## Access-Control-Methods

Question 1 |

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 2 |

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 |

_{t})/Transmission time+Polling delay

(T

_{t}) = 1000bytes/10×10

^{6}bits/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 3 |

500 meters of cable | |

200 meters of cable | |

20 meters of cable | |

50 meters of cable |

_{t}= L / B => 1/ 10

^{7}= 0.1 microsec

Given T

_{p}= 200 m / microsec

In, 1 microsec it covers 200m

Therefore, in 0.1 microsec it is 200 * 0.1 = 20 meters

Question 4 |

A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200 Kbps bandwidth. Find the throughput of the system, if the system (all stations put together) produces 250 frames per second :

49 | |

368 | |

149 | |

151 |

Given data,

Slotted ALOHA transmits = 200 bit frames

Bandwidth = 200 Kbps

System produces = 250 frames per second

Throughput = ?

Step-2:

Pure ALOHA formula S = G * e

^{-2G}

Slotted ALOHA formula S = G * e

^{-G}

Frame transmission time = 200/200 kbps = 1ms

Here,

G = ¼ and S = G * e

^{-G}

= 0.195 (or) 19.5%

System produces = 250 * 0.195

= 48.75