Question

A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. The transmission time of the acknowledgement is negligible. To get a channel efficiency of at least 50 %, the minimum frame size should be

• A. 80 bytes
• B. 80 bits
• C. 160 bytes
• D. 160 bits

Answer 1

The correct option is D 160 bits

Propagation time = 20 m sec
Bandwidth = 4 Kbps
When efficiency is at least 50%, then
$T.T\ge 2P.T$
$\frac{x}{4kbps}\ge 2\times 20msec$
$x\ge 2\times 20\times {10}^{-3}+4\times {10}^{3}bits$
$x\ge 40\times 4bits$
$x\ge 160bits$