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BBF221 FINAL CA - FT
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- 2 marks] Give the subnet mask for the address that has been allocated to the organisation. [1 mark]
- 2 marks] ANSWERS
| QUESTION FIVE
Differentiate between congestion and flow control. [2 marks] Explain any two methods that you would employ to prevent data overrun. [6 mark] Give a well-explained example of replay and outline how you would prevent the problem of replay packets. [2 mark] Explain how duplicate packets can occur in an internet. [2 marks] Using a well-labeled diagram give an example of an internet that can easily experience congestion and outline two methods that can be employed to avoid congestion. [3 marks] An organisation has one thousand five hundred (1 500) hosts to connect to its internet. The organisation has been allocated the network address 197.69.128.0. As an expert in your own right, explain to the organisation what should be done in order for all its hosts to be catered for by the allocated address. [2 marks] Give the subnet mask for the address that has been allocated to the organisation. [1 mark] In its endeavour to make the allocated address cater for its 1 500 hosts, the organisation may alter the allocated address in one way or another. Give the subnet mask that may result due to this alteration. [2 marks] ANSWERS When too many packets rushing to a node or a part of network, the network performance degrades while Flow control refers to a set of procedures used to restrict the amount of data. The sender can send before waiting for acknowledgment. The simplest form of flow control is a stop-and-go system in which a sender waits after transmitting each packet. When the receiver is ready for another packet, the receiver sends a control message, usually a form of acknowledgement. Although stop-and-go protocols prevent overrun, they can cause extremely inefficient use of network bandwidth. To understand why, let us consider what happens on a network that has a packet size of 1000 octets, a throughput capacity of 2 Mbps, and a delay of 50 milliseconds. The network hardware can transport 2 Mbps from one computer to another. However, after transmitting a packet, the sender must wait 100 msec before sending another packet (i.e., 50 msec for the packet to reach the receiver and 50 msec for an acknowledgement to travel back). Thus, the maximum rate at which data can be sent using stop-and-go is one packet every 100 milliseconds. When expressed as a bit rate, the maximum rate that stop-and-go can achieve is, 80,000 bps, which is only 4% of the hardware capacity. To obtain high throughput rates, protocols use a flow control technique known as sliding window. The sender and receiver are programmed to use a fixed window size, which is the maximum amount of data that can be sent before an acknowledgement arrives. For example, the sender and receiver might agree on a window size of four packets. The sender begins with the data to be sent, extracts data to fill the first window, and transmits copies. If reliability is needed, the sender retains a copy in case retransmission is needed. The receiver must have buffer space ready to receive the entire window. When a packet arrives in sequence, the receiver passes the packet to the receiving application and transmits an acknowledgement to the sender. When an acknowledgement arrives, the sender discards its copy of the acknowledged packet and transits the next packet. Replay means that an old, delayed packet affects later communication. To prevent replay, protocols mark each session with a unique ID (e.g., the time the session was established), and require the unique ID to be present in each packet. The protocol software discards any arriving packet that contains an incorrect ID. To avoid replay, an ID must not be reused until a reasonable time has passed (e.g., hours). Malfunctioning hardware can cause packets to be duplicated. Duplication often arises in WANs, but can also occur in LANs. For example, a transceiver malfunction in a LAN that uses CSMA/CD can cause the receiver to sense a valid transmission while the sender senses a collision. As a result, the sender will back off from the collision and retransmit, causing two copies of the frame to reach the receiver. Sequencing solves the problem of duplication. Congestion is a fundamental problem in packet switching systems. To understand why, let us consider a network represented by the graph below. 1 5 3 4 Arrange for packet switches to inform senders when congestion occurs. Use packet loss as an estimate of congestion. i. There is need to subnet the network to accommodate the 1500 hosts. 2n=1500 N=11 Therefore 32-11 = 21. Number of hosts will be 2046 ii. 255.255.255.0 iii. Subnet mask will be 255.255.248.0 Download 290.5 Kb. Share with your friends:
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