Data Link Layer
Short Notes
Responsible for node-to-node delivery, error control, and flow control.
Flow Control
- Stop-and-Wait: Send one frame, wait for ACK.
- Sliding Window: Send multiple frames.
- Go-Back-N (GBN)
- Selective Repeat (SR)
Key Theories & Formulas
1. Efficiency (\(\eta\))
\(\eta = \frac{T_{trans}}{T_{trans} + 2 T_{prop}}\)
- Let \(a = T_{prop} / T_{trans}\).
- Stop-and-Wait: \(\eta = 1 / (1 + 2a)\).
- Sliding Window: \(\eta = W / (1 + 2a)\) (where \(W\) is window size).
2. CSMA/CD (Ethernet)
Used for collision detection in shared channels.
- Minimum Frame Size (\(L\)): \(L \ge 2 \times T_{prop} \times B\) (where \(B\) is bandwidth).
- \(T_{trans} \ge 2 \times T_{prop}\).
Example Problems
Problem: A 1 Mbps link has \(T_{prop} = 25\) ms. Frame size is 1000 bits. Find efficiency of Stop-and-Wait.
- \(T_{trans} = 1000 / 10^6 = 1\) ms.
- \(a = 25 / 1 = 25\).
- \(\eta = 1 / (1 + 2 \times 25) = 1/51 \approx 2\%\).
Hardest GATE Questions
Topic: Window Size and Sequence Numbers Tricky Question (GATE 2011/2013/2016): What is the minimum number of sequence bits for Selective Repeat with window size \(W\)?
- Analysis: \(Max\_Seq\_No \ge 2W\).
- Bits = \(\lceil log_2(2W) \rceil\).
- The "Trap": GBN vs SR window sizes.
- GBN: \(W_{sender} = N-1, W_{receiver} = 1\).
- SR: \(W_{sender} = W_{receiver} = N/2\).
- Hard Aspect: Efficiency with errors.
- \(\eta = (1-p) \times (\eta_{ideal})\) where \(p\) is probability of frame error.
- Complexity: Binary Exponential Backoff in CSMA/CD.
- After \(n\) collisions, wait for \(K \times 512\) bit-times where \(K \in [0, 2^n - 1]\).
- Max \(n\) is 10 (range \([0, 1023]\))