliding window utilization formula
For a sliding window protocol, the efficiency (utilization) is given by:
\[ U = \frac{W \cdot T_t}{T_t + 2T_p} \]
For 100% utilization, \(U = 1\), which means the sender must be able to keep transmitting continuously without waiting for an ACK. This requires:
\[ W \cdot T_t \geq T_t + 2T_p \]
\[ W \geq \frac{T_t + 2T_p}{T_t} \]
Step 5: Substitute the values
\[ W \geq \frac{10 + 200}{10} = \frac{210}{10} = 21 \]
Since \(W\) must be an integer number of frames, the minimum window size that achieves 100% utilization is \(W = 21\).
Step 6: Verify with the given options
Among the options 10, 21, 20, 11, only \(W = 21\) satisfies \(W \geq 21\). A value of 20 would leave the link idle for a brief interval each cycle, so it does not achieve full utilization.
Final Answer: \[ \boxed{W = 21 \text{ frames}} \]
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