Question

Consider the sliding window flow-control protocol operating between a sender and a receiver over a full-duplex error-free link. Assume the following:

• The time taken for processing the data frame by the receiver is negligible.
• The time taken for processing the acknowledgement frame by the sender is negligible.
• The sender has infinite number of frames available for transmission.
• The size of the data frame is 2,000 bits and the size of the acknowledgement frame is 10 bits.
• The link data rate in each direction is 1 Mbps (= 10⁶ bits per second).
• One way propagation delay of the link is 100 milliseconds.

The minimum value of the sender’s window size in terms of the number of frames (rounded to the nearest integer) needed to achieve a link utilization of 50% is  .

Hint:

Large propagation delays significantly increase the required window size to maintain good utilization.

Answer 1

Correct Answer: 50

Step 1: Transmission times.
Given:
Data frame size = 2000 bits
Link rate = 1 Mbps

\[ T_f = \frac{2000}{10^6} = 0.002 \text{ s} \]

Acknowledgement frame size = 10 bits

\[ T_a = \frac{10}{10^6} = 0.00001 \text{ s} \]

Since the ACK transmission time is negligible compared to propagation delay, it is ignored in utilization calculation.

Step 2: Propagation delay.
One-way propagation delay: \[ T_p = 100 \text{ ms} = 0.1 \text{ s} \]

Round-trip propagation delay: \[ 2T_p = 0.2 \text{ s} \]

Step 3: Utilization condition.
For a sliding window protocol, link utilization is: \[ U = \frac{W \times T_f}{T_f + 2T_p} \]

Step 4: Substitute values for 50% utilization.
\[ 0.5 = \frac{W \times 0.002}{0.002 + 0.2} \]

Step 5: Solve for window size.
\[ W = \frac{0.5 \times 0.202}{0.002} \]

\[ W = 50.5 \]

Step 6: Required window size.
Since window size must be an integer and sufficient to achieve at least 50% utilization, we take the next higher integer.

\[ W = 52 \]

Final Answer:
\[ \boxed{52} \]