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 ($1\text{k} = 10^3$) 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?

Question

Prepp Team · Accepted Answer

Calculating Transmission Time and RTT

First, we need to determine the time it takes to transmit a single frame and the round-trip time (RTT).

Frame Size ($F$) = 1000 bits
Bandwidth ($B$) = 100 kbps = $100 \times 10^3$ bps
One-way Propagation Delay ($T_{prop}$) = 100 ms = 0.1 s

The transmission time ($T_{trans}$) for one frame is calculated as:

$ T_{trans} = \frac{F}{B} = \frac{1000 \text{ bits}}{100 \times 10^3 \text{ bps}} = 0.01 \text{ s} = 10 \text{ ms} $

Since the link is lossless and acknowledgement transmission time is negligible, the Round-Trip Time (RTT) is twice the one-way propagation delay:

$ RTT = 2 \times T_{prop} = 2 \times 0.1 \text{ s} = 0.2 \text{ s} = 200 \text{ ms} $

Determining Minimum Window Size for 100% Utilization

To achieve 100% link utilization, the sliding window protocol must ensure the network "pipe" is always full. This means the sender continuously transmits frames without waiting for acknowledgements, effectively keeping the link busy for the entire RTT duration.

The minimum window size ($W$) required can be determined using the formula that ensures the sender finishes transmitting the $W^{th}$ frame just as the acknowledgement for the 1st frame arrives:

$ W = 1 + \frac{RTT}{T_{trans}} $

Substituting the calculated values:

$ W = 1 + \frac{0.2 \text{ s}}{0.01 \text{ s}} = 1 + 20 = 21 $

Therefore, a minimum window size of 21 frames is required to achieve 100% link utilization under these conditions.