GATE CS 2004 Q54 — Data Link Layer: Framing, Error Detection, MAC, Ethernet Bridging

To find the probability that station $A$ wins the second backoff race, we need to track the collision history and the resulting backoff windows for both stations $A$ and $B$.

Step 1: First Collision and Backoff

• Both $A$ and $B$ attempt to transmit their first frames and collide. This is the $1^{\text{st}}$ collision for both.
• Since $A$ wins the backoff race, $A$ successfully transmits its frame.

Step 2: Second Collision

• After $A$'s successful transmission, $A$ attempts to transmit its next frame (since both have a steady queue of frames). This is the first attempt for this new frame.
• $B$ attempts to transmit its current frame (which previously collided).
• They collide again. This collision represents:
  • The $1^{\text{st}}$ collision for $A$'s new frame.
  • The $2^{\text{nd}}$ collision for $B$'s current frame.

Step 3: Second Backoff Race

According to the Binary Exponential Backoff algorithm, the backoff interval is chosen randomly from the range $[0, 2^c - 1]$, where $c$ is the collision count for the frame:

• For station $A$ ($c_A = 1$): The backoff window is $[0, 2^1 - 1] = \{0, 1\}$.
  Thus, $A$ chooses $K_A \in \{0, 1\}$ with equal probability of $\frac{1}{2}$.
• For station $B$ ($c_B = 2$): The backoff window is $[0, 2^2 - 1] = \{0, 1, 2, 3\}$.
  Thus, $B$ chooses $K_B \in \{0, 1, 2, 3\}$ with equal probability of $\frac{1}{4}$.

Step 4: Probability that $A$ wins

Station $A$ wins the race if its chosen backoff slot is strictly less than $B$'s backoff slot ($K_A < K_B$).

The total number of possible outcomes for the pair $(K_A, K_B)$ is $2 \times 4 = 8$ equally likely outcomes.

Let's list the favorable outcomes where $K_A < K_B$:

• If $K_A = 0$, then $K_B$ can be $1, 2, 3$ ($3$ outcomes).
• If $K_A = 1$, then $K_B$ can be $2, 3$ ($2$ outcomes).

Total favorable outcomes = $3 + 2 = 5$.

$$\text{Probability}(A \text{ wins}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{5}{8} = 0.625$$