assignment2_solutions

assignment2_solutions - Solutions: Assignment #1 (Based on...

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Solutions: Assignment #1 (Based on Edition 2 of Textbook) Page 1 of 10 ± 2.28. What is the difference between a physical address, a network address, and a domain name? Solution: The physical address is the unique hardware address that identifies an interface of a machine on a physical network such as a LAN. Physical addresses are used in the data link layer. A network address is a machine’s logical address on a network. The network address is used in the network layer. The network address used on the Internet is the IP address. Domain names are used as an aid to identify hosts and networks in the Internet, since names are easier to remember than numbers. The DNS system is used to translate between domain names and IP addresses. The domain name for the network address 128.100.132.30 is toronto.edu.[1] The physical address is normally hard coded into the device (e.g. the MAC address when you by a Ethernet NIC card, the serial number in cell phones,…). The network address can be software assigned and dynamic. The best examples of network addresses are IP addresses. The domain name is just a word representation of the network address. Note: when you use domain names to specify an address, you must have some device (machine) in the network which converts this domain name (e.g. dea.ece.concordia.ca) to an IP address (dea.ece.concordia.ca == 132.205.9.4) that the routers in the Internet will understand. This device is a domain name server.
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Solutions: Assignment #1 (Based on Edition 2 of Textbook) Page 2 of 10 ± 2.34. Suppose the population of the world is 4 billion, and that there is an average of 1000 communicating devices per person. a) How many bits are required to assign a unique host address to each communicating device? b) Suppose that each device attaches to a single network and that each network on average has 10000 devices. How many bits are required to provide unique network ids to each network? Solution: a) number of devices in the world = 4x10 9 * 1000 = 4x10 12 . Each device is to have a unique binary host address. The number of bits required to accommodate 4x10 12 devices is: log 2 (4x10 12 ) = 41.9 Therefore, 42 bits are required to assign a unique host address to each communicating device. b) We need to determine the number of networks, given that each network has 10000 devices attached. As a result, number of networks = 4x10 12 /10000 = 4x10 8 . If each of these networks is to have a unique network id, then the number of bits require for this network id
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This note was uploaded on 10/09/2009 for the course ENSC 1156 taught by Professor Rhardy during the Spring '07 term at Simon Fraser.

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assignment2_solutions - Solutions: Assignment #1 (Based on...

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