Question

Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4$\mu$s. The minimum frame size is:

• A. 464
• B. 512
• C. 416
• D. 94

Answer 1

The correct option is A 464

$\text{Round trip propagation delay is}2\ast T_p$ minimum frame size of ethernet can be found by using formula
$\text{Frame size}=\left(2T_p\right)\ast \left(Bandwidth\right)$
$=46.4ms\ast 10Mbps$
$=464kbits$