Network Fundamentals with Cisco - Accommodating Hosts in the Student Networks - Assignment Example

Doted Decimal Binary Octets Host 60.61.62.63 00111100 00111101 00111110 00111111 Mask 255.255.255.192 11111111 11111111 11111111 11000000 Question 1 Network 60.61.62.0 00111100 00111101 00111110 00000000 The subnet mask is 255.255.255.192 that borrow 2 bits from the last octet. 2 bits provide 4 sub networks i.e. (22=4). The sub networks follow in the range 0-64-128-192-256. The number of network devices for each subnet is 26-2=64-2= 62 Devices Host IP addresses for the first subnet range from 60.61.62.1- 60.61.62.62 with 60.61.62.63 being the last IP address in that subnet. The Last IP address in a network is usually a broadcast address hence this proves 60.61.62.63 is not a valid IP address to assign to a device on the network. Question 2 Dotted Decimal Host Address Binary Octets The Host Range Broadcast Subnet address 192.168.102.64/27 11000000.10101000.01100110.01000000 192.168.102.65-192.168.102.94 192.168.102.95 Router1 192.168.102.65/27 11000000.10101000.01100110.01000001 In the Range 192.168.102.95 Mask 255.255.255.224 11111111.11111111.11111111.11100000 A 192.168.102.73/27 11000000.10101000.01100110.01001001 In the Range 192.168.102.95 B 192.168.102.81/27 11000000.10101000.01100110.01010001 In the Range 192.168.102.95 Subnet address 192.168.102.96/27 11000000.10101000.01100110.01100000 192.168.102.97-192.168.102.126 192.168.102.127 Router 2 192.168.102.97/27 11000000.10101000.01100110.01100001 In the Range 192.168.102.127 Mask 255.255.255.224 11111111.11111111.11111111.11100000 C 192.168.102.123/27 11000000.10101000.01100110.01111011 In the Range 192.168.102.127 D 192.168.102.148/27 11000000.10101000.01100110.10010100 Not in the Range In the subnet 192.168.102.64/27, host IP addresses range from 192.168.102.65-192.168.102.94 with IP address 192.168.102.95 being the broadcast IP. Router 1, PC A and B are in the range of IP addresses 192.168.102.64/27. The host IP addresses are valid since when converted to binary they have same subnet address. In the subnet 192.168.102.96/27, host IP addresses range from 192.168.102.97-192.168.102.126 with IP address 192.168.102.127 being the broadcast IP address. Router 2 and PC C are in the range of subnet address 192.168.102.96/27. However, D is not in the range of subnet 192.168.102.96/27. Since PC C has invalid IP address assigned, there is no connectivity between PC B to D as the case with A and C. Question 3 Section Min. No.of Hosts Subnet Address Mask Broadcast Address First Host Address Last Host Address Students 8,000 200.200.32.0/19 255.255.224.0 200.200.63.255 200.200.32.1 200.200.63.254 Administration 2,000 200.200.16.0/21 255.255.248.0 200.200.23.255 200.200.16.1 200.200.23.254 Academic staff 400 200.200.18.0/23 255.255.254.0 200.200.19.255 200.200.18.1 200.200.19.254 Computer Lab 4000 200.200.0.0/20 255.255.240.0 200.200.15.255 200.200.0.1 200.200.15.254 Minimum number of Hosts for the 4 Faculties is 14,400 to share a large group of IP addresses starting at 200.200.0.0/18. The IP address 200.200.0.0/18 has 18 bit network portion and 14 bit host portion unused that is 16,382 IP addresses (214-2). Subnet mask is 255.255.192.0. 2 bits have been borrowed from third octet. This result in 4subnets using the formula (2n) where n is the number of bits borrowed. Subnet 1: 0= 00000000 11001000.11001000.00000000.00000000 200.200.0.0/18 Subnet 2: 64=01000000 11001000.11001000.01000000.00000000 200.200.64.0/18 Subnet3: 128=10000000 11001000.11001000.10000000.00000000 200.200.128.0/18 Subnet4: 192=11000000 11001000. 