Question

The minimum frame size required for a CSMA/CD based computer network running at 1 Gbps on a 200 m cable with a link speed of $2\times {10}^8$ m/s is

• A. 125 bytes
• B. 250 bytes
• C. None of these
• D. 500 bytes

Answer 1

The correct option is B 250 bytes

In CSMA/CD :
T. T> $\ge$ 2.P. T. (To detect the collision, frame should be transmitted atleast 2 PT)
$\frac{x}{1\times {10}^{9}bps}\ge 2\times \left(\frac{200m}{2\times {10}^{8}m/sec}\right)$

$x\ge 2\times (\frac{200m}{2\times {10}^{8}m/sec}\times 1\times {10}^{9}bps)$
$x\ge \frac{400\times {10}^9}{2\times {10}^8}bits$
$x\ge 200\times 10bits$
$x\ge \frac{2000}{8}$ bytes
$x\ge 250$ bytes