Access-Control-Methods
Question 1 |
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 μs 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
1 Mbps | |
100/11 Mbps | |
10 Mbps | |
100 Mbps |
(Tt) = 1000bytes/10×106bits/sec = 800μs
Polling delay = 80 μs
Efficiency = 800/800+80 = 800/880 = 10/11
Maximum throughput is
= 10/11 × 10 Mbps
= 100/11 Mbps
Question 2 |
Let us consider a statistical time division multiplexing of packets. The number of sources is 10. In a time unit, a source transmits a packet of 1000 bits. The number of sources sending data for the first 20 time units is 6, 9, 3, 7, 2, 2, 2, 3, 4, 6, 1, 10, 7, 5, 8, 3, 6, 2, 9, 5 respectively. The output capacity of multiplexer is 5000 bits per time unit. Then the average number of backlogged of packets per time unit during the given period is
5 | |
4.45 | |
3.45 | |
0 |
If the no. of packets transmitted is larger than 5 then the extra packets are backlogged. This means gets added to the next number and further backlog is calculated.
Average no. of backlogged packets = 89/20 = 4.45
Question 3 |
0.268 | |
0.468 | |
0.368 | |
0.568 |
Slotted ALOHA transmits=200 bit frames
Bandwidth=200 Kbps
System produces=1000 frames per second
Throughput=?
Step-2: Slotted ALOHA formula S= G*e-G
Frame transmission time=200/200 kbps=1ms
Here, G=1 and S=G*e-G
=0.368 (or) 36.8%
System produces throughput =1000*0.368
=368 frames
Question 4 |
Maximum throughput of slotted ALOHA network is:
18.4%
| |
36.8% | |
50%
| |
35.8%
|
And maximum throughput of pure aloha is 18.4%.