GATE CS 2015 Set 3 Q.16 (CS Q.6): Medium Access Control (ALOHA, CSMA/CD) and Ethernet/Switching

Consider a CSMA/CD network that transmits data at a rate of 100 Mbps ( $10^8$ bits per second) over a 1 km (Kilometer) cable with no repeaters. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable?

8000

10000

16000

20000

Official answer and solution

Official answer: D
From: Official GATE 2015 answer key

Solution

Apply the CSMA/CD collision detection condition relating frame transmission time and round-trip propagation time.

In a CSMA/CD network, to ensure that collisions are always detected before the transmitter finishes transmitting the frame, the frame transmission time $T_t$ must be at least twice the one-way propagation delay $T_p$ :

T_t \ge 2 T_p

Given parameters:

Data rate (bandwidth) $B = 100\text{ Mbps} = 10^8\text{ bits/second}$
Cable distance $d = 1\text{ km}$
Minimum frame size $L = 1250\text{ bytes} = 1250 \times 8\text{ bits} = 10000\text{ bits} = 10^4\text{ bits}$

Compute transmission time:

T_t = \frac{L}{B} = \frac{10^4\text{ bits}}{10^8\text{ bits/sec}} = 10^{-4}\text{ seconds}

Setting $T_t = 2 T_p$ at the minimum frame limit:

10^{-4} = 2 \times \frac{d}{v} = 2 \times \frac{1\text{ km}}{v}

v = \frac{2}{10^{-4}} = 2 \times 10^4 = 20000\text{ km/sec}

Thus, the signal speed in the cable is $20000\text{ km/sec}$ (Option D).

Final answer

Common trap

Using one-way propagation delay instead of round-trip time (2 * Tp), which would yield 10000 km/sec.

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 3
Question: 16 of 65, printed as CS Q.6
Section: Computer Science and Information Technology

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