170280 Tutorial 09 17Sol PDF

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TSN2201 COMPUTER NETWORKS TUTORIAL 9 (Tri 1, PART 1. Two popular approaches to packet switching are the datagram approach and the virtual circuit approach. ANS: T 2. In the datagram approach, each packet is treated dependent to each other. If one packet is dropped, all packets are dropped. ANS: F 3....

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TSN2201 COMPUTER NETWORKS TUTORIAL 9 (Tri 1, 2017/2018) PART A- TRUE/FALSE 1. Two popular approaches to packet switching are the datagram approach and the virtual circuit approach. ANS: T 2. In the datagram approach, each packet is treated dependent to each other. If one packet is dropped, all packets are dropped. ANS: F 3. At the network layer, a global addressing system that uniquely identifies every host and router is necessary for delivery of a packet from network to network. ANS: T 4. In IPv6, there are five classes of IP addresses. ANS: F 5. The portion of the IP address that identifies the network is called the MAC address. ANS: F 6. An IP address is composed of at least two basic parts: the network ID and the host ID. ANS: T 7. Subnetting extends a network by adding another hierarchical level in the IP addressing scheme for IPv4. ANS: T 8. Unicast communication is one source sending a packet to all network destinations. ANS: F 9. Network Address Translation (NAT) allows a private network to use a set of private addresses for internal communication and a set of global Internet addresses for external communication. ANS: T 10. DHCP is a dynamic configuration protocol that utilizes a database of IP addresses to lease them out based on requests to access network resources from clients logging into the network. ANS: T

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PART B – MCQ 1. Which protocol can automatically configure a computer's IP address and subnet mask? a. TCP c. IP b. DHCP d. ARP ANS: B 2. For a Class C network address 192.168.10.0 which of the following subnet masks delivers 30 subnets? a. 255.255.255.252 c. 255.255.255.247 b. 255.255.255.281 d. 255.255.255.248 ANS: D 3. For the Class C network mask for 198.162.128.0, what subnet mask supports up to 14 subnets? a. 255.255.255.252 c. 255.255.255.240 b. 255.255.255.248 d. 255.255.255.224 ANS: C 4. The loopback address for IPv6 is “::1”. What about IPv4? a. 127.0.0.1 c. localhost b. 192.168.0.1 d. 10.0.0.1 ANS: A & C 5. What will happen if you do not set default gateway when you configure TCP/IP for your computer in MMU network. a. you are able to access to any part of MMU network. b. you are only able to communicate with hosts in the same subnet your computer resides in. c. you are unable to communicate with hosts in other subnets. d. you are unable to access Internet. ANS: B, C & D PART C- SHORT ANSWER 1. Describe how a static IP address table might be updated. ANS: Manually via a network administrator. 2. Compare and contrast dynamic IP addressing vs. static IP addressing schemes in a LAN. Which is more efficient and why? ANS: Dynamic, uses DHCP to lease available IP addresses to clients, and automatically updates address tables of border devices and inter-network devices. Static is manually handled by the network administrator, is laborious and takes time. Dynamic IP addressing is the more efficient because you build it once, and keep it updated with the network as changes are made in the network. Static requires that once you build it, you must continuously manually make changes to the IP address tables within the network devices to ensure accuracy and integrity. 3. Given IP address 192.168.5.77 /28. Find i) Subnet address for this IP address. ANS: 192.168.5.64 (use AND operation) ii) The broadcast address for this subnet. ANS: 192.168.5.79 iii) What is the range of useable IP addresses for this subnet.

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ANS: 192.168.5.65-192.168.5.78 (subnet address and broadcast address cannot be assigned to hosts) 4. Identify the five protocols in the network layer. ANS: IP, ICMP, ARP, RARP and IGMP. 5. The IPv4 address space is not enough for current needs. List the short term solutions and a long term solution to solve the problem. ANS: Short term solutions – private addressing scheme, NAT, classless addressing, subnetting. Long Term solution – IPv6 6. Abbreviate the following IPv6 addresses. i) 2001:0DA8:E800:0000:0000:0000:0000:0001 ANS: 2001:DA8:E800::1 ii) FF02:0:0:0:0:0:0:1 ANS: FF02::1 iii) 3FFE:0501:0008:0000:0260:97FF:FE40:EFAB ANS: 3FFE:501:8:0:260:97FF:FE40:EFAB iv) 0:0:0:0:0:0:0:1 ANS: ::1 v) 0:0:0:0:0:0:0:0 ANS: ::

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