GATE CS 2007 Q69 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

To achieve maximum utilization (100% efficiency) in a sliding window protocol, the sender's window size ($W$) must be large enough to allow continuous transmission while waiting for the first acknowledgment.

Step-by-Step Derivation:

1. Transmission Delay ($T_t$):

   The time taken to transmit a frame of size $K$ bits over a channel of capacity $R$ bits/second is:

   $$T_t = \frac{K}{R} \text{ seconds}$$
2. Propagation Delay ($T_p$):

   The time taken for a bit to travel the distance $L$ km with a propagation delay of $t$ seconds/km is:

   $$T_p = L \times t \text{ seconds}$$
3. Minimum Window Size ($W$) for Maximum Utilization:

   To keep the channel busy continuously, the window size must satisfy:

   $$W \ge 1 + 2a$$

   where $a = \frac{T_p}{T_t}$ is the ratio of propagation delay to transmission delay.

   Substituting the values of $T_p$ and $T_t$:

   $$a = \frac{L \cdot t}{\frac{K}{R}} = \frac{L \cdot t \cdot R}{K}$$

   Thus, the minimum window size is:

   $$W = 1 + 2\left(\frac{L \cdot t \cdot R}{K}\right) = \frac{K + 2LtR}{K}$$
4. Minimum Sequence Number Bits:

   The minimum number of bits required in the sequence number field to represent a window of size $W$ is:

   $$\lceil \log_2 W \rceil = \left\lceil \log_2 \left(\frac{2LtR + K}{K}\right) \right\rceil$$

This matches Option C.