11001000.11000000.00000000 200.200.192.0/18 Use first subnet 200.200.0.0/18 We borrow 1 bit more to accommodate at least 8,000 network devices for the students network. Therefore the network prefix is /19. 200.200.0.0/19 Subnet0: 0= 00000000 11001000. 11001000. 00000000.00000000 200.200.0.0/19 Subnet 1: 32= 00100000 11001000.11001000.00100000.00000000 200.200.32.0/19 We borrow 1 Bit more to accommodate 4000 hosts. Network Prefix /20 Subnet1: 0= 00000000 11001000. 11001000. 00000000.00000000 200.200.0.0/20 Subnet2: 16= 0001000 11001000.11001000.00010000.00000000 200.200.16.0/20 We borrow 1 Bit more to accommodate 2000 hosts. Network Prefix /21 Subnet1: 16= 00010000 11001000.11001000.00010000.00000000 200.200.16.0/21 Subnet2: 24= 00011000 11001000. 11001000.00011000.00000000 200.200.24.0/21 We borrow 2 bits to accommodate 400 hosts. Network prefix will be /23 Subnet1: 16= 00010000 11001000.11001000.00010000.00000000 200.200.16.0/23 Subnet2: 18= 00010010 11001000.11001000.00010010.0000000 200.200.18.0/23 Students Network We are to accommodate at least 8000 hosts in the student network. The available IP address to use is 200.200.32.0/19. We have borrowed 1 bit from 200.200.0.0/18 to accommodate 8000 hosts hence the network portion has 19 bits and host portion 13 bits. The network prefix is /19. Subnet mask in Decimal=255.255.224.0 Subnet mask in binary= 11111111.11111111.11100000.00000000 Number of Subnet= 21=2 Subnets Number of Hosts= 213-2=8,190 Hosts per subnet Network 200.200.32.0/19 with one borrowed address we have 2 Subnets starting from (0, 32) 200.200.32.0/19 11001000. 11001000. 00100000.00000000 Network Portion Host Portion Lowest Host Address: The lowest host address is always 1 greater than network address 11001000. 11001000. 00100000.00000000 00000.00000001 All host bits except the last significant address are all 0s:0+0+0+0+0+0+0+1=1. With lowest bit of the host address set to a 1, the address is 200.200.32.1 so, the lowest host address is 200.200.32.1 Broadcast Address All host bits are set to 1. 11001000. 11001000. 00100000.00000000 11111.11111111 This gives 32+16+8+4+2+1. 128+64+32+16+8+4+2+1 ==63.255 ==Broadcast address is 200.200.63.255 Highest Host address Highest host address is 1 less broadcast address. This gives 200.200.63.254 Administration Network Administration needs 2000 hosts. We take the next available IP address which is 200.200.16.0/21. We have borrowed 1 bit for the network portion. Therefore the network prefix is /21. Subnet address is 200.200.16.0 /21. Subnet mask in binary = 11111111.11111111.11111000.0000000 Subnet mask in Decimal= 255.255.248.0 Number of Subnets= 21= 2 Subnets Hosts per Subnet = 211-2= 2,046 Hosts Lowest Host IP address: The lowest host address is always 1 greater than network address 11001000. 11001000. 00010000.00000000 000.00000001 11001000. 11001000. 00010000.00000001 This gives 200.200.16.1 Broadcast address All host bits are set to 1.Therefore, 11001000. 11001000. 00010000.00000000 111.11111111 This gives 16+4+2+1.128+64+32+16+8+4+2+1= 23.255 Broadcast Address is 200.200.23.255 Highest Address Highest host address is 1 less broadcast address. This gives 200.200.23.254 Therefore, the subnet address is 200.200.16.0 /21, subnet mask 255.255.248.0, Broadcast address 200.200.23.255 and lowest and highest IP address being 200.200.16.1 and 200.200.23.254 respectively. Academic Staff Network To accommodate 400 hosts we use IP 200.200.18.0/23 has 2 bits borrowed. Subnet in binary= 11111111.11111111.11111110.00000000 Subnet in Decimal= 255.255.254.0 Subnets=22=4 Number of Hosts per subnet= 29-2= 510 Lowest IP address 11001000. 11001000. 00010010.00000000 0.00000001 11001000. 11001000. 00010010.00000001 =200.200.18.1 Broadcast address: Set all host bits to 1 11001000. 11001000. 