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?

The correct option is 21.

To find the minimum window size $W$ required to achieve 100% link utilization, we need to analyze the timing of the sliding window protocol. Link utilization is 100% when the sender can continuously transmit frames without waiting for acknowledgements. This requires the total transmission time of the frames within the window to be at least equal to the round-trip time (RTT) of the link.

Let's first calculate the transmission time ($T_{\mathrm{tx}}$) of a single frame:
Frame size ($L$) = 1000 bits
Bandwidth ($B$) = 100 kbps = $100\times {10}^3$ bits per second = 100,000 bps
Transmission time is given by the formula:
$T_{\mathrm{tx}}=\frac{L}{B}=\frac{1000}{\mathrm{100,000}}=0.01\text{seconds}=10\text{milliseconds}$

Next, we determine the Round-Trip Time (RTT) of the link. RTT is the time from when the sender starts transmitting a frame to when it receives the corresponding acknowledgement. The problem specifies that processing times and the transmission time of acknowledgements are zero:
One-way propagation delay ($T_p$) = 100 milliseconds
RTT is defined as:
$\text{RTT}=T_{\mathrm{tx}}+2\times T_p$
Substituting the values, we get:
$\text{RTT}=10\text{ms}+2\times 100\text{ms}=210\text{milliseconds}$

To achieve 100% link utilization, the sender must be able to send $W$ frames continuously during one RTT. Therefore, the total transmission time for $W$ frames must satisfy the following inequality:
$W\times T_{\mathrm{tx}}\ge \text{RTT}$
Substituting the calculated values into the inequality:
$W\times 10\text{ms}\ge 210\text{ms}$
Solving for $W$:
$W\ge \frac{210}{10}$
$W\ge 21$

Thus, the minimum size of the window $W$ required to achieve 100% link utilization is 21 frames.