GATE CS 2015 Set 2 Q.18 (CS Q.8): Data Link Layer: Framing, Error Detection and Sliding Window

A link has a transmission speed of $10^6$ bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgment has negligible transmission delay, and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the one-way propagation delay (in milliseconds) is _____.

Your answer

Official answer and solution

Official answer: 12
From: Official GATE 2015 answer key

Solution

Use the standard efficiency formula for Stop-and-Wait protocol $\eta = \dfrac{T_t}{T_t + 2T_p}$ .

Parameters given:

Bandwidth $B = 10^6\text{ bits/sec}$
Packet size $L = 1000\text{ bytes} = 8000\text{ bits}$
Efficiency $\eta = 25\% = 0.25 = \frac{1}{4}$

Transmission time of a data packet:

T_t = \frac{L}{B} = \frac{8000\text{ bits}}{10^6\text{ bits/sec}} = 8 \times 10^{-3}\text{ sec} = 8\text{ ms}.

In Stop-and-Wait with negligible ACK transmission delay and processing delays:

\eta = \frac{T_t}{T_t + 2T_p} = \frac{1}{4}

T_t + 2T_p = 4T_t \implies 2T_p = 3T_t \implies T_p = 1.5 T_t

T_p = 1.5 \times 8\text{ ms} = 12\text{ ms}.

Final answer

Common trap

Forgetting to multiply packet size in bytes by 8 to convert to bits.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 2
Question: 18 of 65, printed as CS Q.8
Section: Computer Science and Information Technology

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