Sliding window protocol: L = 1000 bits, R = 100 Kbps, Tp = 100 ms, Tpm = 0. Find optimal window size?

The correct answer is 21.

To find the optimal window size for the sliding window protocol, we must calculate the window size that allows the sender to transmit continuously without waiting for acknowledgments (ACKs), thereby achieving 100% link utilization. Let's break down the solution step-by-step:

1. Identify the given parameters:
- Packet size ($L$) = 1000 bits
- Transmission rate ($R$) = 100 Kbps = 100,000 bits per second (bps)
- Propagation delay ($T_p$) = 100 ms = 0.1 seconds
- Acknowledgment transmission/processing delay ($T_{\mathrm{pm}}$) = 0 ms

2. Calculate the Transmission Delay ($T_t$):
Transmission delay is the time required to push all of the packet's bits onto the link:

$T_t=\frac{L}{R}$

Substituting the values:

$T_t=\frac{1000}{100000}=0.01\text{seconds}=10\text{ms}$

3. Calculate the total Cycle Time ($T_{\text{cycle}}$):
The total cycle time is the time elapsed from the start of packet transmission until the acknowledgment for that packet is received back at the sender. It includes transmission delay, round-trip propagation delay, and acknowledgment processing time:

$T_{\text{cycle}}=T_t+2·T_p+T_{\mathrm{pm}}$

Substituting the values:

4. Calculate the Optimal Window Size ($W$):
The optimal window size is the number of packets the sender needs to transmit during one full cycle time to keep the channel fully busy:

$W=\frac{T_{\text{cycle}}}{T_t}$

Substituting the calculated times:

$W=\frac{210\text{ms}}{10\text{ms}}=21$

Therefore, the optimal window size is 21.