Question

A broadcast channel has 10 nodes and total capacity of 10 Mbps. It uses polling for medium access. Once a node finishes transmission, there is a polling delay of 80 $\mu$s to poll the next node. Whenever a node is polled, it is allowed to transmit a maximum of 1000 bytes. The maximum throughput of the broadcast channel is

• A. 1 Mbps
• B. 100 Mbps
• C. 100/11 Mbps
• D. 10 Mbps

Answer 1

The correct option is C 100/11 Mbps

Transmission time for one frame
$=\frac{1000B}{10Mbps}=\frac{8000}{10\times {10}^6}$
In polling mechanism, each station involve in polling and one station will be elected to transmit data.
So, total time to transmit frame
$=800\mu sec+80\mu sec$
$=800\mu sec$
Throughput
$=\frac{10\times 800\mu sec}{880\mu \sec }$
$=\frac{800}{88}=\frac{100}{11}$ Mbps