00010010.00000000 1.11111111 =16+2+1.128+64+32+16+8+2+1 =19.255 =200.200.19.255 Highest Host address Highest host address is 1 less broadcast address 200.200.19.254 Therefore, the subnet address is 200.200.18.0/23, subnet mask 255.255.254.0, Broadcast address 200.200.19.255 and lowest and highest address being 200.200.18.1 and 200.200.19.254 respectively. Computer Lab Network We borrow 1 Bit from 200.200.0.0/19 to accommodate 4000 hosts. The network Prefix is /20 200.200.0.0/20. Subnet Mask in Decimal is 255.255.240 Subnet mask in Binary is 11111111.11111111.11110000.00000000 Subnets = 21= 2 Subnets Hosts per Subnet = 212-2= 4,094 Lowest Host address 11001000. 11001000. 00000000.00000000 0000.00000001 11001000. 11001000. 00000000.00000001 The lowest IP address is 200.200.0.1 Broadcast address 11001000. 11001000. 00000000.00000000 (Binary) 1111.11111111 11001000. 11001000. 00001111.11111111 The broadcast address is 200.200.15.255 Highest host address Highest host address is 1 less broadcast address Hence it is 200.200.15.254 Therefore, the subnet address is 200.200.0.0/20, subnet mask 255.255.240, Broadcast address 200.200.15.255 and lowest and highest address being 200.200.0.1 and 200.200.15.254 respectively. Question 4 (a) Source MAC address and Destination MAC address The destination MAC addresses is stored in Ethernet Frame in Destination Address field where 6 bytes are allocated to it and the source MAC address is stored in Source Address field which also has 6 bytes allocated. (b) Source IP address and Destination IP address The destination IP address and the Source IP address is stored in the IP packet Header. Source IP address and destination IP address are usually dealt in Layer 3 of OSI Model (Net-work Layer). (c) Source MAC address, Destination MAC Address, Source IP Address and Destination IP address Leaving the Client Source MAC address: 00D0.AB12.CD71 Destination MAC Address: 0011.ABCD.554C Source IP Address: 172.172.172.1 Destination IP address: 172.172.172.254 Reaching Router B Source MAC address: 00D0.AB12.CD71 Destination MAC Address: 0011.ABCD.554C Source IP Address: 172.172.172.1 Destination IP address: 172.172.172.254 Leaving Router B Source MAC address: No MAC Address Destination MAC Address: No MAC Address Source IP Address: 173.173.173.34 Destination IP address: 173.173.173.33 Reaching Router A Source MAC address: NONE Destination MAC Address: NONE Source IP Address: 173.173.173.34 Destination IP address: 173.173.173.33 Leaving Router A Source MAC address: 0011.ABCD.554A Destination MAC Address: 00D0.A034.127F Source IP Address: 191.168.5.254 Destination IP address: 191.168.5.5 Reaching Server Source MAC address: 0011.ABCD.554A Destination MAC Address: 00D0.A034.127F Source IP Address: 191.168.5.254 Destination IP address: 191.168.5.5 Leaving Server Source MAC address: 00D0.A034.127F Destination MAC Address: 0011.ABCD.554A Source IP Address: 191.168.5.5 Destination IP address: 191.168.5.254 Reaching Router A Source MAC address: 00D0.A034.127F Destination MAC Address: 0011.ABCD.554A Source IP Address: 191.168.5.5 Destination IP address: 191.168.5.254 Leaving Router A Source MAC address: NONE Destination MAC Address: NONE Source IP Address: 173.173.173.33 Destination IP address: 173.173.173.34 Reaching Router B Source MAC address: NONE Destination MAC Address: NONE Source IP Address: 173.173.173.33 Destination IP address: 173.173.173.34 Leaving Router B Source MAC address: 0011.ABCD.554C Destination MAC Address: 00D0.AB12.CD71 Source IP Address: 172.172.172.254 Destination IP address: 172.172.172.1 Reaching Client Source MAC address: 0011.ABCD.554C Destination MAC Address: 00D0.AB12.CD71 Source IP Address: 172.172.172.254 Destination IP address: 172.172.172.1 Read More