HW3_TDC563.docx - TDC 563 Protocols and Techniques Homework#3 1(10 pts The following questions are related to IPv6 addressing a Show the shortest form

HW3_TDC563.docx - TDC 563 Protocols and Techniques...

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TDC 563 Protocols and Techniques Homework #3 1. (10 pts) The following questions are related to IPv6 addressing: a. Show the shortest form of: 0000:0000:1230:0000:00AA:0000:0200:0000 ::1230:0:AA:0:200:0 b. Fully expand: FEF1:0:139::1234 FEF1:0000:0139:0000:0000:0000:0000:1234 c. What is the type of IPv6 address 2001:12:4::1 ? Teredo address. d. What is the type of IPv6 address FE91:139::1234 ? Link Local Address. e. Why is there no broadcast address in IPv6? How can an IPv6 node broadcast to everyone? Broadcast address is not needed in the IPv6. IPv6 uses Unicast, Multicast and Anycast instead of broadcast. Unicast is one to one, multicast is one to many, and anycast is many to many. It is using these types instead of the the broadcast since there is more than one option and other hosts in the network does not have to be bothered by the traffic that is not destant to them. 2. (10 pts) A path consisting of 4 links has the following characteristics: 1 st link: 100 Mbps Ethernet 2 nd link: Gigabit Ethernet (with a delay of 10µs) 3 rd link: Gigabit Ethernet (with a delay of 10µs) 4 th link: 100 Mbps Ethernet Calculate the routing metric for this path for the following routing protocols: a. RIPv2: assuming there are 3 routers between the subnets, the cost is 2 because RIP uses hop count. First hop is from R! to R2, second hop is R2 to R3 and then it reached the destination. b. IGRP: Metric = bandwidth + delay. Delay = (10 + 10) / 10 = 2 Bandwidth = 10 7 / 10 5 = 100 Metric = 100 + 2 = 102. c. EIGRP: EIGRP_metric = 256 * IGRP_metric. = 256 * 102 = 26112. d. OSPF: To reach the B subnet from A. The route will go through Three interfaces. The first interface is 1 Gigabit Ethernet and the cost is 1, the second is also 1 Gigabit Ethernet with the cost of 1 and the last one is 100Mbps with the cost of 1. Total cost is 3. e. Which metric provides the best value? How does this relate to the AD? 3. (10 pts) An AS has 12 BGP speakers. a. (2 pts) How many TCP connections for IBGP are needed (full mesh)? Each router would a point to point connection to each router in the AS and we have 12 BGP speakers. N(N-1)/2, N=node=Routers = 12(12-1)/2 = 66 TCP connections. b. (2 pts) How many TCP connections are needed at a minimum if the concept of confederation is used and the AS is divided into 3 groups of 3, 3, and 6 BGP speakers? May 17, 2017 TDC 563 – HW 3 1
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Using confederation and dividing the BGP speakers into 3 groups would reduce the number of TCP connections. Group 1 = 3 routers. 3(3-1)/2= 3 . Group 2 = 3 routers. 3(3-1)/2= 3 . Group 3 = 6 routers. 6(6-1)/2= 15. since the router are divided into three ASs, each AS should have two TCP connections to the other two routers in the other ASs which adds 3 TCP connections. 3 + 3 + 15 + 3 = 24 TCP connections. c. (2 pts) How many TCP connections are needed if there are three reflectors with 2, 2, and 5 clients each, and that the reflectors are fully meshed? The idea of the reflectors is that the routers in an AS would have only one TCP connection to one router in the same AS, this router is called RR. The RR would have TCP connection with the other RRs in the other ASs.
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