" NETWORK ROUTING ANDSWITCHING |Table of Contents1. Part A ................................................................................................................................................... 22. Part B ................................................................................................................................................... 23. Part C ................................................................................................................................................... 54. Part D ................................................................................................................................................... 65. Part E ................................................................................................................................................... 9References .................................................................................................................................................. 10 1 1. Part A2. Part Ba) 139.145.56.0/22Subnet Mask for /22 is 255.255.252.0111111111 11111111 11111100 00000000, n – 2The host calculation formula is 210= 2 2 = 2X2X2X2X2X2X2 X2X2X2 = 1024The IP range : 126.145.56.0 - 126.145.59.255Total number of addresses including network and broadcast address : 1024b)Department Subnet address Last address MaskMarketing 139.145.56.127 255.255.255.128Accounting 139.145.56.255 255.255.255.1283 Head Office 139.145.57.63 255.255.255.192Cust Services 139.145.57.79 255.255.255.240c) /22 subnet.Total number of address in this subnet is 1024 Binary: 11111111.11111111.11111100.00000000.Used IP Range starting from 139.145.56.0 to 139.245.57.79.IP range in the full subnet is 139.145.56.0 to 139.145.59.2554 Free IP range in this subnet: 139.145.57.80 – 139.145.59.255.IP address in 139.145.58.0 – 255 IP addressesUsed IP’s : 80, Free IP’s : 175 IP addressesIP address in 139.145.59.0 – 255, Free IP addresses: 255Total no of Free IP addresses : 4303. Part Ca) In the task IP datagram size is 5,600 bytes including header. The router receive the packet size of 5,600 bytes. The datagram fragmented asbelow(NETworking, 2013).First fragment? The offset of the first fragment = 0;? Length =1500;? The modified IP header of original has 20 Bytes data. Thus the first fragment offset iscalculated.Second fragment? The offset value of the second fragment = 185;? For 8 fragments the offset value = 185 x 8 = 1480;? It represents the data portion of this fragment starts at 1480 bytes into the original IPdatagram. ? The length = 1500; ? For this fragment an additional IP header is created. Thus the second fragment offset iscalculated.Third fragment? The offset value of third fragment = 370? For 8 fragments, the offset value = 370 x 8 = 2960;? It represents that this fragment contains the data portion with 2960 bytes into the originalIP datagram. ? The total length of this fragment = 1500; ? For this fragment an additional IP header is created. Thus the third fragment can becalculated as above.5 Fourth fragment? The offset value of fourth fragment = 612;? For 8 fragments, the offset value = 612 x 8 = 4896. ? The size of the fragment’s data portion = 4440 bytes;? Its length is 700 bytes; For this fragment an additional IP header is created. It can be doneonly when the original IP datagram is It is only when the last fragment is received that thesize of the original IP datagram can be determined.The offset value of the last fragment (612) provides a data offset that contains 4896 bytes of datainto the original IP datagram. If the data bytes are added in to the last fragments, (704 = 684 - 20),Then totally it provides 5120 bytes, It represents the original IP datagram and its data portion.After that, 20 bytes are added into the IP and it equals to the original IP datagram’s size (4896 +704 + 20 = 5600).b)c) The reason for increasing the overall length by 60 is creating another three additional IP address.One IP address is for each fragment and after the first fragment.4. Part Da) Autonomous routing system:Though Internet, a group of network can be controlled by network administrator who work forcommon. It is based on the policy. It can be also referred as a routing domain. So that this system6 can be as the global unique number.b) Routing protocols The Routing protocols are classified into two types such as dynamic and static routing.The mainly used routing protocols are RIP, EIGRP, OSPF and BGP.Why inter-domain/intra-domain protocols7 The Intra-site routing protocols helps to send the data packets at constant rate. The link betweenback and forth states can be established in the routing table. The intra-site routing protocols helpsto keep the routers count as below 10. c)EIGRP is easy to configure rather than others but this is not work at all time. Because, it does not8 support the other vendor’s equipment so that it uses the Cisco Proprietary. d)e)5. Part E The value can be placed with the short explanation, for (read question 5):Version Number: 6Length of Payload: 128000 bytes9 Next header: 06Hop limit: 07Source Address: 581E:1456:2EFF:0000:0000:0000:0001:0000Destination Address: 581E:1456:2EFF:0000:0000:0000:0001:0000Note: source & destination address are worth 1 mark togetherIPv6 has version number of 6 whereas IPv4 is 4. The size of hop filed is 8bits.the hop limit isbased on the maximum number of links the IPv6 packet traversed. The source and destination IPaddress has 128 bits length. The extension headers and the upper layer PDU together withPayload Length field.Note: source & destination address are worth 1 mark togetherReferencesNETworking. (2013). 1st ed. [Hong Kong]: Regional NET Coordinating Team, NET Section,CDI, EDB.Technet.microsoft.com. (2017). Networking Basics: Network addressing. [online] Available at:https://technet.microsoft.com/en-us/library/cc527495(v=ws.10).aspx [Accessed 17 May2017].10 "