GATE CS 2009 Q58 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

To find the waiting time of the sender, we first determine the value of $I$ and then calculate the timeline of transmission and acknowledgment.

Step 1: Find the number of bits $I$

• Transmission time of a frame ($T_t$) = $\frac{1000\text{ bits}}{10^6\text{ bps}} = 1\text{ ms}$
• Propagation delay ($T_p$) = $25\text{ ms}$

The maximum number of frames that can be in transit (within the link in one direction) is given by the bandwidth-delay product:

To represent these $25$ frames distinctly, we need $I$ bits such that:

Step 2: Calculate the waiting time

The sender window size is $2^I = 32$ frames.

• Transmission of 32 frames: The sender starts transmitting frame $0$ at $t = 0$ and finishes transmitting the $32^{\text{nd}}$ frame (frame $31$) at $t = 32 \times 1\text{ ms} = 32\text{ ms}$.
• Arrival of the first ACK:
  • Frame $0$ transmission finishes at $t = 1\text{ ms}$.
  • It reaches the receiver at $t = 1 + 25 = 26\text{ ms}$.
  • Since acknowledgements are piggybacked, the reverse frame (carrying the ACK) takes $1\text{ ms}$ to transmit, finishing at $t = 26 + 1 = 27\text{ ms}$.
  • The ACK arrives back at the sender at $t = 27 + 25 = 52\text{ ms}$.

The sender must wait from the moment it finishes transmitting the window ($t = 32\text{ ms}$) until it receives the first ACK ($t = 52\text{ ms}$):