home work

Question

A broadcast channel has 10 nodes and total capacity of 16Mbps. It uses polling for medium access. Once a node finishes transmission, there is a polling delay of 100 µseconds to poll the next node. Whenever a node is polled, it is allowed to transmit a maximum of 1500 Bytes. The maximum throughput of broadcast channel is:

8 Mbps

14 Mbps

100/11Mbps

750/85 Mbps

Comments

@Samaa.qudah,

you've to add question source in title

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14 Mbps

Answer 1

Given,

         $B = 16 Mbps$

         $Tpoll = 100μsec$

         $L = 1500 bytes = 12000 bits$

then,

$Ttrans = L/B= 12000𝑏𝑖𝑡𝑠/16×10^6 bps = 3/(4×*10^3)sec=0.75=750μsec$

$Cycle time = Ttrans + Tpoll = 750 + 100 = 850 μsec$

$Utilization = (Useful time ) / (total time) = 750/850=0.8823$

$Throughput = 0.8823 × 16 Mbps = 14.1176 Mbps$