GATE 2007

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 microseconds to poll the next node. Whenever a node is polled, it is allowed to transmit a maximum of 1000 bytes. The maximum throughput of broadcast channel is

(A) 1 Mbps

(B) 100/11 Mbps

(C) 10 Mbps

(D) 100 Mbps

Comments

(B) propagation delay is not given thus it is negligible thus total max throughput =bytes transferred/(transmission + polling delay)

Answer 1

Check this:

http://gateoverflow.in/3509/gate2007-it-64

Comments

Thanks.